= Solution
One conduction <electron> per atom gives number density $n=\rho N_A/M\simeq8.50\times10^{28}\,\mathrm{m^{-3}}$. Filling two spin states per <momentum> state in the three-dimensional <Fermi gas> gives $k_F=(3\pi^2n)^{1/3}$ and
$$
\boxed{E_F=\frac{\hbar^2}{2m_e}(3\pi^2n)^{2/3}\simeq7.05\,\mathrm{eV},\qquad
T_F=E_F/k_B\simeq8.18\times10^4\,\mathrm K.}
$$
The <Fermi temperature> greatly exceeds ordinary <temperatures>, explaining why these conduction <electrons> are degenerate rather than a classical Maxwell gas. The numerical result uses the stipulated free-electron band and bare <electron> <mass>, rather than a detailed copper band structure.
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