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

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.

The eight body-centred cubic neighbour vectors are $(a/\sqrt{3})(\pm 1,\pm 1,\pm 1)$ with independent signs. Their diagonal sums are $8a^2/3$ and the sign-paired off-diagonal sums cancel. HenceThe conventional cubic cell edge is $2a/\sqrt{3}$ in this comparison; using that edge as the hopping distance would spoil the stipulated equal-distance comparison.

The twelve face-centred cubic neighbours are $(a/\sqrt{2})(\pm 1,\pm 1,0)$ and permutations of the zero coordinate. Eight vectors have a nonzero given component, each contributing $a^2/2$, while off-diagonal terms cancel between signs. ThusFor a cubic neighbour shell this can also be checked from ${\sum }_j{r}_{j,\alpha }{r}_{j,\beta }=(z{a}^2/3){\delta }_{\alpha \beta }$ with coordination number $z=12$.