Module 6 : PHYSICS OF SEMICONDUCTOR DEVICES
Lecture 31 : Electron in a Periodic Potential
  Example 1
  The density of free electrons in silver is $ 6\times 10^{28}$ per cubic meter. Calculate the Fermi energy of silver in eV.
  Solution
  The Fermi momentum is given by
 
$\displaystyle k_F = (3\pi^2n)^{1/3} = 1.21\times 10^{10}\ \ {\rm m}^{-1}$
  The Fermi energy
 
$\displaystyle E_F= \frac{\hbar^2k_F^2}{2m} = 9.053\times 10^{-19}\ \ {\rm J}$
  Dividing by electronic charge, the Fermi energy is 5.65 eV.
 
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