= Solution
Let $d=1,2,3$ and take a specimen of length, area or volume $L^d$. One <monovalent atom> per <unit cell> supplies
$$
N_e=(L/a)^d
$$
<Electrons>. At zero <temperature>, the <Pauli exclusion principle> fills the lowest <energy> states, with two <spin> states per spatial <wavevector>. The occupied region is an interval, disk or ball of radius $k_F$. Using the $k$-space <density of states> gives, in one dimension,
$$
\frac La=2\frac L{2\pi}(2k_F),\qquad
\boxed{k_F^{(1)}=\frac\pi{2a},\quad E_F^{(1)}=\frac{\hbar^2\pi^2}{8ma^2}.}
$$
In two dimensions,
$$
\frac{L^2}{a^2}=2\left(\frac L{2\pi}\right)^2\pi k_F^2,\qquad
\boxed{k_F^{(2)}=\frac{\sqrt{2\pi}}a,\quad E_F^{(2)}=\frac{\pi\hbar^2}{ma^2}.}
$$
In three dimensions,
$$
\frac{L^3}{a^3}=2\left(\frac L{2\pi}\right)^3\frac{4\pi k_F^3}3,\qquad
\boxed{k_F^{(3)}=\frac{(3\pi^2)^{1/3}}a,\quad E_F^{(3)}=\frac{\hbar^2(3\pi^2)^{2/3}}{2ma^2}.}
$$
Each <Fermi energy> follows from the <free electron model> <dispersion relation> $E_F=\hbar^2k_F^2/(2m)$. The <Fermi wavevectors> have units $\mathrm{m^{-1}}$ and the <Fermi energies> have units <joules> for $a$ in metres and $m$ in kilograms. No numerical lattice spacing is supplied, so numerical energies in <electronvolts> cannot be uniquely calculated. The <spin> factor of two is necessary here for actual monovalent <crystals>; the spinless auxiliary count of part (a) was not an instruction to remove electron spin from these crystals.
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