HW06Solution

# HW06Solution - PROBLEM 4-4 B.D Cullity and S.R Stock...

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4-4 Derive simplified expressions for F 2 for diamond, including the rules governing observed reflections. This crystal is cubic and contains 8 carbon atoms per unit cell, located in the following positions: 0,0,0 ½ , ½ ,0 ½ ,0, ½ , 0, ½ , ½ ¼ , ¼ , ¼ ¾ , ¾ , ¼ ¾ , ¼ , ¾ ¼ , ¾ , ¾ Solution Begin by summing amplitudes and phases according to the structure factor equation (4-11). Note that because there are 8 atoms in the unit cell, the summation must contain 8 terms. Simplify, Now by trial-and-error determine those values of h , k , and l that make this expression zero. Recall that e n π i = e - n π i = (-1) n where n is any integer, and that 0 is an even integer. F 2 = 0 for mixed indices (even and odd), F 2 = 0 for ( h + k + l ) = odd multiple of 2. Finally, determine the rules governing observed reﬂections. F 2 = 64 f 2 for ( h+k+l ) = even multiple of 2, and F 2 = 32 f 2 for ( h+k+l ) = odd. F = f exp(2 π i [0 + 0 + 0]) + f exp(2 π i [ h/ 2 + k/ 2 + 0]) + f exp(2 π i [ h/ 2 + 0 + l/ 2]) + f exp(2 π i [0 + k/ 2 + l/ 2]) + f exp(2 π i [ h/ 4 + k/ 4 + l/ 4]) + f exp(2 π i [3 h/ 4 + 3 k/ 4 + l/ 4]) + f exp(2 π i [3 h/ 4 + k/ 4 + 3 l/ 4]) + f exp(2 π i [ h/ 4 + 3 k/ 4 + 3 l/ 4]) F = f (1 + exp( π i [ h + k ] + exp( π i [ h + l ] + exp( π i [ k + l ]) + f (exp( π i [ h + k + l 2 ]) + exp( π i [ 3 h + 3 k + l 2 ]) + exp( π i [ 3 h + k + 3 l 2 ]) + exp( π i [ h + 3 k + 3 l 2 ])) PROBLEM 4-4 B.D. Cullity and S.R. Stock, Elements of X-Ray Diffraction, 3 rd Ed., Prentice Hall, (2001) MSE 104 Materials Characterization Professor R. Gronsky page 1 of 1

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4-5 A certain tetragonal crystal has four atoms of the same kind per unit cell, located at 0, ½ , ¼ , ½ ,0, ¼ , ½ ,0, ¾ , and 0, ½ , ¾ , (Do not change axes). a) Derive simplified expressions for F 2 . b) What is the Bravais lattice of this crystal? c) What are the values of F2 for the 100, 002, 111, and 011 reflections? Solution ( a ) The structure factor is calculated by summing over all (four) atoms in the unit cell: Simplifying, When h and k mixed (even and odd), F = 0, so F 2 = 0. When h and k are unmixed, or when l is even, F 2 = 16 f 2 and when l is odd F 2 = 0. F = f e 2 π i [ k/ 2+ l/ 4] + f e 2 π i [ h/ 2+ l/ 4] + f e 2 π i [ h/ 2+3 l/ 4] + f e 2 π i [ k/ 2+3 l/ 4] F = f e π i [ k + l/ 2] + e π i [ h + l/ 2] + e π i [ h +3 l/ 2] + e π i [ k +3 l/ 2] F = f [ e - h π i + e - k π i ] e π i [ h + k + l/ 2] + e π i [ h + k +3 l/ 2] F 2 = [ ± 2] 2 f e π i [ h + k + l/ 2] + f e π i [ h + k +3 l/ 2] ± 2 F 2 = 4 f 2 f e π i [ h + k + l/ 2] + f e π i [ h + k +3 l/ 2] f e - π i [ h + k + l/ 2] + f e - π i [ h + k +3 l/ 2] F 2 = 4 f 2 (1 + 1 + [ e l π i + e - l π i ) PROBLEM 4-5 B.D. Cullity and S.R. Stock, Elements of X-Ray Diffraction, 3 rd Ed., Prentice Hall, (2001) MSE 104 Materials Characterization Professor R. Gronsky page 1 of 2
( b ) As the sketches below show, the eight atoms at fractional locations within the original unit cell can be re-assigned to a new parallelepiped (all unit cells in 3-D must be parallelepipeds) with atoms at the corners. Edge lengths in the base are identical, and the height of the cell is different; consequently, the Bravais lattice is simple tetragonal .

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HW06Solution - PROBLEM 4-4 B.D Cullity and S.R Stock...

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