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HW3solution

# HW3solution - Louisiana State University Mechanical...

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Louisiana State University Mechanical Engineering Department Professor Meletis ME 2733 Fall 2004 Materials of Engineering Homework Assignment #3 Solution 1. First, we can calculate the d spacing for the six given peaks based on d hkl = λ /2.sin θ Peak: #1 #2 #3 #4 #5 #6 2 θ ( º ) 40.64 43.2 50.5 58.82 73.94 74.2 d hkl (Å) 2.22 2.094 1.807 1.570 1.282 1.278 BCC diffraction rule: h+k+l = even number, thus the first three planes in the BCC metal structure that will reflect are: 110, 200, 211 FCC diffraction rule: h, k, l = unmixed numbers (all odd or all even), thus the first three planes in the FCC metal structure that will reflect are: 111, 200, 220 Using the formula d hkl =a/ 2 2 2 l k h + + (where a is the lattice parameter) we can calculate the d spacing ratio with respect to the first reflecting plane that is expected for each one of the above two structures: For BCC: d 110 = α / 2 d 200 = α /2 d 211 = α / 6 and the ratios with respect to d 110 are: d 110 /d 200 =1.414, d 110 /d 211 =1.733 For FCC: d 111 =a/ 3 d 200 =a/2 d 220 =a/ 8 and the ratios with respect to d 111 are: d 111 /d 200 = 1.154, d 111 /d 220 = 1.633 Assuming that the first peak is BCC, check the d spacing ratios: d #1 /d #2 =2.22/2.094=1.06; d #1 /d #3 =2.22/1.807=1.23; d #1 /d #4 =1.414; d #1 /d #5 =2.22/1.282=1.733; d #1 /d #6 =2.22/1.278=1.737

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