= Solution
For a <free-electron Fermi circle in a square lattice>, include two spin states per <wavevector>. A square cell of side $a$ supplies $z/a^2$ <electrons> per unit area; counting the disk of occupied states gives
$$
\frac{z}{a^2}=\frac{2\pi k_F^2}{(2\pi)^2},\qquad k_F=\frac{\sqrt{2\pi z}}a.
$$
The <first Brillouin zone> is $|k_x|,|k_y|\leq\pi/a$. For $z=1$, \b[$k_F=2.507/a$], so the circular <Fermi surface> lies wholly inside this square and encloses half its area. The figure uses the stated free-electron approximation; the lattice symmetry alone does not prescribe an exact circular surface.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2011/paper-1-fermi-circles.png]
{title=Free-electron circles and the square Brillouin zone}
{description=Original state-counting construction. Dashed squares are first-zone boundaries; the right-hand circle is shown in the extended-zone representation.}
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