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Anderson&Anderson12/15/04SolutionsChapter1SOLUTIONSCHAPTER11.1.ShowthatEquation(1.6)followsfromEquation(1.3).Solution:Equation(1.3)isdEP=q24πε0r2dr.IntegratingbothsidesweobtaindEP=q24πε0r2dr=q24πε01r2dr=∫∫∫q24πε0−1r⎡⎣⎢⎤⎦⎥+const=EPTofindtheconstant,weemploytheboundaryconditionthatatr=∞,EP=Evac:20()04PvacqErEconstconstπε−=∞==+=+⋅∞∴const=Evacand204PvacvacqEEEπε−=−=⋅∞,Equation(1.6)1.2.Consideralithiumnucleus,ofcharge+3q.CalculatethefirstthreeelectronenergiesforanelectroninaLi++ion,usingtheBohrmodel.Werepeattheanalysisthatweusedforthehydrogenatom,exceptthatnowthechargeofthenucleusQ1isequalto1933(1.610)qC−=+×.Theresultsofthekeystepsare2122200344QQqFrrπεπε−==203()4PvacqErErπε=−2220304mvqrrπε−=nnmvrn=h()20314nqvnπε⎛⎞=⎜⎟⎝⎠h()20224()3nrnmqπε=h()2422203124KmqEnπε⎛⎞=⎜⎟⎝⎠h()422209124nPnKnvacmqEEEEnπε⎛⎞=+=−⎜⎟⎝⎠hThus()1913.6122vacvacEEeVEeV=−=−Anderson&Anderson22/15/04SolutionsChapter1and()22913.630.62vacvaceVEEEeV=−=−and()32913.613.63vacvaceVEEEeV=−=−1.3.ShowthatEquations(1.12)and(1.13)followfrom(1.8)and(1.11).Equation(1.8)is:222004mvqrrπε−=Multiplybothsidesbyr2anddividebyv:204qmvrvπε=whichfromEquation(1.11)is204qmvrnvπε==hSolvingtherighthandequalityforv:204qvnπε=h(Equation(1.13))Solvingthelefthandequalityforrsubstitutinginv:()22002244nnnnrmvmqmqπεπε===hhhh(Equation(1.12)).1.3.IneachofthepotentialenergydistributionsinFigure1P.1,sketchthemagnitudeanddirectionoftheforceontheelectron.Anderson&Anderson32/15/04SolutionsChapter1Theforceisminusthegradientofthepotentialenergy(Equation(1.2)).1.5.ConsidertheelectronintheenergydiagramofFigure1P.2.TakingtheenergytheelectronhasatPointAasEtotal,ateachoftheindicatedpositions,findthetotalenergy,thekineticenergy,thepotentialenergy,andtheelectron’svelocity.Indicatethedirectionofforce(ifany).Recallthattotalenergyisconserved.Atpoint“D”theelectroncollidesinelasticallywithsomething(perhapsanatominthecrystal).Afterthecollision,theelectron’senergyisequaltoitspotentialenergy,anditskineticenergyiszero.Itstotalenergyismuchlessthanbeforethecollision;wheredidtheextraenergygo?TheelectronatPointAisatrest.ItstotalenergyisEtotal,itskineticenergyandvelocityarezero,anditspotentialenergyisequaltoitstotalenergy.Theforceontheelectronistotheright,becausetheslopeofthepotentialenergyisnegative.AtPointB,thetotalenergyoftheelectronisthesame(byconservationofenergy),butthepotentialenergyisEP(B)=Etotal−5eV.Thekineticenergyisthedifferencebetweenthetotalandthepotentialenergies,soEK(B)=Etotal−EP(B)=Etotal−Etotal−5eV()=5eV.Thevelocityoftheelectronis()196312(5)1.610/21.3310/9.110KeVJeVEvmsmkg−−⋅⋅×===××(or3millionmilesanhour).AtPointC,theelectron’senergyisEtotal,butitspotentialenergyisAnderson&Anderson42/15/04SolutionsChapter1EP(C)=Etotal−10eV.Therefore,thekineticenergyoftheelectronisEK(C)=Etotal−EP(C)=Etotal−Etotal−10eV()=10eV,and()19()6312210(1.610/1.8610/9.110KCEeVeVJvmsmkg−−⋅×===××Thereisnoforceontheelectronbecausetheslopeofthepotentialenergyiszero,buttheelectroncontinuesmovingtotherightbecauseitstillhaskineticenergy.AtPointD,theelectronhasmadeacollision.Itstotalenergyisnow10eVlessthanitwas-theextra10eVofenergywastransferredtotheatomitcollidedwith.ThepotentialenergyisEP(D)=Etotal−10eV.Theelectron’stotalenergyisnowEtotal(new)=Etotal(original)−10eV,whichisthesameasitspotentialenergy.Therefore,thekineticenergyoftheelectroniszero,andsoisthevelocity.Thereisnoforceontheelectronbecausetheslopeofthepotentialenergyiszero.1.6.Findthekineticenergiesineachofthefollowing.Expressallyouranswersinelectronvolts.(a)anelectroninthefirstallowedenergystatesofthehydrogenatom(accordingtotheBohrmodel,Eq.(1.14))Weusetherelation()42220124KmqEnπε=h()()()()()4422222229.1311.61911248.812/1.06342.151813.6KEkgECEEFmEJsEJeVπ−−=−−=−=(b)afreeelectron,initiallyatrestatthebackofacathoderaytubeinyourtelevision,acceleratedthroughapotentialof10kVtostrikethephosphorlayerAnelectronacceleratedthrough10kVacquiresanenergyof10KeV.Sincethiselectronwasinitiallyatrest,itskineticenergyisnow10KeV,or191510,0001.6101.610JJ−−××=×.(c)atiny,driftingdustparticle,ofmass1µgandvelocityaleisurely1mm/s.Weuse2932161110(10/)5103.122KEmvkgmsJkeV−−−==⋅⋅=×=Anderson&Anderson52/15/04SolutionsChapter11.7.Forthefollowingsemiconductormaterials,indicatetowhatdegreeyouexpectcovalentorionicbonding,andwhy:GeGaPInGaAsPHgCdTeGe:ThiselementisinColumnIVoftheperiodictable.Eachatominthecrystalshares4electronswitheachneighboringatom,andinreturn“receives”4electrons.Becauseeveryatomis“giving”and“receiving”thesamenumberofelectrons,thechargeisevenlydistributedbetweenatoms,andthebondingiscovalent.GaP:GalliumisinColumnIIIandphosphorousisinColumnV.Thusinthebonding,indiumhaslesspositivechargethanthephosphorus,sotheelectronsspendsslightlymoretimenearthePatomsthantheGa.Thisbondingislargelycovalaentbutwithaslightlyionicflavor.InGaAsP:IndiumandgalliumareinColumnIIIandarsenicandphosphorousareinColumnV,sothebondinghasthesamecharacterasthatofInP,mostlycovalentbutsomewhationic.HgCdTe:MercuryisinColumnII,whilecadmiumandtelluriumareincolumnVI.ThisII-VIsemiconductormaterialhasalargelyionicbutslightlycovalentbonding.T