{ "metadata": { "name": "", "signature": "sha256:43ad060be6803a5e6c90770bf46ae3612188f9380f800bb70a03161cb97405cb" }, "nbformat": 3, "nbformat_minor": 0, "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": {} } ] }