The classical (Boltzmann) prediction for the heat capacity of a gas of electrons is about 100 times larger than what is actually measured in metals at room temperature. The correct quantum explanation is:
AElectrons in metals move more slowly than the classical model predicts, so they carry less thermal energy
BPauli exclusion prevents all electrons except those within ~k_BT of the Fermi energy from absorbing thermal energy — the vast majority are 'frozen' in filled states with no available nearby states to transition into
CThe Fermi gas model overestimates the number of electrons by including core electrons that don't move freely
DElectrons lose most of their thermal energy to the lattice before it can be measured as heat capacity
In a classical gas, every particle can absorb ~k_BT of thermal energy when temperature rises. In a Fermi gas, Pauli exclusion blocks this for all electrons except those within ~k_BT of the Fermi surface — roughly k_BT/E_F ≈ 1% of electrons at room temperature. Each of these electrons gains ~k_BT, giving C_V ∝ (k_BT/E_F) × Nk_B × T ∝ T, much smaller than the classical Nk_B. The enormous depth of filled states means almost all electrons cannot change their energy at all — there are simply no empty states nearby for them to move into.
Question 2 Multiple Choice
The density of states g(E) ∝ √E in a 3D free-electron gas arises because:
AHigher-energy electrons are more likely to be thermally excited, creating more available states near the top
BThe Pauli exclusion principle forces states to spread out more at higher energies
CStates are uniformly distributed in k-space, and the volume of a spherical shell at radius k grows as k² — which, since E ∝ k², means the number of states per unit energy grows as √E
DElectrons at higher energies have longer de Broglie wavelengths, allowing more standing wave modes
The √E dependence is purely geometric. In k-space, each allowed wavevector occupies the same tiny volume (2π/L)³. The number of states with energy below E equals the volume of a sphere of radius k(E) ∝ √E in k-space, giving N(E) ∝ E^(3/2). The density of states g(E) = dN/dE ∝ E^(1/2) = √E. More states are packed per unit energy at higher energies simply because the spherical shell in k-space has larger area (∝ k² ∝ E). Option D has the wavelength relationship backwards — higher energy means shorter de Broglie wavelength (λ = h/p).
Question 3 True / False
In a Fermi gas at low temperature, the electronic heat capacity is small because only electrons near the Fermi energy can be thermally excited. At absolute zero, this contribution is exactly zero.
TTrue
FFalse
Answer: True
At T = 0, the Fermi-Dirac distribution is a perfect step function: all states below E_F are filled, all above are empty. No electron can change its energy without violating Pauli exclusion (there are no nearby empty states). When T > 0, only electrons within ~k_BT of E_F have access to empty states just above them and can absorb thermal energy. As T → 0, k_BT → 0 and the fraction of excitable electrons vanishes, making C_V → 0. The linear-T electronic heat capacity C_V = γT with γ ∝ g(E_F) reflects this: at T = 0, C_V = 0.
Question 4 True / False
The density of states at the Fermi energy g(E_F) is the key quantity determining the electronic heat capacity of a metal, with C_V proportional to g(E_F) × T.
TTrue
FFalse
Answer: True
g(E_F) determines how many electrons can be thermally excited at a given temperature — it counts the states available within ~k_BT of E_F. The number of excitable electrons is ∝ g(E_F) × k_BT, and each gains ~k_BT energy, giving C_V ∝ g(E_F) k_B² T. The formula g(E_F) = 3N/(2E_F) makes this quantitative. This is why materials with high densities of states at the Fermi level (transition metals, superconductors near their transition temperature) have enhanced heat capacities — and why engineering the density of states via band structure is central to materials design.
Question 5 Short Answer
Why does only a small fraction of electrons in a metal contribute to its heat capacity, and how does the density of states g(E_F) quantify this fraction?
Think about your answer, then reveal below.
Model answer: Pauli exclusion means that for an electron to absorb thermal energy, there must be an empty state at slightly higher energy for it to move into. At low temperature, all states up to E_F are filled and all above are empty. Only electrons within ~k_BT of E_F have nearby empty states and can be thermally excited. The fraction of excitable electrons is ∝ k_BT/E_F — at room temperature in copper (~0.025 eV vs. E_F ~ 7 eV) this is about 0.4%. The density of states g(E_F) counts how many states per unit energy are available at the Fermi surface, so the number of excitable electrons is g(E_F) × k_BT, and C_V = (dU/dT) ∝ g(E_F) k_B² T.
The contrast with a classical gas is stark: classically every electron contributes ~k_B/2 per degree of freedom. In the Fermi gas, only those within k_BT of E_F contribute — roughly 1% at room temperature. This explains why metals have much lower electronic heat capacities than naive classical models predict, a puzzle that was resolved only after Fermi-Dirac statistics were developed.