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Test01_solution

# Test01_solution - 1 Test 1 – Solution Problem 1 The CsCl...

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Unformatted text preview: 1 Test # 1 – Solution Problem # 1 The CsCl structure is simple cubic with two non-equivalent atoms in a unit cell which have coordinates 1 = r and 2 ˆ ˆ ˆ ( / 2)( ) = + + r x y z a . Denoting the atomic form-factors 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 (non-equivalent 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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Test01_solution - 1 Test 1 – Solution Problem 1 The CsCl...

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