6.10 Fermi Energy of the Free-Electron Gas

As the previous section discussed, a system of noninteracting electrons, a free-electron gas, occupies a range of single-particle energies. Now the electrons with the highest single-particle energies are particularly important. The reason is that these electrons have empty single-particle states available at just very slightly higher energy. Therefore, these electrons are easily excited to do useful things, like conduct electricity for example. In contrast, electrons in energy states of lower energy do not have empty states within easy reach. Therefore lower energy electron are essentially stuck in their states; they do not usually contribute to nontrivial electronic effects.

Valence electrons in metals behave qualitatively much like a free-electron gas. For them too, the electrons in the highest energy single-particle states are the critical ones for the metallic properties. Therefore, the highest single-particle energy occupied by electrons in the system ground state has been given a special name; the Fermi energy. In the energy spectrum of the free-electron gas to the right in figure 6.11, the Fermi energy is indicated by a red tick mark on the axis.

Also, the surface that the electrons of highest energy occupy in wave number space is called the Fermi surface. For the free-electron gas the wave number space was illustrated to the left in figure 6.11. The Fermi surface is outlined in red in the figure; it is the spherical outside surface of the occupied region.

One issue that is important for understanding the properties of systems of electrons is the overall magnitude of the Fermi energy. Recall first that for a system of bosons, in the ground state all bosons are in the single-particle state of lowest energy. That state corresponds to the point closest to the origin in wave number space. It has very little energy, even in terms of atomic units of electronic energy. That was illustrated numerically in table 6.1. The lowest single-particle energy is, assuming that the box is cubic

\begin{displaymath}
{\vphantom' E}^{\rm p}_{111} = 3\pi^2 \frac{\hbar^2}{2m_e} \frac{1}{{\cal V}^{2/3}}
\end{displaymath} (6.15)

where $m_{\rm e}$ is the electron mass and ${\cal V}$ the volume of the box.

Unlike for bosons, for electrons only two electrons can go into the lowest energy state. Or in any other spatial state for that matter. And since a macroscopic system has a gigantic number of electrons, it follows that a gigantic number of states must be occupied in wave number space. Therefore the states on the Fermi surface in figure 6.11 are many orders of magnitude further away from the origin than the state of lowest energy. And since the energy is proportional to the square distance from the origin, that means that the Fermi energy is many orders of magnitude larger than the lowest single-particle energy ${\vphantom' E}^{\rm p}_{111}$.

More precisely, the Fermi energy of a free-electron gas can be expressed in terms of the number of electrons per unit volume $I$$\raisebox{.5pt}{$/$}$${\cal V}$ as:

\begin{displaymath}
\fbox{$\displaystyle
{\vphantom' E}^{\rm p}_{\rm{F}} = \...
...\hbar^2}{2m_e}
\left(\frac{I}{{\cal V}}\right)^{2/3}
$} %
\end{displaymath} (6.16)

To check this relationship, integrate the density of states (6.6) given in section 6.3 from zero to the Fermi energy. That gives the total number of occupied states, which equals the number of electrons $I$. Inverting the expression to give the Fermi energy in terms of $I$ produces the result above.

It follows that the Fermi energy is larger than the lowest single-particle energy by the gigantic factor

\begin{displaymath}
\frac{I^{2/3}}{\left(3\pi^2\right)^{1/3}}
\end{displaymath}

It is instructive to put some ballpark number to the Fermi energy. In particular, take the valence electrons in a block of copper as a model. Assuming one valence electron per atom, the electron density $I$$\raisebox{.5pt}{$/$}$${\cal V}$ in the expression for the Fermi energy equals the atom density. That can be estimated to be 8.5 10$\POW9,{28}$ atoms/m$\POW9,{3}$ by dividing the mass density, 9,000 kg/m$\POW9,{3}$, by the molar mass, 63.5 kg/kmol, and then multiplying that by Avogadro’s number, 6.02 10$\POW9,{26}$ particles/kmol. Plugging it in (6.16) then gives a Fermi energy of 7 eV (electron Volt). That is quite a lot of energy, about half the 13.6 eV ionization energy of hydrogen atoms.

The Fermi energy gives the maximum energy that an electron can have. The average energy that they have is comparable but somewhat smaller:

\begin{displaymath}
{\vphantom' E}^{\rm p}_{\rm average} = {\textstyle\frac{3}{5}} {\vphantom' E}^{\rm p}_{\rm {F}}
\end{displaymath} (6.17)

To verify this expression, find the total energy $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\int{\vphantom' E}^{\rm p}{\cal V}{\cal D}{\,\rm d}{\vphantom' E}^{\rm p}$ of the electrons using (6.6) and divide by the number of electrons $I$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\int{\cal V}{\cal D}{\,\rm d}{\vphantom' E}^{\rm p}$. The integration is again over the occupied states, so from zero to the Fermi energy.

For copper, the ballpark average energy is 4.2 eV. To put that in context, consider the equivalent temperature at which classical particles would need to be to have the same average kinetic energy. Multiplying 4.2 eV by $e$$\raisebox{.5pt}{$/$}$$\frac32k_{\rm B}$ gives an equivalent temperature of 33,000 K. That is gigantic even compared to the melting point of copper, 1,356 K. It is all due to the exclusion principle that prevents the electrons from dropping down into the already filled states of lower energy.


Key Points
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The Fermi energy is the highest single-particle energy that a system of electrons at absolute zero temperature will occupy.

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It is normally a very high energy.

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The Fermi surface is the surface that the electrons with the Fermi energy occupy in wave number space.

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The average energy per electron for a free-electron gas is 60% of the Fermi energy.