adding book

1 files changed, 37 insertions, 310 deletions

diff --git a/Engineering_Physics/Chapter4_1.ipynb b/Engineering_Physics/Chapter4_1.ipynb
index c2e7944f..5651b165 100755
--- a/Engineering_Physics/Chapter4_1.ipynb
+++ b/Engineering_Physics/Chapter4_1.ipynb
@@ -1,7 +1,6 @@
 {
  "metadata": {
-  "name": "",
-  "signature": "sha256:bd975238ecc341317a545ed613e73ea0b105f0af115e6d7857237510924e96a0"
+  "name": "Chapter4"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -12,61 +11,25 @@
      "cell_type": "heading",
      "level": 1,
      "metadata": {},
-     "source": [
-      "4: Diffraction"
-     ]
+     "source": "4: Quantum Physics"
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": [
-      "Example number 4.1, Page number 91"
-     ]
+     "source": "Example number 4.1, Page number 117"
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": [
-      " \n",
-      "#importing modules\n",
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "D = 50;    #Distance between source and the screen(cm)\n",
-      "lamda = 6563;    #Wavelength of light of parallel rays(A)\n",
-      "d = 0.385;        #Width of the slit(mm)\n",
-      "n1 = 1;    #Order of diffraction for first minimum\n",
-      "n2 = 5;    #Order of diffraction for fifth minimum\n",
-      "\n",
-      "#Calculation\n",
-      "lamda = lamda*10**-8;        #Wavelength of light of parallel rays(cm)\n",
-      "d = d*10**-1;      #Width of the slit(cm)\n",
-      "#As sin(theta1) = n*lambda/d = x1/D, solving for x1\n",
-      "x1 = n1*lamda*D/d;    #Distance from the centre of the principal maximum to the first minimum(cm)\n",
-      "x1 = x1*10;         #Distance from the centre of the principal maximum to the first minimum(mm)\n",
-      "x1 = math.ceil(x1*10**3)/10**3;     #rounding off the value of x1 to 3 decimals\n",
-      "x2 = n2*lamda*D/d;    #Distance from the centre of the principal maximum to the fifth minimum(cm)\n",
-      "x2 = x2*10;         #Distance from the centre of the principal maximum to the fifth minimum(mm)\n",
-      "x2 = math.ceil(x2*10**3)/10**3;     #rounding off the value of x2 to 3 decimals\n",
-      "\n",
-      "#Result\n",
-      "print \"The Distance from the centre of the principal maximum to the first minimum is\",x1, \"mm\"\n",
-      "print \"The Distance from the centre of the principal maximum to the fifth minimum is\",x2, \"mm\"\n",
-      "\n",
-      "#answer for x2 given in the book is wrong"
-     ],
+     "input": "#importing modules\nimport math\n\n#Variable declaration\nc = 3*10**8;       #velocity of light(m/sec)\nh = 6.62*10**-34;     #planck's constant\nlamda = 1.2;          #wavelength of photon(Angstrom)\ne = 1.6*10**-19;      #conversion factor from J to eV\n\n#Calculation\nlamda = lamda*10**-10;       #wavelength of photon(m)\nE = (h*c)/(lamda*e);         #energy of photon(eV)\nE=math.ceil(E*10)/10;   #rounding off to 1 decimal\np = h/lamda;            #momentum of photon(kg m/s)\n\n#Result\nprint \"energy of the photon is\",E,\"eV\"\nprint \"momentum of the photon is\",p,\"kg m/s\"",
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": [
-        "The Distance from the centre of the principal maximum to the first minimum is 0.853 mm\n",
-        "The Distance from the centre of the principal maximum to the fifth minimum is 4.262 mm\n"
-       ]
+       "text": "energy of the photon is 10343.8 eV\nmomentum of the photon is 5.51666666667e-24 kg m/s\n"
       }
      ],
      "prompt_number": 1
@@ -75,405 +38,169 @@
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": [
-      "Example number 4.2, Page number 91"
-     ]
+     "source": "Example number 4.2, Page number 117"
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": [
-      " \n",
-      "#importing modules\n",
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "D = 0.04;    #Diameter of circular aperture(cm)\n",
-      "f = 20;      #Focal length of convex lens(cm)\n",
-      "lamda = 6000;    #Wavelength of light used(A)\n",
-      "\n",
-      "#Calculation\n",
-      "lamda = lamda*10**-8;    #Wavelength of light used(cm)\n",
-      "#We have sin(theta) = 1.22*lambda/D = theta, for small theta\n",
-      "#For first dark ring\n",
-      "theta = 1.22*lamda/D;    #The half angular width at central maximum(rad)\n",
-      "r1 = theta*f;    #The half width of central maximum for first dark ring(cm)\n",
-      "r1 = r1*10**2;\n",
-      "#We have sin(theta) = 5.136*lambda/(%pi*D) = theta, for small theta\n",
-      "#For second dark ring\n",
-      "theta = 5.136*lamda/(math.pi*D);    #The half angular width at central maximum(rad)\n",
-      "r2 = theta*f;    #The half width of central maximum for second dark ring(cm)\n",
-      "r2 = r2*10**2;\n",
-      "r2 = math.ceil(r2*100)/100;     #rounding off the value of r2 to 2 decimals\n",
-      "\n",
-      "#Result\n",
-      "print \"The radius of first dark ring is\",r1,\"*10**-2 cm\"\n",
-      "print \"The radius of second dark ring is\",r2,\"*10**-2 cm\""
-     ],
+     "input": "#importing modules\nimport math\n\n#Variable declaration\nh = 6.625*10**-34;     #planck's constant\nnew = 900;             #frequency(kHz)\nE1 = 10;               #power radiated(kW)\n\n#Calculation\nE1 = E1*10**3;         #power radiated(W)\nnew = new*10**3;       #frequency(Hz)\nE = h*new;         #energy of photon(J)\nN = E1/E;          #number of photons emitted \n\n#Result\nprint \"number of photons emitted per second is\",N",
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": [
-        "The radius of first dark ring is 3.66 *10**-2 cm\n",
-        "The radius of second dark ring is 4.91 *10**-2 cm\n"
-       ]
+       "text": "number of photons emitted per second is 1.67714884696e+31\n"
       }
      ],
-     "prompt_number": 6
+     "prompt_number": 2
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": [
-      "Example number 4.3, Page number 92"
-     ]
+     "source": "Example number 4.3, Page number 118"
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": [
-      " \n",
-      "#importing modules\n",
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration\n",
-      "n = 2;    #Order of diffraction\n",
-      "lamda = 650;    #Wavelength of light used(nm)\n",
-      "d = 1.2*10**-3;    #Distance between two consecutive slits of grating(cm)\n",
-      "\n",
-      "#Calculation\n",
-      "#We have sin(theta) = n*N*lambda = n*lambda/d, solving for theta\n",
-      "lamda = lamda*10**-9;     #Wavelength of light used(m)\n",
-      "d = d*10**-2;    #Distance between two consecutive slits of grating(m)\n",
-      "a=n*lamda/d;\n",
-      "theta = math.asin(a);     #Angle at which the 650 nm light produces a second order maximum(rad)\n",
-      "theta = theta*57.2957795;     #angle in degrees\n",
-      "theta = math.ceil(theta*10**2)/10**2;     #rounding off the value of theta to 2 decimals\n",
-      "\n",
-      "#Result\n",
-      "print \"The angle at which the light produces a second order maximum is\",theta, \"degrees\""
-     ],
+     "input": "#importing modules\nimport math\n\n#Variable declaration\nc = 3*10**8;       #velocity of light(m/sec)\nh = 6.63*10**-34;     #planck's constant\nlamda = 5893;         #wavelength of photon(Angstrom)\nE1 = 100;             #power of lamp(W) \n\n#Calculation\nlamda = lamda*10**-10;       #wavelength of photon(m)\nE = h*c/lamda;         #energy of photon(J)\nN = E1/E;          #number of photons emitted \n\n#Result\nprint \"number of photons emitted per second is\",N\nprint \"answer given in the book is wrong\"",
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": [
-        "The angle at which the light produces a second order maximum is 6.22 degrees\n"
-       ]
+       "text": "number of photons emitted per second is 2.96279537456e+20\nanswer given in the book is wrong\n"
       }
      ],
-     "prompt_number": 7
+     "prompt_number": 3
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": [
-      "Example number 4.4, Page number 92"
-     ]
+     "source": "Example number 4.4, Page number 118"
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": [
-      " \n",
-      "#importing modules\n",
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration\n",
-      "lamda = 650;    #Wavelength of light used(nm)\n",
-      "N = 6000;    #Number of lines per cm on grating\n",
-      "theta = 90;    #Angle at which the highest spectral order is obtained(degrees)\n",
-      "\n",
-      "#Calculation\n",
-      "theta = theta*0.0174532925;     #Angle at which the highest spectral order is obtained(rad)\n",
-      "#We have sin(theta) = n*N*lambda, solving for n\n",
-      "lamda = lamda*10**-9;    #Wavelength of light used(m)\n",
-      "N = N*10**2;    #Number of lines per m on grating\n",
-      "n = math.sin(theta)/(N*lamda);      #The highest order of spectra with diffraction grating\n",
-      "n = math.ceil(n*10**3)/10**3;     #rounding off the value of theta to 3 decimals\n",
-      "i,d = divmod(n, 1);     #divides the value of n into integer and decimal parts where i is integer\n",
-      "\n",
-      "#Result\n",
-      "print \"value of n is\",n\n",
-      "print \"The highest order of spectra obtained with diffraction grating is\",i\n"
-     ],
+     "input": "#importing modules\nimport math\n\n#Variable declaration\nc = 3*10**8;       #velocity of light(m/sec)\nh = 6.6*10**-34;     #planck's constant\nm0 = 9.1*10**-31;        #mass of photon(kg)\ntheta = 30;              #viewing angle(degrees)\nlamda = 2.8*10**-10;         #wavelength of photon(m)\n\n#Calculation\nx = math.pi/180;       #conversion factor from degrees to radians\ntheta = theta*x;       #viewing angle(radian) \nlamda_dash = (2*h*(math.sin(theta/2))**2/(m0*c))+lamda;        #wavelength of scattered radiation(m)\nlamda_dash = lamda_dash*10**10;           #wavelength of scattered radiation(Angstrom)\nlamda_dash=math.ceil(lamda_dash*10**5)/10**5;   #rounding off to 5 decimals\n\n#Result\nprint \"wavelength of scattered radiation is\",lamda_dash,\"Angstrom\"",
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": [
-        "value of n is 2.565\n",
-        "The highest order of spectra obtained with diffraction grating is 2.0\n"
-       ]
+       "text": "wavelength of scattered radiation is 2.80324 Angstrom\n"
       }
      ],
-     "prompt_number": 8
+     "prompt_number": 4
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": [
-      "Example number 4.5, Page number 92"
-     ]
+     "source": "Example number 4.5, Page number 119"
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": [
-      " \n",
-      "#importing modules\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "N = 4000;    #Number of lines per cm on grating\n",
-      "#For Blue Line\n",
-      "lamda1 = 450;    #Wavelength of blue light(nm)\n",
-      "n1 = 3;    #Order of diffraction spectrum\n",
-      "#For Red Line\n",
-      "lamda2 = 700;    #Wavelength of red light(nm)\n",
-      "n2 = 2;    #Order of diffraction spectrum\n",
-      "\n",
-      "#Calculation\n",
-      "N = N*10**2;    #Number of lines per m on grating\n",
-      "lamda1 = lamda1*10**-9;     #Wavelength of blue light(m)\n",
-      "lamda2 = lamda2*10**-9;     #Wavelength of red light(m)\n",
-      "#We have sin(theta) = n*N*lambda, solving for sin(theta)\n",
-      "sin_theta_3 = n1*N*lamda1;    #Sine of angle at third order diffraction \n",
-      "sin_theta_2 = n2*N*lamda2;    #Sine of angle at second order diffraction\n",
-      "\n",
-      "#Result\n",
-      "print \"Sine of angle at third order diffraction is\",sin_theta_3\n",
-      "print \"Sine of angle at second order diffraction is\",sin_theta_2    \n",
-      "#Check for overlapping\n",
-      "if (sin_theta_2-sin_theta_3)<0.05:\n",
-      "    print \"The two orders overlap\"\n",
-      "else:\n",
-      "     print \"The two orders do not overlap\"   "
-     ],
+     "input": "#importing modules\nimport math\n\n#Variable declaration\nh = 6.6*10**-34;     #planck's constant\nm = 0.040;        #mass of bullet(kg)\nv = 1;            #speed of bullet(km/s)\n\n#Calculation\nv = v*10**3;      #speed of bullet(m/s)\np = m*v;          #momemtun of bullet(kg m/s)\nlamda = h/p;      #deBroglie wavelength(m)\nlamda = lamda*10**10;     #deBroglie wavelength(Angstrom)\n\n#Result\nprint \"deBroglie wavelength is\",lamda,\"Angstrom\"",
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": [
-        "Sine of angle at third order diffraction is 0.54\n",
-        "Sine of angle at second order diffraction is 0.56\n",
-        "The two orders overlap\n"
-       ]
+       "text": "deBroglie wavelength is 1.65e-25 Angstrom\n"
       }
      ],
-     "prompt_number": 9
+     "prompt_number": 5
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": [
-      "Example number 4.6, Page number 93"
-     ]
+     "source": "Example number 4.6, Page number 119"
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": [
-      " \n",
-      "#importing modules\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "n = 1;    #Order of diffraction spectrum\n",
-      "N = 6000;    #Number of lines per cm on diffraction grating\n",
-      "D = 2;    #Distance of screen from the source(m)\n",
-      "lamda1 = 400;    #Wavelength of blue light(nm)\n",
-      "lamda2 = 750;    #Wavelength of blue light(nm)\n",
-      "\n",
-      "#Calculation\n",
-      "N = N*10**2;    #Number of lines per m on grating\n",
-      "lamda1 = lamda1*10**-9;     #Wavelength of blue light(m)\n",
-      "lamda2 = lamda2*10**-9;     #Wavelength of blue light(m)\n",
-      "#We have sin(theta1) = n*N*lamda1, solving for theta1\n",
-      "theta1 = math.asin(n*N*lamda1);    #Angle at first order diffraction for Blue light(rad)\n",
-      "theta1_d = theta1*57.2957795;     #Angle at first order diffraction for Blue light(degrees)\n",
-      "theta2 = math.asin(n*N*lamda2);    #Angle at first order diffraction for Red light(rad)\n",
-      "theta2_d = theta2*57.2957795;      #Angle at first order diffraction for Red light(degrees)\n",
-      "x1 = D*math.tan(theta1);    #Half width position at central maximum for blue color(m)\n",
-      "x2 = D*math.tan(theta2);    #Half width position at central maximum for red color(m)\n",
-      "x = x2-x1;      #width of first order spectrum on the screen(m)\n",
-      "x = x*10**2;    #width of first order spectrum on the screen(cm)\n",
-      "x = math.ceil(x*10**2)/10**2;     #rounding off the value of x to 2 decimals\n",
-      "\n",
-      "#Result\n",
-      "print \"The width of first order spectrum on the screen is\",x, \"cm\""
-     ],
+     "input": "#importing modules\nimport math\n\n#Variable declaration\nn = 1;               #lowest energy state\na = 0.1;             #width of box(nm)\nh = 6.625*10**-34;     #planck's constant\ne = 1.602*10**-19;       #conversion factor from J to eV\nm = 9.11*10**-31;        #mass of particle(kg)\n\n#Calculation\na = a*10**-9;        #width of box(m)\nE = (n**2)*(h**2)/(8*m*(a**2));      #energy of particle(J)\nE_eV = E/e;             #energy of particle(eV)\nE_eV=math.ceil(E_eV*10)/10;   #rounding off to 1 decimal\n\n#Result\nprint \"energy of particle is\",E,\"J or\",E_eV,\"eV\" ",
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": [
-        "The width of first order spectrum on the screen is 51.34 cm\n"
-       ]
+       "text": "energy of particle is 6.02231407794e-18 J or 37.6 eV\n"
       }
      ],
-     "prompt_number": 10
+     "prompt_number": 6
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": [
-      "Example number 4.7, Page number 93"
-     ]
+     "source": "Example number 4.7, Page number 120"
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": [
-      " \n",
-      "#importing modules\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "w = 5;    #Width of the grating(cm)\n",
-      "N = 32;    #Number of lines per mm on grating\n",
-      "lamda = 640;    #Wavelength of light(nm)\n",
-      "n = 2;    #Order of diffraction\n",
-      "\n",
-      "#Calculation\n",
-      "N= N*10;    #Number of lines per cm on grating\n",
-      "N0 = w*N;   #Total number of lines on the grating\n",
-      "d_lambda = lamda/(n*N0);    #Separation between wavelengths(nm)\n",
-      "\n",
-      "#Result\n",
-      "print \"The separation between wavelengths which the grating can just resolve is\",d_lambda, \"nm\""
-     ],
+     "input": "#importing modules\nimport math\n\n#Variable declaration\nn = 1;               #lowest energy state\na = 4;             #width of well(nm)\nh = 6.625*10**-34;     #planck's constant\ne = 1.6025*10**-19;       #conversion factor from J to eV\nm = 9.11*10**-31;        #mass of electron(kg)\n\n#Calculation\na = a*10**-9;        #width of box(m)\nE = (n**2)*(h**2)/(8*m*(a**2));      #energy of particle(J)\nE_eV = E/e;             #energy of particle(eV)\nE_eV=math.ceil(E_eV*10**4)/10**4;   #rounding off to 4 decimals\n\n#Result\nprint \"minimum energy of electron is\",E,\"J or\",E_eV,\"eV\" ",
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": [
-        "The separation between wavelengths which the grating can just resolve is 0.2 nm\n"
-       ]
+       "text": "minimum energy of electron is 3.76394629871e-21 J or 0.0235 eV\n"
       }
      ],
-     "prompt_number": 11
+     "prompt_number": 7
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": [
-      "Example number 4.8, Page number 93"
-     ]
+     "source": "Example number 4.8, Page number 120"
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": [
-      " \n",
-      "#importing modules\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration\n",
-      "lamda = 550;    #Wavelength of light(nm)\n",
-      "D = 3.2;    #Diameter of circular lens(cm)\n",
-      "f = 24;     #Focal length of the lens(cm)  \n",
-      "\n",
-      "#Calculation\n",
-      "lamda = lamda*10**-9;    #Wavelength of light(m)\n",
-      "D = D*10**-2;    #Diameter of circular lens(m)\n",
-      "theta_min = 1.22*lamda/D;    #Minimum angle of resolution provided by the lens(rad)\n",
-      "#As delta_x/f = theta_min, solving for delta_x\n",
-      "f = f*10**-2;    #Focal length of the lens(m) \n",
-      "delta_x = theta_min*f;     #Separation of the centres of the images in the focal plane of lens(m)\n",
-      "delta_x = delta_x*10**6;    #Separation of the centres of the images in the focal plane of lens(micro m)\n",
-      " \n",
-      "#Result\n",
-      "print \"The separation of the centres of the images in the focal plane is\",round(delta_x), \"micro-metre\""
-     ],
+     "input": "#importing modules\nimport math\n\n#Variable declaration\nn1 = 1;               #lowest energy state\nn2 = 6;               #for 6th excited state\na = 0.1;             #width of box(nm)\nh = 6.625*10**-34;     #planck's constant\ne = 1.602*10**-19;       #conversion factor from J to eV\nm = 9.11*10**-31;        #mass of electron(kg)\n\n#Calculation\na = a*10**-9;        #width of box(m)\nE1 = (n1**2)*(h**2)/(8*m*(a**2));      #energy of electron in ground state(J)\nE6 = (n2**2)*(h**2)/(8*m*(a**2));      #energy of electron in excited state(J)\nE = E6-E1;                          #energy required to excite the electron(J)\nE_eV = E/e;             #energy required to excite the electron(eV)\nE_eV=math.ceil(E_eV*10)/10;   #rounding off to 1 decimal\n\n#Result\nprint \"energy required to excite the electron is\",E,\"J or\",E_eV,\"eV\" \nprint \"answer for energy in eV given in the book is wrong\"",
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": [
-        "The separation of the centres of the images in the focal plane is 5.0 micro-metre\n"
-       ]
+       "text": "energy required to excite the electron is 2.10780992728e-16 J or 1315.8 eV\nanswer for energy in eV given in the book is wrong\n"
       }
      ],
-     "prompt_number": 12
+     "prompt_number": 8
     },
     {
      "cell_type": "heading",
      "level": 2,
      "metadata": {},
-     "source": [
-      "Example number 4.9, Page number 94"
-     ]
+     "source": "Example number 4.9, Page number 121"
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": [
-      " \n",
-      "#importing modules\n",
-      "import math\n",
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration\n",
-      "lamda = 550;    #Wavelength of light(nm)\n",
-      "D = 20;     #Diameter of objective of telescope(cm)\n",
-      "d = 6;     #Distance of two points from the objective of telescope(km)\n",
-      "\n",
-      "#Calculation\n",
-      "lamda = lamda*10**-9;    #Wavelength of light(m)\n",
-      "D = D*10**-2;    #Diameter of objective of telescope(m)\n",
-      "d = d*10**3;    #Distance of two points from the objective of telescope(m)\n",
-      "theta = 1.22*lamda/D;    #Angular separation between two points(rad)\n",
-      "x = theta*d;    #Linear separation between two points(m)\n",
-      "x = x*10**3;    #Linear separation between two points(mm)\n",
-      "\n",
-      "#Result\n",
-      "print \"The linear separation between two points is\",x, \"mm\""
-     ],
+     "input": "#importing modules\nimport math\n\n#Variable declaration\nh = 6.625*10**-34;     #planck's constant\nc = 3*10**8;       #velocity of light(m/sec)\nm0 = 9.11*10**-31;        #rest mass of electron(kg)\nphi = 90;              #angle of scattering(degrees)\nx = math.pi/180;       #conversion factor from degrees to radians\n\n#Calculation\nphi = phi*x;           ##angle of scattering(radian)\ndelta_lamda = h*(1-math.cos(phi))/(m0*c);         #change in wavelength(m)\n\n#Result\nprint \"change in wavelength of X-ray photon is\",delta_lamda,\"m\"",
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": [
-        "The linear separation between two points is 20.13 mm\n"
-       ]
+       "text": "change in wavelength of X-ray photon is 2.42407610684e-12 m\n"
       }
      ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [],
-     "language": "python",
-     "metadata": {},
-     "outputs": []
+     "prompt_number": 9
     }
    ],
    "metadata": {}