For $z=2$, \b[$k_F=\sqrt{4\pi}/a=3.545/a$]. The circle encloses one whole zone's area, crosses each face because $k_F>\pi/a$, and misses the corners because $k_F<\sqrt2\pi/a$. In a <reduced Brillouin zone>, fold the portions outside the square into neighbouring band branches. A weak <periodic potential> opens <band gaps> at crossings and deforms these pieces; divalence alone does not establish a <band insulator>.

Solution

Past exam of the physics A course of the University of Cambridge / 2011 / Paper 1 / Section A / 5 / b / Solution

 Solution (source code)