- #1

DrClaude

Mentor

- 7,506

- 3,786

## Homework Statement

Consider a one-dimensional metal wire with one free electron per atom and an atomic spacing of ##d##. Calculate the Fermi temperature.

## Homework Equations

Energy of a particle in a box of length ##L##: ##E_n = \frac{\pi^2 \hbar^2}{2 m L^2} n^2##

1D density of states: ##g(E) = \frac{L}{\pi \hbar} \sqrt{\frac{2m}{\epsilon}}##

Fermi temperature: ##T_\mathrm{F} = \frac{E_\mathrm{F}}{k_\mathrm{B}}##

Fermi-Dirac occupancy: ##\bar{n}_\mathrm{FD} = \left[ e^{\beta (E - \mu)} + 1 \right]^{-1}##

## The Attempt at a Solution

The solution is easy to find by considering the system at ##T=0##, where one simply needs to calculate the energy of the highest occupied level. For ##N## electrons, with ##2## electrons per ##n## state, i.e., ##N= 2 n_\mathrm{max}##,

$$

\begin{align*}

E_\mathrm{F} &= \frac{\pi^2 \hbar^2}{2 m L^2} n_\mathrm{max}^2 \\

&= \frac{\pi^2 \hbar^2}{2 m L^2} \left(\frac{N}{2} \right)^2 \\

&= \frac{\pi^2 \hbar^2}{8 m} \left(\frac{N}{L} \right)^2 \\

&= \frac{\pi^2 \hbar^2}{8 m d^2}

\end{align*}

$$

since ##N/L = 1/d##, so ##T_\mathrm{F} = \frac{\pi^2 \hbar^2}{8 m k_\mathrm{B} d^2} \left(\frac{N}{L} \right)^2 = \frac{\pi^2 \hbar^2}{8 m k_\mathrm{B}}##.

My problem comes when trying an alternate approach, by considering the system at ##T = T_\mathrm{F}##. In this case, one can calculate the average number of electrons using

$$

N = \int_0^\infty \bar{n}_\mathrm{FD} g(E) dE

$$

##\mu(T_\mathrm{F}) = 0##, so the equation becomes

$$

\begin{align*}

N &= \frac{L \sqrt{2m}}{\pi \hbar} \int_0^\infty \frac{E^{-1/2}}{e^{E / (k_\mathrm{B} T_\mathrm{F})} + 1} dE \\

&= \frac{L \sqrt{2m}}{\pi \hbar} \left(1 - \sqrt{2} \right) \sqrt{\pi } \zeta \left(1/2\right) \sqrt{k_\mathrm{B} T_\mathrm{F}}

\end{align*}

$$

from which I recover

$$

T_\mathrm{F} = \frac{\left(2 \sqrt{2}+3\right) \pi \hbar ^2}{2 \zeta^2 \left(1/2\right) k_\mathrm{B} m} \left(\frac{N}{L} \right)^2

$$

Comparing with the approach above, this result is ##\approx 3.5## times bigger.

I am confident of the first result above (and it is the same as I found in some on-line and off-line resources) and would appreciate some help in figuring out what is wrong in the second approach.