= Solution
For an isotropic free-electron gas at zero <temperature>, the <Fermi wavevector> is the <radius> of the occupied region in <wavevector> space. Its <energy> is $E_F=\hbar^2k_F^2/(2m)$. The first <Brillouin zone> is the Wigner-Seitz cell of the <reciprocal lattice>; it contains one distinct crystal momentum per primitive cell and per band, with <spin> counted separately.
For a two-dimensional sample of area $A$, allowed wavevectors have density $A/(2\pi)^2$. Including two <spin> states, filling a circle gives $N=2[A/(2\pi)^2]\pi k_F^2$. Dividing by area therefore yields
$$
\boxed{k_F=\sqrt{2\pi n}.}
$$
The square lattice has one atom per area $a^2$, so for valence $z$ the free-electron density is $z/a^2$. Its <Brillouin zone> has $|k_x|,|k_y|\leq\pi/a$. Thus
$$
\boxed{k_F^{(1)}=\frac{\sqrt{2\pi}}a\simeq8.36\times10^9\,\mathrm{m^{-1}},\qquad k_F^{(2)}=\frac{\sqrt{4\pi}}a\simeq1.18\times10^{10}\,\mathrm{m^{-1}}.}
$$
The monovalent Fermi circle lies inside the square. The divalent circle crosses the faces but misses the corners; its area equals the zone's area. These are the free-electron contours before distortion by the periodic potential.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2004-square-lattice-fermi-circles.png]
{title=Free-electron Fermi circles for one and two electrons per square-lattice cell}
{description=Original reciprocal-space diagrams in units of $\pi/a$. The divalent circle has the same area as the square zone but a different shape, extending beyond the faces while leaving corner regions outside the circle.}
At a Bragg plane, the periodic potential mixes degenerate plane waves differing by a <reciprocal lattice vector>. The two standing-wave combinations have different charge densities relative to the ions and split in <energy>, opening the gap. In a two-wave approximation, the boundary energies are $E_B\pm E_g/2$, with $E_g=2|V_G|$.
The boundary <wavevector> magnitude is $\pi/a$ along $[10]$ and $\sqrt2\pi/a$ along $[11]$. The free-electron boundary energies are therefore
$$
E_X=\frac{\hbar^2\pi^2}{2ma^2}\simeq4.18\,\mathrm{eV},\qquad E_M=2E_X\simeq8.36\,\mathrm{eV}.
$$
Within the stated direction-independent gap picture, the upper band's lowest face-edge <energy> is approximately $E_X+2\,\mathrm{eV}=6.18\,\mathrm{eV}$, while the lower band's corner-edge <energy> is approximately $E_M-2\,\mathrm{eV}=6.36\,\mathrm{eV}$. There is consequently a small overlap, about $0.18\,\mathrm{eV}$, rather than a global insulating gap. The divalent Fermi level lies in that overlap, producing occupied upper-band states and unfilled lower-band states.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2004-square-lattice-band-edge-sketch.png]
{title=An equal local splitting can leave overlap between different zone-edge directions}
{description=Original schematic two-wave dispersions near the $[10]$ and $[11]$ boundaries. The shaded Fermi-level range marks the overlap implied by the question's direction-independent gap approximation; no precise Fermi energy is inferred. These separate directional sketches do not specify a complete two-dimensional periodic potential.}
Thus \b[the intended isotropic-gap approximation predicts a metal], with small <electron> and hole pockets. <Positive direct gaps can coexist with band overlap>. A fully specified square-lattice potential would require simultaneous treatment of the corner degeneracies; a numerical boundary splitting alone does not uniquely determine an arbitrary two-dimensional band structure. The conclusion here uses the given simplified gap picture, rather than <electron> count alone.
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