= Solution
For an undoped neutral <intrinsic semiconductor>, $n=p$. Equating the two expressions from <4/a> and taking logarithms gives
$$
\ln N_c-\frac{E_c-\mu}{k_BT}
=\ln N_v-\frac{\mu-E_v}{k_BT},
$$
and hence the <intrinsic-semiconductor chemical potential>
$$
\boxed{\mu_i(T)=\frac{E_c+E_v}{2}
+\frac{k_BT}{2}\ln\frac{N_v}{N_c}
=\frac{E_c+E_v}{2}+\frac34k_BT\ln\frac{m_h^*}{m_e^*}.}
$$
Therefore $\boxed{\alpha=\tfrac34\ln(m_h^*/m_e^*)}$. The logarithm is <temperature>-independent for fixed parabolic <effective masses>. The <chemical potential> is exactly at midgap only if the two <effective density of states> values coincide; for $m_h^*>m_e^*$ it moves above midgap. Additional valley multiplicities would add their logarithmic contribution.
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