Fermi distribution

by aaaa202
Tags: distribution, fermi
aaaa202 is online now
Nov5-13, 01:07 PM
P: 992
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)?
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hokhani is offline
Nov23-13, 08:08 AM
P: 241
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 [itex]\epsilon_f[/itex] to the number of particles in the region [[itex]\epsilon_f -kT, \epsilon_f +kT [/itex]](which is approximately the area under the triangle) you will get to part 1.

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