 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-3/section-d/43/b/iii/1/solution

F = ω_{0} e^{- Δ E / (k_{B} T)}

and

K = n Q^{2} F / (2 k_{B} T)

. The six simple cubic neighbour vectors are

\pm a \overset{x}{^}

\pm a \overset{y}{^}

\pm a \overset{z}{^}

. Each diagonal direction receives

2 a^{2}

, and every off-diagonal product vanishes. Therefore

σ_{sc} = 2 K a^{2} I = \frac{n Q ^{2} F a ^{2}}{k _{B} T} I .

The result uses

a

as the nearest-neighbour distance specified here, not automatically as the conventional-cell edge for the other lattices.

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