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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.

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.