Fermions that can access 10 distinct energy states; Statistical Physics

[matt_crouch]

Homework Statement

Consider a system made of 4 quantum fermions that can access 10 distinct states respectively with energies:

E_n=n/10 eV with n=1,2,3,4,5,6,7,8,9,10

1) Write the expression for the entropy when the particles can access all states with equal probability

2) Compute the Entropy of the isolated system at energy U =1 eV

3) Compute the entropy of the isolated system at energy 1.1 eV

Homework Equations

Ω=G!/m!(G-m)!
s=k_Bln(Ω)

The Attempt at a Solution

the first question i think is answered basically by the first equation i gave for the statistical weight because that is for indistinguishable particles with multiple occupancy not allowed. I am a little bit stuck on the 2nd and 3rd questions. The probability of finding a particle in the lowest state must be more probable than finding a particle in the highest state but the equation for the statistical weight won't take that into account. if i can be pointed in the right direction that would be awesome

 

[clamtrox]

the first question i think is answered basically by the first equation i gave for the statistical weight because that is for indistinguishable particles with multiple occupancy not allowed. I am a little bit stuck on the 2nd and 3rd questions. The probability of finding a particle in the lowest state must be more probable than finding a particle in the highest state but the equation for the statistical weight won't take that into account. if i can be pointed in the right direction that would be awesome

The physical way of looking at entropy is that it's the logarithm of the number of states corresponding to a given energy. For example, when the total energy is fixed to 1eV, there's only one possible arrangement to get this: 1/10eV + 2/10eV + 3/10eV + 4/10eV = 1eV. Now it becomes just a combinatorics problem, and you only need to figure out how many such configurations there are.

 

[matt_crouch]

ok so because there is only 1 possible arrangement for u=1ev the statistical weight can be calculated used the equation above so

Ω= 4!/4!(1)!
Ω=1

so s=kln(1)

then for the u=1.1 ev so the only possible state will be when 3/10 + 5/10 + 2/10 + 1/10 = 1.1 ev

so again Ω=1

s=kln (1)

 

[clamtrox]

Oops, I forgot about spin. If your fermions have spin 1/2, you can have two of them occupying a state with same n. Maybe you should at least mention this, if not calculate it completely.