= Solution
\b[<spin normalization in two-dimensional free-electron state counting>.] Treat the mobile <electrons> as independent particles in a constant interior potential, with <parabolic dispersion> $\epsilon=\hbar^2k^2/(2m)$, and impose confinement or <periodic boundary conditions>. <electron>–<electron> interactions and the detailed periodic ionic potential are neglected, while Fermi statistics are retained. Each spatial state occupies area $(2\pi)^2/A$ in two-dimensional <wavevector> space. For <spin degeneracy> $g_s$,
$$
\mathcal N(k)=g_s\frac{A\pi k^2}{(2\pi)^2},\qquad
\boxed{g(\epsilon)=\frac{d\mathcal N}{d\epsilon}=\frac{g_sAm}{2\pi\hbar^2}}.
$$
The printed coefficient $Am/(2\pi\hbar^2)$ is the \b[per-spin <density of states>]. Unpolarized spin-one-half <electrons> have $g_s=2$, doubling it. Actual <graphene> near its <Dirac points> has approximately linear dispersion, so its energy-dependent <density of states> is not this parabolic-model result; the plane geometry alone does not establish the free-electron dispersion.<Two-Dimensional Gas of Massless Dirac Fermions in Graphene by Konstantin Novoselov and others>[Abstract: massless Dirac charge carriers rather than a finite-mass parabolic band]{ref}
\b[Occupied disk and <Fermi energy>.] At zero <temperature> fill the lowest states up to a disk of radius $k_F$. State counting gives
$$
\boxed{k_F=\sqrt{\frac{4\pi N}{g_sA}},\qquad
\epsilon_F=\frac{2\pi\hbar^2N}{g_sAm}}.
$$
Thus the physical spin-degenerate result is $\epsilon_F=\pi\hbar^2N/(Am)$. If $N$ instead counts one spin species, the per-spin convention gives $2\pi\hbar^2N/(Am)$. Each occupied <wavevector> can contain one <electron> of each allowed spin, never two in the same complete <one-particle state>.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2013/paper-2-two-dimensional-fermi-gas.png]
{title=Original occupied wavevector disk and Fermi-Dirac occupations for a parabolic two-dimensional electron gas}
\b[Occupation probability and <chemical potential of a two-dimensional ideal Fermi gas>.] A single <fermion> state permits occupation zero or one. In a <grand canonical ensemble> their relative weights are $1$ and $e^{-\beta(\epsilon-\mu)}$. Normalizing gives
$$
\boxed{p(\epsilon)=\frac{e^{-\beta(\epsilon-\mu)}}{1+e^{-\beta(\epsilon-\mu)}}
=\frac1{e^{\beta(\epsilon-\mu)}+1}}.
$$
The <chemical potential> is fixed by the total particle number. Here the total <density of states> $D=g_sAm/(2\pi\hbar^2)$ is constant above the band bottom. Integrating the occupation gives $N=Dk_BT\ln(1+e^{\mu/(k_BT)})$, and therefore
$$
\boxed{\mu=\epsilon_F+k_BT\ln\left(1-e^{-\epsilon_F/(k_BT)}\right)
\longrightarrow\epsilon_F\quad(T\to0)}.
$$
The approach is exponentially close in this ideal two-dimensional constant-density-of-states model; a three-dimensional $T^2$ correction should not be inserted here. A narrow interval of width of order $k_BT$ around $\mu$ is thermally smeared.
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