added books

1 files changed, 156 insertions, 106 deletions

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": {}