Suppose there is a semiconductor with Fermi energy $E_f$ and that there are $N$ bound electron states.(adsbygoogle = window.adsbygoogle || []).push({});

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

where [itex]\mu[/itex] 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 [tex]\bar n = {1\over \exp\beta(E-\mu)+1}[/tex]

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 [itex]N[/itex] bound [itex]e^-[/itex] states. I am not quite so confident as to why [tex]\mu\to E_f, E\to \mu[/tex]

Could someone please explain?

Many thanks.

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# Suppose there is a semiconductor with Fermi energy

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