= Solution
At $300\,\mathrm K$, $k_BT=0.0258520\,\mathrm{eV}$. With the scalar <effective masses> and two-spin convention of <4/a>, the <effective density of states> values are
$$
N_c=3.32684\times10^{24}\,\mathrm{m^{-3}},\qquad
N_v=6.11180\times10^{24}\,\mathrm{m^{-3}}.
$$
The <intrinsic carrier concentration> follows from $n=p=n_i$:
$$
\boxed{p=n_i=\sqrt{N_cN_v}\,e^{-E_g/(2k_BT)}
=2.1405\times10^{15}\,\mathrm{m^{-3}}.}
$$
From <4/b>, $\alpha=(3/4)\ln(0.39/0.26)=0.304099$. Taking $E_v=0$ as a reference,
$$
\boxed{\mu_i-E_v=0.562862\,\mathrm{eV},\qquad
E_c-\mu_i=0.547138\,\mathrm{eV}.}
$$
It lies $7.862\,\mathrm{meV}$ above the middle of the <band gap>, because the hole <effective density of states> is larger. Both band edges are more than twenty $k_BT$ away, validating the <nondegenerate semiconductor> approximation.
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