 Solution

In the nondegenerate conduction band, the Fermi-Dirac distribution reduces to its Boltzmann distribution tail. ThusHere the substitution $x=(ϵ-{ϵ}_c)/(k_{B}T)$ leaves ${\int }_0^{\infty }{x}^{1/2}{e}^{-x}dx=\sqrt{\pi }/2$. For electron holes use the empty-state probability $1-f(ϵ)\simeq e^{(ϵ-\mu )/(k_{B}T)}$, not the conduction-band Boltzmann factor. This givesThe chemical potential cancels in the product. With $E_g={ϵ}_c-{ϵ}_v$,This semiconductor mass-action law holds in thermal equilibrium for nondegenerate carriers, including a doped semiconductor. Only in an intrinsic semiconductor must $n=p=n_i$. The factors $N_c,N_v$ are the effective density of states; the factor $1/2$ in the final expression follows from the two spin factors already included in (a).