Determining hkl planes of a kikuchi pattern?

  • Thread starter Queequeg
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  • #1
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.
  • #2
To obtain the hkl planes of a Kikuchi pattern, you will need to use the Bragg's law equation. This equation states that the angle of reflection (2θ) from a crystalline material is related to the spacing between the atomic planes in the lattice (d) and the wavelength of the incident X-ray radiation (λ):2θ = (2π/λ)*dSince you have the angle of reflection and the lattice parameter for aluminum (a = 4.05 A), you can use this equation to calculate the d-spacing for each of the four bands. Once you have the d-spacings, you can then use the Miller-Bravais indices to obtain the hkl planes for your Kikuchi pattern.

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