Register to reply 
Fermi distribution 
Share this thread: 
#1
Nov513, 01:07 PM

P: 1,005

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 (εkT,ε+kT) is given by:
T/T_{F} (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
Nov2313, 08:08 AM

P: 265

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.



Register to reply 
Related Discussions  
Fermi distribution  Atomic, Solid State, Comp. Physics  1  
Fermi distribution  Quantum Physics  2  
Fermi Dirac Distribution  Quantum Physics  0  
FermiDirac Distribution  Advanced Physics Homework  3  
FermiDirac Distribution  Quantum Physics  1 