Back to the on-screen lesson ·
Filling particle-in-a-box states to the Fermi energy, the Fermi temperature, average energy and degeneracy pressure, the small electronic heat capacity of metals, and white dwarf stars.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to derive and compute the Fermi energy of a free electron gas, find its Fermi temperature, average energy and degeneracy pressure, explain and compute its heat capacity, and apply the model to metals and white dwarfs.
You know the Fermi-Dirac distribution and that the conduction electrons in a metal are far denser than their quantum concentration. From modern physics you know that a particle in a box has standing-wave states with discrete energies. This lesson counts those states to find how high the electrons in a metal must pile up, and what that implies for heat, pressure and stars.
| Term | What it means |
|---|---|
| Degenerate Fermi gas | A gas of fermions at $n \gg n_Q$, filling states from the bottom up to a sharp top. |
| Fermi energy | $E_F = \frac{h^2}{8m}\left(\frac{3N}{\pi V}\right)^{2/3}$, the energy of the highest filled state at zero temperature. |
| Fermi temperature | $T_F = E_F/k$; the gas is degenerate when $T \ll T_F$. |
| Fermi sea | The filled states below $E_F$. |
| Density of states | $g(E)$, the number of states per unit energy; for a free gas in three dimensions it grows as $\sqrt{E}$. |
| Degeneracy pressure | $P = \tfrac{2}{5}(N/V)E_F$, the pressure of a Fermi gas at zero temperature. |
| White dwarf | A dead star about Earth's size with the Sun's mass, held up by electron degeneracy pressure. |
Put $N$ electrons in a box of side $L$. Each is a standing wave, with allowed momenta $p = hn_i/2L$ in each direction for positive integers $n_x$, $n_y$, $n_z$, and energy
$$E = \frac{h^2}{8mL^2}\left(n_x^2 + n_y^2 + n_z^2\right).$$
Each triple $(n_x, n_y, n_z)$ is a point in an "$n$-space", and each point holds two electrons, spin up and spin down. At zero temperature the electrons fill the lowest states, which are the points inside an eighth of a sphere (positive $n$ only) of some radius $n_{\max}$. Counting them:
$$N = 2 \times \frac{1}{8} \times \frac{4}{3}\pi n_{\max}^3 = \frac{\pi n_{\max}^3}{3}.$$
The energy of the top state is the Fermi energy. Solving for $n_{\max}$ and substituting,
$$E_F = \frac{h^2}{8m}\left(\frac{3N}{\pi V}\right)^{2/3}.$$
It depends only on the mass of the particles and their density, not on temperature. For copper's $8.5 \times 10^{28}$ electrons per cubic meter it is $7.0$ eV, which corresponds to a Fermi temperature $T_F = E_F/k$ of $82\,000$ K. At any temperature a metal can survive, $T \ll T_F$: the electron gas is effectively at absolute zero, a Fermi sea filled to a sharp surface, blurred only over a band about $kT$ wide by the Fermi-Dirac distribution.
Integrating over the filled states gives the average energy, $\tfrac{3}{5}E_F$, and the total energy $U = \tfrac{3}{5}NE_F$. Because $E_F \propto V^{-2/3}$, squeezing the gas raises $U$, and $P = -\partial U/\partial V = \tfrac{2}{5}(N/V)E_F$: the degeneracy pressure, present even at absolute zero.
Another way: picture
Picture a stadium where each seat holds two people and seats are sold strictly from the front row back. However cold and still the crowd, the last people in must sit high up in the stands. The Fermi energy is the height of the last filled row. Warming the crowd lets only the people near the top move, into the empty rows just above; everyone deeper in is boxed in by full seats all around.
Another way: steps
Checks. Metals' Fermi energies are a few electronvolts, from about $2$ eV for potassium to $12$ eV for aluminum, and Fermi temperatures tens of thousands of kelvin. Doubling the density raises $E_F$ by $2^{2/3} = 1.59$, not by two. A heavier fermion at the same density has a proportionally smaller $E_F$, which is why the neutrons in a neutron star need a density a billion times a metal's to become degenerate. And the degeneracy pressure of a metal's electrons, hundreds of thousands of atmospheres, is what the attraction to the positive ions balances; it is not an extra pressure the metal exerts on its surroundings.
Heating a metal by $dT$ gives each electron a chance to gain about $kT$ of energy — but only if there is an empty state to move into. An electron deep in the Fermi sea is surrounded by filled states, so it cannot absorb a small amount of energy. Only the electrons within about $kT$ of the Fermi energy, a fraction roughly $kT/E_F$ of them, can respond, and each absorbs about $k$ per kelvin. The careful result is
$$C_{\text{el}} = \frac{\pi^2}{2}Nk\,\frac{kT}{E_F} = \frac{\pi^2}{2}Nk\,\frac{T}{T_F}.$$
For copper at room temperature, $T/T_F = 0.0037$, so the electrons contribute about $0.018R$ per mole, against the $3R$ of the lattice vibrations and the $\tfrac{3}{2}R$ that equipartition would have given a classical electron gas. This solved a long-standing puzzle: the free electrons that make metals conduct seemed to contribute nothing to their heat capacity.
At very low temperature the tables turn. The lattice's heat capacity falls as $T^3$, the electrons' only as $T$, so below a few kelvin the electrons dominate. Plotting $C/T$ against $T^2$ gives a straight line whose intercept is the electronic coefficient $\gamma$ — the standard way physicists measure a metal's density of states at the Fermi energy.
When a star like the Sun exhausts its fuel, its core contracts until the electrons, stripped from their atoms and crushed together, become a degenerate Fermi gas. Their degeneracy pressure, which does not depend on temperature, then holds the star up indefinitely as it cools: a white dwarf. Sirius B has about the Sun's mass in a sphere the size of Earth, a density of about $10^{9}$ kg/m³, and electron concentrations near $10^{36}$ m⁻³, where the Fermi energy is about $0.4$ MeV — close to the electron's rest energy of $0.511$ MeV, so the fastest electrons are nearly relativistic.
That matters. For relativistic electrons the Fermi energy grows only as $n^{1/3}$, and the pressure no longer rises fast enough with compression to resist gravity. Chandrasekhar found in 1930 that no white dwarf can exceed about $1.4$ solar masses. Heavier stellar cores collapse further, into neutron stars held up by neutron degeneracy pressure, or into black holes.
The Fermi energy corresponds to a speed: $\tfrac{1}{2}m_ev_F^2 = E_F$ gives $v_F = \sqrt{2E_F/m_e}$. For copper, $v_F = \sqrt{2 \times 1.13 \times 10^{-18}/9.11 \times 10^{-31}} = 1.57 \times 10^{6}$ m/s, about half a percent of the speed of light. The electrons at the top of the Fermi sea move this fast even at absolute zero. By contrast, the drift speed of electrons carrying an ordinary current in a copper wire is a fraction of a millimeter per second: the current is a tiny net drift superimposed on enormously fast, random motion.
The same physics, one step further, holds up a neutron star. When a stellar core heavier than the Chandrasekhar limit collapses, electrons are squeezed into protons to make neutrons, and the core shrinks until the neutrons themselves form a degenerate Fermi gas. Neutrons are about $1840$ times heavier than electrons, so by $E_F \propto 1/m$ they need far higher densities to become degenerate: a neutron star packs one and a half solar masses into a sphere about $20$ km across, at the density of an atomic nucleus, $2 \times 10^{17}$ kg/m³. Its neutron Fermi energy is tens of MeV. Pulsars, first discovered in 1967, are rotating neutron stars, and their masses, measured from binary orbits, test the equation of state of this densest known matter.
At liquid-helium temperatures, a few kelvin, the heat capacity of a metal has two parts: the lattice vibrations, which by Debye's theory go as $T^3$, and the conduction electrons, which go as $T$. Writing $C = \gamma T + \beta T^3$ and plotting $C/T$ against $T^2$ gives a straight line whose intercept is $\gamma$, the electronic coefficient.
For copper the free-electron model predicts $\gamma = \pi^2R/(2T_F) = 4.93 \times 8.31/81\,700 = 0.50$ mJ/(mol K²). The measured value is $0.69$ mJ/(mol K²), about forty percent larger. The difference is itself informative: the electrons in a real metal interact with the lattice and with each other, which makes them behave as if they were heavier than free electrons, and the ratio $0.69/0.50 = 1.38$ is read directly as that "effective mass" ratio.
The same measurement on the alloys used in superconducting magnets, or in the heavy-fermion compounds studied in university laboratories, finds $\gamma$ hundreds of times larger than for copper — electrons behaving as if a hundred times heavier. At $4$ K the electrons in copper contribute about $2$ mJ/(mol K), comparable with the lattice, which is why these experiments are done where they are.
Sirius B, the faint companion of the brightest star in the night sky, has a mass of about $1.0$ solar mass and a radius of about $5800$ km, a little smaller than Earth. Its mean density is $2 \times 10^{30}/(\tfrac{4}{3}\pi \times (5.8 \times 10^{6})^3) \approx 2.4 \times 10^{9}$ kg/m³ — a teaspoon would weigh about twelve tonnes.
Made mostly of carbon and oxygen, with about one electron per two nucleons, it holds roughly $n = 2.4 \times 10^{9}/(2 \times 1.67 \times 10^{-27}) \approx 7 \times 10^{35}$ electrons per cubic meter. Its interior temperature is about ten million kelvin, yet the quantum concentration of electrons there is only about $8 \times 10^{31}$ m⁻³, so $n/n_Q \approx 10^{4}$: the electrons are degenerate despite the heat. Their Fermi energy, about $0.3$ MeV, corresponds to a Fermi temperature of over three billion kelvin.
Because degeneracy pressure does not depend on temperature, Sirius B will simply cool, over billions of years, at a fixed size, until it fades to a cold black dwarf. The universe is not yet old enough for any white dwarf to have finished cooling.
Classical physics says that as a gas cools its particles slow down, and at absolute zero they stop. For a degenerate Fermi gas this is completely wrong. The exclusion principle allows only two electrons, one of each spin, in each standing-wave state, so the electrons must fill states up to the Fermi energy however cold the metal. At absolute zero the electrons in copper still have an average kinetic energy of $4.2$ eV, and the fastest move at $1.6 \times 10^{6}$ m/s. What cooling removes is only the thin blur of excitations, about $kT$ wide, at the top of the Fermi sea.
A related misconception is that degeneracy pressure is a thermal pressure like that of an ordinary gas. It does not come from temperature at all; it comes from the exclusion principle forcing particles into higher-energy states when they are squeezed together. That is exactly why a white dwarf can cool for billions of years without shrinking: its support never depended on its heat.
Copper has $8.49 \times 10^{28}$ free electrons per cubic meter. Write the Fermi energy.
$E_F = \dfrac{h^2}{8m_e}\left(\dfrac{3n}{\pi}\right)^{2/3}$
From counting standing-wave states.
Evaluate the prefactor.
$\dfrac{h^2}{8m_e} = \dfrac{(6.626 \times 10^{-34})^2}{8 \times 9.109 \times 10^{-31}} = 6.025 \times 10^{-38}\ \text{J m}^2$
Planck's constant squared over eight electron masses.
Evaluate the density factor.
$\left(\dfrac{3 \times 8.49 \times 10^{28}}{\pi}\right)^{2/3} = (8.11 \times 10^{28})^{2/3} = 1.873 \times 10^{19}\ \text{m}^{-2}$
Cube root, then square.
Multiply the factors.
$E_F = 6.025 \times 10^{-38} \times 1.873 \times 10^{19} = 1.129 \times 10^{-18}\ \text{J}$
An energy per electron.
Convert to electronvolts.
$E_F = \dfrac{1.129 \times 10^{-18}}{1.602 \times 10^{-19}} = 7.04\ \text{eV}$
Close to the $7.0$ eV found from experiments on copper.
Find copper's Fermi temperature.
$T_F = \dfrac{E_F}{k} = \dfrac{7.04}{8.617 \times 10^{-5}} = 81\,700\ \text{K}$
The temperature at which $kT$ would reach $E_F$.
Compare with room temperature.
$\dfrac{T}{T_F} = \dfrac{300}{81\,700} = 0.0037$
The electron gas is, in effect, at absolute zero.
Find the average energy per electron.
$\bar{E} = \tfrac{3}{5} \times 7.04 = 4.22\ \text{eV}$
More than a hundred and fifty times $kT$ at room temperature.
Find the electronic heat capacity per mole.
$C_{\text{el}} = \dfrac{\pi^2}{2}R\dfrac{T}{T_F} = 4.93 \times 8.31 \times 0.0037 = 0.15\ \text{J/(mol K)}$
Tiny.
Compare with the classical prediction.
$\tfrac{3}{2}R = 12.5\ \text{J/(mol K)}$
About eighty times more than is measured: the exclusion principle freezes almost every electron.
Find the speed of the fastest electrons.
$v_F = \sqrt{\dfrac{2E_F}{m_e}} = \sqrt{\dfrac{2 \times 1.129 \times 10^{-18}}{9.109 \times 10^{-31}}} = 1.57 \times 10^{6}\ \text{m/s}$
Even at absolute zero.
Write the total energy of the Fermi sea.
$U = \tfrac{3}{5}NE_F, \qquad E_F \propto \left(\dfrac{N}{V}\right)^{2/3}$
Three fifths of $E_F$ per electron.
Write the energy as a function of volume.
$U = CV^{-2/3}$
At fixed $N$ everything else is a constant $C$.
Differentiate for the pressure.
$P = -\dfrac{\partial U}{\partial V} = \tfrac{2}{3}CV^{-5/3} = \tfrac{2}{3}\dfrac{U}{V}$
Pressure is minus the derivative of energy with volume at zero temperature.
Substitute the energy.
$P = \tfrac{2}{3} \times \tfrac{3}{5}\dfrac{N}{V}E_F = \tfrac{2}{5}nE_F$
Two fifths of the density times the Fermi energy.
Evaluate for copper.
$P = 0.4 \times 8.49 \times 10^{28} \times 1.129 \times 10^{-18} = 3.83 \times 10^{10}\ \text{Pa}$
Pascals from inverse cubic meters times joules.
Convert to atmospheres.
$\dfrac{3.83 \times 10^{10}}{1.013 \times 10^{5}} \approx 3.8 \times 10^{5}\ \text{atm}$
Nearly four hundred thousand atmospheres.
Say what holds it in.
$\text{electron-ion attraction}$
The electrons' attraction to the copper ions balances this pressure; it is also why metals resist compression so strongly.
Compare with copper's measured stiffness.
$B_{\text{free electron}} = \tfrac{5}{3}P \approx 6.4 \times 10^{10}\ \text{Pa} \quad \text{vs} \quad B_{\text{measured}} = 1.4 \times 10^{11}\ \text{Pa}$
The Fermi gas alone accounts for a large share of how hard it is to squeeze a metal.
Divide the Fermi energy by Boltzmann's constant.
$T_F = \dfrac{3.24}{8.617 \times 10^{-5}} = 37\,600\ \text{K}$
Still enormous beside room temperature.
Write the average energy.
$\bar{E} = \tfrac{3}{5}E_F$
The average over the filled Fermi sea.
Evaluate the expression.
A metal's conduction electrons have a Fermi energy of $2$ eV. If the metal were squeezed until its electron density was eight times larger, what would the Fermi energy become?
Complete the worked solution: a metal has $8 \times 10^{28}$ free electrons per cubic meter with a Fermi energy of $7$ eV. Find the average energy per electron at zero temperature and the degeneracy pressure.
Take three fifths of the Fermi energy.
$\bar{E} = 0.6 \times 7 =$ u eV
The density of states weights the top of the filled sea.
Convert the Fermi energy to joules.
$E_F = 7 \times 1.602 \times 10^{-19} =$ j $\times 10^{-19}$ J
The pressure must come out in pascals.
Multiply two fifths of the density by the Fermi energy.
$P = 0.4 \times 8 \times 10^{28} \times E_F =$ p $\times 10^{9}$ Pa
$P = \tfrac{2}{5}nE_F$ from differentiating the total energy.
For a free electron gas with Fermi energy $6$ eV and $N$ electrons in volume $V$, match each quantity to its expression.
| $\frac{h^2}{8m}(3N/\pi V)^{2/3}$ | $\tfrac{3}{5}E_F$ | $\tfrac{2}{5}(N/V)E_F$ | $\frac{\pi^2}{2}Nk(kT/E_F)$ | |
|---|---|---|---|---|
| the Fermi energy | ||||
| the average energy per electron at zero temperature | ||||
| the degeneracy pressure | ||||
| the heat capacity at low temperature |
The metal aluminum has a Fermi energy of $11.67$ eV. Fill in its Fermi temperature, to the nearest hundred kelvin, and the ratio $T/T_F$ at $300$ K.
| Fermi temperature (K) | $T/T_F$ at 300 K | |
|---|---|---|
| this metal |
The Fermi energy of aluminum is $11.67$ eV. What is its Fermi temperature, in kelvin?
Answer: K
The metal sodium has $2.65 \times 10^{28}$ conduction electrons per cubic meter. What is its Fermi energy, in eV? Use $h^2/8m_e = 6.025 \times 10^{-38}$ J m² and $1$ eV $= 1.602 \times 10^{-19}$ J.
Answer: eV
Low-temperature calorimetry measures the electronic heat capacity of metals. For aluminum, with Fermi temperature $134900$ K, the free-electron model gives $\gamma = \pi^2R/(2T_F) = 0.30399$ mJ/(mol K²). What is its electronic heat capacity at $4.0$ K, in mJ/(mol K)?
Answer: mJ/(mol K)
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The metal sodium has a Fermi energy of $3.24$ eV. Fill in its Fermi temperature, to the nearest hundred kelvin, and the ratio $T/T_F$ at $300$ K.
| Fermi temperature (K) | $T/T_F$ at 300 K | |
|---|---|---|
| this metal |
You can find the Fermi energy of a metal and what follows from it. Explain to someone why the electrons in a metal move at a million meters per second even at absolute zero.
16. Your turn: sodium's Fermi energy is $3.24$ eV. What is its Fermi temperature and the average energy of its conduction electrons at zero temperature?, step 3
$\bar{E} = 0.6 \times 3.24 = 1.94\ \text{eV}$
Sodium's lower electron density gives a smaller Fermi energy than copper's.