Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fermi distribution

  1. Nov 5, 2013 #1
    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. jcsd
  3. Nov 23, 2013 #2
    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.
    Last edited: Nov 23, 2013
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook