path: root/Engineering_Physics/chapter7.ipynb

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 "worksheets": [

  {

   "cells": [

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Chapter7:WAVE MECHANICS"

     ]

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.1:pg-200"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate de Broglie wavelength \n",

      "v=1.5*10**7 #velocity of proton =(1/20)*velocity of light i.e.3*10**8 in m/s\n",

      "m=1.67*10**-27 #mass of the proton in kg\n",

      "h=6.6*10**-34 #plank's constant \n",

      "lamda=h/(m*v)\n",

      "print \"the de Broglie wavelength is lamda=\",\"{:.3e}\".format(lamda),\"m\" \n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "the de Broglie wavelength is lamda= 2.635e-14 m\n"

       ]

      }

     ],

     "prompt_number": 4

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.2:pg-200"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate de Broglie wavelength\n",

      "#mo*c**2=1.507*10**-10/1.6*10**-19=941.87 Mev\n",

      "#since 12.8 Mev is very small compared to rest mass energy hence relavistic consideration may be ignored\n",

      "m=1.67*10**-27 #mass in kg\n",

      "h=6.62*10**-34 #plank's constant\n",

      "E=12.8*10**6 #energy in Mev\n",

      "lamda=h/math.sqrt(2*m*E*1.6*10**-19)/(1e-10)\n",

      "print \"the de Broglie wavelength is lamda=\",round(lamda,5),\"angstrom\"\n",

      "\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "the de Broglie wavelength is lamda= 8e-05 angstrom\n"

       ]

      }

     ],

     "prompt_number": 11

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.4:pg-201"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate wavelength\n",

      "h=6.6*10**-34 #plank's constant\n",

      "m=9.1*10**-31 #mass of electron in kg\n",

      "E=1.25*10**3 #pottential difference keV\n",

      "lamda=h/math.sqrt(2*m*E*1.6*10**-19)\n",

      "print \"the wavelength is lamda=\",\"{:.2e}\".format(lamda),\"m\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "the wavelength is lamda= 3.46e-11 m\n"

       ]

      }

     ],

     "prompt_number": 6

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.5:pg-201"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate kinetic energy of an electron \n",

      "h=6.63*10**-34 #plank's constant\n",

      "mo=9.1*10**-31 #rest mass of an electron in kg\n",

      "lamda=5896*10**-10  #wavelength in angstrom\n",

      "K=(h**2)/(2*mo*(lamda**2)*1.6*10**-19) \n",

      "print \"kinetic energy of an electron is K=\",\"{:.2e}\".format(K),\"eV\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "kinetic energy of an electron is K= 4.34e-06 eV\n"

       ]

      }

     ],

     "prompt_number": 7

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.6:pg-202"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate the wavelength of an electron of kinetic energy\n",

      "mo=9.1*10**-31 #mass of an electron in kg\n",

      "c=3*10**8 #speed of light in m/s \n",

      "K=1*10**6#kinetic energy in eV\n",

      "h=6.62*10**-34 #planck's constant in J-s\n",

      "#E=moc**2=81.9*10**-15/1.6*10**-19 eV=0.51MeV\n",

      "E=0.51*10**6\n",

      "lamda=(h*c)/(math.sqrt(K*(K+2*E))*1.6*10**-19)\n",

      "print \"wavelength of an electron of kinetic energy is lamda=\",round(lamda,14),\"m\"\n",

      "\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "wavelength of an electron of kinetic energy is lamda= 8.7e-13 m\n"

       ]

      }

     ],

     "prompt_number": 16

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.7:pg-203"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate de Broglie wavelength\n",

      "V=100 #potential difference in volts\n",

      "lamda=12.25/math.sqrt(V)\n",

      "print \"de Broglie wavelength of any electron is lamda=\",lamda,\"angstrom\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "de Broglie wavelength of any electron is lamda= 1.225 angstrom\n"

       ]

      }

     ],

     "prompt_number": 7

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.9:pg-203"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate energy of the neutron\n",

      "h=6.60*10**-34 #plank's constant in J/s\n",

      "m=1.674*10**-27 #mass of the neutron in kg\n",

      "lamda=10**-10 #de Broglie wavelength in m\n",

      "E=(h**2)/(2*m*(lamda**2)*1.6*10**-19)\n",

      "print \"energy of the neutron is E=\",\"{:.2e}\".format(E),\"eV\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "energy of the neutron is E= 8.13e-02 eV\n"

       ]

      }

     ],

     "prompt_number": 8

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.10:pg-204"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate wavelength\n",

      "h=6.6*10**-34 #plank's constant in J/sec\n",

      "m=9.1*10**-31 #mass of electron in kg\n",

      "c=3*10**8 #light speed in m/s\n",

      "lamda=h/(m*c)/(1e-10) # in angstrom\n",

      "print \"wavelength of quantum of radiant energy is lamda=\",round(lamda,4),\"angstrom\"\n",

      "#to calculate number of photons \n",

      "power=12 #power emitted by the lamp =150*(8/100) in watts\n",

      "E=12.0 #energy emitted per second\n",

      "lamda=4500*10.0**-10\n",

      "energy=(h*c)/lamda #energy contained in one photon in J\n",

      "number=E/energy\n",

      "print \"number of photons emitted per sec is number=\",round(number,-16),\"unitless\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "wavelength of quantum of radiant energy is lamda= 0.0242 angstrom\n",

        "number of photons emitted per sec is number= 2.727e+19 unitless\n"

       ]

      }

     ],

     "prompt_number": 22

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.11:pg-209"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate uncertainity in position\n",

      "#actual formula is (delx)min*(delp)max=h/2*math.pi-------------eq(1)\n",

      "#(delp)max=p(momentum of the electron)\n",

      "#mv=mov/math.sqrt(1-(v/c)**2)---------------------eq(2)\n",

      "mo=9*10**-31 #mass of an electron in m/s\n",

      "c=3*10**8 #light speed in m/s\n",

      "v=3*10**7 #velocity in m/s  \n",

      "h=6.6*10**-34 #plank's constant in J/s\n",

      "#from eq(1) and eq(2),we get\n",

      "delxmin=(h*math.sqrt(1-(v/c)**2))/(2*math.pi*mo*v)\n",

      "print \"smallest possible uncertainity in the position of an electron is delxmin=\",round(delxmin/1e-10,4),\"angstrom\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "smallest possible uncertainity in the position of an electron is delxmin= 0.0389 angstrom\n"

       ]

      }

     ],

     "prompt_number": 12

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.12:pg-209"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate minimum uncertainity in the velocity\n",

      "delxmax=10**-8 #maximum uncertainity in position in m\n",

      "h=6.626*10**-34 #planck's constant\n",

      "delpmin=h/(2*math.pi*delxmax) #minimum uncertainity in momentum in kg-m/s**2 \n",

      "m=9*10**-31 #mass of an electron in kg\n",

      "delvmin=delpmin/m\n",

      "print \"minimum uncertainity in the velocity is delvmin=\",\"{:.2e}\".format(delvmin),\"m/s\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "minimum uncertainity in the velocity is delvmin= 1.17e+04 m/s\n"

       ]

      }

     ],

     "prompt_number": 13

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.13:pg-209"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate uncertainity in the momentum of the parcticle \n",

      "h=6.626*10**-34 #planck's constant J-s\n",

      "delx=0.01*10**-2 #uncertainity in position in m\n",

      "delp=h/(2*math.pi*delx)\n",

      "print \"uncertainity in the momentum of the parcticle is delp=\",\"{:.2e}\".format(delp),\"kg-m/s**2\"\n",

      "#to calculate uncertainity in the velocity of an electron\n",

      "m=9*10**-31 #mass of an electron in kg\n",

      "delx=5*10**-10 \n",

      "delv=h/(2*math.pi*m*delx)\n",

      "print \"uncertainity in the velocity of an electron is delv=\",\"{:.3e}\".format(delv),\"m/s\"\n",

      "#to calculate uncertainity in the velocity of alpha particle \n",

      "m=4*1.67*10**-27 #mass of alpha particle in kg\n",

      "delx=5*10**-10\n",

      "delv=h/(2*math.pi*m*delx)\n",

      "print \"uncertainity in the velocity of an electron is delv=\",round(delv,2),\"m/s\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "uncertainity in the momentum of the parcticle is delp= 1.05e-30 kg-m/s**2\n",

        "uncertainity in the velocity of an electron is delv= 2.343e+05 m/s\n",

        "uncertainity in the velocity of an electron is delv= 31.57 m/s\n"

       ]

      }

     ],

     "prompt_number": 15

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.14:pg-210"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate uncertainity in position\n",

      "m=9.11*10**-31 #mass of electron in kg\n",

      "delv=40 #uncertainity in velocity in m/s\n",

      "h=6.6*10**-34 #plank's constant \n",

      "delx=h/(2*math.pi*m*delv)\n",

      "print \"uncertainity in the position of the electron is delx=\",\"{:.2e}\".format(delx),\"m\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "uncertainity in the position of the electron is delx= 2.88e-06 m\n"

       ]

      }

     ],

     "prompt_number": 16

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.15:pg-210"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate uncertainity in frequency\n",

      "#delE*delt=h/2*math.pi----eq(1)\n",

      "#delE=h*delv-----------eq(2)\n",

      "delt=10**-8 #uncertainity in time in s\n",

      "#from eq(1) and eq(2),we get\n",

      "delnu=1/(2*math.pi*delt)\n",

      "print \"minimum uncertainity in the frequency of the photon is delv=\",\"{:.3e}\".format(delnu),\"sec**-1\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "minimum uncertainity in the frequency of the photon is delv= 1.592e+07 sec**-1\n"

       ]

      }

     ],

     "prompt_number": 17

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.16:pg-211"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate uncertainity in the energy\n",

      "h=6.63*10**-34 #plank's constant in J-s\n",

      "delt=2.5*10**-14 #uncertainity in time in s\n",

      "delE=h/(2*math.pi*delt*1.6*10**-19)\n",

      "print \"minimum error with which the energy of the state can be measured is delE=\",round(delE,3),\"ev\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "minimum error with which the energy of the state can be measured is delE= 0.026 ev\n"

       ]

      }

     ],

     "prompt_number": 18

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.17:pg-211"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate time required for the atomic system \n",

      "#delE=h*c*dellamda/lamda**2 -----eq(1)\n",

      "#delE*delt=h/2*math.pi----------eq(2)\n",

      "dellamda=10**-14\n",

      "c=3*10**8\n",

      "lamda=6*10**-7\n",

      "#from eq(1)and eq(2),we get\n",

      "delt=(lamda**2)/(2*math.pi*c*dellamda)\n",

      "print \"time required for the atomic system to retain rotational energy is delt=\",\"{:.1e}\".format(delt),\"s\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "time required for the atomic system to retain rotational energy is delt= 1.9e-08 s\n"

       ]

      }

     ],

     "prompt_number": 19

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.18:pg-211"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate minimum uncertainity in the momentum \n",

      "delxmax=5*10**-14 #uncertainity in position in m\n",

      "h=6.626*10**-34 #plank's constant in Js\n",

      "delpmin=h/(2*math.pi*delxmax)\n",

      "print \"minimum uncertainity in the momentum of the nucleon is delpmin=\",\"{:.2e}\".format(delpmin),\"kg m/s\"\n",

      "m=1.675*10**-27 #mass in kg\n",

      "Emin=(delpmin**2)/(2*m*1.6*10**-19)\n",

      "print \"minimum kinetic energy of the nucleon is Emin=\",round(Emin,2),\"eV\"\n",

      "#the answer is given wrong in the book Emin=0.039 eV\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "minimum uncertainity in the momentum of the nucleon is delpmin= 2.11e-21 kg m/s\n",

        "minimum kinetic energy of the nucleon is Emin= 8299.24 eV\n"

       ]

      }

     ],

     "prompt_number": 9

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.19:pg-212"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate uncertainity in velocity\n",

      "delx=1.1*10**-8 #uncertainity in velocity in m\n",

      "h=6.626*10**-34 #plank's constant\n",

      "m=9.1*10**-31 #mass of electron in kg\n",

      "delv=h/(2*math.pi*m*delx)\n",

      "print \"minimum uncertainity in velocity is delv=\",\"{:.2e}\".format(delv),\"m/s\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "minimum uncertainity in velocity is delv= 1.05e+04 m/s\n"

       ]

      }

     ],

     "prompt_number": 14

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.20:pg-212"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate uncertainity in frequency\n",

      "delt=10**-8 #uncertainity in time\n",

      "delnu=1/(2*math.pi*delt) \n",

      "print \"minimum uncertainity in the frequency of a photon is delnu=\",\"{:.2e}\".format(delnu),\"sec**-1\"\n",

      "#to use the uncertainity principle to place a lower limit on the energy an electron must have if it is to be part of a nucleus\n",

      "delx=5*10**-15 #uncertainity in position\n",

      "delp=h/(2*2*math.pi*delx) #uncertainbity in momentum\n",

      "c=3*10**8 #/speed of light in m/s\n",

      "E=delp*c\n",

      "print \"energy of an electron is E=\",\"{:.2e}\".format(E),\"J\"\n",

      "\n",

      "# the answer is slightlty different due to approximation in textboook\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "minimum uncertainity in the frequency of a photon is delnu= 1.59e+07 sec**-1\n",

        "energy of an electron is E= 3.16e-12 J\n"

       ]

      }

     ],

     "prompt_number": 15

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.22:pg-223"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate probability of finding the particle\n",

      "a=25*10**-10#width in angstrom \n",

      "#wave function of the particle is chi(x)=math.sqrt(2/a)*math.sin(n*math.pi*x/a),for the particle in the least energy state n=1\n",

      "chix=math.sqrt(2/a)*math.sin(math.pi*(a/2)/a)\n",

      "delx=5*10**-10 #interval in angstrom\n",

      "P=delx*chix**2\n",

      "print \"probability of finding the particle is P=\",P,\"unitless\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "probability of finding the particle is P= 0.4 unitless\n"

       ]

      }

     ],

     "prompt_number": 9

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.24:pg-224"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate energy of an electron\n",

      "n=1 #least energy of the particle \n",

      "h=6.63*10**-34 #planck's constant in Js\n",

      "m=9.11*10**-31 #mass of electron in kg\n",

      "a=10**-10 #width in angstrom\n",

      "E=(n**2)*(h**2)/(8*m*(1.602*10**-19)*a**2)\n",

      "print \"energy of an electron moving in one dimension in an infinitely high potential box is E=\",round(E,2),\"eV\"\n",

      "#the answer is given wrong in the book E=5.68 eV\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "energy of an electron moving in one dimension in an infinitely high potential box is E= 37.65 eV\n"

       ]

      }

     ],

     "prompt_number": 16

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.26:pg-225"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate probability\n",

      "x1=0.45 #x1=0.45*L\n",

      "x2=0.55       #x2=0.55*L\n",

      "n=1                     #for ground state \n",

      "#formula is P=integrate('(2/L)*math.sin(n*math.pi*x)**2),'x',x1,x2)\n",

      "from scipy.integrate import quad\n",

      "def integrand(x):\n",

      "    return 2*(math.sin(n*math.pi*x)**2)\n",

      "P1 ,er=quad(integrand,x1,x2)\n",

      "\n",

      "print \"P1=\",round(P1,3),\"unitless\"\n",

      "probability1=P1*100\n",

      "print \"probability for the ground states is probability1 =\",round(probability1,1),\"%\"\n",

      "n=2   #for first excited state\n",

      "P2, er=quad(integrand,x1,x2)\n",

      "print \"P2=\",round(P2,4),\"unitless\"\n",

      "probability2=P2*100 \n",

      "print \"probability for first excited states is probability2=\",round(probability2,2),\"%\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "P1= 0.198 unitless\n",

        "probability for the ground states is probability1 = 19.8 %\n",

        "P2= 0.0065 unitless\n",

        "probability for first excited states is probability2= 0.65 %\n"

       ]

      }

     ],

     "prompt_number": 5

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.28:pg-226"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate energy of a neutron\n",

      "#consider nucleus as a cubical box of size 10**-14m\n",

      "#x=y=z=a=10**-14=l\n",

      "#for neutron to be in the lowest energy state nx=ny=nz=1\n",

      "#formula is E=(math.pi**2*h**2/8*math.pi**2*m)*((nx/lx)**2+(ny/ly)**2+(nz/lz)**2)\n",

      "h=6.626*10**-34  #planck's constant in Js\n",

      "m=1.6*10**-27 #mass in kg\n",

      "l=10**-14 #in m\n",

      "E=(math.pi**2)*(h**2)*3/(4*(math.pi**2)*2*m*(1.6*10**-19)*l**2)\n",

      "print \"lowest energy of a neutron is E=\",round(E/(1e6),2),\"MeV\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "lowest energy of a neutron is E= 6.43 MeV\n"

       ]

      }

     ],

     "prompt_number": 13

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.29:pg-226"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate zero point energy of a linear harmonic oscillator\n",

      "h=6.63*10**-34 #planck's constant in Js\n",

      "nu=50 #frequency in Hz\n",

      "zeropointenergy=(h*nu)/2\n",

      "print \"zeropointenergy=\",\"{:.2e}\".format(zeropointenergy),\"J\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "zeropointenergy= 1.66e-32 J\n"

       ]

      }

     ],

     "prompt_number": 21

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.30:pg-226"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate zero point energy\n",

      "nu=1 #frequency in Hz\n",

      "h=6.63*10**-34 #planck's constant in Js\n",

      "zeropointenergy=(h*nu)/2\n",

      "print \"zeropointenergy=\",zeropointenergy,\"J\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "zeropointenergy= 3.315e-34 J\n"

       ]

      }

     ],

     "prompt_number": 14

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex7.31:pg-226"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "import math\n",

      "#to calculate frequency of vibration\n",

      "En=0.1*1.6*10**-19 #energy of a linear harmonic oscillator in eV\n",

      "n=3.0 #third excited state\n",

      "h=6.63*10**-34 #planck's constant\n",

      "nu=En/((n+(1/2.0))*h)\n",

      "print \"the frequency of vibration is nu=\",round(nu,-9),\"Hz\"\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "the frequency of vibration is nu= 6.895e+12 Hz\n"

       ]

      }

     ],

     "prompt_number": 16

    }

   ],

   "metadata": {}

  }

 ]

}