 Solution

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.