p.259, 8.2110Statistically, the probability for a Cuatom to appear at a lattice point is385pmequal to the percentage (x) of Cudoping. Thus the AuCu alloybelongstoface-centered cubic latticebsystem.Each unit cell contains4lattice points and each of them is aa=385pmstatistic atom (Aui-xCu)AuCu2)The ordered phase belongs to simple tetragonal lattice; each unitcell/lattice point contains AuCu. Au(0,0,0), Cu (1/2,1/2,1/2)3)Cell parameterfortheorderedphase:a' = b' = a/N2 = 272.23 pm,c =a=385 pmThe first observed diffraction is (O01)do01= 1/(h/a')2+(k/a)2+(V/c)2)1/2 = c' = aAccording to Bragg's Law, we have→0=11.5°sin0oo, = 2/2do01 = /2a= 0.2
p.259, 8.21 1) Statistically, the probability for a Cu atom to appear at a lattice point is equal to the percentage (x) of Cu doping. Thus the AuCu alloy belongs to face-centered cubic lattice system. Each unit cell contains 4 lattice points and each of them is a statistic atom (Au1-xCux ). 2) The ordered phase belongs to simple tetragonal lattice; each unit cell/lattice point contains AuCu. Au(0,0,0), Cu (1/2,1/2,1/2); 3) Cell parameter for the ordered phase: a = b = a/2 = 272.23 pm, c = a = 385 pm The first observed diffraction is (001) d001= 1/(h/a) 2+(k/a) 2+(l/c) 2 ) 1/2 = c = a According to Bragg’s Law, we have sin001= /2d001 = /2a= 0.2 = 11.5
p.259, 8.21The randomly doping phase belongsto fcc lattice. Hence its first observablediffraction is (111)385pmdi1i= a/(h2+k2+P)1/2 = a/31/2AccordingtoBragg'sLawDsinQ11 = N2dul385pm= 31/2 2/2a = 0.3464385pm0 = 22.3 °CuAuNote:insomecases,random/multipleorientations/rotationsof structural units(clustersmolecules,orgroups)exist withinasinglecrystal,whichposes difficulty in the structural determination based on Xray diffraction data!
p.259, 8.21 The randomly doping phase belongs to fcc lattice. Hence its first observable diffraction is (111). d111= a/(h2+k2+l2 ) 1/2 = a/31/2 According to Bragg’s Law, sin111 = /2d111 = 31/2/2a = 0.3464 = 22.3 Note: in some cases, random/multiple orientations/rotations of structural units (clusters, molecules, or groups) exist within a single crystal, which poses difficulty in the structural determination based on Xray diffraction data!
Example银为立方晶系，用CuK_射线(2=154.18 pm)作粉末衍射，在hkl类型衍射中，hk奇偶混合的系统消光。衍射线经指标化后，选取333衍射线，Q=78.64，试计算晶胞参数。已知Ag的密度为10.507g?cm-3，相对原子质量为107.87，问晶胞中有几个Ag原子：试写出Ag原子的分数坐标。Answer: when h,k,l are neither all odd nor all even, system absence!→ Cubic F-centredBragg law: 2dhk sin0= dhk N2sin0cell parameter:a=dh×h?+k?+1?=vh?+k?+P/2sino=154.18×/27/2sin(78.64)=408.58pmnAg=NVp/MAg =Noa'p/MAgatoms in a unit cell:=6.022×1023×（408.58×10-10)3×10.507/107.87=4Thus each atom corresponds to one lattice point, atomic(0,0,0) (1/2,1/2,0) (1/2,0,1/2) (0,1/2,1/2)coordinates:
Example: • 银为立方晶系，用CuK射线(=154.18 pm)作粉末衍射，在hkl 类型衍射中，hkl奇偶混合的系统消光。衍射线经指标化后， 选取333 衍射线， = 78.64º ，试计算晶胞参数。已知Ag 的密 度为10.507 g·cm-3，相对原子质量为107.87，问晶胞中有几个 Ag 原子; 试写出Ag 原子的分数坐标。 Answer: when h,k,l are neither all odd nor all even, system absence! Cubic F-centred. Bragg law: 2dhkl sin = dhkl = /2sin cell parameter: atoms in a unit cell: Thus each atom corresponds to one lattice point, atomic coordinates: (0,0,0) (1/2,1/2,0) (1/2,0,1/2) (0,1/2,1/2) . / sin( . ) . pm a dhkl h k l h k l / sin 154 18 27 2 78 64 408 58 2 2 2 2 2 2 2 λ θ 6.022 10 (408.58 10 ) 10.507 /107.87 4 / ~ / ~ 2 3 1 0 3 3 0 0 nAg N V MAg N a MAg
Example金属Mg是由Mg原子按A3型堆积而成，已知Mg的原子半径是160pm，求晶胞参数。Answer:Hexagonal close-packingmode→ a=b =2R=320 pm2R2RThe height of a tetrahedron with an edgelengthof2Rish=2R×V6/3C=2h=4R×/6/3h= 522.6 pm
Example: • 金属Mg是由Mg原子按A3型堆积而成，已知Mg的原子半径是 160 pm，求晶胞参数。 Answer: Hexagonal close-packing mode a = b = 2R = 320 pm, The height of a tetrahedron with an edge length of 2R is h 2R 6 / 3 pm c h R 522.6 2 4 6 / 3 2R 2R h
Example: p.286, 9.9KF crystal-cubic system,MoKα2=70.8 pm, diffraction sin?e:0.0132,0.0256,0.0391,0.0514,0.0644,0.0769,0.102,0.115,0.127, 0.139, ....., 1) plz derive its lattice type and cellparameter; 2)Suppose the F-adopts the simple cubicpackingwith cubic interstices being occupied by K+; Rk =133 pm andR,=136 pm.plz derive the cell parameter.Answer:1) indexing: sin?0 = 1:2:3:4:5:6:8:9:10:11... →> simple cubicBragg law: 2dhk sin@= →dhk= N2sinea=dh×Vh?+k?+1=2/h?+k?+1?/2sin0a/3a70.8×11/(2/0.139)=314pmOr.2/4a?=sin?0/(h?+k?+12）取四条高指标衍射线求该式的平均值得22/4a=0.012716=a=313.9pm
Example: p.286, 9.9 • KF crystal– cubic system, Mo K =70.8 pm, diffraction sin2: 0.0132, 0.0256, 0.0391, 0.0514, 0.0644, 0.0769, 0.102, 0.115, 0.127, 0.139, ., 1) plz derive its lattice type and cell parameter; 2) Suppose the F- adopts the simple cubic packing with cubic interstices being occupied by K+ ; RK =133 pm and RF = 136 pm. plz derive the cell parameter. Answer: 1) indexing: sin2 = 1:2:3:4:5:6:8:9:10:11. simple cubic Bragg law: 2dhkl sin = dhkl = /2sin Or 取四条高指标衍射线求该式的 平均值得 pm a d h k l h k l hkl 70.8 11 /(2 0.139) 314 / 2sin 2 2 2 2 2 2 3a 2a / 4 sin /( ) 2 2 2 2 2 2 a h k l / 4a 0.012716 a 313.9pm 2 2