Question

Show how the average energy in a Fermi gas can be expressed in terms of Ef0 

Answer 1

When we have the 1-D box, then the energy of the electron in the box can be represents as below

$E_n=E_0+\frac{ℏ^2π^2}{2mL^2}n^2$

We know that Fermi level is the energy of the last electron, if we have 2N electron then we can find the mean energy of the electron with the given equation,

$E_{av}=E_0+\frac{2\Sigma_1^NE_n}{2N}$

$\Rightarrow E_{av}=E_0+\frac{2\Sigma_1^N E_o +\frac{ℏ^2π^2}{2mL^2}n^2}{2N}$

we know that $\Sigma n^2=\frac{N(N+1)(2N+1)}{6}=\frac{N^3}{3}+\frac{N^2}{2}+\frac{N}{6}$

If we consider N is large value, then we can ignore the second and third terms.

If we assume that there is lots of electron, then $E_{av}=E_o+\frac{ℏ^2π^2 N^3}{2mL^23N}$

But, Fermi level $E_f=E_o+\frac{ℏ^2π^2 }{2mL^2}N^2$

But if $E_o=0$

$E_{av}=\frac{E_f}{3}$