= Solution
{c}
Assume uncompensated <donor impurities>, negligible intrinsic <electron holes>, and a nondegenerate <conduction band>. Then <charge neutrality> equates the conduction <electron number density> to the ionized-donor <number density>:
$$
n=n_{\rm empty},\qquad n_{\rm full}=N_D-n.
$$
The low-<temperature> condition relative to the <band gap> supports neglect of intrinsic carriers; one also needs $n_i\ll n$ and the <Boltzmann approximation> for the conduction <electrons>. Combining part (b)(i) with the <effective density of states> result gives
$$
\frac{N_D-n}{n}=e^{\beta(\mu-\epsilon_D)},\qquad
n=N_c e^{\beta(\mu-\epsilon_c)}.
$$
Eliminate the <chemical potential> by writing $e^{\beta(\mu-\epsilon_D)}=(n/N_c)e^{\beta(\epsilon_c-\epsilon_D)}$. With the donor binding <energy> $E_i=\epsilon_c-\epsilon_D>0$,
$$
\boxed{\frac{n^2}{N_D-n}=N_c e^{-\beta E_i}
=2\left(\frac{m_e}{2\pi\beta\hbar^2}\right)^{3/2}e^{-\beta E_i}.}
$$
Writing $A(T)=N_c e^{-\beta E_i}$, the physical root of the resulting <quadratic equation> is
$$
n=\frac{\sqrt{A^2+4AN_D}-A}{2}.
$$
This lies between zero and $N_D$. At <semiconductor carrier freeze-out>, $A\ll N_D$, it becomes $n\simeq\sqrt{N_DN_c}\,e^{-E_i/(2k_BT)}$; at essentially complete ionization, $A\gg N_D$, it tends to $N_D$. Including <donor charge-state degeneracy> would replace $A$ by $A\,g_{\rm empty}/g_{\rm full}$.
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