 Solution

Retain the inferred mobilities and assume one ionized donor per added arsenic atom, negligible compensation, the same band gap, and the nondegenerate semiconductor mass-action law. Relative to the intrinsic sample, a conductivity factor $f$ requiresFor complete donor ionization, charge neutrality gives $N_D=n_e-n_h$, soThe host atom density, using $M_{\mathrm{Ge}}\simeq 72.63\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, isThus the required atomic fraction isapproximately one donor per $5.8\times 10^4$ host atoms. The electron density must rise to about $2000n_i$, not $1000n_i$, because the original intrinsic sample conducted through both electrons and holes. This is an estimate under unchanged mobility: impurity scattering, incomplete ionization, compensation and band-edge changes can alter it. Here $n_e/N_c\simeq 0.12$, so the nondegenerate treatment is an approximation rather than an arbitrarily dilute limit.