= Solution
{c}
In a <direct band gap> <semiconductor>, the small <photon momentum> makes optical transitions approximately vertical in <wave vector>. Their <energy> is
$$
\mathcal E(k)=E_c(k)-E_v(k)=E_g+\frac{\hbar^2k^2}{2m_r},
\qquad \boxed{\frac1{m_r}=\frac1{m_c^*}+\frac1{m_v^*}}.
$$
The <joint density of states> per unit volume per transition <energy>, including the supplied spin counting, is
$$
D_J(\mathcal E)=\frac{k^2}{\pi^2}\frac{dk}{d\mathcal E}
=\frac{m_r k}{\pi^2\hbar^2}
=\frac{\sqrt2\,m_r^{3/2}}{\pi^2\hbar^3}\sqrt{\mathcal E-E_g},\qquad \mathcal E>E_g.
$$
For an upper <electron> state of occupation $f_c$ and a lower <electron> state of occupation $f_v$, <stimulated emission> has occupation factor $f_c(1-f_v)$; <absorption> has factor $f_v(1-f_c)$. Their difference is exactly $f_c-f_v$. Multiplication by the <joint density of states> therefore gives
$$
\boxed{g(\omega)\ \propto\ m_r^{3/2}\sqrt{\hbar\omega-E_g}\,[f_c(E_2)-f_v(E_1)].}
$$
The proportionality suppresses the optical transition <matrix element> and other smoothly varying optical prefactors. It assumes parabolic isotropic <energy bands>, an allowed direct transition, and negligible linewidth and many-body corrections. Here $f_v$ counts <electrons>, not <electron holes>.
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