 Solution

Take uncompensated donor impurities and initially approximate them as ionized. Charge neutrality then gives $n-p=N_d$, while the semiconductor mass-action law gives $np=n_i^2$. Solving the quadratic without prematurely discarding minority carriers,Since $N_d\gg n_i$, the hole density isUse the conduction population to locate the chemical potential:It moves toward the conduction band but remains about $10.4k_{B}T$ below it, so the mobile electrons are nondegenerate. Despite the donor binding energy being comparable with $k_{B}T$, the large number of available conduction states makes ionization almost complete at this dilute doping. For a singly occupied two-spin donor, $N_d^{+}/N_d=[1+2e^{(\mu -E_D)/(k_{B}T)}{]}^{-1}$, with $E_D=E_c-0.02584\mathrm{eV}$. The inferred value is $0.99984$, confirming the approximation to much better than one percent. Including this small correction raises the hole density only to about $4.5824\times 10^{10}{\mathrm{m}}^{-3}$.