量子化学课程习题及标准答案

量子化学习题及标准答案Chapter011.Acertainone-particle,one-dimensionalsystemhas/2bmxibteae,whereaandbareconstantsandmistheparticle’smass.Findthepotential-energyfunctionVforthissystem.(Hint:Usethetime-dependentSchrodingerequation.)Solution：As(x,t)isknown,wecanderivethecorrespondingderivatives.222222/42),(),(),(2xmbbmxtxibttxeaetxbmxibtAccordingtotime-dependentSchroedingerequation,222(,)(,)(,)(,)2xtxtVxtxtitmxsubstitutingintothederivatives,weget222),(mxbtxV2.Atacertaininstantoftime,aone-particle,one-dimensionalsystemhasbxxeb/||2/13)/2(,whereb=3.000nm.Ifameasurementofxismadeatthistimeinthesystem,findtheprobabilitythattheresult(a)liesbetween0.9000nmand0.9001nm(treatthisintervalasinfinitesimal);(b)liesbetween0and2nm(usethetableofintegrals,ifnecessary).(c)Forwhatvalueofxistheprobabilitydensityaminimum?(Thereisnoneedtousecalculustoanswerthis.)(d)Verifythatisnormalized.Solution：a)Theprobabilityoffindinganparticleinaspacebetweenxandx+dxisgivenby6/223210*29.32dxexbdxPbxb)0753.02910*20/223dxexbPbxc)Clearly,theminimumofprobabilitydensityisatx=0,wheretheprobabilitydensityvanishes.d)144422330/223/223/2232bbdxexbdxexbdxexbdxPbxbxbx3.Aone-particle,one-dimensionalsystemhasthestatefunction2222/4/16/4/12)/32)((cos)/2)((sincxcxxecatecatwhereaisaconstantandc=2.000Å.Iftheparticle’spositionismeasuredatt=0,estimatetheprobabilitythattheresultwillliebetween2.000Åand2.001Å.Solution：whent=0,thewavefunctionissimplifiedas441610*158.2)32(),(22cxxectxChapter021.Consideranelectroninaone-dimensionalboxoflength2.000Åwiththeleftendoftheboxatx=0.(a)Supposewehaveonemillionofthesesystems,eachinthen=1state,andwemeasurethexcoordinateoftheelectronineachsystem.Abouthowmanytimeswilltheelectronbefoundbetween0.600Åand0.601Å?Considertheintervaltobeinfinitesimal.Hint:Checkwhetheryourcalculatorissettodegreesorradians.(b)Supposewehavealargenumberofthesesystems,eachinthen=1state,andwemeasurethexcoordinateoftheelectronineachsystemandfindtheelectronbetween0.700Åand0.701Åin126ofthemeasurements.Inabouthowmanymeasurementswilltheelectronbefoundbetween1.000Åand1.001Å?Solution:a)Ina1Dbox,theenergyandwave-functionofamicro-systemaregivenby)sin(2,22222xlnlmlnEtherefore,theprobabilitydensityoffindingtheelectronbetween0.600and0.601Åis65510*545.6)(sin242dxxlnlPb)Fromthedefinitionofprobability,theprobabilityoffindinganelectronbetweenxandx+dxisgivenbydxxlnlP)(sin22Asthenumberofmeasurementsoffindingtheelectronbetween0.700and0.701Åisknown,thenumberofsystemis159001.0)0.12*1(sin22*158712158712001.0)7.02*1(sin2212612622PPN2.Whenaparticleofmass9.1*10-28ginacertainone-dimensionalboxgoesfromthen=5leveltothen=2level,itemitsaphotonoffrequency6.0*1014s-1.Findthelengthofthebox.Solution.mlhvlmlhnnmlnnEloweruplowerup923622222222210*78.1110*26646.18)(2)(3.Anelectroninastationarystateofaone-dimensionalboxoflength0.300nmemitsaphotonoffrequency5.05*1015s-1.Findtheinitialandfinalquantumnumbersforthistransition.Solution:2,3588)(2)(222222222222lowerupperloweruploweruplowerupnnhvmlnnhvmlhnnmlnnE4.Fortheparticleinaone-dimensionalboxoflengthl,wecouldhaveputthecoordinateoriginatthecenterofthebox.Findthewavefunctionsandenergylevelsforthischoiceoforigin.Solution:Thewavefunctionforaparticleinaone-dimernsionalboxcanbewrittenas)2()2()(xmEBSinxmEACosxIfthecoordinateoriginisdefinedatthecenterofthebox,theboundaryconditionsaregivenas)2(,0)22()22(0)()1(,0)22()22(0)(22EqlmEBSinlmEACosxEqlmEBSinlmEACosxlxlxCombiningEq1withEq2,weget)4(,0)22()3(,0)22(EqlmEBSinEqlmEACosEq3leadstoA=0,or)22(lmECos=0.Wewilldiscussbothsituationsinthefollowingsection.IfA=0,Bmustbenon-zeronumberotherwisethewavefunctionvanishes.)2(22220)22(0222xlnSinlmlhnEnlmElmESinBIfA≠0))12((28)12()21(220)22(00)22(0)22(0222xlnCoslmlhnEnlmElmECosBlmESinlmECosA5.Foranelectroninacertainrectangularwellwithadepthof20.0eV,thelowestenergylies3.00eVabovethebottomofthewell.Findthewidthofthiswell.Hint:Usetanθ=sinθ/cosθSolution:Fortheparticleinacertainrectangularwell,theEfulfillwith)2sin()2()2cos()(21010lmEVElmEEEVSubstitutingintotheVandE,weget101010011110*64.2)7954.0(10*127.12)7954.0(7954.020202.12)(2)2()2()2(lowestlnmEnlnlmEVEEEVlmETanlmECoslmESinChapter031.IfAˆf(x)=3x2f(x)+2xdf/dx,giveanexpressionforAˆ.Solution：Extractingf(x)fromtheknownequationleadstotheexpressionofAdxdxxA23ˆ22.(a)Showthat(Aˆ+Bˆ)2=(Bˆ+Aˆ)2foranytwooperators.(b)Underwhatconditionsis(Aˆ+Bˆ)2equaltoAˆ2+2AˆBˆ+Bˆ2?Solution:a)222222)ˆˆ()ˆˆ)(ˆˆ(ˆˆˆˆˆˆˆˆˆˆˆˆ)ˆˆ)(ˆˆ()ˆˆ(ABABABABAABBBABBAABABABAb)ABBABBAABABBAABAˆˆˆˆˆˆˆ2ˆˆˆˆˆˆˆ)ˆˆ(22222IfandonlyifAandBcommute,(Aˆ+Bˆ)2equalstoAˆ2+2AˆBˆ+Bˆ23.IfAˆ=d2/dx2andBˆ=x2,find(a)AˆBˆx3;(b)BˆAˆx3;(c)AˆBˆf(x);(d)BˆAˆf(x)Solution:a)3522320ˆˆxxdxdxBAb)3322236ˆˆxxdxdxxABc))()(4)(2)]()(2[)]([)(ˆˆ2222222xfdxdxxfdxdxxfxfdxdxxxfdxdxfxdxdxfBAd))()()(ˆˆ222222xfdxdxxfdxdxxfAB4.Classifytheseope