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1、Chapteronehomework1.(P803-3)Calculatetheatomicradiusincmforthefollowing:(a)BCCmetalwitha0=0.3294nmandoneatomperlatticepoint;and(b)FCCmetalwitha0=4.0862Åandoneatomperlatticepoint.Solution:(a)InBCCstructures,atomstouchalongthebodydiagonal,whichisa0inlength.Therea
2、retwoatomicradiifromthecenteratomandoneatomicradiusfromeachofthecorneratomsonthebodydiagonal,so:=0.14263nm=1.4263cm(b)InFCCstructures,atomstouchalongthefacediagonalofthecube,whichisinlength.Therearefouratomicradiialongthislength—tworadiifromtheface-centeredatom
3、andoneradiusfromeachcorner,so,=1.44447Å=1.44447cm2.(P803-4)determinethecrystalstructureforthefollowing:(a)ametalwitha0=4.9489Å,r=1.75Å,andoneatomperlatticepoint;and(b)ametalwitha0=0.42906nm,r=0.1858nm,andoneatomperlatticepoint.Solution:Weknowtherelationshipsbet
4、weenatomicradiiandlatticeparametersareinBCCandinFCC.(a)1.75soitscrystalstructureisFCC;(b)==0.186nmsoitscrystalstructureisBCC.1.(P803-5)thedensityofpotassium,whichhastheBCCstructureandoneatomperlatticepoint,is0.855g/cm3.theatomicweightofpotassiumis39.09g/mol.Cal
5、culate(a)thelatticeparameter;and(b)theatomicradiusofpotassium.Solution(a)ForaBCCunitcell,therearetwoatomsinperunitcell,atomicmassis39.09g/mol,densityρ=0.855g/cm3Avogadro’snumberNA=6.02atoms/mol0.855g/cm3=Soa==0.53=5.3Å(b)thenr==0.229cm=2.29Å1.(P813-20)determine
6、theindicesforthedirectionsinthecubicunitcellshowninFigure3-32.TheprocedureforfindingtheMillerindicesfordirectionsisasfollows:1.Usingaright-handedcoordinatesystem,determinethecoordinatesoftwopoints,whichlieonthedirection.2.Subtractthecoordinatesofthe“tail”pointf
7、romthecoordinatesofthe“head”pointtoobtainthenumberoflatticeparameterstraveledinthedirectionofeachaxisofthecoordinatesystem.3.Clearfractionsand/orreducetheresultsobtainedfromthesubtractiontolowestintegers.4.Enclosethenumberinsquarebrackets[].Ifanegativesignispro
8、duced,representthenegativesignwithabaroverthenumber.SolutionDirectionA1.Twopointsare0,0,1and1,0,02.0,0,1-1,0,0=-1,0,13.nofractiontoclearorintegerstoreduce4.DirectionB1.Twopo
