Free Electron Model Questions and Answers

Answer: B

The ratio n_B/n_A is 2.

For free electrons in two dimensions the electron concentration is proportional to the area of the Fermi circle, πk_F², so n ∝ k_F².

• Metal A: the circle touches the zone boundary at (π/a, 0), so k_F = π/a.
• Metal B: the circle passes through the corners of the square zone, at (π/a, π/a), so k_F = √2 π/a.

\[ \begin{aligned} \frac{n_{B}}{n_{A}} &= \left(\frac{\sqrt{2}\,\pi/a}{\pi/a}\right)^{2} \\ &= 2 \end{aligned} \]

√2 is the ratio of the Fermi wavevectors, not of the concentrations. 3 and 3√3 would need other radii, and 3√3 a three-dimensional k_F³ scaling.

Answer: A

The resistance is \(\frac{mL}{ne^{2}A\tau}\).

In the Drude model an electron accelerated by a field \(E\) gains drift velocity \(v_{d} = \frac{eE\tau}{m}\) between collisions, so the current density is

\[ \begin{aligned} J &= nev_{d} = \frac{ne^{2}\tau}{m}E \Rightarrow \sigma = \frac{ne^{2}\tau}{m} \\ R &= \frac{L}{\sigma A} = \frac{mL}{ne^{2}A\tau} \end{aligned} \]

The other options carry spurious factors of 2 or 4 that the Drude derivation does not produce: the resistance is simply the length over conductivity times area.

Answer: A

The value is \(n = -\frac{2}{3}\).

In the free electron model the Fermi energy depends on the electron density as \(E_{F} \propto N^{2/3}\) (for a fixed volume), since \(E_{F} = \frac{\hbar^{2}}{2m}(3\pi^{2}N/V)^{2/3}\). Filling only half the electrons gives

\[ E = E_{F}\left(\frac{1}{2}\right)^{2/3} = 2^{-2/3}E_{F} \]

so \(n = -\frac{2}{3}\). The energy must be less than \(E_{F}\), which rules out \(+\frac{2}{3}\); \(-\frac{1}{3}\) and \(-1\) use the wrong power of the density.

Answer: A

The primitive cell has area \( \frac{2\pi}{k_{F}^{2}} \).

In two dimensions each \( k \)-state occupies an area \( (2\pi/L)^{2} \), and each holds two electrons. Filling a Fermi circle of radius \( k_{F} \):

\[ \begin{aligned} N &= 2 \times \frac{\pi k_{F}^{2}}{(2\pi/L)^{2}} = \frac{k_{F}^{2}L^{2}}{2\pi} \\ \frac{L^{2}}{N} &= \frac{2\pi}{k_{F}^{2}} \end{aligned} \]

With one electron per atom, N is also the number of atoms, so \( L^{2}/N \) is the area of one primitive cell. \( \pi/k_{F}^{2} \) forgets the spin factor of 2, and the options containing L grow with the size of the crystal, whereas the area of one cell cannot depend on L.

Answer: B

In two dimensions the density of states is independent of energy.

The number of states with energy below E is proportional to the area of the circle of radius \( k \) in k-space, \( \pi k^{2} \), and \( E = \hbar^{2}k^{2}/2m \) makes \( k^{2} \propto E \). So N(E) grows linearly with E and its derivative is constant:

\[ g(E) = \frac{mA}{\pi\hbar^{2}} \]

for area A including spin. \( E^{1/2} \) is the three-dimensional result and \( E^{-1/2} \) the one-dimensional one.

Answer: D

The Fermi velocity varies as \(n^{1/3}\).

For a free electron gas in three dimensions the Fermi wave vector is fixed by the electron density:

\[ \begin{aligned} k_{F} &= (3\pi^{2}n)^{1/3} \\ v_{F} &= \frac{\hbar k_{F}}{m} \propto n^{1/3} \end{aligned} \]

The Fermi energy, \(E_{F} = \hbar^{2}k_{F}^{2}/2m\), goes as \(n^{2/3}\), which is the trap in option A. The velocity is the square root of an energy, so it goes as \(n^{1/3}\).

Answer: D

The Fermi level is a direct measure of the chemical potential of electrons.

In the Fermi-Dirac distribution \(f(E) = \frac{1}{e^{(E - \mu)/k_{B}T} + 1}\), the Fermi level is the chemical potential \(\mu\): the energy at which the occupation probability is one half, and the energy needed to add one electron to the system. At absolute zero it equals the Fermi energy, the highest occupied level. Two materials in contact exchange electrons until their Fermi levels line up, which is what a common chemical potential means.

The Fermi level is not a current, a temperature or a resistivity, although it shifts slightly with temperature and helps decide how many carriers conduct.

Answer: B

The factor is b = 2.

For a square lattice of side a, the first Brillouin zone is a square with corners at \((\pm\pi/a, \pm\pi/a)\) and side midpoints at \((\pm\pi/a, 0)\) and \((0, \pm\pi/a)\). A free electron's kinetic energy is \(\hbar^{2}k^{2}/2m\), so

\[ \begin{aligned} k_{corner}^{2} &= 2\left(\frac{\pi}{a}\right)^{2} \\ k_{mid}^{2} &= \left(\frac{\pi}{a}\right)^{2} \\ b &= \frac{k_{corner}^{2}}{k_{mid}^{2}} = 2 \end{aligned} \]

\(\sqrt{2}\) is the ratio of the wave vectors, not of the energies; 4 and 8 square that ratio once or twice too often.