Physics Problems & Solutions: Conduction Electrons

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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 – 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

Condensed Matter - Effective mass

Lattice forces affect the motion of electrons in a metallic crystal, so that the relationship between the energy E and wave number k is not the classical equation E = ħ²k²/2m, where m is the electron mass. Instead, it is possible to use an effective mass m* given by which of the following?

A. $m^*=\frac{1}{2}\hbar^2 k \left( \frac{dk}{dE} \right)$

B. $m^*= \frac{\hbar^2 k^2}{\left( \frac{dk}{dE} \right)}$

C. $m^*= \frac{1}{2} \hbar^2 m^2 \left( \frac{d^2k}{dE^2} \right)^{1/3}$

D. $m^*= \frac{\hbar^2}{\left( \frac{d^2E}{dk^2} \right)}$

E. $m^*=\frac{1}{2} \hbar^2 m^2 \left( \frac{d^2E}{dk^2} \right)$

(GR9277 #97)

Solution:

F = m*a

F = dp/dt = ħ dk/dt (momentum, p = ħk)

a = dv_g/dt

Group velocity, v_g= dω/dk

Energy of electron, E = hf = ħω
ω = E/ħ

$v_g=\frac{d\omega}{dk} = \frac{1}{\hbar} \left( \frac{dE}{dk} \right)$

$a=\frac{dv_g}{dt}= \frac{1}{\hbar} \frac{d}{dt}\left( \frac{dE}{dk} \right)\frac{dk}{dk}= \frac{1}{\hbar} \left( \frac{d^2E}{dk^2} \right)\frac{dk}{dt}$

F = m*a

$\hbar \frac{dk}{dt} =m^*\cdot \frac{1}{\hbar} \left( \frac{d^2E}{dk^2} \right)\frac{dk}{dt}$

$m^*= \frac{\hbar^2}{\left( \frac{d^2E}{dk^2} \right)}$