Most probable macrostates of two Einstein solids

[Yuriick]

Homework Statement

Consider two Einstein solid A and B with N_{A} = 200 and N_{B} = 100. Suppose now that the oscillations of solids A and B are different. The energies of oscillators in solid A are measured in units of ε_{A}, so that the energy of solid A is U_{A} = q_{A} ε_{A}. Likewise, for solid B, the energy is U_{B} = q_{B} \epsilon_{B}. Assume that ε_{A} = 2ε_{B}, and that the total energy is 100 ε_{A}. 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: \ln n! = n\ln n - n

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

The Attempt at a Solution

U_{Total} = U_{A} + U_{B} = q_{A} ε_{A} + q_{B} ε_{B} = 100 ε_{A}
or
q_{A} ε_{A} + 2 q_{B} ε_{A} = 100 ε_{A}
q_{A} + 2 q_{B} = 100

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

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

So the we can use (3),
Ω_{A}(q_{A})Ω_{B}(q_{B}) = (\frac{e q_{A}}{N_{A}})^{N_{A}} <br /> (\frac{e(50 - \frac{q_{A}}{2})}{\frac{N_{A}}{2}})^{\frac{N_{A}}{2}}<br /> <br />


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 Ω_{A} and Ω_{B}
Use

S = kln(Ω_{A}) +kln(Ω_{B})

Then we know that the most probable macrostate is when,

\frac{δS}{δq_{A}} = 0

So just solve for q_{A},then use it to solve for q_{B}.

I got
q_{A} = \frac{200}{3}
q_{B} = \frac{50}{3}