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+{
+ "metadata": {
+  "name": "Chapter 4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Quantum Physics"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example number 4.1, Page number 133 "
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#To calculate the change in wavelength\n\n#importing modules\nimport math\n\n#Variable declaration\nh=6.63*10**-34;     #plancks constant in Js\nm0=9.1*10**-31;     #mass of the electron in kg\nc=3*10**8;       #velocity of light in m/s\nphi=135;        #angle of scattering in degrees\nphi=phi*0.0174532925   #converting degrees to radians  \n\n#Calculation\ndelta_lamda=(h*(1-math.cos(phi)))/(m0*c);\n\n#Result\nprint(\"change in wavelength in metres is\",delta_lamda);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "('change in wavelength in metres is', 4.1458307496867315e-12)\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example number 4.2, Page number 134 "
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#To calculate the compton shift, wavelength,energy and angle\n\n#importing modules\nimport math\n\n#Variable declaration\nh=6.63*10**-34;     #plancks constant in Js\nm0=9.1*10**-31;     #mass of the electron in kg\nc=3*10**8;       #velocity of light in m/s\nlamda=2;     #wavelength in angstrom\nlamdaA=lamda*10**-10;      #converting lamda from Angstrom to m\nphi=90;     #angle of scattering in degrees\nphi=phi*0.0174532925   #converting degrees to radians  \n\n#Calculation\ndelta_lamda=(h*(1-math.cos(phi)))/(m0*c);\ndelta_lamda=delta_lamda*10**10;     #converting delta_lamda from m to Angstrom\ndelta_lamda=math.ceil(delta_lamda*10**5)/10**5;   #rounding off to 5 decimals\nlamda_dash=delta_lamda+lamda;\nlamdaA_dash=lamda_dash*10**-10;     #converting lamda_dash from Angstrom to m\n#energy E=h*new-h*new_dash\nE=h*c*((1/lamdaA)-(1/lamdaA_dash));\nEeV=E/(1.602176565*10**-19);      #converting J to eV\nEeV=math.ceil(EeV*10**3)/10**3;   #rounding off to 3 decimals\nnew=c/lamda;\nnew_dash=c/lamda_dash;\ntheta=math.atan((h*new*math.sin(phi))/((h*new)-(h*new_dash*math.cos(phi))));\ntheta=theta*57.2957795;       #converting radians to degrees\n\n#Result\nprint(\"change in compton shift in Angstrom is\",delta_lamda);\nprint(\"wavelength of scattered photons in Angstrom is\",lamda_dash);\nprint(\"energy of recoiling electron in J is\",E);\nprint(\"energy of recoiling electron in eV is\",EeV);\nprint(\"angle at which recoiling electron appears in degrees is\",int(theta));\n\n#answers given in the book are wrong",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "('change in compton shift in Angstrom is', 0.02429)\n('wavelength of scattered photons in Angstrom is', 2.02429)\n('energy of recoiling electron in J is', 1.1933272900621974e-17)\n('energy of recoiling electron in eV is', 74.482)\n('angle at which recoiling electron appears in degrees is', 45)\n"
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example number 4.3, Page number 135"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#To calculate the compton shift, wavelength of the scattered photon\n\n#importing modules\nimport math\n\n#Variable declaration\nh=6.626*10**-34;     #plancks constant in Js\nm0=9.1*10**-31;     #mass of the electron in kg\nc=3*10**8;       #velocity of light in m/s\nphi=60;     #angle of scattering in degrees\nphi=phi*0.0174532925;      #converting degrees to radians\nE=10**6;    #energy of photon in eV\nE=E*1.6*10**-19;    #converting eV into J\n\n#Calculation\ndelta_lamda=(h*(1-math.cos(phi)))/(m0*c);\ndelta_lamda=delta_lamda*10**10;    #converting metre to angstrom\ndelta_lamda=math.ceil(delta_lamda*10**4)/10**4;   #rounding off to 4 decimals\nlamda=(h*c)/E;\nlamdaA=lamda*10**10;     #converting metre to angstrom\nlamda_dash=delta_lamda+lamdaA;\nlamda_dash=math.ceil(lamda_dash*10**3)/10**3;   #rounding off to 3 decimals\n\n#Result\nprint(\"compton shift in angstrom is\",delta_lamda);\nprint(\"energy of incident photon in m\",lamda);\nprint(\"wavelength of scattered photons in angstrom is\",lamda_dash);\n\n#answer for wavelength of scattered photon given in the book is wrong",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "('compton shift in angstrom is', 0.0122)\n('energy of incident photon in m', 1.242375e-12)\n('wavelength of scattered photons in angstrom is', 0.025)\n"
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example number 4.4, Page number 135"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#To calculate the number of photons emitted\n\n#importing modules\nimport math\n\n#Variable declaration\nh=6.626*10**-34;     #plancks constant in Js\nc=3*10**8;       #velocity of light in m/s\nlamda=5893;     #wavelength in angstrom\nP=60;      #output power in Watt\n\n#Calculation\nlamda=lamda*10**-10;    #wavelength in metre\nE=(h*c)/lamda;\nEeV=E/(1.602176565*10**-19);      #converting J to eV\nEeV=math.ceil(EeV*10**4)/10**4;   #rounding off to 4 decimals\nN=P/E;\n\n#Result\nprint(\"energy of photon in J is\",E);\nprint(\"energy of photon in eV is\",EeV);\nprint(\"number of photons emitted per se cond is\",N);\n\n#answer for energy in eV given in the book is wrong",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "('energy of photon in J is', 3.373154590191753e-19)\n('energy of photon in eV is', 2.1054)\n('number of photons emitted per se cond is', 1.7787503773015396e+20)\n"
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example number 4.5, Page number 136"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#To calculate the momentum, energy and mass of a photon\n\n#importing modules\nimport math\n\n#Variable declaration\nh=6.626*10**-34;     #plancks constant in Js\nc=3*10**8;       #velocity of light in m/s\nlamda=10;     #wavelength in angstrom\n\n#Calculation\nlamda=lamda*10**-10;    #wavelength in metre\nE=(h*c)/lamda;\nEeV=E/(1.602176565*10**-19);      #converting J to eV\nEeV=EeV*10**-3;     #converting eV to keV\nEeV=math.ceil(EeV*10**3)/10**3;   #rounding off to 3 decimals\nP=h/lamda;\nM=h/(lamda*c);\n\n#Result\nprint(\"energy of photon in J is\",E);\nprint(\"energy of photon in keV is\",EeV);\nprint(\"momentum in kg m/sec is\",P);\nprint(\"mass of photon in kg is\",M);\n\n#answer for energy of photon in keV given in the book is wrong by 1 decimal",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "('energy of photon in J is', 1.9878e-16)\n('energy of photon in keV is', 1.241)\n('momentum in kg m/sec is', 6.626e-25)\n('mass of photon in kg is', 2.2086666666666664e-33)\n"
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example number 4.6, Page number 136"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#To calculate the wavelength\n\n#importing modules\nimport math\n\n#Variable declaration\nh=6.626*10**-34;     #plancks constant in Js\nm=9.1*10**-31;     #mass of the electron in kg\ne=1.602*10**-19;\nV=1.25;    #potential difference in kV\n\n#Calculation\nV=V*10**3;    #converting kV to V\nlamda=h/math.sqrt(2*m*e*V);\nlamda=lamda*10**10;     #converting metre to angstrom\nlamda=math.ceil(lamda*10**4)/10**4;   #rounding off to 4 decimals\n\n#Result\nprint(\"de Broglie wavelength in angstrom is\",lamda);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "('de Broglie wavelength in angstrom is', 0.3471)\n"
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example number 4.7, Page number 136"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#To calculate the de Broglie wavelength\n\n#importing modules\nimport math\n\n#Variable declaration\nE=45;    #energy of electron in eV\nE=E*1.6*10**-19;     #energy in J\nh=6.626*10**-34;     #plancks constant in Js\nm=9.1*10**-31;     #mass of the electron in kg\n\n#Calculation\nlamda=h/math.sqrt(2*m*E);\nlamda=lamda*10**10;   #converting metres to angstrom\nlamda=math.ceil(lamda*10**4)/10**4;   #rounding off to 4 decimals\n\n#Result\nprint(\"de Broglie wavelength in angstrom is\",lamda);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "('de Broglie wavelength in angstrom is', 1.8305)\n"
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example number 4.8, Page number 137"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#To calculate the de Broglie wavelength\n\n#importing modules\nimport math\n\n#Variable declaration\nv=10**7;      #velocity of electron in m/sec\nh=6.626*10**-34;     #plancks constant in Js\nm=9.1*10**-31;     #mass of the electron in kg\n\n#Calculation\nlamda=h/(m*v);\nlamda=lamda*10**10;   #converting metres to angstrom\nlamda=math.ceil(lamda*10**4)/10**4;   #rounding off to 4 decimals\n\n#Result\nprint(\"de Broglie wavelength in angstrom is\",lamda);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "('de Broglie wavelength in angstrom is', 0.7282)\n"
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example number 4.9, Page number 137"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#To calculate the de Broglie wavelength of alpha particle\n\n#importing modules\nimport math\n\n#Variable declaration\nV=1000;       #potential difference in V\nh=6.626*10**-34;     #plancks constant in Js\nm=1.67*10**-27;     #mass of proton in kg\ne=1.6*10**-19;      #charge of electron in J\n\n#Calculation\nlamda=h/math.sqrt(2*m*e*V);\n\n#Result\nprint(\"de Broglie wavelength of alpha particle in metre is\",lamda);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "('de Broglie wavelength of alpha particle in metre is', 9.063964727801313e-13)\n"
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example number 4.10, Page number 138"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#To calculate the probability of finding the particle\n\n#importing modules\nimport math\n\n#Variable declaration\nL=25;    #width of potential in armstrong\ndelta_x=0.05;   #interval in armstrong\nn=1;   #particle is in its least energy\nx=L/2;   #particle is at the centre\npi=180;   #angle in degrees\n\n#Calculation\npi=pi*0.0174532925;   #angle in radians\nL=L*10**-10;   #width in m\ndelta_x=delta_x*10**-10;   #interval in m\n#probability P = integration of (A**2)*(math.sin(n*pi*x/L))**2*delta_x\n#but A=math.sqrt(2/L)\n#since the particle is in a small interval integration need not be applied\n#therefore P=2*(L**(-1))*(math.sin(n*pi*x/L))**2*delta_x\nP=2*(L**(-1))*((math.sin(n*pi*x/L))**2)*delta_x;\nP=math.ceil(P*10**3)/10**3;   #rounding off to 3 decimals\n\n#Result\nprint(\"probability of finding the particle is\",P);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "('probability of finding the particle is', 0.004)\n"
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example number 4.11, Page number 138"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#To calculate the lowest energy of electron in eV\n\n#importing modules\nimport math\n\n#Variable declaration\nn=1;\nh=6.626*10**-34;     #plancks constant in Js\nm=9.1*10**-31;     #mass of the electron in kg\nL=1;    #width of potential well in angstrom\n\n#Calculation\nL=L*10**-10;     #converting angstrom into metre\nE=((n**2)*h**2)/(8*m*L**2);\nEeV=E/(1.6*10**-19);       #converting J to eV\nEeV=math.ceil(EeV*10**3)/10**3;   #rounding off to 3 decimals\n\n#Result\nprint(\"lowest energy of electron in J is\",E);\nprint(\"lowest energy of electron in eV is\",EeV);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "('lowest energy of electron in J is', 6.030752197802197e-18)\n('lowest energy of electron in eV is', 37.693)\n"
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example number 4.12, Page number 139"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#To calculate the lowest energy of the system\n\n#importing modules\nimport math\n\n#Variable declaration\nn=1;\nh=6.626*10**-34;     #plancks constant in Js\nm=9.1*10**-31;     #mass of the electron in kg\nL=1;    #width of potential well in angstrom\n\n#Calculation\nL=L*10**-10;     #converting angstrom into metre\nE=(2*(n**2)*h**2)/(8*m*L**2);\nE=E/(1.6*10**-19);       #converting J to eV\nE=math.ceil(E*10**3)/10**3;   #rounding off to 3 decimals\n\n#Result\nprint(\"lowest energy of system in eV is\",E);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "('lowest energy of system in eV is', 75.385)\n"
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example number 4.13, Page number 139"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#To calculate the lowest energy of the system and quantum numbers\n\n#importing modules\nimport math\n\n#Variable declaration\nh=6.626*10**-34;     #plancks constant in Js\nm=9.1*10**-31;     #mass of the electron in kg\nL=1;    #width of potential well in angstrom\n\n#Calculation\nL=L*10**-10;     #converting angstrom into metre\n#according to pauli's exclusion principle, 1st electron occupies n1=1 and second electron occupies n2=2\nn1=1;\nn2=2;\nE=((2*(n1**2)*h**2)/(8*m*L**2))+(((n2**2)*h**2)/(8*m*L**2));\nE=E/(1.6*10**-19);       #converting J to eV\nE=math.ceil(E*10**3)/10**3;   #rounding off to 3 decimals\n\n#Result\nprint(\"lowest energy of system in eV is\",E);\nprint(\"quantum numbers are\");\nprint(\"n=1,l=0,mL=0,mS=+1/2\");\nprint(\"n=1,l=0,mL=0,mS=-1/2\");\nprint(\"n=2,l=0,mL=0,mS=+1/2\");",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "('lowest energy of system in eV is', 226.154)\nquantum numbers are\nn=1,l=0,mL=0,mS=+1/2\nn=1,l=0,mL=0,mS=-1/2\nn=2,l=0,mL=0,mS=+1/2\n"
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example number 4.14, Page number 140"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#To calculate the mass of the particle\n\n#Variable declaration\nn=1;\nh=6.626*10**-34;     #plancks constant in Js\nL=100;    #width of potential well in angstrom\n\n#Calculation\nL=L*10**-10;     #converting angstrom into metre\nE=0.025;    #lowest energy in eV\nE=E*(1.6*10**-19);       #converting eV to J\nm=((n**2)*h**2)/(8*E*L**2);\n\n#Result\nprint(\"mass of the particle in kg is\",m);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "('mass of the particle in kg is', 1.3719961249999998e-31)\n"
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example number 4.15, Page number 141"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#To calculate the energy density\n\n#importing modules\nimport math\n\n#Variable declaration\nk=1.38*10**-23;\nT=6000;      #temperature in K\nh=6.626*10**-34;     #plancks constant in Js\nc=3*10**8;       #velocity of light in m/s\nlamda1=450;     #wavelength in nm\nlamda2=460;     #wavelength in nm\n\n#Calculation\nlamda1=lamda1*10**-9;     #converting nm to metre\nlamda2=lamda2*10**-9;     #converting nm to metre\nnew1=c/lamda1;\nnew2=c/lamda2;\nnew=(new1+new2)/2;\nA=math.exp((h*new)/(k*T));\nrho_v=(8*math.pi*h*new**3)/(A*c**3);\n\n#Result\nprint(\"energy density of the black body in J/m^3 is\",rho_v);\n\n#answer given in the book is wrong",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "('energy density of the black body in J/m^3 is', 9.033622836188887e-16)\n"
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file