= Solution
{c}
Equal <effective masses> and equal <spin> and <valley degeneracies> give <electron-hole symmetry>. In the <intrinsic semiconductor>, <charge neutrality> requires equal numbers of <conduction band> <electrons> and <valence band> <holes>. It therefore pins the <chemical potential> to the band touching point, $\mu=0$. The <Fermi-Dirac distribution> gives
$$
N_c=D\int_0^\infty\frac{dE}{e^{E/(k_BT)}+1}
=Dk_BT\ln2,
\qquad
\boxed{N_c(T)=\frac{Am^*k_BT}{\pi\hbar^2}\ln2.}
$$
Measured relative to the filled <valence band> at zero <temperature>, the thermal <internal energy> includes both species. Removing a <valence band> <electron> at $-E$ costs the positive <hole> <energy> $E$, just as adding a <conduction band> <electron> at $E$ does. Hence
$$
\Delta U=2D\int_0^\infty\frac{E\,dE}{e^{E/(k_BT)}+1}
=\frac{\pi^2}{6}D(k_BT)^2.
$$
Differentiate at fixed area and fixed total electronic <particle number>, allowing creation of <electron-hole pairs>:
$$
C_v^{\rm tot}=\frac{\pi^2}{3}Dk_B^2T,\qquad
\boxed{\frac{C_v^{\rm tot}}{N_c}=\frac{\pi^2}{3\ln2}k_B\simeq4.746\,k_B.}
$$
There is no conflict with the <equipartition theorem>. The normalization $N_c$ counts only the excited <conduction band> <electrons>, while the <heat capacity> includes both <electrons> and <holes>; moreover, $N_c\propto T$ is not fixed. In fact $\Delta U/N_c=\pi^2k_BT/(6\ln2)$, and differentiating $N_c(T)$ supplies another equal contribution to the <heat capacity>. Finally, $\mu=0$ gives occupation $1/2$ at the touching point, so the excitation gas is not in the classical dilute limit even though $k_BT$ exceeds the vanishing <band gap>.
Back to article page
