Question 4 of 7: Free-Electron Fermi Gas — Fermi Energy, Speed, Occupation Probability, Band Gap

Question 4: Free-Electron Fermi Gas — Fermi Energy, Speed, Occupation Probability, Band Gap (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Free-electron gas; Cu conduction-electron concentration $N/V=8.45\times10^{22}\ \text{cm}^{-3}$; a copper electron with wavevector $k=2\times10^8\ \text{cm}^{-1}$; $T=300\ \text{K}$; energy $0.12\ \text{eV}$ above the chemical potential $\mu$.

Find. (a) $\varepsilon_F$ (eV); (b) speed $v(k)$; (c) Fermi–Dirac occupation probability $f(\varepsilon)$; (d) origin of the band gap in the nearly-free-electron model.

Approach. (a) invert the free-electron density-of-states relation, eq. (11); (b) use $v=\hbar k/m$; (c) evaluate the Fermi–Dirac function, eq. (10); (d) recall the Bragg-reflection argument for the nearly-free-electron model.

1. Part (a) — Fermi energy of Cu. From eq. (11), $\varepsilon_F=\dfrac{\hbar^2}{2m}\left(\dfrac{3\pi^2N}{V}\right)^{2/3}$. Converting $N/V=8.45\times10^{22}\ \text{cm}^{-3}=8.45\times10^{28}\ \text{m}^{-3}$: $$\varepsilon_F=\frac{(1.055\times10^{-34})^2}{2(9.11\times10^{-31})}\left(3\pi^2\times8.45\times10^{28}\right)^{2/3}\ \text{J}$$ $$\boxed{\varepsilon_F\approx7.03\ \text{eV}}$$ This matches the textbook Fermi energy of copper ($\approx7.0\ \text{eV}$) almost exactly, confirming the free-electron approximation and the given concentration are both realistic (i.e. this is genuine copper data, not an invented number).
2. Part (b) — speed at $k=2\times10^8\ \text{cm}^{-1}$. For a free electron, $E=\hbar^2k^2/2m\Rightarrow v=\hbar k/m$ (group velocity of the parabolic band). Converting $k=2\times10^8\ \text{cm}^{-1}=2\times10^{10}\ \text{m}^{-1}$: $$v=\frac{(1.055\times10^{-34})(2\times10^{10})}{9.11\times10^{-31}}\ \text{m/s}$$ $$\boxed{v\approx2.32\times10^6\ \text{m/s}}$$ of the same order as, though about 1.5 times, the Fermi velocity of Cu ($v_F=\hbar k_F/m\approx1.57\times10^6\ \text{m/s}$, with $k_F=(3\pi^2N/V)^{1/3}=1.36\times10^8\ \text{cm}^{-1}$). This $k$ lies outside the Fermi sphere, so the state (energy $\hbar^2k^2/2m\approx15.2$ eV) is empty in the ground state; the question asks only for its speed.
3. Part (c) — Fermi–Dirac occupation probability. With $\varepsilon-\mu=0.12\ \text{eV}$ and $T=300\ \text{K}$ ($k_BT=0.0259\ \text{eV}$), eq. (10): $$f(\varepsilon)=\frac{1}{\exp\!\left(\dfrac{\varepsilon-\mu}{k_BT}\right)+1}=\frac{1}{\exp(0.12/0.0259)+1}=\frac{1}{\exp(4.63)+1}$$ $$\boxed{f(\varepsilon)\approx0.0095\ (0.95\%)}$$ a small but nonzero probability — states well above $\mu$ are only lightly populated at room temperature, consistent with $0.12\ \text{eV}\gg k_BT$.
4. Part (d) — origin of the band gap (Figure P4c). In the nearly-free-electron model, the periodic ion potential is treated as a small perturbation on the free-electron parabola (Figure P4b). At the zone boundary $k=\pm\pi/a$, the electron wave satisfies the Bragg condition and is Bragg-reflected, so the traveling-wave solutions $e^{\pm i\pi x/a}$ combine into two standing waves, $\psi_+\propto\cos(\pi x/a)$ and $\psi_-\propto\sin(\pi x/a)$. These two standing waves pile electron charge density at different locations relative to the positive ion cores — $\psi_+$ concentrates charge on the ions (lower potential energy) while $\psi_-$ concentrates charge between the ions (higher potential energy) — so the periodic potential shifts their energies apart. This splits the single free-electron energy at $k=\pm\pi/a$ into two distinct values, opening the energy gap shown in Figure P4c; no travelling-wave (propagating) solution exists for energies inside the gap.