Physics Problems & Solutions: Fermi Energy

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Thermal Physics - Conduction Electron

The Fermi temperature of Cu is about 80,000 K. Which of the following is most nearly equal to the average speed of a conduction electron in Cu?

A. 2 × 10⁻² m/s
B. 2 m/s
C. 2 × 10² m/s
D. 2 × 10⁴ m/s
E. 2 × 10⁶ m/s

(GR9277 #23)

Solution:

T_F = 8 × 10⁴ K
m_e = 9.11 × 10⁻³¹ kg
k = 1.38 × 10⁻²³ J/K

Fermi energy = Kinetic energy

$\begin{aligned} &\varepsilon _F = kT_F = \frac{1}{2}mv^2 \rightarrow v=\sqrt{\frac{2kT_F}{m}} \\&v=\sqrt{\frac{2\times 1.38\times 10^{-23}\times 8\times 10^4}{9.11 \times10^{-31} }} \\&v\approx \sqrt{2.76 \times 10^{12}}\approx 2 \times 10^6 \end{aligned}$

Answer: E

Thermal Physics - Conduction Electrons

The mean kinetic energy of electrons in metals at room temperature is usually many times the thermal energy kT. Which of the following can best be used to explain this fact?

A. The energy-time uncertainty relation
B. The Pauli exclusion principle
C. The degeneracy of the energy levels
D. The Born approximation
E. The wave-particle duality

(GR8677 #55)

Solution:

The mean kinetic energy of electrons in metals at room temperature is usually many times the thermal energy kT due to Pauli’s exclusion principle.

See problem GR0177 #76

Answer: B

Thermal Physics - Maxwell-Boltzmann Statistics

Consider a system of N non-interacting particles confined in a volume V at a temperature such that the particles obey classical Boltzmann statistics. If the temperature is lowered to the point at which quantum effects become important, the pressure of the gas may differ depending on whether the particles are fermions or bosons. Let P_F be the pressure exerted by the particles if they are fermions. P_B be the pressure if they are bosons and, and P_C be the pressure the particles would exert if quantum effects are ignored. Which of the following is true?

(GR8677 #88)

Solution:

Ideal gas: $PV=NkT \rightarrow P=\frac{N}{V}kT$

Classical (Maxwell-Boltzmann Statistics): $E=\frac{3}{2}kT$

$P=\frac{N}{V}kT\rightarrow P_C = \frac{2}{3} \frac{N}{V} \frac{3}{2} kT=\frac{2}{3} \frac{NE}{V}$

For Bosons (photon):

$P=\frac{N}{V}kT\rightarrow P_B = \frac{1}{3} \frac{N}{V} 3 kT=\frac{1}{3} \frac{NE}{V}$

→

For Fermion: $E=\frac{3}{5}E_F$
E_F

= Fermi Energy, $E_F \gg kT$

For the same temperature:

Answer: B

Thermal Physics – Conduction Electrons

The mean kinetic energy of the conduction electrons in metals is ordinarily much higher than kT because

A. electrons have many more degrees of freedom than atoms do
B. the electrons and the lattice are not in thermal equilibrium
C. the electrons form a degenerate Fermi gas
D. electrons in metals are highly relativistic
E. electron interact strongly with phonons

(GR0177 #76)

Solution:

The mean kinetic energy of electrons in metals at room temperature is usually many times the thermal energy kT due to Pauli’s exclusion principle.

According to classical physics, the mean thermal energy of the electrons is ³⁄₂kT.

According to quantum physics, the mean energy of electrons (fermion energy) is ³⁄₅E_F
where
$E_F= $ Fermi energy $ = \frac{\pi^2 \hbar^2}{2m_e} \left(\frac{3n}{\pi}\right)^{2/3}$
n = N/a³→ the number of electrons per unit volume

If E_F ≪ kT : electrons are hot (high T) → electrons are non-degenerate → classic (Maxwell-Boltzman) statistic.

In reality, E_F ≫ kT: electrons are cold (room T) → electrons are degenerate → quantum (Fermi-Dirac) statistic.

Answer: C

Thermal Physics - Fermion Energy in 3D and 2D

Show that:

Fermion Energy in 3D: E = (3/5) E_F

Fermion Energy 2D: E = (1/2) E_F

Solution:

click image to enlarge