= 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,
$$
n=\frac{N_d+\sqrt{N_d^2+4n_i^2}}2,\qquad
p=\frac{2n_i^2}{N_d+\sqrt{N_d^2+4n_i^2}}.
$$
Since $N_d\gg n_i$, the hole density is
$$
\boxed{p\simeq\frac{n_i^2}{N_d}
=4.5816\times10^{10}\,\mathrm{m^{-3}},\qquad
n\simeq1.0000\times10^{20}\,\mathrm{m^{-3}}.}
$$
Use the conduction population to locate the <chemical potential>:
$$
\boxed{E_c-\mu=k_BT\ln\frac{N_c}{n}=0.269180\,\mathrm{eV},\qquad
\mu-E_v=0.840820\,\mathrm{eV}.}
$$
It moves toward the <conduction band> but remains about $10.4k_BT$ below it, so the mobile electrons are nondegenerate. Despite the donor binding energy being comparable with $k_BT$, the large number of available conduction states makes ionization almost complete at this dilute doping. For a singly occupied two-spin donor, $N_d^+/N_d=[1+2e^{(\mu-E_D)/(k_BT)}]^{-1}$, 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 $4.5824\times10^{10}\,\mathrm{m^{-3}}$.
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