= Solution
Assume dilute nondegenerate <electrons> and <electron holes>, parabolic three-dimensional band edges with the stated scalar <band effective masses>, <spin> multiplicity two, one effective valley for each edge, and thermal equilibrium. Take the unchanged host <ground state> to have unit activity and neglect bound excitons. The <effective density of states> values are
$$
N_c=2\left(\frac{2\pi m_e^*k_{\rm B}T}{h^2}\right)^{3/2},\qquad
N_v=2\left(\frac{2\pi m_h^*k_{\rm B}T}{h^2}\right)^{3/2}.
$$
For a <Fermi level> $\mu$ between the band edges,
$$
n_e=N_c e^{-(E_c-\mu)/(k_{\rm B}T)},\qquad
n_h=N_v e^{-(\mu-E_v)/(k_{\rm B}T)}.
$$
The factors of $\mu$ cancel on multiplication. With $E_g=E_c-E_v$, the <semiconductor mass-action law> is
$$
\boxed{K_n\equiv n_en_h=N_cN_v e^{-E_g/(k_{\rm B}T)}
=4\left(\frac{2\pi k_{\rm B}T}{h^2}\right)^3(m_e^*m_h^*)^{3/2}
e^{-E_g/(k_{\rm B}T)}.}
$$
This density-form equilibrium constant has units $\mathrm{m^{-6}}$. A dimensionless thermodynamic constant uses a specified reference <number density> $n^\circ$, giving $K=K_n/(n^\circ)^2$. The dot representing the host <ground state> is not a freely translating gas species requiring a further gas partition function. For an <intrinsic semiconductor>, <charge neutrality> gives $n_e=n_h=n_i=\sqrt{K_n}$.
Back to article page
