# Suppose there is a semiconductor with Fermi energy

1. Mar 28, 2013

### c299792458

Suppose there is a semiconductor with Fermi energy $E_f$ and that there are $N$ bound electron states.

I'd like to know why the mean number of excited electrons takes the form $$\bar n(T)={N\over \exp\beta(\mu-E_f)+1}$$

where $\mu$ is the chemical potential.

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I can see that the Fermi Dirac statistics say that for one fermion the mean occupation number is $$\bar n = {1\over \exp\beta(E-\mu)+1}$$
I am not sure however as to how for the semi-conductor it should assume the above form. The N is clearly due to the $N$ bound $e^-$ states. I am not quite so confident as to why $$\mu\to E_f, E\to \mu$$

Could someone please explain?

Many thanks.

2. Mar 30, 2013

### Giga_Man

i think by E, you mean the top of the valence band?
Every electron that is not bound is excited (or makes a "hole" in the valence band and an electron in the conduction band).
There are N bound states, so the number of excited electron is N* (1- fermi_dirac_dist(E))
mathematically, this is the the same as changing the palces of Ef and E, try it.
By the way, using E as the top of the valence band is an approximation which doesn't work if, for example, the semiconductor is very heavily p-type.

Last edited: Mar 30, 2013
3. Apr 2, 2013

### c299792458

Thank you, Giga_Man!

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