 Solution

For one electron per square cell, $n_e=1/a^2$, soNumerically, $\pi /a\simeq 2.17\times 10^{10}{\mathrm{m}}^{-1}$ exceeds $k_F\simeq 1.73\times 10^{10}{\mathrm{m}}^{-1}$, so the fermi circle lies entirely inside the first Brillouin zone.

The state-counting explanation is that a spin-degenerate band has two electron states per primitive cell. One electron per cell therefore occupies half of that band. In the circular free-electron approximation the occupied reciprocal-space area isA disk of that area fits inside this square: its radius is less than the distance to an edge. Half filling alone is not a shape-independent theorem that every possible Fermi surface must avoid every zone boundary; the square geometry and circular free-electron dispersion make the statement true here.