= Solution
There are two <electrons> per <primitive unit cell>; a spin-degenerate <energy band> accommodates exactly that number. A <semiconductor> results when the first band is full and does not overlap the second. In the leading two-wave/four-wave approximation, compare the first-band maximum at $M$ with the second-band minimum at $X$:
$$
\Delta_{\rm ind}=E_{X,2}-E_{M,1}
=(E_X-u)-(2E_X+w+2u)=-E_X-w-3u.
$$
This is an <indirect band gap>; a nonzero local splitting at $X$ alone is insufficient. At the onset,
$$
\boxed{V_{10,\rm crit}=-\frac{E_X+V_{11}}3.}
$$
With $a=0.3\,\mathrm{nm}$, $\hbar^2/(2m_e)=0.03810\,\mathrm{eV\,nm^2}$ gives $E_X\simeq4.178\,\mathrm{eV}$, hence
$$
\boxed{V_{10,\rm crit}\simeq-1.326\,\mathrm{eV}.}
$$
For $V_{10}$ more negative than this value the estimated <indirect band gap> is positive; at equality the bands just touch. This is the onset estimate within the source's <nearly-free electron model> truncation, not an exact critical potential after coupling all <plane waves>.
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