Past exam of the chemistry A course of the University of Cambridge / 2022 / Paper 2 / A10 / a / ii / Solution

In a nondegenerate semiconductor, the Fermi-Dirac distribution reduces near the conduction edge to $f(E)\simeq e^{-(E-E_F)/k_{\mathrm{B}}T}$. The corresponding dilute hole probability near the valence edge givesAn intrinsic, neutral sample has $n=p=n_i$. Multiplying the two expressions therefore yieldsEach excitation creates an electron-hole pair; charge neutrality is why the carrier concentration has half the gap in its activation exponent. The electrical conductivity isThus $\sigma \propto e^{-E_g/(2k_{\mathrm{B}}T)}$ to leading exponential accuracy. The prefactor contains the mobilities and the density-of-states temperature dependence; for fixed parabolic masses, $\sqrt{N_c{N}_v}\propto T^{3/2}$.

Measure conductivity in the intrinsic regime and plot $\ln \sigma$ against $1/T$. If the prefactor is constant or its temperature dependence has been corrected, a slope $m$ givesUse the intrinsic branch rather than a donor freeze-out or extrinsic branch. A temperature-dependent gap or mobility changes the apparent activation energy, so a precision extraction must account for those prefactors.