= Solution
<Semiconductor doping> introduces electrically active impurities. In <gallium arsenide>, <silicon> has one more valence <electron> than <gallium> but one fewer than <arsenic>. <Silicon> on a <gallium> site can therefore supply a shallow donor <electron>, while <silicon> on an <arsenic> site can accept an <electron> and leave a mobile hole. It is an <amphoteric dopant>, so its net effect depends on site occupation and growth conditions.<Thermal equilibrium concentrations of the amphoteric dopant Si and the associated carrier concentrations in GaAs by Chang-Ho Chen and others>[1999 abstract: neutral and ionized donors on Ga sites and acceptors on As sites]{ref}
For isotropic parabolic bands, shift the <energy> origin in the given free-electron <density of states> to the appropriate band edge and use its effective <mass>:
$$
\boxed{g_c(E)=\frac1{2\pi^2}\left(\frac{2m_e}{\hbar^2}\right)^{3/2}\sqrt{E-\epsilon_c},\quad E\geq\epsilon_c,}
$$
and, for valence <electron> states,
$$
\boxed{g_v(E)=\frac1{2\pi^2}\left(\frac{2m_h}{\hbar^2}\right)^{3/2}\sqrt{\epsilon_v-E},\quad E\leq\epsilon_v.}
$$
The stated normalization includes the two <spin> states and assumes one effective valley. In a nondegenerate semiconductor, the conduction occupation is approximately $e^{-(E-\mu)/(k_BT)}$ and the probability of a valence hole is approximately $e^{(E-\mu)/(k_BT)}$. Substituting $x=(E-\epsilon_c)/(k_BT)$ in the first integral, and the corresponding hole <energy> in the second, gives
$$
n=N_ce^{-(\epsilon_c-\mu)/(k_BT)},\qquad p=N_ve^{-(\mu-\epsilon_v)/(k_BT)},
$$
where the <effective density of states> values are $N_c=2[m_ek_BT/(2\pi\hbar^2)]^{3/2}$ and $N_v=2[m_hk_BT/(2\pi\hbar^2)]^{3/2}$. The factor $\sqrt\pi/2$ comes from the supplied Gaussian-weighted integral. Multiplying cancels the shared <thermodynamic equilibrium> <chemical potential> and proves the <semiconductor mass-action law>:
$$
\boxed{np=N_cN_ve^{-E_g/(k_BT)}=4\left(\frac{k_BT}{2\pi\hbar^2}\right)^3(m_em_h)^{3/2}e^{-E_g/(k_BT)}.}
$$
For the given sample $N_D=2\times10^{21}\,\mathrm{m^{-3}}$ and $N_A=10^{21}\,\mathrm{m^{-3}}$. With complete donor ionization and occupied acceptors, <charge neutrality> gives $n-p=N_D-N_A=10^{21}\,\mathrm{m^{-3}}$. At $293\,\mathrm K$ the given masses and gap give
$$
N_c\simeq3.92\times10^{23}\,\mathrm{m^{-3}},\quad N_v\simeq8.56\times10^{24}\,\mathrm{m^{-3}},\quad n_i^2\simeq1.26\times10^{24}\,\mathrm{m^{-6}}.
$$
Solving the <compensated semiconductor> equations, or using $n_i\ll N_D-N_A$, yields
$$
\boxed{n\simeq1.00\times10^{21}\,\mathrm{m^{-3}},\qquad p=\frac{n_i^2}{n}\simeq1.26\times10^3\,\mathrm{m^{-3}}.}
$$
The very small minority density is consistent with the wide gap and the strongly n-type balance. Also $n/N_c\simeq2.6\times10^{-3}\ll1$ and $p/N_v\ll1$, checking the nondegenerate approximation. The acceptor levels are occupied in this n-type regime; complete acceptor ionization in the compensation equation is appropriate to the model.
At low <temperature> <dopant freeze-out> reduces the conduction-electron population; a quantitative curve would require dopant binding energies and occupation factors. Once donors are ionized while intrinsic excitation is negligible, $n\simeq N_D-N_A$ gives an extrinsic plateau. At higher <temperature> intrinsic pairs eventually dominate, so $n\simeq p\simeq n_i(T)\propto T^{3/2}e^{-E_g/(2k_BT)}$ if the gap and masses are treated as constant. Real gap and <mass> changes modify that last expression.
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