 Solution

ID: past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2022/1/a/iii/solution

In elastic scattering, the incoming and outgoing wave vectors have equal magnitude

k_{0} = 2 π / λ

. Their mutual angle is

2 θ_{hk l}

q = k_{f} - k_{i}

therefore has magnitude

q^{2} = 2 k_{0}^{2} [1 - cos (2 θ_{hk l})] = 4 k_{0}^{2} sin^{2} θ_{hk l} .

q = G_{hk l}

G_{hk l} = \frac{4 π}{λ} sin θ_{hk l} .

Equivalently,

G_{hk l} = 2 π / d_{hk l}

turns this relation into Bragg's law,

2 d_{hk l} sin θ_{hk l} = λ

, with the reflection order incorporated into the Miller indices.

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