材料科学与工程基础习题评讲-1

习题讲解第一次作业英文2.6Allowedvaluesforthequantumnumbersofelectronsareasfollows:TherelationshipsbetweennandtheshelldesignationsarenotedinTable2.1.Relativetothesubshells,l＝0correspondstoanssubshelll＝1correspondstoapsubshelll＝2correspondstoadsubshelll＝3correspondstoanfsubshellFortheKshell,thefourquantumnumbersforeachofthetwoelectronsinthe1sstate,intheorderofnlmlms,are100(1/2)and100(-1/2).WritethefourquantumnumbersforalloftheelectronsintheLandMshells,andnotewhichcorrespondtothes,p,anddsubshellsK:s:100(1/2);100(-1/2)L:s:200(1/2);200(-1/2)p:210(1/2);210(-1/2);21-1(1/2);21-1(-1/2);211(1/2);211(-1/2)M:s:300(1/2);300(-1/2)p:310(1/2);310(-1/2);31-1(1/2);31-1(-1/2);311(1/2);311(-1/2)d:320(1/2);320(-1/2);32-1(1/2);32-1(-1/2);321(1/2);321(-1/2);32-2(1/2);32-2(-1/2);322(1/2);322(-1/2)2.7Givetheelectronconfigurationsforthefollowingions:Fe2+,Fe3+,Cu+,Ba2+,Br-,andS2-.SOLUTIONFe2+:1s22s22p63s23p63d6Fe3+:1s22s22p63s23p63d5Cu+:1s22s22p63s23p63d10Ba2+:1s22s22p63s23p63d104s24p64d105s25p6Br-:1s22s22p63s23p63d104s24p6S2-:1s22s22p63s23p62.17(a)Brieflycitethemaindifferencesbetweenionic,covalent,andmetallicbonding.(b)StatethePauliexclusionprinciple.SOLUTION(a)离子键：无方向性球形正、负离子堆垛取决电荷数——电荷平衡体积（离子半径）金属键：无方向性球形正离子较紧密堆垛共价键：有方向性、饱和性，电子云最大重叠(b)原子中的每个电子不可能有完全相同的四个量子数（或运动状态）2.19Computethepercentsioniccharacteroftheinteratomicbondsforthefollowingcompounds:TiO2,ZnTe,CsCl,InSb,andMgCl2.SOLUTION由公式：已知：TiO2,XTi=1.5andXO=3.5ZnTe,已知：XZn=1.6andXTe=2.1，故，%IC=6.05%CsCl,已知：XCs=0.7andXCl=3.0，故：%IC=73.4%InSb,已知：XIn=1.7andXSb=1.9，故：%IC=1.0%MgCl2,已知：XMg=1.2andXCl=3.0故：%IC=55.5%2.24，Onthebasisofthehydrogenbond,explaintheanomalousbehaviorofwaterwhenitfreezes.Thatis,whyistherevolumeexpansionuponsolidification?水冻结时结晶，非球形的水分子规整排列时受氢键方向性和饱和性的更强限制，不能更紧密地堆积，故密度变小，体积增大。2-7影响离子化合物和共价化合物配位数的因素有那些？离子化合物：体积电荷共价化合物：价电子数电子云最大重叠第二次作业2.18Offeranexplanationastowhycovalentlybondedmaterialsaregenerallylessdensethanionicallyormetallicallybondedones.共价键需按键长、键角要求堆垛，相对离子键和金属键较疏松2.21UsingTable2.2,determinethenumberofcovalentbondsthatarepossibleforatomsofthefollowingelements:germanium,phosphorus,selenium,andchlorine.SOLUTIONGe:4P:3Se:2Cl:12-6按照杂化轨道理论，说明下列的键合形式：（1）CO2的分子键合Csp杂化（2）甲烷CH4的分子键合Csp3杂化（3）乙烯C2H4的分子键合Csp2杂化（4）水H2O的分子键合Osp3杂化（5）苯环的分子键合Csp2杂化（6）羰基中C、O间的原子键合Csp2杂化2-10当CN=6时，K+离子的半径为0.133nm(a)当CN=4时，对应负离子半径是多少？(b)当CN=8时，对应负离子半径是多少？若(按K+半径不变)求负离子半径,则：CN=6R=r/0.414=0.133/0.414=0.321nmCN=4R=r/0.225=0.133/0.225=0.591nmCN=8R=r+/0.732=0.133/0.732=0.182nm第三次作业3.48Drawanorthorhombicunitcell,andwithinthatcella[121]directionanda(210)plane.3.50Hereareunitcellsfortwohypotheticalmetals:a.Whataretheindicesforthedirectionsindicatedbythetwovectorsinsketch(a)?bWhataretheindicesforthetwoplanesdrawninsketch(b)?（a）direction1,xyzProjections0ab/2cProjectionsintermsofa,b,andc01/21Reductiontointegers012Enclosure[012]direction2,xyzProjectionsa/2b/2-cProjectionsintermsofa,b,andc1/21/2-1Reductiontointegers11-2Enclosure[112]（b）Plane1，∞：1/2：∞；0：2：0；(020)Plane2，1/2：-1/2：1；2：-2：1；(221)3.51*Withinacubicunitcell,sketchthefollowingdirections:ab（c）[012]（d）[133]（e）[111]（f）[122]（g）[123]（h）[103]3.53Determinetheindicesforthedirectionsshowninthefollowingcubicunitcell:DirectionA:xyz-2/3ab/20c-2/31/20-430[430]DirectionA:DirectionB:xyzxyz-2/3ab/20c2/3a-b2/3c-2/31/202/3-12/3-4302-32[430][232]4DirectionCDirectionDxyzxyz1/3a-b-ca/6b/2-c1/3-1-11/61/2-11-3-313-6[133][136]3.57DeterminetheMillerindicesfortheplanesshowninthefollowingunitcell:planeAxyza/3b/2-c/21/31/2-1/23/12/1-2/1(322)planeB(101)3.58DeterminetheMillerindicesfortheplanesshowninthefollowingunitcell:planeA以（0，1，0）为新原点xyz2/3a-bc/22/3-11/23/2-1/12/13/2-2/24/2(324)planeB(221)3.61*Sketchwithinacubicunitcellthefollowingplanes:a3.61*Sketchwithinacubicunitcellthefollowingplanes:abcdefgh3.62Sketchtheatomicpackingof(a)the(100)planefortheFCCcrystalstructure,and(b)the(111)planefortheBCCcrystalstructure(similartoFigures3.24band3.25b).(a)FCC:(100)plane(b)BCC:(111)plane3.81ThemetaliridiumhasanFCCcrystalstructure.Iftheangleofdiffractionforthe(220)setofplanesoccursat69.22°(first-orderreflection)whenmonochromaticx-radiationhavingawavelengthof0.1542nmisused,compute(a)theinterplanarspacingforthissetofplanes,and(b)theatomicradiusforaniridiumatom.SOLUTION：已知：铱FCC的（220）晶面,2θ=69.22°;λ=0.1542nm；n=1由公式：nλ=2dsinθ和dhkl=a/()故解：（a）d220=0.1542/(2sin34.61°)=0.1357（nm）（b）∵a=d220×=2d220；又FCC的2R=a/=2d220∴R=d220=0.1357（nm）即：其（220）晶面间距为0.1357nm；铱原子半径也为0.1357nm。222lkh22202222222-14计算（a）面心立方金属的原子致密度；（b）面心立方化合物NaCl的离子致密度（离子半径r(Na+)=0.097，r(Cl-)=0.181）；（C）由计算结果，可以引出什么结论？解：（a）∵面心立方金属∴一个晶胞中有四个金属原子，且a=4R/√2有：PF=4(4πR3/3)/a3=16πR3(2√2)/(3)(64R3)=0.74答：面心立方金属的原子致密度为0.74（b）解：∵面心立方化合物NaCl,且:r(Na+)=0.097，r(Cl-)=0.181∴一个晶胞中有四个Na离子、四个Cl离子有：PF=[4(4πr3/3)+4(4πR3/3)]/(2r+2R)3=16π(0.0973+0.1813)/3(8)(0.097+0.181)3=0.64答：面心立方化合物NaCl的离子致密度为0.64（c）结论:(1)同种原子晶体的致密度只与晶胞类型相关,与原子尺寸无关(2)化合物晶体的离子致密度与离子大小相关2-18在体心立方结构晶胞的（100）面上按比例画出该面上的原子以及八面体和四面体间隙。八面体四面体2-24方向为[111]的直线通过1/2,0,1/2点,则在此直线上的另外两点的坐标是什么?解：∵该直线过为:u=1/2v=0w=1/2u=0:v=-1/2,w=0.即过(0,-1/2,0)点u=1:v=1/2,w=1.即过(1,1/2,1)点(0,-½,0)(1,½,1)(½,0,½)2-27一平面与三轴的截距为a=1,b=-2/3,c=2