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The finite-temperature Fermi level lies at the middle of the gap in the stipulated symmetric intrinsic model:Here $E_v$ is the valence-band maximum and $E_c$ the conduction-band minimum. Thermal promotion creates one conduction electron and one electron hole, so charge neutrality requires $n=p$. Equal constant densities of states in broad, symmetric bands give equal populations when the distances $E_c-\mu$ and $\mu -E_v$ are equal.

This is a statement about the chemical potential, not an available electron state in the gap. Unequal band densities of states would shift the intrinsic Fermi level away from exact midgap; the equality follows from the symmetry assumed here.

The smaller $|{\beta }_v|$ gives the valence edge a larger density-of-states prefactor than the conduction edge. If the Fermi level were exactly halfway across the gap, there would therefore be more thermally generated holes than electrons. Charge neutrality in an intrinsic semiconductor requires equal carrier densities, so the chemical potential moves upward: it suppresses holes while increasing electrons.

Explicitly, in the nondegenerate limit,and $n=p$ gives the intrinsic Fermi levelSince $N_v>N_c$, the Fermi level lies in the upper half of the gap at finite temperature.

The supplied inverse-$|\beta |$ density-of-states rule gives the correct ordering. Strictly, an isotropic three-dimensional parabolic band has a prefactor proportional to $(m^{\ast }{)}^{3/2}\propto |\beta {|}^{-3/2}$, rather than $|\beta {|}^{-1}$; this changes the size of the displacement but not its sign. At zero temperature the limiting intrinsic expression approaches midgap; a chemical potential in a completely empty gap is not uniquely fixed by carrier neutrality alone.