added books

1 files changed, 96 insertions, 294 deletions

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": {}