 Solution

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

G_{hk l} = (2 π / a) h^{2} + k^{2} + l^{2}

. Combining this with Bragg's law gives the Bragg angle

sin θ_{hk l} = \frac{λ}{2 a} h^{2} + k^{2} + l^{2} .

Using

a = 0.375 nm

and

λ = 0.154 nm

yields

2 θ_{110} = 33.76 2^{\circ} ≃ 33. 8^{\circ}, 2 θ_{111} = 41.66 6^{\circ} ≃ 41. 7^{\circ} .

These are the full scattering angles; the Bragg angles themselves are half as large.

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