Solution Manual For Solid State Electronic Devices, 6th Edition

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Chapter1SolutionsProb.1.1WhichsemiconductorinTable1-1hasthelargestEg?thesmallest?WhatisthecorrespondingN’HowisthecolumnIIIcomponentrelatedtoEg?largestEg:ZnS,3.6eV1.24A=——=0.344356wmsmallestEg:InSb,0.18eV124A=—"—=6.89oisomAlcompoundsE,>correspondingGacompoundsEg>thecorrespondingIncompoundsEgProb.1.2Findpackingfractionoffccunitcell.nearestatomseparation=REN=3.54Atetrahedralradius=1.77Avolumeofeachatom=23.14A°numberofatomspercube=6-1+8-1=4atoms3packingfraction=BIKA=0.74=74%(54)+StudyXY

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DownloadedfromStudyXY.com®+StudyXYSdYe.o>\|iF’prE\3SStudyAnythingThisContentHasbeenPostedOnStudyXY.comassupplementarylearningmaterial.StudyXYdoesnotendroseanyuniversity,collegeorpublisher.Allmaterialspostedareundertheliabilityofthecontributors.wv8)www.studyxy.com

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Prob.1.3Labelplanes.zz643)|212)«yxwy.XyZzXyZz2342421/21/31/41/21/41/2643212Prob.1.4CalculatedensitiesofSiandGaAs.TheatomicweightsofSi,Ga,andAsare28.1,69.7,and74.9,respectively.Si:a=5.43-10"cm,8atoms/cell8atoms=8=5-10715a(5:43-10%m)domsityS107BIEensity=——&>=233-£RTRTEESGaAs:a=5.65-10°cm,4eachGa,Asatoms/cellEa.SERTaN2(5.65-10%m)densi2.22-10715.(69.7+74.9)-%;533ensity=—ew>7J/mol_5338y6.02-10%2;Es+StudyXY

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Prob.1.5ForInSb,findlatticeconstant,primitivecellvolume,(110)atomicdensity.La=1.44A+136A=2.84a=6.47A3FCCunitcellhas4latticepoints..volumeofprimitivecell===67.7A°areaof(110)plane=2a’.4.242.122densityofInatoms=—=—2=2=3.37-10"1;2aaonsamenumberofSbatoms=3.37-10"1;Prob.1.6Finddensityofscunitcell.nearestatomseparation=2-2.5A=5Anumberofatomspercube=8-1=1atom542-5massofoneatom=———mL=9.](*_£_6.02107omomL9.10%_&density=latom9-107gp=0.072%;(4)’EProb.1.7Draw<110>directionofdiamondlattice.(Pr-—-1LIRs~=)Thisviewistiltedslightlyfrom(110)toshowthe!)ZAae2,alignmentofatoms.Theopenchannelsare\17hiyhexagonalalongthisdirection.[}WN)[OdpoXEXpddI)SUSoyeo:Lagigvanise)Zins+StudyXxy

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Prob.1.8Showbcclatticeasinterpenetratingsclattices.SOVatrlVoipdirgad©ViewdirectionTheshadedpointsareonesclattice.Theopenpointsaretheinterpenetratingsclatticelocateda/2behindtheplaneofthefrontshadedpoints.Prob.1.9:(a)FindnumberofSiatoms/cm’on(100)surface.feelatticewitha=5.4341numberatomsper(100)surface=4-4+1=2atomsa=5.43A5|atomsper(100)surfacearea=————=6.78-10"1;(5.434)(b)FindthenearestneighbordistanceinInP.a4.2?|NERS2feelatticewitha=5.87Anearestneighbordistance=22-288pais+StudyXY

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Prob.1.10FindNaCldensity.Na”:atomicweight23g/mol,radius14CI’:atomicweight35.5g/mol,radius1.84unitcellwitha=2.8Abyhardsphereapproximation.lstomsD34latms.355.LY%Naand%Clatomsperunitcell=FEBa+355=4.86-10756.02107stems;1078density=486-1075=22-L(2.8-10%cm)’J;aThehardsphereapproximationiscomparablewiththemeasured2.172%density.Prob.1.11Findpackingfraction,Batomsperunitvolume,andAatomsperunitarea.2AoOOONote:TheatomsarethesamesizeandtoucheachEootherbythehardsphereapproximation.(5)0.O58radiiofAandBatomsarethen1A4AShOo0numberofAatomsperunitcell=8-1=1aAnumberofBatomsperunitcell=1volumeofatomsperunitcell=1-4.(14)+1-4.(14)°=&A°volumeofunitcell=(4A)°=64A°=Aonackingfraction=2——=—=0.13=13%packing6824°Batomsvolumedensity=te=1.56-10"15numberofAatomson(100)plane=4-1=1Aatoms(100)aerialdensity=~2°%_65.10%1(44)E|+studyxy

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Prob.1.12AFindatoms/cellandnearestneighbordistanceforsc,bce,andfeclattices.OOsc:atoms/cell=8-1=1at©;C}anearestneighbordistance=aava2al!.bee:atoms/cell=8-3+1=2OooOAEA5Oe25nearestneighbordistance=a30]a22O%Oo2abte®®zOOfee:atoms/cell=8-3+6-1=4VacY%AaesiTOLTOLshe?nearestneighbordistance=AO22a“15FlzProb.1.13Drawcubesshowingfour{111}planesandfour{110}planes.<4EeRT{111}pl4I-A.anes7éPViL77V47Eears<=W/L[7LTT727LF707)Ng{110}planes.ERKps:.77/8weld,lfa,PON

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Prob.1.14:Findfractionoccupiedforsc,bee,anddiamondlattices.sc:atoms/cell=8-1=1OOnearestneighbour=a=>radius=:Totes”a4n(a)_ma’]atomspherevolume==)=—OOYTaunitcellvolume=a’©,©)LEfractionoccupied=—==0.5200bec:atoms/cell=8-§+1=2a-%aOOnearestneighbour=el>radius=SlMo]O.3.2%9)atomspherevolume=huda3=wie@,©,3416%a32unitcellvolume=a’:Bb72eli>.2.52fractionoccupied=——10=zh=0.6822adiamond:atoms/cell=4(fcc)+4(offsetfoc)=84/]lattice,..al2nearestneighbour=a3->atomradius=a3O‘=o].’.a37Oatomspherevolume=w==LEE/|JVOaunitcellvolume=a*aoNn3a®4_0g.rN__SN.1287-3fractionoccupied=——=&—=——=0.34ENa16rv)_—

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Prob.1.15CalculatedensitiesofGeandInP.TheatomicweightsofGe,In,andPare72.6,114.8,and31,respectively.Ge:a=5.66-10"cm,8atoms/cell8aloms=SEBa=4.41-1015a(5.6610cm)Lo441-10”15.72.65%;_.density=TIL=532%GaAs:2a=5.87-10"cm,4eachIn,Patoms/cellAo4jesi0ty2(5.87-10%m)1.98-10%1;(114.8+31)%;density=———=—————~"==4.79-£,6.02-10%1;emProb.1.16Sketchdiamondlatticeshowingfouratomsofinterpenetratingfccinunitcell..SS$$aOOOOS>4S(25)&()LuleLg)JENSENdgTORSONoxallolilN5)——SNigLIPS4S(I0.ENOOW—_adOLSS$$5$$S$”(CHE(HY(CHOFullinterpenetratingLatticeFourInterpenetratingAtomsinUnitCell+Studyxy

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Prob.1.17FindAlSb,As.tolatticematchInPandgivebandgap.LatticeconstantsofA1Sb,AlAs,andInPare6.14A,5.66A,and5.874,respectivelyfromAppendixIII.UsingVegard’sLaw,6.14Ax+5.66A-(1-x)=5.87A—»x=0.44AlSbo.44As56latticematchesInPandhasE;~1.9eVfromFigure1-13.FindInyGay..PtolatticematchGaAsandgivebandgap.LatticeconstantofInP,GaP,andGaAsare5.87A,5.454,and5.654,respectivelyfromAppendixIII.UsingVegard’sLaw,5.87Ax+5.45A-(1-x)=5.654—>x=048Ino43Gay52PlatticematchesGaAsandhasE;=2.0eVfromFigure1-13.Prob.1.18FindweightofAs(k;=0.3)addedto1kgSiinCzochralskigrowthfor10”emdoping.atomicweightofAs=74.9%;101.C,=k,-C,=10"1;—»C=—==333.10"1;om03ESassumeAsmaybeneglectedforoverallmeltweightandvolume1000g1=429.2cm’®Si2.33%3.33-10°ZL+429.2cm’=1.43-10'*Asatoms10".&143:107atoms-74955_|g104gAs=18-107kgAs6.02-10%someI~StudyXY

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Chapter2SolutionsProb.2.1(a&b)Sketchavacuumtubedevice.GraphphotocurrentIversusretardingvoltageV.“forseverallightintensities.|hv|light[]intensityFaVovNotethatV,remainssameforallintensities.(¢)Findretardingpotential.A=2440A=0.244pmD=4.09¢VV,=hy-@=120VoHmgy124eVepmoo508eV-4.096V~16V-Aum)0.244pmProb.2.2.ShowthirdBohrpostulateequatestointegernumberofDeBrogliewavesfittingwithin©circumferenceofaBohrcircularorbit.22222ALNRILGIAP—.mq’4me,rT_Amen’nh’4ner’or,_oRBETmi?ofwm?me?aaqmr,’qmr,”mv’mv’,m*v2r,?=n?mvt,=nhp,=nhisthethirdBohrpostulate+StudyXYB:SE—S—

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Prob.2.3(a)FindgenericequationforLyman,Balmer,andPaschenseries.44ap=fooi-—222A32n’e’n’R’ 32n’e’n,’hhe—mq*(n,’-n’)-mq‘(n,’-n’)Lr32’n’n’mn’8e.’n’n,’h?a=8e,’n’n,’h’he_8s'Kcn/n;mg'@,’-n’) mq*nn’a=8(8.85-10™£)*.(6.63-107Js)*2.998-10°=n/n)’9.11-10%'kg-(1.60-10°C)*n,’-n?2.22.2A=91110°m-282—g14.n,"-nyn,-n;n,=1forLyman,2forBalmer,and3forPaschen(b)PlotwavelengthversusnforLyman,Balmer,andPaschenseries.— LyMANSERES[PASCHENSERES|[oTooTn2q|ned)[etme][nT2Tno[n"2(n%29)pironan2s)[2|aT3T7415|[3ToT78113|025|[6126116|"1406|2811|[411616[107|oz|[6138127T1200|10932|[512124|104|040|71#140|410s|does|LYMANLIMIT911AI5[0J70|1|oss|o010|[nT0%Tn24[anrain2a)[e114nmed)PASCHENLIMIT~~8199A8Te15T720|6560|HE2=xTTwe616|a[as|100][7140145123|oes|BALMERLIMIT~~3644AProb.2.4m-q*Showequation2-17correspondstoequation2-3.Thatisshowc¢-R=TERFrom2-17andsolutionto2.3,2.998-10°=1vy=5_29%10:=a1]9.11-10°m.Jemmn,n,From2-3,vymek|5-1|=2.008.10m.1.097.1073L-L|=320-1058LLn,n,no,”mn,n,n,—StudyXY

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Prob.2.5(a)FindApxforAx=IA.hh6.63-10™J5Ap,-Ax=—>Ap,=——=""—_"~=503.107knPTTTAx4mi0'm:(b)FindAtforAE=1eV.15AE-At=a>At=_h_414107eV:s=3.30-10"%4r4m-AE4n-leVProb.2.6Findwavelengthof100eVand12keVelectrons.Commentonelectronmicroscopescomparedtovisiblelightmicroscopes.E=imv’>v=ZEBm\-34r=hh___h__663107T'spo=E7.491.10"F.mpmvV2.Em2.91110%kgFor100eV,A=E*.491.10"J.m=(100eV-1.602-10™2)-4.91-10"F:m=1.23-10™m=1.234For12keV,A=E*.491-107Fm=(1.2-10%V-1.602-10%°2)4.91.10):m=1.12-10"'m=0.1124Theresolutiononavisiblemicroscopeisdependentonthewavelengthofthelightwhichisaround50004;so,themuchsmallerelectronwavelengthsprovidemuchbetterresolution.Prob.2.7Showthat7istheaveragelifetimeinexponentialradioactivedecay.TheprobabilityoffindinganatominthestablestateattimetisN(t)=N,-¢™.Thisisanalogoustotheprobabilityoffindingaparticleatpositionxforfindingtheaverage.feta,OE——CEfeat’0Thismayalsobefoundbymimickingthediffusionlengthcalculation(Equations4-37to4-39).—StudyXYaEESEoo

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Prob.2.8Findtheprobabilityoffindinganelectronatx<0.Istheprobabilityoffindinganelectronatx>0zero-ornon-zero?Istheclassicalprobabilityoffindinganelectronatx>6zeroornon?Theenergybarrieratx=0isinfinite;so,thereiszeroprobabilityoffindinganelectronatx<0(|¢|’=0).However,itispossibleforelectronstotunnelthroughthebarrierat5<x<6;s0,theprobabilityoffindinganelectronatx>6wouldbequantummechanicallygreaterthanzero(|y|*>0)andclassicalmechanicallyzero.-|v=10vHPaBySH|™Bio....050£3LJx(nm,Prob.2.9Find4-p’+2-p;+ESorP(x,y,z,t)=A-e/OFm»2:faLgi0x3y-40(22)A.l10x3y-404,jox(0.2)=_J100-7flAl£TU0RBY-40J(10xs3v4)g©’2|A".gri0x3y-4y(ha)A.el10x3y493,joz(0.2)_=_J-0flAferi10x3y-49,1i(10%133-4)dzTarLgI0xI3y-49(22Jacorerornajor(By==VIF)uy,flAffTIXYAA034)34opre2.pp+E=400%"PE—m9.11-107'kg—StudyXY._—SE_

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Prob.2.10Findtheuncertaintyinposition(Ax)andmomentum(Ap).LW(x,t)=EomIX)ehangfe“Wdx=1ALLJK2%TX(x)=[#"xwax=£[x-sin?(5)os=0.5L(fromproblemnote)0L0LRU2%,(wx2(x)=[§3-x-Wdx==fx-sin®|—|dx=0.28"(fromproblemnote)[4]L0LAx=[(x®)-(x)=Jo.28%(0.5L?=0.17LAp>tom4m-AxLProb.2.11Calculatethefirstthreeenergylevelsfora1odquantumwellwithinfinitewalls.2,23210342I2SS(2253UaSRTe2:m-L8-9.11-10™-(107)E,=6.03-10%T=0.377eVE,=4-0.377eV=1.508eVE;=9-0.377¢V=3.393eVProb.2.12Showschematicofatomwith1s’25°2p"andatomicweight21.Commentonitsreactivity.nucleuswith...8protonsandThisatomischemicallyreactivebecause13neutronstheouter2pshellisnotfull.Itwilltend\2electronsin1stotrytoaddtwoelectronstothatouterwm,i.shell.’ix2electronsin2sTkhd4electronsin2p&=proton’@=neturon.e=electronI~StudyXY

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Chapter3SolutionsProb.3.1CalculatetheapproximatedonorbindingenergyforGads(€,=13.2,m*=0.067m,).FromEquation3-8andAppendixII,at..31,.."194E=—wd4>=_0.067:0.11:107ke)-(1.6-107°C)"O1110ke)1.6.10©>=8.34:107T=52meV8:(c,€,)*h’ 8-(8.85-10™E.13.2)%.(6.63-10™]s)Prob.3.2PlotFermifunctionforEr=1eVandshowtheprobabilityofanoccupiedstateAEaboveErisequaltotheprobabilityofanemptystateAEbelowEpsof(E,+AE)=1-f(E,-AE).usef(E)=—andkT=0.025%eV1+estE(eV]E-Er)/kTf(E10.75-9.65250.999940s0.903.8610|097939|-0.95-1.93050.87330_080.980.77220.68399eo1.000.00000.50000021.020.77220.3160021.051.93050.12669Aonpy1.103.86100.02061’wn:1.259.65250.00006occupationprobabilityaboveE;,=f(E.+AE)=ER1+ekTemptyprobabilitybelowE,=1-f(E,-AB)=1-—T+e-am1ek11L-fBAR)=1-—5=——5=5—=—3={E+E:T+eXT14eXM4]I+4eiTThisshowsthattheprobabilityofanoccupiedstateAEaboveEyisequaltotheprobabilityofanemptystateAEbelowEg.—StudyXY

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