Past exam of the chemistry A course of the University of Cambridge / 2023 / Paper 2 / A10 / a / iv / Solution

Let $D$ be the equal constant density of states per unit volume and energy in each band. In the nondegenerate semiconductor regime, the conduction-band occupation isFor bands broad compared with $k_{B}T$, integrating above the conduction edge givesThe valence-hole integral similarly gives $p=D{k}_{B}T{e}^{-(\mu -E_v)/(k_{B}T)}$. Midgap $\mu$ therefore impliesInserting the carrier densities into the transport expression yieldsOn treating the algebraic and mobility prefactor as approximately constant, the requested leading activation law isThe factor $1/2$ is essential: an intrinsic excitation creates an electron–hole pair, and charge neutrality places the chemical potential halfway across the symmetric gap. The conduction-electron population is governed by $E_c-\mu =E_g/2$, not by replacing this separation with the full gap.

There is a qualification to the proportionality as printed. A literal constant-density-of-states intrinsic semiconductor with temperature-independent mobilities retains the additional factor $T$. Indeed, using the full Fermi-Dirac distribution giveswhich reduces to the expression above when $E_g\gg k_{B}T$. Thus the pure exponential is an activation approximation, not an exact consequence of constant DOS alone. A mobility sum proportional to $T^{-1}$ would cancel the leading algebraic factor.