What is the Multiplicity of a Macrostate in a Two-State Atomic System?

[kate31]

Homework Statement

Consider a hypothetical crystal A consisting of 5 atoms, each of which has only two states, with energies 0 and ε. A microstate is described by the distribution of energy among the individual atoms; a macrostate is described by the total energy of the crystal. For instance, in the microstates with configuration 00ε0ε, the 3rd and 5th atoms have energy ε, the remaining 3 atoms have energy 0, and the crystal is in the macrostate with energy EA = 2ε.

(i) How many microstates of the crystal have energy EA = 2ε? Write down the configurations of 2 of these microstates.
(ii) Explain what is meant by the multiplicity Ω of a macrostate, and confirm that the macrostate with energy 4ε has multiplicity given by
N!/(n!(N-n)!)
where N is the number of atoms in the system and n is the number of atoms excited to the energy level ε.

Homework Equations

for part i) (q+N-1)!/((q!)(N-1)!)=the number of microstates, where q is the units of energy to be distributed amongst N oscillators, and for part ii) the equation N!/(n!(N-n)!)

The Attempt at a Solution

By using these two equations I can get the required answer. The problem that I am having is that I thought that equation i) and equation ii) were basically the same thing. I don't understand the difference between them.. I know they are different because when I put q=2, N=5 into i) i get 15, and if I put n=2 and N=5 into ii) i get 10.
If someone could explain what situation I would use which equation for in general that would be good.

 

[Nate810]

For part i) I got that there are 10 microstates with energy EA = 2ε. The configurations of two of these microstates are ε00ε0 and 0εε00. For part ii) I found that the multiplicity of the macrostate with energy 4ε is 10, which is given by N!/(n!(N-n)!) (where N is the number of atoms in the system and n is the number of atoms excited to the energy level ε).