 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 $X$:This is an indirect band gap; a nonzero local splitting at $X$ alone is insufficient. At the onset,With $a=0.3\mathrm{nm}$, ${\hslash }^2/(2m_e)=0.03810\mathrm{eV}\mathrm{n}{\mathrm{m}}^2$ gives $E_X\simeq 4.178\mathrm{eV}$, henceFor $V_{10}$ 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.