Determining hkl planes of a kikuchi pattern?

  • Thread starter Queequeg
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I am struggling to obtain the hkl planes of a kikuchi pattern. The material is Aluminum with a = 4.05 A. There are 4 bands, the first, b1, is pointing vertically down, the second, b1, is at the 34.5 degree angle from b1, the third, b3, is 65 degrees from b1, and the fourth, b4 is 90 degrees from b1. the image is in reciprocal lattice.

The widths are:
b1 = 9 mm
b2 = 5 mm
b3 = 11 mm
b4 = 6.5 mm

My work so far is that the material is FCC, so the allowed HKL planes are all odd and all even. I reasoned that the ratio of the widths are equal to the ratio of the hkl planes magnitude, or w1/w2 = (h1^2+k1^2+l1^2)^.5 / (h2^2+k2^2+l2^2)^.5. When I tried taking the ratios, I couldn't match any of the h^2+k^2+l^2 ratio values to the width ratios for FCC, however, and am unsure of where to go from there.
 

Answers and Replies

  • #2
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Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 

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