電子材料與器件原理答案_第1頁
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1、SolutionstoPrinciplesofElectronicMaterialsDevices:3rdEdition(13May2005)Chapter11.1ThirdEdition(????2005McGrawHill)Chapter1Note:Thefirstprintinghasafewoddtyposwhichareindicatedinbluebelow.Thesewillbecrectedinthereprint.1.

2、1VirialtheemTheLiatomhasanucleuswitha3epositivegewhichissurroundedbyafull1sshellwithtwoelectronsasinglevalenceelectronintheouter2ssubshell.TheatomicradiusoftheLiatomisabout0.17nm.UsingtheVirialtheemassumingthatthevalence

3、electronseesthenuclear3eshieldedbythetwo1selectronsthatisageofeestimatetheionizationenergyofLi(theenergyrequiredtofreethe2selectron).Comparethisvaluewiththeexperimentalvalueof5.39eV.Supposethattheactualnucleargeseenbythe

4、valenceelectronisnotebutalittlehighersay1.25eduetotheimperfectshieldingprovidedbytheclosed1sshell.WhatwouldbethenewionizationenergyWhatisyourconclusionSolutionFirstweconsiderthecasewhentheoutermostvalenceelectroncanseeag

5、eofe.FromCoulomb’slawwehavethepotentialenergy0000rπεeerπεQQPE4))((421?==m)10)(0.17Fm108.85(4C)10(1.69112219?????=π=1.3541018J8.46eVVirialtheemrelatestheoverallenergytheaveragekiicenergyKEaveragepotentialenergyPEthroughth

6、erelationsKEPEE=PEKE21?=ThususingVirialtheemthetotalenergyiseV46.85.021?==PEE=4.23eVTheionizationenergyistherefe4.23eV.Nowweconsiderthesecondcasewhereelectronthesees1.25eduetoimperfectshielding.AgaintheCoulombicPEbetween

7、e1.25ewillbe000021r4πeer4πQQPEεε))(1.25(?==m)10)(0.17Fm10(854C)10(1.61.259112219??????=π=?1.6921018J?10.58eVThetotalenergyisSolutionstoPrinciplesofElectronicMaterialsDevices:3rdEdition(13May2005)Chapter11.3b.Supposethatw

8、ehavetheatomicfractionniofanelementwithatomicmassMi.Themassoftheelementinthealloywillbetheproductoftheatomicmasswiththeatomicfractioni.e.niMi.MassofthealloyistherefenAMAnBMB…nNMN=MalloyBydefinitiontheweightfractioniswi=m

9、assoftheelementiMassofalloy.Therefe...=CCBBAAAAAMnMnMnMnw...=CCBBAABBBMnMnMnMnwc.TheatomicmassofCdSeare112.41gmol178.96gmol1.SinceoneatomofeachelementisinthecompoundCdSetheatomicfractionnCdnSeare0.5.TheweightfractionofCd

10、inCdSeistherefe111SeSeCdCdCdCdCdmolg96.785.0molg41.1125.0molg41.1125.0???==MnMnMnw=0.58758.7%SimilarlyweightfractionofSeis111SeSeCdCdSeSeSemolg96.785.0molg41.1125.0molg96.785.0???==MnMnMnw=0.412641.3%Consider100gofCdSe.T

11、henthemassofCdweneedisMassofCd=wCdMcompound=0.587100g=58.7g(Cd)MassofSe=wSeMcompound=0.413100g=41.3g(Se)d.Theatomicfractionsoftheconstituentscanbecalculatedusingtherelationsprovedabove.TheatomicmassesofthecomponentsareMS

12、e=78.6gmol1MTe=127.6gmol1MP=30.974gmol1.Applyingtheweighttoatomicfractionconversionequationderivedinpart(a)wefind1111PPTeTeSeSeSeSeSemolg974.3003.0molg6.1272.0molg6.7877.0molg6.7877.0????==MwMwMwMwn∴∴∴∴nSe=0.79479.4%1111

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