# How is the dynamic equilibrium nature of fermi-dirac distribution of particles facilitated?

I read this in Kittel: Introduction to Solid State Physics about deriving that product of electron and hole concentration as independent at a given temperature by the law of mass action.

For this model, it is assumed wherein the electron-hole pairs are generated using photons of black body radiation at rate $A(T)$, with $T$ being temperature, within the semiconductor and the recombination rate of pair at $B(T)np$. At equilibrium

$\frac{dn}{dt} = A(T) - B(T)np=0 \implies np = \frac{A(T)}{B(T)}$

Is this model a specific case of facilitating dynamic equilibrium nature of Fermi-Dirac distribution or the principle way of how fermions interact through photons?

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Probably a little late, but could you clarify this? I am looking at the discussion you reference (Ch.8 of Kittel, p.207 in my edition), and I don't quite get what you are asking. The basic argument in Kittel seems to be that the product of electron and hole concentrations doesn't depend on impurity concentration. Kittel shows this first by using the Fermi-Dirac statistics for electrons and holes (p.205-206). Then he shows that this result is plausible based off of simple kinetics - the product np only depends on rates of production and recombination of electron-hole pairs. – AJK May 8 '13 at 6:21

Photons are modeled as bosons with an integer spin, have a symmetry and can occupy the same quantum state.

All this means, is that they use the Bose-Einsten distribution instead. Where the Bose Einstein distribution gives the average number of Bosons found in an energy state, $\epsilon$.

$$\overline{n}=f_{-}{(\epsilon)}=\frac{1}{\exp(\beta(\epsilon-\mu)-1}$$

$$\mu=\left(\frac{\partial{F}}{\partial{N}}\right)_{T,V}\text{is the chemical potential.}$$

$$\beta = \frac{1}{k_{B}T}$$.

Boltzmann's constant, $k_{B}=1.38 \times 10^{-23}\mathrm{ m^2 kg s^{-2} K^{-1}}$

$T$ is the temperature in Kelvin.

Remember fermions $\frac{1}{2}$ spin and are not symmetrical. They also obey the pauli exclusion principle. Does that sound like photons to you? Their average number is given by the Fermi-Dirac.

$$\overline{n}=f_{+}{(\epsilon)}=\frac{1}{\exp(\beta(\epsilon-\mu)+1}$$

You may already know that stuff but I am guessing you are confused at the blackbody model. The model for blackbody radiation is a cavity.

In this situation energy is quantised such that $\epsilon = nh\upsilon$ with $h=6.64\times 10^{-34}\mathrm{m^2kgs^{-1}}$ which means that the mean number of particles in a given energy state is given by

$$\overline{n}=f_{-}{(\upsilon)}=\frac{1}{\exp(\beta(h\upsilon-\mu)-1}$$

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Hi Magpie, in this case I do really not see what the OP of the question is up to and I do not have access to book he mentions. At a first glance I see nothing wrong in your answer. I see no reason for a downvote and if somebody thinks it does not answer the question or something, he could just leave a comment saying this and write a better answer himself if he thinks he understands the question better... – Dilaton May 8 '13 at 21:04
+1 Who downvoted this? – centralcharge Jul 3 '13 at 9:59