Past exam of the condensed matter physics course of the University of Oxford / 2013 / 2 / a / Solution

Put $C=\alpha q^2/(4\pi {ϵ}_0)$ with $\alpha =2\ln 2$. Relative to a chosen positive ion, successive neighbours have alternating charge. The sum of its interactions with all other ions is $-C/R$. Summing this over all $2N$ ions would count every pair twice, so the total electrostatic energy is $-NC/R$, or $-C/R$ per ion pair.

There are $2N$ nearest-neighbour bonds in a periodic chain of $N$ pairs; boundary corrections are negligible for large $N$. If $A/R^n$ is the energy of each such bond, the repulsive contribution per ion pair is $2A/R^n$. Therefore the alternating-charge chain cohesive energy isThe factor two in the short-range term counts two bonds per ion pair. If a repulsion parameter were instead defined as the whole pair's repulsive coefficient, it would equal $2A$ in this convention; changing that definition must not silently change the equilibrium formula. The absolute zero of energy is separated ions for a genuinely decaying repulsion.