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1、41ATOMICPHYSICSConceptualQuestions41.1.The5dstatehasthehigherenergybecause2En=−13.6eV/ndependsonlyontheprincipalquantumnumbern,and−−13.6eV13.6eV>5422ψ2(,,)xyz41.2.Inthree-dimensionalspace,theprobabilitydensityisthelikelihoodoffindinganelectronina2volumeδVatagivenpositionx,y
2、,z.TheradialprobabilitydensityRnl()risthelikelihoodoffindinganelectronatthedistancerfromtheorigin(whichisusuallydefinedbythepositionofthenucleus).Bothhaveunitsof(length).−341.3.listheorbitalquantumnumber.Itisaninteger:lL=0,1,2,…,istheactualnumericalvalueoftheorbitalangularm
3、omentum,withunitsofJs.ItisrelatedtolbyLl=+(1l).=41.4.Thequantumnumbers=1isthespinofanelectron.Itplaysaroleanalogoustotheorbitalquantumnumberl.2Althoughitisnotaninteger,itsallowedvaluesdifferbytheinteger±1.Sistheactualnumericalvalueofthespinangularmomentum.ItisrelatedtosbySs
4、=+(1s).=41.5.(a)Theup-downdeflectionofatomsinaStern-Gerlachexperimentisalongthez-axis,sothedeflectiondependsonthemagneticmomentμzwhichprovidesinformationaboutthez-componentoftheangularmomentum.Eachpeakrepresentsadifferentvalueofthez-componentoftotalangularmomentum.(b)Orbita
5、langularmomentumnumberslareintegers,andthereareanoddnumber(2l+1)ofz-components.Onecomponenthasm=0andwouldnotbedeflectedatall.However,weseeanevennumberofpeaks,allofwhicharedeflected.Thattellsusthatthetotalangularmomentumcan’tbeaninteger.Thisismorelikeanatomwhosetotalangularm
6、omentumconsistsonlyofitsspin,withS=1/2andwithtwoz-components1/2.mS=±Inthatcase,thedeflectionhad2peaks.Anexperimentwith4deflectionpeakscanbeexplainedifthetotalangularmomentumquantumnumberis3/2,becausethenthez-componentwouldhavethepossiblevalues−−3/2,1/2,1/2,and3/2.Totalangul
7、armomentum3/2canariseifanatomhasbothorbitalandspinangularmomentum.41.6.(a)Yes,thisisthegroundstateofnitrogen(seeFigure41.22).Notethatthepsubshellshave6states,sotheycanacceptthreeelectronswithoutviolatingthePauliexclusionprinciple(seeTable41.2).(b)No,onlytwoelectronsareallow
8、edinthe2sstate.(c)No,the2sstatesarelowerinenergythanthe2pstates,sointhegroundstate
