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Unformatted text preview: 1 Test # 1 – Solution Problem # 1 The CsCl structure is simple cubic with two nonequivalent atoms in a unit cell which have coordinates 1 = r and 2 ˆ ˆ ˆ ( / 2)( ) = + + r x y z a . Denoting the atomic formfactors of the two atoms 1 f and 2 f and taking into account that the reciprocal lattice vector is 2 ˆ ˆ ˆ ( ) h k l a π = + + G x y z , we obtain for the structure factor [ ] 1 1 2 2 1 2 1 2 2 ˆ ˆ ˆ ˆ ˆ ˆ exp( ) exp( ) exp ( ) ( ) 2 exp ( ) a S f i f i f f i h k l a f f i h k l π π & ¡ = ⋅ + ⋅ = + + + ⋅ + + = ¢ £ ¤ ¥ = + + + G G r G r x y z x y z If 1 2 f f ≠ , for any h , k and l one can observe diffraction peaks provided that the diffraction condition is met. The latter is given by 2 2 G = ⋅ k G . It follows from this expression that 2 sin 2 2 G G kG kG k θ = = = k G & , where 2 k π λ = . Since for cubic lattice 2 2 2 2 2 G h k l d a π π = = + + , we find that 2 2 2 sin 2 h k l a λ θ = + + . The allowed values of h , k and l are therefore limited by the existence of the solution of this equation. Thus, we find that there are 12 planes (nonequivalent by symmetry of the lattice) and the associated scattering angles which are { } { } { } { } { } { } { } { } { } { } { } { } 100 16.57 110 23.79 111 29.60 200 34.78 210 39.62 211 44.31 220 53.77 221 58.82 222 81.08 300 58.82 310 64.40 311 71.06 ° ° ° ° ° ° ° ° ° ° ° ° If 1 2 f f = the diffraction peaks corresponding to odd values of ( h + k + l ) vanish, which is the case for the bcc lattice. Therefore, only scattering from the {110}, {200}, {220}, {222}, and {310} planes will be observed. 2 Problem # 2...
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 Spring '08
 WORMER

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