{ "metadata": { "name": "", "signature": "sha256:624622ddf9fdf57adcae6c7032213756e820e3fa0a2228db7dd3864940305f16" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter1:WAVE MECHANICS" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg1:pg-12" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.62*10**-34 #planck constant in joule-sec\n", "m=9.1*10**-31 #mass of electron in kg\n", "q=1.6*10**-19 #charge of electron in coulombs\n", "#V energy of electron in eV\n", "lamda=round(((h/math.sqrt(2*m*q))*10**10),2) \n", "print\"de-Broglie wavelength for an electron of energy V ev is= %s/sqrt(V) Angstrom\"%lamda" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "de-Broglie wavelength for an electron of energy V ev is= 12.27/sqrt(V) Angstrom\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg2:pg-12" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.63*10**-34 #planck constant in joule-sec\n", "m=9.1*10**-31 #mass of electron in kg\n", "q=1.6*10**-19 #charge of electron in coulombs\n", "V=50. #potential difference in volts(given)\n", "lamda=int(((h/math.sqrt(2*m*q))*10**10)*1e2)*1e-2/sqrt(V) #lamda=h/sqrt(2mE)=h/sqrt(2mqV)\n", "print\"de-Broglie wavelength = \", round(lamda,4),\"Angstrom\" " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "de-Broglie wavelength = 1.7367 Angstrom\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg3:pg-12" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.62*10**-34 #planck constant in joule-sec\n", "m=9.1*10**-31 #mass of electron in kg\n", "q=1.6*10**-19 #charge of electron in coulombs\n", "V=100. #potential difference in volts(given)\n", "E=q*V \n", "lamda=round(((h/math.sqrt(2*m*q))*10**10),2)/math.sqrt(V) #lamda=h/sqrt(2mE)=h/sqrt(2mqV)\n", "print\"de-Broglie wavelength =\",round(lamda,3),\"Angstrom\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "de-Broglie wavelength = 1.227 Angstrom\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg4:pg-12" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.6*10**-34 #planck constant in joule-sec\n", "m=9.0*10**-31 #mass of electron in kg\n", "KE=(15*10**3)*(1.6*10**-19) #Kinetic Energy of electron in joule\n", "v=math.sqrt((2*KE)/m) #KE=1/2(mv**2) joule\n", "p=m*v #momentum of electron in Kg-m/sec\n", "lamda=h/p #de-broglie wavelength\n", "print\"de-Broglie wavelength is=\",round(lamda*10**10,1),\"Angstrom\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "de-Broglie wavelength is= 0.1 Angstrom\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg5:pg-13" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.62*10**-34 #planck constant in joule-sec\n", "m=1.67*10**-27 #mass of neutron in Kg\n", "lamda=10**-10 #de-broglie wavelength in meter(given)\n", "v=h/(m*lamda) #since lamda=h/mv \n", "KE=(1./2)*m*v**2 #in joule\n", "KE=KE/(1.6*10**-19) #in eV\n", "print\"Velocity of neutron is= %.2e m/sec\"%v\n", "print\"Kinetic Energy of Neutron is= \",round(KE,3),\"eV\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Velocity of neutron is= 3.96e+03 m/sec\n", "Kinetic Energy of Neutron is= 0.082 eV\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg6:pg-13" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.6*10**-34 #planck constant in joule-sec\n", "m=9.1*10**-31 #mass of electron in kg\n", "q=1.6*10**-19 #charge of electron in coulombs\n", "E=(1.25*10**3)*(1.6*10**-19) #Kinetic energy in joule\n", "lamda=h/math.sqrt(2*m*E)\n", "print\"wavelength is =\",\"{:.2e}\".format(lamda),\"m\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "wavelength is = 3.46e-11 m\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg7:pg-13" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.63*10**-34 #planck constant in joule-sec\n", "mo=1.67*10**-27 #mass of proton in Kg\n", "v=2.0*10**8 #speed of proton in m/sec\n", "c=3*10**8 #speed of light in m/sec\n", "p=(mo*v)/math.sqrt(1-(v/c)**2) #momentum of proton\n", "lamda=h/p\n", "print\"wavelength is =\",\"{:.2e}\".format(lamda*10**10),\"Angstrom\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "wavelength is = 1.48e-05 Angstrom\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg8:pg-14" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.62*10**-34 #planck constant in joule-sec\n", "mp=1.67*10**-27 #mass of proton in Kg\n", "c=3*10**8 #speed of light in m/sec\n", "v=c/20 #speed of proton in m/sec\n", "lamda=h/(mp*v)\n", "print\"de-Broglie wavelength =\",\"{:.3e}\".format(lamda),\"m\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "de-Broglie wavelength = 2.643e-14 m\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg9:pg-14" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.62*10**-34 #planck constant in joule-sec\n", "m=9.1*10**-31 #mass of electron in kg\n", "c=3*10**8 #speed of light in m/sec\n", "lamda=2.0*10**-10#wavelength of electron and photon in meter\n", "#(a) momenta\n", "Pe=h/lamda #momentum of electron in Kg-m/s\n", "Pp=h/lamda #momentum of photon in kg-m/s\n", "print\"(a)Since wavelength of photon is same as that of an electron, therefore their moments is also same i.e \",Pe,\"Kg-m/s\"\n", "\n", "#(b) total energies\n", "me=mo=m\n", "KE=(Pe**2)/(2*me) #Kinetic energy of electron in joule\n", "Re=mo*c**2 #rest energy of electron in joule\n", "Re=Re/(1.6*10**-19) #in eV\n", "Ee=Re/10**6 #total energy of electron in Mev(since K.E. of electron is negligible compared to its rest energy so total energy is equal to the rest energy) \n", "Ep=Pp*c #total energy of photon in joule(since rest energy of photon is zero so its total energy is same as its K.E.) \n", "Ep=(Ep/(1.6*10**-19))*10**-3 #in KeV\n", "print\"(b)Total energy of electron is\",round(Ee,2),\"MeV\",\"\\n Total energy of photon is\",round(Ep,2),\"KeV\"\n", "\n", "#(c) ratio of kinetic energies\n", "Ke=round((KE/(1.6*10**-19)),1)#Kinetic energy of electron in eV\n", "Kp=round(Ep,2)*(10**3) #Kinetic energy of photon in eV\n", "ratio=Ke/Kp\n", "print\"(c)Ratio of kinetic energies=\",\"{:.2e}\".format(ratio)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)Since wavelength of photon is same as that of an electron, therefore their moments is also same i.e 3.31e-24 Kg-m/s\n", "(b)Total energy of electron is 0.51 MeV \n", " Total energy of photon is 6.21 KeV\n", "(c)Ratio of kinetic energies= 6.05e-03\n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg10:pg-15" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.62*10**-34 #planck constant in joule-sec\n", "m=1.67*10**-27 #mass of neutron in Kg\n", "c=3*10**8 #speed of light in m/sec\n", "E=28.8 #Kinetic energy of neutron in eV(given)\n", "E=28.8*1.6*(10**-19)#in joule\n", "Rn=m*c**2 #Rest mass energy of neutron in joule\n", "Rn=(Rn/(1.6*10**-19))/10**6 #in MeV\n", "#since Kinetic energy of neutron under consideration is very small compared to its rest mass energy,the relativistic consideration may be ignored. \n", "lamda=h/math.sqrt(2*m*E)\n", "print\"de-Broglie wavelength=\",round(lamda*10**10,5),\"Angstrom\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "de-Broglie wavelength= 0.05336 Angstrom\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg11:pg-15" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.63*10**-34 #planck constant in joule-sec\n", "m=4*1.67*10**-27 #mass of alpha particle Kg=4*mass of proton\n", "q=2*1.6*10**-19 #for alpha particle q=2*e coulomb\n", "V=200 #potential difference in volts \n", "lamda=h/math.sqrt(2*m*q*V)\n", "print\"de-Broglie wavelength=\",round(lamda*10**10,5),\"Angstrom\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "de-Broglie wavelength= 0.00717 Angstrom\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg12:pg-15" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.63*10**-34 #planck constant in joule-sec\n", "m=9.1*10**-31 #mass of electron in kg\n", "mo=m #for velocity much less than the velocity of light, m=mo\n", "#(a)wavelength for a ball of mass 1.0Kg and v=1.0m/s\n", "v=1.0 #speed of ball in m/sec\n", "mass=1.0 #mass of ball in Kg\n", "lamda=h/(mass*v)\n", "print\"de-Broglie wavelength for a ball of mass 1.0Kg and v=1.0m/s =\",\"{:.2e}\".format(lamda*10**10),\"Angstrom\"\n", "\n", "#(b)wavelength for an electron of mass 9.1*10**-31 Kg and v=10**6 m/sec\n", "v=10**6 #speed of electron in m/sec\n", "lamda=h/(mo*v)\n", "print\"de-Broglie wavelength for an electron of mass 9.1*10**-31Kg and v=10**6m/sec =\",round(lamda*10**10,2),\"Angstrom\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "de-Broglie wavelength for a ball of mass 1.0Kg and v=1.0m/s = 6.63e-24 Angstrom\n", "de-Broglie wavelength for an electron of mass 9.1*10**-31Kg and v=10**6m/sec = 7.29 Angstrom\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg13:pg-16" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.63*10**-34 #planck constant in joule-sec\n", "m=1.67*10**-27 #mass of neutron in Kg\n", "c=3*10**8 #speed of light in m/sec\n", "E=1. #Kinetic energy of neutron in eV\n", "Rn=m*c**2 #rest mass energy of neutron in joule\n", "Rn=(Rn/(1.6*10**-19))/10**6 #in MeV\n", "#Kinetic energy of given neutron 1eV is very small as compared to its rest mass energy,therefore the relativistic consideration may be ignored \n", "E=1*1.6*10**-19 #in joule\n", "lamda=h/math.sqrt(2*m*E)\n", "print\"de-Broglie wavelength=\",round(lamda*10**10,3),\"Angstrom\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "de-Broglie wavelength= 0.287 Angstrom\n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg14:pg-16" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.62*10**-34 #planck constant in joule-sec\n", "m=1.675*10**-27 #mass of neutron in Kg\n", "c=3*10**8 #speed of light in m/sec\n", "E=12.8 #energy of neutron in MeV\n", "Rn=m*c**2 #rest mass energy of neutron in joule\n", "Rn=(Rn/(1.6*10**-19))/10**6 # in MeV\n", "#since the given energy 12.8MeV is very small as compared to the rest mass energy,therefore the relativistic consideration may be ignored \n", "E=E*(10**6)*(1.6*10**-19) # in eV\n", "lamda=h/math.sqrt(2*m*E)\n", "print\"de-Broglie wavelength=\",\"{:.1e}\".format(lamda*10**10),\"Angstrom\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "de-Broglie wavelength= 8.0e-05 Angstrom\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg15:pg-17" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.6*10**-34 #planck constant joule-sec\n", "m=9.1*10**-31 #mass of electron kg\n", "e=1.6*10**-19 #charge of electron in coulomb\n", "lamda=0.40*10**-10#wavelength in meter\n", "V=(h**2)/round(((lamda**2)*2*m*e),72)#lamda=h/sqrt(2mE)=h/sqrt(2meV)\n", "print\"applied voltage=\",round(V,2),\"Volt\" " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "applied voltage= 934.76 Volt\n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg17:pg-17" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.6*10**-34 #planck constant in joule-sec\n", "m=1.67*10**-27 #mass of neutron in Kg\n", "K=8.6*10**-5 #Boltzmann constant in eV/degree\n", "K=K*1.6*10**-19 #in J/K\n", "T=27+273 #temperature in Kelvin\n", "E=K*T #energy of particle\n", "lamda=h/round(math.sqrt(2*m*E),26)\n", "print\"wavelength=\",round(lamda*10**10,3),\"Angstrom\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "wavelength= 1.779 Angstrom\n" ] } ], "prompt_number": 21 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg18:pg-18" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.62*10**-34 #planck constant in joule-sec\n", "m=4.65*10**-26 #mass of nitrogen atom in Kg\n", "T=27+273 #Temperature in Kelvin\n", "K=1.38*10**-23 #Boltzmann constant in J/K\n", "E=(3./2)*K*T #for nitrogen atom E=(3/2)*K*T\n", "lamda=h/math.sqrt(2*m*E) \n", "print\"de-Broglie wavelength=\",round(lamda*10**10,4),\"Angstrom\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "de-Broglie wavelength= 0.2755 Angstrom\n" ] } ], "prompt_number": 22 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg19:pg-18" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.62*10**-34 #planck constant in joule-sec\n", "m=6.7*10**-27 #mass of helium atom in Kg\n", "K=1.38*10**-23 #Boltzmann constant in J/K\n", "T=400 #Temperature in Kelvin\n", "E=(3./2)*K*T #for helium atom E=(3/2)*K*T\n", "lamda=h/math.sqrt(2*m*E)\n", "print\"de-Broglie wavelength=\",round(lamda*10**10,3),\"Angstrom\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "de-Broglie wavelength= 0.628 Angstrom\n" ] } ], "prompt_number": 23 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg20:pg-18" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "E=100*1.6*10**-19 #kinetic energy of electron in joule\n", "m=9.0*10**-31 #mass of electron in Kg\n", "h=6.62*10**-34 #planck constant joule-sec\n", "D=20 #distance of screen from foil in cm\n", "diameter=2.44 #diameter of ring in cm\n", "r=diameter/2 #radius of ring in cm\n", "lamda=h/math.sqrt(2*m*E)\n", "tan_theta=r/D\n", "#for small value of theta tan(theta)=sin(theta)\n", "#According to Bragg's law, 2d(sin(theta))=n*lamda\n", "n=1\n", "sin_theta=tan_theta\n", "d=(n*lamda)/(2*sin_theta)\n", "print\"spacing of the related lattice planes in the metal=\",round(d*10**10,2),\"Angstrom\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "spacing of the related lattice planes in the metal= 10.11 Angstrom\n" ] } ], "prompt_number": 24 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg21:pg-19" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "r=0.53*10**-10 #radius of first Bohr orbit in hydrogen atom in Meter\n", "h=6.6*10**-34 #planck constant in joule-sec\n", "m=9.1*10**-31 #mass of electron in Kg\n", "lamda=2*math.pi*r#since 2*pi*r=n*lamda where n=1 for the first Bohr orbit, So lamda=2*pi*r=h/(m*v)\n", "v=h/(lamda*m)\n", "print\"velocity of electron=\",\"{:.2e}\".format(v),\"m/s\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "velocity of electron= 2.18e+06 m/s\n" ] } ], "prompt_number": 25 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg22:pg-19" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.63*10**-34 #planck constant in joule-sec\n", "mo=9.1*10**-31 #rest mass of electron in Kg\n", "lamda=5896*10**-10#wavelength in meter\n", "#Since lamda=h/(mo*v) and Kinetic energy=(1/2)*mo*v**2 \n", "#therefore on putting v=h/(mo*lamda) in equation of Kinetic energy\n", "K=((h/lamda)**2)/(2*mo) #kinetic energy of electron in joule\n", "K=K/(1.6*10**-19) #in eV\n", "print\"Kinetic energy of electron=\",\"{:.2e}\".format(K),\"eV\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Kinetic energy of electron= 4.34e-06 eV\n" ] } ], "prompt_number": 26 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg23:pg-20" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.63*10**-34 #planck constant in joule-sec\n", "mo=1.67*10**-27 #rest mass of neutron in Kg\n", "lamda=10**-10 #de-broglie wavelength in meter\n", "v=h/(mo*lamda) #velocity of neutron in m/s (since lamda=h/(mo*v))\n", "print\"velocity of neutron=\",\"{:.2e}\".format(v),\"m/s\"\n", "K=(mo*v**2)/2 #kinetic energy of neutron in joule\n", "K=K/(1.6*10**-19) #in eV\n", "print\"kinetic energy of neutron=\",round(K,3),\"eV\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "velocity of neutron= 3.97e+03 m/s\n", "kinetic energy of neutron= 0.082 eV\n" ] } ], "prompt_number": 27 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg25:pg-21" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.62*10**-34 #planck constant in joule-sec\n", "mo=9.1*10**-31 #rest mass of electron in Kg\n", "c=3e8 #speed of light in m/sec\n", "K=1 #Kinetic energy in MeV\n", "Re=(mo*c**2/(1.6*10**-19))/10**6 #rest mass energy of electron in Mev\n", "lamda=h*c/math.sqrt(K*(K+(2*Re)))\n", "print\"Wavelength is %.2e Angstrom\"%(lamda*1e10/(1.6e-19*1e6))\n", "#answer is wrong in book because of calculation mistake" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Wavelength is 8.73e-03 Angstrom\n" ] } ], "prompt_number": 28 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg26:pg-22" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.63*10**-34 #planck constant in joule-sec\n", "mo=9.1*10**-31 #rest mass of electron in Kg(in book it is given wrong in question)\n", "c=3*10**8 #speed of light in m/sec\n", "\n", "#(a)wavelength associated with 1MeV electron\n", "K=1 #kinetic energy of electron in MeV\n", "Re=(mo*c**2/(1.6*10**-19))/10**6 #rest mass energy of electron in Mev\n", "#since given K.E(1MeV)of electron is comparable with its rest mass energy \n", "#therefore relativistic variation of mass with velocity is taken in to account\n", "d=round(math.sqrt(K*(K+(2*Re))),2)*1.6*10**-13 #value of sqrt(K*(K+(2*mo*c**2))) in volt\n", "lamda=h*c/d\n", "print\"wavelength associated with 1MeV electron=\",\"{:.2e}\".format(lamda*10**10),\"Angstrom\"\n", "\n", "#(b)wavelength associated with 1MeV proton\n", "K=1*1.6*10**-13 #kinetic energy of electron in volt\n", "mo=1.67*10**-27 #rest mass of proton in Kg\n", "Rp=(mo*c**2/(1.6*10**-19))/10**6 #rest mass energy of proton in Mev\n", "#since given K.E(1MeV)of proton is much less than its rest mass energy \n", "#therefore relativistic effect can be ignored\n", "lamda=h/math.sqrt(2*mo*K)\n", "print\"wavelength associated with 1MeV proton=\",\"{:.2e}\".format(lamda*10**10),\"Angstrom\"\n", "\n", "#(c)wavelength associated with 1MeV photon\n", "#since rest mass of photon is zero so its rest mass energy is also zero\n", "E=K #Energy of photon is entirely kinetic energy in volt\n", "lamda=h*c/E\n", "print\"wavelength associated with 1MeV photon=\",\"{:.2e}\".format(lamda*10**10),\"Angstrom\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "wavelength associated with 1MeV electron= 8.75e-03 Angstrom\n", "wavelength associated with 1MeV proton= 2.87e-04 Angstrom\n", "wavelength associated with 1MeV photon= 1.24e-02 Angstrom\n" ] } ], "prompt_number": 29 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg27:pg-23" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "V=54. #potential difference in volt\n", "lamda=12.28/math.sqrt(V) #de-Broglie wavelength of electron\n", "phi=50 #angle of scattering in degree\n", "sin_phi=math.sin(math.radians(phi))\n", "D=lamda/sin_phi #according to Bragg's law for normal incidence lamda=D*(sin phi),where D is the distance between two consecutive atoms in the surface layer \n", "print\"Distance between two neighbouring atoms in the surface of Ni-crystal=\",round(D,2),\"Angstrom\"\n", "theta=90-(phi/2) #glancing angle in degree\n", "sin_theta=math.sin(math.radians(theta))\n", "d=lamda/(2*sin_theta) \n", "print\"Distance between successive Bragg's planes of Ni-crystal=\",round(d,2),\"Angstrom\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Distance between two neighbouring atoms in the surface of Ni-crystal= 2.18 Angstrom\n", "Distance between successive Bragg's planes of Ni-crystal= 0.92 Angstrom\n" ] } ], "prompt_number": 30 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg28:pg-24" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "V=54. #potential difference in volt\n", "lamda=12.28/math.sqrt(V) #de-Broglie wavelength of electron\n", "D=2.15 #distance between successive atoms of crystal plane in angstrom\n", "\n", "#for first order diffraction\n", "n=1 \n", "phi=math.asin(n*lamda/D)#Bragg's equation, D*sin(phi)=n*lamda,where phi is angle of scattering \n", "phi=math.degrees(phi)# in degree\n", "#for second order diffraction\n", "n=2\n", "sin_phi_2=n*lamda/D\n", "print\"since sin_phi_2=\",round(sin_phi_2,2),\">1 which is impossible because sin(phi) can never exceed 1.Hence second and third orders can't occur.\"\n", "\n", "#when V is increased from 54volt to 60 volt\n", "V=60.\n", "n=1\n", "lamda=12.28/math.sqrt(V)#de-Broglie wavelength of electron\n", "phi=math.asin(n*lamda/D)\n", "phi=math.degrees(phi)\n", "print\"When accelerating potential were changed from 54volt to 60volt, first order diffracted beams occur at=\",int(round(phi)),\"degree\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "since sin_phi_2= 1.55 >1 which is impossible because sin(phi) can never exceed 1.Hence second and third orders can't occur.\n", "When accelerating potential were changed from 54volt to 60volt, first order diffracted beams occur at= 48 degree\n" ] } ], "prompt_number": 31 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg30:pg-28" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.63*10**-34 #planck constant in joule-sec\n", "mo=9.1*10**-31 #rest mass of electron in kg\n", "c=3*10**8 #speed of light in m/sec\n", "lamda=10**-12 #wavelength in meter\n", "pc=(h*c/lamda)/(1.6*10**-19)\n", "Eo=(mo*c**2)/(1.6*10**-19)#rest energy of electron in eV\n", "E=math.sqrt((pc**2)+(Eo**2)) #total energy of electron eV\n", "v=round(math.sqrt(1-(Eo/E)**2),3)\n", "print\"Group velocity of the de-Broglie waves,Vg = v = %s*c\"%v\n", "Vp=round((1/math.sqrt(1-(Eo/E)**2)),2) \n", "print\"Phase velocity of the de-Broglie waves,Vp= %s*c\"%Vp" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Group velocity of the de-Broglie waves,Vg = v = 0.925*c\n", "Phase velocity of the de-Broglie waves,Vp= 1.08*c\n" ] } ], "prompt_number": 32 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg31:pg-29" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.63*10**-34 #planck constant in joule-sec\n", "mo=9.1*10**-31 #rest mass of electron in kg\n", "c=3*10**8 #speed of light in m/sec\n", "lamda=2.0*10**-12 #wavelength in meter \n", "pc=round((h*c/lamda)/(1.6*10**-16),2)#in KeV\n", "Eo=int((mo*c**2)/(1.6*10**-16))#rest energy of electron in KeV\n", "E=round(math.sqrt((pc**2)+(Eo**2)),2)#total energy of electron KeV\n", "KE=E-Eo\n", "print\"Kinetic energy of electron=\",KE,\"KeV\"\n", "v=round(math.sqrt(1-(Eo/E)**2),4)\n", "print\"Group velocity of the de-Broglie waves,Vg = v = %s*c\"%v\n", "Vp=round((1/math.sqrt(1-(Eo/E)**2)),2)\n", "print\"Phase velocity of the de-Broglie waves,Vp= %s*c\"%Vp" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Kinetic energy of electron= 293.65 KeV\n", "Group velocity of the de-Broglie waves,Vg = v = 0.7725*c\n", "Phase velocity of the de-Broglie waves,Vp= 1.29*c\n" ] } ], "prompt_number": 33 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg33:pg-39" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#From Heisenberg uncertainty relation (delta_x*delta_p)>=h or delta_p>=h/delta_x\n", "h=6.63*10**-34 #planck constant joule-sec\n", "delta_x=0.2*10**-10 #uncertainty in position in meter\n", "delta_p=h/(2*math.pi*delta_x) #uncertainty in momentum \n", "print\"Uncertainty in momentum =\",\"{:.2e}\".format(delta_p),\"Kg-ms-1\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Uncertainty in momentum = 5.28e-24 Kg-ms-1\n" ] } ], "prompt_number": 34 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg34:pg-39" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#From Heisenberg uncertainty relation (delta_x*delta_p)>=h or delta_x>=h/delta_p\n", "h=6.62*10**-34 #planck constant in joule-sec\n", "mo=9.0*10**-31 #rest mass of electron in kg\n", "c=3*10**8 #speed of light in m/sec\n", "v=3.0*10**7 #velocity of electron in m/sec\n", "delta_p=mo*v/math.sqrt(1-(v/c)**2)#maximum uncertainty in momentum\n", "delta_x=h/(2*math.pi*delta_p)#smallest uncertainty in position \n", "print\"Smallest possible uncertainty in position of an electron=\",round(delta_x*10**10,4),\"Angstrom\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Smallest possible uncertainty in position of an electron= 0.0388 Angstrom\n" ] } ], "prompt_number": 35 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg35:pg-40" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#From Heisenberg uncertainty relation (delta_x*delta_p)>=h or delta_p>=h/delta_x\n", "h=1.05*10**-34 #planck constant in joule-sec\n", "m=9.1*10**-31 # mass of electron in kg\n", "delta_x=1.1*10**-8#uncertainty in position in meter\n", "delta_p=h/delta_x #uncertainty in momentum\n", "delta_v=delta_p/m #minimum uncertainty in velocity\n", "print\"minimum uncertainty in velocity of an electron=\",\"{:.2e}\".format(delta_v),\"m/s\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "minimum uncertainty in velocity of an electron= 1.05e+04 m/s\n" ] } ], "prompt_number": 36 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg36:pg-40" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#From Heisenberg uncertainty relation (delta_x*delta_p)>=h or delta_p>=h/delta_x\n", "h=6.62*10**-34 #planck constant in joule-sec\n", "m=9.1*10**-31 #mass of electron in kg\n", "delta_x=10*10**-10#uncertainty in position in meter\n", "delta_p=h/(2*math.pi*delta_x)#uncertainty in momentum in Kg-m/s\n", "delta_v=delta_p/m #uncertainty in velocity of an electron\n", "print\"uncertainty in velocity of an electron=\",\"{:.2e}\".format(delta_v),\"m/s\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "uncertainty in velocity of an electron= 1.16e+05 m/s\n" ] } ], "prompt_number": 37 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg37:pg-40" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#From Heisenberg uncertainty relation (delta_x*delta_p)>=h or delta_x>=h/delta_p\n", "h=6.62*10**-34 #planck constant in joule-sec\n", "m=9.0*10**-31 #mass of electron in kg\n", "v=1.05*10**4 #speed of electron in m/s\n", "p=m*v #momentum of electron in Kg-m/s\n", "delta_p=(0.01/100)*p#uncertainty in momentum(since uncertainty in value of p is 0.01% of its value)\n", "delta_x=h/(2*math.pi*delta_p)#uncertainty in position of electron\n", "print\"uncertainty in the position of electron=\",\"{:.3e}\".format(delta_x),\"m\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "uncertainty in the position of electron= 1.115e-04 m\n" ] } ], "prompt_number": 38 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg38:pg-40" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#From Heisenberg uncertainty relation (delta_x*delta_p)>=h or delta_x>=h/delta_p\n", "h=6.63*10**-34 #planck constant in joule-sec\n", "m=9.0*10**-31 #mass of electron in kg\n", "v=600 #speed of electron in m/s\n", "delta_v=(.005/100)*v#uncertainty in velocity of an electron in m/s\n", "delta_p=m*delta_v#uncertainty in momentum of an electron in Kg-m/s\n", "#value of delta_p is wrong in book\n", "delta_x=h/(2*math.pi*delta_p)#uncertainty in position of electron\n", "print\"uncertainty in position of electron=\",round(delta_x,4),\"m\"#answer is wrong in book" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "uncertainty in position of electron= 0.0039 m\n" ] } ], "prompt_number": 39 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg39:pg-41" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.63*10**-34 #planck constant in joule-sec\n", "m=9*10**-31 #mass of electron in kg\n", "v=6.6*10**4 #speed of electron in m/s\n", "p=m*v #momentum of electron in Kg-m/s\n", "delta_p=(0.01/100)*p#uncertainty in momentum(since uncertainty in value of p is 0.01% of its value)\n", "delta_x=h/(2*math.pi*delta_p)#uncertainty in position of electron(From Heisenberg uncertainty relation)\n", "print\"uncertainty in the position of electron=\",\"{:.2e}\".format(delta_x),\"m\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "uncertainty in the position of electron= 1.78e-05 m\n" ] } ], "prompt_number": 40 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg40:pg-41" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.63*10**-34 #planck constant in joule-sec\n", "m=9.1*10**-31 #mass of electron in kg\n", "r=0.5*10**-10 #radius of hydrogen atom in meter\n", "delta_x=r #uncertainty in position of electron meter\n", "delta_Px=round(h/(2*math.pi*delta_x),25)#uncertainty in momentum of electron in Kg-m/s(From Heisenberg uncertainty relation)\n", "p=delta_Px #momentum of electron(since magnitude of momentum can't be less than that of uncertainty)\n", "KE=((p**2)/(2*m))/(1.6*10**-19)#Kinetic energy in eV\n", "print\"Kinetic energy=\",round(KE,1),\"eV\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Kinetic energy= 15.1 eV\n" ] } ], "prompt_number": 41 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg41:pg-41" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.63*10**-34 #planck constant in joule-sec\n", "m=9.0*10**-31 #mass of electron in kg\n", "v=5.00*10**3 #speed of electron in m/s\n", "p=m*v #momentum of electron in Kg-m/s\n", "delta_p=(0.003/100)*p#uncertainty in momentum(since uncertainty in value of p is 0.003% of its value)\n", "delta_x=h/(2*math.pi*delta_p)#uncertainty in position of electron(From Heisenberg uncertainty relation)\n", "print\"uncertainty in the position of electron=\",\"{:.2e}\".format(delta_x),\"m\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "uncertainty in the position of electron= 7.82e-04 m\n" ] } ], "prompt_number": 42 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg42:pg-41" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.63*10**-34 #planck constant in joule-sec\n", "m=9.1*10**-31 #mass of electron in kg\n", "r=0.53*10**-10 #radius of hydrogen atom in meter\n", "delta_x=r #uncertainty in position of electron in meter\n", "delta_Px=h/(2*math.pi*delta_x)#uncertainty in momentum of electron in Kg-m/s(From Heisenberg uncertainty relation)\n", "p=delta_Px #momentum of electron(since magnitude of momentum can't be less than that of uncertainty)\n", "KE=((p**2)/(2*m))/(1.6*10**-19)#Kinetic energy in eV\n", "print\"Minimum energy=\",round(KE,1),\"eV\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Minimum energy= 13.6 eV\n" ] } ], "prompt_number": 43 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg43:pg-42" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "c=3*10**8 #speed of light in m/sec\n", "dlamda=(10**-4)*(10**-10) #width of spectral line in meter\n", "lamda=5000*10**-10 #wavelength of spectral line in meter\n", "\n", "#From Heisenberg uncertainty relation (delta_E*delta_t)>=h or delta_t>=h/(2*pi*delta_E)\n", "#since E=h*c/lamda so delta_E=(h*c/lamda**2)*dlamda\n", "#putting value of delta_E in Heisenberg uncertainty relation,delta_t=lamda**2/(2*pi*c*dlamda)\n", "delta_t=(lamda**2)/(2*math.pi*c*dlamda)\n", "print\"minimum time required=\",\"{:.3e}\".format(delta_t),\"sec\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "minimum time required= 1.326e-08 sec\n" ] } ], "prompt_number": 44 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg44:pg-42" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.62*10**-34 #planck constant joule-sec\n", "m=9.0*10**-31 #mass of electron kg\n", "delta_x=.01*10**-2#uncertainty in position of a particle in meter\n", "#(a)uncertainty in momentum of particle\n", "delta_p=h/(2*math.pi*delta_x)#From Heisenberg uncertainty relation\n", "print\"uncertainty in momentum of particle=\",\"{:.3e}\".format(delta_p),\"Kg-m/s\"\n", "\n", "#(b)uncertainty in velocity of electron\n", "delta_x=5*10**-10 #uncertainty in position of a electron in meter\n", "delta_p=h/(2*math.pi*delta_x)\n", "delta_v=delta_p/m\n", "print\"uncertainty in velocity of electron=\",\"{:.2e}\".format(delta_v),\"m/s\"\n", "\n", "#(c)uncertainty in velocity of alpha-particle\n", "mp=1.67*10**-27 #mass of proton in Kg\n", "m=4*mp #mass of alpha=particle in Kg\n", "delta_x=5*10**-10 #uncertainty in position of alpha-particle in meter\n", "delta_p=h/(2*math.pi*delta_x)\n", "delta_v=delta_p/m\n", "print\"uncertainty in velocity of alpha-particle=\",round(delta_v,2),\"m/s\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "uncertainty in momentum of particle= 1.054e-30 Kg-m/s\n", "uncertainty in velocity of electron= 2.34e+05 m/s\n", "uncertainty in velocity of alpha-particle= 31.55 m/s\n" ] } ], "prompt_number": 45 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg45:pg-43" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#From Heisenberg uncertainty relation for energy E and time t, (delta_E*delta_t)>=h\n", "#Also E=h*v or delta_E=h*(delta_v)\n", "#putting this value in uncertainty relation, delta_v>=1/(2*pi*delta_t)\n", "delta_t=10**-8 #uncertainty in time in sec\n", "delta_v=1/(2*math.pi*delta_t) \n", "print\"minimum uncertainty in the frequency=\",\"{:.3e}\".format(delta_v),\"sec-1\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "minimum uncertainty in the frequency= 1.592e+07 sec-1\n" ] } ], "prompt_number": 46 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg46:pg-43" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#From Heisenberg uncertainty relation for energy E and time t, (delta_E*delta_t)>=h\n", "#Also E=h*v or delta_E=h*(delta_v)\n", "#putting this value in uncertainty relation, delta_v>=1/(2*pi*delta_t)\n", "delta_t=10**-8 #uncertainty in time in sec\n", "delta_v=1/(2*math.pi*delta_t) \n", "print\"minimum uncertainty in the frequency=\",\"{:.3e}\".format(delta_v),\"sec-1\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "minimum uncertainty in the frequency= 1.592e+07 sec-1\n" ] } ], "prompt_number": 47 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg47:pg-43" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.62*10**-34 #planck constant joule-sec\n", "m=9.0*10**-31 #mass of electron kg\n", "delta_x=1*10**-10#uncertainty in position of a electron in meter\n", "delta_p=h/(2*math.pi*delta_x)#uncertainty in momentum of electron in Kg-m/s(From Heisenberg uncertainty relation) \n", "E=(1*10**3)*(1.6*10**-19)#energy in joule\n", "p=math.sqrt(2*m*E) #momentum in Kg-m/s\n", "percentage=(delta_p/p)*100\n", "print\"Percentage of uncertainty in its momentum=\",round(percentage,1),\"%\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Percentage of uncertainty in its momentum= 6.2 %\n" ] } ], "prompt_number": 48 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg48:pg-43" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.63*10**-34 #planck constant joule-sec\n", "delta_t=2.5*10**-20 #uncertainty in time in sec\n", "delta_E=h/(2*math.pi*delta_t)#From Heisenberg uncertainty relation for energy E and time t\n", "delta_E=delta_E/(1.6*10**-19)\n", "print\"minimum error=\",round(delta_E/10**3,3),\"KeV\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "minimum error= 26.38 KeV\n" ] } ], "prompt_number": 49 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg49:pg-44" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=1.054*10**-34 #planck constant joule-sec\n", "delta_t=10**-12 #uncertainty in time in sec\n", "delta_E=h/(2*delta_t)#uncertainty in energy(From Heisenberg uncertainty relation)\n", "delta_E=delta_E/(1.6*10**-19)\n", "print\"uncertainty in energy of gamma-ray photon emitted=\",\"{:.1e}\".format(delta_E),\"eV\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "uncertainty in energy of gamma-ray photon emitted= 3.3e-04 eV\n" ] } ], "prompt_number": 50 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg50:pg-44" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.62*10**-34 #planck constant joule-sec\n", "m=9.1*10**-31 #mass of electron kg\n", "e=1.6*10**-19 #charge of electron in coulomb\n", "V=1000. #potential difference in volt\n", "delta_V=12 #in volt\n", "#delta_P=m*e*delta_V/sqrt(2*m*e*V)\n", "#From Heisenberg uncertainty relation delta_x*delta_p)>=h or delta_x>=h/delta_p\n", "delta_x=((h*math.sqrt(2.0))/math.sqrt(m*e))*(math.sqrt(V)/delta_V)\n", "print\"uncertainty in position of an electron=\",round(delta_x,11),\"m =\",round(delta_x*1e10,1),\"Angstrom\"\n", "#answer in book is wrong" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "uncertainty in position of an electron= 6.47e-09 m = 64.7 Angstrom\n" ] } ], "prompt_number": 51 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg51:pg-44" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#According to uncertainty principle (delta_x*delta_p)>=h ......eq.1\n", "#momentum,p=h/lamda or p*lamda=h ........eq.2\n", "#on differentiating eq.2 and solving we get delta_p=(h*delta_lamda)/lamda**2,put this value in eq.1\n", "#we get (delta_x*delta_lamda)>=(lamda**2)/2*pi\n", "lamda=10**-10 #wavelength in meter\n", "delta_lamda=lamda*10**-6#since uncertainty in wavelength is given to be in ratio 1/10**6\n", "delta_x=lamda/(2*math.pi*10**-6)#since (delta_x)=(lamda**2)/(2*pi*delta_lamda)\n", "print\"uncertainty in position=\"\"{:.2e}\".format(delta_x),\"m\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "uncertainty in position=1.59e-05 m\n" ] } ], "prompt_number": 52 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg52:pg-54" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.63*10**-34 #planck constant in joule-sec\n", "m=9.11*10**-31 #mass of electron in kg\n", "L=10**-10 #width of box in meter\n", "n=1 #minimum energy of the particle is obtained for n=1\n", "E=(n*h)**2/(8*m*L**2) #in joule\n", "E=round(E,20)/(1.6*10**-19)#in eV\n", "print\"energy of an electron=\",round(E,2),\"eV\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "energy of an electron= 37.69 eV\n" ] } ], "prompt_number": 53 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg54:pg-54" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "delta_x=5*10**-10 #uncertainty in position in meter\n", "L=25*10**-10 #width of box in meter\n", "x=L/2 #at the center of the box x=L/2\n", "n=1 #since the particle is in the state of least energy so n=1\n", "def si(x): #si(x) is the wave function of particle moving in an infinite potential well\n", " return math.sqrt(2/L)*math.sin(n*math.pi*x/L)\n", "p=(math.fabs(si(x))**2)*delta_x\n", "print\"probability =\",round(p,1)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "probability = 0.4\n" ] } ], "prompt_number": 54 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg57:pg-56" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "h=6.63*10**-34 #planck constant in joule-sec\n", "m=9.1*10**-31 #mass of electron in kg\n", "L=2.5*10**-10 #width of box in meter\n", "#first lowest permitted energy value\n", "n=1\n", "E=(n*h)**2/(8*m*L**2) #in joule\n", "E=E/(1.6*10**-19) #in eV\n", "print\"first lowest permitted energy value=\",round(E,2),\"eV\"\n", "#second lowest permitted energy value\n", "n=2\n", "E=round(E,2)*n**2 #in eV\n", "print\"second lowest permitted energy value=\",round(E,2),\"eV\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "first lowest permitted energy value= 6.04 eV\n", "second lowest permitted energy value= 24.16 eV\n" ] } ], "prompt_number": 55 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg58:pg-56" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#five antinodes signify that particle is in 5th quantum state i.e n=5\n", "E5=5**2 #energy of the 5th quantum state\n", "E1=230*1.6*10**-19 #energy of the 1st quantum state in joule\n", "E1=round(E1/E5,20)\n", "h=6.62*10**-34 #planck constant in joule-sec\n", "L=0.2*10**-9 #width of well in meter\n", "#for n=1\n", "m=h**2/(8*E1*L**2) #in Kg(since En=(n*h)**2/(8*m*L**2))\n", "print\"mass of the particle=\",\"{:.2e}\".format(m),\"Kg\"\n", "En=(1*10**3)*(1.6*10**-19)#in joule(given)\n", "n=math.sqrt(En/E1)\n", "print\"since n=\",round(n,2),\" is not an integer. Hence,En=1KeV is not permitted value of energy.\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "mass of the particle= 9.32e-31 Kg\n", "since n= 10.43 is not an integer. Hence,En=1KeV is not permitted value of energy.\n" ] } ], "prompt_number": 56 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg59:pg-57" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from scipy import integrate\n", "import math\n", "x1=0.35 #lower limit\n", "x2=0.45 #upper limit\n", "#Wn=a*x is wave function\n", "p=integrate.quad(lambda x: x**2,x1,x2)\n", "p=round(p[0],4)\n", "print\"Probability= %s*a**2\"%p\n", "X=integrate.quad(lambda x: x**3,0,1)\n", "X=(X[0])\n", "print\"Expectation value of particle's position= %s*a**2\"%X" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Probability= 0.0161*a**2\n", "Expectation value of particle's position= 0.25*a**2\n" ] } ], "prompt_number": 59 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg60:pg-58" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "L = 1 # let unit length\n", "x1=0.45*L #lower limit\n", "x2=0.55*L #upper limit\n", "n=1 #for ground state\n", "p = (1/L)*((x2-(L/(2*math.pi*n) *math.sin(2*x2*math.pi*n/L)))- (x1-(L/(2*math.pi*n) *math.sin(2*x1*math.pi*n/L))))\n", "p_per = p*100 # probability of finding particle in percentage\n", "print(\"Probability of finding particle(ground state)=\"),round(p_per,1),\"%\"\n", "n=2 #for first excited state\n", "p = (1/L)*((x2-(L/(2*math.pi*n) *math.sin(2*x2*math.pi*n/L)))- (x1-(L/(2*math.pi*n) *math.sin(2*x1*math.pi*n/L))))\n", "p_per = p*100 # probability of finding particle in percentage\n", "print(\"Probability of finding particle(first excited state)=\"),round(p_per,2),\"%\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Probability of finding particle(ground state)= 19.8 %\n", "Probability of finding particle(first excited state)= 0.65 %\n" ] } ], "prompt_number": 60 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg61:pg-58" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "L = 1e-8 # length of box in cm\n", "h = 6.626e-34 # Plank constant in joule-sec\n", "m = 9.1e-31 # mass of electron in Kg\n", "E1 = (h)**2/(8*m*(L*1e-2)**2) # Calculation of energy of ground state in Joule\n", "E1_eV = round(E1/1.6e-19 )# Calculation of energy in eV\n", "E2_eV =2**2*E1_eV # Calculation of energy of first excited state in eV\n", "del_E = E2_eV - E1_eV # calculation of difference between first state and ground state\n", "print(\"Energy difference between ground state and first excited state =\"),int(del_E),\"eV\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Energy difference between ground state and first excited state = 114 eV\n" ] } ], "prompt_number": 61 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg62:pg-59" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "a=1 #let\n", "n=2 #for second energy state\n", "x=a/2 #at the center of the box\n", "W2=math.sqrt(2./a)*(math.sin((n*math.pi*x)/a))#wave function of particle in second energy state\n", "print\"probability of finding particle in interval del_x, p=del_x*(W2)**2= \",int(W2)\n", "#probability of finding particle in interval del_x is, p=del_x*(W2)**2" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "probability of finding particle in interval del_x, p=del_x*(W2)**2= 0\n" ] } ], "prompt_number": 62 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg63:pg-59" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "L = 2. # length of box in Angstrom\n", "x1=1.6000 #lower limit in Angstrom\n", "x2=1.6001 #upper limit in Angstrom\n", "n=1 #given\n", "p = (1/L)*((x2-(L/(2*math.pi*n) *math.sin(2*x2*math.pi*n/L)))- (x1-(L/(2*math.pi*n) *math.sin(2*x1*math.pi*n/L))))\n", "print\"Probability of finding particle=\",\"{:.2e}\".format(p)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Probability of finding particle= 3.45e-05\n" ] } ], "prompt_number": 63 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg64:pg-60" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "lamda=1.21*10**-10 #de-Broglie wavelength in meter\n", "L=lamda/2 #length of a loop in meter\n", "#since there are 7 loops between the walls of the box \n", "a=7*L\n", "print\"Distance between the walls=\", a,\"m\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Distance between the walls= 4.235e-10 m\n" ] } ], "prompt_number": 64 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg65:pg-60" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "L = 10**-9 # width of potential well in meter\n", "h = 6.63e-34 # Plank constant in joule-sec\n", "m = 9.1e-31 # mass of electron in Kg\n", "n1=1\n", "n2=2\n", "n3=3\n", "lamda1 = 2*L/n1 # Calculation of wavelength\n", "lamda2 = 2*L/n2 # Calculation of wavelength\n", "lamda3 = 2*L/n3 # Calculation of wavelength\n", "E=h**2/(8*m*L**2) # Calculation of energy in Joule\n", "E=round(E/(1.6*10**-19),2) # Calculation of energy in eV\n", "E1_eV = n1**2*E # Calculation of energy in eV\n", "E2_eV = n2**2*E # Calculation of energy in eV\n", "E3_eV = n3**2*E # Calculation of energy in eV\n", "print\"For first energy state: wavelength in angstrom & Energy in eV=\",int(lamda1*10**10),\",\",round(E1_eV,2)\n", "print\"For second energy state: wavelength in angstrom & Energy in eV=\",int(lamda2*10**10),\",\",round(E2_eV,2)\n", "print\"For third energy state: wavelength in angstrom & Energy in eV=\",round(lamda3*10**10,1),\",\",round(E3_eV,2)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "For first energy state: wavelength in angstrom & Energy in eV= 20 , 0.38\n", "For second energy state: wavelength in angstrom & Energy in eV= 10 , 1.52\n", "For third energy state: wavelength in angstrom & Energy in eV= 6.7 , 3.42\n" ] } ], "prompt_number": 65 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Eg66:pg-60" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "c=3*10**10 #speed of light in cm/sec\n", "del_t=2*10**-7#uncertainty in time in sec\n", "#x is the distance between source and reflecting object\n", "del_x=(c/2)*del_t\n", "print\"Uncertainty in distance=\",int(del_x),\"cm\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Uncertainty in distance= 3000 cm\n" ] } ], "prompt_number": 66 } ], "metadata": {} } ] }