Physics Problems & Solutions: Condensed Matter

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

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

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

Physics Problems & Solutions

Condensed Matter - Effective mass

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