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As mentioned in a previous question, the number of electrons in conduction band in a semiconductor can be computed as follows:

$$N = \int_{E_c}^{+\infty} g_c(E)f(E)dE$$

where $g_c(E)$ is the density of states of electrons in the conduction band and $f(E)$ is the Fermi-Dirac distribution.

But how can an actual number of electrons be computed from a probability function $f(E)$? The Fermi-Dirac distribution gives only the probability that a state with energy $E$ is occupied, it doesn't give the number of electrons occupying that state. So, why is it used anyway?

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$$g(E)=\text{number of states at energy E available to be occupied}$$

$$f(E)=\text{probability that a state with energy E is occupied}$$

so that

$$g(E) \ f(E) = \text{average number of occupied states with energy E} \\ =\text{average number of particles with energy E} = N(E)$$

So that the total number of particles will be given by

$$N=\int N(E) \ d E$$

To understand better why $$g(E) \ f(E)$$ is the average number of occupied states with energy E, think about this example: if I have a group of $g$ people and the probability that a person has blonde hair is $f$, what will be the average number of blonde people?

Answer: $(g \cdot f)$

In this case it is the same: if I have $g$ states and the probability that a state is occupied is $f$, what will be the average number of occupied states? $(g \cdot f)$

(Or maybe your doubt arose because you didn't understand that that was an average number?)

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  • $\begingroup$ First and foremost, thank you. My doubt arose because very often $N$ is said to be not the average number of electrons in the conduction band, but the actual, effective number of electrons. Speaking about your example: I don't actually know how many blonde people are there, but if the probability of having a blonde person is $f$, then on average I'll have $g \cdot f$ blonde people. So $g \cdot f$ is not an exact number, it is an average number. Is it correct? $\endgroup$ – BowPark Jun 29 '16 at 9:07
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    $\begingroup$ @BowPark You're welcome. Yes, it is an average number (en.wikipedia.org/wiki/…) , and I explicitly wrote it. But notice that it will become close to the actual number in the limit in which the number of states is very large. $\endgroup$ – valerio Jun 29 '16 at 9:12

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