 Solution

An isolated spin-degenerate lowest energy band accommodates two electrons per atom when there is one atom per primitive cell. Starting from one per atom, the estimate for filling the band is thereforeThis corresponds to an additional areal carrier count of about $4.77\times 10^{19}{\mathrm{m}}^{-2}$. If a genuine band gap separates the filled first band from the next band, there are no partially occupied band states at low temperature, so ordinary metallic conduction is lost.

Simply touching an edge is insufficient. In the circular approximation first contact would occur at $k_F=\pi /a$, corresponding to $n_e{a}^2=\pi /2\simeq 1.57$ electrons per cell; that is a partly filled band and need not be insulating. Near zone boundaries, the periodic potential reshapes the free-electron contours. The two-electron filling estimate assumes a global gap without overlapping bands, rather than inferring an insulator merely from one local avoided crossing.