= Solution
{c}
Assume equal excess <electron> and <electron hole> <number densities> $N_t$, with no additional doping imbalance. At zero <temperature>, both occupied carrier spheres have the same <Fermi wavevector> $k_F$ because their spin counting is the same:
$$
N_t=\int_0^{k_F}\frac{k^2}{\pi^2}\,dk
=\frac{k_F^3}{3\pi^2},\qquad k_F=(3\pi^2N_t)^{1/3}.
$$
For $k<k_F$, the <conduction band> is occupied and the <valence band> has <electron holes>, so $f_c-f_v=1$. Beyond $k_F$ the occupations reverse and the medium absorbs. Thus the upper edge of the positive-<optical gain> interval is the supremum
$$
\boxed{\hbar\omega_{\max}=E_g+\frac{\hbar^2}{2m_r}(3\pi^2N_t)^{2/3},\qquad
\beta=1+\frac{\hbar^2(3\pi^2N_t)^{2/3}}{2m_rE_g}.}
$$
Equivalently $E_{\rm Fc}=E_g+\hbar^2k_F^2/(2m_c^*)$ and $E_{\rm Fv}=-\hbar^2k_F^2/(2m_v^*)$. With the supplied <effective masses>,
$$
\begin{aligned}
m_r&=0.0587934\,m_e,\quad k_F=3.09367\times10^8\,\mathrm{m}^{-1},\\
\hbar\omega_{\max}-E_g&\simeq0.0620214\,\mathrm{eV},\\
&\boxed{\beta\simeq1.04368,\qquad \hbar\omega_{\max}\simeq1.48202\,\mathrm{eV}.}
\end{aligned}
$$
Using the <vacuum> <wavelength> $\lambda=hc/(\hbar\omega)$ reverses the endpoints:
$$
\boxed{836.59\,\mathrm{nm}<\lambda<873.13\,\mathrm{nm}.}
$$
These boundaries are idealized and open: the lower-<energy> edge has zero <joint density of states>, and the upper edge is the zero-<temperature> occupation discontinuity, not a finite-width gain peak. The calculation gives positive material <optical gain>, not the stronger requirement of exceeding the losses at <laser threshold>.
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