= Solution
{c}
There are three regimes. At the lowest <temperatures>, most <donor impurities> retain their <electrons>; thermal ionization gives <semiconductor carrier freeze-out>. The result from part (b) is
$$
n\simeq\sqrt{N_DN_c}\,e^{-E_i/(2k_BT)}\propto T^{3/4}e^{-E_i/(2k_BT)}.
$$
This produces the approximately straight descending low-<temperature> tail on the <logarithmic scale> plot against $1/T$. The activation <energy> in the exponential is $E_i/2$, because the <chemical potential> adjusts with donor ionization.
At intermediate <temperatures>, the <donor impurities> are essentially fully ionized while intrinsic excitation remains small. <Charge neutrality> gives $n\simeq N_D$, producing the extrinsic plateau, roughly from $20$ to $40\,\mathrm K$ in the figure.
At higher <temperatures>, excitation from the <valence band> to the <conduction band> creates <electron-hole pairs>. The <intrinsic carrier concentration> grows as $n_i\propto T^{3/2}e^{-E_g/(2k_BT)}$ for constant parabolic <band effective masses> and constant <band gap>. With ionized <donor impurities>, <charge neutrality> and the <semiconductor mass-action law> give
$$
n-p=N_D,\qquad np=n_i^2,\qquad
n=\frac{N_D+\sqrt{N_D^2+4n_i^2}}2.
$$
Thus $n\simeq n_i+N_D/2$ when $n_i\gg N_D$. This explains the renewed rapid rise and the eventual intrinsic branch; that branch is governed by the <band gap>, rather than the donor binding <energy>.
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