# Fermi distribution

1. Nov 5, 2013

### aaaa202

My book notes, that at a given temperature the ratio of the total number of particles in the fermi gas to the total number lying within (ε-kT,ε+kT) is given by:
T/TF (1)
And that each of these particles has an energy of ≈kT (2).
I can't see where this comes from? :S Could anyone explain (1) and (2)?

2. Nov 23, 2013

### hokhani

As far as I know, to estimate the heat capacity of metals at T>0, there is a rule of thumb that considers deflection of particles' distribution from that of at T=0 very simple. You can imagine two symmetric triangles with the length equal to KT and width equal to 1/2 near the Fermi energy which is made by interceptions of the two Fermi distributions, one at T=0 and the other at T>0. At T>0 the particles which were in the left triangle (at T=0) would enter to the right triangle and their energy change is approximately KT(the reply to the part 2 of your question). Now if you calculate the ratio of number of all particles which is $\epsilon_f$ to the number of particles in the region [$\epsilon_f -kT, \epsilon_f +kT$](which is approximately the area under the triangle) you will get to part 1.

Last edited: Nov 23, 2013