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

Set $F={\omega }_0{e}^{-\Delta E/(k_{B}T)}$ and $K=n{Q}^{2}F/(2k_{B}T)$. The six simple cubic neighbour vectors are $\pm a\hat{x}$, $\pm a\hat{y}$, $\pm a\hat{z}$. Each diagonal direction receives $2a^2$, and every off-diagonal product vanishes. ThereforeThe result uses $a$ as the nearest-neighbour distance specified here, not automatically as the conventional-cell edge for the other lattices.