Solution
= Solution
There are twelve model conduction <electrons> in the conventional unit cell. Hence $n=12/a^3=3.78495\times10^{29}\ \mathrm m^{-3}$. For a spin-twofold <free electron gas>,
$$
k_F=(3\pi^2n)^{1/3},\qquad E_F=\frac{\hbar^2k_F^2}{2m_e}.
$$
Numerically,
$$
\boxed{k_F=2.23783\ \text{\AA}^{-1}},\qquad
\boxed{E_F=19.0799\ \mathrm{eV}}.
$$
The six-electron count is the stipulated model; treating transition-metal d states as completely free is an approximation.