{
 "metadata": {
  "name": "",
  "signature": "sha256:e581747b76e15afc0096179446c0fbd68c3566f21f4931be3d8fc722fc1225b8"
 },
 "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": [
      "\n",
      "\n",
      "#importing modules\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "h=6.63*10**-34;     #plancks constant in Js\n",
      "m0=9.1*10**-31;     #mass of the electron in kg\n",
      "c=3*10**8;       #velocity of light in m/s\n",
      "phi=135;        #angle of scattering in degrees\n",
      "phi=phi*0.0174532925   #converting degrees to radians  \n",
      "\n",
      "#Calculation\n",
      "delta_lamda=(h*(1-math.cos(phi)))/(m0*c);\n",
      "\n",
      "#Result\n",
      "print(\"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": [
      "\n",
      "\n",
      "#importing modules\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "h=6.63*10**-34;     #plancks constant in Js\n",
      "m0=9.1*10**-31;     #mass of the electron in kg\n",
      "c=3*10**8;       #velocity of light in m/s\n",
      "lamda=2;     #wavelength in angstrom\n",
      "lamdaA=lamda*10**-10;      #converting lamda from Angstrom to m\n",
      "phi=90;     #angle of scattering in degrees\n",
      "phi=phi*0.0174532925   #converting degrees to radians  \n",
      "\n",
      "#Calculation\n",
      "delta_lamda=(h*(1-math.cos(phi)))/(m0*c);\n",
      "delta_lamda=delta_lamda*10**10;     #converting delta_lamda from m to Angstrom\n",
      "delta_lamda=math.ceil(delta_lamda*10**5)/10**5;   #rounding off to 5 decimals\n",
      "lamda_dash=delta_lamda+lamda;\n",
      "lamdaA_dash=lamda_dash*10**-10;     #converting lamda_dash from Angstrom to m\n",
      "#energy E=h*new-h*new_dash\n",
      "E=h*c*((1/lamdaA)-(1/lamdaA_dash));\n",
      "EeV=E/(1.602176565*10**-19);      #converting J to eV\n",
      "EeV=math.ceil(EeV*10**3)/10**3;   #rounding off to 3 decimals\n",
      "new=c/lamda;\n",
      "new_dash=c/lamda_dash;\n",
      "theta=math.atan((h*new*math.sin(phi))/((h*new)-(h*new_dash*math.cos(phi))));\n",
      "theta=theta*57.2957795;       #converting radians to degrees\n",
      "\n",
      "#Result\n",
      "print(\"change in compton shift in Angstrom is\",delta_lamda);\n",
      "print(\"wavelength of scattered photons in Angstrom is\",lamda_dash);\n",
      "print(\"energy of recoiling electron in J is\",E);\n",
      "print(\"energy of recoiling electron in eV is\",EeV);\n",
      "print(\"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": [
      "\n",
      "\n",
      "#importing modules\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "h=6.626*10**-34;     #plancks constant in Js\n",
      "m0=9.1*10**-31;     #mass of the electron in kg\n",
      "c=3*10**8;       #velocity of light in m/s\n",
      "phi=60;     #angle of scattering in degrees\n",
      "phi=phi*0.0174532925;      #converting degrees to radians\n",
      "E=10**6;    #energy of photon in eV\n",
      "E=E*1.6*10**-19;    #converting eV into J\n",
      "\n",
      "#Calculation\n",
      "delta_lamda=(h*(1-math.cos(phi)))/(m0*c);\n",
      "delta_lamda=delta_lamda*10**10;    #converting metre to angstrom\n",
      "delta_lamda=math.ceil(delta_lamda*10**4)/10**4;   #rounding off to 4 decimals\n",
      "lamda=(h*c)/E;\n",
      "lamdaA=lamda*10**10;     #converting metre to angstrom\n",
      "lamda_dash=delta_lamda+lamdaA;\n",
      "lamda_dash=math.ceil(lamda_dash*10**3)/10**3;   #rounding off to 3 decimals\n",
      "\n",
      "#Result\n",
      "print(\"compton shift in angstrom is\",delta_lamda);\n",
      "print(\"energy of incident photon in m\",lamda);\n",
      "print(\"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": [
      "\n",
      "\n",
      "#importing modules\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "h=6.626*10**-34;     #plancks constant in Js\n",
      "c=3*10**8;       #velocity of light in m/s\n",
      "lamda=5893;     #wavelength in angstrom\n",
      "P=60;      #output power in Watt\n",
      "\n",
      "#Calculation\n",
      "lamda=lamda*10**-10;    #wavelength in metre\n",
      "E=(h*c)/lamda;\n",
      "EeV=E/(1.602176565*10**-19);      #converting J to eV\n",
      "EeV=math.ceil(EeV*10**4)/10**4;   #rounding off to 4 decimals\n",
      "N=P/E;\n",
      "\n",
      "#Result\n",
      "print(\"energy of photon in J is\",E);\n",
      "print(\"energy of photon in eV is\",EeV);\n",
      "print(\"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": [
      "\n",
      "\n",
      "#importing modules\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "h=6.626*10**-34;     #plancks constant in Js\n",
      "c=3*10**8;       #velocity of light in m/s\n",
      "lamda=10;     #wavelength in angstrom\n",
      "\n",
      "#Calculation\n",
      "lamda=lamda*10**-10;    #wavelength in metre\n",
      "E=(h*c)/lamda;\n",
      "EeV=E/(1.602176565*10**-19);      #converting J to eV\n",
      "EeV=EeV*10**-3;     #converting eV to keV\n",
      "EeV=math.ceil(EeV*10**3)/10**3;   #rounding off to 3 decimals\n",
      "P=h/lamda;\n",
      "M=h/(lamda*c);\n",
      "\n",
      "#Result\n",
      "print(\"energy of photon in J is\",E);\n",
      "print(\"energy of photon in keV is\",EeV);\n",
      "print(\"momentum in kg m/sec is\",P);\n",
      "print(\"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": [
      "\n",
      "\n",
      "#importing modules\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "h=6.626*10**-34;     #plancks constant in Js\n",
      "m=9.1*10**-31;     #mass of the electron in kg\n",
      "e=1.602*10**-19;\n",
      "V=1.25;    #potential difference in kV\n",
      "\n",
      "#Calculation\n",
      "V=V*10**3;    #converting kV to V\n",
      "lamda=h/math.sqrt(2*m*e*V);\n",
      "lamda=lamda*10**10;     #converting metre to angstrom\n",
      "lamda=math.ceil(lamda*10**4)/10**4;   #rounding off to 4 decimals\n",
      "\n",
      "#Result\n",
      "print(\"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": [
      "\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "E=45;    #energy of electron in eV\n",
      "E=E*1.6*10**-19;     #energy in J\n",
      "h=6.626*10**-34;     #plancks constant in Js\n",
      "m=9.1*10**-31;     #mass of the electron in kg\n",
      "\n",
      "#Calculation\n",
      "lamda=h/math.sqrt(2*m*E);\n",
      "lamda=lamda*10**10;   #converting metres to angstrom\n",
      "lamda=math.ceil(lamda*10**4)/10**4;   #rounding off to 4 decimals\n",
      "\n",
      "#Result\n",
      "print(\"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": [
      "\n",
      "#importing modules\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "v=10**7;      #velocity of electron in m/sec\n",
      "h=6.626*10**-34;     #plancks constant in Js\n",
      "m=9.1*10**-31;     #mass of the electron in kg\n",
      "\n",
      "#Calculation\n",
      "lamda=h/(m*v);\n",
      "lamda=lamda*10**10;   #converting metres to angstrom\n",
      "lamda=math.ceil(lamda*10**4)/10**4;   #rounding off to 4 decimals\n",
      "\n",
      "#Result\n",
      "print(\"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": [
      "\n",
      "\n",
      "#importing modules\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "V=1000;       #potential difference in V\n",
      "h=6.626*10**-34;     #plancks constant in Js\n",
      "m=1.67*10**-27;     #mass of proton in kg\n",
      "e=1.6*10**-19;      #charge of electron in J\n",
      "\n",
      "#Calculation\n",
      "lamda=h/math.sqrt(2*m*e*V);\n",
      "\n",
      "#Result\n",
      "print(\"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": [
      "\n",
      "#importing modules\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "L=25;    #width of potential in armstrong\n",
      "delta_x=0.05;   #interval in armstrong\n",
      "n=1;   #particle is in its least energy\n",
      "x=L/2;   #particle is at the centre\n",
      "pi=180;   #angle in degrees\n",
      "\n",
      "#Calculation\n",
      "pi=pi*0.0174532925;   #angle in radians\n",
      "L=L*10**-10;   #width in m\n",
      "delta_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\n",
      "P=2*(L**(-1))*((math.sin(n*pi*x/L))**2)*delta_x;\n",
      "P=math.ceil(P*10**3)/10**3;   #rounding off to 3 decimals\n",
      "\n",
      "#Result\n",
      "print(\"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": [
      "\n",
      "\n",
      "#importing modules\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "n=1;\n",
      "h=6.626*10**-34;     #plancks constant in Js\n",
      "m=9.1*10**-31;     #mass of the electron in kg\n",
      "L=1;    #width of potential well in angstrom\n",
      "\n",
      "#Calculation\n",
      "L=L*10**-10;     #converting angstrom into metre\n",
      "E=((n**2)*h**2)/(8*m*L**2);\n",
      "EeV=E/(1.6*10**-19);       #converting J to eV\n",
      "EeV=math.ceil(EeV*10**3)/10**3;   #rounding off to 3 decimals\n",
      "\n",
      "#Result\n",
      "print(\"lowest energy of electron in J is\",E);\n",
      "print(\"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": [
      "\n",
      "\n",
      "#importing modules\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "n=1;\n",
      "h=6.626*10**-34;     #plancks constant in Js\n",
      "m=9.1*10**-31;     #mass of the electron in kg\n",
      "L=1;    #width of potential well in angstrom\n",
      "\n",
      "#Calculation\n",
      "L=L*10**-10;     #converting angstrom into metre\n",
      "E=(2*(n**2)*h**2)/(8*m*L**2);\n",
      "E=E/(1.6*10**-19);       #converting J to eV\n",
      "E=math.ceil(E*10**3)/10**3;   #rounding off to 3 decimals\n",
      "\n",
      "#Result\n",
      "print(\"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": [
      "\n",
      "\n",
      "#importing modules\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "h=6.626*10**-34;     #plancks constant in Js\n",
      "m=9.1*10**-31;     #mass of the electron in kg\n",
      "L=1;    #width of potential well in angstrom\n",
      "\n",
      "#Calculation\n",
      "L=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\n",
      "n1=1;\n",
      "n2=2;\n",
      "E=((2*(n1**2)*h**2)/(8*m*L**2))+(((n2**2)*h**2)/(8*m*L**2));\n",
      "E=E/(1.6*10**-19);       #converting J to eV\n",
      "E=math.ceil(E*10**3)/10**3;   #rounding off to 3 decimals\n",
      "\n",
      "#Result\n",
      "print(\"lowest energy of system in eV is\",E);\n",
      "print(\"quantum numbers are\");\n",
      "print(\"n=1,l=0,mL=0,mS=+1/2\");\n",
      "print(\"n=1,l=0,mL=0,mS=-1/2\");\n",
      "print(\"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)\n",
        "quantum numbers are\n",
        "n=1,l=0,mL=0,mS=+1/2\n",
        "n=1,l=0,mL=0,mS=-1/2\n",
        "n=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": [
      "\n",
      "#Variable declaration\n",
      "n=1;\n",
      "h=6.626*10**-34;     #plancks constant in Js\n",
      "L=100;    #width of potential well in angstrom\n",
      "\n",
      "#Calculation\n",
      "L=L*10**-10;     #converting angstrom into metre\n",
      "E=0.025;    #lowest energy in eV\n",
      "E=E*(1.6*10**-19);       #converting eV to J\n",
      "m=((n**2)*h**2)/(8*E*L**2);\n",
      "\n",
      "#Result\n",
      "print(\"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": [
      "\n",
      "\n",
      "#importing modules\n",
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "k=1.38*10**-23;\n",
      "T=6000;      #temperature in K\n",
      "h=6.626*10**-34;     #plancks constant in Js\n",
      "c=3*10**8;       #velocity of light in m/s\n",
      "lamda1=450;     #wavelength in nm\n",
      "lamda2=460;     #wavelength in nm\n",
      "\n",
      "#Calculation\n",
      "lamda1=lamda1*10**-9;     #converting nm to metre\n",
      "lamda2=lamda2*10**-9;     #converting nm to metre\n",
      "new1=c/lamda1;\n",
      "new2=c/lamda2;\n",
      "new=(new1+new2)/2;\n",
      "A=math.exp((h*new)/(k*T));\n",
      "rho_v=(8*math.pi*h*new**3)/(A*c**3);\n",
      "\n",
      "#Result\n",
      "print(\"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": {}
  }
 ]
}