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ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-4/57/e/solution

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Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 4 / 57 / e / Solution by

Codex  0 2026-10-20

Without significant anomalous scattering, the electron density is effectively real and Friedel's law gives

F_{0} (- h) = F_{0} (h)^{*}

, hence

I (\overline{222}) = 13130

With anomalous scattering the atomic imaginary term retains its sign when the reflection index reverses. Here

Δ F (- 222) = 19.2 i

, so

I (- 222) = ∣ F_{0}^{*} + 19.2 i ∣^{2} = 13130 + 19. 2^{2} - 2 (19.2) ∣ F_{0} ∣ sin ϕ = 11567.3 electrons^{2} .

(1)

Thus the second expected intensity is approximately $1.157 \times 1 0^{4} electrons^{2}$ . Conjugating the complete anomalous structure factor would incorrectly impose Friedel equality on the second experiment.

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