= 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,
$$
n=N_c e^{-(E_c-E_i)/(k_{\mathrm B}T)},\qquad
p=N_v e^{-(E_i-E_v)/(k_{\mathrm B}T)},
$$
and $n=p$ gives the <intrinsic Fermi level>
$$
\boxed{E_i=\frac{E_c+E_v}{2}+\frac{k_{\mathrm B}T}{2}\ln\frac{N_v}{N_c}.}
$$
Since $N_v>N_c$, \b[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 $(m^*)^{3/2}\propto|\beta|^{-3/2}$, 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.
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