= Solution
{c}
The <Boltzmann carrier density of a two-dimensional parabolic band> can be derived directly. Use one isotropic parabolic <conduction band>, $E(\mathbf k)=E_c+\hbar^2k^2/(2m_e^*)$, with two spin states per <wave vector>. In the nondegenerate limit the <Fermi-Dirac distribution> reduces to its Boltzmann factor. The number per area is therefore
$$
\begin{aligned}
n&=2\int\frac{d^2k}{(2\pi)^2}\exp\left[-\beta\left(E_c-\mu+\frac{\hbar^2k^2}{2m_e^*}\right)\right]\\
&=\frac{e^{-\beta(E_c-\mu)}}\pi\int_0^\infty k\,e^{-\beta\hbar^2k^2/(2m_e^*)}\,dk\\
&=\frac{m_e^*k_BT}{\pi\hbar^2}e^{-\beta(E_c-\mu)}.
\end{aligned}
$$
The integral uses $\int_0^\infty ke^{-ak^2}dk=1/(2a)$. Thus
$$
\boxed{C=\frac{k_B}{\pi\hbar^2},\qquad a=1,\qquad b=1.}
$$
Equivalently, the two-dimensional <density of states> is constant, $D_c=m_e^*/(\pi\hbar^2)$, and $n=N_c e^{-\beta(E_c-\mu)}$ with $N_c=D_ck_BT$. These expressions include twofold <spin degeneracy> and one valley. With a general total band degeneracy $g_c$, replace the factor $1/\pi$ by $g_c/(2\pi)$; extra valleys or occupied confinement subbands would change the coefficient.
Back to article page
