 Solution

For a single impurity level in a crystal with allowed occupations zero and one and equal occupied/empty statistical weights, the Fermi-Dirac distribution givesSince $n_{\mathrm{full}}=N_D{f}_D$ and $n_{\mathrm{empty}}=N_D(1-f_D)$,The same result follows from the two grand canonical ensemble weights $1$ and $e^{-\beta ({ϵ}_D-\mu )}$.

The equal-statistical-weight convention is necessary for the printed formula when $n_{\mathrm{full}}$ and $n_{\mathrm{empty}}$ count whole donor impurities. More generally, donor charge-state degeneracy gives $n_{\mathrm{full}}/n_{\mathrm{empty}}=(g_{\mathrm{full}}/g_{\mathrm{empty}})e^{\beta (\mu -{ϵ}_D)}$. An ordinary singly occupied donor with two spin orientations and a nondegenerate empty state has ratio $2e^{\beta (\mu -{ϵ}_D)}$; at $\mu ={ϵ}_D$ its occupied-to-empty ratio is two, a concrete counterexample to treating the printed expression as universal. The following part uses the paper's ratio-one convention.