Question

According to the Free Electron Theory, if the temperature of a pure metal is increased, the collision frequency of electrons increases. How does this specifically affect the Fermi energy ($E_F$) of the metal?

• A. It decreases linearly with temperature
• B. It drops to zero at the melting point
• C. It increases significantly due to higher thermal energy
• D. It remains essentially constant

Hint:

The Fermi energy is a property dictated primarily by the density of free electrons in the lattice, which is constant.
Because the Fermi temperature of metals ($T_F = E_F/k_B$) is extremely high (tens of thousands of Kelvin), standard operating temperatures have negligible influence on the Fermi level.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks about the effect of increasing temperature (and the resulting higher electron collision frequency) on the Fermi energy ($E_F$) of a pure metal according to the Free Electron Theory.

Step 2: Key Formula or Approach:

• The Fermi energy at absolute zero temperature ($E_{F0}$) is defined solely by the electron density ($N/V$): \[ E_{F0} = \frac{\hbar^2}{2m} \left( 3\pi^2 \frac{N}{V} \right)^{2/3} \]
• At a finite temperature $T$, the temperature-dependent Fermi energy $E_F(T)$ is given by the Sommerfeld expansion: \[ E_F(T) \approx E_{F0} \left[ 1 - \frac{\pi^2}{12} \left( \frac{k_B T}{E_{F0}} \right)^2 \right] \]

Step 3: Detailed Explanation:

• Order of Magnitude Analysis:
For most typical metals, the Fermi energy $E_{F0}$ is in the range of 2 to 10 eV.
The thermal energy equivalent of 1 eV is approximately $11600 \text{ K}$.
At room temperature ($T \approx 300 \text{ K}$), the thermal energy $k_B T$ is only about $0.025 \text{ eV}$.
Even at the melting point of most metals ($T \approx 1000 \text{ K}$ to $1500 \text{ K}$), $k_B T$ is around $0.1 \text{ eV}$.

• Evaluation of the Correction Term:
The ratio $\frac{k_B T}{E_{F0}}$ is exceptionally small (on the order of $10^{-2}$).
When squared in the Sommerfeld correction, the term $\left(\frac{k_B T}{E_{F0}}\right)^2$ becomes on the order of $10^{-4}$.
Therefore, the decrease in Fermi energy with temperature is extremely tiny (less than $0.1\%$).

• Conclusion:
While collision frequency increases and electrical resistivity rises with temperature due to electron-phonon interactions, the Fermi energy itself remains virtually unchanged.
Thus, for all practical purposes and solid-state calculations, the Fermi energy of a metal remains essentially constant.

Step 4: Final Answer:
The Fermi energy of the metal remains essentially constant as the temperature increases.