20111ee123B_1_EE123B_HW2_Solutions

20111ee123B_1_EE123B_HW2_Solutions - CHAPTER 2 1. The...

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Unformatted text preview: CHAPTER 2 1. The crystal plane with Miller indices bid is a plane defined by the points als'h. 333k. and 33 / f . (a) Two vectors that he in the plane may be taken as mill 7 azf‘k and 31 /ll — fl3 / I." . But each of these vectors gives zero as its scalar product with G : ha] +ka2 + {a3 . so that G must be peipendicular to the plane 111d. (1)) If fl is the unit normal to the p1ane. the interplanar spacing is fl -al/ll. But 11 :Gf I G whence d(lll([) : G -a1/'11|Gl : 2TH I GI. (c) For a siinple cubic lattice G : (lit/3X11}? +k§' + (i). whence d2 7TZ 82 l 3a La 0 2 2 1 l 2. (a) Cell volume al~a7><a3:i— 3 —a 0 ' 2 2 0 0 c 1 a 2 :— 33C 2 i y" i 21 x3 4?: 1 1 b b’ZTE 2 3 * fia a 0 () 1 lalsazxafi fiazc 2 2 0 O c 27': 1 1/ \ 1 "2 (c) Six vectors in the reciprocal lattice are shown as solid lilies. The broken —,+ — — L— - -u 1ines are the perpendicular bisectors at the midpoints. The inscribed hexagon forms the first Bri11ouin Zone. 3. By definition of the pl‘lllllllYE reciprocal lattice rectors VBZ 70103012 Xa3)'(aa X30291 X32) 7 (27.03 I ‘ (81 .62 Xflg)‘ ‘1'? Hal'aZXaS) l : (my me. For the rector identity. see G. A. Korn and '1'. M. Korn. Mathematical handbook for scientists and engineers. McGraw—Hill. 1961. p. 147 4. (a) This follows by forming 2—1 _ 17 exp[7iM(a -Ak)] ‘ 1 iexp[iM(a - A10] 1: 2 I I l— exp[—i(a - A10] '1 — exp[i(a A10] 7 1— cos M(a Ak) 7 sin2 %M(a Ali) licos(a 2.1;) si112%(a-Ak) ' (b) The first zero in 5111 3 N18 occurs for E2 : 2mM. That this is the cou‘ect conSideration tollows tron) . l . l . l sn1M(rth+—E) :sni Terfli cos — M2+cos TENHI Sin — ME. 2 a—/ 2 iv? 2 zero. I — as M11 is an integer -2' : v - + - 5. S(vlv2'u3):f 2 e “Wm” z‘”) Referred to an fcc lattice. the basis of diamond is 000; . Thus in the product iii 444 S(V’1V2V’3) : S(fcc lattice)>< S (basis). we take the lattice structure factor from (48). and for the basis 1 *1 7 11(v1+V‘3+V'3j S (basis) :1+e 3 Now S(fcc) : 0 only if all indices are even or all indices are odd. If all indices are even the stnicttn‘e factor of the basis vanishes unless \‘1 + V2 * \‘3 = 411. Where n is an integer. For example. for the reflection (222) we have S(basis) : l * e'fl’I : O. and this reflection is forbidden. ...
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This note was uploaded on 02/20/2012 for the course EE 123B taught by Professor Dianahuffaker during the Winter '11 term at UCLA.

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20111ee123B_1_EE123B_HW2_Solutions - CHAPTER 2 1. The...

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