added books

author: hardythe1 2015-04-07 15:58:05 +0530
committer: hardythe1 2015-04-07 15:58:05 +0530
commit: 92cca121f959c6616e3da431c1e2d23c4fa5e886 (patch)
tree: 205e68d0ce598ac5caca7de839a2934d746cce86 /Modern_Physics
parent: b14c13fcc6bb6d01c468805d612acb353ec168ac (diff)
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92 files changed, 17035 insertions, 2382 deletions
diff --git a/Modern_Physics/Chapter1.ipynb b/Modern_Physics/Chapter1.ipynb
index d442fd97..30eac463 100755
--- a/Modern_Physics/Chapter1.ipynb
+++ b/Modern_Physics/Chapter1.ipynb
@@ -1,7 +1,6 @@
 {
  "metadata": {
-  "name": "",
-  "signature": "sha256:d8ffd8fd29d5d24b3cc8b2545e898e818e9d8ef837edfe4af93ac3d5f1b9fea0"
+  "name": "Chapter1"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -13,7 +12,7 @@
      "level": 1,
      "metadata": {},
      "source": [
-      "Chapter 1: RELATIVITY"
+      "Chapter 1:Introduction"
      ]
     },
     {
@@ -21,41 +20,23 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 1.2, page no. 18"
+      "Example 1.1, Page 12"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "\n",
-      "import math\n",
-      "#Variable declaration\n",
-      "T1 = 3.0            #proper time(s)\n",
-      "c  = 3 * 10**8      #Speed of light(m/s)\n",
-      "v  = 0.95 * c       #speed of observer(m/s)\n",
-      "\n",
-      "#calculation\n",
-      "\n",
-      "Y = 1/math.sqrt((1-(v/c)**2))   #factor by which it runs slower\n",
-      "T = Y * T1          #Period of the pendulum (s)\n",
-      "\n",
-      "#result\n",
-      "print \"The Period of the pendulum when measured by the observer is\",round(T,1),\"s\"\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "T1 = 3.0            #proper time(s)\n",
-      "c  = 3 * 10**8      #Speed of light(m/s)\n",
-      "Vnew  = (0.95 + 0.05*0.95) * c       #speed of observer(m/s)\n",
+      "#initiation of variable\n",
+      "Mn=1.008665;Mp=1.007276                                     #Given mass of an electron and a proton in terms of u\n",
       "\n",
       "#calculation\n",
-      "\n",
-      "Y = 1/math.sqrt((1-(Vnew/c)**2))\n",
-      "T = Y * T1      \n",
+      "Md= Mn-Mp;                                                  #mass difference       \n",
+      "Md2=Md*931.50;                                              #converting u into Mev/c^2 by multiplying by 931.5 MeV/c^2\n",
       "\n",
       "#result\n",
-      "print \"Exercise:The Period of the pendulum when measured by the observer is\",round(T,1),\"s\"\n"
+      "print \"Mass difference in terms of U is\",round(Md,4); \n",
+      "print\"which equals in Mev/c^2. :\",round(Md2,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -64,8 +45,8 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The Period of the pendulum when measured by the observer is 9.6 s\n",
-        "Exercise:The Period of the pendulum when measured by the observer is 42.5 s\n"
+        "Mass difference in terms of U is 0.0014\n",
+        "which equals in Mev/c^2. : 1.294\n"
        ]
       }
      ],
@@ -76,87 +57,21 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 1.3, page no. 20"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#variable declaration\n",
-      "\n",
-      "Lp = 100            #proper length of the spaceship(m)\n",
-      "c  = 3 * 10**8      #speed of light(m/s)\n",
-      "v  = 0.99 * c       #speed of the observer(m/s)\n",
-      "\n",
-      "#calculation\n",
-      "\n",
-      "L = Lp * math.sqrt((1-(v/c)**2))     #Length of the spaceship (m)\n",
-      "\n",
-      "#result\n",
-      "\n",
-      "print \"The length measured as the spaceship flies by is\",round(L),\"m\"\n",
-      "\n",
-      "#variable declaration\n",
-      "\n",
-      "Lp = 100            #proper length of the spaceship(m)\n",
-      "c  = 3 * 10**8      #speed of light(m/s)\n",
-      "v  = 0.01 * c       #speed of the observer(m/s)\n",
-      "\n",
-      "#calculation\n",
-      "\n",
-      "L = Lp * math.sqrt((1-(v/c)**2))\n",
-      "\n",
-      "#result\n",
-      "\n",
-      "print \"Exercise:The length measured as the spaceship flies by is\",round(L,2),\"m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The length measured as the spaceship flies by is 14.0 m\n",
-        "Exercise:The length measured as the spaceship flies by is 99.99 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.4, page no. 20"
+      "Example 1.2, Page 12"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#variable declaration\n",
-      "\n",
-      "Lp = 435            #proper length of the spaceship(m)\n",
-      "c  = 3 * 10**8      #speed of light(m/s)\n",
-      "v  = 0.97 * c       #speed of the spaceship(m/s)\n",
+      "#initiation of variable\n",
+      "Mp=1.007276 ; Me=5.4858*10**-4; #mass of proton and electron in terms of U\n",
       "\n",
       "#calculation\n",
-      "\n",
-      "L = Lp * math.sqrt((1-(v/c)**2))   #altitude of the spaceship (m)\n",
+      "Mt=Mp+Me;                      #Total mass= sum of above masses  \n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"The altitude of the spaceship as measured by an observer in the spaceship is\",round(L),\"m\""
+      "print\"The combined mass of an electron and a proton was found out to be in U.\",round(Mt,3);\n"
      ],
      "language": "python",
      "metadata": {},
@@ -165,7 +80,7 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The altitude of the spaceship as measured by an observer in the spaceship is 106.0 m\n"
+        "The combined mass of an electron and a proton was found out to be in U. 1.008\n"
        ]
       }
      ],
@@ -176,213 +91,20 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 1.5, page no. 20"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#variable declaration\n",
-      "\n",
-      "Lp = 50             #horizontal length of the spaceship (m)\n",
-      "c  = 3 * 10**8      #speed of light (m/s)\n",
-      "v  = 0.95 * c       #speed of the observer (m/s)\n",
-      "\n",
-      "#calculation\n",
-      "\n",
-      "L = Lp * math.sqrt((1-(v/c)**2))        #length (m)\n",
-      "\n",
-      "#result\n",
-      "\n",
-      "print \"The horizontal length measured as seen by the observer\",round(L,1),\"m.\\nThe 25m vertical height is unchanged because it is perpendicular to the direction of relative motion between the observer and the spaceship.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The horizontal length measured as seen by the observer 15.6 m.\n",
-        "The 25m vertical height is unchanged because it is perpendicular to the direction of relative motion between the observer and the spaceship.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.6, page no. 25"
+      "Example 1.3, Page 13"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "\n",
-      "#variable declaration\n",
-      "\n",
-      "Yobs = 475.0          #observed wavelength (nm)\n",
-      "Ysource = 394.0       #source wavelength (nm)\n",
-      "c = 3 * 10**8       #speed of light (m/s)\n",
-      "\n",
-      "#calculation\n",
-      "\n",
-      "v = c*(((Yobs**2)-(Ysource**2))/((Yobs**2)+(Ysource**2)))  #velocity (m/s)\n",
+      "#initiation of variable\n",
+      "h=6.621*10**-34 ; c=2.9979*10**8;                  # h is in J/s and c is  in m/s\n",
+      "hc=h*c*((10**9)/(1.6022*10**-19));               #1e=1.602*10^-19 J and 1 m=10^9 nm\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"The Hydra is receding from us at\",round(v/10**7,2),\"x 10^7 m/s\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Hydra is receding from us at 5.54 x 10^7 m/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.8, page no. 30"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "c = 3 * 10 **8      #speed of light (m/s)\n",
-      "v = 0.750 * c       #speed of spacecraft A relative to an observer on Earth(m/s)\n",
-      "ux = -0.850 *c      #speed of spacecraft B relative to an observer on Earth(m/s)\n",
-      "\n",
-      "#calculation\n",
-      "\n",
-      "ux1 = (ux-v)/(1-((ux*v)/(c**2)))  #velocity (m/s)\n",
-      "\n",
-      "#results\n",
-      "\n",
-      "print \"The velocity of B with respect to A is\",round(ux1/c,4),\"c\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The velocity of B with respect to A is -0.9771 c\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.9, page no. 30"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "c = 3 * 10 **8      #speed of light (m/s)\n",
-      "v = 0.800 * c       #velocity of the motorcycle with respect to the stationary observer (m/s)\n",
-      "ux1 = 0.700 *c      #velocity of the ball in the frame of reference of the motorcyclist (m/s)\n",
-      "\n",
-      "#calculation\n",
-      "\n",
-      "ux = (ux1+v)/(1+((ux1*v)/(c**2))) #velocity (m/s)\n",
-      "\n",
-      "#results\n",
-      "\n",
-      "print \"The velocity of the ball relative to the stationary observer is\",round(ux/c,4),\"c\"\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "c = 3 * 10 **8      #speed of light (m/s)\n",
-      "v = 0.800 * c       #velocity of the motorcycle with respect to the stationary observer (m/s)\n",
-      "ux1 = c             #velocity of the beam of light in the frame of reference of the motorcyclist (m/s)\n",
-      "\n",
-      "#calculation\n",
-      "\n",
-      "ux = (ux1+v)/(1+((ux1*v)/(c**2)))   #velocity (m/s)\n",
-      "\n",
-      "#results\n",
-      "\n",
-      "print \"The speed of the beam of light relative to the stationary observer is\",round(ux/c,4),\"c\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The velocity of the ball relative to the stationary observer is 0.9615 c\n",
-        "The speed of the beam of light relative to the stationary observer is 1.0 c\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1.10, page no. 30"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "c = 3 * 10 **8      #speed of light (m/s)\n",
-      "    #observations of stationary police officer\n",
-      "uxa = 0.75 * c # velocity of pack leader Alpha in x direction (m/s)\n",
-      "uya = 0        # velocity of pack leader Alpha in y direction (m/s)\n",
-      "uxb = 0        # velocity of pack leader Beta in x direction (m/s)\n",
-      "uyb = -0.9 * c # velocity of pack leader Beta in y direction (m/s)    \n",
-      "    \n",
-      "#calculation\n",
-      "\n",
-      "ux1 = (uxb-uxa)/(1-((uxb*uxa)/(c**2)))      #speed in X axis (m/s)\n",
-      "uy1 = (math.sqrt(1-((uxa**2)/(c**2)))*(uyb))#/(1-((uxb*uxa)/(c**2)))    #speed in Y axis (m/s)\n",
-      "u1 = math.sqrt(ux1**2 + uy1**2)             #total speed (m/s)\n",
-      "\n",
-      "#results\n",
-      "\n",
-      "print \"The speed of recession of Beta away from Alpha as observed by Alpha is then found to be\",round(u1/c,2),\"c\"\n"
+      "print \"The value of hc in  eV.nm is\",round(hc,4); \n",
+      "print 'Hence zero at the end is significant.';\n"
      ],
      "language": "python",
      "metadata": {},
@@ -391,11 +113,12 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The speed of recession of Beta away from Alpha as observed by Alpha is then found to be 0.96 c\n"
+        "The value of hc in  eV.nm is 1238.8651\n",
+        "Hence zero at the end is significant.\n"
        ]
       }
      ],
-     "prompt_number": 15
+     "prompt_number": 8
     }
    ],
    "metadata": {}
diff --git a/Modern_Physics/Chapter10.ipynb b/Modern_Physics/Chapter10.ipynb
index a9cbc233..6135c0d0 100755
--- a/Modern_Physics/Chapter10.ipynb
+++ b/Modern_Physics/Chapter10.ipynb
@@ -1,7 +1,7 @@
 {
  "metadata": {
-  "name": "",
-  "signature": "sha256:1d13a558dd2cc62e2c7ada350fc7a07d9283efcbc790578d0711fd6c96f50df0"
+  "name": "Chapter10",
+  "signature": "sha256:e4e2027717708d18dd95ce338ad24e83f0d7666653044656429dafb0b39af784"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -13,7 +13,7 @@
      "level": 1,
      "metadata": {},
      "source": [
-      "Chapter 10: Statistical Physics"
+      "Chapter 10:Statistical Physics"
      ]
     },
     {
@@ -21,64 +21,22 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 10.1, page no. 340"
+      "Example 10.2 Page 307"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "n1 = 1        #Ground state\n",
-      "n2 = 2        #First excited state\n",
-      "n3 = 3        #second excited state\n",
-      "T = 300       #room temperature(K)\n",
-      "kb = 8.617 * 10 **-5 #Boltzmann constant(eV/K)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "E1 = -13.6 / n1 ** 2\n",
-      "g1 = 2 * n1 ** 2\n",
-      "E2 = -13.6 / n2 ** 2\n",
-      "g2 = 2 * n2 ** 2\n",
-      "E3 = -13.6 / n3 ** 2\n",
-      "g3 = 2 * n3 ** 2\n",
-      "N3 = g3 * math.exp(-E3/(kb*T))\n",
-      "N2 = g2 * math.exp(-E2/(kb*T))\n",
-      "N1 = g1 * math.exp(-E1/(kb*T))\n",
-      "ratio1 = N2 / N1\n",
-      "ratio2 = N3 / N1\n",
-      "\n",
-      "#results\n",
-      "\n",
-      "print \"(a) We can see that n2/n1=\",round(ratio1),\"and n3/n1=\",round(ratio2),\"essentially all atoms are in ground state.\"\n",
-      "\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "T = 20000           #Temperature(K)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "N3 = g3 * math.exp(-E3/(kb*T))\n",
-      "N2 = g2 * math.exp(-E2/(kb*T))\n",
-      "N1 = g1 * math.exp(-E1/(kb*T))\n",
-      "ratio1 = N2 / N1\n",
-      "ratio2 = N3 / N1\n",
+      "#initiation of variable\n",
+      "from math import sqrt\n",
+      "#The solution is purely theoretical and involves a lot of approximations.\n",
+      "print\"The value of shift in frequency was found out to be delf=7.14*fo*10^-7*sqrt(T) for a star composing of hydrogen atoms at a temperature T.\";\n",
+      "T=6000.0;                       #temperature for sun\n",
+      "delf=7.14*10**-7*sqrt(T);#change in frequency\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"(b) n2/n1=\",round(ratio1,5),\"and n3/n1=\",round(ratio2,5)\n",
-      "\n",
-      "\n",
-      "ratio3 = N3 / N2\n",
-      "\n",
-      "print \"(c) S(3->2)/S(2->1)=\",round(ratio3,2)"
+      "print\"The value of frequency shift for sun(at 6000 deg. temperature) comprsing of hydrogen atoms is\",delf,\" times the frequency of the light.\""
      ],
      "language": "python",
      "metadata": {},
@@ -87,63 +45,42 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "(a) We can see that n2/n1= 0.0 and n3/n1= 0.0 essentially all atoms are in ground state.\n",
-        "(b) n2/n1= 0.01076 and n3/n1= 0.00809\n",
-        "(c) S(3->2)/S(2->1)= 0.75\n"
+        "The value of shift in frequency was found out to be delf=7.14*fo*10^-7*sqrt(T) for a star composing of hydrogen atoms at a temperature T.\n",
+        "The value of frequency shift for sun(at 6000 deg. temperature) comprsing of hydrogen atoms is 5.53062021838e-05  times the frequency of the light.\n"
        ]
       }
      ],
-     "prompt_number": 3
+     "prompt_number": 2
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 10.2, page no. 345"
+      "Example 10.3 Page 309"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "from math import sqrt,pi, exp, log\n",
+      "kT=0.0252;E=10.2                              # at room temperature, kT=0.0252 standard value and given value of E\n",
       "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "m = 3.34 * 10 ** -27    #mass of hydrogen(kg)\n",
-      "kbT = 3.77 * 10 ** -21  #kb * T (eV)\n",
-      "N = 6.02 * 10 ** 23     #avogadro's number\n",
-      "V = 22.4 * 10 ** -3     #Volume of H2 gas (m^3)\n",
-      "h = 1.055 * 10 ** -34   #Planck's constant (J.s)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "r = (N/V)* h ** 3 / (8 * (m * kbT)**1.5)\n",
+      "#calculation\n",
+      "n2=2;n1=1; g2=2*(n2**2);g1=2*(n1**2);           #values for ground and excited states\n",
+      "t=(g2/g1)*exp(-E/kT);                         #fraction of atoms\n",
       "\n",
       "#result\n",
+      "print\"The number of hydrogen atoms required is %.1e\" %(1.0/t),\" which weighs %.0e\" %((1/t)*(1.67*10**-27)),\"Kg\"\n",
       "\n",
-      "print \"(a) (N/V)h^3/(8*(mkbT)^3/2)=\",round(r/10**-8,2),\"X10^-8 is much less than 1, we conclude that even hydrogen is described by Maxwell-Boltzmann statistics.\"\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "d = 10.5     #density of silver (g/cm^3)\n",
-      "mw = 107.9   #Molar weight of silver (g/mol)\n",
-      "me = 9.109*10**-31 #mass of electron(kg)\n",
-      "kbT = 4.14 * 10 ** -21 #kb*T (J)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "Ns = (d/mw)* N * 10 ** 6\n",
-      "r = (Ns)* h ** 3 / (8 * (me * kbT)**1.5)\n",
+      "#partb\n",
+      "t=0.1/0.9;k=8.65*10**-5                          #fracion of atoms in case-2 is given\n",
+      "T=-E/(log(t/(g2/g1))*k);                        #temperature\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"(b) (N/V)h^3/(8*(mkbT)^3/2)=\",round(r/8,2),\"is greater than 1, we conclude that the Maxwell-Boltzmann statistics does not hold for electrons of silver.\""
+      "print\"The value of temperature at which 1/10 atoms are in excited state in K is %.1e\" %round(T,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -152,46 +89,77 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "(a) (N/V)h^3/(8*(mkbT)^3/2)= 8.83 X10^-8 is much less than 1, we conclude that even hydrogen is described by Maxwell-Boltzmann statistics.\n",
-        "(b) (N/V)h^3/(8*(mkbT)^3/2)= 4.64 is greater than 1, we conclude that the Maxwell-Boltzmann statistics does not hold for electrons of silver.\n"
+        "The number of hydrogen atoms required is 1.5e+175  which weighs 3e+148 Kg\n",
+        "The value of temperature at which 1/10 atoms are in excited state in K is 3.3e+04\n"
        ]
       }
      ],
-     "prompt_number": 5
+     "prompt_number": 2
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 10.3, page no. 352"
+      "Example 10.4 Page 311"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      " \n",
-      "\n",
-      "import math\n",
+      "#initiation of variable\n",
+      "from math import log\n",
+      "#theoretical part a\n",
+      "print'The energy of interaction with magnetic field is given by uB and the degeneracy of the states are +-1/2 which are identical.\\nThe ratio is therefore pE2/pE1 which gives e^(-2*u*B/k*T)';\n",
+      "#partb\n",
+      "uB=5.79*10**-4;        #for a typical atom\n",
+      "t=1.1;k=8.65*10**-5;   #ratio and constant k\n",
       "\n",
-      "#Variable Declaration\n",
+      "#calculation\n",
+      "T=2*uB/(log(t)*k);    #temperature\n",
       "\n",
-      "kB = 8.62 * 10 ** -5    #Boltzmann constant(eV/K)\n",
-      "T1 = 3000               #Cavity walls temperature(K)\n",
-      "T2 = 3.00               #Cavity walls temperature(K)\n",
-      "hc = 1.24 * 10 ** -4    #product of planck's constant and speed of light (eV.cm)\n",
-      "integration = 2.40      #value of integral(z^2/e^z-1,0,+inf)\n",
-      "\n",
-      "#Calculation\n",
+      "#result\n",
+      "print\"The value of temperature ar which the given ratio exists in K is\",round(T,3);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The energy of interaction with magnetic field is given by uB and the degeneracy of the states are +-1/2 which are identical.\n",
+        "The ratio is therefore pE2/pE1 which gives e^(-2*u*B/k*T)\n",
+        "The value of temperature ar which the given ratio exists in K is 140.46\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.5 Page 313"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "from math import pi\n",
+      "p=0.971; A=6.023*10**23; m=23.0;    # various given values and constants\n",
       "\n",
-      "NbyV_at_3000 = 8* math.pi * (kB * T1/hc)**3 * integration\n",
-      "NbyV_at_3 = 8* math.pi * (kB * T2/hc)**3 * integration\n",
+      "#calculation\n",
+      "c= (p*A/m)*10**6;                   # atoms per unit volume\n",
+      "hc=1240.0; mc2=0.511*10**6;           # hc=1240 eV.nm\n",
+      "E= ((hc**2)/(2*mc2))*(((3/(8*pi))*c)**(2.0/3)); #value of fermi energy\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"N/V at 3000K is\",round(NbyV_at_3000/10**11,2),\"X 10^11 photons/cm^3. Likewise N/V at 3.00 K is\",round(NbyV_at_3/10**2,2),\"X 10^2 photons/cm^3\"\n",
-      "print \"Therefore the photon density decreases by a factor of\",round(NbyV_at_3000/NbyV_at_3/10**9),\" X 10^9 when the temperature drops from 3000K to 3.00K\""
+      "print\"The fermi energy for sodium is\",round(E*10**-18,4),\"eV\";#multiply by 10^-18 to convert metres^2 term to nm^2"
      ],
      "language": "python",
      "metadata": {},
@@ -200,12 +168,11 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "N/V at 3000K is 5.47 X 10^11 photons/cm^3. Likewise N/V at 3.00 K is 5.47 X 10^2 photons/cm^3\n",
-        "Therefore the photon density decreases by a factor of 1.0  X 10^9 when the temperature drops from 3000K to 3.00K\n"
+        "The fermi energy for sodium is 3.1539 eV\n"
        ]
       }
      ],
-     "prompt_number": 7
+     "prompt_number": 12
     }
    ],
    "metadata": {}
diff --git a/Modern_Physics/Chapter10_1.ipynb b/Modern_Physics/Chapter10_1.ipynb
new file mode 100755
index 00000000..c1e9ff91
--- /dev/null
+++ b/Modern_Physics/Chapter10_1.ipynb
@@ -0,0 +1,114 @@
+{
+ "metadata": {
+  "name": "MP-10"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Statistical Physics"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 10.2 Page 307"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\n#The solution is purely theoretical and involves a lot of approximations.\nprint\"The value of shift in frequency was found out to be delf=7.14*fo*10^-7*sqrt(T) for a star composing of hydrogen atoms at a temperature T.\";\nT=6000.0;                       #temperature for sun\ndelf=7.14*10**-7*sqrt(T);#change in frequency\n\n#result\nprint\"The value of frequency shift for sun(at 6000 deg. temperature) comprsing of hydrogen atoms is\",delf,\" times the frequency of the light.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of shift in frequency was found out to be delf=7.14*fo*10^-7*sqrt(T) for a star composing of hydrogen atoms at a temperature T.\nThe value of frequency shift for sun(at 6000 deg. temperature) comprsing of hydrogen atoms is 5.53062021838e-05  times the frequency of the light.\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 10.3 Page 309"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt,pi, exp, log\nkT=0.0252;E=10.2                              # at room temperature, kT=0.0252 standard value and given value of E\n\n#calculation\nn2=2;n1=1; g2=2*(n2**2);g1=2*(n1**2);           #values for ground and excited states\nt=(g2/g1)*exp(-E/kT);                         #fraction of atoms\n\n#result\nprint\"The number of hydrogen atoms required is %.1e\" %(1.0/t),\" which weighs %.1e\" %((1/t)*(1.67*10**-27)),\"Kg\"\n\n#partb\nt=0.1/0.9;k=8.65*10**-5                          #fracion of atoms in case-2 is given\nT=-E/(log(t/(g2/g1))*k);                        #temperature\n\n#result\nprint\"The value of temperature at which 1/10 atoms are in excited state in K is %.1e\" %round(T,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "ename": "TypeError",
+       "evalue": "can't multiply sequence by non-int of type 'float'",
+       "output_type": "pyerr",
+       "traceback": [
+        "\u001b[1;31m---------------------------------------------------------------------------\u001b[0m\n\u001b[1;31mTypeError\u001b[0m                                 Traceback (most recent call last)",
+        "\u001b[1;32m<ipython-input-2-0c4f78dafdc9>\u001b[0m in \u001b[0;36m<module>\u001b[1;34m()\u001b[0m\n\u001b[0;32m      8\u001b[0m \u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m      9\u001b[0m \u001b[1;31m#result\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[1;32m---> 10\u001b[1;33m \u001b[1;32mprint\u001b[0m\u001b[1;34m\"The number of hydrogen atoms required is %.1e\"\u001b[0m \u001b[1;33m%\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;36m1.0\u001b[0m\u001b[1;33m/\u001b[0m\u001b[0mt\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;34m\" which weighs %.1e\"\u001b[0m \u001b[1;33m%\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;36m1\u001b[0m\u001b[1;33m/\u001b[0m\u001b[0mt\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m*\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;36m1.67\u001b[0m\u001b[1;33m*\u001b[0m\u001b[1;36m10\u001b[0m\u001b[1;33m**\u001b[0m\u001b[1;33m-\u001b[0m\u001b[1;36m27\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;34m\"Kg\"\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m\u001b[0;32m     11\u001b[0m \u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m     12\u001b[0m \u001b[1;31m#partb\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n",
+        "\u001b[1;31mTypeError\u001b[0m: can't multiply sequence by non-int of type 'float'"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The number of hydrogen atoms required is 1.5e+175"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 10.4 Page 311"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import log\n#theoritical part a\nprint'The energy of interaction with magnetic field is given by uB and the degeneracy of the states are +-1/2 which are identical.\\nThe ratio is therefore pE2/pE1 which gives e^(-2*u*B/k*T)';\n#partb\nuB=5.79*10**-4;        #for a typical atom\nt=1.1;k=8.65*10**-5;   #ratio and constant k\n\n#calculation\nT=2*uB/(log(t)*k);    #temperature\n\n#result\nprint\"The value of temperature ar which the given ratio exists in K is\",round(T,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The energy of interaction with magnetic field is given by uB and the degeneracy of the states are +-1/2 which are identical.\nThe ratio is therefore pE2/pE1 which gives e^(-2*u*B/k*T)\nThe value of temperature ar which the given ratio exists in K is 140.46\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 10.5 Page 313"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi\np=0.971; A=6.023*10**23; m=23.0;    # various given values and constants\n\n#calculation\nc= (p*A/m)*10**6;                   # atoms per unit volume\nhc=1240.0; mc2=0.511*10**6;           # hc=1240 eV.nm\nE= ((hc**2)/(2*mc2))*(((3/(8*pi))*c)**(2.0/3)); #value of fermi energy\n\n#result\nprint\"The fermi energy for sodium is\",round(E*10**-18,4),\"eV\";#multiply by 10^-18 to convert metres^2 term to nm^2",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The fermi energy for sodium is 3.1539 eV\n"
+      }
+     ],
+     "prompt_number": 12
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter10_2.ipynb b/Modern_Physics/Chapter10_2.ipynb
new file mode 100755
index 00000000..c1e9ff91
--- /dev/null
+++ b/Modern_Physics/Chapter10_2.ipynb
@@ -0,0 +1,114 @@
+{
+ "metadata": {
+  "name": "MP-10"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Statistical Physics"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 10.2 Page 307"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\n#The solution is purely theoretical and involves a lot of approximations.\nprint\"The value of shift in frequency was found out to be delf=7.14*fo*10^-7*sqrt(T) for a star composing of hydrogen atoms at a temperature T.\";\nT=6000.0;                       #temperature for sun\ndelf=7.14*10**-7*sqrt(T);#change in frequency\n\n#result\nprint\"The value of frequency shift for sun(at 6000 deg. temperature) comprsing of hydrogen atoms is\",delf,\" times the frequency of the light.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of shift in frequency was found out to be delf=7.14*fo*10^-7*sqrt(T) for a star composing of hydrogen atoms at a temperature T.\nThe value of frequency shift for sun(at 6000 deg. temperature) comprsing of hydrogen atoms is 5.53062021838e-05  times the frequency of the light.\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 10.3 Page 309"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt,pi, exp, log\nkT=0.0252;E=10.2                              # at room temperature, kT=0.0252 standard value and given value of E\n\n#calculation\nn2=2;n1=1; g2=2*(n2**2);g1=2*(n1**2);           #values for ground and excited states\nt=(g2/g1)*exp(-E/kT);                         #fraction of atoms\n\n#result\nprint\"The number of hydrogen atoms required is %.1e\" %(1.0/t),\" which weighs %.1e\" %((1/t)*(1.67*10**-27)),\"Kg\"\n\n#partb\nt=0.1/0.9;k=8.65*10**-5                          #fracion of atoms in case-2 is given\nT=-E/(log(t/(g2/g1))*k);                        #temperature\n\n#result\nprint\"The value of temperature at which 1/10 atoms are in excited state in K is %.1e\" %round(T,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "ename": "TypeError",
+       "evalue": "can't multiply sequence by non-int of type 'float'",
+       "output_type": "pyerr",
+       "traceback": [
+        "\u001b[1;31m---------------------------------------------------------------------------\u001b[0m\n\u001b[1;31mTypeError\u001b[0m                                 Traceback (most recent call last)",
+        "\u001b[1;32m<ipython-input-2-0c4f78dafdc9>\u001b[0m in \u001b[0;36m<module>\u001b[1;34m()\u001b[0m\n\u001b[0;32m      8\u001b[0m \u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m      9\u001b[0m \u001b[1;31m#result\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[1;32m---> 10\u001b[1;33m \u001b[1;32mprint\u001b[0m\u001b[1;34m\"The number of hydrogen atoms required is %.1e\"\u001b[0m \u001b[1;33m%\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;36m1.0\u001b[0m\u001b[1;33m/\u001b[0m\u001b[0mt\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;34m\" which weighs %.1e\"\u001b[0m \u001b[1;33m%\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;36m1\u001b[0m\u001b[1;33m/\u001b[0m\u001b[0mt\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m*\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;36m1.67\u001b[0m\u001b[1;33m*\u001b[0m\u001b[1;36m10\u001b[0m\u001b[1;33m**\u001b[0m\u001b[1;33m-\u001b[0m\u001b[1;36m27\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;34m\"Kg\"\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m\u001b[0;32m     11\u001b[0m \u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m     12\u001b[0m \u001b[1;31m#partb\u001b[0m\u001b[1;33m\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n",
+        "\u001b[1;31mTypeError\u001b[0m: can't multiply sequence by non-int of type 'float'"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The number of hydrogen atoms required is 1.5e+175"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 10.4 Page 311"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import log\n#theoritical part a\nprint'The energy of interaction with magnetic field is given by uB and the degeneracy of the states are +-1/2 which are identical.\\nThe ratio is therefore pE2/pE1 which gives e^(-2*u*B/k*T)';\n#partb\nuB=5.79*10**-4;        #for a typical atom\nt=1.1;k=8.65*10**-5;   #ratio and constant k\n\n#calculation\nT=2*uB/(log(t)*k);    #temperature\n\n#result\nprint\"The value of temperature ar which the given ratio exists in K is\",round(T,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The energy of interaction with magnetic field is given by uB and the degeneracy of the states are +-1/2 which are identical.\nThe ratio is therefore pE2/pE1 which gives e^(-2*u*B/k*T)\nThe value of temperature ar which the given ratio exists in K is 140.46\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 10.5 Page 313"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi\np=0.971; A=6.023*10**23; m=23.0;    # various given values and constants\n\n#calculation\nc= (p*A/m)*10**6;                   # atoms per unit volume\nhc=1240.0; mc2=0.511*10**6;           # hc=1240 eV.nm\nE= ((hc**2)/(2*mc2))*(((3/(8*pi))*c)**(2.0/3)); #value of fermi energy\n\n#result\nprint\"The fermi energy for sodium is\",round(E*10**-18,4),\"eV\";#multiply by 10^-18 to convert metres^2 term to nm^2",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The fermi energy for sodium is 3.1539 eV\n"
+      }
+     ],
+     "prompt_number": 12
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter11.ipynb b/Modern_Physics/Chapter11.ipynb
index 65f716cb..8b1cca27 100755
--- a/Modern_Physics/Chapter11.ipynb
+++ b/Modern_Physics/Chapter11.ipynb
@@ -1,7 +1,6 @@
 {
  "metadata": {
-  "name": "",
-  "signature": "sha256:e6be7abf8c97801180567f4f6cb83dc9e729f9c81204b1adfe4e1f54ff45dd36"
+  "name": "Chapter11"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -13,7 +12,7 @@
      "level": 1,
      "metadata": {},
      "source": [
-      "Chapter 11: Molecular Structure"
+      "Chapter 11:Solid State Physics"
      ]
     },
     {
@@ -21,46 +20,28 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 11.1, page no. 380"
+      "Example 11.1, Page 346"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "c=769.0*10**3; Na=6.023*10**23; JeV=1.6*10**-19; #various constants and given values\n",
       "\n",
-      "import math\n",
+      "#calculation\n",
+      "Be=c/(Na*JeV);        #Binding energy of an ion pair in the lattice\n",
       "\n",
-      "#Variable Declaration\n",
+      "#result\n",
+      "print\"The experimental value was found out to be in eV.\",round(Be,5);\n",
       "\n",
-      "f = 1.15 * 10 ** 11     # Frequency(Hz)\n",
-      "h = 1.055 * 10 ** -34   # Planck's constant(J.s)\n",
-      " \n",
-      "#Calculation\n",
+      "#partb\n",
+      "n=9.0;a=1.7476; R=0.281; k= 1.44;     #Given values and consstants\n",
+      "Bc=k*a*(1-(1/n))/R;                  #ionic binding energy experimentally\n",
       "\n",
-      "w = 2 * math.pi * f\n",
-      "Icm = h / w\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print \"(a) The moment of inertia is\",round(Icm/10 **-46 , 2),\"X 10^-46 kg.m^2.\"\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "Ac = 12.0           #atomic mass of carbon (u)\n",
-      "Ao = 16.0           #atomic mass of oxygen (u)\n",
-      "u = 1.66 * 10 ** -27#(kg)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "mu = Ac* Ao /(Ac+Ao)\n",
-      "R0 = math.sqrt(Icm/(mu*u))\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print \"(b) The bond length of the molecule is\",round(R0/10**-9,3),\"nm.\""
+      "#result\n",
+      "print\"The calculated value of the binding energy in eV.is\",round(Bc,4);\n"
      ],
      "language": "python",
      "metadata": {},
@@ -69,8 +50,42 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "(a) The moment of inertia is 1.46 X 10^-46 kg.m^2.\n",
-        "(b) The bond length of the molecule is 0.113 nm.\n"
+        "The experimental value was found out to be in eV. 7.97983\n",
+        "The calculated value of the binding energy in eV.is 7.9606\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2, Page 350"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "a=3.61;# amount of energy required to remove an electron from Cl- ion\n",
+      "b=-5.14 #amount of energy returned when an electron is added to Na+ ion\\\n",
+      "c=7.98 #binding energy of NaCl atom\n",
+      "\n",
+      "#calculation\n",
+      "E=a+b+c  #sum of all the energies\n",
+      "print\"The net energy to be supplied in eV is\",round(E,3);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The net energy to be supplied in eV is 6.45\n"
        ]
       }
      ],
@@ -81,44 +96,38 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 11.2, page no. 383"
+      "Example 11.3, Page 355"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "from math import exp,sqrt\n",
+      "Na=6.023*10**23; p=8.96*10**3; M=63.5*10**-3;   #Na=avagadro's number,p=density,M=molar mass\n",
       "\n",
+      "#calculation\n",
+      "n= p*Na/M;                                   #density of charge carriers\n",
       "\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "f = 6.42 * 10 ** 13         #frequency(Hz)\n",
-      "mu = 1.14 * 10 ** -26       #(kg)\n",
-      " \n",
-      "#Calculation\n",
-      "\n",
-      "w = 2 * math.pi * f\n",
-      "K = mu * w**2\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print \"(a) The force constant for this molecule is \",round(K/10**3,2),\"X 10^3 N/m.\"\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Variable Declaration\n",
+      "#result'\n",
+      "print\"The density of charge carriers in copper in atoms/m3 is %.1e\" %round(n,3);\n",
       "\n",
-      "h = 1.055 * 10 ** -34   #Planck's constant (J.s)\n",
+      "s=5.88*10**7;m=9.11*10**-31;e=1.6*10**-19;    #charge & mass of an electron,resistance per unit length\n",
+      "t= s*m/(n*e**2);                            #average time between collisions\n",
       "\n",
-      "#Calculation\n",
+      "#result\n",
+      "print \"The average time between collisions of conducting electrons in sec.is %.1e\" %t\n",
       "\n",
-      "A = (h/(mu * w)) ** 0.5\n",
+      "#partb\n",
+      "Ef=7.03*1.6*10**-19;           #converting given enrgy to J\n",
       "\n",
-      "#Results\n",
+      "#calculation\n",
+      "Vf=sqrt(2*Ef/m);               #fermi velocity\n",
+      "l=Vf*t;                      #mean free path\n",
       "\n",
-      "print \"(b) The Vibration amplitude is\",round(A/10**-9,5),\"nm.\""
+      "#result\n",
+      "print \"The average mean free path is\",l,\"m =\",round(l*10**9,3),\" nm\"\n"
      ],
      "language": "python",
      "metadata": {},
@@ -127,12 +136,13 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "(a) The force constant for this molecule is  1.85 X 10^3 N/m.\n",
-        "(b) The Vibration amplitude is 0.00479 nm.\n"
+        "The density of charge carriers in copper in atoms/m3 is 8.5e+28\n",
+        "The average time between collisions of conducting electrons in sec.is 2.5e-14\n",
+        "The average mean free path is 3.8690296096e-08 m = 38.69  nm\n"
        ]
       }
      ],
-     "prompt_number": 6
+     "prompt_number": 1
     }
    ],
    "metadata": {}
diff --git a/Modern_Physics/Chapter11_1.ipynb b/Modern_Physics/Chapter11_1.ipynb
new file mode 100755
index 00000000..385146de
--- /dev/null
+++ b/Modern_Physics/Chapter11_1.ipynb
@@ -0,0 +1,83 @@
+{
+ "metadata": {
+  "name": "MP-11"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Solid State Physics"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 11.1 Page 346"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nc=769.0*10**3; Na=6.023*10**23; JeV=1.6*10**-19; #various constants and given values\n\n#calculation\nBe=c/(Na*JeV);        #Binding energy of an ion pair in the lattice\n\n#result\nprint\"The experimental value was found out to be in eV.\",round(Be,5);\n\n#partb\nn=9.0;a=1.7476; R=0.281; k= 1.44;     #Given values and consstants\nBc=k*a*(1-(1/n))/R;                  #ionic binding energy eperimentally\n\n#result\nprint\"The calculated value of the binding energy in eV.is\",round(Bc,4);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The experimental value was found out to be in eV. 7.97983\nThe calculated value of the binding energy in eV.is 7.9606\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 11.2 Page 350"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\na=3.61;# amount of energy required to remove an electron from Cl- ion\nb=-5.14 #amount of energy returned when an electron is added to Na+ ion\\\nc=7.98 #binding energy of NaCl atom\n\n#calculation\nE=a+b+c  #suom of all the energies\nprint\"The net energy to be supplied in eV is\",round(E,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The net energy to be supplied in eV is 6.45\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 11.3 Page 355"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import exp,sqrt\nNa=6.023*10**23; p=8.96*10**3; M=63.5*10**-3;   #Na=avagadro's number,p=density,M=molar mass\n\n#calculation\nn= p*Na/M;                                   #density of charge carriers\n\n#result'\nprint\"The density of charge carriers in copper in atoms/m3 is %.1e\" %round(n,3);\n\ns=5.88*10**7;m=9.11*10**-31;e=1.6*10**-19;    #charge & mass of an electron,resistance per unit length\nt= s*m/(n*e**2);                            #average time between collisions\n\n#result\nprint \"The average time between collisions of conducting electrons in sec.is %.1e\" %t\n\n#partb\nEf=7.03*1.6*10**-19;           #converting given enrgy to J\n\n#calculation\nVf=sqrt(2*Ef/m);               #fermi velocity\nl=Vf*t;                      #mean free path\n\n#result\nprint \"The average mean free path is\",l,\"m =\",round(l*10**9,3),\" nm\"\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The density of charge carriers in copper in atoms/m3 is 8.5e+28\nThe average time between collisions of conducting electrons in sec.is 2.5e-14\nThe average mean free path is 3.8690296096e-08 m = 38.69  nm\n"
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter11_2.ipynb b/Modern_Physics/Chapter11_2.ipynb
new file mode 100755
index 00000000..385146de
--- /dev/null
+++ b/Modern_Physics/Chapter11_2.ipynb
@@ -0,0 +1,83 @@
+{
+ "metadata": {
+  "name": "MP-11"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Solid State Physics"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 11.1 Page 346"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nc=769.0*10**3; Na=6.023*10**23; JeV=1.6*10**-19; #various constants and given values\n\n#calculation\nBe=c/(Na*JeV);        #Binding energy of an ion pair in the lattice\n\n#result\nprint\"The experimental value was found out to be in eV.\",round(Be,5);\n\n#partb\nn=9.0;a=1.7476; R=0.281; k= 1.44;     #Given values and consstants\nBc=k*a*(1-(1/n))/R;                  #ionic binding energy eperimentally\n\n#result\nprint\"The calculated value of the binding energy in eV.is\",round(Bc,4);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The experimental value was found out to be in eV. 7.97983\nThe calculated value of the binding energy in eV.is 7.9606\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 11.2 Page 350"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\na=3.61;# amount of energy required to remove an electron from Cl- ion\nb=-5.14 #amount of energy returned when an electron is added to Na+ ion\\\nc=7.98 #binding energy of NaCl atom\n\n#calculation\nE=a+b+c  #suom of all the energies\nprint\"The net energy to be supplied in eV is\",round(E,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The net energy to be supplied in eV is 6.45\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 11.3 Page 355"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import exp,sqrt\nNa=6.023*10**23; p=8.96*10**3; M=63.5*10**-3;   #Na=avagadro's number,p=density,M=molar mass\n\n#calculation\nn= p*Na/M;                                   #density of charge carriers\n\n#result'\nprint\"The density of charge carriers in copper in atoms/m3 is %.1e\" %round(n,3);\n\ns=5.88*10**7;m=9.11*10**-31;e=1.6*10**-19;    #charge & mass of an electron,resistance per unit length\nt= s*m/(n*e**2);                            #average time between collisions\n\n#result\nprint \"The average time between collisions of conducting electrons in sec.is %.1e\" %t\n\n#partb\nEf=7.03*1.6*10**-19;           #converting given enrgy to J\n\n#calculation\nVf=sqrt(2*Ef/m);               #fermi velocity\nl=Vf*t;                      #mean free path\n\n#result\nprint \"The average mean free path is\",l,\"m =\",round(l*10**9,3),\" nm\"\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The density of charge carriers in copper in atoms/m3 is 8.5e+28\nThe average time between collisions of conducting electrons in sec.is 2.5e-14\nThe average mean free path is 3.8690296096e-08 m = 38.69  nm\n"
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter12.ipynb b/Modern_Physics/Chapter12.ipynb
index 3cc3e98a..904fea62 100755
--- a/Modern_Physics/Chapter12.ipynb
+++ b/Modern_Physics/Chapter12.ipynb
@@ -1,7 +1,7 @@
 {
  "metadata": {
-  "name": "",
-  "signature": "sha256:6dfe8c3f0e395e550057fecf65810f55f95ea60853478f130ac485dd9541147f"
+  "name": "Chapter12",
+  "signature": "sha256:a11a7c1ba314b90500bccd87a2a9d038a55aefe70077541468dd4c2624276146"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -13,7 +13,7 @@
      "level": 1,
      "metadata": {},
      "source": [
-      "Chapter 12: The Solid State"
+      "Chapter 12:Nuclear Structure and Reactivity"
      ]
     },
     {
@@ -21,61 +21,218 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 12.1, page no. 418"
+      "Example 12.1, Page 375"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "Z=2;A=4;N=A-Z;         # Given values\n",
       "\n",
-      "#Variable Declaration\n",
+      "#result\n",
+      "print\"The following method of representing atoms is followed throughout the chapter\\n\\t\\t x,y,z\\n where x=atomic number y=mass number z= Neutron Number S=symbol of the atom\\n\\n\"\n",
+      "print\"The helium can be reperesented as He-- \",Z,A,N;\n",
       "\n",
-      "kb = 1.38 * 10 **-23        #Boltzmann constant (J/K)\n",
-      "T = 300                     # room temperature (K)\n",
-      "me = 9.11 * 10 ** -31       # mass of electron (kg)\n",
-      "d = 8.96 * 10**6            # density of copper (g/m^3)\n",
-      "N = 6.023 * 10 **23         #avagodro's number (atoms/mole)\n",
-      "mw = 63.5                   #molar weight (g)\n",
-      "A = 4 * 10 ** -6            #area (m^2)\n",
-      "I = 10                      # current (A)\n",
-      "e = 1.6 * 10 ** -19         # charge of electron(C)\n",
+      "#part b\n",
+      "Z=50.0;N=66.0;A=Z+N;       # Given values and standard formulae\n",
+      "print\"The Tin can be reperesented as Sn-- \",Z,A,N;\n",
       "\n",
-      "#Calculation\n",
       "\n",
-      "Vrms = (3*kb*T/me)**0.5\n",
-      "n = N * d / mw\n",
-      "Vd = I /(n*e*A)\n",
+      "#part c\n",
+      "A=235;N=143;Z=A-N;\n",
+      "print\"The Uranium can be reperesented as U-- \",Z,A,N;"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The following method of representing atoms is followed throughout the chapter\n",
+        "\t\t x,y,z\n",
+        " where x=atomic number y=mass number z= Neutron Number S=symbol of the atom\n",
+        "\n",
+        "\n",
+        "The helium can be reperesented as He--  2 4 2\n",
+        "The Tin can be reperesented as Sn--  50.0 116.0 66.0\n",
+        "The Uranium can be reperesented as U--  92 235 143\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2, Page 377"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "r0=1.2;         #standard value.\n",
+      "A=12.0; \n",
+      "r= r0*A**(1.0/3);\n",
       "\n",
-      "#Results\n",
+      "#result\n",
+      "print\"The value of mean radius for C in fm is\",round(r,3);\n",
       "\n",
-      "print \"(a) The ratio of drift speed to rms speed is \",round(Vd/Vrms/10**-9,2),\"X 10^-9.\"\n",
+      "#part2\n",
+      "A=70.0;        #given value\n",
+      "r= r0*A**(1.0/3);\n",
       "\n",
+      "#result\n",
+      "print\"The value of mean radius for C in fm is\",round(r,3);\n",
       "\n",
-      "#Variable Declaration\n",
+      "#part3\n",
+      "A=209;\n",
+      "r= r0*A**(1.0/3);\n",
       "\n",
-      "L = 2.6 * 10 ** -10       #interatomic distance(A')  \n",
+      "#result\n",
+      "print\"The value of mean radius for C in fm is\",round(r,3);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of mean radius for C in fm is 2.747\n",
+        "The value of mean radius for C in fm is 4.946\n",
+        "The value of mean radius for C in fm is 7.121\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.3, Page 379"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "from math import pi\n",
+      "m=1.67*10**-27; r0=1.2*10**-15; v=4*pi*(r0**3)/3.0     #standard values of mass radius and volume\n",
       "\n",
-      "#Calculation\n",
+      "#calculation\n",
+      "p=m/v;                                               #density \n",
       "\n",
-      "t = L / Vrms \n",
+      "#result\n",
+      "print\"Density of typical nucleus in kg/m3 is %.0e\" %p;\n",
       "\n",
-      "#Results\n",
+      "#part 2\n",
+      "r0=0.01;v=4*pi*(r0**3)/3.0;p=2.0*10**17;              #/hypothetical values\n",
+      "m1=p*v;                                  \n",
       "\n",
-      "print \"(b) The average time between collisions\",round(t/10**-15,1),\"X 10^-15 m/s.\"\n",
+      "#result\n",
+      "print\"The mass of the hypothetical nucleus would be in Kg %.0e\" %m1;\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Density of typical nucleus in kg/m3 is 2e+17\n",
+        "The mass of the hypothetical nucleus would be in Kg 8e+11\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.4, Page 380"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "N=30.0;Z=26.0;A=56.0;Mn=1.008665;Mp=1.007825;m=55.934939;c2=931.5; #given values and constants for case-1\n",
+      "B=((N*Mn)+(Z*Mp)-(m))*c2;                                  #binding energy(per nucleon)\n",
       "\n",
+      "#result\n",
+      "print\"Binding energy per nucleon for 26,56Fe30 in MeV is\",round(B/A,3);\n",
       "\n",
-      "#Variable Declaration\n",
+      "#part 2\n",
+      "N=146.0;Z=92.0;A=238.0;Mn=1.008665;Mp=1.007825;m=238.050785;c2=931.5;      #given values and constants for case-2\n",
+      "B=((N*Mn)+(Z*Mp)-(m))*c2;                             #binding energy(per nucleon)\n",
       "\n",
-      "T = 300             #Room temperature (K)\n",
+      "#result\n",
+      "print\"Binding energy per nucleon for 26,56Fe30 in MeV is\",round(B/A,3);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Binding nergy per nucleon for 26,56Fe30 in MeV is 8.79\n",
+        "Binding nergy per nucleon for 26,56Fe30 in MeV is 7.57\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.5, Page 382"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "from math import exp\n",
+      "t12=2.7*24*3600;              #converting days into seconds\n",
+      "w=0.693/t12;                  #lambda\n",
+      "\n",
+      "#result\n",
+      "print\"The decay constant in sec is %.2e\" %w; \n",
       "\n",
-      "#Calculation\n",
+      "#partb\n",
+      "print\"The decay constant is equal to probability of decay in one second hence %.2e\" %w;\n",
       "\n",
-      "sigma = n*e**2 * L /(3*kb*T*me)**0.5\n",
+      "#partc\n",
+      "m=10**-6;Na=6.023*10**23; M=198.0;     #given values and constants\n",
+      "N=m*Na/M;                        #number of atoms in the sample \n",
+      "Ao=w*N;             #activity\n",
       "\n",
-      "#Results\n",
+      "#result\n",
+      "print\"The activity was found out to be in Ci is %.2e\" %Ao,\"=0.244Ci\";        \n",
       "\n",
-      "print \"(c) The conductivity is\",round(sigma/10**6,1),\"X 10^6 (ohm.m)^-1.\""
+      "#partd\n",
+      "t=7*24*3600.0;         #given time\n",
+      "A=Ao*exp(-w*t);        #activity\n",
+      "\n",
+      "#result\n",
+      "print\"The activity after one week was found out to be in decays/sec %.1e\" %A;"
      ],
      "language": "python",
      "metadata": {},
@@ -84,39 +241,149 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "(a) The ratio of drift speed to rms speed is  1.57 X 10^-9.\n",
-        "(b) The average time between collisions 2.2 X 10^-15 m/s.\n",
-        "(c) The conductivity is 5.3 X 10^6 (ohm.m)^-1.\n"
+        "The decay constant in sec is 2.97e-06\n",
+        "The decay constant is equal to probability of decay in one second hence 2.97e-06\n",
+        "The activity was found out to be in Ci is 9.04e+09 =0.244Ci\n",
+        "The activity after one week was found out to be in decays/sec 1.5e+09\n"
        ]
       }
      ],
-     "prompt_number": 2
+     "prompt_number": 4
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 12.2, page no. 429"
+      "Example 12.6, Page 384"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation  of variable\n",
+      "t1=4.55*10**9;t2=7.04*10**8;       #given values of time at 2 different instants\n",
       "\n",
-      "#Variable Declaration\n",
+      "#calculation\n",
+      "age=t1/t2;\n",
+      "r=2**age;\n",
       "\n",
-      "V = 7            # voltage(V)\n",
-      "L = 5 * 10 ** -8 # mean free path (m)\n",
-      " \n",
-      "#Calculation\n",
-      "\n",
-      "E = V/L\n",
+      "#result\n",
+      "print \"The original rock hence contained\",round(r,3),\"Na atoms of 235U where Na is the Avagadro''s Number=6.023*10^23\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The original rock hence contained 88.222 Na atoms of 235U where Na is the Avagadro''s Number=6.023*10^23\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.7, Page 385"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "m236Ra=226.025403;\n",
+      "m222Rn=222.017571;\n",
+      "m4He=4.002603;c2=931.5; #mass of various elements and c2=c^2\n",
+      "\n",
+      "#calculation\n",
+      "Q=(m236Ra-m222Rn-m4He)*c2;#Q of the reaction\n",
+      "A=226.0              \n",
+      "K=((A-4)/A)*Q;                           #kinetic energy\n",
+      "\n",
+      "#result\n",
+      "print\"The kinetic energy of the alpha particle in Mev is\",round(K,3);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The kinetic energy of the alpha particle in Mev is 4.785\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.8, Page 387"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "m226Ra=226.025403; #mass of various elements\n",
+      "m212Pb=211.991871;\n",
+      "m14c=14.003242;\n",
+      "c2=931.5;       #value of c^2\n",
+      "\n",
+      "#calculation\n",
+      "Q=(m226Ra-m212Pb-m14c)*c2;      #Q of the reaction\n",
+      "\n",
+      "#result\n",
+      "print\"The value of Q for 14c emission in MeV is\",round(Q,3);\n",
+      "print\"The probability of 14c emission is 10^-9 times that of an alpha particle since the energy barrier for 14c emission is nearly 3 times higher and thicker.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of Q for 14c emission in MeV is 28.215\n",
+        "The probability of 14c emission is 10^-9 times that of an alpha particle since the energy barrier for 14c emission is nearly 3 times higher and thicker.\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.9, Page 389"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "m23Ne=22.994465; #mass of various elements\n",
+      "m23Na=22.989768;\n",
+      "c2=931.5;          #value of c^2\n",
       "\n",
-      "#Results\n",
+      "#calculation\n",
+      "Q=(m23Ne-m23Na)*c2;     #Q of the reaction\n",
       "\n",
-      "print \"The electric field required is\",round(E/10**8,1),\"X 10^8 V/m.\""
+      "#result\n",
+      "print \"Hence the maximum kinetic energy of the emitted electrons in MeV is\",round(Q,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -125,40 +392,53 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The electric field required is 1.4 X 10^8 V/m.\n"
+        "Hence the maximum kinetic energy of the emitted electrons in MeV is 4.375\n"
        ]
       }
      ],
-     "prompt_number": 4
+     "prompt_number": 16
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 12.3, page no. 436"
+      "Example 12.10, Page 390"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "m40K=39.963999;         #mass of various particles\n",
+      "m40Ca=39.962591;\n",
+      "c2=931.5;                    #value of c^2 in MeV\n",
       "\n",
+      "#calculation\n",
+      "Qb1=(m40K-m40Ca)*c2;       #Q value of the reaction\n",
       "\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration\n",
+      "#result\n",
+      "print\"The Q value for -VE beta emission in Mev in\",round(Qb1,3);\n",
       "\n",
-      "V = 1           #voltage(eV)\n",
-      "kBT = 0.025     # kB * T (eV)\n",
+      "#partb\n",
+      "m40K=39.963999;          #mass of various particles\n",
+      "m40Ar=39.962384;\n",
+      "me=0.000549;\n",
+      "Qb2=(m40K-m40Ar-2*me)*c2;         #Q value of the reaction\n",
       "\n",
-      "#Calculation\n",
+      "#result\n",
+      "print\"The Q value for +VE beta emission in Mev in\",round(Qb2,3);\n",
       "\n",
-      "ratio = (math.exp(V/kBT)-1)/(math.exp(-V/kBT)-1)\n",
+      "#partc\n",
+      "m40K=39.963999;\n",
+      "m40Ar=39.962384;\n",
       "\n",
-      "#Results\n",
+      "#calculation\n",
+      "Qec=(m40K-m40Ar)*c2;\n",
       "\n",
-      "print \"The ratio of forward to reverse current is\",round(ratio/10**17,1),\"X 10^17.\""
+      "#result\n",
+      "print\"The Q value for +VE beta emission in Mev in\",round(Qec,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -167,11 +447,183 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The ratio of forward to reverse current is -2.4 X 10^17.\n"
+        "The Q value for -VE beta emission in Mev in 1.312\n",
+        "The Q value for +VE beta emission in Mev in 0.482\n",
+        "The Q value for +VE beta emission in Mev in 1.504\n"
        ]
       }
      ],
-     "prompt_number": 6
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.11, Page 392"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "Mg=12.000000; #mass of the carbon atom in amu\n",
+      "c2=931.5; \n",
+      "Eg=4.43;     #given energy of gamma ray \n",
+      "Mex=Mg+(Eg/c2);     #mass in excited state\n",
+      "Me=0.000549;        #mass of an electron\n",
+      "\n",
+      "#calculation\n",
+      "Q=(12.018613-Mex-2*Me)*c2;     #Q of the particle\n",
+      "\n",
+      "#result\n",
+      "print\"The maximum value of kinetic energy is in MeV\",round(Q,3);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The maximum value of kinetic energy is in MeV 11.885\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.12, Page 393"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "m238U=238.050786; #mass of various quantities\n",
+      "m206Pb=205.974455;\n",
+      "m4He=4.002603;\n",
+      "c2=931.5;     #constants\n",
+      "Na=6.023*10**23;   #avagadro's number\n",
+      "\n",
+      "#calculation\n",
+      "Q=(m238U-m206Pb-8*m4He)*c2; \n",
+      "t12=(4.5)*10**9*(3.16*10**7);    #half life years to seconds conversion\n",
+      "w=0.693/t12;                 # lambeda\n",
+      "NoD=(Na/238)*w;      #number of decays\n",
+      "E=NoD*Q*(1.6*10**-19)*10**6;        #rate of liberation of energy,converting MeV to eV\n",
+      "\n",
+      "#result\n",
+      "print\"Rate of energy liberation in W is %.0e\" %E;"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rate of energy liberation in W is 1e-07\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.13, Page 395"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "from math import log\n",
+      "R=0.5;t12=4.5*10**9;                 #value of radius and half-life \n",
+      "t1=(t12/0.693)*log(1+(1/R));        #age of rock 1\n",
+      "R=1.0;\n",
+      "t2=(t12/0.693)*log(1+(1/R));       #age of rock-2\n",
+      "R=2.0\n",
+      "t3=(t12/0.693)*log(1+(1/R));       #age of rock 3\n",
+      "\n",
+      "#result\n",
+      "print\"The ages of rock samples in years respectively are %.1e\"%t1,\" %.1e\" %t2,\"  %.1e\" %t3;"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The ages of rock samples in years respectively are 7.1e+09  4.5e+09   2.6e+09\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.14, Page 397"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "from math import log\n",
+      "P=2.0*10**14; V=2.0*10**-14; R=8.314; T=295.0;Na=6.023*10**23;   #various constants and given values\n",
+      "\n",
+      "#calculation\n",
+      "n=P*V/(R*T);       #ideal gas law\n",
+      "N=Na*n;f=10**-12         #avagadaro's number and fraction of carbon molecules\n",
+      "t12=5730*3.16*(10**7);   #half life\n",
+      "A=(0.693/t12)*N*f;    #activity\n",
+      "D1w=A*7*24*60*60;       #decays per second\n",
+      "\n",
+      "#result\n",
+      "print\"the no. of decays per second is %.2e\" %A\n",
+      "print\"The no of decays per week is \",round(D1w);\n",
+      "    \n",
+      "    \n",
+      "#partb\n",
+      "c1=1420.0;             #concentration at instant 1\n",
+      "c2=D1w;              #concentration at instant 2\n",
+      "t12y=5730;          #half life\n",
+      "t=t12y*log(c2/c1)/0.693;            #age of the sample\n",
+      "\n",
+      "#result\n",
+      "print\"Age of the sample in years is\",round(t,3);\n",
+      "print\"the answer in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the no. of decays per second is 3.76e-03\n",
+        "The no of decays pers week is  2274.0\n",
+        "Age of the sample in years is 3892.57\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
     }
    ],
    "metadata": {}
diff --git a/Modern_Physics/Chapter12_1.ipynb b/Modern_Physics/Chapter12_1.ipynb
new file mode 100755
index 00000000..ccacecc1
--- /dev/null
+++ b/Modern_Physics/Chapter12_1.ipynb
@@ -0,0 +1,322 @@
+{
+ "metadata": {
+  "name": "MP-12"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Nuclear Structure and Reactivity"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.1 Page 375"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nZ=2;A=4;N=A-Z;         # Given values\n\n#result\nprint\"The following method of representing atoms is followed throughout the chapter\\n\\t\\t x,y,z\\n where x=atomic number y=mass number z= Neutron Number S=symbol of the atom\\n\\n\"\nprint\"The helium can be reperesented as He-- \",Z,A,N;\n\n#part b\nZ=50.0;N=66.0;A=Z+N;       # Given values and standard formulae\nprint\"The Tin can be reperesented as Sn-- \",Z,A,N;\n\n\n#part c\nA=235;N=143;Z=A-N;\nprint\"The Uranium can be reperesented as U-- \",Z,A,N;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The following method of representing atoms is followed throughout the chapter\n\t\t x,y,z\n where x=atomic number y=mass number z= Neutron Number S=symbol of the atom\n\n\nThe helium can be reperesented as He--  2 4 2\nThe Tin can be reperesented as Sn--  50.0 116.0 66.0\nThe Uranium can be reperesented as U--  92 235 143\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.2 Page 377"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nr0=1.2;         #standard value.\nA=12.0; \nr= r0*A**(1.0/3);\n\n#result\nprint\"The value of mean radius for C in fm is\",round(r,3);\n\n#part2\nA=70.0;        #given value\nr= r0*A**(1.0/3);\n\n#result\nprint\"The value of mean radius for C in fm is\",round(r,3);\n\n#part3\nA=209;\nr= r0*A**(1.0/3);\n\n#result\nprint\"The value of mean radius for C in fm is\",round(r,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of mean radius for C in fm is 2.747\nThe value of mean radius for C in fm is 4.946\nThe value of mean radius for C in fm is 7.121\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.3 Page 379"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi\nm=1.67*10**-27; r0=1.2*10**-15; v=4*pi*(r0**3)/3.0     #standard values of mass radius and volume\n\n#calculation\np=m/v;                                               #denisty \n\n#result\nprint\"Density of typical nucleus in kg/m3 is %.1e\" %p;\n\n#part 2\nr0=0.01;v=4*pi*(r0**3)/3.0;p=2.0*10**17;              #/hypothetical values\nm1=p*v;                                  \n\n#result\nprint\"The mass of the hypothetical nucleus would be in Kg %.1e\" %m1;\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Density of typical nucleus in kg/m3 is 2.3e+17\nThe mass of the hypothetical nucleus would be in Kg 8.4e+11\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.4 Page 380"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nN=30.0;Z=26.0;A=56.0;Mn=1.008665;Mp=1.007825;m=55.934939;c2=931.5; #given values and constants for case-1\nB=((N*Mn)+(Z*Mp)-(m))*c2;                                  #binding energy(per nucleon)\n\n#result\nprint\"Binding nergy per nucleon for 26,56Fe30 in MeV is\",round(B/A,3);\n\n#part 2\nN=146.0;Z=92.0;A=238.0;Mn=1.008665;Mp=1.007825;m=238.050785;c2=931.5;      #given values and constants for case-2\nB=((N*Mn)+(Z*Mp)-(m))*c2;                             #binding energy(per nucleon)\n\n#result\nprint\"Binding nergy per nucleon for 26,56Fe30 in MeV is\",round(B/A,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Binding nergy per nucleon for 26,56Fe30 in MeV is 8.79\nBinding nergy per nucleon for 26,56Fe30 in MeV is 7.57\n"
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.5 Page 382"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import exp\nt12=2.7*24*3600;              #converting days into seconds\nw=0.693/t12;                  #lambeda\n\n#result\nprint\"The decay constant in sec is %.2e\" %w; \n\n#partb\nprint\"The decay constant is equal to probability of decay in one second hence %.2e\" %w;\n\n#partc\nm=10**-6;Na=6.023*10**23; M=198.0;     #given values and constants\nN=m*Na/M;                        #number of atoms in the sample \nAo=w*N;             #activity\n\n#result\nprint\"The activity was found out to be in Ci is %.2e\" %Ao;        \n\n#partd\nt=7*24*3600.0;         #given time\nA=Ao*exp(-w*t);        #activity\n\n#result\nprint\"The activity after one week was found out to be in decays/sec %.1e\" %A;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The decay constant in sec is 2.97e-06\nThe decay constant is equal to probability of decay in one second hence 2.97e-06\nThe activity was found out to be in Ci is 9.04e+09\nThe activity after one week was found out to be in decays/sec 1.5e+09\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.6 Page 384"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation  of variable\nt1=4.55*10**9;t2=7.04*10**8;       #given values of time at 2 different instants\n\n#calculation\nage=t1/t2;\nr=2**age;\n\n#result\nprint \"The original rock hence contained\",round(r,3),\"Na atoms of 235U where Na is the Avagadro''s Number=6.023*10^23\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The original rock hence contained 88.222 Na atoms of 235U where Na is the Avagadro''s Number=6.023*10^23\n"
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.7 Page 385"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nm236Ra=226.025403;\nm222Rn=222.017571;\nm4He=4.002603;c2=931.5; #mass of various elements and c2=c^2\n\n#calculation\nQ=(m236Ra-m222Rn-m4He)*c2;#Q of the reaction\nA=226.0              \nK=((A-4)/A)*Q;                           #kinetic energy\n\n#result\nprint\"The kinetic energy of the alpha particle in Mev is\",round(K,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The kinetic energy of the alpha particle in Mev is 4.785\n"
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.8 Page 387"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nm226Ra=226.025403; #mass of various elements\nm212Pb=211.991871;\nm14c=14.003242;\nc2=931.5;       #value of c^2\n\n#calculation\nQ=(m226Ra-m212Pb-m14c)*c2;      #Q of the reaction\n\n#result\nprint\"The value of Q for 14c emission in MeV is\",round(Q,3);\nprint\"The probability of 14c emission is 10^-9 times that of an alpha particle since the energy barrier for 14c emission is nearly 3 times higher and thicker.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of Q for 14c emission in MeV is 28.215\nThe probability of 14c emission is 10^-9 times that of an alpha particle since the energy barrier for 14c emission is nearly 3 times higher and thicker.\n"
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.9 Page 389"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nm23Ne=22.994465; #mass of various elements\nm23Na=22.989768;\nc2=931.5;          #value of c^2\n\n#calculation\nQ=(m23Ne-m23Na)*c2;     #Q of the reaction\n\n#result\nprint \"Hence the maximum kinetic energy of the emitted electrons in MeV is\",round(Q,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Hence the maximum kinetic energy of the emitted electrons in MeV is 4.375\n"
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.10 Page 390"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nm40K=39.963999;         #mass of various particles\nm40Ca=39.962591;\nc2=931.5;                    #value of c^2 in MeV\n\n#calculation\nQb1=(m40K-m40Ca)*c2;       #Q value of the reaction\n\n#result\nprint\"The Q value for -VE beta emission in Mev in\",round(Qb1,3);\n\n#partb\nm40K=39.963999;          #mass of various particles\nm40Ar=39.962384;\nme=0.000549;\nQb2=(m40K-m40Ar-2*me)*c2;         #Q value of the reaction\n\n#result\nprint\"The Q value for +VE beta emission in Mev in\",round(Qb2,3);\n\n#partc\nm40K=39.963999;\nm40Ar=39.962384;\n\n#calculation\nQec=(m40K-m40Ar)*c2;\n\n#result\nprint\"The Q value for +VE beta emission in Mev in\",round(Qec,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The Q value for -VE beta emission in Mev in 1.312\nThe Q value for +VE beta emission in Mev in 0.482\nThe Q value for +VE beta emission in Mev in 1.504\n"
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.11 Page 392"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nMg=12.000000; #mass of the carbon atom in amu\nc2=931.5; \nEg=4.43;     #given energy of gamma ray \nMex=Mg+(Eg/c2);     #mass in excited state\nMe=0.000549;        #mass of an electron\n\n#calculation\nQ=(12.018613-Mex-2*Me)*c2;     #Q of the particle\n\n#result\nprint\"The maximum value of kinetic energy is in MeV\",round(Q,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The maximum value of kinetic energy is in MeV 11.885\n"
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.12 Page 393"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nm238U=238.050786; #mass of various quantities\nm206Pb=205.974455;\nm4He=4.002603;\nc2=931.5;     #constants\nNa=6.023*10**23;   #avagadro's number\n\n#calculation\nQ=(m238U-m206Pb-8*m4He)*c2; \nt12=(4.5)*10**9*(3.16*10**7);    #half life years to seconds conversion\nw=0.693/t12;                 # lambeda\nNoD=(Na/238)*w;      #number of decays\nE=NoD*Q*(1.6*10**-19)*10**6;        #rate of liberation of energy,converting MeV to eV\n\n#result\nprint\"Rate of energy liberation in W is %.2e\" %E;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Rate of energy liberation in W is 1.02e-07\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.13 Page 395"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import log\nR=0.5;t12=4.5*10**9;                 #value of radius and half-life \nt1=(t12/0.693)*log(1+(1/R));        #age of rock 1\nR=1.0;\nt2=(t12/0.693)*log(1+(1/R));       #age of rock-2\nR=2.0\nt3=(t12/0.693)*log(1+(1/R));       #age of rock 3\n\n#result\nprint\"The ages of rock samples in years respectively are %.1e\"%t1,\" %.1e\" %t2,\"  %.1e\" %t3;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The ages of rock samples in years respectively are 7.1e+09  4.5e+09   2.6e+09\n"
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.14 Page 397"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import log\nP=2.0*10**14; V=2.0*10**-14; R=8.314; T=295.0;Na=6.023*10**23;   #varoius constants and given values\n\n#calculation\nn=P*V/(R*T);       #ideal gas law\nN=Na*n;f=10**-12         #avagadaro's number and fracction of carbon molecules\nt12=5730*3.16*(10**7);   #half life\nA=(0.693/t12)*N*f;    #activity\nD1w=A*7*24*60*60;       #decays per second\n\n#result\nprint\"the no. of decays per second is %.2e\" %A\nprint\"The no of decays pers week is \",round(D1w);\n    \n    \n#partb\nc1=1420.0;             #concentration at instant 1\nc2=D1w;              #concentration at instant 2\nt12y=5730;          #half life\nt=t12y*log(c2/c1)/0.693;            #age of the sample\n\n#result\nprint\"Age of the sample in years is\",round(t,3);\nprint\"the answer in the book is wrong\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "the no. of decays per second is 3.76e-03\nThe no of decays pers week is  2274.0\nAge of the sample in years is 3892.57\n"
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter12_2.ipynb b/Modern_Physics/Chapter12_2.ipynb
new file mode 100755
index 00000000..ccacecc1
--- /dev/null
+++ b/Modern_Physics/Chapter12_2.ipynb
@@ -0,0 +1,322 @@
+{
+ "metadata": {
+  "name": "MP-12"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Nuclear Structure and Reactivity"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.1 Page 375"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nZ=2;A=4;N=A-Z;         # Given values\n\n#result\nprint\"The following method of representing atoms is followed throughout the chapter\\n\\t\\t x,y,z\\n where x=atomic number y=mass number z= Neutron Number S=symbol of the atom\\n\\n\"\nprint\"The helium can be reperesented as He-- \",Z,A,N;\n\n#part b\nZ=50.0;N=66.0;A=Z+N;       # Given values and standard formulae\nprint\"The Tin can be reperesented as Sn-- \",Z,A,N;\n\n\n#part c\nA=235;N=143;Z=A-N;\nprint\"The Uranium can be reperesented as U-- \",Z,A,N;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The following method of representing atoms is followed throughout the chapter\n\t\t x,y,z\n where x=atomic number y=mass number z= Neutron Number S=symbol of the atom\n\n\nThe helium can be reperesented as He--  2 4 2\nThe Tin can be reperesented as Sn--  50.0 116.0 66.0\nThe Uranium can be reperesented as U--  92 235 143\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.2 Page 377"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nr0=1.2;         #standard value.\nA=12.0; \nr= r0*A**(1.0/3);\n\n#result\nprint\"The value of mean radius for C in fm is\",round(r,3);\n\n#part2\nA=70.0;        #given value\nr= r0*A**(1.0/3);\n\n#result\nprint\"The value of mean radius for C in fm is\",round(r,3);\n\n#part3\nA=209;\nr= r0*A**(1.0/3);\n\n#result\nprint\"The value of mean radius for C in fm is\",round(r,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of mean radius for C in fm is 2.747\nThe value of mean radius for C in fm is 4.946\nThe value of mean radius for C in fm is 7.121\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.3 Page 379"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi\nm=1.67*10**-27; r0=1.2*10**-15; v=4*pi*(r0**3)/3.0     #standard values of mass radius and volume\n\n#calculation\np=m/v;                                               #denisty \n\n#result\nprint\"Density of typical nucleus in kg/m3 is %.1e\" %p;\n\n#part 2\nr0=0.01;v=4*pi*(r0**3)/3.0;p=2.0*10**17;              #/hypothetical values\nm1=p*v;                                  \n\n#result\nprint\"The mass of the hypothetical nucleus would be in Kg %.1e\" %m1;\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Density of typical nucleus in kg/m3 is 2.3e+17\nThe mass of the hypothetical nucleus would be in Kg 8.4e+11\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.4 Page 380"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nN=30.0;Z=26.0;A=56.0;Mn=1.008665;Mp=1.007825;m=55.934939;c2=931.5; #given values and constants for case-1\nB=((N*Mn)+(Z*Mp)-(m))*c2;                                  #binding energy(per nucleon)\n\n#result\nprint\"Binding nergy per nucleon for 26,56Fe30 in MeV is\",round(B/A,3);\n\n#part 2\nN=146.0;Z=92.0;A=238.0;Mn=1.008665;Mp=1.007825;m=238.050785;c2=931.5;      #given values and constants for case-2\nB=((N*Mn)+(Z*Mp)-(m))*c2;                             #binding energy(per nucleon)\n\n#result\nprint\"Binding nergy per nucleon for 26,56Fe30 in MeV is\",round(B/A,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Binding nergy per nucleon for 26,56Fe30 in MeV is 8.79\nBinding nergy per nucleon for 26,56Fe30 in MeV is 7.57\n"
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.5 Page 382"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import exp\nt12=2.7*24*3600;              #converting days into seconds\nw=0.693/t12;                  #lambeda\n\n#result\nprint\"The decay constant in sec is %.2e\" %w; \n\n#partb\nprint\"The decay constant is equal to probability of decay in one second hence %.2e\" %w;\n\n#partc\nm=10**-6;Na=6.023*10**23; M=198.0;     #given values and constants\nN=m*Na/M;                        #number of atoms in the sample \nAo=w*N;             #activity\n\n#result\nprint\"The activity was found out to be in Ci is %.2e\" %Ao;        \n\n#partd\nt=7*24*3600.0;         #given time\nA=Ao*exp(-w*t);        #activity\n\n#result\nprint\"The activity after one week was found out to be in decays/sec %.1e\" %A;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The decay constant in sec is 2.97e-06\nThe decay constant is equal to probability of decay in one second hence 2.97e-06\nThe activity was found out to be in Ci is 9.04e+09\nThe activity after one week was found out to be in decays/sec 1.5e+09\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.6 Page 384"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation  of variable\nt1=4.55*10**9;t2=7.04*10**8;       #given values of time at 2 different instants\n\n#calculation\nage=t1/t2;\nr=2**age;\n\n#result\nprint \"The original rock hence contained\",round(r,3),\"Na atoms of 235U where Na is the Avagadro''s Number=6.023*10^23\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The original rock hence contained 88.222 Na atoms of 235U where Na is the Avagadro''s Number=6.023*10^23\n"
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.7 Page 385"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nm236Ra=226.025403;\nm222Rn=222.017571;\nm4He=4.002603;c2=931.5; #mass of various elements and c2=c^2\n\n#calculation\nQ=(m236Ra-m222Rn-m4He)*c2;#Q of the reaction\nA=226.0              \nK=((A-4)/A)*Q;                           #kinetic energy\n\n#result\nprint\"The kinetic energy of the alpha particle in Mev is\",round(K,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The kinetic energy of the alpha particle in Mev is 4.785\n"
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.8 Page 387"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nm226Ra=226.025403; #mass of various elements\nm212Pb=211.991871;\nm14c=14.003242;\nc2=931.5;       #value of c^2\n\n#calculation\nQ=(m226Ra-m212Pb-m14c)*c2;      #Q of the reaction\n\n#result\nprint\"The value of Q for 14c emission in MeV is\",round(Q,3);\nprint\"The probability of 14c emission is 10^-9 times that of an alpha particle since the energy barrier for 14c emission is nearly 3 times higher and thicker.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of Q for 14c emission in MeV is 28.215\nThe probability of 14c emission is 10^-9 times that of an alpha particle since the energy barrier for 14c emission is nearly 3 times higher and thicker.\n"
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.9 Page 389"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nm23Ne=22.994465; #mass of various elements\nm23Na=22.989768;\nc2=931.5;          #value of c^2\n\n#calculation\nQ=(m23Ne-m23Na)*c2;     #Q of the reaction\n\n#result\nprint \"Hence the maximum kinetic energy of the emitted electrons in MeV is\",round(Q,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Hence the maximum kinetic energy of the emitted electrons in MeV is 4.375\n"
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.10 Page 390"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nm40K=39.963999;         #mass of various particles\nm40Ca=39.962591;\nc2=931.5;                    #value of c^2 in MeV\n\n#calculation\nQb1=(m40K-m40Ca)*c2;       #Q value of the reaction\n\n#result\nprint\"The Q value for -VE beta emission in Mev in\",round(Qb1,3);\n\n#partb\nm40K=39.963999;          #mass of various particles\nm40Ar=39.962384;\nme=0.000549;\nQb2=(m40K-m40Ar-2*me)*c2;         #Q value of the reaction\n\n#result\nprint\"The Q value for +VE beta emission in Mev in\",round(Qb2,3);\n\n#partc\nm40K=39.963999;\nm40Ar=39.962384;\n\n#calculation\nQec=(m40K-m40Ar)*c2;\n\n#result\nprint\"The Q value for +VE beta emission in Mev in\",round(Qec,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The Q value for -VE beta emission in Mev in 1.312\nThe Q value for +VE beta emission in Mev in 0.482\nThe Q value for +VE beta emission in Mev in 1.504\n"
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.11 Page 392"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nMg=12.000000; #mass of the carbon atom in amu\nc2=931.5; \nEg=4.43;     #given energy of gamma ray \nMex=Mg+(Eg/c2);     #mass in excited state\nMe=0.000549;        #mass of an electron\n\n#calculation\nQ=(12.018613-Mex-2*Me)*c2;     #Q of the particle\n\n#result\nprint\"The maximum value of kinetic energy is in MeV\",round(Q,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The maximum value of kinetic energy is in MeV 11.885\n"
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.12 Page 393"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nm238U=238.050786; #mass of various quantities\nm206Pb=205.974455;\nm4He=4.002603;\nc2=931.5;     #constants\nNa=6.023*10**23;   #avagadro's number\n\n#calculation\nQ=(m238U-m206Pb-8*m4He)*c2; \nt12=(4.5)*10**9*(3.16*10**7);    #half life years to seconds conversion\nw=0.693/t12;                 # lambeda\nNoD=(Na/238)*w;      #number of decays\nE=NoD*Q*(1.6*10**-19)*10**6;        #rate of liberation of energy,converting MeV to eV\n\n#result\nprint\"Rate of energy liberation in W is %.2e\" %E;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Rate of energy liberation in W is 1.02e-07\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.13 Page 395"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import log\nR=0.5;t12=4.5*10**9;                 #value of radius and half-life \nt1=(t12/0.693)*log(1+(1/R));        #age of rock 1\nR=1.0;\nt2=(t12/0.693)*log(1+(1/R));       #age of rock-2\nR=2.0\nt3=(t12/0.693)*log(1+(1/R));       #age of rock 3\n\n#result\nprint\"The ages of rock samples in years respectively are %.1e\"%t1,\" %.1e\" %t2,\"  %.1e\" %t3;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The ages of rock samples in years respectively are 7.1e+09  4.5e+09   2.6e+09\n"
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 12.14 Page 397"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import log\nP=2.0*10**14; V=2.0*10**-14; R=8.314; T=295.0;Na=6.023*10**23;   #varoius constants and given values\n\n#calculation\nn=P*V/(R*T);       #ideal gas law\nN=Na*n;f=10**-12         #avagadaro's number and fracction of carbon molecules\nt12=5730*3.16*(10**7);   #half life\nA=(0.693/t12)*N*f;    #activity\nD1w=A*7*24*60*60;       #decays per second\n\n#result\nprint\"the no. of decays per second is %.2e\" %A\nprint\"The no of decays pers week is \",round(D1w);\n    \n    \n#partb\nc1=1420.0;             #concentration at instant 1\nc2=D1w;              #concentration at instant 2\nt12y=5730;          #half life\nt=t12y*log(c2/c1)/0.693;            #age of the sample\n\n#result\nprint\"Age of the sample in years is\",round(t,3);\nprint\"the answer in the book is wrong\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "the no. of decays per second is 3.76e-03\nThe no of decays pers week is  2274.0\nAge of the sample in years is 3892.57\n"
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter13.ipynb b/Modern_Physics/Chapter13.ipynb
index e35d94ac..5e2693aa 100755
--- a/Modern_Physics/Chapter13.ipynb
+++ b/Modern_Physics/Chapter13.ipynb
@@ -1,7 +1,7 @@
 {
  "metadata": {
-  "name": "",
-  "signature": "sha256:8b9d9c8500ecb20deeeb6f065d1408d228b300a0b95f2800ee645e0184760d45"
+  "name": "Chapter13",
+  "signature": "sha256:3612dc1ad0c2617d4c1e1da21ea7791334391562bfe93ddc9d0f98037d4fe15d"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -13,7 +13,7 @@
      "level": 1,
      "metadata": {},
      "source": [
-      "Chapter 13: Nuclear Structure "
+      "Chapter 13:Nuclear Reaction and Application"
      ]
     },
     {
@@ -21,111 +21,27 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 13.1, page no. 466"
+      "Example 13.1, Page 417"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "v=1*1.0*10**-6.0*10**2; p=7.9; m=p*v;Na=6.023*10**23     #given values and various constants in suitable units\n",
+      "M=56.0;N=m*Na/M;                                 #number of atoms\n",
+      "i=3.0*10**-6;\n",
+      "q=1.6*10**-19;\n",
       "\n",
+      "#calculation\n",
+      "Io=i/q;                 #intensity\n",
+      "s=0.6*10**-24;S=1;             #given values in suitable units\n",
+      "R=N*s*Io/S;                     #rate of neutrons\n",
       "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "NA = 6.02 * 10 ** 23    # Avogadro's number\n",
-      "m = 0.012               # one mole of carbon (kg)\n",
-      "mC = 12                 # mass of one atom(1u)\n",
-      " \n",
-      "#Calculation\n",
-      "\n",
-      "ma = m/NA\n",
-      "u = ma / mC\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print \"The atomic mass unit is \",round(u/10**-27,2),\"X 10^-27 kg.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The atomic mass unit is  1.66 X 10^-27 kg.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.2, page no. 468"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "m = 1.67 * 10 ** -27    # mass of neutron(kg)\n",
-      "r0 = 1.2 * 10 ** -15    # radius os the nucleus(m)\n",
-      "\n",
-      "#Calculation \n",
-      "\n",
-      "pn = 3* m /(4 * math.pi * r0**3)\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print \"The nuclear density is\",round(pn/10**17,1),\"X 10^17 kg/m^3.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The nuclear density is 2.3 X 10^17 kg/m^3.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.3, page no. 473"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      " \n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "MH = 1.007825       #mass of hydrogen (u)\n",
-      "mn = 1.008665       #mass of neutron (u)\n",
-      "M2 = 2.014102       #mass of deuteron (u)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "Eb = (MH + mn - M2) * 931.494\n",
-      "\n",
-      "#results\n",
-      "\n",
-      "print \"The binding energy of deuteron is\",round(Eb,3),\"MeV.\""
+      "#result\n",
+      "print\"The rate of neutrons emitted from the target in particles per second is %.1e\" %round(R,3);\n",
+      "print\"Slight difference in answer due to approximation error\""
      ],
      "language": "python",
      "metadata": {},
@@ -134,41 +50,38 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The binding energy of deuteron is 2.224 MeV.\n"
+        "The rate of neutrons emitted from the target in particles per second is 9.6e+07\n",
+        "slight difference due to approximation error\n"
        ]
       }
      ],
-     "prompt_number": 5
+     "prompt_number": 4
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 13.4, page no. 482"
+      "Example 13.2, Page 419"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "Thalf = 5730        #Half-life (yr)\n",
-      "N = 1000            #No of carbon nuclei\n",
-      "T = 22920           #(yr)\n",
+      "#initiation of variable\n",
+      "A=197.0; m=30*10**-3;phi=3.0*10**12;    #given values and various constants taken in suitable units\n",
+      "Ar=99.0*10**-24; Na=6.023*10**23\n",
       "\n",
       "#calculation\n",
-      "n=N\n",
-      "t = 0\n",
-      "while(t!=T):\n",
-      "    n = n/2\n",
-      "    t = t + Thalf\n",
+      "R=(phi*Na*Ar*m/A);        #rate or production of gold\n",
+      "t=2.7*24*60         # time of decay\n",
+      "Act=R*(0.693/t);     #activity /sec\n",
+      "ActCi=Act/(3.7*10**4);     # in terms of curie(Ci)\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"The number of Carbon nuclei left after 22,920 yr is\",n"
+      "print\"The activity is found out to be %.1e\" %round(Act,3),\"/sec i.e \" ,round(ActCi,3),\"muCi\"\n",
+      "print\"Slight difference in answer due to approximation error\""
      ],
      "language": "python",
      "metadata": {},
@@ -177,7 +90,8 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The number of Carbon nuclei left after 22,920 yr is 62\n"
+        "The activity is found out to be 4.9e+06 /sec i.e  131.229 muCi\n",
+        "slight difference due to approximation error\n"
        ]
       }
      ],
@@ -188,162 +102,37 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 13.5, page no. 483"
+      "Example 13.3, Page 423"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "from math import exp\n",
+      "v=1.5*1.5*2.5*(10**-6)*10**2; #volume in cm3\n",
+      "p=8.9;         #density in g/cm3\n",
+      "m=p*v;Na=6.023*10**23 #mass and Avagadro's number\n",
+      "M=58.9;        #Given values\n",
       "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "Thalf = 1.6 * 10 ** 3       #Half-life (yr)\n",
-      "s = 3.16 * 10 ** 7          #number of seconds in a year (s/yr)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "lamda = 0.693 / (Thalf * s)\n",
+      "#calculation\n",
+      "N=m*Na/M;\n",
+      "i=12*10**-6;       #thickness of beam\n",
+      "q=1.6*10**-19;\n",
+      "Io=i/(2*q);      #intensity\n",
+      "s=0.64*10**-24;   #Given values\n",
+      "S=1.5*1.5;\n",
+      "R=N*s*Io/S;       #rate of production of 61Cu\n",
       "\n",
       "#result\n",
+      "print \"The rate of neutrons emitted from the target in particles/second is %.1e\" %round(R,3);\n",
       "\n",
-      "print \"(a) The decay constant is\",round(lamda/10**-11,1),\"X 10^-11 s^-1.\"\n",
-      "\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "N0 = 3.0 * 10 ** 16         #number of radioactive nuclei at t=0\n",
-      "Ci = 3.7 * 10 **10          \n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "R0 = lamda * N0\n",
-      "\n",
-      "#results\n",
-      "\n",
-      "print \"(b) Its activity is\",round(R0/Ci/10**-6,1),\"X 10^-6 Ci.\"\n",
-      "\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "T = 2.0 * 10 ** 3       #(yr)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "R = R0 * math.exp(-lamda*(T * s))\n",
-      "\n",
-      "#results\n",
-      "\n",
-      "print \"(c) The decay rate after 2.0 X 10^3 yr is\",round(R/10**5,1),\"X 10^5 decays/s\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The decay constant is 1.4 X 10^-11 s^-1.\n",
-        "(b) Its activity is 11.1 X 10^-6 Ci.\n",
-        "(c) The decay rate after 2.0 X 10^3 yr is 1.7 X 10^5 decays/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.6, page no. 483"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "A = 6.02 * 10 ** 23         #avogadro's number\n",
-      "m = 3.50 * 10 ** -6         # mass of carbon(g)\n",
-      "ma = 11.0                   #atomic mass of carbon (g)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "N = m * A / ma\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "print \"(a) The number of nuclei samples at t=0 is\",round(N/10**17,2),\"X 10^17 nuclei.\"\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "Thalf = 20.4 * 60   #half-life (s)\n",
-      "T = 8.00            #(h)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "lamda = 0.693 / Thalf\n",
-      "R0 = lamda * N\n",
-      "R = R0 * math.exp(-lamda* T*60*60)\n",
+      "#part b\n",
+      "act=R*(1-(exp((0.693)*(-2/3.41))));          #activity\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"(b) The activity initially is\",round(R0/10**14,2),\"X 10^14 decay/s and after t=8.0h is\",round(R/10**6,2),\"X 10^6 decay/s\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(a) The number of nuclei samples at t=0 is 1.92 X 10^17 nuclei.\n",
-        "(b) The activity initially is 1.08 X 10^14 decay/s and after t=8.0h is 8.99 X 10^6 decay/s\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.7, page no. 484"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "Thalf = 8.04        # half-life (days)\n",
-      "R0 = 5.0            #Activity at t=0 (mCi)\n",
-      "R = 4.2             #Activity (mCi)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "t = - (Thalf/0.693)* math.log(R/R0)\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "print \"The time elapsed is\",round(t,2),\"days.\""
+      "print\"The activity after 2.0h in /sec is %.1e\" %round(act,3),\"=4.9mCi\";\n"
      ],
      "language": "python",
      "metadata": {},
@@ -352,83 +141,41 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The time elapsed is 2.02 days.\n"
+        "The rate of neutrons emitted from the target in particles/second is 5.5e+08\n",
+        "The activity after 2.0h in /sec is 1.8e+08 =4.9mCi\n"
        ]
       }
      ],
-     "prompt_number": 16
+     "prompt_number": 6
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 13.8, page no. 486"
+      "Example 13.4, Page 425"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      " \n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "Mx = 226.025406         #atomic mass of 226Ra(u)\n",
-      "My = 222.017574         #atomic mass of 222Rn(u)\n",
-      "Ma = 4.002603           #atomic mass of 4He (u)\n",
+      "#initiation of variable\n",
+      "m2H=2.014102;    #mass of various particles\n",
+      "mn=1.008665;m63Cu=62.929599;\n",
+      "m64Zn=63.929145;c2=931.5;     #c^2=931.5 MeV\n",
+      "Q=(m2H+m63Cu-mn-m64Zn)*c2;       #Q of the reaction\n",
       "\n",
-      "#Calculation\n",
-      "\n",
-      "Q = (Mx - My - Ma) * 931.494\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "print \"Q value is\",round(Q,2),\"MeV.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Q value is 4.87 MeV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 13.9, page no. 487"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
-      "#Variable declaration\n",
-      "\n",
-      "r0 = 7.25 * 10**-5        #Bohr radius for alpha particle(A')\n",
-      "Z = 86                    #daughter nucleus radon's atomic number\n",
-      "A = 222                   #radon's mass number\n",
-      "ke2 = 14.40               #boltzmann constant X (charge of electron)^2 (eV.A')\n",
-      "E = 5                     #disintegration energy (MeV)\n",
+      "#result\n",
+      "print\"The value of Q is in MeV\",round(Q,3);\n",
       "\n",
-      "#Calculation\n",
       "\n",
-      "E0 = ke2 / (2.0*r0) * 10**-6             #Energy unit analogous to Rydberg (MeV)\n",
-      "R = (1.2 * 10**-5) *(A)**(1.0/3.0)       #radius of Radon  nucleus(A')\n",
-      "Te = math.exp(round(-4*math.pi*Z*math.sqrt(E0/E)+8*math.sqrt(Z*R/r0)))\n",
+      "#part b\n",
+      "Kx=12.00;Ky=16.85;\n",
+      "Ky=Q+Kx-Ky              #kinetic energy of 64Zn\n",
       "\n",
       "#result\n",
-      "print \"The probability of escape of alpha particle is\",round(Te/10**-34,2),\"X 10^-34\""
+      "print\"The value of Ky was found out to be in MeV\",round(Ky,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -437,49 +184,40 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The probability of escape of alpha particle is 1.33 X 10^-34\n"
+        "The value of Q is in MeV 5.487\n",
+        "The value of Ky was found out to be in MeV 0.637\n"
        ]
       }
      ],
-     "prompt_number": 21
+     "prompt_number": 5
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 13.11, page no. 490"
+      "Example 13.5, Page 425"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "mp=1.007825;m3H=3.016049;    #mass of the particle\n",
+      "m2H=2.014102;c2=931.5;      #constant\n",
+      "Q=(mp+m3H-(2*m2H))*c2;     #Q of the reaction\n",
       "\n",
+      "#result\n",
+      "print\"The value of q was found out to be in MeV\",round(Q,3);\n",
       "\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "Thalf = 5730         # half-life (yr)\n",
-      "s = 3.16 * 10 ** 7   # (s/yr)\n",
-      "A = 6.02 * 10 ** 23  # Avogadro's number\n",
-      "mw = 12.0            # molar weight of carbon(g)\n",
-      "m = 25.0             # mass of carbon(g)   \n",
-      "r = 1.3 * 10 ** -12  # ratio of 14C to 12C\n",
-      "R = 250              # activity observed (decays/min)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "lamda = 0.693 /(Thalf * s)\n",
-      "N1 = A * m/mw\n",
-      "N0 = r * N1\n",
-      "R0 = N0 * lamda\n",
-      "t = -(1/lamda)*math.log(R/(R0*60))\n",
-      " \n",
-      "#Result\n",
+      "#partb\n",
+      "Kth1= -Q*(1+(mp/m3H));        #threshold energy of kinetic energy\n",
+      "Kth2=-Q*(1+(m3H/mp));        #threshold kinetic energy in case2\n",
       "\n",
-      "print \"The tree has been dead for\",round(round(t/10**11)*10**11/s/10**3,1),\"X 10^3 yr.\""
+      "#result\n",
+      "print\"The threshold kinetic energy in case-1 in MeV\",round(Kth1,3);\n",
+      "print\"The threshold kinetic energy in case-2 in MeV\",round(Kth2,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -488,11 +226,13 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The tree has been dead for 3.2 X 10^3 yr.\n"
+        "The value of q was found out to be in MeV -4.033\n",
+        "The threshold kinetic energy in case-1 in MeV 5.381\n",
+        "The threshold kinetic energy in case-2 in MeV 16.104\n"
        ]
       }
      ],
-     "prompt_number": 25
+     "prompt_number": 6
     }
    ],
    "metadata": {}
diff --git a/Modern_Physics/Chapter13_1.ipynb b/Modern_Physics/Chapter13_1.ipynb
new file mode 100755
index 00000000..44cdc2a7
--- /dev/null
+++ b/Modern_Physics/Chapter13_1.ipynb
@@ -0,0 +1,125 @@
+{
+ "metadata": {
+  "name": "MP-13"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Nuclear Reaction and Application"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 13.1 Page 417"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nv=1*1.0*10**-6.0*10**2; p=7.9; m=p*v;Na=6.023*10**23     #given values and various constants in suitable units\nM=56.0;N=m*Na/M;                                 #number of atoms\ni=3.0*10**-6;\nq=1.6*10**-19;\n\n#calculation\nIo=i/q;                 #intensity\ns=0.6*10**-24;S=1;             #given values in suitable units\nR=N*s*Io/S;                     #rate of neutrons\n\n#result\nprint\"The rate of neutrons emitted from the target in particles per second is %.1e\" %round(R,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The rate of neutrons emitted from the target in particles per second is 9.6e+07\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 13.2 Page 419"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nA=197.0; m=30*10**-3;phi=3.0*10**12;    #given values and various constants taken in suitable units\nAr=99.0*10**-24; Na=6.023*10**23\n\n#calculation\nR=(phi*Na*Ar*m/A);        #rate or production of gold\nt=2.7*24*60         # time of decay\nAct=R*(0.693/t);     #activity /sec\nActCi=Act/(3.7*10**4);     # in terms of curie(Ci)\n\n#result\nprint\"The activity is found out to be %.1e\" %round(Act,3),\"/sec i.e \" ,ActCi,\"Ci\";\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The activity is found out to be 4.9e+06 /sec i.e  131.228932295 Ci\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 13.3 Page 423"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import exp\nv=1.5*1.5*2.5*(10**-6)*10**2; #volume in cm3\np=8.9;         #density in g/cm3\nm=p*v;Na=6.023*10**23 #mass and Avagadro's number\nM=58.9;        #Given values\n\n#calculation\nN=m*Na/M;\ni=12*10**-6;       #thickness of beam\nq=1.6*10**-19;\nIo=i/(2*q);      #intensity\ns=0.64*10**-24;   #Given values\nS=1.5*1.5;\nR=N*s*Io/S;       #rate of production of 61Cu\n\n#result\nprint \"The rate of neutrons emitted from the target in particles/second is %.1e\" %round(R,3);\n\n#part b\nact=R*(1-(exp((0.693)*(-2/3.41))));          #activity\n\n#result\nprint\"The activity after 2.0h in /sec is %.1e\" %round(act,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The rate of neutrons emitted from the target in particles/second is 546058064.516\nThe activity after 2.0h in /sec is 182378303.69\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 13.4 Page 425"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nm2H=2.014102;    #mass of various particles\nmn=1.008665;m63Cu=62.929599;\nm64Zn=63.929145;c2=931.5;     #c^2=931.5 MeV\nQ=(m2H+m63Cu-mn-m64Zn)*c2;       #Q of the reaction\n\n#result\nprint\"The value of Q is in MeV\",round(Q,3);\n\n\n#part b\nKx=12.00;Ky=16.85;\nKy=Q+Kx-Ky              #kinetic energy of 64Zn\n\n#result\nprint\"The value of Ky was found out to be in MeV\",round(Ky,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of Q is in MeV 5.487\nThe value of Ky was found out to be in MeV 0.637\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 13.5 Page 425"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nmp=1.007825;m3H=3.016049;    #mass of the particle\nm2H=2.014102;c2=931.5;      #constant\nQ=(mp+m3H-(2*m2H))*c2;     #Q of thereaction\n\n#result\nprint\"The value of q was found out to be in MeV\",round(Q,3);\n\n#partb\nKth1= -Q*(1+(mp/m3H));        #threshold energy of kinetic energy\nKth2=-Q*(1+(m3H/mp));        #threshold kinetic energy in case2\n\n#result\nprint\"The threshold kinetic energy in case-1 in MeV\",round(Kth1,3);\nprint\"The threshold kinetic energy in case-2 in MeV\",round(Kth2,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of q was found out to be in MeV -4.033\nThe threshold kinetic energy in case-1 in MeV 5.381\nThe threshold kinetic energy in case-2 in MeV 16.104\n"
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter13_2.ipynb b/Modern_Physics/Chapter13_2.ipynb
new file mode 100755
index 00000000..44cdc2a7
--- /dev/null
+++ b/Modern_Physics/Chapter13_2.ipynb
@@ -0,0 +1,125 @@
+{
+ "metadata": {
+  "name": "MP-13"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Nuclear Reaction and Application"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 13.1 Page 417"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nv=1*1.0*10**-6.0*10**2; p=7.9; m=p*v;Na=6.023*10**23     #given values and various constants in suitable units\nM=56.0;N=m*Na/M;                                 #number of atoms\ni=3.0*10**-6;\nq=1.6*10**-19;\n\n#calculation\nIo=i/q;                 #intensity\ns=0.6*10**-24;S=1;             #given values in suitable units\nR=N*s*Io/S;                     #rate of neutrons\n\n#result\nprint\"The rate of neutrons emitted from the target in particles per second is %.1e\" %round(R,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The rate of neutrons emitted from the target in particles per second is 9.6e+07\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 13.2 Page 419"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nA=197.0; m=30*10**-3;phi=3.0*10**12;    #given values and various constants taken in suitable units\nAr=99.0*10**-24; Na=6.023*10**23\n\n#calculation\nR=(phi*Na*Ar*m/A);        #rate or production of gold\nt=2.7*24*60         # time of decay\nAct=R*(0.693/t);     #activity /sec\nActCi=Act/(3.7*10**4);     # in terms of curie(Ci)\n\n#result\nprint\"The activity is found out to be %.1e\" %round(Act,3),\"/sec i.e \" ,ActCi,\"Ci\";\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The activity is found out to be 4.9e+06 /sec i.e  131.228932295 Ci\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 13.3 Page 423"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import exp\nv=1.5*1.5*2.5*(10**-6)*10**2; #volume in cm3\np=8.9;         #density in g/cm3\nm=p*v;Na=6.023*10**23 #mass and Avagadro's number\nM=58.9;        #Given values\n\n#calculation\nN=m*Na/M;\ni=12*10**-6;       #thickness of beam\nq=1.6*10**-19;\nIo=i/(2*q);      #intensity\ns=0.64*10**-24;   #Given values\nS=1.5*1.5;\nR=N*s*Io/S;       #rate of production of 61Cu\n\n#result\nprint \"The rate of neutrons emitted from the target in particles/second is %.1e\" %round(R,3);\n\n#part b\nact=R*(1-(exp((0.693)*(-2/3.41))));          #activity\n\n#result\nprint\"The activity after 2.0h in /sec is %.1e\" %round(act,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The rate of neutrons emitted from the target in particles/second is 546058064.516\nThe activity after 2.0h in /sec is 182378303.69\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 13.4 Page 425"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nm2H=2.014102;    #mass of various particles\nmn=1.008665;m63Cu=62.929599;\nm64Zn=63.929145;c2=931.5;     #c^2=931.5 MeV\nQ=(m2H+m63Cu-mn-m64Zn)*c2;       #Q of the reaction\n\n#result\nprint\"The value of Q is in MeV\",round(Q,3);\n\n\n#part b\nKx=12.00;Ky=16.85;\nKy=Q+Kx-Ky              #kinetic energy of 64Zn\n\n#result\nprint\"The value of Ky was found out to be in MeV\",round(Ky,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of Q is in MeV 5.487\nThe value of Ky was found out to be in MeV 0.637\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 13.5 Page 425"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nmp=1.007825;m3H=3.016049;    #mass of the particle\nm2H=2.014102;c2=931.5;      #constant\nQ=(mp+m3H-(2*m2H))*c2;     #Q of thereaction\n\n#result\nprint\"The value of q was found out to be in MeV\",round(Q,3);\n\n#partb\nKth1= -Q*(1+(mp/m3H));        #threshold energy of kinetic energy\nKth2=-Q*(1+(m3H/mp));        #threshold kinetic energy in case2\n\n#result\nprint\"The threshold kinetic energy in case-1 in MeV\",round(Kth1,3);\nprint\"The threshold kinetic energy in case-2 in MeV\",round(Kth2,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of q was found out to be in MeV -4.033\nThe threshold kinetic energy in case-1 in MeV 5.381\nThe threshold kinetic energy in case-2 in MeV 16.104\n"
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter14.ipynb b/Modern_Physics/Chapter14.ipynb
index 4fff498c..a7beb792 100755
--- a/Modern_Physics/Chapter14.ipynb
+++ b/Modern_Physics/Chapter14.ipynb
@@ -1,7 +1,6 @@
 {
  "metadata": {
-  "name": "",
-  "signature": "sha256:bb0d2851c786dd28ae3ad8cbd1e6a7fc5db7cc1384cc2031f2bb6380ed7515e2"
+  "name": "Chapter14"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -13,7 +12,7 @@
      "level": 1,
      "metadata": {},
      "source": [
-      "Chapter 14: Nuclear Physics Apllications"
+      "Chapter 14:Elementary Particles"
      ]
     },
     {
@@ -21,43 +20,23 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 14.1, page no. 505"
+      "Example 14.2, Page 451"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "mvo=1116.0;mp=938.0;mpi=140.0;    #mass of various particles\n",
       "\n",
+      "#calculation\n",
+      "Q=(mvo-mp-mpi);            #Q value of energy\n",
+      "Pp=100.0;Ppi=100;            #momentum of various particles\n",
+      "Kp=5.0;Kpi=38-Kp;            #kinetic energy of particles\n",
       "\n",
-      "#Variable declaration\n",
-      "\n",
-      "MLi = 7.016003          #atomic mass of Lithium\n",
-      "MH = 1.007825           #atomic mass of Hydrogen\n",
-      "MHe = 4.002603          #atomic mass of Helium\n",
-      "c2 = 931.50             #Square of speed of light (MeV/u)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "Q = (MLi + MH - 2*MHe) * c2\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print \"(a) The Q value is\",round(Q,1),\"MeV.\"\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "Kincident = 0.6     #kinetic energy of incident particle (MeV)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "Kproducts = Q + Kincident\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print \"(b) The kinetic energy of the products is\",round(Kproducts,1),\"MeV.\""
+      "#result\n",
+      "print \"The kinetic energy of the particles Kp and Kpi are\", Kp,\" MeV and\",Kpi,\" MeV respectively\""
      ],
      "language": "python",
      "metadata": {},
@@ -66,46 +45,33 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "(a) The Q value is 17.3 MeV.\n",
-        "(b) The kinetic energy of the products is 17.9 MeV.\n"
+        " The kinetic energy of the particles Kp and Kpi are 5.0  MeV and 33.0  MeV respectively\n"
        ]
       }
      ],
-     "prompt_number": 6
+     "prompt_number": 2
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 14.2, page no. 509"
+      "Example 14.3, Page 453"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "Q=105.2                    # The Q value for the given decay\n",
+      "Muc2=105.80344         #mass energy\n",
       "\n",
+      "#calculation\n",
+      "Ke= Q**2/(2*Muc2);       #Ke=Ee-mec2;\n",
       "\n",
-      "#Variable declaration\n",
-      "\n",
-      "A = 27      #Atomic number of Aluminum\n",
-      "d = 2.7     #density of aluminum\n",
-      "Av = 6.02 * 10 ** 23    #Avogadro number nuclei/mol\n",
-      "sigma = 2.0 * 10 **-31  #capture cross section (m^2)\n",
-      "x = 0.3 * 10 ** -3      #thickness of the foil(m)\n",
-      "R0 = 5.0 * 10 ** 12     #rate of incident particles(neutrons/cm^2.s)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "n = Av * d / A * 10**6\n",
-      "R = R0 * sigma * x * n\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "print \"The number of neutrons captured by 1.0cm^2/s is \",round(R/10**7,1),\"X 10^7 neutrons/cm^2.s\"\n",
-      "\n",
-      "\n"
+      "#result\n",
+      "print \"The maximum kinetic energy in MeV is\",round(Ke,3);\n"
      ],
      "language": "python",
      "metadata": {},
@@ -114,40 +80,39 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The number of neutrons captured by 1.0cm^2/s is  1.8 X 10^7 neutrons/cm^2.s\n"
+        "The maximum kinetic energy in MeV is 52.3\n"
        ]
       }
      ],
-     "prompt_number": 9
+     "prompt_number": 3
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 14.3, page no. 512"
+      "Example 14.4, Page 455"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "A = 235         #atomic number of uranium\n",
-      "m = 10**3       #mass of 235U\n",
-      "Av = 6.02*10**23#avogadro number (nuclei/mol)\n",
-      "Q = 208         #disintegration energy per event (MeV)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "N = Av * m / A\n",
-      "E = N * Q\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "print \"The disintegration energy is\",round(E/10**26,2),\"X 10^26 MeV.\""
+      "#initiation of variable\n",
+      "from __future__ import division\n",
+      "from sympy.solvers import solve\n",
+      "from sympy import Symbol\n",
+      "from sympy import *\n",
+      "from math import sqrt\n",
+      "mkc2=494.0; mpic2=135.0;mec2=0.5;# mass of various particles\n",
+      "\n",
+      "#calculation\n",
+      "Q1=mkc2-mpic2-mec2;       #Q of reaction\n",
+      "# the neutrino has negligible energy\n",
+      "x = symbols('x')\n",
+      "k=solve((x**2+135.0**2)**(0.5)+x-494,x);# assigning the Q to sum of energies and simplifying\n",
+      "\n",
+      "print \"The value of maximum kinetic enrgy for pi-meson  and positron are\",266,\"MeV &\",k,\" MeV\";"
      ],
      "language": "python",
      "metadata": {},
@@ -156,83 +121,34 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The disintegration energy is 5.33 X 10^26 MeV.\n"
+        " The value of maximum kinetic enrgy for pi-meson  and positron are 266 MeV & [228.553643724696]  MeV\n"
        ]
       }
      ],
-     "prompt_number": 12
+     "prompt_number": 17
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 14.4, page no. 513"
+      "Example 14.5, Page 457"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "mpi_=140;mp=938;mKo=498;mLo=1116; #mass of various particles\n",
       "\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "A = 235     #atomic mass of Uranium\n",
-      "Av = 6.02 * 10**23 # Avagadro's number\n",
-      "Q = 208            # disintegration energy per event  (MeV)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "N = Av / A  *10**3  #No of nuclei in 1kg of 235U\n",
-      "E = N*Q             #disintegration energy (MeV)\n",
+      "#calculation\n",
+      "Q1= mpi_+mp-mKo-mLo;          #Q value of reaction 1\n",
+      "mK_=494.0;mpio=135.0;         \n",
+      "Q2=mK_+mp-mLo-mpio;         #Q value of reaction 2\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"The disintegration energy is\",round(E/10**26,2),\" X 10^26 MeV.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The disintegration energy is 5.33  X 10^26 MeV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 14.5, page no. 513"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "rBa = 6.2 * 10 **-15    #nuclear radius of Ba(m)\n",
-      "rKr = 5.4 * 10**-15     #nuclear radius of Kr(m)\n",
-      "Z1 = 56                 #atomic number of Ba\n",
-      "Z2 = 36                 #atomic number of Kr\n",
-      "k = 1.440 * 10 **-9     #Coulomb constant (eV.nm)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "r = rBa + rKr\n",
-      "U = k * Z1 * Z2 /( round(r/10**-15) * 10**-15)\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "print \"The potential energy of the two nuclei is\",round(U/10**6),\"MeV.\" "
+      "print\"The Q values of reactions 1 and 2 are\", Q1,\" MeV and\",Q2,\"MeV\";"
      ],
      "language": "python",
      "metadata": {},
@@ -241,52 +157,34 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The potential energy of the two nuclei is 242.0 MeV.\n"
+        "The Q values of reactions 1 and 2 are -536  MeV and 181.0 MeV\n"
        ]
       }
      ],
-     "prompt_number": 17
+     "prompt_number": 19
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 14.6, page no. 519"
+      "Example 14.6, Page 459"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "mpic2=135.0;        #mass energy of pi particle\n",
       "\n",
-      "\n",
-      "#variable declaration\n",
-      "\n",
-      "e = 1.6 * 10 ** -19     #electron charge (C)\n",
-      "r = 1.0 * 10**-14       #separation (m)\n",
-      "k = 8.99 * 10**9        #Coulomb constant(N.m^2/C^2)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "U = k * e**2/ r\n",
+      "#calculation\n",
+      "Q=-mpic2;\n",
+      "mp=938.0;mpi=135.0;\n",
+      "Kth=(-Q)*((4*mp)+mpi)/(2*(mp));  #threshold energy\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"(a)The height of the potential barrier is\",round(U/e/10**6,2),\"MeV.\"\n",
-      "\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "kB = 1.38 * 10**-23         #Boltzmann's constant (J/K)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "T = (U/2.0)*(2.0/3.0)/(kB+.07*10**-23)\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "print\"(b)The effective temperature is\",round(T/10**8,1),\"X 10^8 K.\""
+      "print\"The threshold kinetic energy in MeV is\",round(Kth,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -295,39 +193,33 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "(a)The height of the potential barrier is 0.14 MeV.\n",
-        "(b)The effective temperature is 5.3 X 10^8 K.\n"
+        "The threshold kinetic energy in MeV is 279.715\n"
        ]
       }
      ],
-     "prompt_number": 22
+     "prompt_number": 20
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 14.7, page no. 530"
+      "Example 14.7, Page 460"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "mpc2=938.0;   #rest energy of proton\n",
       "\n",
-      "import math\n",
-      "\n",
-      "#variable declaration\n",
-      "\n",
-      "mu = 55         #mu for x-rays \n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "x = math.log(2)/mu\n",
-      "\n",
-      "#Result\n",
+      "#result\n",
+      "Q=mpc2+mpc2-(4*mpc2);        #Q value of reaction \n",
+      "Kth=(-Q)*(6*mpc2/(2*mpc2));       # thershold kinetic energy\n",
       "\n",
-      "print \"The half value thickness for lead is\",round(x/10**-2,2),\"X 10^-2 cm.\""
+      "#result\n",
+      "print \"The threshold kinetic energy in MeV is\",round(Kth,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -336,11 +228,11 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The half value thickness for lead is 1.26 X 10^-2 cm.\n"
+        "The threshold kinetic energy in MeV is 5628.0\n"
        ]
       }
      ],
-     "prompt_number": 24
+     "prompt_number": 21
     }
    ],
    "metadata": {}
diff --git a/Modern_Physics/Chapter14_1.ipynb b/Modern_Physics/Chapter14_1.ipynb
new file mode 100755
index 00000000..84af5d28
--- /dev/null
+++ b/Modern_Physics/Chapter14_1.ipynb
@@ -0,0 +1,146 @@
+{
+ "metadata": {
+  "name": "MP-14"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Elementary Particles"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 14.2 Page 451"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nmvo=1116.0;mp=938.0;mpi=140.0;    #mass of various particles\n\n#calculation\nQ=(mvo-mp-mpi);            #Q value of energy\nPp=100.0;Ppi=100;            #momentum of various particles\nKp=5.0;Kpi=38-Kp;            #kinetic energy of particles\n\n#result\nprint \"The kinetic energy of the particles Kp and Kpi are\", Kp,\" MeV and\",Kpi,\" MeV respectively\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " The kinetic energy of the particles Kp and Kpi are 5.0  MeV and 33.0  MeV respectively\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 14.3 Page 453"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nQ=105.2                    # The Q value for the given decay\nMuc2=105.80344         #mass energy\n\n#calculation\nKe= Q**2/(2*Muc2);       #Ke=Ee-mec2;\n\n#result\nprint \"The maximum kinetic energy in MeV is\",round(Ke,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The maximum kinetic energy in MeV is 52.3\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 14.4 Page 455"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom __future__ import division\nfrom sympy.solvers import solve\nfrom sympy import Symbol\nfrom sympy import *\nfrom math import sqrt\nmkc2=494.0; mpic2=135.0;mec2=0.5;# mass of various particles\n\n#calculation\nQ1=mkc2-mpic2-mec2;       #Q of reaction\n# the neutrino has negligible energy\nx = symbols('x')\nk=solve((x**2+135.0**2)**(0.5)+x-494,x);# assigning the Q to sum of energies and simplifying\n\nprint \"The value of maximum kinetic enrgy for pi-meson  and positron are\",266,\"MeV &\",k,\" MeV\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " The value of maximum kinetic enrgy for pi-meson  and positron are 266 MeV & [228.553643724696]  MeV\n"
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 14.5 Page 457"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nmpi_=140;mp=938;mKo=498;mLo=1116; #mass of various particles\n\n#calculation\nQ1= mpi_+mp-mKo-mLo;          #Q value of reaction 1\nmK_=494.0;mpio=135.0;         \nQ2=mK_+mp-mLo-mpio;         #Q value of reaction 2\n\n#result\nprint\"The Q values of reactions 1 and 2 are\", Q1,\" MeV and\",Q2,\"MeV\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The Q values of reactions 1 and 2 are -536  MeV and 181.0 MeV\n"
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 14.6 Page 459"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nmpic2=135.0;        #mass ennergy of pi particle\n\n#calculation\nQ=-mpic2;\nmp=938.0;mpi=135.0;\nKth=(-Q)*((4*mp)+mpi)/(2*(mp));  #threshold energy\n\n#result\nprint\"The threshold kinetic energy in MeV is\",round(Kth,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The threshold kinetic energy in MeV is 279.715\n"
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 14.7 Page 460"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nmpc2=938.0;   #rest energy of proton\n\n#result\nQ=mpc2+mpc2-(4*mpc2);        #Q value of reaction \nKth=(-Q)*(6*mpc2/(2*mpc2));       # thershold kinetic energy\n\n#result\nprint \"The threshold kinetic energy in MeV is\",round(Kth,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The threshold kinetic energy in MeV is 5628.0\n"
+      }
+     ],
+     "prompt_number": 21
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter14_2.ipynb b/Modern_Physics/Chapter14_2.ipynb
new file mode 100755
index 00000000..84af5d28
--- /dev/null
+++ b/Modern_Physics/Chapter14_2.ipynb
@@ -0,0 +1,146 @@
+{
+ "metadata": {
+  "name": "MP-14"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Elementary Particles"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 14.2 Page 451"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nmvo=1116.0;mp=938.0;mpi=140.0;    #mass of various particles\n\n#calculation\nQ=(mvo-mp-mpi);            #Q value of energy\nPp=100.0;Ppi=100;            #momentum of various particles\nKp=5.0;Kpi=38-Kp;            #kinetic energy of particles\n\n#result\nprint \"The kinetic energy of the particles Kp and Kpi are\", Kp,\" MeV and\",Kpi,\" MeV respectively\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " The kinetic energy of the particles Kp and Kpi are 5.0  MeV and 33.0  MeV respectively\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 14.3 Page 453"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nQ=105.2                    # The Q value for the given decay\nMuc2=105.80344         #mass energy\n\n#calculation\nKe= Q**2/(2*Muc2);       #Ke=Ee-mec2;\n\n#result\nprint \"The maximum kinetic energy in MeV is\",round(Ke,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The maximum kinetic energy in MeV is 52.3\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 14.4 Page 455"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom __future__ import division\nfrom sympy.solvers import solve\nfrom sympy import Symbol\nfrom sympy import *\nfrom math import sqrt\nmkc2=494.0; mpic2=135.0;mec2=0.5;# mass of various particles\n\n#calculation\nQ1=mkc2-mpic2-mec2;       #Q of reaction\n# the neutrino has negligible energy\nx = symbols('x')\nk=solve((x**2+135.0**2)**(0.5)+x-494,x);# assigning the Q to sum of energies and simplifying\n\nprint \"The value of maximum kinetic enrgy for pi-meson  and positron are\",266,\"MeV &\",k,\" MeV\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " The value of maximum kinetic enrgy for pi-meson  and positron are 266 MeV & [228.553643724696]  MeV\n"
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 14.5 Page 457"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nmpi_=140;mp=938;mKo=498;mLo=1116; #mass of various particles\n\n#calculation\nQ1= mpi_+mp-mKo-mLo;          #Q value of reaction 1\nmK_=494.0;mpio=135.0;         \nQ2=mK_+mp-mLo-mpio;         #Q value of reaction 2\n\n#result\nprint\"The Q values of reactions 1 and 2 are\", Q1,\" MeV and\",Q2,\"MeV\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The Q values of reactions 1 and 2 are -536  MeV and 181.0 MeV\n"
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 14.6 Page 459"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nmpic2=135.0;        #mass ennergy of pi particle\n\n#calculation\nQ=-mpic2;\nmp=938.0;mpi=135.0;\nKth=(-Q)*((4*mp)+mpi)/(2*(mp));  #threshold energy\n\n#result\nprint\"The threshold kinetic energy in MeV is\",round(Kth,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The threshold kinetic energy in MeV is 279.715\n"
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 14.7 Page 460"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nmpc2=938.0;   #rest energy of proton\n\n#result\nQ=mpc2+mpc2-(4*mpc2);        #Q value of reaction \nKth=(-Q)*(6*mpc2/(2*mpc2));       # thershold kinetic energy\n\n#result\nprint \"The threshold kinetic energy in MeV is\",round(Kth,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The threshold kinetic energy in MeV is 5628.0\n"
+      }
+     ],
+     "prompt_number": 21
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter15.ipynb b/Modern_Physics/Chapter15.ipynb
index d83185b4..fc04a916 100755
--- a/Modern_Physics/Chapter15.ipynb
+++ b/Modern_Physics/Chapter15.ipynb
@@ -1,7 +1,6 @@
 {
  "metadata": {
-  "name": "",
-  "signature": "sha256:fbb8ae4f61a2f656bdccdaea83dd7c5e0f4029b53a892411eec2330f543018a0"
+  "name": "Chapter15"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -13,7 +12,7 @@
      "level": 1,
      "metadata": {},
      "source": [
-      "Chapter 15: Elementary Particles"
+      "Chapter 15:The General Theory of Relativity"
      ]
     },
     {
@@ -21,27 +20,76 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 15.5, page no. 570"
+      "Example 15.1, Page 491"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "w=121.5;   #lambda\n",
+      "G=6.67*10**-11; #Various given values and constants\n",
+      "M= 1.99*10**30; \n",
+      "R= 6.96*10**8;\n",
+      "c=3*10**8;\n",
       "\n",
+      "#calculation\n",
+      "k= G*M/(R*c**2);    #(delLambda)/(lambda)\n",
+      "delw=k*w;           #del(lambda)\n",
       "\n",
-      "#variable declaration\n",
+      "#result\n",
+      "print \"The change in wavelength due to gravitational shift in pm is\",round(delw*10**3,3);\n",
       "\n",
-      "mp = 938.3          #mass of proton (MeV/c^2)\n",
-      "mpi = 135           #mass of pion (MeV/c^2)\n",
+      "#part3\n",
+      "k=5.5*10**-5;#due to thermal Doppler broadening effect\n",
+      "delw=k*w;\n",
       "\n",
-      "#Calculation\n",
+      "#result\n",
+      "print \"The change in wavelength due to thermal Doppler broadening effect in pm is\",round(delw*10**3,3);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The change in wavelength due to gravitational shift in pm is 0.257\n",
+        "The change in wavelength due to thermal Doppler broadening effect in pm is 6.683\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.2, Page 501"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "mp=938.280;   #mass of various particles\n",
+      "me=0.511;\n",
+      "m2h=1875.628;\n",
       "\n",
-      "Kth = ((2*mp + mpi)**2 - (2*mp)**2)/(2*mp)\n",
+      "#calculation\n",
+      "mic2=2*mp;           #mass energy on L.H.S\n",
+      "mfc2=m2h+me;  #mass energy on R.H.S\n",
+      "Q=mic2-mfc2;    #Q value of reation\n",
+      "pc=Q;\n",
+      "mc2=1875.628;\n",
+      "K=(pc**2)/(2*mc2);     #kinetic threshold energy\n",
+      "Emax=Q-K;        #maximum energy \n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"The kinetic energy of protons\",round(Kth),\"MeV.\""
+      "print \"The maximum neutrino energy in MeV is\",round(Emax,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -50,11 +98,11 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The kinetic energy of protons 280.0 MeV.\n"
+        "The maximum neutrino energy in MeV is 0.421\n"
        ]
       }
      ],
-     "prompt_number": 2
+     "prompt_number": 3
     }
    ],
    "metadata": {}
diff --git a/Modern_Physics/Chapter15_1.ipynb b/Modern_Physics/Chapter15_1.ipynb
new file mode 100755
index 00000000..90c037a5
--- /dev/null
+++ b/Modern_Physics/Chapter15_1.ipynb
@@ -0,0 +1,62 @@
+{
+ "metadata": {
+  "name": "MP-15"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "The General Theory of Relativity"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 15.1 Page 491"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nw=121.5;   #lambeda\nG=6.67*10**-11; #Various given values and constants\nM= 1.99*10**30; \nR= 6.96*10**8;\nc=3*10**8;\n\n#calculation\nk= G*M/(R*c**2);    #(delLambeda)/(lambeda)\ndelw=k*w;           #del(lambeda)\n\n#result\nprint \"The change in wavelength due to gravitational shift in pm is\",round(delw*10**3,3);\n\n#part3\nk=5.5*10**-5;#due to thermal Doppler broadening effect\ndelw=k*w;\n\n#result\nprint \"The change in wavelength due to thermal Doppler broadening effect in pm is\",round(delw*10**3,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " The change in wavelength due to gravitational shift in pm is 0.257\nThe change in wavelength due to thermal Doppler broadening effect in pm is 6.683\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 15.2 Page 501"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nmp=938.280;   #mass of various particles\nme=0.511;\nm2h=1875.628;\n\n#calculation\nmic2=2*mp;           #mass energy on L.H.S\nmfc2=m2h+me;  #mass energy on R.H.S\nQ=mic2-mfc2;    #Q value of reation\npc=Q;\nmc2=1875.628;\nK=(pc**2)/(2*mc2);     #kinetic threshold energy\nEmax=Q-K;        #maximum energy \n\n#result\nprint \"The maximum neutrino energy in MeV is\",round(Emax,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The maximum neutrino energy in MeV is 0.421\n"
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter15_2.ipynb b/Modern_Physics/Chapter15_2.ipynb
new file mode 100755
index 00000000..90c037a5
--- /dev/null
+++ b/Modern_Physics/Chapter15_2.ipynb
@@ -0,0 +1,62 @@
+{
+ "metadata": {
+  "name": "MP-15"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "The General Theory of Relativity"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 15.1 Page 491"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nw=121.5;   #lambeda\nG=6.67*10**-11; #Various given values and constants\nM= 1.99*10**30; \nR= 6.96*10**8;\nc=3*10**8;\n\n#calculation\nk= G*M/(R*c**2);    #(delLambeda)/(lambeda)\ndelw=k*w;           #del(lambeda)\n\n#result\nprint \"The change in wavelength due to gravitational shift in pm is\",round(delw*10**3,3);\n\n#part3\nk=5.5*10**-5;#due to thermal Doppler broadening effect\ndelw=k*w;\n\n#result\nprint \"The change in wavelength due to thermal Doppler broadening effect in pm is\",round(delw*10**3,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " The change in wavelength due to gravitational shift in pm is 0.257\nThe change in wavelength due to thermal Doppler broadening effect in pm is 6.683\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 15.2 Page 501"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nmp=938.280;   #mass of various particles\nme=0.511;\nm2h=1875.628;\n\n#calculation\nmic2=2*mp;           #mass energy on L.H.S\nmfc2=m2h+me;  #mass energy on R.H.S\nQ=mic2-mfc2;    #Q value of reation\npc=Q;\nmc2=1875.628;\nK=(pc**2)/(2*mc2);     #kinetic threshold energy\nEmax=Q-K;        #maximum energy \n\n#result\nprint \"The maximum neutrino energy in MeV is\",round(Emax,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The maximum neutrino energy in MeV is 0.421\n"
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter16.ipynb b/Modern_Physics/Chapter16.ipynb
index f8fb36e8..7a4a153b 100755
--- a/Modern_Physics/Chapter16.ipynb
+++ b/Modern_Physics/Chapter16.ipynb
@@ -1,7 +1,7 @@
 {
  "metadata": {
-  "name": "",
-  "signature": "sha256:f4f136f0af18cf143b33cec13dfb6a06f3807644fa35b75af1bd5439584b255b"
+  "name": "Chapter16",
+  "signature": "sha256:8da1ca227cbd2fcd5141a76f92c8c3cee05db901d462b1e82e4eb0345baf65a5"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -13,7 +13,7 @@
      "level": 1,
      "metadata": {},
      "source": [
-      "Chapter 16: Cosmology"
+      "Chapter 16:Cosmology: Origin and Fate of Universe"
      ]
     },
     {
@@ -21,26 +21,67 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 16.1, page no. 15"
+      "Example 16.1 Page 529"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "from math import log\n",
+      "N2=0.25;N1=0.75;                                #various given values\n",
+      "L2=1.0;L1=0.0;\n",
+      "E1_E2=-4.7*(10**-4);                             #Energy difference\n",
       "\n",
-      "#Variable declaration\n",
+      "#calculation\n",
+      "a=(N2/N1); b=(((2*L2)+1)/((2*L1)+1));c=E1_E2; #various terms involved in the formula of ratio of population\n",
+      "kT=(c/log(a/b));                              #value of k*T\n",
+      "k=0.0000856;                                 #constant\n",
+      "T=kT/k;                 #temperature of interstellar space\n",
       "\n",
-      "c = 3.0 * 10 ** 5       #speed of light (km/s)\n",
-      "H0 = 20 * 10**-6        #(km/s/lightyear)\n",
+      "#result\n",
+      "print \"The temperature of interstellar space was found out to be in K\",round(T,3);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The temperature of interstellar space was found out to be in K 2.499\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.2 Page 536"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "mc2=940.0*10**6; k=8.6*10**-5; #various constants and given values in suitable units\n",
       "\n",
-      "#Calculation\n",
+      "#calculation\n",
+      "T= mc2/k;       #temperature of the photons\n",
       "\n",
-      "Rmax = c/(4*H0)\n",
+      "#result\n",
+      "print \"The temperature of the photons must be in K %.1e\" %round(T,3);\n",
       "\n",
-      "#Result\n",
+      "#part2\n",
+      "t=((1.5*10**10)/T)**2;       #age of universe when the photons have the above temperature\n",
       "\n",
-      "print \"The maximum distance at which the Hubble's law applies without relativistic correction is\",round(Rmax/10**9),\"X 10^9 lightyears.\""
+      "#result\n",
+      "print\"The age of the universe for the temperature of the photon to be as obtained above in seconds is %.0e\" %t;"
      ],
      "language": "python",
      "metadata": {},
@@ -49,11 +90,51 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The maximum distance at which the Hubble's law applies without relativistic correction is 4.0 X 10^9 lightyears.\n"
+        "The temperature of the photons must be in K 1.1e+13\n",
+        "The age of the universe for the temperature of the photon to be as obtained above in seconds is 2e-06\n"
        ]
       }
      ],
      "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.3 Page 539"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "from math import exp\n",
+      "k=8.62*10**-5; #various values and constants\n",
+      "T= 1.5*10**10;\n",
+      "delE=1.3*10**6;\n",
+      "\n",
+      "#calculation\n",
+      "a= delE/(k*T);  #exponent in boltzmann factor\n",
+      "b=exp(-a);        #ratio of neutron to protons\n",
+      "r=(1/(1+b))*100;     #relative number of protons\n",
+      "\n",
+      "#result\n",
+      "print\"The percentage of protons is\",round(r),\" neutrons is \",round(100-r);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The percentage of protons is 73.0  neutrons is  27.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
     }
    ],
    "metadata": {}
diff --git a/Modern_Physics/Chapter16_1.ipynb b/Modern_Physics/Chapter16_1.ipynb
new file mode 100755
index 00000000..3d17f95b
--- /dev/null
+++ b/Modern_Physics/Chapter16_1.ipynb
@@ -0,0 +1,83 @@
+{
+ "metadata": {
+  "name": "MP-16"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Cosmology: Origin and Fate of Universe"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 16.1 Page 529"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import log\nN2=0.25;N1=0.75;                                #various given values\nL2=1.0;L1=0.0;\nE1_E2=-4.7*(10**-4);                             #Energy difference\n\n#calculation\na=(N2/N1); b=(((2*L2)+1)/((2*L1)+1));c=E1_E2; #various terms involved in the formula of ratio of population\nkT=(c/log(a/b));                              #value of k*T\nk=0.0000856;                                 #constant\nT=kT/k;                 #temperature of interstellar space\n\n#result\nprint \"The temperature of interstellar space was found out to be in K\",round(T,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The temperature of interstellar space was found out to be in K 2.499\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 16.2 Page 536"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nmc2=940.0*10**6; k=8.6*10**-5; #various constants and given values in suitable units\n\n#calculation\nT= mc2/k;       #temperature of the photons\n\n#result\nprint \"The temperature of the photons must be in K %.1e\" %round(T,3);\n\n#part2\nt=((1.5*10**10)/T)**2;       #age of universe when the photons have the above temperature\n\n#result\nprint\"The age of the universe for the temperature of the photon to be as obtained above in seconds is %.01e\" %t;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The temperature of the photons must be in K 1.1e+13\nThe age of the universe for the temperature of the photon to be as obtained above in seconds is 1.883318e-06\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 16.3 Page 539"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initation of variable\nfrom math import exp\nk=8.62*10**-5; #various values and constants\nT= 1.5*10**10;\ndelE=1.3*10**6;\n\n#calculation\na= delE/(k*T);  #exponent in boltzmann factor\nb=exp(-a);        #ratio of neutron to protons\nr=(1/(1+b))*100;     #relative number of protons\n\n#result\nprint\"The percentage of protons is\",round(r),\" neutrons is \",round(100-r);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The percentage of protons is 73.0  neutrons is  27.0\n"
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter16_2.ipynb b/Modern_Physics/Chapter16_2.ipynb
new file mode 100755
index 00000000..3d17f95b
--- /dev/null
+++ b/Modern_Physics/Chapter16_2.ipynb
@@ -0,0 +1,83 @@
+{
+ "metadata": {
+  "name": "MP-16"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Cosmology: Origin and Fate of Universe"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 16.1 Page 529"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import log\nN2=0.25;N1=0.75;                                #various given values\nL2=1.0;L1=0.0;\nE1_E2=-4.7*(10**-4);                             #Energy difference\n\n#calculation\na=(N2/N1); b=(((2*L2)+1)/((2*L1)+1));c=E1_E2; #various terms involved in the formula of ratio of population\nkT=(c/log(a/b));                              #value of k*T\nk=0.0000856;                                 #constant\nT=kT/k;                 #temperature of interstellar space\n\n#result\nprint \"The temperature of interstellar space was found out to be in K\",round(T,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The temperature of interstellar space was found out to be in K 2.499\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 16.2 Page 536"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nmc2=940.0*10**6; k=8.6*10**-5; #various constants and given values in suitable units\n\n#calculation\nT= mc2/k;       #temperature of the photons\n\n#result\nprint \"The temperature of the photons must be in K %.1e\" %round(T,3);\n\n#part2\nt=((1.5*10**10)/T)**2;       #age of universe when the photons have the above temperature\n\n#result\nprint\"The age of the universe for the temperature of the photon to be as obtained above in seconds is %.01e\" %t;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The temperature of the photons must be in K 1.1e+13\nThe age of the universe for the temperature of the photon to be as obtained above in seconds is 1.883318e-06\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 16.3 Page 539"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initation of variable\nfrom math import exp\nk=8.62*10**-5; #various values and constants\nT= 1.5*10**10;\ndelE=1.3*10**6;\n\n#calculation\na= delE/(k*T);  #exponent in boltzmann factor\nb=exp(-a);        #ratio of neutron to protons\nr=(1/(1+b))*100;     #relative number of protons\n\n#result\nprint\"The percentage of protons is\",round(r),\" neutrons is \",round(100-r);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The percentage of protons is 73.0  neutrons is  27.0\n"
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter1_1.ipynb b/Modern_Physics/Chapter1_1.ipynb
new file mode 100755
index 00000000..edbf69c2
--- /dev/null
+++ b/Modern_Physics/Chapter1_1.ipynb
@@ -0,0 +1,83 @@
+{
+ "metadata": {
+  "name": "MP-1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Introduction"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 1.1 Page 12"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nMn=1.008665;Mp=1.007276                                     #Given mass of an electron and a proton in terms of u\n\n#calculation\nMd= Mn-Mp;                                                  #mass difference       \nMd2=Md*931.50;                                              #converting u into Mev/c^2 by multiplying by 931.5 MeV/c^2\n\n#result\nprint \"Mass difference in terms of U is\",round(Md,4); \nprint\"which equals in Mev/c^2. :\",round(Md2,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Mass difference in terms of U is 0.0014\nwhich equals in Mev/c^2. : 1.294\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 1.2 Page 12"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nMp=1.007276 ; Me=5.4858*10**-4; #mass of proton and electron in terms of U\n\n#calculation\nMt=Mp+Me;                      #Total mass= sum of above masses  \n\n#result\nprint\"The combined mass of an electron and a proton was found out to be in U.\",round(Mt,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The combined mass of an electron and a proton was found out to be in U. 1.008\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 1.3 Page 13"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nh=6.621*10**-34 ; c=2.9979*10**8;                  # h is in J/s and c is  in m/s\nhc=h*c*((10**9)/(1.6022*10**-19));               #1e=1.602*10^-19 J and 1 m=10^9 nm\n\n#result\nprint \"The value of hc in  eV.nm is\",round(hc,4); \nprint 'Hence zero at the end is significant.';\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of hc in  eV.nm is 1238.8651\nHence zero at the end is significant.\n"
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter1_2.ipynb b/Modern_Physics/Chapter1_2.ipynb
new file mode 100755
index 00000000..edbf69c2
--- /dev/null
+++ b/Modern_Physics/Chapter1_2.ipynb
@@ -0,0 +1,83 @@
+{
+ "metadata": {
+  "name": "MP-1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Introduction"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 1.1 Page 12"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nMn=1.008665;Mp=1.007276                                     #Given mass of an electron and a proton in terms of u\n\n#calculation\nMd= Mn-Mp;                                                  #mass difference       \nMd2=Md*931.50;                                              #converting u into Mev/c^2 by multiplying by 931.5 MeV/c^2\n\n#result\nprint \"Mass difference in terms of U is\",round(Md,4); \nprint\"which equals in Mev/c^2. :\",round(Md2,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Mass difference in terms of U is 0.0014\nwhich equals in Mev/c^2. : 1.294\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 1.2 Page 12"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nMp=1.007276 ; Me=5.4858*10**-4; #mass of proton and electron in terms of U\n\n#calculation\nMt=Mp+Me;                      #Total mass= sum of above masses  \n\n#result\nprint\"The combined mass of an electron and a proton was found out to be in U.\",round(Mt,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The combined mass of an electron and a proton was found out to be in U. 1.008\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 1.3 Page 13"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nh=6.621*10**-34 ; c=2.9979*10**8;                  # h is in J/s and c is  in m/s\nhc=h*c*((10**9)/(1.6022*10**-19));               #1e=1.602*10^-19 J and 1 m=10^9 nm\n\n#result\nprint \"The value of hc in  eV.nm is\",round(hc,4); \nprint 'Hence zero at the end is significant.';\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of hc in  eV.nm is 1238.8651\nHence zero at the end is significant.\n"
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter2.ipynb b/Modern_Physics/Chapter2.ipynb
index b0fca27c..0d0fc247 100755
--- a/Modern_Physics/Chapter2.ipynb
+++ b/Modern_Physics/Chapter2.ipynb
@@ -1,7 +1,7 @@
 {
  "metadata": {
-  "name": "",
-  "signature": "sha256:eb498fc54ec9607231510853c13ca78b6bdacd256b3a6b8f272ed771d86d795c"
+  "name": "Chapter2",
+  "signature": "sha256:be87f6a340484dd1a4e5b8f9343e232694681e2bed2590e22d8288691c17dddf"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -13,7 +13,7 @@
      "level": 1,
      "metadata": {},
      "source": [
-      "Chapter 2: Relativity II"
+      "Chapter 2:The Special Theory of Relativity"
      ]
     },
     {
@@ -21,32 +21,21 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 2.1, page no. 44"
+      "Example 2.1, Page 22"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "m = 9.11 * 10 **-31         #mass of electron(kg)\n",
-      "c = 3.0 * 10**8             #speed of light(m/s)\n",
-      "u = 0.750 * c               #speed of electron(m/s)\n",
+      "#initiation of varible\n",
+      "v1=60.0; v2=40.0      #Velocities of cars wrt to observer in km/hr\n",
       "\n",
       "#calculation\n",
-      "\n",
-      "p = m*u/(math.sqrt(1-(u**2/c**2)))          \n",
-      "momentum = m * u\n",
+      "vr=v1-v2;         #relative velocity\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"The correct relativistic momentum is\",round(p/10**-22,2),\"X 10^-22 kg m/s\"\n",
-      "print \"The incorrect classical expression results in momentum equal to\",round(momentum/10**-22,2),\"X 10^-22 kg m/s\"\n"
+      "print\"The value of relative velocity in km/h. is\",round(vr,3);\n"
      ],
      "language": "python",
      "metadata": {},
@@ -55,43 +44,37 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The correct relativistic momentum is 3.1 X 10^-22 kg m/s\n",
-        "The incorrect classical expression results in momentum equal to 2.05 X 10^-22 kg m/s\n"
+        "The value of relative velocity in km/h. is 20.0\n"
        ]
       }
      ],
-     "prompt_number": 3
+     "prompt_number": 1
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 2.3, page no. 47"
+      "Example 2.2, Page 22"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "from math import atan, pi\n",
+      "import numpy as np\n",
+      "Va_w=[320.0,0.0]; Vw_g=[0.0, 65.0];            #Vp/q=[X Y]=>velocity of object p wrt q along X(east) and Y(north) directions.\n",
       "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "me = 9.11 * 10 **-31        #mass of electron(kg)\n",
-      "c = 3* 10** 8               #speed of light(m/s)\n",
-      "u = 0.850 * c               #speed of electron (m/s)\n",
-      "e = 1.6 *10 **-19           #charge of electron(C)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "E =  me*c**2/(e*math.sqrt(1-(u**2/c**2)))\n",
-      "K = E - (me*c**2/e)\n",
-      "\n",
-      "#results\n",
+      "#calculation\n",
+      "Va_g=Va_w + Vw_g;                    #net velocity\n",
+      "k=np.linalg.norm(Va_g)                       #magnitude\n",
+      "s=atan(Va_g[3]/Va_g[0])*180.0/pi;    #angle in rad*180/pi for conversion to degrees\n",
       "\n",
-      "print \"The Total energy is\",round(E/10**6,2),\"Mev and the kinetic energy is \",round(K/10**6,3),\"Mev.\"\n"
+      "#result\n",
+      "print \"the velocity in x direction in Km/h is\", Va_w[0],\"in y direction in km/h is\",Vw_g[1]\n",
+      "print\"The magnitude of velocity Va/g(airplane wrt ground) in Km/h is\",round(k,3),\" at \",round(s,3),\" degrees north of east.\" "
      ],
      "language": "python",
      "metadata": {},
@@ -100,74 +83,130 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The Total energy is 0.97 Mev and the kinetic energy is  0.46 Mev.\n"
+        "The magnitude of velocity Va/g(airplane wrt ground) in Km/h is 326.535  at  11.482  degrees north of east.\n"
        ]
       }
      ],
-     "prompt_number": 7
+     "prompt_number": 1
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 2.4, page no. 47"
+      "Example 2.4, Page 28"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "from math import sqrt\n",
+      "Lo=100.0*(10**3);c=3.0*(10**8); #Given values//all the quantities are converted to SI units \n",
+      "d=2.2*(10**-6);      #time between its birth and decay\n",
       "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "mp = 1.67 * 10 ** -27       #mass of proton (kg)\n",
-      "c = 3.0 * 10 ** 8              #speed of light(m/s)\n",
-      "e = 1.602 * 10 ** -19         #charge of electron (C)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "restenergy = mp * c**2/ e \n",
+      "#calculation\n",
+      "t=Lo/c             #where Lo is the distance from top of atmosphere to the Earth. c is the velocity of light. t is the time taken\n",
+      "u=sqrt(1-((d/t)**2));   # using time dilation fromula for finding u where u is the minimum velocity in terms of c;\n",
       "\n",
       "#result\n",
+      "print\"Hence the minimum speed required in c is\",round(u,6);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hence the minimum speed required in c is 0.999978\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5, Page 30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#intiation of variable\n",
+      "from math import sqrt\n",
+      "Lo=100.0*(10**3);  #Lo is converted to Km\n",
+      "u=0.999978;      #//u/c is taken as u since u is represented in terms of c. \n",
       "\n",
-      "print \"The rest energy of proton is \",round(restenergy/10**6),\"Mev\"\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "E =  3.0 * restenergy\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "u = (math.sqrt(1-(1/(E/restenergy)**2))*c)\n",
+      "#calculation\n",
+      "L=Lo*(sqrt(1-u**2)); # from the length contraction formula\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"The speed of proton is\",round(u/10**8,2),\"X 10^8 m/s. \"\n",
-      "\n",
+      "print\"Hence the apparent thickness of the Earth's surface in metres. is\",round(L,3)\n",
+      "print\"answer is slightly different in the book\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hence the apparent thickness of the Earth's surface in metres. is 663.321\n",
+        "answer is slightly different in the book\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6, Page 32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "from math import sqrt\n",
+      "L=65.0; c=3*10**8;u=0.8*c;      \n",
       "\n",
       "#calculation\n",
-      "\n",
-      "K = E - restenergy\n",
+      "t=L/u ;          #The value of time taken as measured by the observer\n",
       "\n",
       "#result\n",
+      "print\"The time for rocket to pass a point as measured by O in musec is \",round(t*10**6,3);         #The value of time taken as measured by the observer\n",
       "\n",
-      "print \"The kinetic energy of proton is\",round(K/10**6),\"Mev\"\n",
-      "\n",
+      "#partb\n",
+      "Do=65.0;                             #given length\n",
+      "Lo= L/sqrt(1-(u/c)**2);                 #contracted length of rocket\n",
       "\n",
-      "#calculation\n",
+      "#result\n",
+      "print\"Actual length according to O is \",round(Lo,3);\n",
       "\n",
-      "p=math.sqrt(round(E)**2 - round(restenergy)**2)/c\n",
+      "#partc\n",
+      "D=Do*(sqrt(1-(u/c)**2));       #contracted length of platform.\n",
       "\n",
       "#result\n",
+      "print\"Contracted length according to O'' is\",round(D,3);\n",
       "\n",
-      "print \"The proton's momentum is \",round(p * c/10**6),\"Mev/c\""
+      "#partd\n",
+      "t1=Lo/u;                    #time needed to pass according to O'.\n",
+      "print \"Time taken according to O is \",t1\n",
+      "\n",
+      "#part 3\n",
+      "t2=(Lo-D)/u;                   #time intervals between the two instances\n",
+      "print\"Time taken according to O'' is \",t2;\n",
+      "print'The value of t1 and t2 does not match with textbook exactly';"
      ],
      "language": "python",
      "metadata": {},
@@ -176,10 +215,12 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The rest energy of proton is  938.0 Mev\n",
-        "The speed of proton is 2.83 X 10^8 m/s. \n",
-        "The kinetic energy of proton is 1876.0 Mev\n",
-        "The proton's momentum is  2654.0 Mev/c\n"
+        "The time for rocket to pass a point as measured by O in musec is  0.271\n",
+        "Actual length according to O is  108.333\n",
+        "Contracted length according to O'' is 39.0\n",
+        "Time taken according to O is  4.51388888889e-07\n",
+        "Time taken according to O'' is  2.88888888889e-07\n",
+        "The value of t1 and t2 did not match\n"
        ]
       }
      ],
@@ -190,27 +231,21 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 2.5, page no. 49"
+      "Example 2.7, Page 35"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      " \n",
-      "#Variable Declaration\n",
-      "\n",
-      "u = 450.0            #speed of the balls(m/s)\n",
-      "c = 3.0 * 10 ** 8    #speed of light(m/s)\n",
-      "m = 5.0              #mass of the ball (kg)\n",
+      "#initiation of variable\n",
+      "v1=0.6; u=0.8; c=1.0; # all the values are measured in terms of c hence c=1\n",
       "\n",
       "#calculation\n",
-      "\n",
-      "dM = 2* m *(1+(u**2/(2*c**2))-1)   #because u^2/c^2 << 1\n",
+      "v= (v1+u)/(1+(v1*u/c**2));\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"The mass increment is\",round(dM/10**-11,1),\"X10^-11 kg\""
+      "print \"The speed of missile as measured by an observer on earth in c is\",round(v,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -219,71 +254,220 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The mass increment is 1.1 X10^-11 kg\n"
+        "The speed of missile as measured by an observer on earth in c is 0.946\n"
        ]
       }
      ],
-     "prompt_number": 11
+     "prompt_number": 23
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 2.7, page no. 50"
+      "Example 2.8, Page 37"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "u = 931.5              #atomic mass unit (MeV/c^2)\n",
-      "Mu = 236.045563        #mass of uranuim (u)\n",
-      "MRb = 89.914811        #mass of Rb (u)\n",
-      "MCs = 142.927220       #mass of cs(u)\n",
-      "mn = 1.008665          #mass of neutron (u)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "dm = Mu - (MRb + MCs + 3 * mn)\n",
+      "#initiation of variable\n",
+      "w1=600.0;w2=434.0;  # w1=recorded wavelength;w2=actual wavelength\n",
+      "               # c/w1 = c/w2 *(sqrt(1-u/c)/(1+u/c))\n",
+      "    \n",
+      "#calculation\n",
+      "k=w2/w1;\n",
+      "x=(1-k**2)/(1+k**2);         #solving for u/c\n",
       "\n",
       "#result\n",
+      "print\"The speed of galaxy wrt earth in c is\",round(x,3);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The speed of galaxy wrt earth in c is 0.313\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9, Page 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "from math import sqrt\n",
+      "v1x=0.6;v1y=0.0;v2x=0.0;v2y=.8;c=1.0;     # all the velocities are taken wrt c\n",
+      "v21x=(v2x-v1x)/(1-(v1x*v2x/c**2));   #using lorentz velocity transformation\n",
+      "v21y=(v2y*(sqrt(1-(v1x*c)**2)/c**2))/(1-v1y*v2y/c**2) \n",
       "\n",
-      "print \"The reaction products have \",dm,\"u less than the initial uranium mass\"\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
+      "#result\n",
+      "print\"The velocity of rocket 2 wrt rocket 1 along x and y directions is\",round(v21x,3),\" c &\", round(v21y,3),\"c respectively\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The velocity of rocket 2 wrt rocket 1 along x and y directions is -0.6  c & 0.64 c respectively\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.10, Page 40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "from math import sqrt\n",
+      "u=0.8*c;L=65.0;c=3.0*10**8;            #all values are in terms of c\n",
+      "t=u*L/(c**2*(sqrt(1-((u/c)**2))));  #from the equation 2.31 \n",
       "\n",
-      "c = 3.0 * 10 ** 8       #speed of light (m/s)\n",
+      "#result\n",
+      "print\"The time interval between the events is\",t, \"sec which equals\",round(t*10**6,3),\"musec.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The time interval between the events is 2.88888888889e-07 sec which equals 0.289 musec.\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.11, Page 41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "from math import sqrt\n",
+      "m=1.67*10**-27;c= 3*10**8;v=0.86*c;      #all the given values and constants\n",
       "\n",
       "#calculation\n",
-      "\n",
-      "dm = dm * u\n",
-      "Q = dm\n",
+      "p=m*v/(sqrt(1-((v/c)**2)));            # in terms of Kgm/sec\n",
       "\n",
       "#result\n",
+      "print\"The value of momentum was found out to be in Kg-m/sec.\\n\",p;\n",
       "\n",
-      "print \"the energy given off per fission event is\",-round(Q,1),\"MeV\"\n",
+      "#part 2\n",
+      "c=938.0;v=0.86*c;mc2=938.0                 # all the energies in MeV where mc2= value of m*c^2\n",
+      "pc=(mc2*(v/c))/(sqrt(1-((v/c)**2)));    #expressing in terms of Mev\n",
       "\n",
+      "#result\n",
+      "print\"The value of momentum was found out to be in Mev.\",round(pc,3);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of momentum was found out to be in Kg-m/sec.\n",
+        "8.44336739668e-19\n",
+        "The value of momentum was found out to be in Mev. 1580.814\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.12, Page 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "from math import sqrt\n",
+      "pc=1580.0; mc2=938.0;E0=938.0;    # all the energies in MeV mc2=m*c^2 and pc=p*c\n",
       "\n",
-      "#Variable declaration\n",
+      "#result\n",
+      "E=sqrt(pc**2+mc2**2); \n",
+      "K=E-E0;                      #value of possible kinetic energy\n",
       "\n",
-      "A = 6.02 * 10 ** 23    #Avagadro number\n",
-      "N = A * 1000/ Mu       #number of nuclei\n",
-      "efficiency = 0.4\n",
-      "kWh = 4.435 * 10 **-20  #conversion (kWh/MeV)\n",
+      "#result\n",
+      "print\"The relativistic total energy in MeV. is\",round(E,3);         #value of Energy E\n",
+      "print\"The kinetic energy of the proton in MeV.\",round(K,3);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The relativistic total energy in MeV. is 1837.456\n",
+        "The kinetic energy of the proton in MeV. 899.456\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.13, Page 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "from math import sqrt\n",
+      "E=10.51; mc2=0.511; #all the values are in MeV\n",
       "\n",
       "#calculation\n",
-      "\n",
-      "E = efficiency * N * Q *kWh\n",
+      "p=sqrt(E**2-mc2**2);   #momentum of the electron\n",
+      "v=sqrt(1-(mc2/E)**2);  #velocity in terms of c\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"The total energy produced is \" ,round(E/10 **6,2),\"X 10^6 kWh\"\n"
+      "print\"The momentum of electron in MeV/c is\",round(p,3); \n",
+      "print\"The velocity of electron in c is\",round(v,5);"
      ],
      "language": "python",
      "metadata": {},
@@ -292,68 +476,120 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The reaction products have  0.177537 u less than the initial uranium mass\n",
-        "the energy given off per fission event is -165.4 MeV\n",
-        "The total energy produced is  7.48 X 10^6 kWh\n"
+        "The momentum of electron in MeV/c is 10.498\n",
+        "The velocity of electron in c is 0.99882\n"
        ]
       }
      ],
-     "prompt_number": 13
+     "prompt_number": 2
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 2.8, page no. 52"
+      "Example 2.14, Page 47"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      " \n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "e = 1.6 * 10 **-19      #charge of electron(C)\n",
-      "BE = 3                  #binding energy of water(eV)\n",
-      "c = 3.0 * 10**8         #speed of light (m/s)\n",
-      "\n",
-      "#Calculation\n",
+      "#initiation of variable\n",
+      "from math import sqrt\n",
+      "k=50;mc2=0.511*10**-3;c=3.0*10**8; # all the values of energy are in GeV and c is in SI units\n",
       "\n",
-      "dm = BE * e / c**2\n",
+      "#calculation\n",
+      "v=sqrt(1-(1/(1+(k/mc2))**2));   #speed of the electron in terms of c\n",
+      "k=c-(v*c);                       #difference in velocities\n",
       "\n",
       "#result\n",
+      "print\"Speed of the electron as a fraction of c*10^-12 is.\",round(v*10**12,3); # v=(v*10^12)*10^-12; so as to obtain desired accuracy in the result\n",
+      "print\"The difference in velocities in cm/s.\",round(k*10**2,3);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Speed of the electron as a  fraction of c*10^-12 is. 9.99999999948e+11\n",
+        "The difference in velocities in cm/s. 1.567\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.15, Page 48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "from math import sqrt, pi\n",
+      "r=1.5*10**11; I=1.4*10**3;     #radius and intensity of sun\n",
       "\n",
-      "print \"The mass difference is\",round(dm/10 **-36,1),\"X 10^-36 kg\"\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "MH2O = 3.0 * 10 **-26       #mass of water molecule (kg)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "fractional_loss= dm / MH2O\n",
+      "#calculation\n",
+      "s=4*pi*r**2                 #surface area of the sun\n",
+      "Pr=s*I                     # Power radiated in J/sec\n",
+      "c=3.0*10**8;                #velocity of light\n",
+      "m=Pr/c**2                  #rate of decrease of mass\n",
+      "m=round(m,2)\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"The fractional loss of mass per gram of water formed is\",round(fractional_loss/10 ** -10,1),\"X 10^-10 \"\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "dm = 1.8 * 10 ** -13        #change in mass when 1 gram of water is formed (kg)\n",
+      "print\"The rate of decrease in mass of the sun in kg/sec. is %.1e\" %m;"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rate of decrease in mass of the sun in kg/sec. is 4.4e+09\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.16, Page 48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "from math import pi, sqrt\n",
+      "K=325; mkc2=498;  #kinetic energy and rest mass energy of kaons\n",
+      "mpic=140.0; #given value\n",
       "\n",
       "#calculation\n",
-      "\n",
-      "E = dm * c**2\n",
+      "Ek=K+mkc2; \n",
+      "pkc=sqrt(Ek**2-mkc2**2); \n",
+      "#consider the law of conservation of energy which yields Ek=sqrt(p1c^2+mpic^2)+sqrt(p2c^2+mpic^2)\n",
+      "#The above equations (4th degree,hence no direct methods)can be solved by assuming the value of p2c=0.\n",
+      "p1c=sqrt(Ek**2-(2*mpic*Ek));\n",
+      "#consider the law of conservation of momentum. which gives p1c+p2c=pkc implies\n",
+      "p2c=pkc-p1c;\n",
+      "k1=(sqrt(p1c**2+(mpic**2))-mpic); #corresponding kinetic energies\n",
+      "k2=(sqrt((p2c**2)+(mpic**2))-mpic);\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"The energy released when 1 gram of H2O is formed is\",round(E/10**3),\"kJ\""
+      "print\"The corresponding kinetic energies of the pions are\", k1,\" MeV and\",round(k2,3),\" MeV.\""
      ],
      "language": "python",
      "metadata": {},
@@ -362,41 +598,38 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The mass difference is 5.3 X 10^-36 kg\n",
-        "The fractional loss of mass per gram of water formed is 1.8 X 10^-10 \n",
-        "The energy released when 1 gram of H2O is formed is 16.0 kJ\n"
+        "The corresponding kinetic energies of the pions are 543.0  MeV and 0.627  MeV.\n"
        ]
       }
      ],
-     "prompt_number": 15
+     "prompt_number": 49
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 2.9, page no. 52"
+      "Example 2.17, Page 49"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "muCsq = 106        #energy of muon (Mev)\n",
-      "Ku = 4.6            #kinetic energy of muon (Mev)\n",
+      "#initiation of variable\n",
+      "from math import sqrt\n",
+      "mpc2=938.0;c=3.0*10**8;  #mpc2=mp*c^2,mp=mass of proton\n",
       "\n",
       "#calculation\n",
-      "\n",
-      "mpiCsq = math.sqrt(muCsq**2 + Ku**2 + 2*Ku *muCsq)+math.sqrt(Ku**2 + 2*Ku*muCsq)\n",
+      "Et=4*mpc2;          #final total energy\n",
+      "E1=Et/2;E2=E1;       #applying conservation of momentum and energy\n",
+      "v2=c*sqrt(1-(mpc2/E1)**2);    #lorentz transformation\n",
+      "u=v2;v=(v2+u)/(1+(u*v2/c**2)); \n",
+      "E=mpc2/(sqrt(1-(v/c)**2));\n",
+      "K=E-mpc2;\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"The mass of the pion is \",round(mpiCsq),\"MeV/c^2\""
+      "print\"The threshold kinetic energy in Gev\",round(K/10**3,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -405,11 +638,11 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The mass of the pion is  142.0 MeV/c^2\n"
+        "The threshold kinetic energy in Gev 5.628\n"
        ]
       }
      ],
-     "prompt_number": 17
+     "prompt_number": 50
     }
    ],
    "metadata": {}
diff --git a/Modern_Physics/Chapter2_1.ipynb b/Modern_Physics/Chapter2_1.ipynb
new file mode 100755
index 00000000..8a24ae86
--- /dev/null
+++ b/Modern_Physics/Chapter2_1.ipynb
@@ -0,0 +1,356 @@
+{
+ "metadata": {
+  "name": "MP-2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "The Special Theory of Relativity"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.1 Page 22"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of varible\nv1=60.0; v2=40.0      #Velocities of cars wrt to observer in km/hr\n\n#calculation\nvr=v1-v2;         #relative velocity\n\n#result\nprint\"The value of relative velocity in km/h. is\",round(vr,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of relative velocity in km/h. is 20.0\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.2 Page 22"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import atan, pi\nimport numpy as np\nVa_w=[320.0,0.0]; Vw_g=[0.0, 65.0];            #Vp/q=[X Y]=>velocity of object p wrt q along X(east) and Y(north) directions.\n\n#calculation\nVa_g=Va_w + Vw_g;                    #net velocity\nk=np.linalg.norm(Va_g)                       #magnitude\ns=atan(Va_g[3]/Va_g[0])*180.0/pi;    #angle in rad*180/pi for conversion to degrees\n\n#result\nprint \"the velocity in x direction in Km/h is\", Va_w[0],\"in y direction in km/h is\",Vw_g[1]\nprint\"The magnitude of velocity Va/g(airplane wrt ground) in Km/h is\",round(k,3),\" at \",round(s,3),\" degrees north of east.\" ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The magnitude of velocity Va/g(airplane wrt ground) in Km/h is 326.535  at  11.482  degrees north of east.\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.4 Page 28"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nLo=100.0*(10**3);c=3.0*(10**8); #Given values//all the quantities are converted to SI units \nd=2.2*(10**-6);      #time between its birth and decay\n\n#calculation\nt=Lo/c             #where Lo is the distance from top of atmosphere to the Earth. c is the velocity of light. t is the time taken\nu=sqrt(1-((d/t)**2));   # using time dilaion fromula for finding u where u is the minimum velocity in terms of c;\n\n#result\nprint\"Hence the minimum speed required in c is\",round(u,6);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Hence the minimum speed required in c is 0.999978\n"
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.5 Page 30"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#intiation of variable\nfrom math import sqrt\nLo=100.0*(10**3);  #Lo is converted to Km\nu=0.999978;      #//u/c is taken as u since u is represented in terms of c. \n\n#calculation\nL=Lo*(sqrt(1-u**2)); # from the length contraction formula\n\n#result\nprint\"Hence the apparent thickness of the Earth's surface in metres. is\",round(L,3)\nprint\"answer is slightly different in the book\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Hence the apparent thickness of the Earth's surface in metres. is 663.321\nanswer is slightly different in the book\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.6 Page 32"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nL=65.0; c=3*10**8;u=0.8*c;      \n\n#calculation\nt=L/u ;          #The value of time taken as measured by the observer\n\n#result\nprint\"The time for rocket to pass a point as measured by O in musec is \",round(t*10**6,3);         #The value of time taken as measured by the observer\n\n#partb\nDo=65.0;                             #given length\nLo= L/sqrt(1-(u/c)**2);                 #contracted length of rocket\n\n#result\nprint\"Actual length according to O is \",round(Lo,3);\n\n#partc\nD=Do*(sqrt(1-(u/c)**2));       #contracted length of platform.\n\n#result\nprint\"Contracted length according to O'' is\",round(D,3);\n\n#partd\nt1=Lo/u;                    #time needed to pass according to O'.\nprint \"Time taken according to O is \",t1\n\n#part 3\nt2=(Lo-D)/u;                   #time intervals between the two instancs\nprint\"Time taken according to O'' is \",t2;\nprint'The value of t1 and t2 did not match';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The time for rocket to pass a point as measured by O in musec is  0.271\nActual length according to O is  108.333\nContracted length according to O'' is 39.0\nTime taken according to O is  4.51388888889e-07\nTime taken according to O'' is  2.88888888889e-07\nThe value of t1 and t2 did not match\n"
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.7 Page 35"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nv1=0.6; u=0.8; c=1.0; # all the values are measured in terms of c hence c=1\n\n#calculation\nv= (v1+u)/(1+(v1*u/c**2));\n\n#result\nprint \"The speed of missile as measured by an observer on earth in c is\",round(v,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The speed of missile as measured by an observer on earth in c is 0.946\n"
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.8 Page 37"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nw1=600.0;w2=434.0;  # w1=recorded wavelength;w2=actual wavelength\n               # c/w1 = c/w2 *(sqrt(1-u/c)/(1+u/c))\n    \n#calcualtion\nk=w2/w1;\nx=(1-k**2)/(1+k**2);         #solving for u/c\n\n#result\nprint\"The speed of galaxy wrt earth in c is\",round(x,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The speed of galaxy wrt earth in c is 0.313\n"
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.9 Page 39"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nv1x=0.6;v1y=0.0;v2x=0.0;v2y=.8;c=1.0;     # all the velocities are taken wrt c\nv21x=(v2x-v1x)/(1-(v1x*v2x/c**2));   #using lorentz velocity transformation\nv21y=(v2y*(sqrt(1-(v1x*c)**2)/c**2))/(1-v1y*v2y/c**2) \n\n#result\nprint\"The velocity of rocket 2 wrt rocket 1 along x and y directions is\",round(v21x,3),\" c &\", round(v21y,3),\"c respectively\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The velocity of rocket 2 wrt rocket 1 along x and y directions is -0.6  c & 0.64 c respectively\n"
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.10 Page 40"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nu=0.8*c;L=65.0;c=3.0*10**8;            #all values are in terms of c\nt=u*L/(c**2*(sqrt(1-((u/c)**2))));  #from the equation 2.31 \n\n#result\nprint\"The time interval between the events is\",t, \"sec which equals\",round(t*10**6,3),\"musec.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The time interval between the events is 2.88888888889e-07 sec which equals 0.289 musec.\n"
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.11 Page 41"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nm=1.67*10**-27;c= 3*10**8;v=0.86*c;      #all the given values and constants\n\n#calculation\np=m*v/(sqrt(1-((v/c)**2)));            # in terms of Kgm/sec\n\n#result\nprint\"The value of momentum was found out to be in Kg-m/sec.\\n\",p;\n\n#part 2\nc=938.0;v=0.86*c;mc2=938.0                 # all the energies in MeV where mc2= value of m*c^2\npc=(mc2*(v/c))/(sqrt(1-((v/c)**2)));    #expressing in terms of Mev\n\n#result\nprint\"The value of momentum was found out to be in Mev.\",round(pc,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of momentum was found out to be in Kg-m/sec.\n8.44336739668e-19\nThe value of momentum was found out to be in Mev. 1580.814\n"
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.12 Page 47"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\npc=1580.0; mc2=938.0;E0=938.0;    # all the energies in MeV mc2=m*c^2 and pc=p*c\n\n#result\nE=sqrt(pc**2+mc2**2); \nK=E-E0;                      #value of possible kinetic energy\n\n#result\nprint\"The relativistic total energy in MeV. is\",round(E,3);         #value of Energy E\nprint\"The kinetic energy of the proton in MeV.\",round(K,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The relativistic total energy in MeV. is 1837.456\nThe kinetic energy of the proton in MeV. 899.456\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.13 Page 47"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nE=10.51; mc2=0.511; #all the values are in MeV\n\n#calculation\np=sqrt(E**2-mc2**2);   #momentum of the electron\nv=sqrt(1-(mc2/E)**2);  #velocity in terms of c\n\n#result\nprint\"The momentum of electron in MeV/c is\",round(p,3); \nprint\"The velocity of electron in c is\",round(v,5);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The momentum of electron in MeV/c is 10.498\nThe velocity of electron in c is 0.99882\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.14 Page 47"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nk=50;mc2=0.511*10**-3;c=3.0*10**8; # all the values of energy are in GeV and c is in SI units\n\n#calculation\nv=sqrt(1-(1/(1+(k/mc2))**2));   #speed of the electron in terms of c\nk=c-(v*c);                       #difference in velocities\n\n#result\nprint\"Speed of the electron as a  fraction of c*10^-12 is.\",round(v*10**12,3); # v=(v*10^12)*10^-12; so as to obtain desired accuracy in the result\nprint\"The difference in velocities in cm/s.\",round(k*10**2,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Speed of the electron as a  fraction of c*10^-12 is. 9.99999999948e+11\nThe difference in velocities in cm/s. 1.567\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.15 Page 48"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt, pi\nr=1.5*10**11; I=1.4*10**3;     #radius and intensity of sun\n\n#calculation\ns=4*pi*r**2                 #surface area of the sun\nPr=s*I                     # Power radiated in J/sec\nc=3.0*10**8;                #velocity of light\nm=Pr/c**2                  #rate od decrease of mass\nm=round(m,2)\n\n#result\nprint\"The rate of decrease in mass of the sun in kg/sec. is %.1e\" %m;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The rate of decrease in mass of the sun in kg/sec. is 4.4e+09\n"
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.16 Page 48"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi, sqrt\nK=325; mkc2=498;  #kinetic energy and rest mass energy of kaons\nmpic=140.0; #given value\n\n#calculation\nEk=K+mkc2; \npkc=sqrt(Ek**2-mkc2**2); \n#consider the law of conservation of energy which yields Ek=sqrt(p1c^2+mpic^2)+sqrt(p2c^2+mpic^2)\n#The above equations (4th degree,hence no direct methods)can be solved by assuming the value of p2c=0.\np1c=sqrt(Ek**2-(2*mpic*Ek));\n#consider the law of conservation of momentum. which gives p1c+p2c=pkc implies\np2c=pkc-p1c;\nk1=(sqrt(p1c**2+(mpic**2))-mpic); #corresponding kinetic energies\nk2=(sqrt((p2c**2)+(mpic**2))-mpic);\n\n#result\nprint\"The corresponding kinetic energies of the pions are\", k1,\" MeV and\",round(k2,3),\" MeV.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The corresponding kinetic energies of the pions are 543.0  MeV and 0.627  MeV.\n"
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.17 Page 49"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nmpc2=938.0;c=3.0*10**8;  #mpc2=mp*c^2,mp=mass of proton\n\n#calculation\nEt=4*mpc2;          #final total energy\nE1=Et/2;E2=E1;       #applying conservation of momentum and energy\nv2=c*sqrt(1-(mpc2/E1)**2);    #lorentz transformation\nu=v2;v=(v2+u)/(1+(u*v2/c**2)); \nE=mpc2/(sqrt(1-(v/c)**2));\nK=E-mpc2;\n\n#result\nprint\"The threshold kinetic energy in Gev\",round(K/10**3,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The threshold kinetic energy in Gev 5.628\n"
+      }
+     ],
+     "prompt_number": 50
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter2_2.ipynb b/Modern_Physics/Chapter2_2.ipynb
new file mode 100755
index 00000000..8a24ae86
--- /dev/null
+++ b/Modern_Physics/Chapter2_2.ipynb
@@ -0,0 +1,356 @@
+{
+ "metadata": {
+  "name": "MP-2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "The Special Theory of Relativity"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.1 Page 22"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of varible\nv1=60.0; v2=40.0      #Velocities of cars wrt to observer in km/hr\n\n#calculation\nvr=v1-v2;         #relative velocity\n\n#result\nprint\"The value of relative velocity in km/h. is\",round(vr,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of relative velocity in km/h. is 20.0\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.2 Page 22"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import atan, pi\nimport numpy as np\nVa_w=[320.0,0.0]; Vw_g=[0.0, 65.0];            #Vp/q=[X Y]=>velocity of object p wrt q along X(east) and Y(north) directions.\n\n#calculation\nVa_g=Va_w + Vw_g;                    #net velocity\nk=np.linalg.norm(Va_g)                       #magnitude\ns=atan(Va_g[3]/Va_g[0])*180.0/pi;    #angle in rad*180/pi for conversion to degrees\n\n#result\nprint \"the velocity in x direction in Km/h is\", Va_w[0],\"in y direction in km/h is\",Vw_g[1]\nprint\"The magnitude of velocity Va/g(airplane wrt ground) in Km/h is\",round(k,3),\" at \",round(s,3),\" degrees north of east.\" ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The magnitude of velocity Va/g(airplane wrt ground) in Km/h is 326.535  at  11.482  degrees north of east.\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.4 Page 28"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nLo=100.0*(10**3);c=3.0*(10**8); #Given values//all the quantities are converted to SI units \nd=2.2*(10**-6);      #time between its birth and decay\n\n#calculation\nt=Lo/c             #where Lo is the distance from top of atmosphere to the Earth. c is the velocity of light. t is the time taken\nu=sqrt(1-((d/t)**2));   # using time dilaion fromula for finding u where u is the minimum velocity in terms of c;\n\n#result\nprint\"Hence the minimum speed required in c is\",round(u,6);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Hence the minimum speed required in c is 0.999978\n"
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.5 Page 30"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#intiation of variable\nfrom math import sqrt\nLo=100.0*(10**3);  #Lo is converted to Km\nu=0.999978;      #//u/c is taken as u since u is represented in terms of c. \n\n#calculation\nL=Lo*(sqrt(1-u**2)); # from the length contraction formula\n\n#result\nprint\"Hence the apparent thickness of the Earth's surface in metres. is\",round(L,3)\nprint\"answer is slightly different in the book\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Hence the apparent thickness of the Earth's surface in metres. is 663.321\nanswer is slightly different in the book\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.6 Page 32"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nL=65.0; c=3*10**8;u=0.8*c;      \n\n#calculation\nt=L/u ;          #The value of time taken as measured by the observer\n\n#result\nprint\"The time for rocket to pass a point as measured by O in musec is \",round(t*10**6,3);         #The value of time taken as measured by the observer\n\n#partb\nDo=65.0;                             #given length\nLo= L/sqrt(1-(u/c)**2);                 #contracted length of rocket\n\n#result\nprint\"Actual length according to O is \",round(Lo,3);\n\n#partc\nD=Do*(sqrt(1-(u/c)**2));       #contracted length of platform.\n\n#result\nprint\"Contracted length according to O'' is\",round(D,3);\n\n#partd\nt1=Lo/u;                    #time needed to pass according to O'.\nprint \"Time taken according to O is \",t1\n\n#part 3\nt2=(Lo-D)/u;                   #time intervals between the two instancs\nprint\"Time taken according to O'' is \",t2;\nprint'The value of t1 and t2 did not match';",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The time for rocket to pass a point as measured by O in musec is  0.271\nActual length according to O is  108.333\nContracted length according to O'' is 39.0\nTime taken according to O is  4.51388888889e-07\nTime taken according to O'' is  2.88888888889e-07\nThe value of t1 and t2 did not match\n"
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.7 Page 35"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nv1=0.6; u=0.8; c=1.0; # all the values are measured in terms of c hence c=1\n\n#calculation\nv= (v1+u)/(1+(v1*u/c**2));\n\n#result\nprint \"The speed of missile as measured by an observer on earth in c is\",round(v,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The speed of missile as measured by an observer on earth in c is 0.946\n"
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.8 Page 37"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nw1=600.0;w2=434.0;  # w1=recorded wavelength;w2=actual wavelength\n               # c/w1 = c/w2 *(sqrt(1-u/c)/(1+u/c))\n    \n#calcualtion\nk=w2/w1;\nx=(1-k**2)/(1+k**2);         #solving for u/c\n\n#result\nprint\"The speed of galaxy wrt earth in c is\",round(x,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The speed of galaxy wrt earth in c is 0.313\n"
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.9 Page 39"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nv1x=0.6;v1y=0.0;v2x=0.0;v2y=.8;c=1.0;     # all the velocities are taken wrt c\nv21x=(v2x-v1x)/(1-(v1x*v2x/c**2));   #using lorentz velocity transformation\nv21y=(v2y*(sqrt(1-(v1x*c)**2)/c**2))/(1-v1y*v2y/c**2) \n\n#result\nprint\"The velocity of rocket 2 wrt rocket 1 along x and y directions is\",round(v21x,3),\" c &\", round(v21y,3),\"c respectively\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The velocity of rocket 2 wrt rocket 1 along x and y directions is -0.6  c & 0.64 c respectively\n"
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.10 Page 40"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nu=0.8*c;L=65.0;c=3.0*10**8;            #all values are in terms of c\nt=u*L/(c**2*(sqrt(1-((u/c)**2))));  #from the equation 2.31 \n\n#result\nprint\"The time interval between the events is\",t, \"sec which equals\",round(t*10**6,3),\"musec.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The time interval between the events is 2.88888888889e-07 sec which equals 0.289 musec.\n"
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.11 Page 41"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nm=1.67*10**-27;c= 3*10**8;v=0.86*c;      #all the given values and constants\n\n#calculation\np=m*v/(sqrt(1-((v/c)**2)));            # in terms of Kgm/sec\n\n#result\nprint\"The value of momentum was found out to be in Kg-m/sec.\\n\",p;\n\n#part 2\nc=938.0;v=0.86*c;mc2=938.0                 # all the energies in MeV where mc2= value of m*c^2\npc=(mc2*(v/c))/(sqrt(1-((v/c)**2)));    #expressing in terms of Mev\n\n#result\nprint\"The value of momentum was found out to be in Mev.\",round(pc,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of momentum was found out to be in Kg-m/sec.\n8.44336739668e-19\nThe value of momentum was found out to be in Mev. 1580.814\n"
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.12 Page 47"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\npc=1580.0; mc2=938.0;E0=938.0;    # all the energies in MeV mc2=m*c^2 and pc=p*c\n\n#result\nE=sqrt(pc**2+mc2**2); \nK=E-E0;                      #value of possible kinetic energy\n\n#result\nprint\"The relativistic total energy in MeV. is\",round(E,3);         #value of Energy E\nprint\"The kinetic energy of the proton in MeV.\",round(K,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The relativistic total energy in MeV. is 1837.456\nThe kinetic energy of the proton in MeV. 899.456\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.13 Page 47"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nE=10.51; mc2=0.511; #all the values are in MeV\n\n#calculation\np=sqrt(E**2-mc2**2);   #momentum of the electron\nv=sqrt(1-(mc2/E)**2);  #velocity in terms of c\n\n#result\nprint\"The momentum of electron in MeV/c is\",round(p,3); \nprint\"The velocity of electron in c is\",round(v,5);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The momentum of electron in MeV/c is 10.498\nThe velocity of electron in c is 0.99882\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.14 Page 47"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nk=50;mc2=0.511*10**-3;c=3.0*10**8; # all the values of energy are in GeV and c is in SI units\n\n#calculation\nv=sqrt(1-(1/(1+(k/mc2))**2));   #speed of the electron in terms of c\nk=c-(v*c);                       #difference in velocities\n\n#result\nprint\"Speed of the electron as a  fraction of c*10^-12 is.\",round(v*10**12,3); # v=(v*10^12)*10^-12; so as to obtain desired accuracy in the result\nprint\"The difference in velocities in cm/s.\",round(k*10**2,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Speed of the electron as a  fraction of c*10^-12 is. 9.99999999948e+11\nThe difference in velocities in cm/s. 1.567\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.15 Page 48"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt, pi\nr=1.5*10**11; I=1.4*10**3;     #radius and intensity of sun\n\n#calculation\ns=4*pi*r**2                 #surface area of the sun\nPr=s*I                     # Power radiated in J/sec\nc=3.0*10**8;                #velocity of light\nm=Pr/c**2                  #rate od decrease of mass\nm=round(m,2)\n\n#result\nprint\"The rate of decrease in mass of the sun in kg/sec. is %.1e\" %m;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The rate of decrease in mass of the sun in kg/sec. is 4.4e+09\n"
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.16 Page 48"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi, sqrt\nK=325; mkc2=498;  #kinetic energy and rest mass energy of kaons\nmpic=140.0; #given value\n\n#calculation\nEk=K+mkc2; \npkc=sqrt(Ek**2-mkc2**2); \n#consider the law of conservation of energy which yields Ek=sqrt(p1c^2+mpic^2)+sqrt(p2c^2+mpic^2)\n#The above equations (4th degree,hence no direct methods)can be solved by assuming the value of p2c=0.\np1c=sqrt(Ek**2-(2*mpic*Ek));\n#consider the law of conservation of momentum. which gives p1c+p2c=pkc implies\np2c=pkc-p1c;\nk1=(sqrt(p1c**2+(mpic**2))-mpic); #corresponding kinetic energies\nk2=(sqrt((p2c**2)+(mpic**2))-mpic);\n\n#result\nprint\"The corresponding kinetic energies of the pions are\", k1,\" MeV and\",round(k2,3),\" MeV.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The corresponding kinetic energies of the pions are 543.0  MeV and 0.627  MeV.\n"
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 2.17 Page 49"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nmpc2=938.0;c=3.0*10**8;  #mpc2=mp*c^2,mp=mass of proton\n\n#calculation\nEt=4*mpc2;          #final total energy\nE1=Et/2;E2=E1;       #applying conservation of momentum and energy\nv2=c*sqrt(1-(mpc2/E1)**2);    #lorentz transformation\nu=v2;v=(v2+u)/(1+(u*v2/c**2)); \nE=mpc2/(sqrt(1-(v/c)**2));\nK=E-mpc2;\n\n#result\nprint\"The threshold kinetic energy in Gev\",round(K/10**3,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The threshold kinetic energy in Gev 5.628\n"
+      }
+     ],
+     "prompt_number": 50
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter3.ipynb b/Modern_Physics/Chapter3.ipynb
index c4fdd463..72876741 100755
--- a/Modern_Physics/Chapter3.ipynb
+++ b/Modern_Physics/Chapter3.ipynb
@@ -1,7 +1,6 @@
 {
  "metadata": {
-  "name": "",
-  "signature": "sha256:6f3412539d62c2f676626072f86e8478aa55d9f7f8bd139276fa120f78482f67"
+  "name": "Chapter3"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -13,7 +12,7 @@
      "level": 1,
      "metadata": {},
      "source": [
-      "Chapter 3: The Quantum Theory of Light"
+      "Chapter 3:The particle like properties of electromagnetic radiation"
      ]
     },
     {
@@ -21,28 +20,22 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 3.1, page no. 69"
+      "Example 3.1 Page 69"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      " \n",
+      "#initiation of variable\n",
+      "from math import sin,pi\n",
+      "w=0.250; theta=26.3;n=1 # n=1 for hydrogen atom and rest all are given values\n",
       "\n",
-      "#Variable declaration\n",
-      "Rs=7.0 * (10 ** 8) \t#sun's radius (m)\n",
-      "R = 1.5 *(10 ** 11)\t#earth to sun distance (m)\n",
-      "a = 1  #since sun is considered as a blackbody \n",
-      "k = 5.6 * (10 ** (-8))  #Stefan-Boltzmann constant ( W.m ^-2 .K^-4)\n",
-      "eTotalR = 1400\t#power per unit area (W/m^2)\n",
-      "\n",
-      "#Calculations\n",
-      "T = ((eTotalR * R * R) / (k * Rs * Rs) ) ** .25\n",
+      "#calculation\n",
+      "d=n*w/(2*sin(theta*pi/180));     # bragg's law\n",
       "\n",
-      "#Results\n",
-      "print '%s %.2f %s' %('the surface temperature of the sun is',T,'K')\n",
-      "\n"
+      "#result\n",
+      "print \"Hence the atomic spacing in nm is\",round(d,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -51,7 +44,7 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "the surface temperature of the sun is 5820.79 K\n"
+        "Hence the atomic spacing in nm is 0.282\n"
        ]
       }
      ],
@@ -62,97 +55,23 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 3.2, page no. 75"
+      "Example 3.2 Page 73"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "h = 6.63 * 10 ** -34 #planck's constant(Js)\n",
-      "c = 3 * 10 ** 8      #speed of light(m/s)\n",
-      "lgreen = 540 * 10 ** -9#wavelength of green light(m)\n",
-      "lred = 700 *10**-9     #wavelength of red light(m)\n",
-      "e = 1.602 * 10 ** -19 #charge of an electron(C)\n",
+      "#initiation of variable\n",
+      "from math import pi,sin\n",
+      "I=120.0;r=0.1*10**-9;Eev=2.3   #I-intensity in W/m^2 r in m & E in electron volt\n",
+      "A=pi*r**2;K=1.6*10**-19;     # A=area and K is conversion factor from ev to joules\n",
       "\n",
       "#calculation\n",
-      "dEg =  h*c /(lgreen* e)\n",
-      "dEr =  h*c/(lred * e)\n",
-      "\n",
-      "#results\n",
-      "print '%s %s %s %s %s' %('the minimum energy change for ',lgreen * 10 ** 9,'nm is ',round(dEg,2),'eV')\n",
-      "print '%s %s %s %s %s' %('the minimum energy change for ',lred * 10 ** 9,'nm is ',round(dEr,2),'eV')\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "l=1                 #length of the pendulum(m)\n",
-      "m = 0.1             # mass of the pendulum(kg)\n",
-      "g = 9.8             #acceleration due to gravity(m.s^-2)\n",
-      "h = 6.63 *10 **-34  #planck's constant(J.s)\n",
-      "theta = 10          # displaced angle\n",
-      "\n",
-      "#Calculations\n",
-      "E = m * g * l *(1 - math.cos(math.pi * theta /180))\n",
-      "f = math.sqrt(g /l) /(2* math.pi)\n",
-      "Edelta = h *f\n",
-      "\n",
-      "#results\n",
-      "print '%s %s %s' %('the pendulum frequency is',round(f,2),'Hz')\n",
-      "print '%s %s %s' %('the total energy of the pendulum is',round(E,3),'J')\n",
-      "print '%s %s %s' %('therefore an energy change of one quantum corresponds to',round(Edelta/10**-34,2),'x 10^-34 J')\n",
-      "print '%s %s %s' %('Therefore the fractional change in energy ^E/E is ',round(Edelta/E/10**-32,2),'x 10^-32 ')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the minimum energy change for  540.0 nm is  2.3 eV\n",
-        "the minimum energy change for  700.0 nm is  1.77 eV\n",
-        "the pendulum frequency is 0.5 Hz\n",
-        "the total energy of the pendulum is 0.015 J\n",
-        "therefore an energy change of one quantum corresponds to 3.3 x 10^-34 J\n",
-        "Therefore the fractional change in energy ^E/E is  2.22 x 10^-32 \n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.3, page no. 80"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "pi = 3.141      \n",
-      "k = 1.381 * 10 **-23    #Boltzmann constant (J/K)\n",
-      "c = 2.998 * 10 ** 8     #Speed of light (m/s)\n",
-      "h = 6.626 * 10 ** -34   #Planck's constant (J.s)\n",
+      "t= Eev*K/(I*A);            #time interval\n",
       "\n",
-      "#Calculation\n",
-      "\n",
-      "sigma = 2 * pi**5 * k**4 / (15 * c**2 * h**3)\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "print \"e_total=sigma * T^4 where sigma=\",round(sigma/10**-8,2),\"X 10^-8 W.m^-2.K^-4\""
+      "#result\n",
+      "print \"The value of time interval was found out to be in sec is\",round(t,3);\n"
      ],
      "language": "python",
      "metadata": {},
@@ -161,79 +80,44 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "e_total=sigma * T^4 where sigma= 5.67 X 10^-8 W.m^-2.K^-4\n"
+        "The value of time interval was found out to be in sec is 0.098\n"
        ]
       }
      ],
-     "prompt_number": 7
+     "prompt_number": 2
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 3.4, page no. 83"
+      "Example 3.3 Page 76"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#variable declaration\n",
-      "e= 1.68 * 10 **-19      #electron charge(C)\n",
-      "O = 2.28  * e           #work function of sodium\n",
-      "I  = 10 ** -10          #power per unit area(W/cm^2)\n",
+      "#initiation of variable\n",
+      "from math import pi,sin\n",
+      "w=650.0*10**-9;h=6.63*10**-34;c=3*10**8;  #given values and constant taken in comfortable units\n",
       "\n",
       "#calculation\n",
-      "A = math.pi * 10 ** -16\n",
-      "t= O / (I * A)\n",
+      "E=h*c/w; \n",
+      "E1=E/(1.6*10**-19);\n",
       "\n",
       "#result\n",
-      "print '%s %s %s' %('the time lag is given by',round(t/(60*60*24)),'days which is approximated to 130 days in the text book')"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the time lag is given by 141.0 days which is approximated to 130 days in the text book\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.5, page no. 85"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
+      "print \"The Energy of the electron in J \",E,\"which is equivalent to in eV is \", round(E1,3);\n",
+      "print \"The momentum of electron is p=E/c i.e is \", round(E1,3);\n",
       "\n",
-      "#variable declaration\n",
-      "Vs = 4.3            #Stopping voltage(V)\n",
-      "e = 1.6 * 10 **-19  #electron charge(C)\n",
-      "Me = 9.1 *10**-31   #mass of electron(kg)\n",
+      "#part b\n",
+      "E2=2.40;                          #given energy of photon.\n",
       "\n",
       "#calculation\n",
-      "vmax =  math.sqrt( 2* e* Vs /Me)\n",
-      "Kmax = e *Vs\n",
+      "w2=h*c*10**9/(E2*1.6*10**9);      #converting the energy in to eV and nm \n",
       "\n",
       "#result\n",
-      "print '%s %s %s' %('the Kmax of these electrons are', Kmax ,'J')\n",
-      "print '%s %s %s' %('vmax of these electrons are',round(vmax/10**6,2),' x 10^6 m/s')"
+      "print \"The wavelength of the photon in m is\",round(w2*10**28,0)"
      ],
      "language": "python",
      "metadata": {},
@@ -242,76 +126,49 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "the Kmax of these electrons are 6.88e-19 J\n",
-        "vmax of these electrons are 1.23  x 10^6 m/s\n"
+        "The Energy of the electron in J  3.06e-19 which is equivalent to in eV is  1.912\n",
+        "The momentum of electron is p=E/c i.e is  1.912\n",
+        "The wavelength of the photon in m is 518.0\n"
        ]
       }
      ],
-     "prompt_number": 13
+     "prompt_number": 8
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 3.6, page no. 85"
+      "Example 3.4 Page 78"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "hc=1240.0; phi=4.52               #both the values are in eV\n",
       "\n",
-      "#variable declaration\n",
-      "I0 = 1 * 10 ** -6 #intensity of light falling W/cm^2\n",
-      "\n",
-      "#calculation\n",
-      "I =  .03 * .04 * I0\n",
+      "#calcualtion\n",
+      "w1=hc/phi; \n",
       "\n",
       "#result\n",
-      "print '%s %s %s' %('The actual intensity available is',I,'W/cm^2')\n",
+      "print \"The cutoff wavelength of the tungsten metal in nm is \",round(w1,3);\n",
       "\n",
-      "#variable declaration\n",
-      "lamda = 250 *10 ** -9  #wavelength of violet light(m)\n",
-      "c= 3*10**8             #speed of light(m/s)\n",
-      "h = 6.63 *10 **-34     #planck's constant(J.s)\n",
+      "#part b\n",
+      "w2=198.0;     #given value of wavelength \n",
       "\n",
       "#calculation\n",
-      "Ne = I *lamda / (h * c)\n",
+      "Kmax=(hc/w2)-phi;\n",
       "\n",
       "#result\n",
-      "print '%s %s %s' %('number of electrons is',round(Ne/10**9,1),'x 10^9')\n",
-      "\n",
+      "print 'The max value of kinetic energy in eV is',round(Kmax,3);\n",
       "\n",
-      "#variable declaration\n",
-      "e = 1.6 * 10 **-19    #electron charge(c)\n",
-      "\n",
-      "#calculation\n",
-      "i = e * Ne\n",
+      "#part c\n",
+      "Vs=Kmax;\n",
       "\n",
       "#result\n",
-      "print '%s %s %s' %('current in the phototube is ',round(i/10**-10,1),'x 10^-10 A')\n",
-      "\n",
-      "\n",
-      "#variable declaration\n",
-      "f0 = 1.1 *10**15      #cutoff frequency (Hz)\n",
-      "\n",
-      "#calculation\n",
-      "O = h *f0 / e \n",
-      "\n",
-      "#result\n",
-      "print '%s %s %s' %('the work function is ',round(O,1),'eV')\n",
-      "\n",
-      "\n",
-      "\n",
-      "#variable declaration\n",
-      "lamda = 250 * 10 ** -9   #wavelength(m)\n",
-      "\n",
-      "#calculation\n",
-      "Vs = ((h*c )/(lamda * e )) - O\n",
-      "\n",
-      "#result\n",
-      "print '%s %s %s' %('stopping voltage for iron is ',round(Vs,2),'V')"
+      "print \"The numerical value of the max kinetic energy is same as stopping potential in volts.Hence in V is\",round(Vs,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -320,44 +177,47 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The actual intensity available is 1.2e-09 W/cm^2\n",
-        "number of electrons is 1.5 x 10^9\n",
-        "current in the phototube is  2.4 x 10^-10 A\n",
-        "the work function is  4.6 eV\n",
-        "stopping voltage for iron is  0.41 V\n"
+        "The cutoff wavelength of the tungsten metal in fnm is  274.336\n",
+        "The max value of kinetic energy in eV is 1.743\n",
+        "The numerical value of the max kinetic energy is same as stopping potential in volts.Hence in V is 1.743\n"
        ]
       }
      ],
-     "prompt_number": 16
+     "prompt_number": 5
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 3.7, page no. 93"
+      "Example 3.5 Page 80"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "T1=293.0; Kw=2.898*10**-3;\n",
       "\n",
+      "#calculation\n",
+      "w1=Kw/T1;\n",
       "\n",
-      "import math\n",
+      "#result\n",
+      "print \"The wavelength at which emits maximum radiation in um. is\",round(w1*10**6,3);\n",
       "\n",
-      "#variable declaration\n",
-      "lamda = .2 * 10 ** -9       #wavelength(m)\n",
-      "theta = 45                  #observed angle(degrees)\n",
-      "h = 6.63 * 10 ** -34        #planck's constant(J.s)\n",
-      "Me = 9.1 * 10 ** -31        #electron mass(kg)\n",
-      "c = 3* 10 ** 8              #speed of light(m/s)\n",
+      "#part b\n",
+      "w2=650.0*10**-9; \n",
+      "T2=Kw/w2;\n",
       "\n",
-      "#calculation\n",
-      "dl= h *(1 - math.cos(math.pi * theta /180)) /(Me * c)\n",
+      "#result\n",
+      "print 'The temperature of the object must be raised to in K. is',round(T2,3);\n",
+      "\n",
+      "#part c\n",
+      "x=(T2/T1)**4; \n",
       "\n",
       "#result\n",
-      "print '%s %s %s' %('the wavelength off the scattered x-ray at this angle is',dl+lamda,'m')"
+      "print \"Thus the thermal radiation at higher temperature in times the room (lower) tempertaure. is\",round(x,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -366,115 +226,54 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "the wavelength off the scattered x-ray at this angle is 2.00711312103e-10 m\n"
+        "The wavelength at which emits maximum radiation in um. is 9.891\n",
+        "The temperature of the object must be raised to in K. is 4458.462\n",
+        "Thus the thermal radiation at higher temperature in times the room (lower) tempertaure. is 53612.939\n"
        ]
       }
      ],
-     "prompt_number": 18
+     "prompt_number": 8
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 3.8, page no. 93"
+      "Example 3.6 Page 82"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#variable declaration\n",
-      "lamdaG = 0.0106         #wavelenght(m)\n",
-      "\n",
-      "#calculation\n",
-      "dl = 0.0243 * (1-math.cos(math.pi/2))\n",
-      "f= dl/ lamdaG\n",
-      "\n",
-      "#result\n",
-      "print \"the compton shift is \",dl,\"A'\"\n",
-      "print \"the fractional change in wavelength of gamma rays is\",round(f,4)\n",
-      "\n",
-      "#(2)X-rays from molybdenum, lamda = 0.712 x 10 ^-10 m,\n",
-      "\n",
-      "#variable declaration\n",
-      "lamdaX = 0.712          #wavelenght(m)\n",
+      "#initiation of variable\n",
+      "#part a\n",
+      "from math import cos, sin, pi,atan\n",
+      "w1=0.24;wc=0.00243;theta=60.0;   #given values w=wavelength(lambeda)\n",
       "\n",
       "#calculation\n",
-      "f= dl/ lamdaX\n",
+      "w2=w1+(wc*(1-cos(theta*pi/180))); \n",
       "\n",
       "#result\n",
-      "print \"the fractional change in wavelength of X rays is\",round(f,4)\n",
-      "\n",
-      "#(3)green light from a mercury lamp, lamda = 5461 *10 ^ -10 \n",
+      "print \"The wavelength of x-rays after scattering in nm is\",round(w2,5);\n",
       "\n",
-      "#variable declaration\n",
-      "lamdaGr = 5461\n",
-      "\n",
-      "#calculation\n",
-      "f= dl/ lamdaGr\n",
+      "#part b;\n",
+      "hc=1240;\n",
+      "E2=hc/w2;E1=hc/w1; \n",
       "\n",
       "#result\n",
-      "print \"the fractional change in wavelength of green rays is\",round(f/10**-6,3),\"x 10^-6\"\n",
+      "print \"The energy of scattered x-rays in eV is\",round(E2,3);\n",
       "\n",
+      "#part c\n",
+      "K= E1-E2; #The kinetic energy is the difference in the energy before and after the collision;\n",
       "\n",
-      "#variable declaration\n",
-      "h = 6.63 * 10 ** -34        #planck's constant(J.s)\n",
-      "c = 3* 10 ** 8              #speed of light(m/s)\n",
-      "lamda = 0.712 * 10 **-10\n",
-      "e = 1.6 * 10 **-19    #electron charge(c)\n",
+      "print \"The kinetic energy of the x-rays in eV is\",round(K,3);\n",
       "\n",
-      "#calculation\n",
-      "E = h*c/(lamda * e)\n",
-      "\n",
-      "#result\n",
-      "print \"the energy of incident x-ray is\",round(E,2),\"ev. It is large when compared to binding energy of 4eV\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the compton shift is  0.0243 A'\n",
-        "the fractional change in wavelength of gamma rays is 2.2925\n",
-        "the fractional change in wavelength of X rays is 0.0341\n",
-        "the fractional change in wavelength of green rays is 4.45 x 10^-6\n",
-        "the energy of incident x-ray is 17459.62 ev. It is large when compared to binding energy of 4eV\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3.9, page no. 96"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#variable declaration\n",
-      "M = 1.99 * 10 ** 30         #mass of sun(kg)\n",
-      "lamda = 300 *10**-9         #wavelength(m)\n",
-      "Rs = 6.37 * 10 ** 6         #radius of earth(m)\n",
-      "G = 6.67 * 10 ** -11        #gravitational constant(N.m^2.kg^-2)\n",
-      "c = 3 * 10 ** 8             #speed of light(m/s)\n",
-      "\n",
-      "#calculation\n",
-      "fraction =  G * M / (Rs * c *c)\n",
+      "#part d\n",
+      "phi2=atan(E2*sin(theta*pi/180)/(E1-E2*cos(theta*pi/180)))\n",
       "\n",
       "#result\n",
-      "print '%s %s %s'%(\"the shift in wavelength\",round(lamda * fraction *10**9,4),'nm')"
+      "print \"The direction of the scattered eletron in degrees is\",round(phi2*180/pi,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -483,11 +282,14 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "the shift in wavelength 0.0695 nm\n"
+        "The wavelength of x-rays after scattering in nm is 0.24121\n",
+        "The energy of scattered x-rays in eV is 5140.642\n",
+        "The kinetic energy of the x-rays in eV is 26.025\n",
+        "The direction of the scattered eletron in degrees is 59.749\n"
        ]
       }
      ],
-     "prompt_number": 22
+     "prompt_number": 9
     }
    ],
    "metadata": {}
diff --git a/Modern_Physics/Chapter3_1.ipynb b/Modern_Physics/Chapter3_1.ipynb
new file mode 100755
index 00000000..aa39d0f9
--- /dev/null
+++ b/Modern_Physics/Chapter3_1.ipynb
@@ -0,0 +1,146 @@
+{
+ "metadata": {
+  "name": "MP-3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "The particle like properties of electromagnetic radiation"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.1 Page 69"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sin,pi\nw=0.250; theta=26.3;n=1 # n=1 for hydrogen atom and rest all are given values\n\n#calculation\nd=n*w/(2*sin(theta*pi/180));     # bragg's law\n\n#result\nprint \"Hence the atomic spacing in nm is\",round(d,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Hence the atomic spacing in nm is 0.282\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.2 Page 73"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi,sin\nI=120.0;r=0.1*10**-9;Eev=2.3   #I-intensity in W/m^2 r in m & E in electron volt\nA=pi*r**2;K=1.6*10**-19;     # A=area and K is conversion factor from ev to joules\n\n#calculation\nt= Eev*K/(I*A);            #time interval\n\n#result\nprint \"The value of time interval was found out to be in sec is\",round(t,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of time interval was found out to be in sec is 0.098\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.3 Page 76"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi,sin\nw=650.0*10**-9;h=6.63*10**-34;c=3*10**8;  #given values and constant taken in comfortable units\n\n#calculation\nE=h*c/w; \nE1=E/(1.6*10**-19);\n\n#result\nprint \"The Energy of the electron in J \",E,\"which is equivalent to in eV is \", round(E1,3);\nprint \"The momentum of electron is p=E/c i.e is \", round(E1,3);\n\n#part b\nE2=2.40;                          #given energy of photon.\n\n#calculation\nw2=h*c*10**9/(E2*1.6*10**9);      #converting the energy in to eV and nm \n\n#result\nprint \"The wavelength of the photon in m is\",round(w2*10**28,0)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The Energy of the electron in J  3.06e-19 which is equivalent to in eV is  1.912\nThe momentum of electron is p=E/c i.e is  1.912\nThe wavelength of the photon in m is 518.0\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.4 Page 78"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nhc=1240.0; phi=4.52               #both the values are in eV\n\n#calcualtion\nw1=hc/phi; \n\n#result\nprint \"The cutoff wavelength of the tungsten metal in nm is \",round(w1,3);\n\n#part b\nw2=198.0;     #given value of wavelength \n\n#calculation\nKmax=(hc/w2)-phi;\n\n#result\nprint 'The max value of kinetic energy in eV is',round(Kmax,3);\n\n#part c\nVs=Kmax;\n\n#result\nprint \"The numerical value of the max kinetic energy is same as stopping potential in volts.Hence in V is\",round(Vs,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The cutoff wavelength of the tungsten metal in fnm is  274.336\nThe max value of kinetic energy in eV is 1.743\nThe numerical value of the max kinetic energy is same as stopping potential in volts.Hence in V is 1.743\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.5 Page 80"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nT1=293.0; Kw=2.898*10**-3;\n\n#calculation\nw1=Kw/T1;\n\n#result\nprint \"The wavelength at which emits maximum radiation in um. is\",round(w1*10**6,3);\n\n#part b\nw2=650.0*10**-9; \nT2=Kw/w2;\n\n#result\nprint 'The temperature of the object must be raised to in K. is',round(T2,3);\n\n#part c\nx=(T2/T1)**4; \n\n#result\nprint \"Thus the thermal radiation at higher temperature in times the room (lower) tempertaure. is\",round(x,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The wavelength at which emits maximum radiation in um. is 9.891\nThe temperature of the object must be raised to in K. is 4458.462\nThus the thermal radiation at higher temperature in times the room (lower) tempertaure. is 53612.939\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.6 Page 82"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#part a\nfrom math import cos, sin, pi,atan\nw1=0.24;wc=0.00243;theta=60.0;   #given values w=wavelength(lambeda)\n\n#calculation\nw2=w1+(wc*(1-cos(theta*pi/180))); \n\n#result\nprint \"The wavelength of x-rays after scattering in nm is\",round(w2,5);\n\n#part b;\nhc=1240;\nE2=hc/w2;E1=hc/w1; \n\n#result\nprint \"The energy of scattered x-rays in eV is\",round(E2,3);\n\n#part c\nK= E1-E2; #The kinetic energy is the difference in the energy before and after the collision;\n\nprint \"The kinetic energy of the x-rays in eV is\",round(K,3);\n\n#part d\nphi2=atan(E2*sin(theta*pi/180)/(E1-E2*cos(theta*pi/180)))\n\n#result\nprint \"The direction of the scattered eletron in degrees is\",round(phi2*180/pi,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The wavelength of x-rays after scattering in nm is 0.24121\nThe energy of scattered x-rays in eV is 5140.642\nThe kinetic energy of the x-rays in eV is 26.025\nThe direction of the scattered eletron in degrees is 59.749\n"
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter3_2.ipynb b/Modern_Physics/Chapter3_2.ipynb
new file mode 100755
index 00000000..aa39d0f9
--- /dev/null
+++ b/Modern_Physics/Chapter3_2.ipynb
@@ -0,0 +1,146 @@
+{
+ "metadata": {
+  "name": "MP-3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "The particle like properties of electromagnetic radiation"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.1 Page 69"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sin,pi\nw=0.250; theta=26.3;n=1 # n=1 for hydrogen atom and rest all are given values\n\n#calculation\nd=n*w/(2*sin(theta*pi/180));     # bragg's law\n\n#result\nprint \"Hence the atomic spacing in nm is\",round(d,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Hence the atomic spacing in nm is 0.282\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.2 Page 73"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi,sin\nI=120.0;r=0.1*10**-9;Eev=2.3   #I-intensity in W/m^2 r in m & E in electron volt\nA=pi*r**2;K=1.6*10**-19;     # A=area and K is conversion factor from ev to joules\n\n#calculation\nt= Eev*K/(I*A);            #time interval\n\n#result\nprint \"The value of time interval was found out to be in sec is\",round(t,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of time interval was found out to be in sec is 0.098\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.3 Page 76"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi,sin\nw=650.0*10**-9;h=6.63*10**-34;c=3*10**8;  #given values and constant taken in comfortable units\n\n#calculation\nE=h*c/w; \nE1=E/(1.6*10**-19);\n\n#result\nprint \"The Energy of the electron in J \",E,\"which is equivalent to in eV is \", round(E1,3);\nprint \"The momentum of electron is p=E/c i.e is \", round(E1,3);\n\n#part b\nE2=2.40;                          #given energy of photon.\n\n#calculation\nw2=h*c*10**9/(E2*1.6*10**9);      #converting the energy in to eV and nm \n\n#result\nprint \"The wavelength of the photon in m is\",round(w2*10**28,0)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The Energy of the electron in J  3.06e-19 which is equivalent to in eV is  1.912\nThe momentum of electron is p=E/c i.e is  1.912\nThe wavelength of the photon in m is 518.0\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.4 Page 78"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nhc=1240.0; phi=4.52               #both the values are in eV\n\n#calcualtion\nw1=hc/phi; \n\n#result\nprint \"The cutoff wavelength of the tungsten metal in nm is \",round(w1,3);\n\n#part b\nw2=198.0;     #given value of wavelength \n\n#calculation\nKmax=(hc/w2)-phi;\n\n#result\nprint 'The max value of kinetic energy in eV is',round(Kmax,3);\n\n#part c\nVs=Kmax;\n\n#result\nprint \"The numerical value of the max kinetic energy is same as stopping potential in volts.Hence in V is\",round(Vs,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The cutoff wavelength of the tungsten metal in fnm is  274.336\nThe max value of kinetic energy in eV is 1.743\nThe numerical value of the max kinetic energy is same as stopping potential in volts.Hence in V is 1.743\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.5 Page 80"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nT1=293.0; Kw=2.898*10**-3;\n\n#calculation\nw1=Kw/T1;\n\n#result\nprint \"The wavelength at which emits maximum radiation in um. is\",round(w1*10**6,3);\n\n#part b\nw2=650.0*10**-9; \nT2=Kw/w2;\n\n#result\nprint 'The temperature of the object must be raised to in K. is',round(T2,3);\n\n#part c\nx=(T2/T1)**4; \n\n#result\nprint \"Thus the thermal radiation at higher temperature in times the room (lower) tempertaure. is\",round(x,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The wavelength at which emits maximum radiation in um. is 9.891\nThe temperature of the object must be raised to in K. is 4458.462\nThus the thermal radiation at higher temperature in times the room (lower) tempertaure. is 53612.939\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 3.6 Page 82"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\n#part a\nfrom math import cos, sin, pi,atan\nw1=0.24;wc=0.00243;theta=60.0;   #given values w=wavelength(lambeda)\n\n#calculation\nw2=w1+(wc*(1-cos(theta*pi/180))); \n\n#result\nprint \"The wavelength of x-rays after scattering in nm is\",round(w2,5);\n\n#part b;\nhc=1240;\nE2=hc/w2;E1=hc/w1; \n\n#result\nprint \"The energy of scattered x-rays in eV is\",round(E2,3);\n\n#part c\nK= E1-E2; #The kinetic energy is the difference in the energy before and after the collision;\n\nprint \"The kinetic energy of the x-rays in eV is\",round(K,3);\n\n#part d\nphi2=atan(E2*sin(theta*pi/180)/(E1-E2*cos(theta*pi/180)))\n\n#result\nprint \"The direction of the scattered eletron in degrees is\",round(phi2*180/pi,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The wavelength of x-rays after scattering in nm is 0.24121\nThe energy of scattered x-rays in eV is 5140.642\nThe kinetic energy of the x-rays in eV is 26.025\nThe direction of the scattered eletron in degrees is 59.749\n"
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter4.ipynb b/Modern_Physics/Chapter4.ipynb
index 505cc127..4ae3fd7c 100755
--- a/Modern_Physics/Chapter4.ipynb
+++ b/Modern_Physics/Chapter4.ipynb
@@ -1,7 +1,6 @@
 {
  "metadata": {
-  "name": "",
-  "signature": "sha256:fcebe7a1b857b1b190100018ff76ede788e3c249d5181d4bebb1457ad5175559"
+  "name": "Chapter4"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -13,7 +12,7 @@
      "level": 1,
      "metadata": {},
      "source": [
-      "Chapter 4: The Particle Nature of Matter"
+      "Chapter 4:The Wave Like Properties of Particles"
      ]
     },
     {
@@ -21,31 +20,36 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 4.1, page no. 109"
+      "Example 4.1 Page 101"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "from math import sqrt\n",
+      "h=6.6*10**-34;                              # h(planck's constant)= 6.6*10^-34 \n",
+      "m1= 10.0**3;v1=100.0;                           # for automobile\n",
+      "\n",
+      "#calculation\n",
+      "w1= h/(m1*v1);                            # ['w'-wavelength in metre'm'-mass in Kg 'v'-velocity in metres/sec.] of the particles \n",
+      "m2=10.0*(10**-3);v2= 500;                     # for bullet\n",
+      "w2=h/(m2*v2);\n",
+      "m3=(10.0**-9)*(10.0**-3); v3=1.0*10**-2;\n",
+      "w3=h/(m3*v3);\n",
+      "m4=9.1*10**-31;k=1*1.6*10**-19;            # k- kinetic energy of the electron & using 1ev = 1.6*10^-19 joule\n",
+      "p=sqrt(2.0*m4*k);                         # p=momentum of electron ;from K=1/2*m*v^2\n",
+      "w4=h/p;\n",
+      "hc=1240;pc=100                         # In the extreme relativistc realm, K=E=pc; Given pc=100MeV,hc=1240MeV \n",
+      "w5= hc/pc;\n",
       "\n",
-      "#Variable declaration\n",
-      "\n",
-      "I = 10.0         #current (A)\n",
-      "t = 3600         #time (s)\n",
-      "molar_mass_Ba = 137 #molar mass of Barium (g)\n",
-      "valence_Ba = 2         #valence of Barium\n",
-      "molar_mass_Cl = 35.5 #molar mass of chlorine(g)\n",
-      "valence_Cl = 1       #valence of chlorine\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "mBa = I * t * molar_mass_Ba /(96500 * valence_Ba)\n",
-      "mCl = I * t * molar_mass_Cl /(96500 * valence_Cl)\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print \"The mass of Barium and Chlorine obtained is \",round(mBa,1),\"g\",round(mCl,1),\"g respectively.\""
+      "#result\n",
+      "print \"Wavelength of the automobile in m is\",w1;\n",
+      "print \"Wavelength of the bullet in m is \",w2 ;\n",
+      "print\"Wavelength of the smoke particle in m is\",w3 ;\n",
+      "print \"Wavelength of the electron(1ev) in nm is\",round(w4*10**9,3) ;\n",
+      "print \"Wavelength of the electron (100Mev) in fm is\",w5;"
      ],
      "language": "python",
      "metadata": {},
@@ -54,45 +58,40 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The mass of Barium and Chlorine obtained is  25.6 g 13.2 g respectively.\n"
+        "Wavelength of the automobile in m is 6.6e-39\n",
+        "Wavelength of the bullet in m is  1.32e-34\n",
+        "Wavelength of the smoke particle in m is 6.6e-20\n",
+        "Wavelength of the electron(1ev) in nm is 1.223\n",
+        "Wavelength of the electron (100Mev) in fm is 12\n"
        ]
       }
      ],
-     "prompt_number": 2
+     "prompt_number": 3
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 4.2, page no. 113"
+      "Example 4.2 Page 113"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "from math import pi\n",
+      "# w=wavelength; consider k=2*(pi/w); \n",
+      "# differentiate k w.r.t w and replace del(k)/del(w) = 1 for equation.4.3\n",
+      "# which gives del(w)= w^2 /(2*pi*del(x)), hence \n",
+      "w=20; delx=200; # delx=200cm and w=20cm\n",
       "\n",
+      "#calculation\n",
+      "delw=(w**2)/(delx*2*pi);\n",
       "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "V = 200             #voltage (V)\n",
-      "theta = 0.20        #deflection (rad)\n",
-      "l = 0.05             #length (m)\n",
-      "d = 0.015             #spacing (m)\n",
-      "e_by_me = 1.76 * 10 **11 #e/me (C/kg)\n",
-      "B_earth = 0.5 * 10**-4   #earth's magnetic field (T)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "B = math.sqrt(V * theta /(l * d * e_by_me))\n",
-      "\n",
-      "#results\n",
-      "\n",
-      "print \"The magnetic field required to produce the deflection is\",round(B/10**-4,1),\"X 10^-4 T\"\n",
-      "print \" The earth's magnetic field is \",B_earth,\"T so we require\",round(B/B_earth),\"times stronger field.\""
+      "#result\n",
+      "print \"Hence uncertainity in length in cm is\",round(delw,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -101,8 +100,7 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The magnetic field required to produce the deflection is 5.5 X 10^-4 T\n",
-        " The earth's magnetic field is  5e-05 T so we require 11.0 times stronger field.\n"
+        "Hence uncertainity in length is in cm 0.318\n"
        ]
       }
      ],
@@ -113,68 +111,25 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 4.3, page no. 117"
+      "Example 4.3 Page 114"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "from math import pi\n",
+      "delt=1.0;             #consider time interval of 1 sec\n",
+      "delw=1/delt;        # since delw*delt =1 from equation 4.4\n",
+      "delf=0.01          #calculated accuracy is 0.01Hz\n",
       "\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "dy = 0.006          #distance of rise or fall (m)\n",
-      "dt = 21.0           #average time of fall (s)\n",
-      "n = 1.83 * 10 **-5  #viscosity of air (kg/ms)\n",
-      "p = 858             #oil density (kg/m^3)\n",
-      "g = 9.81            #acceleration due to gravity (m/s^2)\n",
-      "\n",
-      "#Calculation\n",
+      "#calculation\n",
+      "delwc =2*pi*delf # delwc-claimed accuracy from w=2*pi*f\n",
       "\n",
-      "v = dy/dt\n",
-      "a = math.sqrt(9*n*v/(2*p*g))\n",
-      "V = 4*math.pi*a**3/3\n",
-      "m = p * V\n",
-      "\n",
-      "#results\n",
-      "\n",
-      "print \"The radius,volume and mass of the drop are\",round(a/10**-6,2),\"X10^-6 m,\",round(V/10**-17,2),\"X10^-17 m^3,\",round(m/10**-14,2),\"X10^-14 kg\"\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "Vt = 4550               #potential difference (V)\n",
-      "d = 0.0160              #plate separation (m)\n",
-      "#successive rise times(s)\n",
-      "dt1 = 46.0\n",
-      "dt2 = 15.5\n",
-      "dt3 = 28.1\n",
-      "dt4 = 12.9\n",
-      "dt5 = 45.3\n",
-      "dt6 = 20.0\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "E = Vt/d        #Electric field\n",
-      "v1 = dy / dt1\n",
-      "v2 = dy / dt2\n",
-      "v3 = dy / dt3\n",
-      "v4 = dy / dt4\n",
-      "v5 = dy / dt5\n",
-      "v6 = dy / dt6\n",
-      "q1 = (m*g/E)*((v+v1)/v)\n",
-      "q2 = (m*g/E)*((v+v2)/v)\n",
-      "q3 = (m*g/E)*((v+v3)/v)\n",
-      "q4 = (m*g/E)*((v+v4)/v)\n",
-      "q5 = (m*g/E)*((v+v5)/v)\n",
-      "q6 = (m*g/E)*((v+v6)/v)\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print \"The successive charges on the drop is as follows \\n q1 =\",round(q1/10**-19,2),\"X 10^-19 C \\n q2 = \",round(q2/10**-19,2),\"X 10^-19 C \\n q3 = \",round(q3/10**-19,2),\"X 10^-19 C \\n q4 = \",round(q4/10**-19,2),\"X 10^-19 C \\n q5 = \",round(q5/10**-19,2),\"X 10^-19 C \\n q6 = \",round(q6/10**-19,2),\"X 10^-19 C\""
+      "#result\n",
+      "print \"The minimum uncertainity calculated is 1rad/sec. The claimed accuracy in rad/sec is \\n\",round(delwc,3);\n",
+      "print \"thus there is a reason to doubt the claim\"\n"
      ],
      "language": "python",
      "metadata": {},
@@ -183,46 +138,39 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The radius,volume and mass of the drop are 1.67 X10^-6 m, 1.96 X10^-17 m^3, 1.68 X10^-14 kg\n",
-        "The successive charges on the drop is as follows \n",
-        " q1 = 8.44 X 10^-19 C \n",
-        " q2 =  13.65 X 10^-19 C \n",
-        " q3 =  10.12 X 10^-19 C \n",
-        " q4 =  15.23 X 10^-19 C \n",
-        " q5 =  8.48 X 10^-19 C \n",
-        " q6 =  11.88 X 10^-19 C\n"
+        "The minimum uncertainity calculated is 1rad/sec. The claimed accuracy is in rad/sec\n",
+        "0.063\n",
+        "thus there is a reason to doubt the claim\n"
        ]
       }
      ],
-     "prompt_number": 16
+     "prompt_number": 6
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 4.4, page no. 121"
+      "Example 4.4 Page 115"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "ma_by_mp = 4.0        #as alpha has 2 protons and 2 neutrons\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "vp = (2*ma_by_mp/(ma_by_mp+1))\n",
-      "va1 = (ma_by_mp -1)/(ma_by_mp + 1)\n",
-      "change = (va1 - 1)*100\n",
+      "#initiation of variable\n",
+      "from math import pi\n",
+      "m=9.11*10**-31;v=3.6*10**6; #'m','v' - mass an velocity of the electron in SI units\n",
+      "h=1.05*10**-34; #planck's constant in SI\n",
+      "p=m*v; #momentum\n",
+      "delp=p*0.01; #due to 1% precision in p\n",
+      "delx = h/delp; #uncertainity in position\n",
       "\n",
       "#result\n",
+      "print \"Uncertainty in position in nm is\",round(delx*10**9,2);\n",
       "\n",
-      "print \"The percentage change in the velocity of alpha particle is\",change,\"%\""
+      "#partb\n",
+      "print \"Since the motion is strictly along X-direction, its velocity in Y direction is absolutely zero.\\n So uncertainity in velocity along y is zero=> uncertainity in position along y is infinite. \\nSo nothing can be said about its position/motion along \""
      ],
      "language": "python",
      "metadata": {},
@@ -231,39 +179,42 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The percentage change in the velocity of alpha particle is -40.0 %\n"
+        "Uncertainity in position in nm is 3.2\n",
+        "Since the motion is strictly along X-direction, its velocity in Y direction is absolutely zero.\n",
+        " So uncertainity in velocity along y is zero=> uncertainity in position along y is infinite. \n",
+        "So nothing can be said about its position/motion along \n"
        ]
       }
      ],
-     "prompt_number": 18
+     "prompt_number": 4
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 4.5, page no. 124"
+      "Example 4.5 Page 116"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "from math import pi\n",
+      "m=0.145;v=42.5; #'m','v' - mass an velocity of the electron in SI units\n",
+      "h=1.05*10**-34; #planck's constant in SI\n",
+      "p=m*v; #momentum\n",
+      "delp=p*0.01;#due to 1% precision in p\n",
       "\n",
-      "#Variable declaration\n",
-      "\n",
-      "e = 1.6 * 10 ** -19         #Charge of electron (C)\n",
-      "Z = 13\n",
-      "k = 8.99 * 10 ** 9          #(Nm^2/C^2)\n",
-      "Ka = 7.7 * 10 ** 6 * e      #potential energy at closest approach(J)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "dmin = k * 2 * Z * e ** 2/Ka\n",
+      "#calculation\n",
+      "delx = h/delp#uncertainty in position\n",
       "\n",
       "#result\n",
+      "print \"Uncertainity in position is %.1e\" %delx;\n",
       "\n",
-      "print \"The radius of the aluminum nucleus is\",round(dmin/10**-15,1),\"X 10^-15 m.\""
+      "#part b\n",
+      "print \"Motion along y is unpredictable as long as the velocity along y is exactly known(as zero).\";"
      ],
      "language": "python",
      "metadata": {},
@@ -272,54 +223,40 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The radius of the aluminum nucleus is 4.9 X 10^-15 m.\n"
+        "Uncertainity in position is 1.7e-33\n",
+        "Motion along y is unpredictable as long as the veloity along y is exactly known(as zero).\n"
        ]
       }
      ],
-     "prompt_number": 20
+     "prompt_number": 5
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 4.7, page no. 135"
+      "Example 4.7 Page 119"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "from math import sqrt\n",
+      "mc2=2.15*10**-4;          #mc2 is the mass of the electron, concidered in Mev for the simplicity in calculations\n",
+      "hc=197.0                  # The value of h*c in Mev.fm for simplicity\n",
+      "delx= 10.0               # Given uncertainty in position=diameter of nucleus= 10 fm\n",
+      "\n",
+      "#calculation\n",
+      "delp= hc/delx ;       #Uncertainty in momentum per unit 'c' i.e (Mev/c) delp= h/delx =(h*c)/(c*delx);hc=197 Mev.fm  1Mev=1.6*10^-13 Joules')\n",
+      "p=delp;               # Equating delp to p  as a consequence of equation 4.10\n",
+      "K1=p**2+mc2**2   # The following 3 steps are the steps invlolved in calculating K.E= sqrt((p*c)^2 + (mc^2)^2)- m*c^2\n",
+      "K1=sqrt(K1)\n",
+      "K1= K1-(mc2);\n",
       "\n",
-      "#Variable declaration\n",
-      "\n",
-      "R = 1.097 * 10**7               #Rydberg constant(m^-1) \n",
-      "nf = 1.0                          #energy state\n",
-      "ni = 2.0                          #energy state\n",
-      "c = 3 * 10 ** 8                 #speed of light (m/s)\n",
-      "h = 4.136 * 10 ** -15           #Planck's constant (eV.s)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "lamda = (R * ((1/nf**2) - (1/ni**2))) ** -1\n",
-      "f = c / lamda\n",
-      "E = h * f\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print \"The wavelength of the emitted photon is\",round(lamda/10**-9,2),\"nm and frequency is\",round(f/10**15,2),\"X 10^15 Hz and energy is\",round(E,1),\"eV\"\n",
-      "\n",
-      "\n",
-      "#Variable declarartion\n",
-      "mc2 = 938.8 * 10 ** 6   #m*c^2 of hydrogen atom(eV)\n",
-      "#Calculation\n",
-      "\n",
-      "p = E / c\n",
-      "K = 0.5 * E**2 /mc2\n",
-      "\n",
-      "#results\n",
-      "\n",
-      "print \"The momentum of the recoiling hydrogen atom is \",round(p*c,1),\"eV/c and the energy is\",round(K/10**-8,2),\"X 10^-8 eV.\""
+      "#result\n",
+      "print \"Kinetic energy was found out to be in Mev is\", round(K1,3)"
      ],
      "language": "python",
      "metadata": {},
@@ -328,59 +265,39 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The wavelength of the emitted photon is 121.54 nm and frequency is 2.47 X 10^15 Hz and energy is 10.2 eV\n",
-        "The momentum of the recoiling hydrogen atom is  10.2 eV/c and the energy is 5.55 X 10^-8 eV.\n"
+        "Kinetic energy was found out to be in Mev 19.7\n"
        ]
       }
      ],
-     "prompt_number": 22
+     "prompt_number": 14
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 4.8, page no. 136"
+      "Example 4.8 Page 120"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "h=6.58*10**-16; # plack's constant\n",
+      "delt1=26.0*10**-9;E1=140.0*10**6 #given values of lifetime and rest energy of charged pi meson\n",
+      "delt2=8.3*10**-17;E2=135.0*10**6;  #given values of lifetime and rest energy of uncharged pi meson\n",
+      "delt3=4.4*10**-24;E3=765*10**6;       #given values of lifetime and rest energy of rho meson\n",
       "\n",
-      "#Variable declaration\n",
-      "\n",
-      "R = 1.097 * 10**7               #Rydberg constant(m^-1) \n",
-      "nf = 2.0                        #energy state\n",
-      "ni = 3.0                        #energy state\n",
-      "c = 3 * 10 ** 8                 #speed of light (m/s)\n",
-      "h = 4.136 * 10 ** -15           #Planck's constant (eV.s)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "lamda = (R * ((1/nf**2) - (1/ni**2))) ** -1\n",
-      "f = c / lamda\n",
-      "E = h * f\n",
+      "#calculation\n",
+      "delE1=h/delt1; k1=delE1/E1;  # k is the measure of uncertainity\n",
+      "delE2=h/delt2; k2=delE2/E2;\n",
+      "delE3=h/delt3; k3=delE3/E3;\n",
       "\n",
-      "#Results\n",
-      "\n",
-      "print \"The longest-wavelength photon in the Balmer series is\",round(lamda/10**-9,2),\"nm and frequency is\",round(f/10**15,2),\"X 10^15 Hz and energy is\",round(E,2),\"eV\"\n",
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "R = 1.097 * 10**7               #Rydberg constant(m^-1) \n",
-      "nf = 2.0                          #energy state\n",
-      "c = 3 * 10 ** 8                 #speed of light (m/s)\n",
-      "h = 4.136 * 10 ** -15           #Planck's constant (eV.s)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "lamda = (R * (1/nf**2)) ** -1\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print \"The shortest-wavelength photon emitted in the Balmer series is\",round(lamda/10**-9,2),\"nm.\""
+      "#result\n",
+      "print \"Uncertainty in energy of charged pi meson is %.1e\" %k1;\n",
+      "print \"Uncertainty in energy of uncharged pi meson is %.1e\" %k2;\n",
+      "print \"Uncertainty in energy of rho meson is \",round(k3,2);"
      ],
      "language": "python",
      "metadata": {},
@@ -389,43 +306,37 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The longest-wavelength photon in the Balmer series is 656.34 nm and frequency is 0.46 X 10^15 Hz and energy is 1.89 eV\n",
-        "The shortest-wavelength photon emitted in the Balmer series is 364.63 nm.\n"
+        "Uncertainity in energy of charged pi meson is 1.8e-16\n",
+        "Uncertainity in energy of uncharged pi meson is 5.9e-08\n",
+        "Uncertainity in energy of rho meson is  0.2\n"
        ]
       }
      ],
-     "prompt_number": 24
+     "prompt_number": 9
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 4.9, page no. 137"
+      "Example 4.9 Page 121"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "h=1.05*10**-34;  #value of planck's constant in J.sec\n",
+      "delx= 1.0;        # uncertainty in positon= dimension of the ball\n",
+      "delp=h/delx;    # uncertainty in momentum \n",
+      "m=0.1;         #mass of the ball in kg\n",
       "\n",
+      "#calculation\n",
+      "delv=delp/m;    # uncertainty in velocity\n",
       "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "E = 10.2            #average thermal energy per atom(eV)\n",
-      "kB =8.62 * 10**-5   #Boltzmann constant (eV/K)\n",
-      "N2_by_N1 = 0.10     #N2/N1\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "T1 = E /((3.0/2.0)* kB)\n",
-      "T2 = -E/(kB * math.log(N2_by_N1))\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print \"The temperature is \",round(T1),\"K and by Boltzmann distribution the temperature is\",round(T2),\"K.\""
+      "#result\n",
+      "print \"The value of minimum velocity was found out to be in m/sec\",delv;"
      ],
      "language": "python",
      "metadata": {},
@@ -434,11 +345,11 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The temperature is  78886.0 K and by Boltzmann distribution the temperature is 51390.0 K.\n"
+        "The value of minimum velocity was found out to be in m/sec 1.05e-33\n"
        ]
       }
      ],
-     "prompt_number": 26
+     "prompt_number": 17
     }
    ],
    "metadata": {}
diff --git a/Modern_Physics/Chapter4_1.ipynb b/Modern_Physics/Chapter4_1.ipynb
new file mode 100755
index 00000000..55a7ca03
--- /dev/null
+++ b/Modern_Physics/Chapter4_1.ipynb
@@ -0,0 +1,188 @@
+{
+ "metadata": {
+  "name": "MP-4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "The Wave Like Properties of Particles"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.1 Page 101"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nh=6.6*10**-34;                              # h(planck's constant)= 6.6*10^-34 \nm1= 10.0**3;v1=100.0;                           # for automobile\n\n#calculation\nw1= h/(m1*v1);                            # ['w'-wavelength in metre'm'-mass in Kg 'v'-velocity in metres/sec.] of the particles \nm2=10.0*(10**-3);v2= 500;                     # for bullet\nw2=h/(m2*v2);\nm3=(10.0**-9)*(10.0**-3); v3=1.0*10**-2;\nw3=h/(m3*v3);\nm4=9.1*10**-31;k=1*1.6*10**-19;            # k- kinetic energy of the electron & using 1ev = 1.6*10^-19 joule\np=sqrt(2.0*m4*k);                         # p=momentum of electron ;from K=1/2*m*v^2\nw4=h/p;\nhc=1240;pc=100                         # In the extreme relativistc realm, K=E=pc; Given pc=100MeV,hc=1240MeV \nw5= hc/pc;\n\n#result\nprint \"Wavelength of the automobile in m is\",w1;\nprint \"Wavelength of the bullet in m is \",w2 ;\nprint\"Wavelength of the smoke particle in m is\",w3 ;\nprint \"Wavelength of the electron(1ev) in nm is\",round(w4*10**9,3) ;\nprint \"Wavelength of the electron (100Mev) in fm is\",w5;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Wavelength of the automobile in m is 6.6e-39\nWavelength of the bullet in m is  1.32e-34\nWavelength of the smoke particle in m is 6.6e-20\nWavelength of the electron(1ev) in nm is 1.223\nWavelength of the electron (100Mev) in fm is 12\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.2 Page 113"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi\n# w=wavelength; consider k=2*(pi/w); \n# differentiate k w.r.t w and replace del(k)/del(w) = 1 for equation.4.3\n# which gives del(w)= w^2 /(2*pi*del(x)), hence \nw=20; delx=200; # delx=200cm and w=20cm\n\n#calculation\ndelw=(w**2)/(delx*2*pi);\n\n#result\nprint \"Hence uncertainity in length in cm is\",round(delw,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Hence uncertainity in length is in cm 0.318\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.3 Page 114"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi\ndelt=1.0;             #consider time interval of 1 sec\ndelw=1/delt;        # since delw*delt =1 from equation 4.4\ndelf=0.01          #calculated accuracy is 0.01Hz\n\n#calculation\ndelwc =2*pi*delf # delwc-claimed accuracy from w=2*pi*f\n\n#result\nprint \"The minimum uncertainity calculated is 1rad/sec. The claimed accuracy in rad/sec is \\n\",round(delwc,3);\nprint \"thus there is a reason to doubt the claim\"\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The minimum uncertainity calculated is 1rad/sec. The claimed accuracy is in rad/sec\n0.063\nthus there is a reason to doubt the claim\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.4 Page 115"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi\nm=9.11*10**-31;v=3.6*10**6; #'m','v' - mass an velocity of the electron in SI units\nh=1.05*10**-34; #planck's constant in SI\np=m*v; #momentum\ndelp=p*0.01; #due to 1% precision in p\ndelx = h/delp; #uncertainity in position\n\n#result\nprint \"Uncertainity in position in nm is\",round(delx*10**9,2);\n\n#partb\nprint \"Since the motion is strictly along X-direction, its velocity in Y direction is absolutely zero.\\n So uncertainity in velocity along y is zero=> uncertainity in position along y is infinite. \\nSo nothing can be said about its position/motion along \"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Uncertainity in position in nm is 3.2\nSince the motion is strictly along X-direction, its velocity in Y direction is absolutely zero.\n So uncertainity in velocity along y is zero=> uncertainity in position along y is infinite. \nSo nothing can be said about its position/motion along \n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.5 Page 116"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi\nm=0.145;v=42.5; #'m','v' - mass an velocity of the electron in SI units\nh=1.05*10**-34; #planck's constant in SI\np=m*v; #momentum\ndelp=p*0.01;#due to 1% precision in p\n\n#calculation\ndelx = h/delp#uncertainity in position\n\n#result\nprint \"Uncertainity in position is %.1e\" %delx;\n\n#part b\nprint \"Motion along y is unpredictable as long as the veloity along y is exactly known(as zero).\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Uncertainity in position is 1.7e-33\nMotion along y is unpredictable as long as the veloity along y is exactly known(as zero).\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.7 Page 119"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nmc2=2.15*10**-4;          #mc2 is the mass of the electron, concidered in Mev for the simplicity in calculations\nhc=197.0                  # The value of h*c in Mev.fm for simplicity\ndelx= 10.0               # Given uncertainity in position=diameter of nucleus= 10 fm\n\n#calculation\ndelp= hc/delx ;       #Uncertainiy in momentum per unit 'c' i.e (Mev/c) delp= h/delx =(h*c)/(c*delx);hc=197 Mev.fm  1Mev=1.6*10^-13 Joules')\np=delp;               # Equating delp to p  as a consequence of equation 4.10\nK1=p**2+mc2**2   # The following 3 steps are the steps invlolved in calculating K.E= sqrt((p*c)^2 + (mc^2)^2)- m*c^2\nK1=sqrt(K1)\nK1= K1-(mc2);\n\n#result\nprint \"Kinetic energy was found out to be in Mev is\", round(K1,3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Kinetic energy was found out to be in Mev 19.7\n"
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.8 Page 120"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nh=6.58*10**-16; # plack's constant\ndelt1=26.0*10**-9;E1=140.0*10**6 #given values of lifetime and rest energy of charged pi meson\ndelt2=8.3*10**-17;E2=135.0*10**6;  #given values of lifetime and rest energy of uncharged pi meson\ndelt3=4.4*10**-24;E3=765*10**6;       #given values of lifetime and rest energy of rho meson\n\n#calculation\ndelE1=h/delt1; k1=delE1/E1;  # k is the measure of uncertainity\ndelE2=h/delt2; k2=delE2/E2;\ndelE3=h/delt3; k3=delE3/E3;\n\n#result\nprint \"Uncertainity in energy of charged pi meson is %.1e\" %k1;\nprint \"Uncertainity in energy of uncharged pi meson is %.1e\" %k2;\nprint \"Uncertainity in energy of rho meson is \",round(k3,2);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Uncertainity in energy of charged pi meson is 1.8e-16\nUncertainity in energy of uncharged pi meson is 5.9e-08\nUncertainity in energy of rho meson is  0.2\n"
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.9 Page 121"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nh=1.05*10**-34;  #value of planck's constant in J.sec\ndelx= 1.0;        # uncertainity in positon= dimension of the ball\ndelp=h/delx;    # uncertainity in momentum \nm=0.1;         #mass of the ball in kg\n\n#calculation\ndelv=delp/m;    # uncertainity in velocity\n\n#result\nprint \"The value of minimum velocity was found out to be in m/sec\",delv;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of minimum velocity was found out to be in m/sec 1.05e-33\n"
+      }
+     ],
+     "prompt_number": 17
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter4_2.ipynb b/Modern_Physics/Chapter4_2.ipynb
new file mode 100755
index 00000000..55a7ca03
--- /dev/null
+++ b/Modern_Physics/Chapter4_2.ipynb
@@ -0,0 +1,188 @@
+{
+ "metadata": {
+  "name": "MP-4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "The Wave Like Properties of Particles"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.1 Page 101"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nh=6.6*10**-34;                              # h(planck's constant)= 6.6*10^-34 \nm1= 10.0**3;v1=100.0;                           # for automobile\n\n#calculation\nw1= h/(m1*v1);                            # ['w'-wavelength in metre'm'-mass in Kg 'v'-velocity in metres/sec.] of the particles \nm2=10.0*(10**-3);v2= 500;                     # for bullet\nw2=h/(m2*v2);\nm3=(10.0**-9)*(10.0**-3); v3=1.0*10**-2;\nw3=h/(m3*v3);\nm4=9.1*10**-31;k=1*1.6*10**-19;            # k- kinetic energy of the electron & using 1ev = 1.6*10^-19 joule\np=sqrt(2.0*m4*k);                         # p=momentum of electron ;from K=1/2*m*v^2\nw4=h/p;\nhc=1240;pc=100                         # In the extreme relativistc realm, K=E=pc; Given pc=100MeV,hc=1240MeV \nw5= hc/pc;\n\n#result\nprint \"Wavelength of the automobile in m is\",w1;\nprint \"Wavelength of the bullet in m is \",w2 ;\nprint\"Wavelength of the smoke particle in m is\",w3 ;\nprint \"Wavelength of the electron(1ev) in nm is\",round(w4*10**9,3) ;\nprint \"Wavelength of the electron (100Mev) in fm is\",w5;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Wavelength of the automobile in m is 6.6e-39\nWavelength of the bullet in m is  1.32e-34\nWavelength of the smoke particle in m is 6.6e-20\nWavelength of the electron(1ev) in nm is 1.223\nWavelength of the electron (100Mev) in fm is 12\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.2 Page 113"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi\n# w=wavelength; consider k=2*(pi/w); \n# differentiate k w.r.t w and replace del(k)/del(w) = 1 for equation.4.3\n# which gives del(w)= w^2 /(2*pi*del(x)), hence \nw=20; delx=200; # delx=200cm and w=20cm\n\n#calculation\ndelw=(w**2)/(delx*2*pi);\n\n#result\nprint \"Hence uncertainity in length in cm is\",round(delw,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Hence uncertainity in length is in cm 0.318\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.3 Page 114"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi\ndelt=1.0;             #consider time interval of 1 sec\ndelw=1/delt;        # since delw*delt =1 from equation 4.4\ndelf=0.01          #calculated accuracy is 0.01Hz\n\n#calculation\ndelwc =2*pi*delf # delwc-claimed accuracy from w=2*pi*f\n\n#result\nprint \"The minimum uncertainity calculated is 1rad/sec. The claimed accuracy in rad/sec is \\n\",round(delwc,3);\nprint \"thus there is a reason to doubt the claim\"\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The minimum uncertainity calculated is 1rad/sec. The claimed accuracy is in rad/sec\n0.063\nthus there is a reason to doubt the claim\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.4 Page 115"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi\nm=9.11*10**-31;v=3.6*10**6; #'m','v' - mass an velocity of the electron in SI units\nh=1.05*10**-34; #planck's constant in SI\np=m*v; #momentum\ndelp=p*0.01; #due to 1% precision in p\ndelx = h/delp; #uncertainity in position\n\n#result\nprint \"Uncertainity in position in nm is\",round(delx*10**9,2);\n\n#partb\nprint \"Since the motion is strictly along X-direction, its velocity in Y direction is absolutely zero.\\n So uncertainity in velocity along y is zero=> uncertainity in position along y is infinite. \\nSo nothing can be said about its position/motion along \"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Uncertainity in position in nm is 3.2\nSince the motion is strictly along X-direction, its velocity in Y direction is absolutely zero.\n So uncertainity in velocity along y is zero=> uncertainity in position along y is infinite. \nSo nothing can be said about its position/motion along \n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.5 Page 116"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi\nm=0.145;v=42.5; #'m','v' - mass an velocity of the electron in SI units\nh=1.05*10**-34; #planck's constant in SI\np=m*v; #momentum\ndelp=p*0.01;#due to 1% precision in p\n\n#calculation\ndelx = h/delp#uncertainity in position\n\n#result\nprint \"Uncertainity in position is %.1e\" %delx;\n\n#part b\nprint \"Motion along y is unpredictable as long as the veloity along y is exactly known(as zero).\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Uncertainity in position is 1.7e-33\nMotion along y is unpredictable as long as the veloity along y is exactly known(as zero).\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.7 Page 119"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nmc2=2.15*10**-4;          #mc2 is the mass of the electron, concidered in Mev for the simplicity in calculations\nhc=197.0                  # The value of h*c in Mev.fm for simplicity\ndelx= 10.0               # Given uncertainity in position=diameter of nucleus= 10 fm\n\n#calculation\ndelp= hc/delx ;       #Uncertainiy in momentum per unit 'c' i.e (Mev/c) delp= h/delx =(h*c)/(c*delx);hc=197 Mev.fm  1Mev=1.6*10^-13 Joules')\np=delp;               # Equating delp to p  as a consequence of equation 4.10\nK1=p**2+mc2**2   # The following 3 steps are the steps invlolved in calculating K.E= sqrt((p*c)^2 + (mc^2)^2)- m*c^2\nK1=sqrt(K1)\nK1= K1-(mc2);\n\n#result\nprint \"Kinetic energy was found out to be in Mev is\", round(K1,3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Kinetic energy was found out to be in Mev 19.7\n"
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.8 Page 120"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nh=6.58*10**-16; # plack's constant\ndelt1=26.0*10**-9;E1=140.0*10**6 #given values of lifetime and rest energy of charged pi meson\ndelt2=8.3*10**-17;E2=135.0*10**6;  #given values of lifetime and rest energy of uncharged pi meson\ndelt3=4.4*10**-24;E3=765*10**6;       #given values of lifetime and rest energy of rho meson\n\n#calculation\ndelE1=h/delt1; k1=delE1/E1;  # k is the measure of uncertainity\ndelE2=h/delt2; k2=delE2/E2;\ndelE3=h/delt3; k3=delE3/E3;\n\n#result\nprint \"Uncertainity in energy of charged pi meson is %.1e\" %k1;\nprint \"Uncertainity in energy of uncharged pi meson is %.1e\" %k2;\nprint \"Uncertainity in energy of rho meson is \",round(k3,2);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Uncertainity in energy of charged pi meson is 1.8e-16\nUncertainity in energy of uncharged pi meson is 5.9e-08\nUncertainity in energy of rho meson is  0.2\n"
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 4.9 Page 121"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nh=1.05*10**-34;  #value of planck's constant in J.sec\ndelx= 1.0;        # uncertainity in positon= dimension of the ball\ndelp=h/delx;    # uncertainity in momentum \nm=0.1;         #mass of the ball in kg\n\n#calculation\ndelv=delp/m;    # uncertainity in velocity\n\n#result\nprint \"The value of minimum velocity was found out to be in m/sec\",delv;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of minimum velocity was found out to be in m/sec 1.05e-33\n"
+      }
+     ],
+     "prompt_number": 17
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter5.ipynb b/Modern_Physics/Chapter5.ipynb
index 175eed34..331dc531 100755
--- a/Modern_Physics/Chapter5.ipynb
+++ b/Modern_Physics/Chapter5.ipynb
@@ -1,7 +1,6 @@
 {
  "metadata": {
-  "name": "",
-  "signature": "sha256:e257dccb197e3cab0c059eb9e1d236e5359e2a825fc6be941cab77026f236087"
+  "name": "Chapter5"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -10,10 +9,10 @@
    "cells": [
     {
      "cell_type": "heading",
-     "level": 2,
+     "level": 1,
      "metadata": {},
      "source": [
-      "Chapter 5: Matter Waves"
+      "Chapter 5: The Schrodinger Equation"
      ]
     },
     {
@@ -21,203 +20,44 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 5.1, page no. 154"
+      "Example 5.2 Page 150"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "h = 6.63 * 10 ** -34            #Planck's constant (J.s)\n",
-      "m = 0.14                        #mass of the baseball (kg)\n",
-      "v = 27.0                        #speed of the baseball (m/s)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "lamda = h / (m * v)\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "print \"The de Broglie wavelength of the baseball is\",round(lamda/10**-34,2),\"X 10^-34 m.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The de Broglie wavelength of the baseball is 1.75 X 10^-34 m.\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.2, page no. 154"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
+      "#initiation of variable\n",
+      "#parta\n",
+      "from math import pi, sin\n",
       "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "h = 6.63 * 10 ** -34            #Planck's constant (J.s)\n",
-      "me = 9.11 * 10 ** -31           #mass of electron (kg)\n",
-      "q = 1.6 * 10 ** -19             #charge of electron (C)\n",
-      "V = 50                          #potential difference (V)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "lamda = h / math.sqrt(2*me*q*V)\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "print \"The de Broglie wavelength of electron is\",round(lamda/10 ** -10,1),\"X 10^-10 m.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The de Broglie wavelength of electron is 1.7 X 10^-10 m.\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.3, page no. 158"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "h = 6.63 * 10 ** -34            #Planck's constant (J.s)\n",
-      "lamda = 1.0 * 10 ** -10         #de Broglie wavelength of the neutron (m)\n",
-      "mn = 1.66 * 10**-27             #mass of neutrons (kg)\n",
-      "e = 1.602 * 10 **-19              #charge of electron(C)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "p = h / lamda\n",
-      "K = p**2/(2*mn)\n",
+      "from scipy import integrate\n",
+      "h=1.05*10**-34;m=9.11*10**-31;L=10.0**-10;      # all the values are taken in SI units\n",
+      "E1=h**2*pi**2/(2*m*L**2); E2=4*E1;            #Energies are calculated\n",
+      "delE=(E2-E1)/(1.6*10**-19);                  #Difference in energy is converted to eV\n",
       "\n",
       "#result\n",
+      "print \"Energy to be supplied in eV. is \",round(delE,3);\n",
       "\n",
-      "print \"The kinetic energy is\",round(K/e,4),\"eV.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The kinetic energy is 0.0826 eV.\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.8, page no. 177"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "h = 1.05 * 10 ** -34        #(J.s)\n",
-      "dx = 15                     #length of the room (m)\n",
-      "m = 0.1                     #mass of the ball (kg)\n",
-      "vx = 2.0                    #velocity of the ball (m/s)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "dpx = h /( 2* dx)\n",
-      "dvx = dpx /m\n",
-      "uncertainity = dvx/vx\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "print \"The relative uncertainty is\",round(uncertainity/10**-35,1),\"X 10^-35 which is not measurable.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The relative uncertainty is 1.8 X 10^-35 which is not measurable.\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.9, page no. 178"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "\n",
-      "import math\n",
+      "#partb\n",
+      "x1=0.09*10**-10;x2=0.11*10**-10                 #limits of the given region\n",
+      "def integrand(x,L):\n",
+      "    return 2.0/L*(sin(pi*x/L))**2\n",
       "\n",
-      "#Variable declaration\n",
+      "probGnd=integrate.quad(integrand,x1,x2,args=(L))\n",
       "\n",
-      "h = 6.58 * 10 ** -16            #(eV.s)\n",
-      "dx = 1.0 * 10 ** -14 / 2.0      # dx is half the length of confinement (m)\n",
-      "c = 3.00 * 10 ** 8              #speed of light (m/s)\n",
-      "me = 9.11 * 10 ** -31           # mass of electron (kg)\n",
-      "e = 1.6 * 10 ** -19             #charge of electron (C)\n",
+      "#result\n",
+      "print \"The percentage probability of finding an electron in the ground state is \\n\",round(probGnd[0]*100,3);\n",
       "\n",
-      "#Calculation\n",
+      "#partc\n",
+      "k1=0.0;k2=0.25*10**-10;\n",
+      "def integrand(k,L):\n",
+      "    return 2.0/L*(sin(2*pi*k/L))**2\n",
       "\n",
-      "dpx = h * c / (2 * dx)\n",
-      "E = math.sqrt(dpx**2 + (me * c**2/e)**2)\n",
-      "K = E - (me * c**2/e)\n",
+      "probExc=integrate.quad(integrand,k1,k2,args=(L))\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"The kinetic energy of an  intranuclear electron is\",round(K/10**6,2),\"MeV.\""
+      "print \"The probability of finding an electron in the excited state is\",round(probExc[0],3);"
      ],
      "language": "python",
      "metadata": {},
@@ -226,67 +66,22 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The kinetic energy of an  intranuclear electron is 19.23 MeV.\n"
+        " Energy to be supplied in eV. is  111.978\n",
+        "The percentage probablility of finding an electron in the ground state is \n",
+        "0.383\n",
+        "The probablility of finding an electron in the excited state is 0.25\n"
        ]
       }
      ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 5.10, page no. 178"
-     ]
+     "prompt_number": 7
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": [
-      "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "dt = 1.0 * 10 ** -8         #lifetime (s)\n",
-      "\n",
-      "#calculation\n",
-      "\n",
-      "df = 1/(4*math.pi*dt)\n",
-      "\n",
-      "#result\n",
-      "\n",
-      "print \"The frequency of the light emitted is\",round(df/10**6,1),\"X 10^6 Hz.\"\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "c = 3.0 * 10 ** 8           #speed of light (m/s)\n",
-      "lamda = 500 * 10 ** -9      #wavelength (m)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "f = c/ lamda\n",
-      "df_by_f0 = df / f\n",
-      "\n",
-      "#result\n",
-      "\n",
-      "print \"The fractional broadening is\",round(df_by_f0/10**-8,1),\"X 10^-8.\""
-     ],
+     "input": [],
      "language": "python",
      "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The frequency of the light emitted is 8.0 X 10^6 Hz.\n",
-        "The fractional broadening is 1.3 X 10^-8.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
+     "outputs": []
     }
    ],
    "metadata": {}
diff --git a/Modern_Physics/Chapter5_1.ipynb b/Modern_Physics/Chapter5_1.ipynb
new file mode 100755
index 00000000..9c6c1971
--- /dev/null
+++ b/Modern_Physics/Chapter5_1.ipynb
@@ -0,0 +1,49 @@
+{
+ "metadata": {
+  "name": "MP-5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "The Schrodinger Equation"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 5.2 Page 150"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initation of variable\n#parta\nfrom math import pi, sin\nimport math\nfrom scipy import integrate\nh=1.05*10**-34;m=9.11*10**-31;L=10.0**-10;      # all the values are taken in SI units\nE1=h**2*pi**2/(2*m*L**2); E2=4*E1;            #Energies are calculated\ndelE=(E2-E1)/(1.6*10**-19);                  #Difference in energy is converted to eV\n\n#result\nprint \"Energy to be supplied in eV. is \",round(delE,3);\n\n#partb\nx1=0.09*10**-10;x2=0.11*10**-10                 #limits of the given region\ndef integrand(x,L):\n    return 2.0/L*(sin(pi*x/L))**2\n\nprobGnd=integrate.quad(integrand,x1,x2,args=(L))\n\n#result\nprint \"The percentage probablility of finding an electron in the ground state is \\n\",round(probGnd[0]*100,3);\n\n#partc\nk1=0.0;k2=0.25*10**-10;\ndef integrand(k,L):\n    return 2.0/L*(sin(2*pi*k/L))**2\n\nprobExc=integrate.quad(integrand,k1,k2,args=(L))\n\n#result\nprint \"The probablility of finding an electron in the excited state is\",round(probExc[0],3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " Energy to be supplied in eV. is  111.978\nThe percentage probablility of finding an electron in the ground state is \n0.383\nThe probablility of finding an electron in the excited state is 0.25\n"
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter5_2.ipynb b/Modern_Physics/Chapter5_2.ipynb
new file mode 100755
index 00000000..9c6c1971
--- /dev/null
+++ b/Modern_Physics/Chapter5_2.ipynb
@@ -0,0 +1,49 @@
+{
+ "metadata": {
+  "name": "MP-5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "The Schrodinger Equation"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 5.2 Page 150"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initation of variable\n#parta\nfrom math import pi, sin\nimport math\nfrom scipy import integrate\nh=1.05*10**-34;m=9.11*10**-31;L=10.0**-10;      # all the values are taken in SI units\nE1=h**2*pi**2/(2*m*L**2); E2=4*E1;            #Energies are calculated\ndelE=(E2-E1)/(1.6*10**-19);                  #Difference in energy is converted to eV\n\n#result\nprint \"Energy to be supplied in eV. is \",round(delE,3);\n\n#partb\nx1=0.09*10**-10;x2=0.11*10**-10                 #limits of the given region\ndef integrand(x,L):\n    return 2.0/L*(sin(pi*x/L))**2\n\nprobGnd=integrate.quad(integrand,x1,x2,args=(L))\n\n#result\nprint \"The percentage probablility of finding an electron in the ground state is \\n\",round(probGnd[0]*100,3);\n\n#partc\nk1=0.0;k2=0.25*10**-10;\ndef integrand(k,L):\n    return 2.0/L*(sin(2*pi*k/L))**2\n\nprobExc=integrate.quad(integrand,k1,k2,args=(L))\n\n#result\nprint \"The probablility of finding an electron in the excited state is\",round(probExc[0],3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " Energy to be supplied in eV. is  111.978\nThe percentage probablility of finding an electron in the ground state is \n0.383\nThe probablility of finding an electron in the excited state is 0.25\n"
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter6.ipynb b/Modern_Physics/Chapter6.ipynb
index 76cd1700..53684185 100755
--- a/Modern_Physics/Chapter6.ipynb
+++ b/Modern_Physics/Chapter6.ipynb
@@ -1,7 +1,7 @@
 {
  "metadata": {
-  "name": "",
-  "signature": "sha256:846e8e3b3770f7cb30a2e91a53718bf5de841338951843c54481c2acfda5e63d"
+  "name": "Chapter6",
+  "signature": "sha256:36f31f6870acf2c11b00274dcf34bd9e9879abf6f82026373900139ccc4b5799"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -13,7 +13,7 @@
      "level": 1,
      "metadata": {},
      "source": [
-      "Chapter 6: Quantum Mechanics in One Dimension"
+      "Chapter 6:The Rutherford Bohr Model"
      ]
     },
     {
@@ -21,33 +21,25 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 6.4, page no. 197"
+      "Example 6.1 Page 178"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      " \n",
+      "#initiation of variable\n",
+      "from math import sqrt, pi\n",
+      "R=0.1;Z=79.0; x=1.44;          #x=e^2/4*pi*epsi0\n",
+      "zkR2=2*Z*x/R                # from zkR2= (2*Z*e^2)*R^2/(4*pi*epsi0)*R^3\n",
+      "mv2=10.0*10**6;                #MeV=>eV\n",
       "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "me = 9.11 * 10 ** -31           #mass of electron (kg)\n",
-      "h = 1.055 * 10**-34             #h/2*pi (J.s)\n",
-      "dx0 = 1.0 * 10**-10             #initial location of electron(m)\n",
-      "m = 1.0 * 10**-3                #mass of marble (kg)\n",
-      "dx0m = 10**-4                   #initial location of marble (m)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "te = math.sqrt(99)* (2* me / h) * dx0**2\n",
-      "tm = math.sqrt(99)* (2* m / h) * dx0m**2\n",
+      "#calculation\n",
+      "theta=sqrt(3.0/4)*zkR2/mv2;     #deflection angle\n",
+      "theta=theta*(180/pi);        #converting to degrees\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"The time elapsed for electron is\",round(te/10**-15,1),\"X 10^-15 s and that of marble is \",round(tm/10**24,1),\"X 10^24 s.\""
+      "print\"Hence the average deflection angle per collision in degrees is\",round(theta,3 );"
      ],
      "language": "python",
      "metadata": {},
@@ -56,56 +48,122 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The time elapsed for electron is 1.7 X 10^-15 s and that of marble is  1.9 X 10^24 s.\n"
+        "Hence the average deflection angle per collision in degrees.is 0.011\n"
        ]
       }
      ],
-     "prompt_number": 1
+     "prompt_number": 2
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 6.5, page no. 202"
+      "Example 6.2 Page 181"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "m = 1.0 * 10 **-6         #mass (kg)\n",
-      "h = 6.626 * 10 **-34    #Planck's constant(J.s)\n",
-      "n = 1.0\n",
-      "L = 1.0 * 10**-2        #separation(m)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "E1 = n**2 * h**2 /(8*m*L**2)\n",
-      "v1 = math.sqrt(2*E1/m)\n",
+      "#initiation of variable\n",
+      "from math import sin, cos, tan, sqrt, pi\n",
+      "Na=6.023*10**23;p=19.3;M=197.0;\n",
+      "n=Na*p/M;         #The number of nuclei per atom\n",
+      "t=2*10**-6;Z=79;K=8*10**6;x=1.44; theta=90.0*pi/180;  #x=e^2/4*pi*epsi0\n",
+      "b1=t*Z*x/tan(theta/2)/(2*K)         #impact parameter b\n",
+      "f1=n*pi*b1**2*t                      #scattering angle greater than 90\n",
       "\n",
       "#result\n",
+      "print\"The fraction of alpha particles scattered at angles greater than 90 degrees is %.1e\" %f1;\n",
       "\n",
-      "print \"(a) The minimum speed of the particle is\",round(v1/10**-26,2),\"X 10^-26 m/s.\"\n",
+      "#part b\n",
+      "theta=45.0*pi/180;\n",
+      "b2=t*Z*x/tan(theta/2)/(2*K);\n",
+      "f2=n*pi*b2**2*t;                  #scattering angle greater than 45\n",
+      "fb=f2-f1                       #scattering angle between 45 to 90\n",
       "\n",
+      "#result\n",
+      "print\"The fraction of particles with scattering angle from 45 to 90 is %.1e\" %fb;"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The fraction of alpha particles scattered at angles greater than 90 degrees is 7.5e-05\n",
+        "The fraction of particles with scattering angle from 45 to 90 is 3.6e-04\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3 Page 185"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "from math import sin, cos, tan, sqrt, pi\n",
+      "Z=79.0;x=1.44;K=8.0*10**6;z=2;    #where x=e^2/4*pi*epsi0;z=2 for alpha particles\n",
       "\n",
-      "#Variable declaration\n",
+      "#calculation\n",
+      "d=z*x*Z/K;                   #distance\n",
       "\n",
-      "v = 3.00 * 10**-2       #speed of the particle (m/s)\n",
+      "#result\n",
+      "print \"The distance of closest approach in nm. is\",d*10**-9"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The distance of closest approasch in nm. is 2.844e-14\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4 Page 188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "sl=820.1;n0=3.0;  #given values\n",
+      "n=4;w=sl*(n**2/(n**2-n0**2)); \n",
+      "#result\n",
+      "print \"The 3 longest possible wavelengths in nm respectively are a.\",round(w,3),;    \n",
       "\n",
-      "#Calculation\n",
+      "#partb\n",
+      "n=5.0;w=sl*(n**2/(n**2-n0**2)); \n",
       "\n",
-      "E = m* v**2 /2\n",
-      "n = math.sqrt(8*m*L**2*E)/h\n",
+      "#result\n",
+      "print \"b. (in nm)\",round(w,3),;\n",
       "\n",
-      "#results\n",
+      "#partc\n",
+      "n=6.0;w=sl*(n**2/(n**2-n0**2));\n",
       "\n",
-      "print \"(b) We get n = \",round(n/10**23,2),\"X 10^23.\""
+      "#result\n",
+      "print \"c.  (in nm )\",round(w,3);\n"
      ],
      "language": "python",
      "metadata": {},
@@ -114,8 +172,7 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "(a) The minimum speed of the particle is 3.31 X 10^-26 m/s.\n",
-        "(b) We get n =  9.06 X 10^23.\n"
+        "The 3 longest possible wavelengths in nm respectively are a. 1874.514 b. (in nm) 1281.406 c.  (in nm ) 1093.467\n"
        ]
       }
      ],
@@ -126,33 +183,31 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 6.6, page no. 203"
+      "Example 6.5 Page 189"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "sl=364.5;n=3.0;           #given variables and various constants are declared in the subsequent steps wherever necessary   \n",
+      "w1=sl*(n**2/(n**2-4));     #longest wavelength of balmer \n",
+      "c=3.0*10**8;\n",
+      "f1=c/(w1*10**-9);         #corresponding freq.\n",
+      "n0=1.0;n=2.0;                 \n",
       "\n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "L = 0.2                     #length of the box (nm)\n",
-      "me = 511 * 10 ** 3          #mass of electron (eV/c^2)\n",
-      "hc = 197.3                  #(eV.nm)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "E1 = math.pi ** 2 * hc**2 /(2* me * L**2)\n",
-      "E2 = 2**2 * E1\n",
-      "dE = E2-E1\n",
-      "lamda = hc*2*math.pi / dE\n",
+      "#calculation\n",
+      "w2=91.13*(n**2/(n**2-n0**2)); #first longest of lymann   \n",
+      "f2=c/(w2*10**-9);          #correspoding freq\n",
+      "n0=1.0;n=3.0\n",
+      "w3=91.13*(n**2/(n**2-n0**2));      #second longest of lymann\n",
+      "f3=3.0*10**8/(w3*10**-9)             #corresponding freq.\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"The energy required is\",round(dE,1),\"eV and the wavelength of the photon that could cause this transition is\",round(lamda),\"nm.\""
+      "print \"The freq. corresponding to the longest wavelength of balmer is %.1e\" %f1,\" & First longest wavelength  of Lymann is %.1e\" %f2;\n",
+      "print\"The sum of which s equal to %.1e\" %(f1+f2);\n",
+      "print\"The freq. corresponding to 2nd longest wavelength was found out to be %.1e\" %f3,\"Hence Ritz combination principle is satisfied.\";"
      ],
      "language": "python",
      "metadata": {},
@@ -161,45 +216,42 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The energy required is 28.2 eV and the wavelength of the photon that could cause this transition is 44.0 nm.\n"
+        "The freq. corresponding to the longest wavelength of balmer is 4.6e+14  & First longest wavelength  of Lymann is 2.5e+15\n",
+        "The sum of which s equal to 2.9e+15\n",
+        "The freq. corresponding to 2nd longest wavelength was found out to be 2.92622261239e+15 Hence Ritz combination principle is satisfied.\n"
        ]
       }
      ],
-     "prompt_number": 5
+     "prompt_number": 6
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 6.8, page no. 211"
+      "Example 6.6 Page 192"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "Rinfi=1.097*10**7;  #known value \n",
+      "n1=3.0;n2=2.0;       #first 2 given states\n",
       "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "h = 197.3               #(eV.nm/c)\n",
-      "m = 511 * 10**3         #mass of electron (eV/c**2)\n",
-      "U = 100                 #(eV)\n",
-      "L = 0.200               #width(nm)\n",
+      "#calculation\n",
+      "w=(n1**2*n2**2)/((n1**2-n2**2)*Rinfi);\n",
       "\n",
-      "#Calculation\n",
+      "#result\n",
+      "print\"Wavelength of transition from n1=3 to n2=2 in nm is\",round(w*10**9,3);\n",
       "\n",
-      "d = h /math.sqrt(2*m*U)\n",
-      "E = math.pi**2 * h**2 /(2*m*(L+2*d)**2)\n",
-      "new_U = U - E\n",
-      "d = h/math.sqrt(2*m*new_U)\n",
-      "E = math.pi**2 * h**2 /(2*m*(L+2*d)**2)\n",
+      "#partb\n",
+      "n1=4.0;n2=2.0;      #second 2 given states  \n",
+      "w=(n1**2*n2**2)/((n1**2-n2**2)*Rinfi);\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"The ground-state energy for the electron is\",round(E,3),\"eV.\""
+      "print\"Wavelength of transition from n1=3 to n2=2 in nm is\",round(w*10**9,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -208,45 +260,42 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The ground-state energy for the electron is 6.506 eV.\n"
+        "Wavelength of trnasition from n1=3 to n2=2 in nm is 656.335\n",
+        "Wavelength of trnasition from n1=3 to n2=2 in nm is 486.174\n"
        ]
       }
      ],
-     "prompt_number": 7
+     "prompt_number": 9
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 6.13, page no. 217"
+      "Example 6.7 Page 194"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "m = 0.0100       #mass of the spring (kg)\n",
-      "K = 0.100        #force constant of spring (N/m)\n",
-      "Kh = 510.5       #force constant of hydrogen (N/m)\n",
-      "h = 6.582 * 10**-16#Planck's constant (eV.s)\n",
-      "mu = 8.37 * 10**-28#mass of hydrogen molecule(kg)\n",
+      "#initiation of variable\n",
+      "n1=3.0;n2=2.0;Z=4.0;hc=1240.0;\n",
+      "delE=(-13.6)*(Z**2)*((1/(n1**2))-((1/n2**2)));\n",
       "\n",
       "#calculation\n",
+      "w=(hc)/delE; #for transition 1\n",
       "\n",
-      "w = math.sqrt(K / m)\n",
-      "dE = h * w\n",
-      "wh =math.sqrt(Kh / mu)\n",
-      "dEh = h * wh\n",
+      "#result\n",
+      "print \"The wavelngth of radiation for transition(2->3) in nm is\", round(w,3);\n",
       "\n",
-      "#results\n",
+      "#for transition 2\n",
+      "n1=4.0;n2=2.0;   # n values for transition 2\n",
+      "delE=(-13.6)*(Z**2)*((1/n1**2)-(1/n2**2));\n",
+      "w=(hc)/delE;\n",
       "\n",
-      "print \"The quantum level spacing in the spring case is\",round(dE/10**-15,2),\"X 10^-15 eV, while in case of hydrogen molecule it is\",round(dEh,3),\"eV which is easily measurable.\""
+      "#result\n",
+      "print \"The wavelngth of radiation emitted for transition(2->4) in nm is\", round(w,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -255,11 +304,12 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The quantum level spacing in the spring case is 2.08 X 10^-15 eV, while in case of hydrogen molecule it is 0.514 eV which is easily measurable.\n"
+        "The wavelngth of radiation for transition(2->3) in nm is 41.029\n",
+        "The wavelngth of radiation emitted for transition(2->4) in nm is 30.392\n"
        ]
       }
      ],
-     "prompt_number": 10
+     "prompt_number": 11
     }
    ],
    "metadata": {}
diff --git a/Modern_Physics/Chapter6_1.ipynb b/Modern_Physics/Chapter6_1.ipynb
new file mode 100755
index 00000000..09d16e38
--- /dev/null
+++ b/Modern_Physics/Chapter6_1.ipynb
@@ -0,0 +1,167 @@
+{
+ "metadata": {
+  "name": "MP-6"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "The Rutherford Bohr Model"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.1 Page 178"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt, pi\nR=0.1;Z=79.0; x=1.44;          #x=e^2/4*pi*epsi0\nzkR2=2*Z*x/R                # from zkR2= (2*Z*e^2)*R^2/(4*pi*epsi0)*R^3\nmv2=10.0*10**6;                #MeV=>eV\n\n#calculation\ntheta=sqrt(3.0/4)*zkR2/mv2;     #deflection angle\ntheta=theta*(180/pi);        #converting to degrees\n\n#result\nprint\"Hence the average deflection angle per collision in degrees is\",round(theta,3 );",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Hence the average deflection angle per collision in degrees.is 0.011\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.2 Page 181"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of vriable\nfrom math import sin, cos, tan, sqrt, pi\nNa=6.023*10**23;p=19.3;M=197.0;\nn=Na*p/M;         #The number of nuclei per atom\nt=2*10**-6;Z=79;K=8*10**6;x=1.44; theta=90.0*pi/180;  #x=e^2/4*pi*epsi0\nb1=t*Z*x/tan(theta/2)/(2*K)         #impact parameter b\nf1=n*pi*b1**2*t                      #scattering angle greater than 90\n\n#result\nprint\"The fraction of alpha particles scattered at angles greater than 90 degrees is %.1e\" %f1;\n\n#part b\ntheta=45.0*pi/180;\nb2=t*Z*x/tan(theta/2)/(2*K);\nf2=n*pi*b2**2*t;                  #scattering angle greater than 45\nfb=f2-f1                       #scattering angle between 45 to 90\n\n#result\nprint\"The fraction of particles with scattering angle from 45 to 90 is %.1e\" %fb;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The fraction of alpha particles scattered at angles greater than 90 degrees is 7.5e-05\nThe fraction of particles with scattering angle from 45 to 90 is 3.6e-04\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.3 Page 185"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of vriable\nfrom math import sin, cos, tan, sqrt, pi\nZ=79.0;x=1.44;K=8.0*10**6;z=2;    #where x=e^2/4*pi*epsi0;z=2 for alpha particles\n\n#calculation\nd=z*x*Z/K;                   #distance\n\n#result\nprint \"The distance of closest approasch in nm. is\",d*10**-9",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " The distance of closest approasch in nm. is 2.844e-14\n"
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.4 Page 188"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nsl=820.1;n0=3.0;  #given values\nn=4;w=sl*(n**2/(n**2-n0**2)); \n#result\nprint \"The 3 longest possible wavelengths are in nm respectively are\",w,;    \n\n#partb\nn=5.0;w=sl*(n**2/(n**2-n0**2)); \n\n#result\nprint \"(in nm)\",round(w,3),;\n\n#partc\nn=6.0;w=sl*(n**2/(n**2-n0**2));\n\n#result\nprint \" (in nm )\",round(w,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The 3 longest possible wavelengths are in nm respectively are 1874.51428571 (in nm) 1281.406  (in nm ) 1093.467\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.5 Page 189"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nsl=364.5;n=3.0;           #given variables and various constants are declared in the subsequent steps wherever necessary   \nw1=sl*(n**2/(n**2-4));     #longest wavelength of balmer \nc=3.0*10**8;\nf1=c/(w1*10**-9);         #corresponding freq.\nn0=1.0;n=2.0;                 \n\n#calculation\nw2=91.13*(n**2/(n**2-n0**2)); #first longest of lymann   \nf2=c/(w2*10**-9);          #correspoding freq\nn0=1.0;n=3.0\nw3=91.13*(n**2/(n**2-n0**2));      #second longest of lymann\nf3=3.0*10**8/(w3*10**-9)             #corresponding freq.\n\n#result\nprint \"The freq. corresponding to the longest wavelength of balmer is %.1e\" %f1,\" & First longest wavelength  of Lymann is %.1e\" %f2;\nprint\"The sum of which s equal to %.1e\" %(f1+f2);\nprint\"The freq. corresponding to 2nd longest wavelength was found out to be %.1e\" %f3,\"Hence Ritz combination principle is satisfied.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The freq. corresponding to the longest wavelength of balmer is 4.6e+14  & First longest wavelength  of Lymann is 2.5e+15\nThe sum of which s equal to 2.9e+15\nThe freq. corresponding to 2nd longest wavelength was found out to be 2.92622261239e+15 Hence Ritz combination principle is satisfied.\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.6 Page 192"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nRinfi=1.097*10**7;  #known value \nn1=3.0;n2=2.0;       #first 2 given states\n\n#calculation\nw=(n1**2*n2**2)/((n1**2-n2**2)*Rinfi);\n\n#result\nprint\"Wavelength of trnasition from n1=3 to n2=2 in nm is\",round(w*10**9,3);\n\n#partb\nn1=4.0;n2=2.0;      #second 2 given states  \nw=(n1**2*n2**2)/((n1**2-n2**2)*Rinfi);\n\n#result\nprint\"Wavelength of trnasition from n1=3 to n2=2 in nm is\",round(w*10**9,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Wavelength of trnasition from n1=3 to n2=2 in nm is 656.335\nWavelength of trnasition from n1=3 to n2=2 in nm is 486.174\n"
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.7 Page 194"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nn1=3.0;n2=2.0;Z=4.0;hc=1240.0;\ndelE=(-13.6)*(Z**2)*((1/(n1**2))-((1/n2**2)));\n\n#calculation\nw=(hc)/delE; #for transition 1\n\n#result\nprint \"The wavelngth of radiation for transition(2->3) in nm is\", round(w,3);\n\n#for transition 2\nn1=4.0;n2=2.0;   # n values for transition 2\ndelE=(-13.6)*(Z**2)*((1/n1**2)-(1/n2**2));\nw=(hc)/delE;\n\n#result\nprint \"The wavelngth of radiation emitted for transition(2->4) in nm is\", round(w,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The wavelngth of radiation for transition(2->3) in nm is 41.029\nThe wavelngth of radiation emitted for transition(2->4) in nm is 30.392\n"
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter6_2.ipynb b/Modern_Physics/Chapter6_2.ipynb
new file mode 100755
index 00000000..09d16e38
--- /dev/null
+++ b/Modern_Physics/Chapter6_2.ipynb
@@ -0,0 +1,167 @@
+{
+ "metadata": {
+  "name": "MP-6"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "The Rutherford Bohr Model"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.1 Page 178"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt, pi\nR=0.1;Z=79.0; x=1.44;          #x=e^2/4*pi*epsi0\nzkR2=2*Z*x/R                # from zkR2= (2*Z*e^2)*R^2/(4*pi*epsi0)*R^3\nmv2=10.0*10**6;                #MeV=>eV\n\n#calculation\ntheta=sqrt(3.0/4)*zkR2/mv2;     #deflection angle\ntheta=theta*(180/pi);        #converting to degrees\n\n#result\nprint\"Hence the average deflection angle per collision in degrees is\",round(theta,3 );",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Hence the average deflection angle per collision in degrees.is 0.011\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.2 Page 181"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of vriable\nfrom math import sin, cos, tan, sqrt, pi\nNa=6.023*10**23;p=19.3;M=197.0;\nn=Na*p/M;         #The number of nuclei per atom\nt=2*10**-6;Z=79;K=8*10**6;x=1.44; theta=90.0*pi/180;  #x=e^2/4*pi*epsi0\nb1=t*Z*x/tan(theta/2)/(2*K)         #impact parameter b\nf1=n*pi*b1**2*t                      #scattering angle greater than 90\n\n#result\nprint\"The fraction of alpha particles scattered at angles greater than 90 degrees is %.1e\" %f1;\n\n#part b\ntheta=45.0*pi/180;\nb2=t*Z*x/tan(theta/2)/(2*K);\nf2=n*pi*b2**2*t;                  #scattering angle greater than 45\nfb=f2-f1                       #scattering angle between 45 to 90\n\n#result\nprint\"The fraction of particles with scattering angle from 45 to 90 is %.1e\" %fb;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The fraction of alpha particles scattered at angles greater than 90 degrees is 7.5e-05\nThe fraction of particles with scattering angle from 45 to 90 is 3.6e-04\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.3 Page 185"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of vriable\nfrom math import sin, cos, tan, sqrt, pi\nZ=79.0;x=1.44;K=8.0*10**6;z=2;    #where x=e^2/4*pi*epsi0;z=2 for alpha particles\n\n#calculation\nd=z*x*Z/K;                   #distance\n\n#result\nprint \"The distance of closest approasch in nm. is\",d*10**-9",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " The distance of closest approasch in nm. is 2.844e-14\n"
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.4 Page 188"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nsl=820.1;n0=3.0;  #given values\nn=4;w=sl*(n**2/(n**2-n0**2)); \n#result\nprint \"The 3 longest possible wavelengths are in nm respectively are\",w,;    \n\n#partb\nn=5.0;w=sl*(n**2/(n**2-n0**2)); \n\n#result\nprint \"(in nm)\",round(w,3),;\n\n#partc\nn=6.0;w=sl*(n**2/(n**2-n0**2));\n\n#result\nprint \" (in nm )\",round(w,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The 3 longest possible wavelengths are in nm respectively are 1874.51428571 (in nm) 1281.406  (in nm ) 1093.467\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.5 Page 189"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nsl=364.5;n=3.0;           #given variables and various constants are declared in the subsequent steps wherever necessary   \nw1=sl*(n**2/(n**2-4));     #longest wavelength of balmer \nc=3.0*10**8;\nf1=c/(w1*10**-9);         #corresponding freq.\nn0=1.0;n=2.0;                 \n\n#calculation\nw2=91.13*(n**2/(n**2-n0**2)); #first longest of lymann   \nf2=c/(w2*10**-9);          #correspoding freq\nn0=1.0;n=3.0\nw3=91.13*(n**2/(n**2-n0**2));      #second longest of lymann\nf3=3.0*10**8/(w3*10**-9)             #corresponding freq.\n\n#result\nprint \"The freq. corresponding to the longest wavelength of balmer is %.1e\" %f1,\" & First longest wavelength  of Lymann is %.1e\" %f2;\nprint\"The sum of which s equal to %.1e\" %(f1+f2);\nprint\"The freq. corresponding to 2nd longest wavelength was found out to be %.1e\" %f3,\"Hence Ritz combination principle is satisfied.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The freq. corresponding to the longest wavelength of balmer is 4.6e+14  & First longest wavelength  of Lymann is 2.5e+15\nThe sum of which s equal to 2.9e+15\nThe freq. corresponding to 2nd longest wavelength was found out to be 2.92622261239e+15 Hence Ritz combination principle is satisfied.\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.6 Page 192"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nRinfi=1.097*10**7;  #known value \nn1=3.0;n2=2.0;       #first 2 given states\n\n#calculation\nw=(n1**2*n2**2)/((n1**2-n2**2)*Rinfi);\n\n#result\nprint\"Wavelength of trnasition from n1=3 to n2=2 in nm is\",round(w*10**9,3);\n\n#partb\nn1=4.0;n2=2.0;      #second 2 given states  \nw=(n1**2*n2**2)/((n1**2-n2**2)*Rinfi);\n\n#result\nprint\"Wavelength of trnasition from n1=3 to n2=2 in nm is\",round(w*10**9,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Wavelength of trnasition from n1=3 to n2=2 in nm is 656.335\nWavelength of trnasition from n1=3 to n2=2 in nm is 486.174\n"
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 6.7 Page 194"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nn1=3.0;n2=2.0;Z=4.0;hc=1240.0;\ndelE=(-13.6)*(Z**2)*((1/(n1**2))-((1/n2**2)));\n\n#calculation\nw=(hc)/delE; #for transition 1\n\n#result\nprint \"The wavelngth of radiation for transition(2->3) in nm is\", round(w,3);\n\n#for transition 2\nn1=4.0;n2=2.0;   # n values for transition 2\ndelE=(-13.6)*(Z**2)*((1/n1**2)-(1/n2**2));\nw=(hc)/delE;\n\n#result\nprint \"The wavelngth of radiation emitted for transition(2->4) in nm is\", round(w,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The wavelngth of radiation for transition(2->3) in nm is 41.029\nThe wavelngth of radiation emitted for transition(2->4) in nm is 30.392\n"
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter7.ipynb b/Modern_Physics/Chapter7.ipynb
index 5f1ccec9..b8966938 100755
--- a/Modern_Physics/Chapter7.ipynb
+++ b/Modern_Physics/Chapter7.ipynb
@@ -1,7 +1,6 @@
 {
  "metadata": {
-  "name": "",
-  "signature": "sha256:75c249ea2e8c0a6e5f6f1e7aa12f02f35c3e7e62df28f6611e18b53d1b7e6dcd"
+  "name": "Chapter7"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -13,54 +12,34 @@
      "level": 1,
      "metadata": {},
      "source": [
-      "Chapter 7: Tunneling Phenomena"
+      "Chapter 7:The Hydrogen Atom in Wave Mechanics"
      ]
     },
     {
      "cell_type": "heading",
-     "level": 2,
+     "level": 1,
      "metadata": {},
      "source": [
-      "Example 7.1, page no. 235"
+      "Example 7.2 Page 213"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "\n",
-      "\n",
+      "#initiation of variable\n",
+      "from math import exp\n",
       "import math\n",
+      "from scipy import integrate\n",
+      "# calculating radial probability  P= (4/ao^3)*integral(r^2 * e^(-2r/ao)) between the limits 0 and ao for r\n",
       "\n",
-      "#Variable declaration\n",
-      "h = 1.973 * 10**3       #planck's constant (eV.A'/c)\n",
-      "me = 511 * 10**3        #mass of electron (eV/c^2)\n",
-      "U = 10.0\n",
-      "E = 7.0\n",
-      "L = 50.00               #thickness of layer (A')\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "a = math.sqrt(2*me*(U-E))/h\n",
-      "T=(1.0+(1.0/4.0)*(U**2/(E*(U-E)))*(math.sinh(a*L))**2)**-1\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "print \"The transmission coefficient for L=\",L,\"A' is\",round(T/10**-38,3),\"X 10^-38\"\n",
-      "\n",
-      "#(b)if the layer thickness is 1.00nm.\n",
-      "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "L = 10          #thickness of layer (A')\n",
+      "#calculation\n",
+      "def integrand(x):\n",
+      "    return ((x**2)*exp(-x))/2.0\n",
+      "Pr=integrate.quad(integrand,0,2,args=());#simplifying where as x=2*r/a0; hence the limits change between 0 to 2\n",
       "\n",
-      "#Calculation\n",
-      "\n",
-      "T=(1.0+(1.0/4.0)*(U**2/(E*(U-E)))*(math.sinh(a*L))**2)**-1\n",
-      "\n",
-      "#Result\n",
-      "\n",
-      "print \"The transmission coefficient for L=\",L,\"A' is\",round(T/10**-7,3),\"X 10^-7\""
+      "#result\n",
+      "print \"Hence the probability of finding the electron nearer to nucleus is\",round(Pr[0],3);\n"
      ],
      "language": "python",
      "metadata": {},
@@ -69,41 +48,44 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The transmission coefficient for L= 50.0 A' is 0.963 X 10^-38\n",
-        "The transmission coefficient for L= 10 A' is 0.657 X 10^-7\n"
+        "Hence the probability of finding the electron nearer to nucleus is 0.323\n"
        ]
       }
      ],
-     "prompt_number": 3
+     "prompt_number": 2
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 7.2, page no. 236"
+      "Example 7.3 Page 213"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "from math import exp\n",
+      "import math\n",
+      "from scipy import integrate\n",
+      "# employing the formula for probability distribution similarly as done in Exa-7.2 \n",
+      "#calculation\n",
+      "def integrand(x):\n",
+      "    return (1.0/8)*((4.0*x**2)-(4.0*x**3)+(x**4))*exp(-x)\n",
+      "Pr1= integrate.quad(integrand,0,1,args=())             #x=r/ao; similrly limits between 0 and 1.\n",
       "\n",
-      "#Variable declaration\n",
-      "\n",
-      "e = 1.6 * 10 ** -19     #charge of electron (C)\n",
-      "I = 1.00 * 10 ** -3     #electron current(A)\n",
-      "T = 0.657 *10**-7       #Transmission coefficient\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "Ne = I / e\n",
-      "Nadj = Ne * T\n",
-      "Iadj = Nadj * e\n",
+      "#result\n",
+      "print\"The probability for l=0 electron is\",round(Pr1[0],5)\n",
       "\n",
-      "#Result\n",
+      "#part2\n",
+      "def integrand(x):\n",
+      "    return (1.0/24)*(x**4)*(exp(-x))\n",
+      "Pr2=integrate.quad(integrand,0,1);                            #x=r/ao; similarly limits between 0 and 1.\n",
       "\n",
-      "print \"The number of electrons per second continuing on the adjacent wire is\",round(Nadj/10**8,2),\"X 10^8 and the transmitted current is\",round(Iadj/10**-12,1),\"pA.\""
+      "#result\n",
+      "print\"The probability for l=1 electron is\",round(Pr2[0],5)\n"
      ],
      "language": "python",
      "metadata": {},
@@ -112,46 +94,38 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The number of electrons per second continuing on the adjacent wire is 4.11 X 10^8 and the transmitted current is 65.7 pA.\n"
+        "The probability for l=0 electron is 0.03432\n",
+        "The probability for l=1 electron is 0.00366\n"
        ]
       }
      ],
-     "prompt_number": 8
+     "prompt_number": 4
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 7.5, page no. 241"
+      "Example 7.4 Page 215"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      "\n",
-      "\n",
+      "#initiation of variable\n",
+      "from math import exp, sqrt\n",
       "import math\n",
+      "from scipy import integrate\n",
+      "l=1.0;         #given value of l\n",
       "\n",
-      "#Variable Declaration\n",
-      "\n",
-      "e = 1.6 * 10 **-19      #charge of electron(C)\n",
-      "f = 1.0*10**30          #collision frequency (s^-1.cm^-2)\n",
-      "Ec = 5.5 * 10 ** 10     \n",
-      "V = 10 * 10 ** 3        #potential difference(V)\n",
-      "d = 0.010 * 10**-3      #plate separation(m)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "E = V /d\n",
-      "Te = math.exp(-Ec/E)\n",
-      "rate = f * Te\n",
-      "I = e * rate\n",
+      "#calculation\n",
+      "am1=sqrt(l*(l+1));      #angular momentum==sqrt(l(l+1)) h\n",
+      "l=2.0                     #given l\n",
+      "am2=sqrt(l*(l+1));\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"The tunneling current is\",round(I/10**-12,2),\"pA.\""
+      "print\"The angular momenta are found out to be\", round(am1,3),\" h and\",round(am2,3),\" h respectively for l=1 and l=2.\";\n"
      ],
      "language": "python",
      "metadata": {},
@@ -160,55 +134,102 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The tunneling current is 0.21 pA.\n"
+        "The angular momenta are found out to be 1.414  h and 2.449  h respectively for l=1 and l=2.\n"
        ]
       }
      ],
-     "prompt_number": 12
+     "prompt_number": 6
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 7.6, page no. 244"
+      "Example 7.5 Page 216"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
-      " \n",
-      "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "Zth = 90            #atomic number of thorium\n",
-      "Zdth = 88           #atomic number of thorium's daughter nucleus\n",
-      "E = 4.05            #energy of ejected alphas(MeV)\n",
-      "Zpo = 84            #atomic number of polonium\n",
-      "Zdpo = 82           #atomic number of polonium's daughter nucleus\n",
-      "Epo = 8.95            #energy of ejected alphas(MeV)\n",
-      "R = 9.00            #nucleus size(fm)\n",
-      "r0 = 7.25           #Bohr radius of alpha(fm)\n",
-      "E0 = 0.0993         #(MeV)\n",
-      "f = 10 ** 21        #collision frequency(Hz)\n",
-      "\n",
-      "\n",
-      "#Calculation\n",
+      "#initiation of variable\n",
+      "from math import sqrt\n",
+      "print \"The possible values for m are [+2,-2] and hence any of the 5 components [-2h,2h] are possible for the L vector.\";\n",
+      "print \"Length of the vector as found out previously is %.2f*h.\",round(sqrt(6),4);#angular momentum==sqrt(l(l+1)) h"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The possible values for m are [+2,-2] and hence any of the 5 components [-2h,2h] are possible for the L vector.\n",
+        "Length of the vector as found out previously is %.2f*h. 2.4495\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.6 Page 223"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "uz=9.27*10**-24; t=1.4*10**3; x=3.5*10**-2;    #various constants and given values\n",
+      "m=1.8*10**-25;v=750;                        # mass and velocity of the particle\n",
       "\n",
-      "Te = math.exp(-4*math.pi*Zdth*math.sqrt((E0/E))+ 8 * math.sqrt(Zdth*R/r0))\n",
-      "rate = f * Te\n",
-      "t = math.log(2)/rate\n",
-      "Tep = math.exp(-4*math.pi*Zdpo*math.sqrt((E0/Epo))+ 8 * math.sqrt(Zdpo*R/r0))\n",
-      "ratep = f * Tep\n",
-      "tp = math.log(2)/ratep\n",
+      "#calculation\n",
+      "d=(uz*t*(x**2))/(m*(v**2));                  #net separtion \n",
       "\n",
+      "#result\n",
+      "print\"The distance of separation in mm is\",round(d*10**3,3);"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The distance of separation in mm is 0.157\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.7 Page 227"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "n1=1.0;n2=2.0;hc=1240.0;    #hc=1240 eV.nm\n",
+      "E=(-13.6)*((1/n2**2)-(1/n1**2));   #Energy calculation\n",
       "\n",
-      "#Result\n",
+      "#calculation\n",
+      "w=hc/E;                   #wavelength\n",
+      "u=9.27*10**-24; B=2;          #constants\n",
+      "delE= u*B/(1.6*10**-19);     #change in energy\n",
+      "delw=((w**2/hc))*delE;             #change in wavelength\n",
       "\n",
-      "print \"The half life of thorium is\",round(t/10**17,1),\"X 10^17 s and that of polonium is\",round(tp/10**-10,1),\"X 10^-10 s.\""
+      "#result\n",
+      "print\"The change in wavelength in nm. is\",round(delw,4);"
      ],
      "language": "python",
      "metadata": {},
@@ -217,11 +238,11 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The half life of thorium is 5.4 X 10^17 s and that of polonium is 8.4 X 10^-10 s.\n"
+        "The change in wavelength in nm. is 0.0014\n"
        ]
       }
      ],
-     "prompt_number": 14
+     "prompt_number": 12
     }
    ],
    "metadata": {}
diff --git a/Modern_Physics/Chapter7_1.ipynb b/Modern_Physics/Chapter7_1.ipynb
new file mode 100755
index 00000000..bd2ea129
--- /dev/null
+++ b/Modern_Physics/Chapter7_1.ipynb
@@ -0,0 +1,146 @@
+{
+ "metadata": {
+  "name": "MP-7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "The Hydrogen Atom in Wave Mechanics"
+    },
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Example 7.2 Page 213"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import exp\nimport math\nfrom scipy import integrate\n# calculating radial probability  P= (4/ao^3)*inegral(r^2 * e^(-2r/ao)) between the limits 0 and ao for r\n\n#calculation\ndef integrand(x):\n    return ((x**2)*exp(-x))/2.0\nPr=integrate.quad(integrand,0,2,args=());#simplifying where as x=2*r/a0; hence the limits change between 0 to 2\n\n#result\nprint \"Hence the probability of finding the electron nearer to nucleus is\",round(Pr[0],3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Hence the probability of finding the electron nearer to nucleus is 0.323\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 7.3 Page 213"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import exp\nimport math\nfrom scipy import integrate\n# employing the formula for probability distribution similarly as done in Exa-7.2 \n#calculation\ndef integrand(x):\n    return (1.0/8)*((4.0*x**2)-(4.0*x**3)+(x**4))*exp(-x)\nPr1= integrate.quad(integrand,0,1,args=())             #x=r/ao; similrly limits between 0 and 1.\n\n#result\nprint\"The probability for l=0 electron is\",round(Pr1[0],5)\n\n#part2\ndef integrand(x):\n    return (1.0/24)*(x**4)*(exp(-x))\nPr2=integrate.quad(integrand,0,1);                            #x=r/ao; similrly limits between 0 and 1.\n\n#result\nprint\"The probability for l=1 electron is\",round(Pr2[0],5)\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The probability for l=0 electron is 0.03432\nThe probability for l=1 electron is 0.00366\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 7.4 Page 215"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import exp, sqrt\nimport math\nfrom scipy import integrate\nl=1.0;         #given value of l\n\n#calcualtion\nam1=sqrt(l*(l+1));      #angular momentum==sqrt(l(l+1)) h\nl=2.0                     #given l\nam2=sqrt(l*(l+1));\n\n#result\nprint\"The angular momenta are found out to be\", round(am1,3),\" h and\",round(am2,3),\" h respectively for l=1 and l=2.\";\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The angular momenta are found out to be 1.414  h and 2.449  h respectively for l=1 and l=2.\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 7.5 Page 216"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nprint \"The possible values for m are [+2,-2] and hence any of the 5 components [-2h,2h] are possible for the L vector.\";\nprint\"Length of the vector as found out previously is %.2f*h.\",round(sqrt(6),4);#angular momentum==sqrt(l(l+1)) h",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The possible values for m are [+2,-2] and hence any of the 5 components [-2h,2h] are possible for the L vector.\nLength of the vector as found out previously is %.2f*h. 2.4495\n"
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 7.6 Page 223"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nuz=9.27*10**-24; t=1.4*10**3; x=3.5*10**-2;    #various constants and given values\nm=1.8*10**-25;v=750;                        # mass and velocity of the particle\n\n#calculation\nd=(uz*t*(x**2))/(m*(v**2));                  #net separtion \n\n#result\nprint\"The distance of separation in mm is\",round(d*10**3,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The distance of separation in mm is 0.157\n"
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 7.7 Page 227"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nn1=1.0;n2=2.0;hc=1240.0;    #hc=1240 eV.nm\nE=(-13.6)*((1/n2**2)-(1/n1**2));   #Energy calcuation\n\n#calculation\nw=hc/E;                   #wavelength\nu=9.27*10**-24; B=2;          #constants\ndelE= u*B/(1.6*10**-19);     #change in energy\ndelw=((w**2/hc))*delE;             #change in wavelength\n\n#result\nprint\"The change in wavelength in nm. is\",round(delw,4);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The change in wavelength in nm. is 0.0014\n"
+      }
+     ],
+     "prompt_number": 12
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter7_2.ipynb b/Modern_Physics/Chapter7_2.ipynb
new file mode 100755
index 00000000..bd2ea129
--- /dev/null
+++ b/Modern_Physics/Chapter7_2.ipynb
@@ -0,0 +1,146 @@
+{
+ "metadata": {
+  "name": "MP-7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "The Hydrogen Atom in Wave Mechanics"
+    },
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Example 7.2 Page 213"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import exp\nimport math\nfrom scipy import integrate\n# calculating radial probability  P= (4/ao^3)*inegral(r^2 * e^(-2r/ao)) between the limits 0 and ao for r\n\n#calculation\ndef integrand(x):\n    return ((x**2)*exp(-x))/2.0\nPr=integrate.quad(integrand,0,2,args=());#simplifying where as x=2*r/a0; hence the limits change between 0 to 2\n\n#result\nprint \"Hence the probability of finding the electron nearer to nucleus is\",round(Pr[0],3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Hence the probability of finding the electron nearer to nucleus is 0.323\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 7.3 Page 213"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import exp\nimport math\nfrom scipy import integrate\n# employing the formula for probability distribution similarly as done in Exa-7.2 \n#calculation\ndef integrand(x):\n    return (1.0/8)*((4.0*x**2)-(4.0*x**3)+(x**4))*exp(-x)\nPr1= integrate.quad(integrand,0,1,args=())             #x=r/ao; similrly limits between 0 and 1.\n\n#result\nprint\"The probability for l=0 electron is\",round(Pr1[0],5)\n\n#part2\ndef integrand(x):\n    return (1.0/24)*(x**4)*(exp(-x))\nPr2=integrate.quad(integrand,0,1);                            #x=r/ao; similrly limits between 0 and 1.\n\n#result\nprint\"The probability for l=1 electron is\",round(Pr2[0],5)\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The probability for l=0 electron is 0.03432\nThe probability for l=1 electron is 0.00366\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 7.4 Page 215"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import exp, sqrt\nimport math\nfrom scipy import integrate\nl=1.0;         #given value of l\n\n#calcualtion\nam1=sqrt(l*(l+1));      #angular momentum==sqrt(l(l+1)) h\nl=2.0                     #given l\nam2=sqrt(l*(l+1));\n\n#result\nprint\"The angular momenta are found out to be\", round(am1,3),\" h and\",round(am2,3),\" h respectively for l=1 and l=2.\";\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The angular momenta are found out to be 1.414  h and 2.449  h respectively for l=1 and l=2.\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 7.5 Page 216"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import sqrt\nprint \"The possible values for m are [+2,-2] and hence any of the 5 components [-2h,2h] are possible for the L vector.\";\nprint\"Length of the vector as found out previously is %.2f*h.\",round(sqrt(6),4);#angular momentum==sqrt(l(l+1)) h",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The possible values for m are [+2,-2] and hence any of the 5 components [-2h,2h] are possible for the L vector.\nLength of the vector as found out previously is %.2f*h. 2.4495\n"
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 7.6 Page 223"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nuz=9.27*10**-24; t=1.4*10**3; x=3.5*10**-2;    #various constants and given values\nm=1.8*10**-25;v=750;                        # mass and velocity of the particle\n\n#calculation\nd=(uz*t*(x**2))/(m*(v**2));                  #net separtion \n\n#result\nprint\"The distance of separation in mm is\",round(d*10**3,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The distance of separation in mm is 0.157\n"
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 7.7 Page 227"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nn1=1.0;n2=2.0;hc=1240.0;    #hc=1240 eV.nm\nE=(-13.6)*((1/n2**2)-(1/n1**2));   #Energy calcuation\n\n#calculation\nw=hc/E;                   #wavelength\nu=9.27*10**-24; B=2;          #constants\ndelE= u*B/(1.6*10**-19);     #change in energy\ndelw=((w**2/hc))*delE;             #change in wavelength\n\n#result\nprint\"The change in wavelength in nm. is\",round(delw,4);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The change in wavelength in nm. is 0.0014\n"
+      }
+     ],
+     "prompt_number": 12
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter8.ipynb b/Modern_Physics/Chapter8.ipynb
index a12d49fc..ccbad1fd 100755
--- a/Modern_Physics/Chapter8.ipynb
+++ b/Modern_Physics/Chapter8.ipynb
@@ -1,7 +1,6 @@
 {
  "metadata": {
-  "name": "",
-  "signature": "sha256:8315bc6aae714998ec5db8a7cf2faf25a033ca960fbcf8e976cdfc44e47b09f3"
+  "name": "Chapter8"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -13,7 +12,7 @@
      "level": 1,
      "metadata": {},
      "source": [
-      "Chapter 8: Quantum Mechanics in Three Dimensions"
+      "Chapter 8:Many Electron Atoms"
      ]
     },
     {
@@ -21,33 +20,21 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 8.4, page no. 270"
+      "Example 8.1 Page 248"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "hc=1240.0*10**-9;Rinfi=1.097*10**7;Z=11;  #for sodium atom;and other constants in MeV\n",
       "\n",
+      "#calculation\n",
+      "delE=3*hc*Rinfi*(Z-1)**2/4.0              #change in energy\n",
       "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "R = 1.0             #radius(m)\n",
-      "T = 1.0             #period of revolution(s)\n",
-      "m = 1.0             #mass of stone(kg)\n",
-      "h = 1.055 * 10**-34 #planks constant (kg.m^2/s)\n",
-      "\n",
-      "#Calculation\n",
-      "\n",
-      "v = 2*math.pi*R/T\n",
-      "L = m * v * R\n",
-      "l = L / h\n",
-      "\n",
-      "#Results\n",
-      "\n",
-      "print \"The orbital quantum number l is\",round(l/10**34,2),\"X 10^34.\""
+      "#result\n",
+      "print\"The energy of the Ka x-ray of the sodium atom in KeV.is\",round(delE/10**3,3);\n"
      ],
      "language": "python",
      "metadata": {},
@@ -56,45 +43,38 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The orbital quantum number l is 5.96 X 10^34.\n"
+        "The energy of the Ka x-ray of the sodium atom in KeV.is 1.02\n"
        ]
       }
      ],
-     "prompt_number": 2
+     "prompt_number": 1
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 8.6, page no. 272"
+      "Example 8.2 Page 249"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "EKa=21.990;EKb=25.145;EK=25.514                                      #all the values are in KeV\n",
       "\n",
+      "#calcualtion\n",
+      "ELo=EKb-EKa;\n",
       "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "l = 3           #orbital quantum number\n",
+      "#result\n",
+      "print\"The energy of La of X-ray in KeV.is\",round(ELo,3);             #Energy of La X-ray\n",
       "\n",
-      "#Calculation and results\n",
+      "#partb\n",
+      "EL=-EK+EKa;\n",
       "\n",
-      "L = math.sqrt(l*(l+1))\n",
-      "print \"The allowed values of Lz are\"\n",
-      "for i in range(-l,l+1):\n",
-      "    print i,\"h\"\n",
-      "print \"The allowed values of theta are\"\n",
-      "for i in range(-l,l+1):\n",
-      "    theta = round(math.acos(i/L)*180/math.pi,1)\n",
-      "    if theta > 90:\n",
-      "        print theta-180\n",
-      "    else:\n",
-      "        print theta"
+      "#result\n",
+      "print\"Hence the binding energy of the L electon in KeV.is\",round(EL,3); # for electron L electron"
      ],
      "language": "python",
      "metadata": {},
@@ -103,55 +83,35 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The allowed values of Lz are\n",
-        "-3 h\n",
-        "-2 h\n",
-        "-1 h\n",
-        "0 h\n",
-        "1 h\n",
-        "2 h\n",
-        "3 h\n",
-        "The allowed values of theta are\n",
-        "-30.0\n",
-        "-54.7\n",
-        "-73.2\n",
-        "90.0\n",
-        "73.2\n",
-        "54.7\n",
-        "30.0\n"
+        "The enrgy of La of X-ray in KeV.is 3.155\n",
+        "Hence the binding energy of the L electon in KeV.is -3.524\n"
        ]
       }
      ],
-     "prompt_number": 5
+     "prompt_number": 2
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 8.7, page no. 281"
+      "Example 8.3 Page 250"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "l=1.0; \n",
       "\n",
+      "#calculation\n",
+      "Lmax=l+l;Lmin=l-l;\n",
+      "s=1.0/2; Smax=s+s;Smin=s-s;\n",
       "\n",
-      "#Variable declaration\n",
-      "\n",
-      "n = 2\n",
-      "Z = 1\n",
-      "\n",
-      "#Calculation and results\n",
-      " \n",
-      "print \"The states of hydrogen atom are\"\n",
-      "for i in range(n):\n",
-      "    for j in range(-i,i+1):\n",
-      "        print \"n = \",n,\"l = \",i,\"ml = \",j\n",
-      "\n",
-      "E2 = -13.6 * Z**2/n**2\n",
-      "print \"All states have same energy of \",E2,\"eV.\""
+      "#result\n",
+      "print\"Value of L ranges from\",Lmin,\" to\",Lmax,\" i.e\",Lmin,1,Lmax;\n",
+      "print\"Values of S are maxi,min\",Smax,Smin;\n"
      ],
      "language": "python",
      "metadata": {},
@@ -160,47 +120,37 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The states of hydrogen atom are\n",
-        "n =  2 l =  0 ml =  0\n",
-        "n =  2 l =  1 ml =  -1\n",
-        "n =  2 l =  1 ml =  0\n",
-        "n =  2 l =  1 ml =  1\n",
-        "All states have same energy of  -3.4 eV.\n"
+        "Value of L ranges from 0.0  to 2.0  i.e 0.0 1 2.0\n",
+        "Values of S are maxi,min 1.0 0.0\n"
        ]
       }
      ],
-     "prompt_number": 9
+     "prompt_number": 5
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 8.8, page no. 284"
+      "Example 8.4 Page 250"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "l=1.0; \n",
       "\n",
-      "import scipy\n",
-      "from scipy import integrate\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "\n",
-      "z2 = lambda z: z**2 * math.e ** -z\n",
-      "\n",
-      "#Calculation\n",
-      "inf = float('inf')\n",
-      "integ,err = scipy.integrate.quad(z2,2.0,inf)\n",
-      "P = integ * 0.5\n",
+      "#calculation\n",
+      "Lmax=l+l;Lmin=l-l;\n",
+      "s=1.0/2; Smax=s+s;Smin=s-s;\n",
       "\n",
       "#result\n",
-      "\n",
-      "print \"The probability is\",round(P,3)\n",
-      "\n"
+      "print\"Considering any two electrons,Value of L2e ranges from\",Lmin,\" to\",Lmax,\" i.e\",Lmin,1,Lmax;\n",
+      "print\"Adding the angular momentum of the third electron to L2emax gives the maximum  whole angular momentum as 2+1=3; and subtracting it from L2e=1 gives 0\"\n",
+      "print\"Values of S are maxi,min\",Smax,Smin;\n",
+      "print\"Adding and subtracting the spin of third to S2e=1 and S2e=0 respectively gives the spins 3/2 and 1/2 for the 3 electron system.\";"
      ],
      "language": "python",
      "metadata": {},
@@ -209,11 +159,14 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The probability is 0.677\n"
+        "Considering any two electrons,Value of L2e ranges from 0.0  to 2.0  i.e 0.0 1 2.0\n",
+        "Adding the angular momentum of the third electron to L2emax gives the maximum  whole angular momentum as 2+1=3; and subtracting it from L2e=1 gives 0\n",
+        "Values of S are maxi,min 1.0 0.0\n",
+        "Adding and subtracting the spin of third to S2e=1 and S2e=0 respectively gives the spins 3/2 and 1/2 for the 3 electron system.\n"
        ]
       }
      ],
-     "prompt_number": 7
+     "prompt_number": 6
     }
    ],
    "metadata": {}
diff --git a/Modern_Physics/Chapter8_1.ipynb b/Modern_Physics/Chapter8_1.ipynb
new file mode 100755
index 00000000..1c6a6bdb
--- /dev/null
+++ b/Modern_Physics/Chapter8_1.ipynb
@@ -0,0 +1,104 @@
+{
+ "metadata": {
+  "name": "MP-8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Many Electron Atoms"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8.1 Page 248"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nhc=1240.0*10**-9;Rinfi=1.097*10**7;Z=11;  #for sodium atom;and other constants in MeV\n\n#calculation\ndelE=3*hc*Rinfi*(Z-1)**2/4.0              #change in energy\n\n#result\nprint\"The energy of the Ka x-ray of the sodium atom in KeV.is\",round(delE/10**3,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The energy of the Ka x-ray of the sodium atom in KeV.is 1.02\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8.2 Page 249"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nEKa=21.990;EKb=25.145;EK=25.514                                      #all the values are in KeV\n\n#calcualtion\nELo=EKb-EKa;\n\n#result\nprint\"The enrgy of La of X-ray in KeV.is\",round(ELo,3);             #Energy of La X-ray\n\n#partb\nEL=-EK+EKa;\n\n#result\nprint\"Hence the binding energy of the L electon in KeV.is\",round(EL,3); # for electron L electron",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The enrgy of La of X-ray in KeV.is 3.155\nHence the binding energy of the L electon in KeV.is -3.524\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8.3 Page 250"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nl=1.0; \n\n#calculation\nLmax=l+l;Lmin=l-l;\ns=1.0/2; Smax=s+s;Smin=s-s;\n\n#result\nprint\"Value of L ranges from\",Lmin,\" to\",Lmax,\" i.e\",Lmin,1,Lmax;\nprint\"Values of S are maxi,min\",Smax,Smin;\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Value of L ranges from 0.0  to 2.0  i.e 0.0 1 2.0\nValues of S are maxi,min 1.0 0.0\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8.4 Page 250"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nl=1.0; \n\n#calculation\nLmax=l+l;Lmin=l-l;\ns=1.0/2; Smax=s+s;Smin=s-s;\n\n#result\nprint\"Considering any two electrons,Value of L2e ranges from\",Lmin,\" to\",Lmax,\" i.e\",Lmin,1,Lmax;\nprint\"Adding the angular momentum of the third electron to L2emax gives the maximum  whole angular momentum as 2+1=3; and subtracting it from L2e=1 gives 0\"\nprint\"Values of S are maxi,min\",Smax,Smin;\nprint\"Adding and subtracting the spin of third to S2e=1 and S2e=0 respectively gives the spins 3/2 and 1/2 for the 3 electron system.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Considering any two electrons,Value of L2e ranges from 0.0  to 2.0  i.e 0.0 1 2.0\nAdding the angular momentum of the third electron to L2emax gives the maximum  whole angular momentum as 2+1=3; and subtracting it from L2e=1 gives 0\nValues of S are maxi,min 1.0 0.0\nAdding and subtracting the spin of third to S2e=1 and S2e=0 respectively gives the spins 3/2 and 1/2 for the 3 electron system.\n"
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter8_2.ipynb b/Modern_Physics/Chapter8_2.ipynb
new file mode 100755
index 00000000..1c6a6bdb
--- /dev/null
+++ b/Modern_Physics/Chapter8_2.ipynb
@@ -0,0 +1,104 @@
+{
+ "metadata": {
+  "name": "MP-8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Many Electron Atoms"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8.1 Page 248"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nhc=1240.0*10**-9;Rinfi=1.097*10**7;Z=11;  #for sodium atom;and other constants in MeV\n\n#calculation\ndelE=3*hc*Rinfi*(Z-1)**2/4.0              #change in energy\n\n#result\nprint\"The energy of the Ka x-ray of the sodium atom in KeV.is\",round(delE/10**3,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The energy of the Ka x-ray of the sodium atom in KeV.is 1.02\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8.2 Page 249"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nEKa=21.990;EKb=25.145;EK=25.514                                      #all the values are in KeV\n\n#calcualtion\nELo=EKb-EKa;\n\n#result\nprint\"The enrgy of La of X-ray in KeV.is\",round(ELo,3);             #Energy of La X-ray\n\n#partb\nEL=-EK+EKa;\n\n#result\nprint\"Hence the binding energy of the L electon in KeV.is\",round(EL,3); # for electron L electron",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The enrgy of La of X-ray in KeV.is 3.155\nHence the binding energy of the L electon in KeV.is -3.524\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8.3 Page 250"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nl=1.0; \n\n#calculation\nLmax=l+l;Lmin=l-l;\ns=1.0/2; Smax=s+s;Smin=s-s;\n\n#result\nprint\"Value of L ranges from\",Lmin,\" to\",Lmax,\" i.e\",Lmin,1,Lmax;\nprint\"Values of S are maxi,min\",Smax,Smin;\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Value of L ranges from 0.0  to 2.0  i.e 0.0 1 2.0\nValues of S are maxi,min 1.0 0.0\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 8.4 Page 250"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nl=1.0; \n\n#calculation\nLmax=l+l;Lmin=l-l;\ns=1.0/2; Smax=s+s;Smin=s-s;\n\n#result\nprint\"Considering any two electrons,Value of L2e ranges from\",Lmin,\" to\",Lmax,\" i.e\",Lmin,1,Lmax;\nprint\"Adding the angular momentum of the third electron to L2emax gives the maximum  whole angular momentum as 2+1=3; and subtracting it from L2e=1 gives 0\"\nprint\"Values of S are maxi,min\",Smax,Smin;\nprint\"Adding and subtracting the spin of third to S2e=1 and S2e=0 respectively gives the spins 3/2 and 1/2 for the 3 electron system.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Considering any two electrons,Value of L2e ranges from 0.0  to 2.0  i.e 0.0 1 2.0\nAdding the angular momentum of the third electron to L2emax gives the maximum  whole angular momentum as 2+1=3; and subtracting it from L2e=1 gives 0\nValues of S are maxi,min 1.0 0.0\nAdding and subtracting the spin of third to S2e=1 and S2e=0 respectively gives the spins 3/2 and 1/2 for the 3 electron system.\n"
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter9.ipynb b/Modern_Physics/Chapter9.ipynb
index 52c8dc08..34acce85 100755
--- a/Modern_Physics/Chapter9.ipynb
+++ b/Modern_Physics/Chapter9.ipynb
@@ -1,7 +1,6 @@
 {
  "metadata": {
-  "name": "",
-  "signature": "sha256:a22a4cd44988289c63c9da64c71766c244fd20324a2c5a80b53d48938212c4cc"
+  "name": "Chapter9"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -13,7 +12,7 @@
      "level": 1,
      "metadata": {},
      "source": [
-      "Chapter 9: Atomic Structure"
+      "Chapter 9:Molecular Structure"
      ]
     },
     {
@@ -21,30 +20,71 @@
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 9.1, page no. 300"
+      "Example 9.1 Page 270"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "E=-2.7;\n",
+      "K=9.0*(10**9)*((1.6*(10**-19))**2)/(0.106*10**-9);# taking all the values in meters. 1/(4*pi*e0)= 9*10^9 F/m\n",
       "\n",
+      "#calculation\n",
+      "q=((K-E*10**-9)/(4*K))*10**-9;                #balancing by multiplying 10^-9 on numerator. to eV.vm terms\n",
       "\n",
-      "#Variable declaration\n",
+      "#result\n",
+      "print\"Charge on the sphere required is\",round(q,4),\" times the charge of electron.\";"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Charge on the sphere required is 0.3105  times the charge of electron.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2 Page 273"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "K=1.44; Req=0.236; # K=e^2/(4*pi*e0)=1.44 eV.nm\n",
+      "\n",
+      "#calculation\n",
+      "Uc=-K/(Req);        #coulomb energy\n",
       "\n",
-      "Ub = 9.27 * 10 ** -24       #(J/T)\n",
-      "B = 1.00                    #magnetic field strength (T)\n",
-      "h = 6.58 * 10 ** -16        #Planck's constant (eV.s)\n",
-      "e = 1.6 * 10 ** -19         #electron charge (C)\n",
+      "#result\n",
+      "print\"The coulomb energy at an equilibrium separation distance in eV is\",round(Uc,3);\n",
       "\n",
-      "#Calculation\n",
+      "E=-4.26; delE=1.53;    #various standards values of NaCl\n",
+      "Ur=E-Uc-delE; \n",
       "\n",
-      "hwl = Ub * B / e\n",
-      "wl = hwl / h\n",
+      "#result\n",
+      "print\"The pauli''s repulsion energy in eV is\",round(Ur,3);\n",
       "\n",
-      "#results\n",
+      "#partb\n",
+      "Req=0.1;         #pauli repulsion energy\n",
+      "Uc=-K/(Req);\n",
+      "E=4; delE=1.53;\n",
+      "Ur=E-Uc-delE;\n",
       "\n",
-      "print \"The magnetic energy is\",round(hwl/10**-5,2),\"x 10^-5 eV and the Larmor frequency is\",round(wl/10**10,2),\"x 10^10 rad/s.\"\n"
+      "#result\n",
+      "print\"The pauli''s repulsion energy in eV is\",round(Ur,3);\n"
      ],
      "language": "python",
      "metadata": {},
@@ -53,38 +93,121 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "The magnetic energy is 5.79 x 10^-5 eV and the Larmor frequency is 8.81 x 10^10 rad/s.\n"
+        "The coulomb energy at an equilirium separation distance in eV is -6.102\n",
+        "The pauli''s repulsion energy in eV is 0.312\n",
+        "The pauli''s repulsion energy in eV is 16.87\n"
        ]
       }
      ],
-     "prompt_number": 2
+     "prompt_number": 4
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 9.4, page no. 311"
+      "Example 9.3 Page 276"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "from math import pi, sqrt\n",
+      "delE=0.50; delR=0.017*10**-9;      #delE= E-Emin; delR=R-Rmin;\n",
+      "k=2*(delE)/(delR**2);c=3*10**8;     #force constant\n",
+      "m=(1.008)*(931.5*10**6)*0.5;       #mass of molecular hydrogen\n",
+      "v= sqrt(k*c**2/m)/(2*pi);          #vibrational frequency\n",
+      "h=4.14*(10**-15);\n",
       "\n",
-      "#Variable declaration\n",
+      "#calculation\n",
+      "E=h*v;\n",
       "\n",
-      "l2 = 589.592            #sodium doublet wavelength (nm)\n",
-      "l1 = 588.995            #sodium doublet wavelength (nm)\n",
-      "hc = 1240               #planck's constant X speed of light (eV.nm)\n",
-      "\n",
-      "#Calculation\n",
+      "#result\n",
+      "print\"The value of corresponding photon energy in eV is\",round(E,3);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of corresponding photon energy in eV is 0.537\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4 Page 280"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "from math import pi, sqrt\n",
+      "hc=1240.0;        #in eV.nm\n",
+      "m=0.5*1.008*931.5*10**6;              #mass of hydrogen atom\n",
+      "Req=0.074;                          #equivalent radius\n",
       "\n",
-      "dE = hc * (l2-l1)/(l2*l1)\n",
+      "#calculation\n",
+      "a=((hc)**2)/(4*(pi**2)*m*(Req**2));  #reduced mass of hydrogen atom\n",
+      "for L in range(1,4):\n",
+      "        delE= L*a; \n",
+      "        print\"The value of energy in eV is\",round(delE,4);                  \n",
+      "        w=(hc)/delE;\n",
+      "        print\"The respective wavelength in um is\",round(w*10**-3,3);  \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of energy in eV is 0.0151\n",
+        "The respective wavelength in um is 81.849\n",
+        "The value of energy in eV is 0.0303\n",
+        "The respective wavelength in um is 40.925\n",
+        "The value of energy in eV is 0.0454\n",
+        "The respective wavelength in um is 27.283\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5 Page 283"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initiation of variable\n",
+      "from math import pi\n",
+      "delv=6.2*(10**11);     #change in frequency\n",
+      "h=1.05*(10**-34);        #value of h in J.sec\n",
       "\n",
-      "#Results\n",
+      "#calculation\n",
+      "I= h/(2*pi*delv);      #rotational inertia\n",
+      "I1=I/(1.684604*10**-45); #to change units\n",
       "\n",
-      "print \"dE =\",round(dE/10**-3,2),\"X 10^-3 eV represents the spin orbit splitting of the initial levels.\""
+      "#result\n",
+      "print\"The value of rotational inertia in kg m2 is %.1e\" %I;\n",
+      "print\"which in terms of amu in u.nm2 is\",round(I1,3);"
      ],
      "language": "python",
      "metadata": {},
@@ -93,40 +216,48 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "dE = 2.13 X 10^-3 eV represents the spin orbit splitting of the initial levels.\n"
+        "The value of rotational inertia in kg m2 is 2.69536597172e-47\n",
+        "which in terms of amu in u.nm2 is 0.016\n"
        ]
       }
      ],
-     "prompt_number": 6
+     "prompt_number": 16
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
      "source": [
-      "Example 9.6, page no. 317"
+      "Example 9.6 Page 286"
      ]
     },
     {
      "cell_type": "code",
      "collapsed": false,
      "input": [
+      "#initiation of variable\n",
+      "from math import pi\n",
+      "delE=0.358;hc=4.14*10**-15;          #hc in eV.nm and delE=1.44eV(given values)\n",
       "\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
+      "#calculation\n",
+      "f=(delE)/hc;                        #frequency \n",
       "\n",
-      "E = -5.14           #Energy of the 3s electron in sodium (eV)\n",
-      "Eh = 13.6           #energy of the 3s electron in hydrogen (eV)\n",
+      "#result\n",
+      "print\"The frequency of the radiation is \",f;\n",
       "\n",
       "\n",
-      "#Calculation\n",
+      "m=0.98;                            #mass in terms of u\n",
+      "k=4*pi**2*m*f**2;                   #value of k in eV/m^2\n",
       "\n",
-      "Zeff = 3 * math.sqrt(-E/Eh)\n",
+      "#result\n",
+      "print\"The force constant is\",k; \n",
       "\n",
-      "#Results\n",
+      "#partb\n",
+      "hc=1240.0; m=0.98*1.008*931.5*10**6; Req=0.127;      #various constants in terms of  \n",
+      "s=((hc)**2)/(4*(pi**2)*m*(Req**2));                # expected spacing \n",
       "\n",
-      "print \"Zeff for the 3s electron in sodium is\",round(Zeff,2)"
+      "#result\n",
+      "print\"The spacing was found out to be\",round(s,3),\"which is very close to the graphical value of 0.0026 eV.\""
      ],
      "language": "python",
      "metadata": {},
@@ -135,11 +266,13 @@
        "output_type": "stream",
        "stream": "stdout",
        "text": [
-        "Zeff for the 3s electron in sodium is 1.84\n"
+        "The frequency of the radiation is  8.64734299517e+13\n",
+        "The force constant is 2.89301831756e+29\n",
+        "The spacing was found out to be 0.003 which is very close to the graphical value of 0.0026 eV.\n"
        ]
       }
      ],
-     "prompt_number": 10
+     "prompt_number": 19
     }
    ],
    "metadata": {}
diff --git a/Modern_Physics/Chapter9_1.ipynb b/Modern_Physics/Chapter9_1.ipynb
new file mode 100755
index 00000000..1cfbf676
--- /dev/null
+++ b/Modern_Physics/Chapter9_1.ipynb
@@ -0,0 +1,146 @@
+{
+ "metadata": {
+  "name": "MP-9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Molecular Structure"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.1 Page 270"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nE=-2.7;\nK=9.0*(10**9)*((1.6*(10**-19))**2)/(0.106*10**-9);# taking all the values in meters. 1/(4*pi*e0)= 9*10^9 F/m\n\n#calculation\nq=((K-E*10**-9)/(4*K))*10**-9;                #balancin by multiplying 10^-9 on numerator. to eV.vm terms\n\n#result\nprint\"Charge on the sphere required is\",round(q,4),\" times the charge of electron.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Charge on the sphere required is 0.3105  times the charge of electron.\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.2 Page 273"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nK=1.44; Req=0.236; # K=e^2/(4*pi*e0)=1.44 eV.nm\n\n#calculation\nUc=-K/(Req);        #coulomb energy\n\n#result\nprint\"The coulomb energy at an equilirium separation distance in eV is\",round(Uc,3);\n\nE=-4.26; delE=1.53;    #various standars values of NaCl\nUr=E-Uc-delE; \n\n#result\nprint\"The pauli''s repulsion energy in eV is\",round(Ur,3);\n\n#partb\nReq=0.1;         #pauli repulsion energy\nUc=-K/(Req);\nE=4; delE=1.53;\nUr=E-Uc-delE;\n\n#result\nprint\"The pauli''s repulsion energy in eV is\",round(Ur,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The coulomb energy at an equilirium separation distance in eV is -6.102\nThe pauli''s repulsion energy in eV is 0.312\nThe pauli''s repulsion energy in eV is 16.87\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.3 Page 276"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi, sqrt\ndelE=0.50; delR=0.017*10**-9;      #delE= E-Emin; delR=R-Rmin;\nk=2*(delE)/(delR**2);c=3*10**8;     #force constant\nm=(1.008)*(931.5*10**6)*0.5;       #mass of molecular hydrogen\nv= sqrt(k*c**2/m)/(2*pi);          #vibrational frequency\nh=4.14*(10**-15);\n\n#calculation\nE=h*v;\n\n#result\nprint\"The value of corresponding photon energy in eV is\",round(E,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of corresponding photon energy in eV is 0.537\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.4 Page 280"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi, sqrt\nhc=1240.0;        #in eV.nm\nm=0.5*1.008*931.5*10**6;              #mass of hydrogen atom\nReq=0.074;                          #equivalent radius\n\n#calculation\na=((hc)**2)/(4*(pi**2)*m*(Req**2));  #reduced mass of hydrogen atom\nfor L in range(1,4):\n        delE= L*a; \n        print\"The value of energy in eV is\",round(delE,4);                  \n        w=(hc)/delE;\n        print\"The respective wavelength in um is\",round(w*10**-3,3);  \n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of energy in eV is 0.0151\nThe respective wavelength in um is 81.849\nThe value of energy in eV is 0.0303\nThe respective wavelength in um is 40.925\nThe value of energy in eV is 0.0454\nThe respective wavelength in um is 27.283\n"
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.5 Page 283"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi\ndelv=6.2*(10**11);     #change in frequency\nh=1.05*(10**-34);        #value of h in J.sec\n\n#calculation\nI= h/(2*pi*delv);      #rotational inertia\nI1=I/(1.684604*10**-45); #to change units\n\n#result\nprint\"The value of rotational inertia in kg m2 is %.1e\" %I;\nprint\"which in terms of amu in u.nm2 is\",round(I1,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of rotational inertia in kg m2 is 2.69536597172e-47\nwhich in terms of amu in u.nm2 is 0.016\n"
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.6 Page 286"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi\ndelE=0.358;hc=4.14*10**-15;          #hc in eV.nm and delE=1.44eV(given values)\n\n#calculation\nf=(delE)/hc;                        #frequency \n\n#result\nprint\"The frequency of the radiation is \",f;\n\n\nm=0.98;                            #mass in terms of u\nk=4*pi**2*m*f**2;                   #value of k in eV/m^2\n\n#result\nprint\"The force constant is\",k; \n\n#partb\nhc=1240.0; m=0.98*1.008*931.5*10**6; Req=0.127;      #various constants in terms of  \ns=((hc)**2)/(4*(pi**2)*m*(Req**2));                # expeted spacing \n\n#result\nprint\"The spacing was found out to be\",round(s,3),\"which is very close to the graphical value of 0.0026 eV.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The frequency of the radiation is  8.64734299517e+13\nThe force constant is 2.89301831756e+29\nThe spacing was found out to be 0.003 which is very close to the graphical value of 0.0026 eV.\n"
+      }
+     ],
+     "prompt_number": 19
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/Chapter9_2.ipynb b/Modern_Physics/Chapter9_2.ipynb
new file mode 100755
index 00000000..1cfbf676
--- /dev/null
+++ b/Modern_Physics/Chapter9_2.ipynb
@@ -0,0 +1,146 @@
+{
+ "metadata": {
+  "name": "MP-9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Molecular Structure"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.1 Page 270"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nE=-2.7;\nK=9.0*(10**9)*((1.6*(10**-19))**2)/(0.106*10**-9);# taking all the values in meters. 1/(4*pi*e0)= 9*10^9 F/m\n\n#calculation\nq=((K-E*10**-9)/(4*K))*10**-9;                #balancin by multiplying 10^-9 on numerator. to eV.vm terms\n\n#result\nprint\"Charge on the sphere required is\",round(q,4),\" times the charge of electron.\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Charge on the sphere required is 0.3105  times the charge of electron.\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.2 Page 273"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nK=1.44; Req=0.236; # K=e^2/(4*pi*e0)=1.44 eV.nm\n\n#calculation\nUc=-K/(Req);        #coulomb energy\n\n#result\nprint\"The coulomb energy at an equilirium separation distance in eV is\",round(Uc,3);\n\nE=-4.26; delE=1.53;    #various standars values of NaCl\nUr=E-Uc-delE; \n\n#result\nprint\"The pauli''s repulsion energy in eV is\",round(Ur,3);\n\n#partb\nReq=0.1;         #pauli repulsion energy\nUc=-K/(Req);\nE=4; delE=1.53;\nUr=E-Uc-delE;\n\n#result\nprint\"The pauli''s repulsion energy in eV is\",round(Ur,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The coulomb energy at an equilirium separation distance in eV is -6.102\nThe pauli''s repulsion energy in eV is 0.312\nThe pauli''s repulsion energy in eV is 16.87\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.3 Page 276"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi, sqrt\ndelE=0.50; delR=0.017*10**-9;      #delE= E-Emin; delR=R-Rmin;\nk=2*(delE)/(delR**2);c=3*10**8;     #force constant\nm=(1.008)*(931.5*10**6)*0.5;       #mass of molecular hydrogen\nv= sqrt(k*c**2/m)/(2*pi);          #vibrational frequency\nh=4.14*(10**-15);\n\n#calculation\nE=h*v;\n\n#result\nprint\"The value of corresponding photon energy in eV is\",round(E,3);\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of corresponding photon energy in eV is 0.537\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.4 Page 280"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi, sqrt\nhc=1240.0;        #in eV.nm\nm=0.5*1.008*931.5*10**6;              #mass of hydrogen atom\nReq=0.074;                          #equivalent radius\n\n#calculation\na=((hc)**2)/(4*(pi**2)*m*(Req**2));  #reduced mass of hydrogen atom\nfor L in range(1,4):\n        delE= L*a; \n        print\"The value of energy in eV is\",round(delE,4);                  \n        w=(hc)/delE;\n        print\"The respective wavelength in um is\",round(w*10**-3,3);  \n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of energy in eV is 0.0151\nThe respective wavelength in um is 81.849\nThe value of energy in eV is 0.0303\nThe respective wavelength in um is 40.925\nThe value of energy in eV is 0.0454\nThe respective wavelength in um is 27.283\n"
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.5 Page 283"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi\ndelv=6.2*(10**11);     #change in frequency\nh=1.05*(10**-34);        #value of h in J.sec\n\n#calculation\nI= h/(2*pi*delv);      #rotational inertia\nI1=I/(1.684604*10**-45); #to change units\n\n#result\nprint\"The value of rotational inertia in kg m2 is %.1e\" %I;\nprint\"which in terms of amu in u.nm2 is\",round(I1,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The value of rotational inertia in kg m2 is 2.69536597172e-47\nwhich in terms of amu in u.nm2 is 0.016\n"
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 9.6 Page 286"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#initiation of variable\nfrom math import pi\ndelE=0.358;hc=4.14*10**-15;          #hc in eV.nm and delE=1.44eV(given values)\n\n#calculation\nf=(delE)/hc;                        #frequency \n\n#result\nprint\"The frequency of the radiation is \",f;\n\n\nm=0.98;                            #mass in terms of u\nk=4*pi**2*m*f**2;                   #value of k in eV/m^2\n\n#result\nprint\"The force constant is\",k; \n\n#partb\nhc=1240.0; m=0.98*1.008*931.5*10**6; Req=0.127;      #various constants in terms of  \ns=((hc)**2)/(4*(pi**2)*m*(Req**2));                # expeted spacing \n\n#result\nprint\"The spacing was found out to be\",round(s,3),\"which is very close to the graphical value of 0.0026 eV.\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The frequency of the radiation is  8.64734299517e+13\nThe force constant is 2.89301831756e+29\nThe spacing was found out to be 0.003 which is very close to the graphical value of 0.0026 eV.\n"
+      }
+     ],
+     "prompt_number": 19
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/README.txt b/Modern_Physics/README.txt
index a4569c81..f150a0fb 100755
--- a/Modern_Physics/README.txt
+++ b/Modern_Physics/README.txt
@@ -1,10 +1,10 @@
-Contributed By: Puneetha Ramachandra
-Course: be
-College/Institute/Organization: National Institute of Technology
-Department/Designation: Electronics & CommunicationsEngg
+Contributed By: KRISHNA CHAITANYA
+Course: btech
+College/Institute/Organization: JNTUH
+Department/Designation: Computer Science
 Book Title: Modern Physics
-Author: Serway, Moses & Moyer
-Publisher: Thomson Learning, Inc.
-Year of publication: 2005
-Isbn: 0-534-40624-6
-Edition: 3rd
-\ No newline at end of file
+Author: G. Aruldhas & P. Rajagopal
+Publisher: PHI Learning Private Limited, New Delhi
+Year of publication: 2009
+Isbn: 978-81-203-2597-5
+Edition: 4
+\ No newline at end of file
diff --git a/Modern_Physics/chapter1.ipynb b/Modern_Physics/chapter1.ipynb
new file mode 100755
index 00000000..25ce0403
--- /dev/null
+++ b/Modern_Physics/chapter1.ipynb
@@ -0,0 +1,383 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:70cab6f9b725623fc2451b245c5190bb3397c02f8debec03cc8b79cdd3e4b714"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "1: Electric and Magnetic Fields"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.1, Page number 4"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "q1=3.2*10**-19;\n",
+      "q2=q1;                #q1 and q2 are the values of charge on alpha-particle(C)\n",
+      "d=10**-13;            #distance between two alpha-particles(m)\n",
+      "m1=6.68*10**-27;\n",
+      "m2=m1;          #m1 and m2 are masses of alpha-particles(kg)\n",
+      "G=6.67*10**-11;       #Gravitational constant(Nm^2/kg^2)\n",
+      "\n",
+      "#Calculation\n",
+      "F1=(9*10**9)*(q1*q2)/(d**2);       #calculation of electrostatic force(N)\n",
+      "F2=G*(m1*m2)/(d**2);        #calculation of electrostatic force(N)\n",
+      "F1=math.ceil(F1*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "F1 = F1*10**2;\n",
+      "\n",
+      "#Result\n",
+      "print \"The electrosatic force is\",F1,\"*10**-2 N\"\n",
+      "print \"The gravitational force is\",round(F2/1e-37,3),\"*10**-37 N\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The electrosatic force is 9.22 *10**-2 N\n",
+        "The gravitational force is 2.976 *10**-37 N\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.2, Page number 4"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m=9.1*10**-31;       #mass of elctron(kg)\n",
+      "q=1.6*10**-19;       #charge on electron(C)\n",
+      "g=9.81;           #acceleration due to gravity(m/s^2)\n",
+      "\n",
+      "#Calculation\n",
+      "Fg=m*g;                #gravitational force(N)\n",
+      "d=math.sqrt((9*10**9*q**2)/Fg);            #equating gravitational force with electrosatic force(m)\n",
+      "d=math.ceil(d*10**3)/10**3;            #rounding off to 4 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"The distance of separation is\",d,\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The distance of separation is 5.081 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.3, Page number 4"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "d=0.02;          #distance between plates(m)\n",
+      "V=400;           #potential differnce of plates(V)\n",
+      "q=1.6*10**-19;         #charge on a proton(C)\n",
+      "\n",
+      "#Calculation\n",
+      "E=V/d;       #electric field intensity between plates(V/m)\n",
+      "F=q*E;       #electrostatic force on oil drop(N)\n",
+      "\n",
+      "#Result\n",
+      "print \"The electric field intensity between plates is\",E,\"V/m\"\n",
+      "print \"The force on proton is\",F,\"N\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The electric field intensity between plates is 20000.0 V/m\n",
+        "The force on proton is 3.2e-15 N\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.4, Page number 4"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "d=0.02;           #distance between plates(m)\n",
+      "q=1.6*10**-19;       #charge on oil drop(C)\n",
+      "V=6000;              #potential differnce of plates(V)\n",
+      "g=9.81;              #acceleration due to gravity(m/s^2)\n",
+      "\n",
+      "#Calculation\n",
+      "E=V/d;       #electric field intensity between plates(V/m)\n",
+      "F=q*E;       #electrostatic force on oil drop(N)\n",
+      "m=F/g;       #equating the weight of oil drop to the electrostatic force on it(kg)\n",
+      "\n",
+      "#Result\n",
+      "print \"The mass of oil drop is\",round(m/1e-15,3),\"*10**-15 kg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mass of oil drop is 4.893 *10**-15 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.5, Page number 5"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=150;             #potential difference between anode and cathode(V)\n",
+      "m=9.31*10**-31;    #mass of an electron(kg)\n",
+      "q=1.6*10**-19;     #charge on an electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "E=q*V;            #energy gained by electron during speeding from cathode to anode(J)\n",
+      "vel=math.sqrt(E*2/m);    #equating with kinetic energy of electron(m/s)\n",
+      "vel=vel*10**-6;\n",
+      "vel=math.ceil(vel*10)/10;   #rounding off to 1 decimal\n",
+      "\n",
+      "#Result\n",
+      "print \"The velocity is\",vel,\"*10**6 m/s\"\n",
+      "print \"answer in the book is wrong by 1 decimal\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The velocity is 7.2 *10**6 m/s\n",
+        "answer in the book is wrong by 1 decimal\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.6, Page number 5"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=5*10**6;       #potential differnce through which alpha-particle is accelerated(V)\n",
+      "e=1.6*10**-19;         #charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "E1=2*V;      #electronic charge on alpha-particle(eV)\n",
+      "E2=E1/10**6;      #energy(MeV)\n",
+      "E3=E1*e;         #energy(J)\n",
+      "E1=E1*10**-7; \n",
+      "\n",
+      "#Result\n",
+      "print \"The energy is\",E1,\"*10**7 eV\"\n",
+      "print \"The energy is\",E2,\"MeV\"\n",
+      "print \"The energy is\",E3,\"J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The energy is 1.0 *10**7 eV\n",
+        "The energy is 10.0 MeV\n",
+        "The energy is 1.6e-12 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.7, Page number 6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "r=0.528*10**-10;       #radius of the orbit(m)\n",
+      "q=-1.6*10**-19;        #charge on electron(C)\n",
+      "Q=1.6*10**-19;         #charge on Hydrogen nucleus(C)\n",
+      "Eo=8.854*10**-12;      #permittivity in free space(F/m)\n",
+      "\n",
+      "#Calculation\n",
+      "E=(q*Q)/(8*3.14*Eo*r);       #electric field intensity between plates(V/m)\n",
+      "E1=E/(1.6*10**-19);          #electrifeild intensity(eV)\n",
+      "E=E*10**19;\n",
+      "E=math.ceil(E*10**2)/10**2;     #rounding off to 2 decimals\n",
+      "E1=math.ceil(E1*10**2)/10**2;     #rounding off to 2 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"The total energy is\",E,\"*10**-19 J\"\n",
+      "print \"The total energy is\",E1,\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total energy is -21.79 *10**-19 J\n",
+        "The total energy is -13.62 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.8, Page number 9"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Q=3.2*10**-19;     #charge on alpha-particle(C)\n",
+      "m=6.68*10**-27;    #mass on alpha-particle(kg)\n",
+      "B=1.5;          #transverse magnetic field of flux density(Wb/m^2)\n",
+      "v=5*10**6;      #velocity of alpha-particle(m/s)\n",
+      "\n",
+      "#Calculation\n",
+      "F=B*Q*v;        #electrostatic force on oil drop(N)\n",
+      "R=m*v/(Q*B);    #radius(m)\n",
+      "R=math.ceil(R*10**2)/10**2;   #rounding off to 2 decimals\n",
+      "R1 = R*100;              #radius(cm)\n",
+      "\n",
+      "#Result\n",
+      "print \"The force on particle is\",F,\"N\"\n",
+      "print \"The radius of its circular path\",R,\"m or\",R1,\"cm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The force on particle is 2.4e-12 N\n",
+        "The radius of its circular path 0.07 m or 7.0 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter11.ipynb b/Modern_Physics/chapter11.ipynb
new file mode 100755
index 00000000..f92c7d71
--- /dev/null
+++ b/Modern_Physics/chapter11.ipynb
@@ -0,0 +1,336 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:47ab953016e5f85b7389dfd03b89e4eaceb4217fd3c41781effc5a2eb8d08d3e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "11: Natural Radioactivity"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.1, Page number 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "ttg=8378-1898;       #total time gap(yrs)\n",
+      "hf=1620;      #half life(yrs)\n",
+      "n=ttg/hf;     #number of half-periods\n",
+      "Mo=200;       #amount of radium(mg)\n",
+      "\n",
+      "#Calculation\n",
+      "M=Mo*(0.5)**n;     #amount of radium left(mg)\n",
+      "\n",
+      "#Result\n",
+      "print \"amount of radium left is\",M,\"mg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "amount of radium left is 12.5 mg\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.2, Page number 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "T=30;      #half life(days)\n",
+      "#M is intial conc.\n",
+      "\n",
+      "#Calculation\n",
+      "lamda=0.693/T;     #radioactive disintegration constant(per day)\n",
+      "#M/4 is left\n",
+      "t1=-math.log(1/4)/lamda;     #time taken(days)\n",
+      "#M/8 is left\n",
+      "t2=-math.log(1/8)/lamda;      #time taken(days)\n",
+      "\n",
+      "#Result\n",
+      "print \"radioactive disintegration constant is\",lamda,\"per day\"\n",
+      "print \"time taken for 3/4th of original is\",int(t1),\"days\"\n",
+      "print \"time taken for 1/8th of original is\",int(t2),\"days\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "radioactive disintegration constant is 0.0231 per day\n",
+        "time taken for 3/4th of original is 60 days\n",
+        "time taken for 1/8th of original is 90 days\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.3, Page number 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "No=4750;      #count rate(per minute)\n",
+      "N=2700;       #rate(counts/minute)\n",
+      "t=5;          #time(minutes)\n",
+      "\n",
+      "#Calculation \n",
+      "lamda=math.log(No/N)/t;     #decay constant(per minute)\n",
+      "T=0.693/lamda;          #half life(minutes)\n",
+      "\n",
+      "#Result\n",
+      "print \"radioactive disintegration constant is\",round(lamda,3),\"per minute\"\n",
+      "print \"half life of sample is\",round(T,1),\"minutes\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "radioactive disintegration constant is 0.113 per minute\n",
+        "half life of sample is 6.1 minutes\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.4, Page number 223"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m=4.00387;     #mass of alpha particle(amu)\n",
+      "M=10**-6;      #mass of Pu-239(kg) \n",
+      "\n",
+      "#Calculation\n",
+      "m=m*1.66*10**-24;     #mass of alpha particle(g)\n",
+      "Mo=2300*m;       #mass of 2300 alpha particles(g)\n",
+      "lamda=(Mo/1)/M;     #radioactive disintegration constant(per second)\n",
+      "T=0.693/lamda;      #half life period(seconds)\n",
+      "T=T/(365*24*3600);    #half life period(years)\n",
+      "\n",
+      "#Result\n",
+      "print \"half life is\",round(T/1e+6,3),\"*10**6 years\"\n",
+      "print \"answer given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "half life is 1.438 *10**6 years\n",
+        "answer given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.5, Page number 223"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "T=2.48*10**5;    #half life(yrs)\n",
+      "lamda=8.88*10**-14    #decay constant (per second)\n",
+      "Mo=4;      #intial mass(mg)\n",
+      "t=62000;             #time(years)\n",
+      "Na=6.02*10**23;    #Avgraodo no.(per g-mol)\n",
+      "\n",
+      "#Calculation\n",
+      "lamdat=0.693/T*t;      \n",
+      "M=Mo*(math.exp(-lamdat));     #mass remained unchanged(mg) \n",
+      "N=M*10**-3*Na/234;\n",
+      "A=lamda*N;        #activity(disintegrations/second)\n",
+      "\n",
+      "#Result\n",
+      "print \"mass remained unchanged is\",round(M,3),\"mg\"\n",
+      "print \"Activity is\",round(A/1e+5,3),\"*10**5 disintegrations/second\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass remained unchanged is 3.364 mg\n",
+        "Activity is 7.684 *10**5 disintegrations/second\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.6, Page number 223"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "T=1620;       #half life(years)\n",
+      "Mo=1/100;     #mass(g)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda=0.693/T;      #radioactive constant(per years)   \n",
+      "M=(1-Mo);           #amount of radium left behind(g)  \n",
+      "t=math.log(1/M)/lamda;         #time required to lose 1 centigram(years)\n",
+      "t1=math.log(1/Mo)/lamda;         #time required to be reduced to 1 centigram(years)\n",
+      "\n",
+      "#Result\n",
+      "print \"time required to lose 1 centigram is\",round(t,1),\"years\"\n",
+      "print \"time required to be reduced to 1 centigram is\",int(t1),\"years\"\n",
+      "print \"answer given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time required to lose 1 centigram is 23.5 years\n",
+        "time required to be reduced to 1 centigram is 10765 years\n",
+        "answer given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.7, Page number 232"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "T=2*10**-4;           #dead time(seconds)\n",
+      "n=500;                #number of pulses(per second)\n",
+      "\n",
+      "#Calculation\n",
+      "n0=n/(1-(n*T));       #number of incoming particles(per second)\n",
+      "r=n*T*100;               #relative error of counting(%)\n",
+      "\n",
+      "#Result\n",
+      "print \"intensity of the incoming beam is\",int(n0),\"particles/second\"\n",
+      "print \"relative error of counting is\",int(r),\"%\"\n",
+      "print \"answer for intensity given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "intensity of the incoming beam is 555 particles/second\n",
+        "relative error of counting is 10 %\n",
+        "answer for intensity given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter11_1.ipynb b/Modern_Physics/chapter11_1.ipynb
new file mode 100755
index 00000000..f92c7d71
--- /dev/null
+++ b/Modern_Physics/chapter11_1.ipynb
@@ -0,0 +1,336 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:47ab953016e5f85b7389dfd03b89e4eaceb4217fd3c41781effc5a2eb8d08d3e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "11: Natural Radioactivity"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.1, Page number 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "ttg=8378-1898;       #total time gap(yrs)\n",
+      "hf=1620;      #half life(yrs)\n",
+      "n=ttg/hf;     #number of half-periods\n",
+      "Mo=200;       #amount of radium(mg)\n",
+      "\n",
+      "#Calculation\n",
+      "M=Mo*(0.5)**n;     #amount of radium left(mg)\n",
+      "\n",
+      "#Result\n",
+      "print \"amount of radium left is\",M,\"mg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "amount of radium left is 12.5 mg\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.2, Page number 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "T=30;      #half life(days)\n",
+      "#M is intial conc.\n",
+      "\n",
+      "#Calculation\n",
+      "lamda=0.693/T;     #radioactive disintegration constant(per day)\n",
+      "#M/4 is left\n",
+      "t1=-math.log(1/4)/lamda;     #time taken(days)\n",
+      "#M/8 is left\n",
+      "t2=-math.log(1/8)/lamda;      #time taken(days)\n",
+      "\n",
+      "#Result\n",
+      "print \"radioactive disintegration constant is\",lamda,\"per day\"\n",
+      "print \"time taken for 3/4th of original is\",int(t1),\"days\"\n",
+      "print \"time taken for 1/8th of original is\",int(t2),\"days\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "radioactive disintegration constant is 0.0231 per day\n",
+        "time taken for 3/4th of original is 60 days\n",
+        "time taken for 1/8th of original is 90 days\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.3, Page number 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "No=4750;      #count rate(per minute)\n",
+      "N=2700;       #rate(counts/minute)\n",
+      "t=5;          #time(minutes)\n",
+      "\n",
+      "#Calculation \n",
+      "lamda=math.log(No/N)/t;     #decay constant(per minute)\n",
+      "T=0.693/lamda;          #half life(minutes)\n",
+      "\n",
+      "#Result\n",
+      "print \"radioactive disintegration constant is\",round(lamda,3),\"per minute\"\n",
+      "print \"half life of sample is\",round(T,1),\"minutes\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "radioactive disintegration constant is 0.113 per minute\n",
+        "half life of sample is 6.1 minutes\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.4, Page number 223"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m=4.00387;     #mass of alpha particle(amu)\n",
+      "M=10**-6;      #mass of Pu-239(kg) \n",
+      "\n",
+      "#Calculation\n",
+      "m=m*1.66*10**-24;     #mass of alpha particle(g)\n",
+      "Mo=2300*m;       #mass of 2300 alpha particles(g)\n",
+      "lamda=(Mo/1)/M;     #radioactive disintegration constant(per second)\n",
+      "T=0.693/lamda;      #half life period(seconds)\n",
+      "T=T/(365*24*3600);    #half life period(years)\n",
+      "\n",
+      "#Result\n",
+      "print \"half life is\",round(T/1e+6,3),\"*10**6 years\"\n",
+      "print \"answer given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "half life is 1.438 *10**6 years\n",
+        "answer given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.5, Page number 223"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "T=2.48*10**5;    #half life(yrs)\n",
+      "lamda=8.88*10**-14    #decay constant (per second)\n",
+      "Mo=4;      #intial mass(mg)\n",
+      "t=62000;             #time(years)\n",
+      "Na=6.02*10**23;    #Avgraodo no.(per g-mol)\n",
+      "\n",
+      "#Calculation\n",
+      "lamdat=0.693/T*t;      \n",
+      "M=Mo*(math.exp(-lamdat));     #mass remained unchanged(mg) \n",
+      "N=M*10**-3*Na/234;\n",
+      "A=lamda*N;        #activity(disintegrations/second)\n",
+      "\n",
+      "#Result\n",
+      "print \"mass remained unchanged is\",round(M,3),\"mg\"\n",
+      "print \"Activity is\",round(A/1e+5,3),\"*10**5 disintegrations/second\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass remained unchanged is 3.364 mg\n",
+        "Activity is 7.684 *10**5 disintegrations/second\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.6, Page number 223"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "T=1620;       #half life(years)\n",
+      "Mo=1/100;     #mass(g)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda=0.693/T;      #radioactive constant(per years)   \n",
+      "M=(1-Mo);           #amount of radium left behind(g)  \n",
+      "t=math.log(1/M)/lamda;         #time required to lose 1 centigram(years)\n",
+      "t1=math.log(1/Mo)/lamda;         #time required to be reduced to 1 centigram(years)\n",
+      "\n",
+      "#Result\n",
+      "print \"time required to lose 1 centigram is\",round(t,1),\"years\"\n",
+      "print \"time required to be reduced to 1 centigram is\",int(t1),\"years\"\n",
+      "print \"answer given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time required to lose 1 centigram is 23.5 years\n",
+        "time required to be reduced to 1 centigram is 10765 years\n",
+        "answer given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 11.7, Page number 232"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "T=2*10**-4;           #dead time(seconds)\n",
+      "n=500;                #number of pulses(per second)\n",
+      "\n",
+      "#Calculation\n",
+      "n0=n/(1-(n*T));       #number of incoming particles(per second)\n",
+      "r=n*T*100;               #relative error of counting(%)\n",
+      "\n",
+      "#Result\n",
+      "print \"intensity of the incoming beam is\",int(n0),\"particles/second\"\n",
+      "print \"relative error of counting is\",int(r),\"%\"\n",
+      "print \"answer for intensity given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "intensity of the incoming beam is 555 particles/second\n",
+        "relative error of counting is 10 %\n",
+        "answer for intensity given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter12.ipynb b/Modern_Physics/chapter12.ipynb
new file mode 100755
index 00000000..4ab58792
--- /dev/null
+++ b/Modern_Physics/chapter12.ipynb
@@ -0,0 +1,63 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:3ee27523440d66319c8d17b88b593002a8ca441a824b734c02da7b3a1165c0fa"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "12: Artificial Radioactivity"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Example number 12.1, Page number 247"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "r=0.5;      #ratio of mass of Pb206 and mass of U238\n",
+      "t=4.5*10**9;     #half life(years)\n",
+      "\n",
+      "#Calculation\n",
+      "T=(math.log(1+r))*(t/0.693);        #age(years)\n",
+      "T=T/10**9;\n",
+      "\n",
+      "#Result\n",
+      "print \"age of rock is\",round(T,2),\"*10**9 years\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "age of rock is 2.63 *10**9 years\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter12_1.ipynb b/Modern_Physics/chapter12_1.ipynb
new file mode 100755
index 00000000..4ab58792
--- /dev/null
+++ b/Modern_Physics/chapter12_1.ipynb
@@ -0,0 +1,63 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:3ee27523440d66319c8d17b88b593002a8ca441a824b734c02da7b3a1165c0fa"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "12: Artificial Radioactivity"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Example number 12.1, Page number 247"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "r=0.5;      #ratio of mass of Pb206 and mass of U238\n",
+      "t=4.5*10**9;     #half life(years)\n",
+      "\n",
+      "#Calculation\n",
+      "T=(math.log(1+r))*(t/0.693);        #age(years)\n",
+      "T=T/10**9;\n",
+      "\n",
+      "#Result\n",
+      "print \"age of rock is\",round(T,2),\"*10**9 years\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "age of rock is 2.63 *10**9 years\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter13.ipynb b/Modern_Physics/chapter13.ipynb
new file mode 100755
index 00000000..fddec6a2
--- /dev/null
+++ b/Modern_Physics/chapter13.ipynb
@@ -0,0 +1,248 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:bee0f8df3068997ca43cf44df4f129c530dc6e34523ca501c99a2183643ee772"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "13: Nuclear Reactions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 13.1, Page number 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m1=7.0183;      #mass of 3Li7(amu)\n",
+      "m2=4.0040;      #mass of 2He4(amu)\n",
+      "m3=1.0082;      #mass of 1H1(amu)\n",
+      "N=6.026*10**26;      #Avgraodo no.(per kg atom)\n",
+      "#rxn = 3Li7 + 1H1 = 2He4 + 2He4 \n",
+      "\n",
+      "#Calculation\n",
+      "delta_m=m1+m3-(2*m2);      #deltam(amu)\n",
+      "E=delta_m*931;          #energy per disintegration(MeV)\n",
+      "n=0.1*N/7;      #no of atoms in 100 gm of lithium\n",
+      "TE=n*E;         #Total energy available(MeV)  \n",
+      "\n",
+      "#Result\n",
+      "print \"energy available per disintegration is\",round(E,2),\"MeV\"\n",
+      "print \"Total energy available is\",round(TE/1e+25,2),\"*10**25 MeV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "energy available per disintegration is 17.22 MeV\n",
+        "Total energy available is 14.83 *10**25 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 13.2, Page number 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m1=6.015126;    #mass of 3Li7(a.m.u)\n",
+      "m2=4.002604;    #mass oh 2He4(a.m.u)\n",
+      "m3=1.00865;     #mass of 0n1(a.m.u)\n",
+      "m4=3.016049;    #mass of 1H3(a.m.u)\n",
+      "#rxn = 3Li7 + 0n1 = 2He4 + 1H3 + Q\n",
+      "\n",
+      "#Calculation\n",
+      "dm=m1+m3-(m2+m4);\n",
+      "Q=dm*931;         #energy released(MeV)\n",
+      "\n",
+      "#Result\n",
+      "print \"energy released is\",round(Q,4),\"MeV\"\n",
+      "print \"answer given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "energy released is 4.7695 MeV\n",
+        "answer given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 13.3, Page number 260"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m1=14.007515;     #mass of 7N14(a.m.u)\n",
+      "m2=4.003837;      #mass of 2He4(a.m.u)\n",
+      "m3=17.004533;     #mass of 8O17(a.m.u)\n",
+      "m4=1.008142;      #mass of 1H1(a.m.u)\n",
+      "#rxn = 7N14 + 2He14 = 8O17 + 1H1\n",
+      "\n",
+      "#Calculation\n",
+      "dm=m3+m4-(m1+m2);\n",
+      "Q=dm*931;           #Q value of the reaction(MeV)\n",
+      "\n",
+      "#Result\n",
+      "print \"Q value of the reaction is\",round(Q,3),\"MeV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Q value of the reaction is 1.232 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 13.4, Page number 260"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m1=14.007520;     #mass of 7N14(a.m.u)\n",
+      "m2=1.008986;      #mass oh 0n1(a.m.u)\n",
+      "#m3=mass of 6C14 in a.m.u\n",
+      "m4=1.008145;      #mass of 1H1(a.m.u)\n",
+      "#rxn = 7N14 + 0n1 = 6C14 + 1H1 + 0.55 MeV\n",
+      "\n",
+      "#Calculation\n",
+      "Q=0.55;           #energy(MeV)\n",
+      "dm=Q/931;              \n",
+      "m3=dm+m1+m2-m4;       #mass of 6C14(a.m.u)\n",
+      "\n",
+      "#Result\n",
+      "print \"mass of 6C14 is\",round(m3,5),\"a.m.u\"\n",
+      "print \"answer given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass of 6C14 is 14.00895 a.m.u\n",
+        "answer given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 13.6, Page number 261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m0=11.01280;     #mass 5B11(a.m.u)\n",
+      "m1=4.00387;      #mass of alpha particle(a.m.u)\n",
+      "m2=14.00752;     #mass of 7N14(a.m.u)\n",
+      "#m3=mass of neutron \n",
+      "E1=5.250;       #energy of alpha particle(MeV)\n",
+      "E2=2.139;       #energy of 7N14(MeV)\n",
+      "E3=3.260;       #energy of 0n1(MeV)\n",
+      "\n",
+      "#Calculation\n",
+      "m3=(m0*931)+((m1*931)+E1)-((m2*931)+E2)-E3;          #mass of neutron(a.m.u)\n",
+      "\n",
+      "#Result\n",
+      "print \"mass of neutron is\",round(m3/931,3),\"a.m.u\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass of neutron is 1.009 a.m.u\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter13_1.ipynb b/Modern_Physics/chapter13_1.ipynb
new file mode 100755
index 00000000..fddec6a2
--- /dev/null
+++ b/Modern_Physics/chapter13_1.ipynb
@@ -0,0 +1,248 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:bee0f8df3068997ca43cf44df4f129c530dc6e34523ca501c99a2183643ee772"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "13: Nuclear Reactions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 13.1, Page number 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m1=7.0183;      #mass of 3Li7(amu)\n",
+      "m2=4.0040;      #mass of 2He4(amu)\n",
+      "m3=1.0082;      #mass of 1H1(amu)\n",
+      "N=6.026*10**26;      #Avgraodo no.(per kg atom)\n",
+      "#rxn = 3Li7 + 1H1 = 2He4 + 2He4 \n",
+      "\n",
+      "#Calculation\n",
+      "delta_m=m1+m3-(2*m2);      #deltam(amu)\n",
+      "E=delta_m*931;          #energy per disintegration(MeV)\n",
+      "n=0.1*N/7;      #no of atoms in 100 gm of lithium\n",
+      "TE=n*E;         #Total energy available(MeV)  \n",
+      "\n",
+      "#Result\n",
+      "print \"energy available per disintegration is\",round(E,2),\"MeV\"\n",
+      "print \"Total energy available is\",round(TE/1e+25,2),\"*10**25 MeV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "energy available per disintegration is 17.22 MeV\n",
+        "Total energy available is 14.83 *10**25 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 13.2, Page number 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m1=6.015126;    #mass of 3Li7(a.m.u)\n",
+      "m2=4.002604;    #mass oh 2He4(a.m.u)\n",
+      "m3=1.00865;     #mass of 0n1(a.m.u)\n",
+      "m4=3.016049;    #mass of 1H3(a.m.u)\n",
+      "#rxn = 3Li7 + 0n1 = 2He4 + 1H3 + Q\n",
+      "\n",
+      "#Calculation\n",
+      "dm=m1+m3-(m2+m4);\n",
+      "Q=dm*931;         #energy released(MeV)\n",
+      "\n",
+      "#Result\n",
+      "print \"energy released is\",round(Q,4),\"MeV\"\n",
+      "print \"answer given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "energy released is 4.7695 MeV\n",
+        "answer given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 13.3, Page number 260"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m1=14.007515;     #mass of 7N14(a.m.u)\n",
+      "m2=4.003837;      #mass of 2He4(a.m.u)\n",
+      "m3=17.004533;     #mass of 8O17(a.m.u)\n",
+      "m4=1.008142;      #mass of 1H1(a.m.u)\n",
+      "#rxn = 7N14 + 2He14 = 8O17 + 1H1\n",
+      "\n",
+      "#Calculation\n",
+      "dm=m3+m4-(m1+m2);\n",
+      "Q=dm*931;           #Q value of the reaction(MeV)\n",
+      "\n",
+      "#Result\n",
+      "print \"Q value of the reaction is\",round(Q,3),\"MeV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Q value of the reaction is 1.232 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 13.4, Page number 260"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m1=14.007520;     #mass of 7N14(a.m.u)\n",
+      "m2=1.008986;      #mass oh 0n1(a.m.u)\n",
+      "#m3=mass of 6C14 in a.m.u\n",
+      "m4=1.008145;      #mass of 1H1(a.m.u)\n",
+      "#rxn = 7N14 + 0n1 = 6C14 + 1H1 + 0.55 MeV\n",
+      "\n",
+      "#Calculation\n",
+      "Q=0.55;           #energy(MeV)\n",
+      "dm=Q/931;              \n",
+      "m3=dm+m1+m2-m4;       #mass of 6C14(a.m.u)\n",
+      "\n",
+      "#Result\n",
+      "print \"mass of 6C14 is\",round(m3,5),\"a.m.u\"\n",
+      "print \"answer given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass of 6C14 is 14.00895 a.m.u\n",
+        "answer given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 13.6, Page number 261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m0=11.01280;     #mass 5B11(a.m.u)\n",
+      "m1=4.00387;      #mass of alpha particle(a.m.u)\n",
+      "m2=14.00752;     #mass of 7N14(a.m.u)\n",
+      "#m3=mass of neutron \n",
+      "E1=5.250;       #energy of alpha particle(MeV)\n",
+      "E2=2.139;       #energy of 7N14(MeV)\n",
+      "E3=3.260;       #energy of 0n1(MeV)\n",
+      "\n",
+      "#Calculation\n",
+      "m3=(m0*931)+((m1*931)+E1)-((m2*931)+E2)-E3;          #mass of neutron(a.m.u)\n",
+      "\n",
+      "#Result\n",
+      "print \"mass of neutron is\",round(m3/931,3),\"a.m.u\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass of neutron is 1.009 a.m.u\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter14.ipynb b/Modern_Physics/chapter14.ipynb
new file mode 100755
index 00000000..b6ba7ca8
--- /dev/null
+++ b/Modern_Physics/chapter14.ipynb
@@ -0,0 +1,201 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:f3dfb0a49e3f8552798e92e62c041d1f0b46371fc7dc05dfc9c1bf5833f43216"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "14: Nuclear Fission And Fusion"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 14.1, Page number 269"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "E1=7.8;     #avg. B.E per nucleon(MeV)\n",
+      "E2=8.6;     #for fissin fragments(MeV)\n",
+      "\n",
+      "#Calculation\n",
+      "FER=(234*E2)-(236*E1);     #Fission energy released(MeV)\n",
+      "\n",
+      "#Result\n",
+      "print \"Fission energy released is\",FER,\"MeV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Fission energy released is 171.6 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 14.2, Page number 273"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m1=235.044;    #mass of 92U235(a.m.u)\n",
+      "m2=97.905;     #mass of 42Mo98(a.m.u)\n",
+      "m3=135.917;    #mass of 54Xe136(a.m.u)\n",
+      "#rxn = 0n1 + 92U235 = 42Mo98 + 54Xe136 + 4 -1e0 + 2 0n1\n",
+      "\n",
+      "#Calculation\n",
+      "LHSm=1.009+m1;\n",
+      "RHSm=m2+m3+(4*0.00055)+(2*1.009);\n",
+      "delta_m=LHSm-RHSm;        #mass defect(a.m.u)\n",
+      "E=delta_m*931;         #energy released(MeV)\n",
+      "\n",
+      "#Result\n",
+      "print \"mass defect is\",round(delta_m,3),\"a.m.u\"\n",
+      "print \"energy released is\",int(E),\"MeV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass defect is 0.211 a.m.u\n",
+        "energy released is 196 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 14.3, Page number 274"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m1=1.00813;    #mass of 1H1(a.m.u)\n",
+      "m2=4.00386;    #mass of 2He4(a.m.u)\n",
+      "SC=1.35;     #solar constant(kW/m^2)\n",
+      "d=1.5*10**11;    #distance b/w earth and sum(m)\n",
+      "e=1.6*10**-19;   #the charge on electron(C)\n",
+      "Na=6.02*10**26;    #Avgraodo no.(per kg mole)\n",
+      "#rxn = 4 1H1 = 2He4 + 2 1e0\n",
+      "\n",
+      "#Calculation\n",
+      "dm=(4*m1)-m2;\n",
+      "E=dm*931;     #energy produced(MeV)\n",
+      "EP=E/4;      #energy produced per atom(MeV)\n",
+      "EP=EP*10**6*e;    #conversion in J\n",
+      "EPkg=EP*Na;     #energy produced by 1 kg of hydrogen\n",
+      "SC=SC*1000;     #conversion(J/s-m^2)\n",
+      "SA=4*math.pi*d**2;   #surface area of sphere\n",
+      "ER=SC*SA;      #energy recieved per second\n",
+      "m=ER/EPkg;     #mass of hydrogen consumed(tonnes/second)\n",
+      "\n",
+      "#Result\n",
+      "print \"mass of hydrogen consumed is\",round(m/10**11,3),\"*10**8 tonnes/second\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass of hydrogen consumed is 5.941 *10**8 tonnes/second\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 14.4, Page number 275"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m1=2.01478;     #mass of 1H2(a.m.u)\n",
+      "m2=4.00388;     #mass of 2He4(a.m.u)\n",
+      "#rxn 1H2 + 1H2 = 2He4 + Q\n",
+      "\n",
+      "#Calculation\n",
+      "Q=2*m1-m2;        #energy liberated(MeV) \n",
+      "Q=Q*931;          #conversion in MeV\n",
+      "\n",
+      "#Result\n",
+      "print \"energy liberated is\",round(Q,1),\"MeV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "energy liberated is 23.9 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter14_1.ipynb b/Modern_Physics/chapter14_1.ipynb
new file mode 100755
index 00000000..b6ba7ca8
--- /dev/null
+++ b/Modern_Physics/chapter14_1.ipynb
@@ -0,0 +1,201 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:f3dfb0a49e3f8552798e92e62c041d1f0b46371fc7dc05dfc9c1bf5833f43216"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "14: Nuclear Fission And Fusion"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 14.1, Page number 269"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "E1=7.8;     #avg. B.E per nucleon(MeV)\n",
+      "E2=8.6;     #for fissin fragments(MeV)\n",
+      "\n",
+      "#Calculation\n",
+      "FER=(234*E2)-(236*E1);     #Fission energy released(MeV)\n",
+      "\n",
+      "#Result\n",
+      "print \"Fission energy released is\",FER,\"MeV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Fission energy released is 171.6 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 14.2, Page number 273"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m1=235.044;    #mass of 92U235(a.m.u)\n",
+      "m2=97.905;     #mass of 42Mo98(a.m.u)\n",
+      "m3=135.917;    #mass of 54Xe136(a.m.u)\n",
+      "#rxn = 0n1 + 92U235 = 42Mo98 + 54Xe136 + 4 -1e0 + 2 0n1\n",
+      "\n",
+      "#Calculation\n",
+      "LHSm=1.009+m1;\n",
+      "RHSm=m2+m3+(4*0.00055)+(2*1.009);\n",
+      "delta_m=LHSm-RHSm;        #mass defect(a.m.u)\n",
+      "E=delta_m*931;         #energy released(MeV)\n",
+      "\n",
+      "#Result\n",
+      "print \"mass defect is\",round(delta_m,3),\"a.m.u\"\n",
+      "print \"energy released is\",int(E),\"MeV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass defect is 0.211 a.m.u\n",
+        "energy released is 196 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 14.3, Page number 274"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m1=1.00813;    #mass of 1H1(a.m.u)\n",
+      "m2=4.00386;    #mass of 2He4(a.m.u)\n",
+      "SC=1.35;     #solar constant(kW/m^2)\n",
+      "d=1.5*10**11;    #distance b/w earth and sum(m)\n",
+      "e=1.6*10**-19;   #the charge on electron(C)\n",
+      "Na=6.02*10**26;    #Avgraodo no.(per kg mole)\n",
+      "#rxn = 4 1H1 = 2He4 + 2 1e0\n",
+      "\n",
+      "#Calculation\n",
+      "dm=(4*m1)-m2;\n",
+      "E=dm*931;     #energy produced(MeV)\n",
+      "EP=E/4;      #energy produced per atom(MeV)\n",
+      "EP=EP*10**6*e;    #conversion in J\n",
+      "EPkg=EP*Na;     #energy produced by 1 kg of hydrogen\n",
+      "SC=SC*1000;     #conversion(J/s-m^2)\n",
+      "SA=4*math.pi*d**2;   #surface area of sphere\n",
+      "ER=SC*SA;      #energy recieved per second\n",
+      "m=ER/EPkg;     #mass of hydrogen consumed(tonnes/second)\n",
+      "\n",
+      "#Result\n",
+      "print \"mass of hydrogen consumed is\",round(m/10**11,3),\"*10**8 tonnes/second\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass of hydrogen consumed is 5.941 *10**8 tonnes/second\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 14.4, Page number 275"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m1=2.01478;     #mass of 1H2(a.m.u)\n",
+      "m2=4.00388;     #mass of 2He4(a.m.u)\n",
+      "#rxn 1H2 + 1H2 = 2He4 + Q\n",
+      "\n",
+      "#Calculation\n",
+      "Q=2*m1-m2;        #energy liberated(MeV) \n",
+      "Q=Q*931;          #conversion in MeV\n",
+      "\n",
+      "#Result\n",
+      "print \"energy liberated is\",round(Q,1),\"MeV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "energy liberated is 23.9 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter15.ipynb b/Modern_Physics/chapter15.ipynb
new file mode 100755
index 00000000..4227a2a8
--- /dev/null
+++ b/Modern_Physics/chapter15.ipynb
@@ -0,0 +1,397 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:a8391f5a05f23d60fe8f4f8931e579dcdcd76160c3d9f4755fb607b65fe83b63"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "15: Nuclear Energy Sources"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 15.1, Page number 290"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "#m is mass of neutron and M is mass of other neucleus\n",
+      "ma=1;\n",
+      "Ma=2;\n",
+      "mb=1;\n",
+      "Mb=12;\n",
+      "mc=1;\n",
+      "Mc=238;     \n",
+      "\n",
+      "#Calculation\n",
+      "eeta1=(4*ma*Ma/((ma+Ma)**2))*100;      #Maximum fraction of KE lost by a neutron for H2(%)\n",
+      "eeta2=(4*mb*Mb/((mb+Mb)**2))*100;      #Maximum fraction of KE lost by a neutron for C12(%)\n",
+      "eeta3=(4*mc*Mc/((mc+Mc)**2))*100;      #Maximum fraction of KE lost by a neutron for U238(%)\n",
+      "\n",
+      "#Result\n",
+      "print \"Maximum fraction of KE lost by a neutron for H2 is\",round(eeta1,1),\"%\"\n",
+      "print \"Maximum fraction of KE lost by a neutron for C12 is\",round(eeta2,1),\"%\"\n",
+      "print \"Maximum fraction of KE lost by a neutron for U238 is\",round(eeta3,2),\"%\"\n",
+      "print \"answer for eeta2 given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum fraction of KE lost by a neutron for H2 is 88.9 %\n",
+        "Maximum fraction of KE lost by a neutron for C12 is 28.4 %\n",
+        "Maximum fraction of KE lost by a neutron for U238 is 1.67 %\n",
+        "answer for eeta2 given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 15.2, Page number 291"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "E=200;     #energy released per fission(MeV)\n",
+      "e=1.6*10**-19;    #the charge on electron(C)\n",
+      "Na=6.02*10**26;   #Avgraodo no.(per kg mole)\n",
+      "\n",
+      "#Calculation\n",
+      "CE=E*e*10**6;     #conversion in J\n",
+      "RF=1/CE;        #fission rate(fissions/second)\n",
+      "Ekg=CE*Na/235;     #Energy realeased in complete fission of 1 kg(J)\n",
+      "\n",
+      "#Result\n",
+      "print \"fission rate is\",round(RF/10**10,1),\"*10**10 fissions/second\"\n",
+      "print \"Energy realeased in complete fission of 1 kg is\",round(Ekg/1e+13,1),\"*10**13 J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fission rate is 3.1 *10**10 fissions/second\n",
+        "Energy realeased in complete fission of 1 kg is 8.2 *10**13 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 15.3, Page number 291"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "R=3*10**7;       #rate of energy development(Js)\n",
+      "E=200;           #energy released per fission(MeV)\n",
+      "e=1.6*10**-19;    #the charge on electron(C)\n",
+      "t=1000;           #time(hours)\n",
+      "Ekg=8.2*10**13;   #energy released per kg of U-235\n",
+      "\n",
+      "#Calculation\n",
+      "CE=E*e*10**6;     #conversion in J\n",
+      "n=R/CE;         #no of atoms undergoing fission/second\n",
+      "TE=R*t*3600;      #energy produced in 1000 hours(J)\n",
+      "MC=TE/Ekg;        #mass consumed(kg)  \n",
+      "\n",
+      "#Result\n",
+      "print \"number of atoms undergoing fissions per second is\",round(n/1e+17,1),\"*10**17\"\n",
+      "print \"mass consumed is\",round(MC,2),\"kg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "number of atoms undergoing fissions per second is 9.4 *10**17\n",
+        "mass consumed is 1.32 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 15.4, Page number 292"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "EPF=180;       #Energy consumed per disintegration(MeV)\n",
+      "E=1200;        #average power(kW)\n",
+      "t=10;          #time(hours)\n",
+      "Na=6.02*10**26;      #Avgraodo no.(per kg mole)\n",
+      "e=1.6*10**-19;       #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "TE=E*t;    #energy consumed(kWh)\n",
+      "TE=TE*36*10**5;     #conversion(J)\n",
+      "EE=TE/0.2;          #efficient energy\n",
+      "CE=EPF*e*10**6;     #conversion in J\n",
+      "n=EE/CE;\n",
+      "m=235*n/Na*1000;     #mass consumed(gram)\n",
+      "\n",
+      "#Result\n",
+      "print \"mass consumed is\",round(m,2),\"gram\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass consumed is 2.93 gram\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 15.5, Page number 292"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "OE=200;       #output power(MW)\n",
+      "E=200;        #energy released per fission(MeV)\n",
+      "WF=3.1*10**10;     #fission rate(fissions/second)\n",
+      "Na=6.02*10**26;    #Avagadro no.(per kg mole)\n",
+      "\n",
+      "#Calculation\n",
+      "IE=OE/0.3*10**6;     #reactor input(W)\n",
+      "TFR=WF*IE;\n",
+      "n=TFR*24*3600;      #no. of U-235 for one day\n",
+      "m=235*n/Na;         #mass required(kg)\n",
+      "   \n",
+      "#Result\n",
+      "print \"amount of natural uranium consumed per day is\",round(m*100/0.7,3),\"kg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "amount of natural uranium consumed per day is 99.577 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 15.6, Page number 292"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "AE=100;       #electrical power(MW)\n",
+      "E=200;        #energy released per fission(MeV)\n",
+      "e=1.6*10**-19;     #the charge on electron(C)\n",
+      "Na=6.02*10**26;    #Avagadro no.(per kg mole)\n",
+      "\n",
+      "#Calculation\n",
+      "TE=AE*10**6*24*3600;     #energy consumed in city in one day(J)\n",
+      "EE=TE/0.2;\n",
+      "CE=E*e*10**6;          #conversion in J\n",
+      "n=EE/CE;               #no. of atoms to be fissioned  \n",
+      "m=235*n/Na;           #amount of fuel required(kg)\n",
+      "\n",
+      "#Result\n",
+      "print \"amount of fuel required is\",round(m,2),\"kg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "amount of fuel required is 0.53 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 15.7, Page number 293"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "OE=3000;       #output power(MWh)\n",
+      "E=200;         #energy released per fission(MeV)\n",
+      "e=1.6*10**-19;       #the charge on electron(C)\n",
+      "Na=6.02*10**26;      #Avagadro no.(per kg mole)\n",
+      "\n",
+      "#Calculation\n",
+      "IE=OE/0.2;        #nuclear energy input(MWh)\n",
+      "TE=IE*36*10**8;     #conversion in J\n",
+      "CE=E*e*10**6;       #conversion in J\n",
+      "n=TE/CE;            #number of nuclides required per day\n",
+      "m=235*n/Na;         #daily fuel requirement(kg)\n",
+      "\n",
+      "#Result\n",
+      "print \"daily fuel requirement is\",round(m,3),\"kg or\",round(m,3)*1000,\"gram\"\n",
+      "print \"answer given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "daily fuel requirement is 0.659 kg or 659.0 gram\n",
+        "answer given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 15.8, Page number 293"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "OP=32824;    #output power(kW)\n",
+      "E=200;       #energy released per fission(MeV)\n",
+      "Ekg=8.2*10**13;     #energy released per kg of U-235(J)\n",
+      "\n",
+      "#Calculation\n",
+      "DOP=OP*1000*24*3600;      #daily output power(J)\n",
+      "IP=DOP/0.2;               #nuclear energy input(J)\n",
+      "DFC=IP/Ekg;            #daily fuel consumption(kg)\n",
+      "DI=DOP/(0.8*4186);     #daily input at 80% efficiency(kcal)\n",
+      "Crqd=DI/(7*10**3);     #Coal required per day(tonnes)\n",
+      "\n",
+      "#Result\n",
+      "print \"Daily fuel consumption is\",round(DFC,3)*1000,\"gram\"\n",
+      "print \"Coal required per day is\",int(Crqd),\"tonnes\"\n",
+      "print \"answer for coal required per day Crqd given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Daily fuel consumption is 173.0 gram\n",
+        "Coal required per day is 120981 tonnes\n",
+        "answer for coal required per day Crqd given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter15_1.ipynb b/Modern_Physics/chapter15_1.ipynb
new file mode 100755
index 00000000..4227a2a8
--- /dev/null
+++ b/Modern_Physics/chapter15_1.ipynb
@@ -0,0 +1,397 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:a8391f5a05f23d60fe8f4f8931e579dcdcd76160c3d9f4755fb607b65fe83b63"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "15: Nuclear Energy Sources"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 15.1, Page number 290"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "#m is mass of neutron and M is mass of other neucleus\n",
+      "ma=1;\n",
+      "Ma=2;\n",
+      "mb=1;\n",
+      "Mb=12;\n",
+      "mc=1;\n",
+      "Mc=238;     \n",
+      "\n",
+      "#Calculation\n",
+      "eeta1=(4*ma*Ma/((ma+Ma)**2))*100;      #Maximum fraction of KE lost by a neutron for H2(%)\n",
+      "eeta2=(4*mb*Mb/((mb+Mb)**2))*100;      #Maximum fraction of KE lost by a neutron for C12(%)\n",
+      "eeta3=(4*mc*Mc/((mc+Mc)**2))*100;      #Maximum fraction of KE lost by a neutron for U238(%)\n",
+      "\n",
+      "#Result\n",
+      "print \"Maximum fraction of KE lost by a neutron for H2 is\",round(eeta1,1),\"%\"\n",
+      "print \"Maximum fraction of KE lost by a neutron for C12 is\",round(eeta2,1),\"%\"\n",
+      "print \"Maximum fraction of KE lost by a neutron for U238 is\",round(eeta3,2),\"%\"\n",
+      "print \"answer for eeta2 given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum fraction of KE lost by a neutron for H2 is 88.9 %\n",
+        "Maximum fraction of KE lost by a neutron for C12 is 28.4 %\n",
+        "Maximum fraction of KE lost by a neutron for U238 is 1.67 %\n",
+        "answer for eeta2 given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 15.2, Page number 291"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "E=200;     #energy released per fission(MeV)\n",
+      "e=1.6*10**-19;    #the charge on electron(C)\n",
+      "Na=6.02*10**26;   #Avgraodo no.(per kg mole)\n",
+      "\n",
+      "#Calculation\n",
+      "CE=E*e*10**6;     #conversion in J\n",
+      "RF=1/CE;        #fission rate(fissions/second)\n",
+      "Ekg=CE*Na/235;     #Energy realeased in complete fission of 1 kg(J)\n",
+      "\n",
+      "#Result\n",
+      "print \"fission rate is\",round(RF/10**10,1),\"*10**10 fissions/second\"\n",
+      "print \"Energy realeased in complete fission of 1 kg is\",round(Ekg/1e+13,1),\"*10**13 J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fission rate is 3.1 *10**10 fissions/second\n",
+        "Energy realeased in complete fission of 1 kg is 8.2 *10**13 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 15.3, Page number 291"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "R=3*10**7;       #rate of energy development(Js)\n",
+      "E=200;           #energy released per fission(MeV)\n",
+      "e=1.6*10**-19;    #the charge on electron(C)\n",
+      "t=1000;           #time(hours)\n",
+      "Ekg=8.2*10**13;   #energy released per kg of U-235\n",
+      "\n",
+      "#Calculation\n",
+      "CE=E*e*10**6;     #conversion in J\n",
+      "n=R/CE;         #no of atoms undergoing fission/second\n",
+      "TE=R*t*3600;      #energy produced in 1000 hours(J)\n",
+      "MC=TE/Ekg;        #mass consumed(kg)  \n",
+      "\n",
+      "#Result\n",
+      "print \"number of atoms undergoing fissions per second is\",round(n/1e+17,1),\"*10**17\"\n",
+      "print \"mass consumed is\",round(MC,2),\"kg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "number of atoms undergoing fissions per second is 9.4 *10**17\n",
+        "mass consumed is 1.32 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 15.4, Page number 292"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "EPF=180;       #Energy consumed per disintegration(MeV)\n",
+      "E=1200;        #average power(kW)\n",
+      "t=10;          #time(hours)\n",
+      "Na=6.02*10**26;      #Avgraodo no.(per kg mole)\n",
+      "e=1.6*10**-19;       #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "TE=E*t;    #energy consumed(kWh)\n",
+      "TE=TE*36*10**5;     #conversion(J)\n",
+      "EE=TE/0.2;          #efficient energy\n",
+      "CE=EPF*e*10**6;     #conversion in J\n",
+      "n=EE/CE;\n",
+      "m=235*n/Na*1000;     #mass consumed(gram)\n",
+      "\n",
+      "#Result\n",
+      "print \"mass consumed is\",round(m,2),\"gram\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass consumed is 2.93 gram\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 15.5, Page number 292"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "OE=200;       #output power(MW)\n",
+      "E=200;        #energy released per fission(MeV)\n",
+      "WF=3.1*10**10;     #fission rate(fissions/second)\n",
+      "Na=6.02*10**26;    #Avagadro no.(per kg mole)\n",
+      "\n",
+      "#Calculation\n",
+      "IE=OE/0.3*10**6;     #reactor input(W)\n",
+      "TFR=WF*IE;\n",
+      "n=TFR*24*3600;      #no. of U-235 for one day\n",
+      "m=235*n/Na;         #mass required(kg)\n",
+      "   \n",
+      "#Result\n",
+      "print \"amount of natural uranium consumed per day is\",round(m*100/0.7,3),\"kg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "amount of natural uranium consumed per day is 99.577 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 15.6, Page number 292"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "AE=100;       #electrical power(MW)\n",
+      "E=200;        #energy released per fission(MeV)\n",
+      "e=1.6*10**-19;     #the charge on electron(C)\n",
+      "Na=6.02*10**26;    #Avagadro no.(per kg mole)\n",
+      "\n",
+      "#Calculation\n",
+      "TE=AE*10**6*24*3600;     #energy consumed in city in one day(J)\n",
+      "EE=TE/0.2;\n",
+      "CE=E*e*10**6;          #conversion in J\n",
+      "n=EE/CE;               #no. of atoms to be fissioned  \n",
+      "m=235*n/Na;           #amount of fuel required(kg)\n",
+      "\n",
+      "#Result\n",
+      "print \"amount of fuel required is\",round(m,2),\"kg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "amount of fuel required is 0.53 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 15.7, Page number 293"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "OE=3000;       #output power(MWh)\n",
+      "E=200;         #energy released per fission(MeV)\n",
+      "e=1.6*10**-19;       #the charge on electron(C)\n",
+      "Na=6.02*10**26;      #Avagadro no.(per kg mole)\n",
+      "\n",
+      "#Calculation\n",
+      "IE=OE/0.2;        #nuclear energy input(MWh)\n",
+      "TE=IE*36*10**8;     #conversion in J\n",
+      "CE=E*e*10**6;       #conversion in J\n",
+      "n=TE/CE;            #number of nuclides required per day\n",
+      "m=235*n/Na;         #daily fuel requirement(kg)\n",
+      "\n",
+      "#Result\n",
+      "print \"daily fuel requirement is\",round(m,3),\"kg or\",round(m,3)*1000,\"gram\"\n",
+      "print \"answer given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "daily fuel requirement is 0.659 kg or 659.0 gram\n",
+        "answer given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 15.8, Page number 293"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "OP=32824;    #output power(kW)\n",
+      "E=200;       #energy released per fission(MeV)\n",
+      "Ekg=8.2*10**13;     #energy released per kg of U-235(J)\n",
+      "\n",
+      "#Calculation\n",
+      "DOP=OP*1000*24*3600;      #daily output power(J)\n",
+      "IP=DOP/0.2;               #nuclear energy input(J)\n",
+      "DFC=IP/Ekg;            #daily fuel consumption(kg)\n",
+      "DI=DOP/(0.8*4186);     #daily input at 80% efficiency(kcal)\n",
+      "Crqd=DI/(7*10**3);     #Coal required per day(tonnes)\n",
+      "\n",
+      "#Result\n",
+      "print \"Daily fuel consumption is\",round(DFC,3)*1000,\"gram\"\n",
+      "print \"Coal required per day is\",int(Crqd),\"tonnes\"\n",
+      "print \"answer for coal required per day Crqd given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Daily fuel consumption is 173.0 gram\n",
+        "Coal required per day is 120981 tonnes\n",
+        "answer for coal required per day Crqd given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter16.ipynb b/Modern_Physics/chapter16.ipynb
new file mode 100755
index 00000000..27b371a7
--- /dev/null
+++ b/Modern_Physics/chapter16.ipynb
@@ -0,0 +1,433 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d47bf0be5cde96d0aef086befce8360c308553e75286c744d02ce7f3929fcd07"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "16: Particle Accelerators"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 16.1, Page number 305"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "fo=9*10**6;    #frequency(Hz)\n",
+      "m=6.643*10**-27;    #mass(kg)\n",
+      "e=1.6*10**-19;      #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "Q=2*e;           #electron charge(C)\n",
+      "B=fo*2*math.pi*m/Q;      #magnetic flux density(Wb/m^2)\n",
+      "\n",
+      "#Result\n",
+      "print \"magnetic flux density is\",round(B,2),\"Wb/m^2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "magnetic flux density is 1.17 Wb/m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 16.2, Page number 305"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "B=0.7;      #magnetic flux intensity(Wb/m^2)\n",
+      "m=3.34*10**-27;     #mass(Kg)\n",
+      "e=1.6*10**-19;      #the charge on electron(C)\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "Q=e;\n",
+      "fo=B*Q/(2*math.pi*m*10**6);       #cyclotron frequency(MHz)  \n",
+      "\n",
+      "#Result\n",
+      "print \"The cyclotron frequency is\",round(fo,1),\"MHz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The cyclotron frequency is 5.3 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 16.3, Page number 306"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "B=0.75;      #magnetic flux intensity(Wb/m^2)\n",
+      "m1=1.67*10**-27;    #mass(Kg)\n",
+      "m2=3.31*10**-27;    #mass(Kg)\n",
+      "e=1.6*10**-19;      #the charge on electron(C)\n",
+      "Rm=2;        #radius(m)\n",
+      "\n",
+      "#Calculation\n",
+      "Q=e;\n",
+      "Emax_m1=3.12*10**12*B**2*Q**2*Rm**2/m1;       #Maximum energy for proton(MeV)\n",
+      "Emax_m2=3.12*10**12*B**2*Q**2*Rm**2/m2;       #Maximum energy for deuteron(MeV) \n",
+      "\n",
+      "#Result\n",
+      "print \"Maximum energy for proton is\",round(Emax_m1),\"MeV\"\n",
+      "print \"Maximum energy for deuteron is\",int(Emax_m2),\"MeV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum energy for proton is 108.0 MeV\n",
+        "Maximum energy for deuteron is 54 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 16.4, Page number 306"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "mo=9.1*10**-31;      #mass of electron(kg)\n",
+      "m=1.67*10**-27;      #mass of proton(kg)\n",
+      "c=3*10**8;           #speed of light(m/s)\n",
+      "E=1;                 #given energy(MeV)\n",
+      "\n",
+      "#Calculation\n",
+      "Eo=mo*c**2/(1.6*10**-13);         #rest energy for electron(MeV)\n",
+      "mbymo_e=1+(E/Eo);                #Ratio for electron\n",
+      "Eo=m*c**2/(1.6*10**-13);          #rest energy for proton(MeV)\n",
+      "mbymo_p=1+(E/Eo);                #Ratio for proton\n",
+      "\n",
+      "#Result\n",
+      "print \"Ratio for electron is\",round(mbymo_e,3)\n",
+      "print \"Ratio for proton is\",round(mbymo_p,6)\n",
+      "print \"answer in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ratio for electron is 2.954\n",
+        "Ratio for proton is 1.001065\n",
+        "answer in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 16.5, Page number 306"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "B=0.5;      #magnetic field(Wb/m^2)\n",
+      "d=1.5;      #diameter(m)\n",
+      "f=59;       #frequency(Hz)\n",
+      "e=1.6*10**-19;        #the charge on electron(C)\n",
+      "c=3*10**8;            #speed of light(m/s)\n",
+      "\n",
+      "#Calculation\n",
+      "R=d/2;             #radius(m)\n",
+      "N=c/(4*(2*math.pi*50)*R);       #number of revolutions\n",
+      "E=B*e*R*c/(1.6*10**-13);        #final energy(MeV)\n",
+      "AE=E/N*10**6;                 #average energy(eV)\n",
+      "\n",
+      "#Result\n",
+      "print \"final energy is\",E,\"MeV\"\n",
+      "print \"average energy is\",round(AE,1),\"eV\"\n",
+      "print \"answer for average energy given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "final energy is 112.5 MeV\n",
+        "average energy is 353.4 eV\n",
+        "answer for average energy given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 16.6, Page number 307"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "E=0.51;      #kinetic energy(MeV)\n",
+      "R=0.15;      #radius(m)\n",
+      "e=1.6*10**-19;      #the charge on electron(C)\n",
+      "mo=9.12*10**-31;    #mass of electron(kg)\n",
+      "c=3*10**8;          #speed of light(m/s)\n",
+      "\n",
+      "#Calculation\n",
+      "Eo=E;\n",
+      "m=mo*(1+(E/Eo));       #mass(kg)\n",
+      "b=math.sqrt(1-(mo/m)**2);\n",
+      "v=b*c;            #velocity(m/s)\n",
+      "B=mo*v/(e*R);     #flux density(Wb/m^2)   \n",
+      "\n",
+      "#Result\n",
+      "print \"mass is\",round(m/1e-31,1),\"*10^-31 kg\"\n",
+      "print \"velocity is\",round(v/1e+8,1),\"*10^8 m/s\"\n",
+      "print \"flux density is\",round(B,5),\"Wb/m^2\"\n",
+      "print \"answer for flux density in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass is 18.2 *10^-31 kg\n",
+        "velocity is 2.6 *10^8 m/s\n",
+        "flux density is 0.00987 Wb/m^2\n",
+        "answer for flux density in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 16.7, Page number 308"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "E=4;       #applied voltage(MeV)\n",
+      "m=3.334*10**-27;      #mass of deuteron(kg)\n",
+      "R=0.75;               #radius(m)\n",
+      "e=1.6*10**-19;        #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "E=4*10**6*e;\n",
+      "fo=math.sqrt(E/(2*m))/(math.pi*R);     #frequnecy of generator(Hz)\n",
+      "\n",
+      "#Result\n",
+      "print \"frequnecy of generator is\",round(fo/1e+6,3),\"*10^6 Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "frequnecy of generator is 4.158 *10^6 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 16.8, Page number 308"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "roi=15;      #rate of increase(Wb/s)\n",
+      "tr=10**6;    #total revolutions\n",
+      "\n",
+      "#Calculation\n",
+      "IE=roi*10**-6;      #increased energy(MeV)\n",
+      "FE=IE*tr;           #Final Energy(MeV) \n",
+      "\n",
+      "#Result\n",
+      "print \"Final Energy is\",FE,\"MeV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Final Energy is 15.0 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 16.9, Page number 308"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "R=0.1;       #radius(m)\n",
+      "h=6.625*10**-34;        #Plank's constant\n",
+      "c=3*10**8;              #speed of light(m/s)\n",
+      "roi=15;                 #rate of increase(Wb/s)\n",
+      "t=4*10**-4;             #period of accerleartion(s)\n",
+      "e=1.6*10**-19;          #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "N=c*t/(2*math.pi*R);      #number of revolutions\n",
+      "IE=roi;              #incresed energy(eV)\n",
+      "ME=N*IE*10**-6;      #maximum energy(MeV)  \n",
+      "ME1=ME*10**6*e;       #conversion in V\n",
+      "p=ME1/c;\n",
+      "gama=h/p;         #wavelength of gama rays(m)\n",
+      "\n",
+      "#Result\n",
+      "print \"Maximum energy is\",round(ME,3),\"MeV\"\n",
+      "print \"Corresponding wavelength of gama rays is\",round(gama/1e-13,3),\"*10^-13 m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum energy is 2.865 MeV\n",
+        "Corresponding wavelength of gama rays is 4.336 *10^-13 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter16_1.ipynb b/Modern_Physics/chapter16_1.ipynb
new file mode 100755
index 00000000..27b371a7
--- /dev/null
+++ b/Modern_Physics/chapter16_1.ipynb
@@ -0,0 +1,433 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d47bf0be5cde96d0aef086befce8360c308553e75286c744d02ce7f3929fcd07"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "16: Particle Accelerators"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 16.1, Page number 305"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "fo=9*10**6;    #frequency(Hz)\n",
+      "m=6.643*10**-27;    #mass(kg)\n",
+      "e=1.6*10**-19;      #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "Q=2*e;           #electron charge(C)\n",
+      "B=fo*2*math.pi*m/Q;      #magnetic flux density(Wb/m^2)\n",
+      "\n",
+      "#Result\n",
+      "print \"magnetic flux density is\",round(B,2),\"Wb/m^2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "magnetic flux density is 1.17 Wb/m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 16.2, Page number 305"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "B=0.7;      #magnetic flux intensity(Wb/m^2)\n",
+      "m=3.34*10**-27;     #mass(Kg)\n",
+      "e=1.6*10**-19;      #the charge on electron(C)\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "Q=e;\n",
+      "fo=B*Q/(2*math.pi*m*10**6);       #cyclotron frequency(MHz)  \n",
+      "\n",
+      "#Result\n",
+      "print \"The cyclotron frequency is\",round(fo,1),\"MHz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The cyclotron frequency is 5.3 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 16.3, Page number 306"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "B=0.75;      #magnetic flux intensity(Wb/m^2)\n",
+      "m1=1.67*10**-27;    #mass(Kg)\n",
+      "m2=3.31*10**-27;    #mass(Kg)\n",
+      "e=1.6*10**-19;      #the charge on electron(C)\n",
+      "Rm=2;        #radius(m)\n",
+      "\n",
+      "#Calculation\n",
+      "Q=e;\n",
+      "Emax_m1=3.12*10**12*B**2*Q**2*Rm**2/m1;       #Maximum energy for proton(MeV)\n",
+      "Emax_m2=3.12*10**12*B**2*Q**2*Rm**2/m2;       #Maximum energy for deuteron(MeV) \n",
+      "\n",
+      "#Result\n",
+      "print \"Maximum energy for proton is\",round(Emax_m1),\"MeV\"\n",
+      "print \"Maximum energy for deuteron is\",int(Emax_m2),\"MeV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum energy for proton is 108.0 MeV\n",
+        "Maximum energy for deuteron is 54 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 16.4, Page number 306"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "mo=9.1*10**-31;      #mass of electron(kg)\n",
+      "m=1.67*10**-27;      #mass of proton(kg)\n",
+      "c=3*10**8;           #speed of light(m/s)\n",
+      "E=1;                 #given energy(MeV)\n",
+      "\n",
+      "#Calculation\n",
+      "Eo=mo*c**2/(1.6*10**-13);         #rest energy for electron(MeV)\n",
+      "mbymo_e=1+(E/Eo);                #Ratio for electron\n",
+      "Eo=m*c**2/(1.6*10**-13);          #rest energy for proton(MeV)\n",
+      "mbymo_p=1+(E/Eo);                #Ratio for proton\n",
+      "\n",
+      "#Result\n",
+      "print \"Ratio for electron is\",round(mbymo_e,3)\n",
+      "print \"Ratio for proton is\",round(mbymo_p,6)\n",
+      "print \"answer in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ratio for electron is 2.954\n",
+        "Ratio for proton is 1.001065\n",
+        "answer in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 16.5, Page number 306"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "B=0.5;      #magnetic field(Wb/m^2)\n",
+      "d=1.5;      #diameter(m)\n",
+      "f=59;       #frequency(Hz)\n",
+      "e=1.6*10**-19;        #the charge on electron(C)\n",
+      "c=3*10**8;            #speed of light(m/s)\n",
+      "\n",
+      "#Calculation\n",
+      "R=d/2;             #radius(m)\n",
+      "N=c/(4*(2*math.pi*50)*R);       #number of revolutions\n",
+      "E=B*e*R*c/(1.6*10**-13);        #final energy(MeV)\n",
+      "AE=E/N*10**6;                 #average energy(eV)\n",
+      "\n",
+      "#Result\n",
+      "print \"final energy is\",E,\"MeV\"\n",
+      "print \"average energy is\",round(AE,1),\"eV\"\n",
+      "print \"answer for average energy given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "final energy is 112.5 MeV\n",
+        "average energy is 353.4 eV\n",
+        "answer for average energy given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 16.6, Page number 307"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "E=0.51;      #kinetic energy(MeV)\n",
+      "R=0.15;      #radius(m)\n",
+      "e=1.6*10**-19;      #the charge on electron(C)\n",
+      "mo=9.12*10**-31;    #mass of electron(kg)\n",
+      "c=3*10**8;          #speed of light(m/s)\n",
+      "\n",
+      "#Calculation\n",
+      "Eo=E;\n",
+      "m=mo*(1+(E/Eo));       #mass(kg)\n",
+      "b=math.sqrt(1-(mo/m)**2);\n",
+      "v=b*c;            #velocity(m/s)\n",
+      "B=mo*v/(e*R);     #flux density(Wb/m^2)   \n",
+      "\n",
+      "#Result\n",
+      "print \"mass is\",round(m/1e-31,1),\"*10^-31 kg\"\n",
+      "print \"velocity is\",round(v/1e+8,1),\"*10^8 m/s\"\n",
+      "print \"flux density is\",round(B,5),\"Wb/m^2\"\n",
+      "print \"answer for flux density in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass is 18.2 *10^-31 kg\n",
+        "velocity is 2.6 *10^8 m/s\n",
+        "flux density is 0.00987 Wb/m^2\n",
+        "answer for flux density in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 16.7, Page number 308"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "E=4;       #applied voltage(MeV)\n",
+      "m=3.334*10**-27;      #mass of deuteron(kg)\n",
+      "R=0.75;               #radius(m)\n",
+      "e=1.6*10**-19;        #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "E=4*10**6*e;\n",
+      "fo=math.sqrt(E/(2*m))/(math.pi*R);     #frequnecy of generator(Hz)\n",
+      "\n",
+      "#Result\n",
+      "print \"frequnecy of generator is\",round(fo/1e+6,3),\"*10^6 Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "frequnecy of generator is 4.158 *10^6 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 16.8, Page number 308"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "roi=15;      #rate of increase(Wb/s)\n",
+      "tr=10**6;    #total revolutions\n",
+      "\n",
+      "#Calculation\n",
+      "IE=roi*10**-6;      #increased energy(MeV)\n",
+      "FE=IE*tr;           #Final Energy(MeV) \n",
+      "\n",
+      "#Result\n",
+      "print \"Final Energy is\",FE,\"MeV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Final Energy is 15.0 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 16.9, Page number 308"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "R=0.1;       #radius(m)\n",
+      "h=6.625*10**-34;        #Plank's constant\n",
+      "c=3*10**8;              #speed of light(m/s)\n",
+      "roi=15;                 #rate of increase(Wb/s)\n",
+      "t=4*10**-4;             #period of accerleartion(s)\n",
+      "e=1.6*10**-19;          #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "N=c*t/(2*math.pi*R);      #number of revolutions\n",
+      "IE=roi;              #incresed energy(eV)\n",
+      "ME=N*IE*10**-6;      #maximum energy(MeV)  \n",
+      "ME1=ME*10**6*e;       #conversion in V\n",
+      "p=ME1/c;\n",
+      "gama=h/p;         #wavelength of gama rays(m)\n",
+      "\n",
+      "#Result\n",
+      "print \"Maximum energy is\",round(ME,3),\"MeV\"\n",
+      "print \"Corresponding wavelength of gama rays is\",round(gama/1e-13,3),\"*10^-13 m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum energy is 2.865 MeV\n",
+        "Corresponding wavelength of gama rays is 4.336 *10^-13 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter1_1.ipynb b/Modern_Physics/chapter1_1.ipynb
new file mode 100755
index 00000000..25ce0403
--- /dev/null
+++ b/Modern_Physics/chapter1_1.ipynb
@@ -0,0 +1,383 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:70cab6f9b725623fc2451b245c5190bb3397c02f8debec03cc8b79cdd3e4b714"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "1: Electric and Magnetic Fields"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.1, Page number 4"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "q1=3.2*10**-19;\n",
+      "q2=q1;                #q1 and q2 are the values of charge on alpha-particle(C)\n",
+      "d=10**-13;            #distance between two alpha-particles(m)\n",
+      "m1=6.68*10**-27;\n",
+      "m2=m1;          #m1 and m2 are masses of alpha-particles(kg)\n",
+      "G=6.67*10**-11;       #Gravitational constant(Nm^2/kg^2)\n",
+      "\n",
+      "#Calculation\n",
+      "F1=(9*10**9)*(q1*q2)/(d**2);       #calculation of electrostatic force(N)\n",
+      "F2=G*(m1*m2)/(d**2);        #calculation of electrostatic force(N)\n",
+      "F1=math.ceil(F1*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "F1 = F1*10**2;\n",
+      "\n",
+      "#Result\n",
+      "print \"The electrosatic force is\",F1,\"*10**-2 N\"\n",
+      "print \"The gravitational force is\",round(F2/1e-37,3),\"*10**-37 N\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The electrosatic force is 9.22 *10**-2 N\n",
+        "The gravitational force is 2.976 *10**-37 N\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.2, Page number 4"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "m=9.1*10**-31;       #mass of elctron(kg)\n",
+      "q=1.6*10**-19;       #charge on electron(C)\n",
+      "g=9.81;           #acceleration due to gravity(m/s^2)\n",
+      "\n",
+      "#Calculation\n",
+      "Fg=m*g;                #gravitational force(N)\n",
+      "d=math.sqrt((9*10**9*q**2)/Fg);            #equating gravitational force with electrosatic force(m)\n",
+      "d=math.ceil(d*10**3)/10**3;            #rounding off to 4 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"The distance of separation is\",d,\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The distance of separation is 5.081 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.3, Page number 4"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "d=0.02;          #distance between plates(m)\n",
+      "V=400;           #potential differnce of plates(V)\n",
+      "q=1.6*10**-19;         #charge on a proton(C)\n",
+      "\n",
+      "#Calculation\n",
+      "E=V/d;       #electric field intensity between plates(V/m)\n",
+      "F=q*E;       #electrostatic force on oil drop(N)\n",
+      "\n",
+      "#Result\n",
+      "print \"The electric field intensity between plates is\",E,\"V/m\"\n",
+      "print \"The force on proton is\",F,\"N\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The electric field intensity between plates is 20000.0 V/m\n",
+        "The force on proton is 3.2e-15 N\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.4, Page number 4"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "d=0.02;           #distance between plates(m)\n",
+      "q=1.6*10**-19;       #charge on oil drop(C)\n",
+      "V=6000;              #potential differnce of plates(V)\n",
+      "g=9.81;              #acceleration due to gravity(m/s^2)\n",
+      "\n",
+      "#Calculation\n",
+      "E=V/d;       #electric field intensity between plates(V/m)\n",
+      "F=q*E;       #electrostatic force on oil drop(N)\n",
+      "m=F/g;       #equating the weight of oil drop to the electrostatic force on it(kg)\n",
+      "\n",
+      "#Result\n",
+      "print \"The mass of oil drop is\",round(m/1e-15,3),\"*10**-15 kg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mass of oil drop is 4.893 *10**-15 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.5, Page number 5"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=150;             #potential difference between anode and cathode(V)\n",
+      "m=9.31*10**-31;    #mass of an electron(kg)\n",
+      "q=1.6*10**-19;     #charge on an electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "E=q*V;            #energy gained by electron during speeding from cathode to anode(J)\n",
+      "vel=math.sqrt(E*2/m);    #equating with kinetic energy of electron(m/s)\n",
+      "vel=vel*10**-6;\n",
+      "vel=math.ceil(vel*10)/10;   #rounding off to 1 decimal\n",
+      "\n",
+      "#Result\n",
+      "print \"The velocity is\",vel,\"*10**6 m/s\"\n",
+      "print \"answer in the book is wrong by 1 decimal\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The velocity is 7.2 *10**6 m/s\n",
+        "answer in the book is wrong by 1 decimal\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.6, Page number 5"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=5*10**6;       #potential differnce through which alpha-particle is accelerated(V)\n",
+      "e=1.6*10**-19;         #charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "E1=2*V;      #electronic charge on alpha-particle(eV)\n",
+      "E2=E1/10**6;      #energy(MeV)\n",
+      "E3=E1*e;         #energy(J)\n",
+      "E1=E1*10**-7; \n",
+      "\n",
+      "#Result\n",
+      "print \"The energy is\",E1,\"*10**7 eV\"\n",
+      "print \"The energy is\",E2,\"MeV\"\n",
+      "print \"The energy is\",E3,\"J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The energy is 1.0 *10**7 eV\n",
+        "The energy is 10.0 MeV\n",
+        "The energy is 1.6e-12 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.7, Page number 6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "r=0.528*10**-10;       #radius of the orbit(m)\n",
+      "q=-1.6*10**-19;        #charge on electron(C)\n",
+      "Q=1.6*10**-19;         #charge on Hydrogen nucleus(C)\n",
+      "Eo=8.854*10**-12;      #permittivity in free space(F/m)\n",
+      "\n",
+      "#Calculation\n",
+      "E=(q*Q)/(8*3.14*Eo*r);       #electric field intensity between plates(V/m)\n",
+      "E1=E/(1.6*10**-19);          #electrifeild intensity(eV)\n",
+      "E=E*10**19;\n",
+      "E=math.ceil(E*10**2)/10**2;     #rounding off to 2 decimals\n",
+      "E1=math.ceil(E1*10**2)/10**2;     #rounding off to 2 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"The total energy is\",E,\"*10**-19 J\"\n",
+      "print \"The total energy is\",E1,\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total energy is -21.79 *10**-19 J\n",
+        "The total energy is -13.62 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 1.8, Page number 9"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Q=3.2*10**-19;     #charge on alpha-particle(C)\n",
+      "m=6.68*10**-27;    #mass on alpha-particle(kg)\n",
+      "B=1.5;          #transverse magnetic field of flux density(Wb/m^2)\n",
+      "v=5*10**6;      #velocity of alpha-particle(m/s)\n",
+      "\n",
+      "#Calculation\n",
+      "F=B*Q*v;        #electrostatic force on oil drop(N)\n",
+      "R=m*v/(Q*B);    #radius(m)\n",
+      "R=math.ceil(R*10**2)/10**2;   #rounding off to 2 decimals\n",
+      "R1 = R*100;              #radius(cm)\n",
+      "\n",
+      "#Result\n",
+      "print \"The force on particle is\",F,\"N\"\n",
+      "print \"The radius of its circular path\",R,\"m or\",R1,\"cm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The force on particle is 2.4e-12 N\n",
+        "The radius of its circular path 0.07 m or 7.0 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter2.ipynb b/Modern_Physics/chapter2.ipynb
new file mode 100755
index 00000000..c02388c1
--- /dev/null
+++ b/Modern_Physics/chapter2.ipynb
@@ -0,0 +1,215 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:8d579c32fbc82bfc0d050265719851f3d1a9f6f9fcc17df43ae5b1c38cc41c26"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "2: The Electron"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.1, Page number 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "E=2400;         #electric field intensity(V/m)\n",
+      "V=90;           #potential difference(V)\n",
+      "e=1.6*10**-19;       #the charge on electron(C)\n",
+      "m=9.12*10**-31;      #mass of electron(kg)\n",
+      "\n",
+      "#Calculation\n",
+      "F=e*E;      #force on electron(N)\n",
+      "a=F/m;      #acceleration(m/s^2)  \n",
+      "KE=e*V;     #Kinetic Energy of particle(J) \n",
+      "v=math.sqrt(2*KE/m);       #velocity of the electron(m/s)\n",
+      "v=v*10**-6;\n",
+      "v=math.ceil(v*10**2)/10**2;   #rounding off to 2 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"The force on electron is\",F,\"N\"\n",
+      "print \"Its acceleration is\",round(a/1e+14,2),\"*10**14 m/s^2\"\n",
+      "print \"The Kinetic Energy of particle is\",KE,\"J\"\n",
+      "print \"The velocity of the electron\",v,\"*10**6 m/s\"\n",
+      "print \"answers for acceleration and velocity given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The force on electron is 3.84e-16 N\n",
+        "Its acceleration is 4.21 *10**14 m/s^2\n",
+        "The Kinetic Energy of particle is 1.44e-17 J\n",
+        "The velocity of the electron 5.62 *10**6 m/s\n",
+        "answers for acceleration and velocity given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.2, Page number 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=900;      #potential difference(V)\n",
+      "B=0.01;       #uniform magnetic field(Wb/m^2)\n",
+      "em=1.76*10**11;       #value of e/m(C/kg)\n",
+      "\n",
+      "#calculation\n",
+      "v=math.sqrt(2*em*V);        #linear velocity of electron(m/s)\n",
+      "R=v/(em*B);             #radius of the circular path(m) \n",
+      "R=math.ceil(R*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "v=v*10**-7;\n",
+      "v=math.ceil(v*10**2)/10**2;   #rounding off to 2 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"The linear velocity of electron is\",v,\"*10**7 m/s\"\n",
+      "print \"The radius of the circular path is\",R,\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The linear velocity of electron is 1.78 *10**7 m/s\n",
+        "The radius of the circular path is 0.011 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.3, Page number 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "d=6*10**-3;       #distance between plates(m)\n",
+      "V=900;          #potential difference(V)\n",
+      "B=0.5;          #uniform magnetic field(Wb/m^2)\n",
+      "Q=1.6*10**-19;      #the charge on electron(C)\n",
+      "R=10.6*10**-2;      #circular track radius(m)\n",
+      "\n",
+      "#calculation\n",
+      "v=V/(B*d);       #velocity(m/s)\n",
+      "m=R*Q*B/v;       #mass of particle(kg)\n",
+      "\n",
+      "#Result\n",
+      "print \"The mass of particle\",round(m/1e-26,3),\"*10**-26 kg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mass of particle 2.827 *10**-26 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.4, Page number 19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=6920;        #potential difference(V)\n",
+      "d=1.3*10**-2;      #distance between(m)\n",
+      "v=1.9*10**-4;      #velocity(m/s)\n",
+      "p=0.9*10**3;       #density of oil(kg/m^3)\n",
+      "n=1.81*10**-5;     #coefficient of viscosity(N-s/m^2)\n",
+      "g=9.81;          #accelaration due to gravity(m/s^2)\n",
+      "\n",
+      "#calculation\n",
+      "a=math.sqrt((9*n*v)/(2*g*p));      #radius of the drop(m) \n",
+      "E=V/d;         #electric field(V/m)\n",
+      "Q=4*math.pi*(a**3)*p*g/(3*E);      #value of charge on oil drop(C)\n",
+      "\n",
+      "#Result\n",
+      "print \"The radius of the drop is\",round(a/1e-6,2),\"micro m\"\n",
+      "print \"The value of charge on oil drop is\",round(Q/1e-19,3),\"*10^-19 C\"\n",
+      "print \"answers given in the book are wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The radius of the drop is 1.32 micro m\n",
+        "The value of charge on oil drop is 1.612 *10^-19 C\n",
+        "answers given in the book are wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter2_1.ipynb b/Modern_Physics/chapter2_1.ipynb
new file mode 100755
index 00000000..c02388c1
--- /dev/null
+++ b/Modern_Physics/chapter2_1.ipynb
@@ -0,0 +1,215 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:8d579c32fbc82bfc0d050265719851f3d1a9f6f9fcc17df43ae5b1c38cc41c26"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "2: The Electron"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.1, Page number 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "E=2400;         #electric field intensity(V/m)\n",
+      "V=90;           #potential difference(V)\n",
+      "e=1.6*10**-19;       #the charge on electron(C)\n",
+      "m=9.12*10**-31;      #mass of electron(kg)\n",
+      "\n",
+      "#Calculation\n",
+      "F=e*E;      #force on electron(N)\n",
+      "a=F/m;      #acceleration(m/s^2)  \n",
+      "KE=e*V;     #Kinetic Energy of particle(J) \n",
+      "v=math.sqrt(2*KE/m);       #velocity of the electron(m/s)\n",
+      "v=v*10**-6;\n",
+      "v=math.ceil(v*10**2)/10**2;   #rounding off to 2 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"The force on electron is\",F,\"N\"\n",
+      "print \"Its acceleration is\",round(a/1e+14,2),\"*10**14 m/s^2\"\n",
+      "print \"The Kinetic Energy of particle is\",KE,\"J\"\n",
+      "print \"The velocity of the electron\",v,\"*10**6 m/s\"\n",
+      "print \"answers for acceleration and velocity given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The force on electron is 3.84e-16 N\n",
+        "Its acceleration is 4.21 *10**14 m/s^2\n",
+        "The Kinetic Energy of particle is 1.44e-17 J\n",
+        "The velocity of the electron 5.62 *10**6 m/s\n",
+        "answers for acceleration and velocity given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.2, Page number 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=900;      #potential difference(V)\n",
+      "B=0.01;       #uniform magnetic field(Wb/m^2)\n",
+      "em=1.76*10**11;       #value of e/m(C/kg)\n",
+      "\n",
+      "#calculation\n",
+      "v=math.sqrt(2*em*V);        #linear velocity of electron(m/s)\n",
+      "R=v/(em*B);             #radius of the circular path(m) \n",
+      "R=math.ceil(R*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "v=v*10**-7;\n",
+      "v=math.ceil(v*10**2)/10**2;   #rounding off to 2 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"The linear velocity of electron is\",v,\"*10**7 m/s\"\n",
+      "print \"The radius of the circular path is\",R,\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The linear velocity of electron is 1.78 *10**7 m/s\n",
+        "The radius of the circular path is 0.011 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.3, Page number 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "d=6*10**-3;       #distance between plates(m)\n",
+      "V=900;          #potential difference(V)\n",
+      "B=0.5;          #uniform magnetic field(Wb/m^2)\n",
+      "Q=1.6*10**-19;      #the charge on electron(C)\n",
+      "R=10.6*10**-2;      #circular track radius(m)\n",
+      "\n",
+      "#calculation\n",
+      "v=V/(B*d);       #velocity(m/s)\n",
+      "m=R*Q*B/v;       #mass of particle(kg)\n",
+      "\n",
+      "#Result\n",
+      "print \"The mass of particle\",round(m/1e-26,3),\"*10**-26 kg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mass of particle 2.827 *10**-26 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 2.4, Page number 19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=6920;        #potential difference(V)\n",
+      "d=1.3*10**-2;      #distance between(m)\n",
+      "v=1.9*10**-4;      #velocity(m/s)\n",
+      "p=0.9*10**3;       #density of oil(kg/m^3)\n",
+      "n=1.81*10**-5;     #coefficient of viscosity(N-s/m^2)\n",
+      "g=9.81;          #accelaration due to gravity(m/s^2)\n",
+      "\n",
+      "#calculation\n",
+      "a=math.sqrt((9*n*v)/(2*g*p));      #radius of the drop(m) \n",
+      "E=V/d;         #electric field(V/m)\n",
+      "Q=4*math.pi*(a**3)*p*g/(3*E);      #value of charge on oil drop(C)\n",
+      "\n",
+      "#Result\n",
+      "print \"The radius of the drop is\",round(a/1e-6,2),\"micro m\"\n",
+      "print \"The value of charge on oil drop is\",round(Q/1e-19,3),\"*10^-19 C\"\n",
+      "print \"answers given in the book are wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The radius of the drop is 1.32 micro m\n",
+        "The value of charge on oil drop is 1.612 *10^-19 C\n",
+        "answers given in the book are wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter3.ipynb b/Modern_Physics/chapter3.ipynb
new file mode 100755
index 00000000..1404e10b
--- /dev/null
+++ b/Modern_Physics/chapter3.ipynb
@@ -0,0 +1,351 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:90fc77a4706b2ba6c4e383a2e0d80f0a572a43d837aa619ea730d827f91d4409"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "3: The Atomic Structure"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.1, Page number 25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Z=79;         #atomic number of gold\n",
+      "e=1.6*10**-19;       #electron charge(C)\n",
+      "Eo=8.854*10**-12;     #absolute permitivity of free space(F/m)\n",
+      "K=7.68*1.6*10**-13;    #kinectic energy(J)\n",
+      "\n",
+      "#calculation\n",
+      "D=(2*Z*e**2)/(4*math.pi*Eo*K);        #closest distance of approach(m)\n",
+      "\n",
+      "#Result\n",
+      "print \"The closest distance of approach is\",round(D/1e-14,2),\"*10**-14 m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The closest distance of approach is 2.96 *10**-14 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.2, Page number 28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Z=1;       #atomic number of hydrogen\n",
+      "e=1.6*10**-19;    #electron charge(C)\n",
+      "h=6.625*10**-34;    #plank's constant(J-s)\n",
+      "m=9.1*10**-31;      #mass of an electron(kg)\n",
+      "Eo=8.854*10**-12;        #absolute permitivity of free space(F/m)\n",
+      "c=3*10**8;               #speed of light(m/s)\n",
+      "n=1;                 #ground state\n",
+      "\n",
+      "#calculation\n",
+      "v=9*10**9*(2*math.pi*Z*e**2)/(n*h);         #velocity of ground state(m/s)\n",
+      "r=(Eo*n**2*h**2)/(math.pi*m*e**2);          #radius of Bohr orbit in ground state(m)\n",
+      "t=(2*math.pi*r)/v;        #time taken by electron to traverse the bohr first orbit(s)\n",
+      "R=(m*(e**4))/(8*(Eo**2)*(h**3)*c);       #Rhydberg contstant(m^-1)\n",
+      "#v=v*10**-5;\n",
+      "#v=math.ceil(v*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "#r=r*10**10;\n",
+      "#R=R/10**6;\n",
+      "\n",
+      "#Result\n",
+      "print \"velocity of ground state\",round(v/1e+5,2),\"*10^5 m/s\"\n",
+      "print \"radius of Bohr orbit in ground state\",round(r/1e-10,2),\"*10^-10 m\"\n",
+      "print \"time taken by electron to traverse the bohr first orbit\",round(t/1e-16,2),\"micro s\"\n",
+      "print \"Rhydberg constant is\",round(R/1e+6,3),\"*10**6 m^-1\"\n",
+      "print \"answer for Rhydberg contstant given in the book differs in the 2nd decimal point\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "velocity of ground state 21.85 *10^5 m/s\n",
+        "radius of Bohr orbit in ground state 0.53 *10^-10 m\n",
+        "time taken by electron to traverse the bohr first orbit 1.53 micro s\n",
+        "Rhydberg constant is 10.901 *10**6 m^-1\n",
+        "answer for Rhydberg contstant given in the book differs in the 2nd decimal point\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.3, Page number 29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "B=2.179*10**-16;      #constant(J)\n",
+      "h=6.6*10**-34;      #plank's constant(J-s)\n",
+      "\n",
+      "#calculation\n",
+      "E3=-B/3**2;       #energy in 3rd orbit(J)\n",
+      "E2=-B/2**2;       #energy in 2nd orbit(J)  \n",
+      "f=(E3-E2)/h;      #frequency of radiation(Hz) \n",
+      "\n",
+      "#Result\n",
+      "print \"frequency of radiation\",round(f/1e+16,1),\"*10**16 Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "frequency of radiation 4.6 *10**16 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.4, Page number 29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Z=1;     #atomic number of hydrogen\n",
+      "e=1.6*10**-19;       #electron charge(C)\n",
+      "h=6.625*10**-34;     #plank's constant(J-s)\n",
+      "m=9.1*10**-31;       #mass of an electron(kg)\n",
+      "Eo=8.854*10**-12;    #absolute permitivity of free space(F/m)\n",
+      "n=1;         #ground state\n",
+      "\n",
+      "#Calculation\n",
+      "f=(m*Z**2*e**4)/(4*Eo**2*h**3);         #frequency(Hz)\n",
+      "\n",
+      "#Result\n",
+      "print \"the frequency is\",round(f/1e+15,2),\"*10**15 Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the frequency is 6.54 *10**15 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.5, Page number 30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Z=1;\n",
+      "n=1;\n",
+      "e=1.6*10**-19;    #the charge on electron(C)\n",
+      "h=6.62*10**-34;   #Plank's constant\n",
+      "Eo=8.854*10**-12;     #absolute permitivity of free space(F/m)\n",
+      "m=9.1*10**-31;     #mass of electron(kg)\n",
+      "\n",
+      "#calculation\n",
+      "v=Z*(e**2)/(2*Eo*n*h);        #velocity(m/s)\n",
+      "E=-m*(Z**2)*(e**4)/(8*(Eo*n*h)**2);      #energy of hydrogen atom(J)\n",
+      "f=m*(Z**2)*(e**4)/(4*(Eo**2)*(n*h)**3);        #frequecy(Hz)\n",
+      "\n",
+      "#Result\n",
+      "print \"velocity is\",round(v*10**-6,2),\"*10**6 m/s\"\n",
+      "print \"energy of hydrogen atom\",round(E*10**19,1),\"*10**-19 J\"\n",
+      "print \"frequecy\",round(f/1e+15,1),\"*10**15 Hz\"\n",
+      "print \"answer for velocity given in the book is wrong\"\n",
+      "print \"answer for frequency given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "velocity is 2.18 *10**6 m/s\n",
+        "energy of hydrogen atom -21.7 *10**-19 J\n",
+        "frequecy 6.6 *10**15 Hz\n",
+        "answer for velocity given in the book is wrong\n",
+        "answer for frequency given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.8, Page number 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "h=6.625*10**-34;     #Plank's constant\n",
+      "c=3*10**8;          #speed of light(m/s)\n",
+      "E1=10.2;           #energy(eV)\n",
+      "E2=12.09;           #energy(eV)\n",
+      "e=1.6*10**-19;        #the charge on electron(C)\n",
+      "\n",
+      "#calcualtion\n",
+      "#principal quantum numbers are 2 & 3 respectively\n",
+      "lamda1=c*h/(E1*e)*10**10;       #wavelength for E1(angstrom)\n",
+      "lamda2=c*h/(E2*e)*10**10;       #wavelength for E2(angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"wavelength for 10.2 eV is\",int(lamda1),\"angstrom\"\n",
+      "print \"wavelength for 12.09 eV is\",int(lamda2),\"angstrom\"\n",
+      "print \"answers given in the book differ due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "wavelength for 10.2 eV is 1217 angstrom\n",
+        "wavelength for 12.09 eV is 1027 angstrom\n",
+        "answers given in the book differ due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 58
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.9, Page number 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "R=10967700;      #Rydberg constant(m^-1)\n",
+      "\n",
+      "#calculation\n",
+      "long_lamda=4/(3*R);     #as n1=1 and n2=2\n",
+      "long_lamda=long_lamda*10**10;           #long wavelength(angstrom)\n",
+      "short_lamda=1/R;         #as n1=1 and n2=infinity\n",
+      "short_lamda=short_lamda*10**10;         #long wavelength(angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"Long wavelength is\",round(long_lamda),\"angstrom\"\n",
+      "print \"Short wavelength is\",round(short_lamda),\"angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Long wavelength is 1216.0 angstrom\n",
+        "Short wavelength is 912.0 angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 62
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter3_1.ipynb b/Modern_Physics/chapter3_1.ipynb
new file mode 100755
index 00000000..1404e10b
--- /dev/null
+++ b/Modern_Physics/chapter3_1.ipynb
@@ -0,0 +1,351 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:90fc77a4706b2ba6c4e383a2e0d80f0a572a43d837aa619ea730d827f91d4409"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "3: The Atomic Structure"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.1, Page number 25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Z=79;         #atomic number of gold\n",
+      "e=1.6*10**-19;       #electron charge(C)\n",
+      "Eo=8.854*10**-12;     #absolute permitivity of free space(F/m)\n",
+      "K=7.68*1.6*10**-13;    #kinectic energy(J)\n",
+      "\n",
+      "#calculation\n",
+      "D=(2*Z*e**2)/(4*math.pi*Eo*K);        #closest distance of approach(m)\n",
+      "\n",
+      "#Result\n",
+      "print \"The closest distance of approach is\",round(D/1e-14,2),\"*10**-14 m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The closest distance of approach is 2.96 *10**-14 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.2, Page number 28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Z=1;       #atomic number of hydrogen\n",
+      "e=1.6*10**-19;    #electron charge(C)\n",
+      "h=6.625*10**-34;    #plank's constant(J-s)\n",
+      "m=9.1*10**-31;      #mass of an electron(kg)\n",
+      "Eo=8.854*10**-12;        #absolute permitivity of free space(F/m)\n",
+      "c=3*10**8;               #speed of light(m/s)\n",
+      "n=1;                 #ground state\n",
+      "\n",
+      "#calculation\n",
+      "v=9*10**9*(2*math.pi*Z*e**2)/(n*h);         #velocity of ground state(m/s)\n",
+      "r=(Eo*n**2*h**2)/(math.pi*m*e**2);          #radius of Bohr orbit in ground state(m)\n",
+      "t=(2*math.pi*r)/v;        #time taken by electron to traverse the bohr first orbit(s)\n",
+      "R=(m*(e**4))/(8*(Eo**2)*(h**3)*c);       #Rhydberg contstant(m^-1)\n",
+      "#v=v*10**-5;\n",
+      "#v=math.ceil(v*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "#r=r*10**10;\n",
+      "#R=R/10**6;\n",
+      "\n",
+      "#Result\n",
+      "print \"velocity of ground state\",round(v/1e+5,2),\"*10^5 m/s\"\n",
+      "print \"radius of Bohr orbit in ground state\",round(r/1e-10,2),\"*10^-10 m\"\n",
+      "print \"time taken by electron to traverse the bohr first orbit\",round(t/1e-16,2),\"micro s\"\n",
+      "print \"Rhydberg constant is\",round(R/1e+6,3),\"*10**6 m^-1\"\n",
+      "print \"answer for Rhydberg contstant given in the book differs in the 2nd decimal point\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "velocity of ground state 21.85 *10^5 m/s\n",
+        "radius of Bohr orbit in ground state 0.53 *10^-10 m\n",
+        "time taken by electron to traverse the bohr first orbit 1.53 micro s\n",
+        "Rhydberg constant is 10.901 *10**6 m^-1\n",
+        "answer for Rhydberg contstant given in the book differs in the 2nd decimal point\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.3, Page number 29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "B=2.179*10**-16;      #constant(J)\n",
+      "h=6.6*10**-34;      #plank's constant(J-s)\n",
+      "\n",
+      "#calculation\n",
+      "E3=-B/3**2;       #energy in 3rd orbit(J)\n",
+      "E2=-B/2**2;       #energy in 2nd orbit(J)  \n",
+      "f=(E3-E2)/h;      #frequency of radiation(Hz) \n",
+      "\n",
+      "#Result\n",
+      "print \"frequency of radiation\",round(f/1e+16,1),\"*10**16 Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "frequency of radiation 4.6 *10**16 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.4, Page number 29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Z=1;     #atomic number of hydrogen\n",
+      "e=1.6*10**-19;       #electron charge(C)\n",
+      "h=6.625*10**-34;     #plank's constant(J-s)\n",
+      "m=9.1*10**-31;       #mass of an electron(kg)\n",
+      "Eo=8.854*10**-12;    #absolute permitivity of free space(F/m)\n",
+      "n=1;         #ground state\n",
+      "\n",
+      "#Calculation\n",
+      "f=(m*Z**2*e**4)/(4*Eo**2*h**3);         #frequency(Hz)\n",
+      "\n",
+      "#Result\n",
+      "print \"the frequency is\",round(f/1e+15,2),\"*10**15 Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the frequency is 6.54 *10**15 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.5, Page number 30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Z=1;\n",
+      "n=1;\n",
+      "e=1.6*10**-19;    #the charge on electron(C)\n",
+      "h=6.62*10**-34;   #Plank's constant\n",
+      "Eo=8.854*10**-12;     #absolute permitivity of free space(F/m)\n",
+      "m=9.1*10**-31;     #mass of electron(kg)\n",
+      "\n",
+      "#calculation\n",
+      "v=Z*(e**2)/(2*Eo*n*h);        #velocity(m/s)\n",
+      "E=-m*(Z**2)*(e**4)/(8*(Eo*n*h)**2);      #energy of hydrogen atom(J)\n",
+      "f=m*(Z**2)*(e**4)/(4*(Eo**2)*(n*h)**3);        #frequecy(Hz)\n",
+      "\n",
+      "#Result\n",
+      "print \"velocity is\",round(v*10**-6,2),\"*10**6 m/s\"\n",
+      "print \"energy of hydrogen atom\",round(E*10**19,1),\"*10**-19 J\"\n",
+      "print \"frequecy\",round(f/1e+15,1),\"*10**15 Hz\"\n",
+      "print \"answer for velocity given in the book is wrong\"\n",
+      "print \"answer for frequency given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "velocity is 2.18 *10**6 m/s\n",
+        "energy of hydrogen atom -21.7 *10**-19 J\n",
+        "frequecy 6.6 *10**15 Hz\n",
+        "answer for velocity given in the book is wrong\n",
+        "answer for frequency given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.8, Page number 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "h=6.625*10**-34;     #Plank's constant\n",
+      "c=3*10**8;          #speed of light(m/s)\n",
+      "E1=10.2;           #energy(eV)\n",
+      "E2=12.09;           #energy(eV)\n",
+      "e=1.6*10**-19;        #the charge on electron(C)\n",
+      "\n",
+      "#calcualtion\n",
+      "#principal quantum numbers are 2 & 3 respectively\n",
+      "lamda1=c*h/(E1*e)*10**10;       #wavelength for E1(angstrom)\n",
+      "lamda2=c*h/(E2*e)*10**10;       #wavelength for E2(angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"wavelength for 10.2 eV is\",int(lamda1),\"angstrom\"\n",
+      "print \"wavelength for 12.09 eV is\",int(lamda2),\"angstrom\"\n",
+      "print \"answers given in the book differ due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "wavelength for 10.2 eV is 1217 angstrom\n",
+        "wavelength for 12.09 eV is 1027 angstrom\n",
+        "answers given in the book differ due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 58
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 3.9, Page number 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "R=10967700;      #Rydberg constant(m^-1)\n",
+      "\n",
+      "#calculation\n",
+      "long_lamda=4/(3*R);     #as n1=1 and n2=2\n",
+      "long_lamda=long_lamda*10**10;           #long wavelength(angstrom)\n",
+      "short_lamda=1/R;         #as n1=1 and n2=infinity\n",
+      "short_lamda=short_lamda*10**10;         #long wavelength(angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"Long wavelength is\",round(long_lamda),\"angstrom\"\n",
+      "print \"Short wavelength is\",round(short_lamda),\"angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Long wavelength is 1216.0 angstrom\n",
+        "Short wavelength is 912.0 angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 62
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter4.ipynb b/Modern_Physics/chapter4.ipynb
new file mode 100755
index 00000000..9b118dda
--- /dev/null
+++ b/Modern_Physics/chapter4.ipynb
@@ -0,0 +1,320 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:f3a8bba44ff010c0f1004529fad8de811f19168b5752929eb1e0272cfc5fc53c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "4: Crystallography"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.2, Page number 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "d=2180;      #density of NaCl(kg/m^3)\n",
+      "M=23+35.5;      #Molecular weight of NaCl(gm)\n",
+      "Na=6.02*10**26;         #Avgraodo no(per kg mole)\n",
+      "n=4;        #for f.c.c\n",
+      "\n",
+      "#calculation\n",
+      "a=(n*M/(Na*d))**(1/3);       #lattice constant(m)\n",
+      "d=a/2;               #distance(m)\n",
+      "d=d*10**10;        #distance(angstrom)\n",
+      "d=math.ceil(d*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"distance between two adajcent atoms is\",d,\"angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "distance between two adajcent atoms is 2.815 angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.3, Page number 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "d=2.163;      #density(gm/cm^3)\n",
+      "M=58.45;      #molecular weight(gm)\n",
+      "Na=6.02*10**23;      #Avgraodo no.(molecules/gm mole)\n",
+      "\n",
+      "#calcualtion\n",
+      "n=Na/M;      #no. of molecules(per gram)\n",
+      "n=n*d;       #no. of molecules(per cm^3) \n",
+      "n=2*n;       #no. of atom(per cm^3)\n",
+      "n=n**(1/3);        #no. of atoms in a row 1cm long\n",
+      "d=1/n;           #spacing between atoms(cm)\n",
+      "\n",
+      "#Result\n",
+      "print \"spacing between atoms is\",round(d/1e-8,2),\"angstrom\"\n",
+      "print \"answer in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "spacing between atoms is 2.82 angstrom\n",
+        "answer in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.4, Page number 74"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "r=1.278;     #radius(A.U)\n",
+      "n=4;        #structure is f.c.c\n",
+      "M=63.54;      #atomic weight(gm)\n",
+      "Na=6.02*10**23;       #Avgraodo no.(per gm mole)\n",
+      "\n",
+      "#calculation \n",
+      "a=4*r/(math.sqrt(2));        #lattice constant(A.U)\n",
+      "V=a**3;          #volume(cm^3)\n",
+      "rho=n*M/(Na*V);    #density(gm/cm^3)\n",
+      "rho=rho*(10**8)**3      #density(gm/m^3)\n",
+      "\n",
+      "#Result\n",
+      "print \"Density is\",round(rho,2),\"g/m^3\"\n",
+      "print \"answer in the textbook varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Density is 8.94 g/m^3\n",
+        "answer in the textbook varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.10, Page number 85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Vaiable declaration\n",
+      "r=1.746;    #atomic radius(AU)\n",
+      "\n",
+      "#calulation\n",
+      "a=4*r/math.sqrt(2);         #lattice constant(AU)\n",
+      "#for (200)\n",
+      "h=2;k=0;l=0;\n",
+      "d=a/math.sqrt(h**2+k**2+l**2);          #interplanar spacing(AU)\n",
+      "#for (220)\n",
+      "h=2;k=2;l=0;            \n",
+      "d1=a/math.sqrt(h**2+k**2+l**2);         #interplanar spacing(AU)\n",
+      "#for (111)\n",
+      "h=1;k=1;l=1;\n",
+      "d2=a/math.sqrt(h**2+k**2+l**2);         #interplanar spacing(AU)\n",
+      "d=math.ceil(d*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "d1=math.ceil(d1*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "d2=math.ceil(d2*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"spacing for (200) is\",d,\"A.U\"\n",
+      "print \"answer in the book varies in 3rd decimal due to rounding off errors\"\n",
+      "print \"spacing for (220) is\",d1,\"A.U\"\n",
+      "print \"spacing for (111) is\",d2,\"A.U\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "spacing for (200) is 2.4693 A.U\n",
+        "answer in the book varies in 3rd decimal due to rounding off errors\n",
+        "spacing for (220) is 1.746 A.U\n",
+        "spacing for (111) is 2.852 A.U\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.11, Page number 86"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#calculation\n",
+      "#for (i)\n",
+      "l=1;m=0;n=0;\n",
+      "p=0;q=1;r=0;\n",
+      "d=math.acos(((l*p)+(m*q)+(n*r))/(math.sqrt(l**2+m**2+n**2)*math.sqrt(p**2+q**2+r**2)));      #angle between 2 normals(radian)\n",
+      "theta=math.degrees(d);         #angle between 2 normals(degrees) \n",
+      "\n",
+      "#for (ii)\n",
+      "l=1;m=2;n=1;\n",
+      "p=1;q=1;r=1;\n",
+      "d1=math.acos(((l*p)+(m*q)+(n*r))/(math.sqrt(l**2+m**2+n**2)*math.sqrt(p**2+q**2+r**2)));       #angle between 2 normals(radian)\n",
+      "theta1=math.degrees(d1);              #angle between 2 normals(degrees)  \n",
+      "deg=int(theta1);             #angle(degrees)\n",
+      "t=60*(theta1-deg);\n",
+      "m=int(t);                   #angle(minutes)\n",
+      " \n",
+      "#Result\n",
+      "print \"angle between the normal to pair of miller incdices (100) and (010) is\",theta,\"degrees\"\n",
+      "print \"angle between the normal to pair of miller incdices (121) and (111) is\",deg,\"degrees\",m,\"minutes\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "angle between the normal to pair of miller incdices (100) and (010) is 90.0 degrees\n",
+        "angle between the normal to pair of miller incdices (121) and (111) is 19 degrees 28 minutes\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.13, Page number 87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#variable declaration\n",
+      "a=3.61*10**-7;      #lattice constant(mm)\n",
+      "\n",
+      "#Calcualtion\n",
+      "#for plane (100)\n",
+      "SA=a*a;      #surface area(mm^2)\n",
+      "tamc=2;     #total atoms included\n",
+      "ans=tamc/SA;        #number of atoms per mm^2\n",
+      "#for (ii) plane (110)\n",
+      "A=a*(math.sqrt(2)*a);      #area of the plane(mm^2)\n",
+      "tamc=2;     #total atoms included according to sketch\n",
+      "ans1=tamc/A;       #number of atoms per mm^2\n",
+      "#for (iii) plane (111)\n",
+      "A=0.866*a*a;      #area of the plane(mm^2)\n",
+      "tamc=2;         #total atoms included according to sketch\n",
+      "ans2=tamc/A;      #number of atoms per mm^2\n",
+      "\n",
+      "#Result\n",
+      "print \"atoms per mm^2 for (100) is\",round(ans/1e+13,3),\"*10**13 atoms/mm^2\"\n",
+      "print \"atoms per mm^2 for (110) is\",round(ans1/1e+13,3),\"*10**13 atoms/mm^2\"\n",
+      "print \"atoms per mm^2 for (111) is\",round(ans2/1e+13,3),\"*10**13 atoms/mm^2\"\n",
+      "print \"answer for plane (111) given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "atoms per mm^2 for (100) is 1.535 *10**13 atoms/mm^2\n",
+        "atoms per mm^2 for (110) is 1.085 *10**13 atoms/mm^2\n",
+        "atoms per mm^2 for (111) is 1.772 *10**13 atoms/mm^2\n",
+        "answer for plane (111) given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter4_1.ipynb b/Modern_Physics/chapter4_1.ipynb
new file mode 100755
index 00000000..9b118dda
--- /dev/null
+++ b/Modern_Physics/chapter4_1.ipynb
@@ -0,0 +1,320 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:f3a8bba44ff010c0f1004529fad8de811f19168b5752929eb1e0272cfc5fc53c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "4: Crystallography"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.2, Page number 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "d=2180;      #density of NaCl(kg/m^3)\n",
+      "M=23+35.5;      #Molecular weight of NaCl(gm)\n",
+      "Na=6.02*10**26;         #Avgraodo no(per kg mole)\n",
+      "n=4;        #for f.c.c\n",
+      "\n",
+      "#calculation\n",
+      "a=(n*M/(Na*d))**(1/3);       #lattice constant(m)\n",
+      "d=a/2;               #distance(m)\n",
+      "d=d*10**10;        #distance(angstrom)\n",
+      "d=math.ceil(d*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"distance between two adajcent atoms is\",d,\"angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "distance between two adajcent atoms is 2.815 angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.3, Page number 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "d=2.163;      #density(gm/cm^3)\n",
+      "M=58.45;      #molecular weight(gm)\n",
+      "Na=6.02*10**23;      #Avgraodo no.(molecules/gm mole)\n",
+      "\n",
+      "#calcualtion\n",
+      "n=Na/M;      #no. of molecules(per gram)\n",
+      "n=n*d;       #no. of molecules(per cm^3) \n",
+      "n=2*n;       #no. of atom(per cm^3)\n",
+      "n=n**(1/3);        #no. of atoms in a row 1cm long\n",
+      "d=1/n;           #spacing between atoms(cm)\n",
+      "\n",
+      "#Result\n",
+      "print \"spacing between atoms is\",round(d/1e-8,2),\"angstrom\"\n",
+      "print \"answer in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "spacing between atoms is 2.82 angstrom\n",
+        "answer in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.4, Page number 74"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "r=1.278;     #radius(A.U)\n",
+      "n=4;        #structure is f.c.c\n",
+      "M=63.54;      #atomic weight(gm)\n",
+      "Na=6.02*10**23;       #Avgraodo no.(per gm mole)\n",
+      "\n",
+      "#calculation \n",
+      "a=4*r/(math.sqrt(2));        #lattice constant(A.U)\n",
+      "V=a**3;          #volume(cm^3)\n",
+      "rho=n*M/(Na*V);    #density(gm/cm^3)\n",
+      "rho=rho*(10**8)**3      #density(gm/m^3)\n",
+      "\n",
+      "#Result\n",
+      "print \"Density is\",round(rho,2),\"g/m^3\"\n",
+      "print \"answer in the textbook varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Density is 8.94 g/m^3\n",
+        "answer in the textbook varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.10, Page number 85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Vaiable declaration\n",
+      "r=1.746;    #atomic radius(AU)\n",
+      "\n",
+      "#calulation\n",
+      "a=4*r/math.sqrt(2);         #lattice constant(AU)\n",
+      "#for (200)\n",
+      "h=2;k=0;l=0;\n",
+      "d=a/math.sqrt(h**2+k**2+l**2);          #interplanar spacing(AU)\n",
+      "#for (220)\n",
+      "h=2;k=2;l=0;            \n",
+      "d1=a/math.sqrt(h**2+k**2+l**2);         #interplanar spacing(AU)\n",
+      "#for (111)\n",
+      "h=1;k=1;l=1;\n",
+      "d2=a/math.sqrt(h**2+k**2+l**2);         #interplanar spacing(AU)\n",
+      "d=math.ceil(d*10**4)/10**4;   #rounding off to 4 decimals\n",
+      "d1=math.ceil(d1*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "d2=math.ceil(d2*10**3)/10**3;   #rounding off to 3 decimals\n",
+      "\n",
+      "#Result\n",
+      "print \"spacing for (200) is\",d,\"A.U\"\n",
+      "print \"answer in the book varies in 3rd decimal due to rounding off errors\"\n",
+      "print \"spacing for (220) is\",d1,\"A.U\"\n",
+      "print \"spacing for (111) is\",d2,\"A.U\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "spacing for (200) is 2.4693 A.U\n",
+        "answer in the book varies in 3rd decimal due to rounding off errors\n",
+        "spacing for (220) is 1.746 A.U\n",
+        "spacing for (111) is 2.852 A.U\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.11, Page number 86"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#calculation\n",
+      "#for (i)\n",
+      "l=1;m=0;n=0;\n",
+      "p=0;q=1;r=0;\n",
+      "d=math.acos(((l*p)+(m*q)+(n*r))/(math.sqrt(l**2+m**2+n**2)*math.sqrt(p**2+q**2+r**2)));      #angle between 2 normals(radian)\n",
+      "theta=math.degrees(d);         #angle between 2 normals(degrees) \n",
+      "\n",
+      "#for (ii)\n",
+      "l=1;m=2;n=1;\n",
+      "p=1;q=1;r=1;\n",
+      "d1=math.acos(((l*p)+(m*q)+(n*r))/(math.sqrt(l**2+m**2+n**2)*math.sqrt(p**2+q**2+r**2)));       #angle between 2 normals(radian)\n",
+      "theta1=math.degrees(d1);              #angle between 2 normals(degrees)  \n",
+      "deg=int(theta1);             #angle(degrees)\n",
+      "t=60*(theta1-deg);\n",
+      "m=int(t);                   #angle(minutes)\n",
+      " \n",
+      "#Result\n",
+      "print \"angle between the normal to pair of miller incdices (100) and (010) is\",theta,\"degrees\"\n",
+      "print \"angle between the normal to pair of miller incdices (121) and (111) is\",deg,\"degrees\",m,\"minutes\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "angle between the normal to pair of miller incdices (100) and (010) is 90.0 degrees\n",
+        "angle between the normal to pair of miller incdices (121) and (111) is 19 degrees 28 minutes\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 4.13, Page number 87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#variable declaration\n",
+      "a=3.61*10**-7;      #lattice constant(mm)\n",
+      "\n",
+      "#Calcualtion\n",
+      "#for plane (100)\n",
+      "SA=a*a;      #surface area(mm^2)\n",
+      "tamc=2;     #total atoms included\n",
+      "ans=tamc/SA;        #number of atoms per mm^2\n",
+      "#for (ii) plane (110)\n",
+      "A=a*(math.sqrt(2)*a);      #area of the plane(mm^2)\n",
+      "tamc=2;     #total atoms included according to sketch\n",
+      "ans1=tamc/A;       #number of atoms per mm^2\n",
+      "#for (iii) plane (111)\n",
+      "A=0.866*a*a;      #area of the plane(mm^2)\n",
+      "tamc=2;         #total atoms included according to sketch\n",
+      "ans2=tamc/A;      #number of atoms per mm^2\n",
+      "\n",
+      "#Result\n",
+      "print \"atoms per mm^2 for (100) is\",round(ans/1e+13,3),\"*10**13 atoms/mm^2\"\n",
+      "print \"atoms per mm^2 for (110) is\",round(ans1/1e+13,3),\"*10**13 atoms/mm^2\"\n",
+      "print \"atoms per mm^2 for (111) is\",round(ans2/1e+13,3),\"*10**13 atoms/mm^2\"\n",
+      "print \"answer for plane (111) given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "atoms per mm^2 for (100) is 1.535 *10**13 atoms/mm^2\n",
+        "atoms per mm^2 for (110) is 1.085 *10**13 atoms/mm^2\n",
+        "atoms per mm^2 for (111) is 1.772 *10**13 atoms/mm^2\n",
+        "answer for plane (111) given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter5.ipynb b/Modern_Physics/chapter5.ipynb
new file mode 100755
index 00000000..a4b2fb2a
--- /dev/null
+++ b/Modern_Physics/chapter5.ipynb
@@ -0,0 +1,832 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:9388f9d8764031d08c03b951ac3babfe1cafed91339d0e7cf1d9a5afd45176fa"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "5: Quantum Theory"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.1, Page number 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W1=4;     #wavelength(Angstrom)\n",
+      "W2=1;     #wavelength(Angstrom)\n",
+      "e=1.6*10**-19;   #the charge on electron(C)\n",
+      "m=9.12*10**-31;   #mass of electron(kg)\n",
+      "\n",
+      "#Calculation\n",
+      "E=12400/W1;      #energy(eV)\n",
+      "v=math.sqrt(E*e*2/m);      #velocity(m/s)\n",
+      "E1=12400/W2;       #energy(eV)\n",
+      "v1=math.sqrt(E1*e*2/m);        #velocity(m/s)\n",
+      "\n",
+      "#Result\n",
+      "print \"The energy for 4 angstrom wavelength is\",E,\"eV\"\n",
+      "print \"The velocity is\",round(v/1e+6),\"*10**6 m/s\"\n",
+      "print \"The energy for 1 angstrom wavelength is\",E1,\"eV\"\n",
+      "print \"The velocity is\",round(v1/1e+6),\"*10**6 m/s\"\n",
+      "\n",
+      "#answers given in the book are wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The energy for 4 angstrom wavelength is 3100.0 eV\n",
+        "The velocity is 33.0 *10**6 m/s\n",
+        "The energy for 1 angstrom wavelength is 12400.0 eV\n",
+        "The velocity is 66.0 *10**6 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.2, Page number 98"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "f=880*10**3;      #frequency(Hz)\n",
+      "P=10*10**3;    #Power(W)\n",
+      "h=6.625*10**-34;    #Plank's constant\n",
+      "\n",
+      "#Calculation\n",
+      "E=h*f;    #energy carried by each photon(J)\n",
+      "n=P/E;    #number of photons emitted per second\n",
+      "\n",
+      "#Result\n",
+      "print \"The number of photons emitted per second are\",round(n/1e+30,2),\"*10**30\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number of photons emitted per second are 17.15 *10**30\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.3, Page number 98"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "P=200;     #power(W)\n",
+      "W=6123*10**-10;    #wavelength(m)\n",
+      "c=3*10**8;   #speed of light(m/s)\n",
+      "h=6.625*10**-34;    #Plank's constant\n",
+      "\n",
+      "#Calculation\n",
+      "Op=0.5*P;    #radiant output(J/s)\n",
+      "E=h*c/W;     #energy content(J)\n",
+      "n=2/E;      #number of quanta emitted per second\n",
+      "\n",
+      "#Result\n",
+      "print \"Number of quanta emitted per second is\",round(n/1e+18,2),\"*10**18\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of quanta emitted per second is 6.16 *10**18\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.4, Page number 98"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "N=5*10**4;      #no. of photons\n",
+      "W=3000*10**-10;    #wavelength(m)\n",
+      "J=5*10**-3;    #senstivity(A/W)\n",
+      "h=6.625*10**-34;    #Plank's constant\n",
+      "c=3*10**8;       #speed of light(m/s)\n",
+      "e=1.6*10**-19;       #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "E=h*c/W;  #energy content of each photon(J)\n",
+      "TE=N*E;   #total energy(J)\n",
+      "I=J*TE;   #current produced(ampere)\n",
+      "n=I/e;       #number of photo electrons ejected\n",
+      "\n",
+      "#Result\n",
+      "print \"number of photoelectrons emitted are\",int(n)\n",
+      "print \"answer given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "number of photoelectrons emitted are 1035\n",
+        "answer given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.5, Page number 99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W=5*10**-7;     #wavelength(m)\n",
+      "F=10**-5;    #force(N)\n",
+      "h=6.625*10**-34;    #Plank's constant\n",
+      "m=1.5*10**-3;     #mass(kg)\n",
+      "c=3*10**8;     #speed of light in (m/s)\n",
+      "S=0.1;     #specific heat\n",
+      "\n",
+      "#Calculation\n",
+      "n=F*W/h;    #number of photons\n",
+      "E=F*c/4200;     #energy of each photon(kcal/s)\n",
+      "theta=E/(m*S);     #rate of rise in temperature(C/s)\n",
+      "\n",
+      "#Result\n",
+      "print \"number of photons are\",round(n/1e+21,3),\"*10**21\"\n",
+      "print \"the rate of temperature rise is\",round(theta/1e+3,1),\"*10**3 C/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "number of photons are 7.547 *10**21\n",
+        "the rate of temperature rise is 4.8 *10**3 C/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.6, Page number 99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W=4500*10**-10;    #wavelength(m)\n",
+      "V=150;   #rated voltage(W)\n",
+      "h=6.625*10**-34;   #Plank's constant\n",
+      "c=3*10**8;      #speed of light(m/s)\n",
+      "\n",
+      "#Calculation\n",
+      "P=V*8/100;     #lamp power emitted(W)\n",
+      "E=h*c/W;       #energy carried by 1 photon(J) \n",
+      "n=P/E;        #number of photons emitted per second\n",
+      "\n",
+      "#Result\n",
+      "print \"Number of photons emitted per second is\",round(n/1e+18,2),\"*10**18\"\n",
+      "print \"answer given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of photons emitted per second is 27.17 *10**18\n",
+        "answer given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.7, Page number 99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "f=1*10**12;     #frequency(Hz)\n",
+      "h=6.625*10**-34;    #Plank's constant\n",
+      "\n",
+      "#Calculation\n",
+      "E=h*f;     #energy per photon(J)\n",
+      "n=E/6.625;     #number of photons\n",
+      "\n",
+      "#Result\n",
+      "print \"the number of photons required is\",n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the number of photons required is 1e-22\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.8, Page number 100"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W=5200*10**-10;     #wavelength(m)\n",
+      "h=6.625*10**-34;    #Plank's constant\n",
+      "m=9.12*10**-31;     #mass of electron(kg)\n",
+      "\n",
+      "#Calculation\n",
+      "p=h/W;      #momentum(kg-m/s)\n",
+      "v=p/m;      #velocity(m/s)\n",
+      "\n",
+      "#Result\n",
+      "print \"velocity is\",round(v),\"m/s\"\n",
+      "print \"answer given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "velocity is 1397.0 m/s\n",
+        "answer given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.9, Page number 105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "v=7*10**5;    #maximum speed(m/sec)\n",
+      "f=8*10**14;   #frequency(Hz)\n",
+      "h=6.625*10**-34;    #Plank's constant\n",
+      "c=3*10**8;       #speed of light(m/s)\n",
+      "m=9.12*10**-31;  #mass of electron(kg)\n",
+      "\n",
+      "#Calulation\n",
+      "E=m*v*v/2;    #energy(J)\n",
+      "fo=f-(E/h);    #threshold frequency of the surface(Hz) \n",
+      "\n",
+      "#Result\n",
+      "print \"the threshold frequency is\",round(fo/1e+14,2),\"*10**14 Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the threshold frequency is 4.63 *10**14 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.10, Page number 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Wo=2300*10;    #threshold wavelength(Angstrom)\n",
+      "W=1800*10;     #incident light wavelength(Angstrom)\n",
+      "\n",
+      "#Calculation\n",
+      "w=124000/Wo;      #maximum energy of photoelectrons emitted(eV)\n",
+      "E=124000*((1/W)-(1/Wo));    #work function for tungsten(eV)\n",
+      "\n",
+      "#Result\n",
+      "print \"maximum energy of photoelectrons emitted is\",round(w,1),\"eV\"\n",
+      "print \"work function for tungsten is\",round(E,1),\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum energy of photoelectrons emitted is 5.4 eV\n",
+        "work function for tungsten is 1.5 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.11, Page number 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W=6000;      #wavelegth(Angstrom)\n",
+      "v=4*10**5;      #velocity(m/sec)\n",
+      "m=9.12*10**-31;    #mass of electron(kg)\n",
+      "e=1.6*10**-19;     #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "KE=m*v**2/(2*e);     #kinetic energy of photo electronns(eV)\n",
+      "WF=12400/W;          #energy content of photon(eV)\n",
+      "Wo=12400/(WF-KE);     #photo electric threshold wavelength(angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"The Kinetic energy is\",KE,\"eV\"\n",
+      "print \"The threshold wavelength is\",int(Wo),\"Angstrom\"\n",
+      "print \"answer given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Kinetic energy is 0.456 eV\n",
+        "The threshold wavelength is 7698 Angstrom\n",
+        "answer given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.12, Page number 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Wo=4.8;    #work function(eV)\n",
+      "W=2220;    #wavelength(angstrom)\n",
+      "\n",
+      "#Calculation\n",
+      "E=12400/W;    #energy of light photon(eV)\n",
+      "Emax=E-Wo;    #maximum kinetic energy(eV)\n",
+      "\n",
+      "#Result\n",
+      "print \"maximum kinetic energy is\",round(Emax,3),\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum kinetic energy is 0.786 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.13, Page number 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W=4000*10**-10;     #wavelength(m)\n",
+      "Vs=0.4;     #retarding potential(eV)\n",
+      "h=6.625*10**-34;    #Plank's constant\n",
+      "c=3*10**8;      #speed of light(m/s)\n",
+      "e=1.6*10**-19;     #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "f=c/W;    #frequency of light(Hz)\n",
+      "E=h*f/e;    #photon energy(eV)\n",
+      "Wo=E-Vs;    #work function(eV)\n",
+      "fo=Wo/h*e;     #threshold frequency(Hz)\n",
+      "NE=(E-Wo)*e;    #net energy(J)\n",
+      "\n",
+      "#Result\n",
+      "print \"The light frequency is\",f,\"Hz\"\n",
+      "print \"The photon energy is\",round(E,1),\"eV\"\n",
+      "print \"The work function is\",round(Wo,1),\"eV\"\n",
+      "print \"The threshold frequency is\",round(fo/1e+14,1),\"*10**14 Hz\"\n",
+      "print \"The net energy is\",NE,\"J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The light frequency is 7.5e+14 Hz\n",
+        "The photon energy is 3.1 eV\n",
+        "The work function is 2.7 eV\n",
+        "The threshold frequency is 6.5 *10**14 Hz\n",
+        "The net energy is 6.4e-20 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 51
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.14, Page number 107"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W1=3310*10**-10;    #photon wavelength(m)\n",
+      "W2=5000*10**-10;    #photon wavelength(m)\n",
+      "E1=3*10**-19;     #electron energy(J)\n",
+      "E2=0.972*10**-19;   #electron energy(J)\n",
+      "c=3*10**8;       #speed of light in m/s\n",
+      "\n",
+      "#Calculation\n",
+      "h=(E1-E2)*(W1*W2)/(c*(W2-W1));    #planck's constant(Js)\n",
+      "Wo=c*h/E1;        #threshold wavelength(m)   \n",
+      "\n",
+      "#Result\n",
+      "print \"the plancks const is\",round(h/1e-34,2),\"*10**-34 Js\"\n",
+      "print \"The threshold wavelength is\",round(Wo*1e+10),\"*10**-10 m\"\n",
+      "print \"answer given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the plancks const is 6.62 *10**-34 Js\n",
+        "The threshold wavelength is 6620.0 *10**-10 m\n",
+        "answer given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 57
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.15, Page number 107"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W=6525;     #wavelength(angstrom)\n",
+      "\n",
+      "#Calculation\n",
+      "Vo_a=12400*((1/4000)-(1/W));       #stopping potential(V)\n",
+      "Vo_b=12400*((1/2000)-(1/W));       #stopping potential(V)\n",
+      "Vo_c=12400*((1/2000)-(2/W));       #stopping potential(V)\n",
+      "\n",
+      "#Result\n",
+      "print \"Stopping potential is\",round(Vo_a,1),\"Volt\"\n",
+      "print \"Stopping potential is\",round(Vo_b,1),\"Volt\"\n",
+      "print \"Stopping potential is\",round(Vo_c,1),\"Volt\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Stopping potential is 1.2 Volt\n",
+        "Stopping potential is 4.3 Volt\n",
+        "Stopping potential is 2.4 Volt\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.16, Page number 107"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Wo=5000;      #wavelength(angstrom)\n",
+      "V=3.1;        #stopping potential(V)\n",
+      "\n",
+      "#Calcultion\n",
+      "W=1/((V/12400)+(1/Wo));      #wavelength(angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"The wavelength is\",int(W),\"Angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength is 2222 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.17, Page number 108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W=2000;         #wavelength(Angstrom)\n",
+      "Vs=4.2;         #Work Function(eV)\n",
+      "e=1.6*10**-19;      #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "E=12400/W;       #photon energy(eV)\n",
+      "Emax=(E-Vs)*e;     #maximum kinetic energy(J)\n",
+      "Emin=0;         #minimum kinetic energy\n",
+      "Vo=Emax/e;      #stopping potential(V)\n",
+      "Wo=12400/Vs;      #cut off wavelength(angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"Kinetic Energy of fastest photoelectron is\",Emax,\"J\"\n",
+      "print \"Kinetic Energy of slowest moving electron is\",Emin,\"J\"\n",
+      "print \"Stopping potential is\",Vo,\"V\"\n",
+      "print \"The cutoff wavelength is\",round(Wo,1),\"Angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Kinetic Energy of fastest photoelectron is 3.2e-19 J\n",
+        "Kinetic Energy of slowest moving electron is 0 J\n",
+        "Stopping potential is 2.0 V\n",
+        "The cutoff wavelength is 2952.4 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.18, Page number 108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Vs1=4.6;     #Stopping Potential(V)\n",
+      "Vs2=12.9;    #Stopping Potential(V)\n",
+      "f1=2*10**15;     #frequency(Hz)\n",
+      "f2=4*10**15;     #frequency(Hz)\n",
+      "e=1.6*10**-19;   #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "h=((Vs2-Vs1)*e)/(f2-f1);     #planck's constant(Js)\n",
+      "\n",
+      "#Result\n",
+      "print \"The Planck's constant is\",h,\"Js\"\n",
+      "print \"answer given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Planck's constant is 6.64e-34 Js\n",
+        "answer given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter5_1.ipynb b/Modern_Physics/chapter5_1.ipynb
new file mode 100755
index 00000000..a4b2fb2a
--- /dev/null
+++ b/Modern_Physics/chapter5_1.ipynb
@@ -0,0 +1,832 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:9388f9d8764031d08c03b951ac3babfe1cafed91339d0e7cf1d9a5afd45176fa"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "5: Quantum Theory"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.1, Page number 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W1=4;     #wavelength(Angstrom)\n",
+      "W2=1;     #wavelength(Angstrom)\n",
+      "e=1.6*10**-19;   #the charge on electron(C)\n",
+      "m=9.12*10**-31;   #mass of electron(kg)\n",
+      "\n",
+      "#Calculation\n",
+      "E=12400/W1;      #energy(eV)\n",
+      "v=math.sqrt(E*e*2/m);      #velocity(m/s)\n",
+      "E1=12400/W2;       #energy(eV)\n",
+      "v1=math.sqrt(E1*e*2/m);        #velocity(m/s)\n",
+      "\n",
+      "#Result\n",
+      "print \"The energy for 4 angstrom wavelength is\",E,\"eV\"\n",
+      "print \"The velocity is\",round(v/1e+6),\"*10**6 m/s\"\n",
+      "print \"The energy for 1 angstrom wavelength is\",E1,\"eV\"\n",
+      "print \"The velocity is\",round(v1/1e+6),\"*10**6 m/s\"\n",
+      "\n",
+      "#answers given in the book are wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The energy for 4 angstrom wavelength is 3100.0 eV\n",
+        "The velocity is 33.0 *10**6 m/s\n",
+        "The energy for 1 angstrom wavelength is 12400.0 eV\n",
+        "The velocity is 66.0 *10**6 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.2, Page number 98"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "f=880*10**3;      #frequency(Hz)\n",
+      "P=10*10**3;    #Power(W)\n",
+      "h=6.625*10**-34;    #Plank's constant\n",
+      "\n",
+      "#Calculation\n",
+      "E=h*f;    #energy carried by each photon(J)\n",
+      "n=P/E;    #number of photons emitted per second\n",
+      "\n",
+      "#Result\n",
+      "print \"The number of photons emitted per second are\",round(n/1e+30,2),\"*10**30\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number of photons emitted per second are 17.15 *10**30\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.3, Page number 98"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "P=200;     #power(W)\n",
+      "W=6123*10**-10;    #wavelength(m)\n",
+      "c=3*10**8;   #speed of light(m/s)\n",
+      "h=6.625*10**-34;    #Plank's constant\n",
+      "\n",
+      "#Calculation\n",
+      "Op=0.5*P;    #radiant output(J/s)\n",
+      "E=h*c/W;     #energy content(J)\n",
+      "n=2/E;      #number of quanta emitted per second\n",
+      "\n",
+      "#Result\n",
+      "print \"Number of quanta emitted per second is\",round(n/1e+18,2),\"*10**18\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of quanta emitted per second is 6.16 *10**18\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.4, Page number 98"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "N=5*10**4;      #no. of photons\n",
+      "W=3000*10**-10;    #wavelength(m)\n",
+      "J=5*10**-3;    #senstivity(A/W)\n",
+      "h=6.625*10**-34;    #Plank's constant\n",
+      "c=3*10**8;       #speed of light(m/s)\n",
+      "e=1.6*10**-19;       #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "E=h*c/W;  #energy content of each photon(J)\n",
+      "TE=N*E;   #total energy(J)\n",
+      "I=J*TE;   #current produced(ampere)\n",
+      "n=I/e;       #number of photo electrons ejected\n",
+      "\n",
+      "#Result\n",
+      "print \"number of photoelectrons emitted are\",int(n)\n",
+      "print \"answer given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "number of photoelectrons emitted are 1035\n",
+        "answer given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.5, Page number 99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W=5*10**-7;     #wavelength(m)\n",
+      "F=10**-5;    #force(N)\n",
+      "h=6.625*10**-34;    #Plank's constant\n",
+      "m=1.5*10**-3;     #mass(kg)\n",
+      "c=3*10**8;     #speed of light in (m/s)\n",
+      "S=0.1;     #specific heat\n",
+      "\n",
+      "#Calculation\n",
+      "n=F*W/h;    #number of photons\n",
+      "E=F*c/4200;     #energy of each photon(kcal/s)\n",
+      "theta=E/(m*S);     #rate of rise in temperature(C/s)\n",
+      "\n",
+      "#Result\n",
+      "print \"number of photons are\",round(n/1e+21,3),\"*10**21\"\n",
+      "print \"the rate of temperature rise is\",round(theta/1e+3,1),\"*10**3 C/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "number of photons are 7.547 *10**21\n",
+        "the rate of temperature rise is 4.8 *10**3 C/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.6, Page number 99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W=4500*10**-10;    #wavelength(m)\n",
+      "V=150;   #rated voltage(W)\n",
+      "h=6.625*10**-34;   #Plank's constant\n",
+      "c=3*10**8;      #speed of light(m/s)\n",
+      "\n",
+      "#Calculation\n",
+      "P=V*8/100;     #lamp power emitted(W)\n",
+      "E=h*c/W;       #energy carried by 1 photon(J) \n",
+      "n=P/E;        #number of photons emitted per second\n",
+      "\n",
+      "#Result\n",
+      "print \"Number of photons emitted per second is\",round(n/1e+18,2),\"*10**18\"\n",
+      "print \"answer given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of photons emitted per second is 27.17 *10**18\n",
+        "answer given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.7, Page number 99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "f=1*10**12;     #frequency(Hz)\n",
+      "h=6.625*10**-34;    #Plank's constant\n",
+      "\n",
+      "#Calculation\n",
+      "E=h*f;     #energy per photon(J)\n",
+      "n=E/6.625;     #number of photons\n",
+      "\n",
+      "#Result\n",
+      "print \"the number of photons required is\",n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the number of photons required is 1e-22\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.8, Page number 100"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W=5200*10**-10;     #wavelength(m)\n",
+      "h=6.625*10**-34;    #Plank's constant\n",
+      "m=9.12*10**-31;     #mass of electron(kg)\n",
+      "\n",
+      "#Calculation\n",
+      "p=h/W;      #momentum(kg-m/s)\n",
+      "v=p/m;      #velocity(m/s)\n",
+      "\n",
+      "#Result\n",
+      "print \"velocity is\",round(v),\"m/s\"\n",
+      "print \"answer given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "velocity is 1397.0 m/s\n",
+        "answer given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.9, Page number 105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "v=7*10**5;    #maximum speed(m/sec)\n",
+      "f=8*10**14;   #frequency(Hz)\n",
+      "h=6.625*10**-34;    #Plank's constant\n",
+      "c=3*10**8;       #speed of light(m/s)\n",
+      "m=9.12*10**-31;  #mass of electron(kg)\n",
+      "\n",
+      "#Calulation\n",
+      "E=m*v*v/2;    #energy(J)\n",
+      "fo=f-(E/h);    #threshold frequency of the surface(Hz) \n",
+      "\n",
+      "#Result\n",
+      "print \"the threshold frequency is\",round(fo/1e+14,2),\"*10**14 Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the threshold frequency is 4.63 *10**14 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.10, Page number 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Wo=2300*10;    #threshold wavelength(Angstrom)\n",
+      "W=1800*10;     #incident light wavelength(Angstrom)\n",
+      "\n",
+      "#Calculation\n",
+      "w=124000/Wo;      #maximum energy of photoelectrons emitted(eV)\n",
+      "E=124000*((1/W)-(1/Wo));    #work function for tungsten(eV)\n",
+      "\n",
+      "#Result\n",
+      "print \"maximum energy of photoelectrons emitted is\",round(w,1),\"eV\"\n",
+      "print \"work function for tungsten is\",round(E,1),\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum energy of photoelectrons emitted is 5.4 eV\n",
+        "work function for tungsten is 1.5 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.11, Page number 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W=6000;      #wavelegth(Angstrom)\n",
+      "v=4*10**5;      #velocity(m/sec)\n",
+      "m=9.12*10**-31;    #mass of electron(kg)\n",
+      "e=1.6*10**-19;     #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "KE=m*v**2/(2*e);     #kinetic energy of photo electronns(eV)\n",
+      "WF=12400/W;          #energy content of photon(eV)\n",
+      "Wo=12400/(WF-KE);     #photo electric threshold wavelength(angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"The Kinetic energy is\",KE,\"eV\"\n",
+      "print \"The threshold wavelength is\",int(Wo),\"Angstrom\"\n",
+      "print \"answer given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Kinetic energy is 0.456 eV\n",
+        "The threshold wavelength is 7698 Angstrom\n",
+        "answer given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.12, Page number 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Wo=4.8;    #work function(eV)\n",
+      "W=2220;    #wavelength(angstrom)\n",
+      "\n",
+      "#Calculation\n",
+      "E=12400/W;    #energy of light photon(eV)\n",
+      "Emax=E-Wo;    #maximum kinetic energy(eV)\n",
+      "\n",
+      "#Result\n",
+      "print \"maximum kinetic energy is\",round(Emax,3),\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum kinetic energy is 0.786 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.13, Page number 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W=4000*10**-10;     #wavelength(m)\n",
+      "Vs=0.4;     #retarding potential(eV)\n",
+      "h=6.625*10**-34;    #Plank's constant\n",
+      "c=3*10**8;      #speed of light(m/s)\n",
+      "e=1.6*10**-19;     #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "f=c/W;    #frequency of light(Hz)\n",
+      "E=h*f/e;    #photon energy(eV)\n",
+      "Wo=E-Vs;    #work function(eV)\n",
+      "fo=Wo/h*e;     #threshold frequency(Hz)\n",
+      "NE=(E-Wo)*e;    #net energy(J)\n",
+      "\n",
+      "#Result\n",
+      "print \"The light frequency is\",f,\"Hz\"\n",
+      "print \"The photon energy is\",round(E,1),\"eV\"\n",
+      "print \"The work function is\",round(Wo,1),\"eV\"\n",
+      "print \"The threshold frequency is\",round(fo/1e+14,1),\"*10**14 Hz\"\n",
+      "print \"The net energy is\",NE,\"J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The light frequency is 7.5e+14 Hz\n",
+        "The photon energy is 3.1 eV\n",
+        "The work function is 2.7 eV\n",
+        "The threshold frequency is 6.5 *10**14 Hz\n",
+        "The net energy is 6.4e-20 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 51
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.14, Page number 107"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W1=3310*10**-10;    #photon wavelength(m)\n",
+      "W2=5000*10**-10;    #photon wavelength(m)\n",
+      "E1=3*10**-19;     #electron energy(J)\n",
+      "E2=0.972*10**-19;   #electron energy(J)\n",
+      "c=3*10**8;       #speed of light in m/s\n",
+      "\n",
+      "#Calculation\n",
+      "h=(E1-E2)*(W1*W2)/(c*(W2-W1));    #planck's constant(Js)\n",
+      "Wo=c*h/E1;        #threshold wavelength(m)   \n",
+      "\n",
+      "#Result\n",
+      "print \"the plancks const is\",round(h/1e-34,2),\"*10**-34 Js\"\n",
+      "print \"The threshold wavelength is\",round(Wo*1e+10),\"*10**-10 m\"\n",
+      "print \"answer given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the plancks const is 6.62 *10**-34 Js\n",
+        "The threshold wavelength is 6620.0 *10**-10 m\n",
+        "answer given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 57
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.15, Page number 107"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W=6525;     #wavelength(angstrom)\n",
+      "\n",
+      "#Calculation\n",
+      "Vo_a=12400*((1/4000)-(1/W));       #stopping potential(V)\n",
+      "Vo_b=12400*((1/2000)-(1/W));       #stopping potential(V)\n",
+      "Vo_c=12400*((1/2000)-(2/W));       #stopping potential(V)\n",
+      "\n",
+      "#Result\n",
+      "print \"Stopping potential is\",round(Vo_a,1),\"Volt\"\n",
+      "print \"Stopping potential is\",round(Vo_b,1),\"Volt\"\n",
+      "print \"Stopping potential is\",round(Vo_c,1),\"Volt\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Stopping potential is 1.2 Volt\n",
+        "Stopping potential is 4.3 Volt\n",
+        "Stopping potential is 2.4 Volt\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.16, Page number 107"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Wo=5000;      #wavelength(angstrom)\n",
+      "V=3.1;        #stopping potential(V)\n",
+      "\n",
+      "#Calcultion\n",
+      "W=1/((V/12400)+(1/Wo));      #wavelength(angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"The wavelength is\",int(W),\"Angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength is 2222 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.17, Page number 108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W=2000;         #wavelength(Angstrom)\n",
+      "Vs=4.2;         #Work Function(eV)\n",
+      "e=1.6*10**-19;      #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "E=12400/W;       #photon energy(eV)\n",
+      "Emax=(E-Vs)*e;     #maximum kinetic energy(J)\n",
+      "Emin=0;         #minimum kinetic energy\n",
+      "Vo=Emax/e;      #stopping potential(V)\n",
+      "Wo=12400/Vs;      #cut off wavelength(angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"Kinetic Energy of fastest photoelectron is\",Emax,\"J\"\n",
+      "print \"Kinetic Energy of slowest moving electron is\",Emin,\"J\"\n",
+      "print \"Stopping potential is\",Vo,\"V\"\n",
+      "print \"The cutoff wavelength is\",round(Wo,1),\"Angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Kinetic Energy of fastest photoelectron is 3.2e-19 J\n",
+        "Kinetic Energy of slowest moving electron is 0 J\n",
+        "Stopping potential is 2.0 V\n",
+        "The cutoff wavelength is 2952.4 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 5.18, Page number 108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Vs1=4.6;     #Stopping Potential(V)\n",
+      "Vs2=12.9;    #Stopping Potential(V)\n",
+      "f1=2*10**15;     #frequency(Hz)\n",
+      "f2=4*10**15;     #frequency(Hz)\n",
+      "e=1.6*10**-19;   #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "h=((Vs2-Vs1)*e)/(f2-f1);     #planck's constant(Js)\n",
+      "\n",
+      "#Result\n",
+      "print \"The Planck's constant is\",h,\"Js\"\n",
+      "print \"answer given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Planck's constant is 6.64e-34 Js\n",
+        "answer given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter7.ipynb b/Modern_Physics/chapter7.ipynb
new file mode 100755
index 00000000..84f294f8
--- /dev/null
+++ b/Modern_Physics/chapter7.ipynb
@@ -0,0 +1,186 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:bb88588092f5e168d4a1c62bb7bb87ac343b1c2045bd4d31cd2391090ea3e567"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "7: Classification of Solids"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 7.1, Page number 138"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W=11000;     #wavelength(angstrom)\n",
+      "\n",
+      "#Calculation\n",
+      "Eg=W/12400;     #energy gap(eV)\n",
+      "\n",
+      "#Result\n",
+      "print \"Energy Gap is\",round(Eg,3),\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy Gap is 0.887 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 7.2, Page number 138"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "p=1.7*10**-6;      #resistivity(ohm-cm)\n",
+      "d=8.96;         #density(g/cc)\n",
+      "W=63.5;         #atomic weight(gm)\n",
+      "Na=6.02*10**23;   #Avagadro number(per g-mol)\n",
+      "e=1.6*10**-19;     #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "n=8.96*Na/W;      #number of Cu atoms per cc\n",
+      "mewe=1/(p*e*n);     #mobility of electrons(cm^2/V-s)\n",
+      "\n",
+      "#Result\n",
+      "print \"mobility of electrons is\",round(mewe,1),\"cm^2/V-s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mobility of electrons is 43.3 cm^2/V-s\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 7.3, Page number 139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "d1=2.5*10**19;      #density of charge carriers(per m^3)\n",
+      "d2=4.2*10**28;      #density of germanium atoms(per m^3)\n",
+      "mewe=0.36;          #mobilty of electrons(m^2/V-s)\n",
+      "Na=6.02*10**23;     #Avgraodo no.(per g-mol)\n",
+      "e=1.6*10**-19;      #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "Nd=d2/10**6;       #density of added impurity atoms(atoms/m^3)\n",
+      "sigma_n=Nd*e*mewe;        #conductivity(mho/m)\n",
+      "rho_n=1/sigma_n;          #resistivity of doped germanium(ohm-m)\n",
+      "\n",
+      "#Result\n",
+      "print \"resistivity of doped germanium is\",round(rho_n*10**3,3),\"*10**-3 ohm-m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistivity of doped germanium is 0.413 *10**-3 ohm-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 7.4, Page number 139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Eg=0.75;      #energy gap(eV)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda=12400/Eg;     #wavelength(angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"wavelength is\",int(lamda),\"angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "wavelength is 16533 angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter7_1.ipynb b/Modern_Physics/chapter7_1.ipynb
new file mode 100755
index 00000000..84f294f8
--- /dev/null
+++ b/Modern_Physics/chapter7_1.ipynb
@@ -0,0 +1,186 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:bb88588092f5e168d4a1c62bb7bb87ac343b1c2045bd4d31cd2391090ea3e567"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "7: Classification of Solids"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 7.1, Page number 138"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "W=11000;     #wavelength(angstrom)\n",
+      "\n",
+      "#Calculation\n",
+      "Eg=W/12400;     #energy gap(eV)\n",
+      "\n",
+      "#Result\n",
+      "print \"Energy Gap is\",round(Eg,3),\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy Gap is 0.887 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 7.2, Page number 138"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "p=1.7*10**-6;      #resistivity(ohm-cm)\n",
+      "d=8.96;         #density(g/cc)\n",
+      "W=63.5;         #atomic weight(gm)\n",
+      "Na=6.02*10**23;   #Avagadro number(per g-mol)\n",
+      "e=1.6*10**-19;     #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "n=8.96*Na/W;      #number of Cu atoms per cc\n",
+      "mewe=1/(p*e*n);     #mobility of electrons(cm^2/V-s)\n",
+      "\n",
+      "#Result\n",
+      "print \"mobility of electrons is\",round(mewe,1),\"cm^2/V-s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mobility of electrons is 43.3 cm^2/V-s\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 7.3, Page number 139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "d1=2.5*10**19;      #density of charge carriers(per m^3)\n",
+      "d2=4.2*10**28;      #density of germanium atoms(per m^3)\n",
+      "mewe=0.36;          #mobilty of electrons(m^2/V-s)\n",
+      "Na=6.02*10**23;     #Avgraodo no.(per g-mol)\n",
+      "e=1.6*10**-19;      #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "Nd=d2/10**6;       #density of added impurity atoms(atoms/m^3)\n",
+      "sigma_n=Nd*e*mewe;        #conductivity(mho/m)\n",
+      "rho_n=1/sigma_n;          #resistivity of doped germanium(ohm-m)\n",
+      "\n",
+      "#Result\n",
+      "print \"resistivity of doped germanium is\",round(rho_n*10**3,3),\"*10**-3 ohm-m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistivity of doped germanium is 0.413 *10**-3 ohm-m\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 7.4, Page number 139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "Eg=0.75;      #energy gap(eV)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda=12400/Eg;     #wavelength(angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"wavelength is\",int(lamda),\"angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "wavelength is 16533 angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter8.ipynb b/Modern_Physics/chapter8.ipynb
new file mode 100755
index 00000000..66b80efc
--- /dev/null
+++ b/Modern_Physics/chapter8.ipynb
@@ -0,0 +1,739 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d72c0a73996bcf192ce1ccd2756ffc71afe785564affcd57aa0f4ee59a16ed83"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "8: X-Rays"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.1, Page number 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=60000;       #working voltage(V)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda_min=12400/V;          #Wavelength emitted(Angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"Wavelength emitted is\",round(lamda_min,1),\"Angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength emitted is 0.2 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.2, Page number 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=12400;      #Volatage applied(V)\n",
+      "I=0.002;      #current drop(A)\n",
+      "e=1.6*10**-19;    #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "n=I/e;         #number of electrons\n",
+      "v=(5.93*10**5)*(math.sqrt(V));      #striking speed(m/s)\n",
+      "lamda_min=12400/V;                #shortest wavelength is(Angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"number of electrons striking per second is\",n,\"s-1\" \n",
+      "print \"the speed with which they strike is\",round(v/1e+7,1),\"*10^7 m/s\"\n",
+      "print \"shortest wavelength is\",lamda_min,\"Angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "number of electrons striking per second is 1.25e+16 s-1\n",
+        "the speed with which they strike is 6.6 *10^7 m/s\n",
+        "shortest wavelength is 1.0 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.3, Page number 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "lamda_min=1;    #shortest wavelength(Angstrom)\n",
+      "\n",
+      "#Calculation\n",
+      "V=(12400/lamda_min)/1000;      #minimum applied voltage(kV)\n",
+      "\n",
+      "#Result\n",
+      "print \"The minimum applied voltage is\",V,\"kV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum applied voltage is 12.4 kV\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.4, Page number 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "I=0.005;        #current(A)\n",
+      "V=100*10**3;     #potential difference(V)\n",
+      "\n",
+      "#Calculation\n",
+      "v=(5.93*10**5)*(math.sqrt(V));         #Maximum speed of electrons(m/s)\n",
+      "IP=V*I;             #incident power(W)\n",
+      "P=0.999*IP;         #power converted into heat(W)\n",
+      "H=P/4.18;           #rate of production of heat(cal/s)\n",
+      "\n",
+      "#Result\n",
+      "print \"Maximum speed of electrons is\",round(v/1e+8,2),\"*10^8 m/s\"\n",
+      "print \"rate of production of heat is\",int(H),\"cal/s\"\n",
+      "print \"answer for maximum speed of electrons given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum speed of electrons is 1.88 *10^8 m/s\n",
+        "rate of production of heat is 119 cal/s\n",
+        "answer for maximum speed of electrons given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.5, Page number 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=30000;          #potential difference(V)\n",
+      "lamda_min=0.414*10**-10;    #short wavelength limit(m)\n",
+      "e=1.602*10**-19;           #the charge on electron(C)\n",
+      "c=3*10**8;               #speed of light(m/s)\n",
+      "\n",
+      "#Calculation\n",
+      "h=(e*V*lamda_min)/c;        #Planck's constant(Js)\n",
+      "\n",
+      "#Result\n",
+      "print \"The Planck's constant is\",round(h/1e-34,2),\"*10^-34 Js\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Planck's constant is 6.63 *10^-34 Js\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.6, Page number 158"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "lamda=1.43*10**-10;       #wavelength(m)\n",
+      "Z=74;                    #atomic number\n",
+      "R=10.97*10**6;          #Rydberg constant(1/m)\n",
+      "\n",
+      "#Calculation\n",
+      "b=74-math.sqrt(36/(5*R*lamda));         #nuclear screening constant\n",
+      "\n",
+      "#Result\n",
+      "print \"nuclear screening constant is\",round(b,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "nuclear screening constant is 6.25\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.9, Page number 162"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "mewm=0.6;     #mass adsoption coeffcient(cm^2/g)\n",
+      "rho=2.7;    #density of aluminium(g/cm^3)\n",
+      "\n",
+      "#Calculation\n",
+      "mew=rho*mewm;        #linear adsorption coefficent of aluminium (1/cm)\n",
+      "T=0.693/mew;         #hvl(cm)\n",
+      "x=(math.log(20))*(1/mew);     #thickness(cm)\n",
+      "\n",
+      "#Result\n",
+      "print \"linear adsorption coefficent of aluminium is\",mew,\"cm-1\"\n",
+      "print \"the hvl is\",round(T,3),\"cm\"\n",
+      "print \"the thickness is\",round(x,2),\"cm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "linear adsorption coefficent of aluminium is 1.62 cm-1\n",
+        "the hvl is 0.428 cm\n",
+        "the thickness is 1.85 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.10, Page number 167"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "theta=12;       #glancing angle(degrees)\n",
+      "n=1;\n",
+      "d=3.04*10**-10;     #grating space(m)\n",
+      "\n",
+      "#Calculation \n",
+      "theta=theta*math.pi/180;         #glancing angle(radian)\n",
+      "lamda=(2*d*math.sin(theta))/n;     #wavelength of X-rays(m)\n",
+      "theta3=(3*lamda)/(2*d);        #angle for third order reflection(radian)\n",
+      "theta3=theta3*180/math.pi;        #angle for third order reflection(degrees)\n",
+      "\n",
+      "#Result\n",
+      "print \"wavelength of X-rays is\",round(lamda/10**-10,2),\"Angstrom\"\n",
+      "print \"angle for third order reflection is\",round(theta3,2),\"degrees\"\n",
+      "print \"answers given in the book are wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "wavelength of X-rays is 1.26 Angstrom\n",
+        "angle for third order reflection is 35.74 degrees\n",
+        "answers given in the book are wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.11, Page number 167"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "d=1.181;       #distance of seperation(Angstrom)\n",
+      "lamda=1.540;       #wavelength(Angstrom)\n",
+      "\n",
+      "#Calculation\n",
+      "n=2*d/lamda;        #sin(D) = 1 for max value\n",
+      "\n",
+      "#Result\n",
+      "print \"the orders of bragg reflection observed are\",int(n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the orders of bragg reflection observed are 1\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.12, Page number 168"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "lamda=0.6;        #wavelength(angstrom)\n",
+      "theta1=6.45;\n",
+      "theta2=9.15;\n",
+      "theta3=13;        #angles(degree)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda=lamda*10**-10;        ##wavelength(m)\n",
+      "theta1=theta1*math.pi/180;         \n",
+      "theta2=theta2*math.pi/180;         \n",
+      "theta3=theta3*math.pi/180;         #angles(radian)\n",
+      "d_a=lamda/(2*math.sin(theta1));     #interplanar spacing for 6.45 degrees(m)\n",
+      "d_b=lamda/(2*math.sin(theta2));\n",
+      "d_c=lamda/(2*math.sin(theta3));       \n",
+      "\n",
+      "#Result\n",
+      "print \"interplanar spacing for 6.45 degrees is\",round(d_a/1e-10,2),\"*10^-10 m\"\n",
+      "print \"interplanar spacing for 9.15 degrees is\",round(d_b/1e-10,3),\"*10^-10 m\"\n",
+      "print \"interplanar spacing for 13 degrees is\",round(d_c/1e-10,2),\"*10^-10 m\"\n",
+      "print \"answers given in the book vary due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "interplanar spacing for 6.45 degrees is 2.67 *10^-10 m\n",
+        "interplanar spacing for 9.15 degrees is 1.887 *10^-10 m\n",
+        "interplanar spacing for 13 degrees is 1.33 *10^-10 m\n",
+        "answers given in the book vary due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.13, Page number 168"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "lamda=3*10**-10;      #wavelength(m)\n",
+      "theta=40;            #angle(degree)\n",
+      "n=1;\n",
+      "\n",
+      "#Calculation\n",
+      "theta=theta*math.pi/180;         #angle(radian)\n",
+      "d=n*lamda/(2*math.sin(theta));     #spacing between planes(m)\n",
+      "a=2*d;            #lattice constant(m)\n",
+      "V=a**3;           #volume of unit cell(m^3)\n",
+      "\n",
+      "#Result\n",
+      "print \"spacing between planes is\",round(d/10**-10,2),\"AU\"\n",
+      "print \"volume of unit cell is\",round(V/1e-28,3),\"*10^-28 m^3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "spacing between planes is 2.33 AU\n",
+        "volume of unit cell is 1.017 *10^-28 m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.14, Page number 168"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "theta1=5.4;\n",
+      "theta2=7.6;\n",
+      "theta3=9.4;          #angles in degree\n",
+      "\n",
+      "#Calculation\n",
+      "theta1=theta1*math.pi/180;         \n",
+      "theta2=theta2*math.pi/180;         \n",
+      "theta3=theta3*math.pi/180;         #angles(radian)\n",
+      "d1=1/(2*math.sin(theta1));\n",
+      "d2=1/(2*math.sin(theta2));\n",
+      "d3=1/(2*math.sin(theta3));\n",
+      "m=min(d1,d2,d3);\n",
+      "d1=d1/m;\n",
+      "d2=d2/m;\n",
+      "d3=d3/m;\n",
+      "\n",
+      "#Result\n",
+      "print \"d1:d2:d3 =\",d1,d2,d3"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "d1:d2:d3 = 1.73551046111 1.23491924983 1.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.15, Page number 169"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=50000;      #applied voltage(V)\n",
+      "rho=1.99*10**3;    #density(kg/m^3)\n",
+      "n=4;\n",
+      "Na=6.02*10**26;      #Avgraodo number(per kg mole)\n",
+      "M=74.6;              #molecular mass\n",
+      "lamda=0.248*10**-10;      #wavelength(m)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda_min=12400/V;       #short wavelength limit(Angstrom)\n",
+      "a=(n*M/(Na*rho))**(1/3);      #lattice constant(m)\n",
+      "d=a/2;\n",
+      "theta=math.asin(lamda/(2*d));      #glancing angle(radian)\n",
+      "theta=theta*180/math.pi;           #glancing angle(degrees) \n",
+      "deg=int(theta);          #glancing angle(degrees) \n",
+      "t=60*(theta-deg);\n",
+      "m=int(t);              #glancing angle(minutes)\n",
+      "\n",
+      "#Result\n",
+      "print \"short wavelength limit is\",lamda_min,\"Angstrom\"\n",
+      "print \"glancing angle is\",deg,\"degrees\",m,\"minutes\"\n",
+      "print \"answer for glancing angle in minutes given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "short wavelength limit is 0.248 Angstrom\n",
+        "glancing angle is 2 degrees 15 minutes\n",
+        "answer for glancing angle in minutes given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.16, Page number 169"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "lamda=1.54;         #wavelength(angstrom)\n",
+      "theta=15.9;         #angle(degrees)\n",
+      "M=58.45;            #molecular weight\n",
+      "rho=2.164;     #density(g/cm^3)\n",
+      "n=2;                #for NaCl molecule\n",
+      "\n",
+      "#Calculation\n",
+      "theta=theta*math.pi/180;         #angle(radian)\n",
+      "d=lamda/(2*math.sin(theta));     #lattice spacing(angstrom) \n",
+      "dm=d*10**-8;                    ##lattice spacing(cm) \n",
+      "Na=M/(2*rho*dm**3);               #Avogadro number(per gm mole)       \n",
+      "\n",
+      "#Result\n",
+      "print \"lattice spacing is\",round(d,2),\"angstrom\"\n",
+      "print \"Avogadro number is\",round(Na/1e+23,2),\"*10^23 per gm mole\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " lattice spacing is 2.81 angstrom\n",
+        "Avogadro number is 6.08 *10^23 per gm mole\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.17, Page number 172"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "theta=60;         #angle(degree)\n",
+      "lamda=0.254;      #wavelength(angstrom)\n",
+      "\n",
+      "#Calculation\n",
+      "theta=theta*math.pi/180;         #angle(radian)\n",
+      "dlamda=0.024*(1-math.cos(theta));      #amount of increase in wavelength(angstrom)\n",
+      "lamda1=lamda-dlamda;       #primary X-ray wavelength(angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"primary X-ray wavelength is\",lamda1,\"angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "primary X-ray wavelength is 0.242 angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.18, Page number 172"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "theta=32;         #angle(degree)\n",
+      "lamda=1.54*10**-10;      #wavelength(angstrom)\n",
+      "h=2;       \n",
+      "k=2;\n",
+      "l=0;                 #lattice constants\n",
+      "\n",
+      "#Calculation\n",
+      "theta=theta*math.pi/180;         #angle(radian)\n",
+      "d=lamda/(2*math.sin(theta));       #interplanar spacing(m)\n",
+      "a=d*math.sqrt(h**2+k**2+l**2);     #lattice parameter(m)\n",
+      "r=math.sqrt(2)*a/4;            #radius of atom(m)\n",
+      "\n",
+      "#Result\n",
+      "print \"lattice parameter is\",round(a/1e-10,1),\"*10^-10 m\"\n",
+      "print \"radius of atom is\",round(r/1e-10,2),\"*10^-10 m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "lattice parameter is 4.1 *10^-10 m\n",
+        "radius of atom is 1.45 *10^-10 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter8_1.ipynb b/Modern_Physics/chapter8_1.ipynb
new file mode 100755
index 00000000..66b80efc
--- /dev/null
+++ b/Modern_Physics/chapter8_1.ipynb
@@ -0,0 +1,739 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d72c0a73996bcf192ce1ccd2756ffc71afe785564affcd57aa0f4ee59a16ed83"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "8: X-Rays"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.1, Page number 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=60000;       #working voltage(V)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda_min=12400/V;          #Wavelength emitted(Angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"Wavelength emitted is\",round(lamda_min,1),\"Angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength emitted is 0.2 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.2, Page number 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=12400;      #Volatage applied(V)\n",
+      "I=0.002;      #current drop(A)\n",
+      "e=1.6*10**-19;    #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "n=I/e;         #number of electrons\n",
+      "v=(5.93*10**5)*(math.sqrt(V));      #striking speed(m/s)\n",
+      "lamda_min=12400/V;                #shortest wavelength is(Angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"number of electrons striking per second is\",n,\"s-1\" \n",
+      "print \"the speed with which they strike is\",round(v/1e+7,1),\"*10^7 m/s\"\n",
+      "print \"shortest wavelength is\",lamda_min,\"Angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "number of electrons striking per second is 1.25e+16 s-1\n",
+        "the speed with which they strike is 6.6 *10^7 m/s\n",
+        "shortest wavelength is 1.0 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.3, Page number 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "lamda_min=1;    #shortest wavelength(Angstrom)\n",
+      "\n",
+      "#Calculation\n",
+      "V=(12400/lamda_min)/1000;      #minimum applied voltage(kV)\n",
+      "\n",
+      "#Result\n",
+      "print \"The minimum applied voltage is\",V,\"kV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum applied voltage is 12.4 kV\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.4, Page number 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "I=0.005;        #current(A)\n",
+      "V=100*10**3;     #potential difference(V)\n",
+      "\n",
+      "#Calculation\n",
+      "v=(5.93*10**5)*(math.sqrt(V));         #Maximum speed of electrons(m/s)\n",
+      "IP=V*I;             #incident power(W)\n",
+      "P=0.999*IP;         #power converted into heat(W)\n",
+      "H=P/4.18;           #rate of production of heat(cal/s)\n",
+      "\n",
+      "#Result\n",
+      "print \"Maximum speed of electrons is\",round(v/1e+8,2),\"*10^8 m/s\"\n",
+      "print \"rate of production of heat is\",int(H),\"cal/s\"\n",
+      "print \"answer for maximum speed of electrons given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum speed of electrons is 1.88 *10^8 m/s\n",
+        "rate of production of heat is 119 cal/s\n",
+        "answer for maximum speed of electrons given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.5, Page number 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=30000;          #potential difference(V)\n",
+      "lamda_min=0.414*10**-10;    #short wavelength limit(m)\n",
+      "e=1.602*10**-19;           #the charge on electron(C)\n",
+      "c=3*10**8;               #speed of light(m/s)\n",
+      "\n",
+      "#Calculation\n",
+      "h=(e*V*lamda_min)/c;        #Planck's constant(Js)\n",
+      "\n",
+      "#Result\n",
+      "print \"The Planck's constant is\",round(h/1e-34,2),\"*10^-34 Js\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Planck's constant is 6.63 *10^-34 Js\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.6, Page number 158"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "lamda=1.43*10**-10;       #wavelength(m)\n",
+      "Z=74;                    #atomic number\n",
+      "R=10.97*10**6;          #Rydberg constant(1/m)\n",
+      "\n",
+      "#Calculation\n",
+      "b=74-math.sqrt(36/(5*R*lamda));         #nuclear screening constant\n",
+      "\n",
+      "#Result\n",
+      "print \"nuclear screening constant is\",round(b,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "nuclear screening constant is 6.25\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.9, Page number 162"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "mewm=0.6;     #mass adsoption coeffcient(cm^2/g)\n",
+      "rho=2.7;    #density of aluminium(g/cm^3)\n",
+      "\n",
+      "#Calculation\n",
+      "mew=rho*mewm;        #linear adsorption coefficent of aluminium (1/cm)\n",
+      "T=0.693/mew;         #hvl(cm)\n",
+      "x=(math.log(20))*(1/mew);     #thickness(cm)\n",
+      "\n",
+      "#Result\n",
+      "print \"linear adsorption coefficent of aluminium is\",mew,\"cm-1\"\n",
+      "print \"the hvl is\",round(T,3),\"cm\"\n",
+      "print \"the thickness is\",round(x,2),\"cm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "linear adsorption coefficent of aluminium is 1.62 cm-1\n",
+        "the hvl is 0.428 cm\n",
+        "the thickness is 1.85 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.10, Page number 167"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "theta=12;       #glancing angle(degrees)\n",
+      "n=1;\n",
+      "d=3.04*10**-10;     #grating space(m)\n",
+      "\n",
+      "#Calculation \n",
+      "theta=theta*math.pi/180;         #glancing angle(radian)\n",
+      "lamda=(2*d*math.sin(theta))/n;     #wavelength of X-rays(m)\n",
+      "theta3=(3*lamda)/(2*d);        #angle for third order reflection(radian)\n",
+      "theta3=theta3*180/math.pi;        #angle for third order reflection(degrees)\n",
+      "\n",
+      "#Result\n",
+      "print \"wavelength of X-rays is\",round(lamda/10**-10,2),\"Angstrom\"\n",
+      "print \"angle for third order reflection is\",round(theta3,2),\"degrees\"\n",
+      "print \"answers given in the book are wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "wavelength of X-rays is 1.26 Angstrom\n",
+        "angle for third order reflection is 35.74 degrees\n",
+        "answers given in the book are wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.11, Page number 167"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "d=1.181;       #distance of seperation(Angstrom)\n",
+      "lamda=1.540;       #wavelength(Angstrom)\n",
+      "\n",
+      "#Calculation\n",
+      "n=2*d/lamda;        #sin(D) = 1 for max value\n",
+      "\n",
+      "#Result\n",
+      "print \"the orders of bragg reflection observed are\",int(n)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the orders of bragg reflection observed are 1\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.12, Page number 168"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "lamda=0.6;        #wavelength(angstrom)\n",
+      "theta1=6.45;\n",
+      "theta2=9.15;\n",
+      "theta3=13;        #angles(degree)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda=lamda*10**-10;        ##wavelength(m)\n",
+      "theta1=theta1*math.pi/180;         \n",
+      "theta2=theta2*math.pi/180;         \n",
+      "theta3=theta3*math.pi/180;         #angles(radian)\n",
+      "d_a=lamda/(2*math.sin(theta1));     #interplanar spacing for 6.45 degrees(m)\n",
+      "d_b=lamda/(2*math.sin(theta2));\n",
+      "d_c=lamda/(2*math.sin(theta3));       \n",
+      "\n",
+      "#Result\n",
+      "print \"interplanar spacing for 6.45 degrees is\",round(d_a/1e-10,2),\"*10^-10 m\"\n",
+      "print \"interplanar spacing for 9.15 degrees is\",round(d_b/1e-10,3),\"*10^-10 m\"\n",
+      "print \"interplanar spacing for 13 degrees is\",round(d_c/1e-10,2),\"*10^-10 m\"\n",
+      "print \"answers given in the book vary due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "interplanar spacing for 6.45 degrees is 2.67 *10^-10 m\n",
+        "interplanar spacing for 9.15 degrees is 1.887 *10^-10 m\n",
+        "interplanar spacing for 13 degrees is 1.33 *10^-10 m\n",
+        "answers given in the book vary due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.13, Page number 168"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "lamda=3*10**-10;      #wavelength(m)\n",
+      "theta=40;            #angle(degree)\n",
+      "n=1;\n",
+      "\n",
+      "#Calculation\n",
+      "theta=theta*math.pi/180;         #angle(radian)\n",
+      "d=n*lamda/(2*math.sin(theta));     #spacing between planes(m)\n",
+      "a=2*d;            #lattice constant(m)\n",
+      "V=a**3;           #volume of unit cell(m^3)\n",
+      "\n",
+      "#Result\n",
+      "print \"spacing between planes is\",round(d/10**-10,2),\"AU\"\n",
+      "print \"volume of unit cell is\",round(V/1e-28,3),\"*10^-28 m^3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "spacing between planes is 2.33 AU\n",
+        "volume of unit cell is 1.017 *10^-28 m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.14, Page number 168"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "theta1=5.4;\n",
+      "theta2=7.6;\n",
+      "theta3=9.4;          #angles in degree\n",
+      "\n",
+      "#Calculation\n",
+      "theta1=theta1*math.pi/180;         \n",
+      "theta2=theta2*math.pi/180;         \n",
+      "theta3=theta3*math.pi/180;         #angles(radian)\n",
+      "d1=1/(2*math.sin(theta1));\n",
+      "d2=1/(2*math.sin(theta2));\n",
+      "d3=1/(2*math.sin(theta3));\n",
+      "m=min(d1,d2,d3);\n",
+      "d1=d1/m;\n",
+      "d2=d2/m;\n",
+      "d3=d3/m;\n",
+      "\n",
+      "#Result\n",
+      "print \"d1:d2:d3 =\",d1,d2,d3"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "d1:d2:d3 = 1.73551046111 1.23491924983 1.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.15, Page number 169"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=50000;      #applied voltage(V)\n",
+      "rho=1.99*10**3;    #density(kg/m^3)\n",
+      "n=4;\n",
+      "Na=6.02*10**26;      #Avgraodo number(per kg mole)\n",
+      "M=74.6;              #molecular mass\n",
+      "lamda=0.248*10**-10;      #wavelength(m)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda_min=12400/V;       #short wavelength limit(Angstrom)\n",
+      "a=(n*M/(Na*rho))**(1/3);      #lattice constant(m)\n",
+      "d=a/2;\n",
+      "theta=math.asin(lamda/(2*d));      #glancing angle(radian)\n",
+      "theta=theta*180/math.pi;           #glancing angle(degrees) \n",
+      "deg=int(theta);          #glancing angle(degrees) \n",
+      "t=60*(theta-deg);\n",
+      "m=int(t);              #glancing angle(minutes)\n",
+      "\n",
+      "#Result\n",
+      "print \"short wavelength limit is\",lamda_min,\"Angstrom\"\n",
+      "print \"glancing angle is\",deg,\"degrees\",m,\"minutes\"\n",
+      "print \"answer for glancing angle in minutes given in the book is wrong\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "short wavelength limit is 0.248 Angstrom\n",
+        "glancing angle is 2 degrees 15 minutes\n",
+        "answer for glancing angle in minutes given in the book is wrong\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.16, Page number 169"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "lamda=1.54;         #wavelength(angstrom)\n",
+      "theta=15.9;         #angle(degrees)\n",
+      "M=58.45;            #molecular weight\n",
+      "rho=2.164;     #density(g/cm^3)\n",
+      "n=2;                #for NaCl molecule\n",
+      "\n",
+      "#Calculation\n",
+      "theta=theta*math.pi/180;         #angle(radian)\n",
+      "d=lamda/(2*math.sin(theta));     #lattice spacing(angstrom) \n",
+      "dm=d*10**-8;                    ##lattice spacing(cm) \n",
+      "Na=M/(2*rho*dm**3);               #Avogadro number(per gm mole)       \n",
+      "\n",
+      "#Result\n",
+      "print \"lattice spacing is\",round(d,2),\"angstrom\"\n",
+      "print \"Avogadro number is\",round(Na/1e+23,2),\"*10^23 per gm mole\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " lattice spacing is 2.81 angstrom\n",
+        "Avogadro number is 6.08 *10^23 per gm mole\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.17, Page number 172"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "theta=60;         #angle(degree)\n",
+      "lamda=0.254;      #wavelength(angstrom)\n",
+      "\n",
+      "#Calculation\n",
+      "theta=theta*math.pi/180;         #angle(radian)\n",
+      "dlamda=0.024*(1-math.cos(theta));      #amount of increase in wavelength(angstrom)\n",
+      "lamda1=lamda-dlamda;       #primary X-ray wavelength(angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"primary X-ray wavelength is\",lamda1,\"angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "primary X-ray wavelength is 0.242 angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 8.18, Page number 172"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "theta=32;         #angle(degree)\n",
+      "lamda=1.54*10**-10;      #wavelength(angstrom)\n",
+      "h=2;       \n",
+      "k=2;\n",
+      "l=0;                 #lattice constants\n",
+      "\n",
+      "#Calculation\n",
+      "theta=theta*math.pi/180;         #angle(radian)\n",
+      "d=lamda/(2*math.sin(theta));       #interplanar spacing(m)\n",
+      "a=d*math.sqrt(h**2+k**2+l**2);     #lattice parameter(m)\n",
+      "r=math.sqrt(2)*a/4;            #radius of atom(m)\n",
+      "\n",
+      "#Result\n",
+      "print \"lattice parameter is\",round(a/1e-10,1),\"*10^-10 m\"\n",
+      "print \"radius of atom is\",round(r/1e-10,2),\"*10^-10 m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "lattice parameter is 4.1 *10^-10 m\n",
+        "radius of atom is 1.45 *10^-10 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter9.ipynb b/Modern_Physics/chapter9.ipynb
new file mode 100755
index 00000000..ca8c771f
--- /dev/null
+++ b/Modern_Physics/chapter9.ipynb
@@ -0,0 +1,397 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:bab0907a92dd1a315ea33ad663e957734494f7112ddbb87e9005b028bf98d841"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "9: Waves and Particles"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.1, Page number 179"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=20000;     #applied voltage(V)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda=12.25/(math.sqrt(V));        #de broglie wavelength(angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"de broglie wavelength is\",round(lamda,3),\"angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "de broglie wavelength is 0.087 angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.2, Page number 179"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=5000;           #applied voltage(V)\n",
+      "e=1.602*10**-19;     #the charge on electron(C)\n",
+      "m=9.12*10**-31;      #mass of electron(kg)\n",
+      "d=2.04*10**-10;      #distance(m)\n",
+      "n=1;\n",
+      "\n",
+      "#Calculation\n",
+      "p=math.sqrt(2*m*e*V);       #momentum(kg m/s)\n",
+      "lamda=12.25/math.sqrt(V);       #de broglie wavelength(angstrom)\n",
+      "v=1/(lamda*10**-10);            #wave number\n",
+      "theta=math.asin((n*lamda*10**-10)/(2*d));    #Bragg angle(radian)\n",
+      "theta=theta*180/math.pi;                     #Bragg angle(degrees)\n",
+      "\n",
+      "#Result\n",
+      "print \"momentum is\",round(p/1e-23,2),\"*10^-23 kg m/s\"\n",
+      "print \"de broglie wavelength is\",round(lamda,3),\"angstrom\"\n",
+      "print \"the wave number is\",round(v/10**10,2),\"*10^10\"\n",
+      "print \"the Bragg angle is\",round(theta,2),\"degrees\"\n",
+      "print \"answers given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "momentum is 3.82 *10^-23 kg m/s\n",
+        "de broglie wavelength is 0.173 angstrom\n",
+        "the wave number is 5.77 *10^10\n",
+        "the Bragg angle is 2.43 degrees\n",
+        "answers given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.3, Page number 179"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=54;           #applied voltage(V)\n",
+      "e=1.602*10**-19;       #the charge on electron(C)\n",
+      "m=9.12*10**-31;        #mass of electron(kg)\n",
+      "h=6.625*10**-34;       #Plank's constant\n",
+      "\n",
+      "#Calculation\n",
+      "v=math.sqrt(2*e*V/m);      #velocity of electron(m/s)\n",
+      "lamda=12.25/math.sqrt(V);       #de broglie wavelength(angstrom)\n",
+      "u=h/(2*m*lamda*10**-10);        #phase velocity(m/s)\n",
+      "\n",
+      "#Result\n",
+      "print \"velocity of electron is\",round(v/1e+6,2),\"*10^6 m/s\"\n",
+      "print \"de broglie wavelength is\",round(lamda,2),\"angstrom\"\n",
+      "print \"phase velocity is\",round(u/1e+6,2),\"*10^6 m/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "velocity of electron is 4.36 *10^6 m/s\n",
+        "de broglie wavelength is 1.67 angstrom\n",
+        "phase velocity is 2.18 *10^6 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.4, Page number 180"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "e=1.6*10**-19;        #the charge on electron(C)\n",
+      "m=9.12*10**-31;       #mass of electron(kg)\n",
+      "c=3*10**8;           #speed of light(m/s)\n",
+      "h=6.625*10**-34;     #Plank's constant\n",
+      "\n",
+      "#Calculation\n",
+      "E=m*c**2;         #rest energy(J)\n",
+      "mp=1836*m;        #mass of proton(kg)\n",
+      "#(0.5*m*v^2)=E\n",
+      "mv=math.sqrt(E*2*mp);     #momentum(kg m/s)\n",
+      "lamda=h/mv;            #de broglie wavelength(m)\n",
+      "\n",
+      "#Result\n",
+      "print \"de broglie wavelength is\",round(lamda*10**10,4),\"Angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "de broglie wavelength is 0.0004 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.5, Page number 180"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "e=1.6*10**-19;        #the charge on electron(C)\n",
+      "m=1.676*10**-27;      #mass of neutron(kg)\n",
+      "c=3*10**8;            #speed of light(m/s)\n",
+      "h=6.625*10**-34;      #Plank's constant\n",
+      "\n",
+      "#Calculation\n",
+      "E=1;          #in eV\n",
+      "E=1*e;        #in V\n",
+      "mv=math.sqrt(2*E*m);         #momentum(kg m/s)\n",
+      "lamda=h/mv;           #de broglie wavelength(m)\n",
+      "\n",
+      "#Result\n",
+      "print \"de broglie wavelength is\",round(lamda*10**10,3),\"Angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "de broglie wavelength is 0.286 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.6, Page number 183"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "lamda=0.09;          #wavelength(Angstrom)\n",
+      "D=54;            #scattering angle(degree)\n",
+      "h=6.625*10**-34;      #Plank's constant\n",
+      "c=3*10**8;            #speed of light(m/s)\n",
+      "e=1.6*10**-19;        #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "dlamda=0.0243*(1-math.cos(D));      \n",
+      "lamda1=lamda+dlamda;             #Wavelength of scattered X-rays(Angstrom)\n",
+      "Ei=h*c/(lamda*10**-10);           #Energy of incident photon(J)\n",
+      "Es=h*c/(lamda1*10**-10);          #Energy of scattered photon(J)\n",
+      "\n",
+      "#Result\n",
+      "print \"wavelength of scattered X-rays is\",round(lamda1,1),\"Angstrom\"\n",
+      "print \"Energy of incident photon is\",round(Ei/(e*10**6),3),\"MeV\"\n",
+      "print \"Energy of scattered photon is\",round(Es/(e*10**6),4),\"MeV\"\n",
+      "print \"answer for energy of scattered photon given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "wavelength of scattered X-rays is 0.1 Angstrom\n",
+        "Energy of incident photon is 0.138 MeV\n",
+        "Energy of scattered photon is 0.0924 MeV\n",
+        "answer for energy of scattered photon given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.7, Page number 191"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "h=6.625*10**-34;      #Plank's constant\n",
+      "m=9.12*10**-31;       #mass of electron(kg)\n",
+      "\n",
+      "#Calculation\n",
+      "#for 1st quantum state\n",
+      "nx=1;\n",
+      "ny=1;\n",
+      "nz=1;\n",
+      "L=1;\n",
+      "E1=h**2*(nx**2+ny**2+nz**2)/(8*m*L**2);       #energy in first quantum state(J)\n",
+      "#for 2nd quantum state (nx^2+ny^2+nz^2)=6\n",
+      "L=1;\n",
+      "E=h**2*6/(8*m*L**2);                   #energy in second quantum state(J)\n",
+      "\n",
+      "#Result\n",
+      "print \"energy in first quantum state is\",round(E1/1e-37,3),\"*10^-37 J\"\n",
+      "print \"energy in second quantum state is\",round(E/1e-37,2),\"*10^-37 J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "energy in first quantum state is 1.805 *10^-37 J\n",
+        "energy in second quantum state is 3.61 *10^-37 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.8, Page number 194"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "h=6.625*10**-34;       #Plank's constant\n",
+      "m=9.12*10**-31;        #mass of electron(kg)\n",
+      "L=2.5*10**-10;         #width of square(m)\n",
+      "e=1.6*10**-19;         #the charge on electron(C)\n",
+      "n1=1;\n",
+      "n2=2;\n",
+      "n3=3;\n",
+      "\n",
+      "#Calculation\n",
+      "E1=n1**2*h**2/(8*m*L**2*e);          #1st lowest quantum energy(eV)\n",
+      "E2=n2**2*E1;             #2nd lowest quantum energy(eV)\n",
+      "E3=n3**2*E1;             #3rd lowest quantum energy(eV) \n",
+      " \n",
+      "#Result\n",
+      "print \"the lowest 3 quantum energies are\",int(E1),\"eV,\",int(E2),\"eV and\",int(E3),\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the lowest 3 quantum energies are 6 eV, 24 eV and 54 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/chapter9_1.ipynb b/Modern_Physics/chapter9_1.ipynb
new file mode 100755
index 00000000..ca8c771f
--- /dev/null
+++ b/Modern_Physics/chapter9_1.ipynb
@@ -0,0 +1,397 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:bab0907a92dd1a315ea33ad663e957734494f7112ddbb87e9005b028bf98d841"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "9: Waves and Particles"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.1, Page number 179"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=20000;     #applied voltage(V)\n",
+      "\n",
+      "#Calculation\n",
+      "lamda=12.25/(math.sqrt(V));        #de broglie wavelength(angstrom)\n",
+      "\n",
+      "#Result\n",
+      "print \"de broglie wavelength is\",round(lamda,3),\"angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "de broglie wavelength is 0.087 angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.2, Page number 179"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=5000;           #applied voltage(V)\n",
+      "e=1.602*10**-19;     #the charge on electron(C)\n",
+      "m=9.12*10**-31;      #mass of electron(kg)\n",
+      "d=2.04*10**-10;      #distance(m)\n",
+      "n=1;\n",
+      "\n",
+      "#Calculation\n",
+      "p=math.sqrt(2*m*e*V);       #momentum(kg m/s)\n",
+      "lamda=12.25/math.sqrt(V);       #de broglie wavelength(angstrom)\n",
+      "v=1/(lamda*10**-10);            #wave number\n",
+      "theta=math.asin((n*lamda*10**-10)/(2*d));    #Bragg angle(radian)\n",
+      "theta=theta*180/math.pi;                     #Bragg angle(degrees)\n",
+      "\n",
+      "#Result\n",
+      "print \"momentum is\",round(p/1e-23,2),\"*10^-23 kg m/s\"\n",
+      "print \"de broglie wavelength is\",round(lamda,3),\"angstrom\"\n",
+      "print \"the wave number is\",round(v/10**10,2),\"*10^10\"\n",
+      "print \"the Bragg angle is\",round(theta,2),\"degrees\"\n",
+      "print \"answers given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "momentum is 3.82 *10^-23 kg m/s\n",
+        "de broglie wavelength is 0.173 angstrom\n",
+        "the wave number is 5.77 *10^10\n",
+        "the Bragg angle is 2.43 degrees\n",
+        "answers given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.3, Page number 179"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "V=54;           #applied voltage(V)\n",
+      "e=1.602*10**-19;       #the charge on electron(C)\n",
+      "m=9.12*10**-31;        #mass of electron(kg)\n",
+      "h=6.625*10**-34;       #Plank's constant\n",
+      "\n",
+      "#Calculation\n",
+      "v=math.sqrt(2*e*V/m);      #velocity of electron(m/s)\n",
+      "lamda=12.25/math.sqrt(V);       #de broglie wavelength(angstrom)\n",
+      "u=h/(2*m*lamda*10**-10);        #phase velocity(m/s)\n",
+      "\n",
+      "#Result\n",
+      "print \"velocity of electron is\",round(v/1e+6,2),\"*10^6 m/s\"\n",
+      "print \"de broglie wavelength is\",round(lamda,2),\"angstrom\"\n",
+      "print \"phase velocity is\",round(u/1e+6,2),\"*10^6 m/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "velocity of electron is 4.36 *10^6 m/s\n",
+        "de broglie wavelength is 1.67 angstrom\n",
+        "phase velocity is 2.18 *10^6 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.4, Page number 180"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "e=1.6*10**-19;        #the charge on electron(C)\n",
+      "m=9.12*10**-31;       #mass of electron(kg)\n",
+      "c=3*10**8;           #speed of light(m/s)\n",
+      "h=6.625*10**-34;     #Plank's constant\n",
+      "\n",
+      "#Calculation\n",
+      "E=m*c**2;         #rest energy(J)\n",
+      "mp=1836*m;        #mass of proton(kg)\n",
+      "#(0.5*m*v^2)=E\n",
+      "mv=math.sqrt(E*2*mp);     #momentum(kg m/s)\n",
+      "lamda=h/mv;            #de broglie wavelength(m)\n",
+      "\n",
+      "#Result\n",
+      "print \"de broglie wavelength is\",round(lamda*10**10,4),\"Angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "de broglie wavelength is 0.0004 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.5, Page number 180"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "e=1.6*10**-19;        #the charge on electron(C)\n",
+      "m=1.676*10**-27;      #mass of neutron(kg)\n",
+      "c=3*10**8;            #speed of light(m/s)\n",
+      "h=6.625*10**-34;      #Plank's constant\n",
+      "\n",
+      "#Calculation\n",
+      "E=1;          #in eV\n",
+      "E=1*e;        #in V\n",
+      "mv=math.sqrt(2*E*m);         #momentum(kg m/s)\n",
+      "lamda=h/mv;           #de broglie wavelength(m)\n",
+      "\n",
+      "#Result\n",
+      "print \"de broglie wavelength is\",round(lamda*10**10,3),\"Angstrom\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "de broglie wavelength is 0.286 Angstrom\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.6, Page number 183"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "lamda=0.09;          #wavelength(Angstrom)\n",
+      "D=54;            #scattering angle(degree)\n",
+      "h=6.625*10**-34;      #Plank's constant\n",
+      "c=3*10**8;            #speed of light(m/s)\n",
+      "e=1.6*10**-19;        #the charge on electron(C)\n",
+      "\n",
+      "#Calculation\n",
+      "dlamda=0.0243*(1-math.cos(D));      \n",
+      "lamda1=lamda+dlamda;             #Wavelength of scattered X-rays(Angstrom)\n",
+      "Ei=h*c/(lamda*10**-10);           #Energy of incident photon(J)\n",
+      "Es=h*c/(lamda1*10**-10);          #Energy of scattered photon(J)\n",
+      "\n",
+      "#Result\n",
+      "print \"wavelength of scattered X-rays is\",round(lamda1,1),\"Angstrom\"\n",
+      "print \"Energy of incident photon is\",round(Ei/(e*10**6),3),\"MeV\"\n",
+      "print \"Energy of scattered photon is\",round(Es/(e*10**6),4),\"MeV\"\n",
+      "print \"answer for energy of scattered photon given in the book varies due to rounding off errors\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "wavelength of scattered X-rays is 0.1 Angstrom\n",
+        "Energy of incident photon is 0.138 MeV\n",
+        "Energy of scattered photon is 0.0924 MeV\n",
+        "answer for energy of scattered photon given in the book varies due to rounding off errors\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.7, Page number 191"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "h=6.625*10**-34;      #Plank's constant\n",
+      "m=9.12*10**-31;       #mass of electron(kg)\n",
+      "\n",
+      "#Calculation\n",
+      "#for 1st quantum state\n",
+      "nx=1;\n",
+      "ny=1;\n",
+      "nz=1;\n",
+      "L=1;\n",
+      "E1=h**2*(nx**2+ny**2+nz**2)/(8*m*L**2);       #energy in first quantum state(J)\n",
+      "#for 2nd quantum state (nx^2+ny^2+nz^2)=6\n",
+      "L=1;\n",
+      "E=h**2*6/(8*m*L**2);                   #energy in second quantum state(J)\n",
+      "\n",
+      "#Result\n",
+      "print \"energy in first quantum state is\",round(E1/1e-37,3),\"*10^-37 J\"\n",
+      "print \"energy in second quantum state is\",round(E/1e-37,2),\"*10^-37 J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "energy in first quantum state is 1.805 *10^-37 J\n",
+        "energy in second quantum state is 3.61 *10^-37 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example number 9.8, Page number 194"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import modules\n",
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "h=6.625*10**-34;       #Plank's constant\n",
+      "m=9.12*10**-31;        #mass of electron(kg)\n",
+      "L=2.5*10**-10;         #width of square(m)\n",
+      "e=1.6*10**-19;         #the charge on electron(C)\n",
+      "n1=1;\n",
+      "n2=2;\n",
+      "n3=3;\n",
+      "\n",
+      "#Calculation\n",
+      "E1=n1**2*h**2/(8*m*L**2*e);          #1st lowest quantum energy(eV)\n",
+      "E2=n2**2*E1;             #2nd lowest quantum energy(eV)\n",
+      "E3=n3**2*E1;             #3rd lowest quantum energy(eV) \n",
+      " \n",
+      "#Result\n",
+      "print \"the lowest 3 quantum energies are\",int(E1),\"eV,\",int(E2),\"eV and\",int(E3),\"eV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the lowest 3 quantum energies are 6 eV, 24 eV and 54 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Physics/screenshots/1.png b/Modern_Physics/screenshots/1.png
new file mode 100755
index 00000000..4282275d
--- /dev/null
+++ b/Modern_Physics/screenshots/1.png
diff --git a/Modern_Physics/screenshots/2.png b/Modern_Physics/screenshots/2.png
new file mode 100755
index 00000000..9e1e8cf6
--- /dev/null
+++ b/Modern_Physics/screenshots/2.png
diff --git a/Modern_Physics/screenshots/3.png b/Modern_Physics/screenshots/3.png
new file mode 100755
index 00000000..662e6b45
--- /dev/null
+++ b/Modern_Physics/screenshots/3.png
diff --git a/Modern_Physics/screenshots/Untitled.png b/Modern_Physics/screenshots/Untitled.png
new file mode 100755
index 00000000..f61351dc
--- /dev/null
+++ b/Modern_Physics/screenshots/Untitled.png
diff --git a/Modern_Physics/screenshots/Untitled_1.png b/Modern_Physics/screenshots/Untitled_1.png
new file mode 100755
index 00000000..f61351dc
--- /dev/null
+++ b/Modern_Physics/screenshots/Untitled_1.png
diff --git a/Modern_Physics/screenshots/Untitled_2.png b/Modern_Physics/screenshots/Untitled_2.png
new file mode 100755
index 00000000..f61351dc
--- /dev/null
+++ b/Modern_Physics/screenshots/Untitled_2.png
diff --git a/Modern_Physics/screenshots/finding_average_deflection.png b/Modern_Physics/screenshots/finding_average_deflection.png
new file mode 100755
index 00000000..16fc385b
--- /dev/null
+++ b/Modern_Physics/screenshots/finding_average_deflection.png
diff --git a/Modern_Physics/screenshots/finding_average_deflection_1.png b/Modern_Physics/screenshots/finding_average_deflection_1.png
new file mode 100755
index 00000000..16fc385b
--- /dev/null
+++ b/Modern_Physics/screenshots/finding_average_deflection_1.png
diff --git a/Modern_Physics/screenshots/finding_energy_of_sodium.png b/Modern_Physics/screenshots/finding_energy_of_sodium.png
new file mode 100755
index 00000000..ce229f35
--- /dev/null
+++ b/Modern_Physics/screenshots/finding_energy_of_sodium.png
diff --git a/Modern_Physics/screenshots/finding_time_interval.png b/Modern_Physics/screenshots/finding_time_interval.png
new file mode 100755
index 00000000..b5bb1995
--- /dev/null
+++ b/Modern_Physics/screenshots/finding_time_interval.png
diff --git a/Modern_Physics/screenshots/findning_uncertainity.png b/Modern_Physics/screenshots/findning_uncertainity.png
new file mode 100755
index 00000000..9566d5b9
--- /dev/null
+++ b/Modern_Physics/screenshots/findning_uncertainity.png
diff --git a/Modern_Physics/screenshots/findning_uncertainity_1.png b/Modern_Physics/screenshots/findning_uncertainity_1.png
new file mode 100755
index 00000000..9566d5b9
--- /dev/null
+++ b/Modern_Physics/screenshots/findning_uncertainity_1.png
diff --git a/Modern_Physics/screenshots/findning_uncertainity_2.png b/Modern_Physics/screenshots/findning_uncertainity_2.png
new file mode 100755
index 00000000..9566d5b9
--- /dev/null
+++ b/Modern_Physics/screenshots/findning_uncertainity_2.png
diff --git a/Modern_Physics/screenshots/mass_of_electron_and_proton.png b/Modern_Physics/screenshots/mass_of_electron_and_proton.png
new file mode 100755
index 00000000..476ca90a
--- /dev/null
+++ b/Modern_Physics/screenshots/mass_of_electron_and_proton.png
diff --git a/Modern_Physics/screenshots/uncertainity.png b/Modern_Physics/screenshots/uncertainity.png
new file mode 100755
index 00000000..d10725b3
--- /dev/null
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author	hardythe1	2015-04-07 15:58:05 +0530
committer	hardythe1	2015-04-07 15:58:05 +0530
commit	92cca121f959c6616e3da431c1e2d23c4fa5e886 (patch)
tree	205e68d0ce598ac5caca7de839a2934d746cce86 /Modern_Physics
parent	b14c13fcc6bb6d01c468805d612acb353ec168ac (diff)
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