= Fermi energy of a model electron gas
{c}
Choose $n=10^{28}\,\mathrm{m^{-3}}$ for a uniform noninteracting <Electron> model with two equally populated <spin> states. This is an original worked example, not a measured <number density> assigned to an actual material. Using $m_e\simeq9.10938\times10^{-31}\,\mathrm{kg}$, $\hbar\simeq1.05457\times10^{-34}\,\mathrm{J\,s}$, and $k_B=1.380649\times10^{-23}\,\mathrm{J\,K^{-1}}$ gives
$$
\begin{aligned}
k_F&=(3\pi^2n)^{1/3}\simeq6.6651\times10^9\,\mathrm{m^{-1}},\\
p_F&=\hbar k_F\simeq7.0288\times10^{-25}\,\mathrm{kg\,m\,s^{-1}},\\
E_F&=\frac{p_F^2}{2m_e}\simeq2.7117\times10^{-19}\,\mathrm J\simeq1.6925\,\mathrm{eV},\\
T_F&=\frac{E_F}{k_B}\simeq1.9641\times10^4\,\mathrm K.
\end{aligned}
$$
The conversion uses $1\,\mathrm{eV}=1.602176634\times10^{-19}\,\mathrm J$: an <electronvolt> is an <energy> unit, while a <joule> is its <SI> counterpart. The dimensions provide a quick check: $n^{1/3}$ is inverse <length>, $\hbar n^{1/3}$ has units of <momentum>, $p_F^2/m$ has units of <energy>, and $E_F/k_B$ has units of <temperature>. The ratio $p_F/(m_ec)\simeq0.002574$ verifies the <nonrelativistic limit> for this model, where $c$ is the <speed of light>.
Increasing this chosen <number density> eightfold doubles the <Fermi momentum> and quadruples the <Fermi energy>, giving approximately $6.7701\,\mathrm{eV}$. The exponent $2/3$ has a geometric origin: the number of occupied states grows as the cube of a <momentum> radius, but <kinetic energy> of <nonrelativistic particles> grows as its square. This makes <Fermi energy> a useful bridge between a count of particles and a mechanical <energy> scale.
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