The number density of conduction electrons in metal A is twice that of metal B. How does the Fermi energy of A compare to that of B?
AIt is twice as large, since twice as many electrons need to be accommodated
BIt is about 1.59 times larger, since E_F ∝ n^{2/3} and 2^{2/3} ≈ 1.587
CIt is the same, since Fermi energy depends on temperature, which is zero for both
DIt is four times larger, since packing twice the electrons requires twice the momentum in each dimension
The Fermi energy formula E_F = (ℏ²/2m)(3π²n)^{2/3} gives E_F ∝ n^{2/3}. Doubling n gives 2^{2/3} ≈ 1.587, not 2. Option A would be true if E_F were proportional to n; option C is wrong because E_F is explicitly independent of temperature; option D would require a linear momentum dependence, but the kinetic energy goes as p²/2m.
Question 2 Multiple Choice
A white dwarf star is supported against gravitational collapse. At its core temperature of ~10⁷ K, the Fermi temperature of the electrons is ~10¹⁰ K. What primarily provides the supporting pressure?
AThermal pressure from hot electrons colliding with the star's walls
BRadiation pressure from photons trapped in the dense core
CDegeneracy pressure arising from the Pauli exclusion principle, independent of temperature
DElectrostatic repulsion between negatively charged electrons
Because T ≪ T_F, the electrons are deeply quantum degenerate — the vast majority cannot be thermally excited above the Fermi sea. The pressure P = (2/5)nE_F is a quantum mechanical result with no temperature dependence; it persists even as the star cools toward absolute zero. This is why white dwarfs don't collapse: the degeneracy pressure is not thermal and cannot be 'turned off' by cooling.
Question 3 True / False
At T=0, the ground-state energy per particle of an ideal Fermi gas equals (3/5)E_F, which is about 60% of the Fermi energy.
TTrue
FFalse
Answer: True
This follows directly from integrating ε × g(ε) from 0 to E_F: U₀ = ∫₀^{E_F} ε g(ε) dε = (3/5)NE_F, giving an average energy per particle of (3/5)E_F. The result is not zero because the Pauli exclusion principle forces fermions to fill all states from the ground state up to E_F — they cannot all sit at the lowest energy level.
Question 4 True / False
At T=0, most fermions in an ideal Fermi gas occupy the single lowest-energy quantum state.
TTrue
FFalse
Answer: False
This is the key misconception to avoid. The Pauli exclusion principle forbids two identical fermions from occupying the same quantum state. At T=0, fermions fill states one by one from the ground state upward, forming the Fermi sea: every state with energy below E_F is exactly filled, every state above is exactly empty. The zero-temperature state is not a Bose-Einstein condensate — it is a fully filled band up to the Fermi energy, which is why the ground-state energy is large and nonzero.
Question 5 Short Answer
Why does the Fermi energy of a metal depend on electron number density but not on temperature?
Think about your answer, then reveal below.
Model answer: The Fermi energy is determined by counting: you fill available quantum states (each holding at most one electron per spin state, by the Pauli principle) until all N electrons are accommodated. This counting depends only on how many electrons there are per unit volume (the density n) and on the density of states g(ε), which is set by mass and the box geometry. Temperature controls how sharply the occupation function cuts off at E_F, but at T=0 the cutoff is a perfect step function and E_F is exactly the energy of the last filled state — a purely combinatorial, not thermal, quantity.
The classical ideal gas has energy that grows linearly with T because particles can explore all energies freely. In the Fermi gas, the Pauli principle forces a fixed 'stacking' of electrons into states regardless of temperature. As long as k_BT ≪ E_F (true for metals at all laboratory temperatures), the Fermi energy is essentially temperature-independent. This is why the electronic heat capacity of metals is far smaller than the classical prediction — most electrons are frozen deep in the Fermi sea and cannot absorb thermal energy.