Fermi level and conductivity

Fermi energy

If you operate at zero temperature, $T = 0$ K and fill the energy-states of a system according to the Pauli-exclusion-principle, the Fermi energy is the boundary at which all lower states are full and all higher states are empty. At $T = 0$ this boundary is a sharp line.

For example, say you have ladder with five steps which you have to “fill” with ten electrons. Due to the Pauli-exclusion principle, each step can only take two electrons. Now you fill up the ladder: 2 electrons in the first step, the next two in the second step and so forth until you put the last two electrons in the 5th step. The energy at this step (the 5th step) is your Fermi energy.

Metals have Fermi energies of several electron-volts (eVs). (Cu: 7 eV, Al: 11 eV) For comparison, the thermal energy at room temperature is about $k_B T \sim $ 0.025 eV.

Connection to the conductivity

First you must know that only electrons with an energy close to the Fermi energy can participate to the conduction process. Why? I mentioned earlier that at $T = 0$, the Fermi energy is a sharp line. At $T > 0$, this sharp line gets “washed out” and you get something like this:

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This means that instead of only full and empty states, you now have half empty states above and below the Fermi energy. This is turn means that you are now able to excite electrons in higher energy states, which you need to do if you want to accelerate them in a direction; i.e., if you want to do work on them.

But in consequence of the small contributions of the thermal and electric energies (thermal $ \sim $ 0.025 eV, electric less than that), you are only able to excite electrons VERY CLOSE to the Fermi energy ($E_F$). So only electrons close to $E_F$ will contribute to the conduction.

With this $E_F$ you can associate a velocity, the Fermi velocity: $$ v_F = \sqrt{2 E_F / m} $$

Now let's turn to the conductivity:

The conductivity $\sigma$ is defined as $$ \sigma = \frac{n e^2 \tau}{m} $$ where $n$ is the number of electrons, $e$ is the electron charge, $\tau$ is the time between two collisions and $m$ is the mass of an electron.

One can obtain $\tau$ from $\mathscr{l}$, the mean free path between two collisions, given as $\mathscr{l} = v_F \tau$ or inversely, $\tau = \mathscr{l}/v_F$. During this time, the electron is accelerated. A LARGE $v_F$ will therefore result in a SHORT acceleration time and LESS gain of speed for the electron in a given direction.

Finally you can write the conductivity as: $$ \sigma = \frac{n e^2 \mathscr{l}}{m v_F} $$

Remember that $v_F$, the Fermi velocity, is directly related to the Fermi Energy $E_F$.

As an instance, consider this table of Fermi energies: Copper has a LOWER Fermi energy (7 eV) than aluminum (11 eV), so it has a LOWER Fermi velocity; hence, the time between two collisions ($\tau$) is LONGER than that in aluminum. This in turn means that the electron has MORE time to accelerate in a given direction which finally explains why copper is a better conductor.