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1、rdSolutionstoPrinciplesofElectronicMaterialsandDevices:3Edition(13May2005)Chapter1ThirdEdition(2005McGraw-Hill)Chapter1Note:Thefirstprintinghasafewoddtypos,whichareindicatedinbluebelow.Thesewillbecorrectedinthereprint.1.1VirialtheoremTheLiatomhasanucleuswitha+3epositivecharge,whichissurro
2、undedbyafull1sshellwithtwoelectrons,andasinglevalenceelectronintheouter2ssubshell.TheatomicradiusoftheLiatomisabout0.17nm.UsingtheVirialtheorem,andassumingthatthevalenceelectronseesthenuclear+3eshieldedbythetwo1selectrons,thatis,anetchargeof+e,estimatetheionizationenergyofLi(theenergyrequi
3、redtofreethe2selectron).Comparethisvaluewiththeexperimentalvalueof5.39eV.Supposethattheactualnuclearchargeseenbythevalenceelectronisnot+ebutalittlehigher,say+1.25e,duetotheimperfectshieldingprovidedbytheclosed1sshell.Whatwouldbethenewionizationenergy?Whatisyourconclusion?SolutionFirstwecon
4、siderthecasewhentheoutermostvalenceelectroncanseeanetchargeof+e.FromCoulomb’slawwehavethepotentialenergyQQ(+e)(−e)12PE==4πεr4πεr0000−192(1.6×10C)-18=−=1.354×10Jor-8.46eV−12−1−94π(8.85×10Fm)(0.17×10m)Virialtheoremrelatestheoverallenergy,theaveragekineticenergyKE,andaveragepotentialenergyPEt
5、hroughtherelations1E=PE+KEandKE=−PE2ThususingVirialtheorem,thetotalenergyis1E=PE=0.5×−8.46eV=-4.23eV2Theionizationenergyistherefore4.23eV.Nowweconsiderthesecondcasewhereelectronthesees+1.25eduetoimperfectshielding.AgaintheCoulombicPEbetween+eand+1.25ewillbeQ1Q2(+1.25e)(−e)PE==4πεr4πεr0000−
6、1921.25⋅(1.6×10C)-18=−=−1.692×10Jor−10.58eV−12−1−94π(85×10Fm)(0.17×10m)Thetotalenergyis,1.1rdSolutionstoPrinciplesofElectronicMaterialsandDevices:3Edition(13May2005)Chapter11E=PE=5.29eV2Theionizationenergy,consideringimperfectshielding,is5.29eV.Thisvalueisincloseragreementwiththeexperiment
7、alvalue.Hencethesecondassumptionseemstobemorerealistic.____________________________________________________________________________________1.2Atomicmassandmolarfractionsa.ConsideramulticomponentalloycontainingNelements.Ifw1,w2,...,wNaretheweightfractionsofcomp
