Most probable macrostates of two Einstein solids

[Yuriick]

Homework Statement

Consider two Einstein solid A and B with [itex]N_{A} = 200[/itex] and [itex]N_{B} = 100[/itex]. Suppose now that the oscillations of solids A and B are different. The energies of oscillators in solid A are measured in units of [itex]ε_{A}[/itex], so that the energy of solid A is [itex]U_{A} = q_{A} ε_{A}[/itex]. Likewise, for solid B, the energy is [itex]U_{B} = q_{B} \epsilon_{B}[/itex]. Assume that [itex]ε_{A}[/itex] = 2[itex]ε_{B}[/itex], and that the total energy is 100 [itex]ε_{A}[/itex]. Compute most probable macrostate

(Hint: to do this, you should write multiplicity/entropy of each solid and will have to and if/where one needs to take into account the fact that the units of energy of the two solids are different).

Homework Equations

(1) Sterlings approximation: [itex]\ln n! = n\ln n - n [/itex]

(2) Multiplicity: [itex]Ω = \frac{(q+N-1)!}{(q)!(N-1)!} [/itex]
There another forumla that says
(3) [itex]Ω = (\frac{eq}{N})^{N} [/itex]

The Attempt at a Solution

[itex]U_{Total} = U_{A} + U_{B} = q_{A} ε_{A} + q_{B} ε_{B} = 100 ε_{A}[/itex]
or
[itex]q_{A} ε_{A} + 2 q_{B} ε_{A} = 100 ε_{A}[/itex]
[itex]q_{A} + 2 q_{B} = 100 [/itex]

This means that for each [itex]Ω_{A}(q_{A}) [/itex] microstates of energy [itex]q_{A}[/itex], there are [itex]Ω_{B}(50 -\frac{q_{A}}{2})[/itex] microstates accesible to B.

The total microstate for partition of energy [itex] q_{A}, q_{B} [/itex] is
[itex]Ω_{A}(q_{A})Ω_{B}(q_{B})[/itex]

So the we can use (3),
[itex]Ω_{A}(q_{A})Ω_{B}(q_{B}) = (\frac{e q_{A}}{N_{A}})^{N_{A}}
(\frac{e(50 - \frac{q_{A}}{2})}{\frac{N_{A}}{2}})^{\frac{N_{A}}{2}}

[/itex]


I have no idea if what I'm doing is right. Any insight would be really appreciated, thanks.

 

[Yuriick]

Figured it out. For anyone who finds this thread:
Once you have [itex]Ω_{A}[/itex] and [itex]Ω_{B}[/itex]
Use

[itex]S = kln(Ω_{A}) +kln(Ω_{B})[/itex]

Then we know that the most probable macrostate is when,

[itex] \frac{δS}{δq_{A}} = 0[/itex]

So just solve for [itex]q_{A}[/itex],then use it to solve for [itex]q_{B}[/itex].

I got
[itex]q_{A} = \frac{200}{3}[/itex]
[itex]q_{B} = \frac{50}{3}[/itex]