= Solution
{c}
The boundary of the first <Brillouin zone> is $|k|=\pi/a$. The unperturbed <free electron model> assigns the <kinetic energy>
$$
\boxed{E_{\rm ZB}=\frac{\hbar^2\pi^2}{2m_ea^2},}
$$
where $m_e$ is the bare <electron mass> and $\hbar$ is the <reduced Planck constant>. This is the free <kinetic energy> at the <Brillouin zone> boundary, not the already split <energy band> edge. A constant mean <potential energy> may be absorbed into the choice of zero of <energy>.
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