= Solution
In a <nondegenerate semiconductor>, the <Fermi-Dirac distribution> reduces near the conduction edge to $f(E)\simeq e^{-(E-E_F)/k_{\rm B}T}$. The corresponding dilute hole probability near the valence edge gives
$$
n=N_c e^{-(E_c-E_F)/k_{\rm B}T},\qquad
p=N_v e^{-(E_F-E_v)/k_{\rm B}T}.
$$
An intrinsic, neutral sample has $n=p=n_i$. Multiplying the two expressions therefore yields
$$
\boxed{n_i=\sqrt{N_cN_v}\,e^{-E_g/(2k_{\rm B}T)}.}
$$
Each excitation creates an <electron-hole pair>; <charge neutrality> is why the carrier concentration has half the gap in its activation exponent. The <electrical conductivity> is
$$
\sigma=e(n\mu_e+p\mu_h)
=e(\mu_e+\mu_h)\sqrt{N_cN_v}\,e^{-E_g/(2k_{\rm B}T)}.
$$
Thus \b[$\sigma\propto e^{-E_g/(2k_{\rm B}T)}$ to leading exponential accuracy]. The prefactor contains the mobilities and the density-of-states <temperature> dependence; for fixed parabolic masses, $\sqrt{N_cN_v}\propto T^{3/2}$.
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 $m$ gives
$$
\boxed{E_g=-2k_{\rm B}m.}
$$
Use 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.
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