固体物理1-7讲习题参考答案

3.1+−=++−=−+=)(2)(2)(2321kjiaakjiaakjiaaa)(2)(21111114212)(223321jibjiakjiaaaab+=+=−−⋅=×Ω=πππ)(2)(2132kjbaab+=×Ω=π)(2)(2213kibaab+=×Ω=πabπ4=aπ43.2(2π)3/ΩΩ)(321aaa×⋅=Ω*12312122()[(bbbbbaa)]πΩ=⋅×=⋅××ΩCBABCACBA⋅⋅−⋅⋅=××)()()(ijjiaaπδ2=⋅Ω=⋅Ω⋅⋅=Ω311*822πππab3.3Kh=h1b1+h2b2+h3b302122=−⋅hhKKkhK0)21(=⋅−hhKKk02122=−⋅hhKKk3.43.5Γ,H,N,P,∆,Σ,Λ,F100111110Γ(000)H(0120)∆(0140)N(14140)Σ(18180)P(141414)Λ(181818)F(183818)4.1Kh=h1b1+h2b2+h3b302122=−⋅hhKKk(h1h2h3)hK0)21(=⋅−hhKKk02122=−⋅hhKKkhK0K0KnKh=02Kdπ=02cos22=⋅−⋅dnπϕλπϕkhKϕπθ−=2λθnd=sin2Bragg4.2KCl,K+Cl-KClKClfff+−==4K4ClK()0,0,0)0,21,21()21,0,21()21,21,0(Cl)0,0,21([K]coord.8()()()1(1)[1]hjiKriHiHKiHLiKLHKLjjFfefeeeeππππ−⋅−−+−+−===++++∑+20HKLHKLIF==iHiiHK,,HLKL+++,,HKL,,HKLKClff+=−)X(111)()HKL′′′()(222HKLHKL′′′=(222)HKL′′′,,HKL4.3X(HKL)G=H2+K2+L21,2,3,4,5,6,8,9,10,11,12,…2,4,6,8,10,12,…3,4,8,11,12,…3,8,11,…10,,HKLHKL(001)(011)(111)(002)(012)(112)(222)(003)(122)(013)(113)G=H2+K2+L21234568910112()[1]iHKLHKLFfeπ−++=+H+K+L=G=1,3,5,92,4,6,8BCCG∴=3()()()[1]iHKiHLiKLHKLFfeeeπππ−+−+−+=+++H,K,L3,4,8,11FCCG=4()2[1]iHKLHKLFfeπ−++=+⋅[]H,K,L(H+K+L)/2iH,K,LiiH,K,L44.4a310()()exp(2/)nraraπ−=−00222016/(4)hKhfKa=+10!nxnnxedxαα∞−+=∫20sin4()hhKhKrfnrrdrKrπ∞=∫13100()()exp(2/)nraraπ−=−1030003000030004exp(2/)sin()41exp(2/)[exp()exp()]22[exp(2/)exp(2/)]hKhhhhhhhhfrraKrdrKarraiKriKrdrKairiKrradrriKrradriKaππ∞∞∞∞=−=−−−=−−−−∫∫∫∫10!nxnnxedxαα∞−+=∫2230002220222[()()]16(4)hKhhhfiKiKaaaKa−−=−−+=+hiK5.12/n;n=1,2,3,4,6la1122la+ABB’B’ABBAA’A’''BAnAB=n''(12cos)BAABθ=−12cosnθ−=cos:11θ−+∼1,0,1,2,3n=−0,60,90,120,180ooooθ=o164321.11122000000εεε1xxxyxzyxyyyzzxzyzzεεεεεεεεεxxxxyyxzzxDDEDεεε=x180oxxxxyyyxzzzxDDDDDDεεε′′′=−=−−−Ex180DD′=⇒0yxzxεε==yz18000xyzyxzyzεεεε====zE60o13ˆˆ22yzEEeE=+ey00000xxyyzzεε0ε01232xyyyzzzDDEDεε=180001313224431332244xyyzyyzzyzyyzzDDDDDDDεεεε′′′=−+=−+++zEDD′=yyzzεε=122000000εεε2T1'TTεε−=εε′=x180T100010001=−−()ε=0000xxyyyzzyzzεεεεεy180T100010001−=−()ε=000000xxyyzzεεεz120T1000cos60sin600sin60cos60=−yyzzεε=2.2NaClNaCl2()nAtBUNrr′=−+00rUr∂=∂210nrq−∝200qUr∝2qq→1104nr−−′=0r0ar∝1004nnWUW−′′=−=2.3()mnUrrrαβ=−+1r02W34m=2,n=10,r0=3A,W=4eVα,β10110mnrUmnrrrαβ++∂=−=∂10nmnrmβα−=2U(r)1()()2totalUrNUr=0001()/()(1)22totalmmWUrNUrrnα=−=−=−3Nc3VcNr=22()totaldUKVdV=00211()18mnrddrcdrrdrrrαβ=−+0220118mnrdcrdrrrαβ−+=0001(1)(1)18mnmmnncrrrαβ++=−+30()18mmnmcrα+−=4290α=eV⋅A2,eV51.1810β=×⋅A102.4sp3zxyabczxyzxyzxyabc,,abca()2aaxy=+−z()2abxy=−++z()2acxy=−+z3/2abca===,,abacbcθθθ11cos()()33ababxyzxyzabθ==+−−++=−ii11cos()()33acacxyzxyzacθ==+−−+=−ii11cos()()33bcbcxyzxyzbcθ==−++−+=−ii10928'abacbcθθθ===6.1mzdγα=∑(n1n2n3)(001)(011)(111)(002)(012)(112)(022)(122)(222)m61286242412248z+-+-+--+-γ1111/21/21/21/41/41/8d12345689121.752α=7.1r6r1rEP1P1/r32P2=αE(α)E=P1P2/r3=αP12/r62.6Lennard-JonesNeLennard-Jones1261261(4)[]2UNAArrσσε=−N00rUr∂=∂1612062ArAσ=260121()(4)24AWUrNAε=−=0.0031eVε=1212.13FCCA=2.74Aσ=614.45FCCA=129.11BCCA=612.25BCCA=0.957BCCFCCWW=2.7H2501016erg2.96Å0.751kJ/molH2fccLennard-Jones1261261()(4)[()()2UrNAArr]σσε=−A12=12.13A6=14.45=5010-16erg=2.96AN=6.0221023/mol00126(())()2[12.13()14.45()]rrdUrdNdrdRrrσσε=−03.16orA=r0U23161260102.962.96()26.0210/5010[12.13()14.45()]3.163.162.5510/2.55/UrmolergergmolKJmol−=×××××−≈−×≈−H22.55KJ/mol0.751KJ/molH2