Structure.homework__6-answer

Structure.homework__6-answer - 000 0 0 0 F= f c e 2 i(0) +f...

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MTLE-2100 Structure of Engineering Materials, Homework #6 --Answer 1. For Aluminum (fcc) compare the x-ray diffraction intensities for the following planes when Cu K α radiation ( λ = 1.54A) was used. Consider only the square of the structure factor, | F | 2 . Give the relative intensity setting the highest intensity 100. 1/d 2 = (h 2 +k 2 +l 2 )/a 2 . a = 4.0497 Å . Use the atomic scattering factor shown in Appendix 10 of Cullity-Stock and appropriate interpolation. hkl mixed integers F 2 =0 unmixed integers F 2 =16f 2 F 2 h 2 +k 2 +l 2 (1/2d)=sin θ / λ f F 2 I/I max (100) 0 0 0 (110) 0 0 0 (111) 16f 2 3 0.2138( Å -1 ) 8.80 1239 100 (200) 16f 2 4 0.2468 8.35 1116 90 (210) 0 0 0 (220) 16f 2 8 0.3492 7.20 829 67 (222) 16f 2 12 0.4277 6.30 635 51 (300) 0 0 0 (310) 0 0 0 (311) 16f 2 11 0.4095 6.5 676 55 (321) 0 0 0 (331) 16f 2 19 0.5382 5.1 416 34 (333) 16f 2 27 0.6415 4.2 282 23 (400) 16f 2 16 0.4939 5.5 484 39
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2. Derive simplified expressions for F 2 for diamond, including the rules governing observed reflections. The crystal is cubic and contains 8 carbon atoms per unit cell, located in the following positions:
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Unformatted text preview: 000 0 0 0 F= f c e 2 i(0) +f c e 2 i(h/2+k/2) +f c e 2 i(h/2+l/2) +f c e 2 i(k/2+l/2) +f c e 2 i(h/4+k/4+l/4) +f c e 2 i(3h/4+3k/4+l/4) +f c e 2 i(3h/4+k/4+3l/4) + f c e 2 i(h/4+3k/4+3l/4) = f c [1 + e i(h+k) + e i(h+l) + e i(k+l) ] +f c e i(h/2+k/2+l/2) [1+ e i(h+k) + e i(h+l) + e i(k+l) ] = f c [1 +e i(h+k) +e i(h+l) +e i(k+l) ][1+e ( i/2)(h+k+l) ] When h,k,l are mixed, F=0 When h,k,l are unmixed F = 4f c [1+e ( i/2)(h+k+l) ] F 2 =16f c 2 [1+e ( i/2)(h+k+l) ] [1+e (- i/2)(h+k+l) ] =16f c 2 [2+2cos( /2)(h+k+l)] When h,k,l are all odd F 2 =32f c 2 When h,k,l are all even For h+k+l =4n F 2 =64f c 2 For h+k+l=4n +2 F 2 =0...
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Structure.homework__6-answer - 000 0 0 0 F= f c e 2 i(0) +f...

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