What Are the Calculations Behind Electron-Phonon Scattering in Copper?

  • Thread starter Thread starter unscientific
  • Start date Start date
  • Tags Tags
    Debye Scattering
unscientific
Messages
1,728
Reaction score
13

Homework Statement

(a) Find fermi temperature and debye temperature. Calculate them for copper.
(b) Show the scattering wave relation
(c) What does ##\lambda## mean?

2014_B6_Q2.png

Homework Equations

The Attempt at a Solution



Part(a)
The fermi temperature and debye temperature is given by:
T_F = \frac{\hbar^2 (3n \pi^2)^{\frac{2}{3}}}{2m_e k_B}
\theta_D = \hbar (6 \pi^2 n)^{\frac{1}{3}} \frac{c}{k_B}

For copper: ##a = 3.5 \times 10^{-10} m##, ##\theta_D = 231 K##, ##\T_F = 5.5 \times 10^4 K##.

Part(b)
k^{'} = (1-\delta)k_F
E^{'} = (1-\delta)^2E_F

I suppose the phonon gains energy by scattering, so ##E_{ph} = \Delta E = E^{'} - E_F##.
E_{ph}= E^{'} - E_F = E_F \left( 1 - (1-\delta)^2 \right)
k_{ph} = \left(1 - (1-\delta)^2 \right)^{\frac{1}{2}} k_F
k_{ph} \approx \left( 1 - \frac{1}{2} (1-\delta)^2 \right) k_F
\frac{k_{ph}}{k_F} \approx \frac{1}{2}(1 + 2\delta)

Substituting in, LHS
= \frac{1}{2} \frac{1 + 2\delta}{2\delta} \frac{1}{k_F}
= \frac{1}{2}(1 + \frac{1}{2\delta}) \frac{1}{k_F}
\approx \frac{1}{4\delta k_F}

How is this ##\approx \lambda##?

Part(c)
Not sure what this "wavelength" means.
 
Physics news on Phys.org
Would appreciate any help on this problem, many thanks in advance!
 
bumpp
 
bumpp
 
Ok, I got this question done. Key is to use the Bragg Condition: ##\vec k^{'} + \vec k_{ph} = \vec k + \vec G##.
 
Back
Top