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.