= Solution
{c}
For a single <impurity level in a crystal> with allowed occupations zero and one and equal occupied/empty statistical weights, the <Fermi-Dirac distribution> gives
$$
f_D=\frac1{e^{\beta(\epsilon_D-\mu)}+1}.
$$
Since $n_{\rm full}=N_D f_D$ and $n_{\rm empty}=N_D(1-f_D)$,
$$
\boxed{\frac{n_{\rm full}}{n_{\rm empty}}=\frac{f_D}{1-f_D}
=e^{\beta(\mu-\epsilon_D)}.}
$$
The same result follows from the two <grand canonical ensemble> weights $1$ and $e^{-\beta(\epsilon_D-\mu)}$.
The equal-statistical-weight convention is necessary for the printed formula when $n_{\rm full}$ and $n_{\rm empty}$ count whole <donor impurities>. More generally, <donor charge-state degeneracy> gives $n_{\rm full}/n_{\rm empty}=(g_{\rm full}/g_{\rm empty})e^{\beta(\mu-\epsilon_D)}$. An ordinary singly occupied donor with two <spin> orientations and a nondegenerate empty state has ratio $2e^{\beta(\mu-\epsilon_D)}$; at $\mu=\epsilon_D$ 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.
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