PX284 - M2 - fermi energy

putting $N$ electrons into a metal one by one
the energy levels fill up one by one because of the pauli exclusion principle until no more electrons are left
the highest energy reached is called the fermi energy, $E_{F}$
this is seen from $f_{F D} (E)$ as $R \to 0 :$

f_{F D} = {\begin{cases} 1 & E < E_{F} \\ 0 & E > E_{F} \end{cases}

E_{F} \equiv μ (T = 0)

to evaluate $E_{F} :$

N = \int_{0}^{\infty} g (E) f_{F D} (E) d E = \int_{0}^{E_{F}} g (E) d E = \frac{V}{2 π^{2}} {(\frac{2 m}{ℏ})}^{3 / 2} \frac{E_{F}^{3 / 2}}{3 / 2}

therefore, the fermi energy is:

\begin{matrix} (1) & E_{F} = \frac{ℏ^{2}}{2 m} (3 π^{2} n)^{\frac{2}{3}} \end{matrix}

where, $n = N / V$ is the carrier density

alternatively:

E_{F} = \frac{ℏ^{2} k_{F}^{2}}{2 m}

where, $k_{F}$ is the fermi wavevector

the fermi surface separates filled from empty states
it is a sphere of radius, $k_{F}$ , in the $k$ -space

\begin{aligned} N & = \frac{volume of fermi surface}{volume of one state} \\ = \frac{\frac{4}{3} π k_{F}^{3}}{\frac{1}{2} {(\frac{2 π}{L})}^{3}} \\ ⟹ k_{F}^{3} & = 3 π^{2} n \\ E_{F} & = \frac{ℏ^{2}}{2 m} (3 π^{2} n)^{2 / 3} \end{aligned}

eg: copper, $n = 8.5 \times 10^{28}$ m $^{- 3}$ , $E_{F} = 7.0$ eV, $T_{F} = E_{F} / k_{B} = 80 000$ K
this is much higher than room temperature, so the low temperature limit approximation holds