How Is Entropy Calculated in Different Macrostates of Einstein Solids?

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The discussion focuses on calculating the entropy of two Einstein solids, with specific values for the number of oscillators and total energy. The most likely macrostate is identified as having solid A with 60 units of energy and solid B with 40 units, rather than an equal distribution. It is emphasized that the number of oscillators affects the energy distribution, meaning ΩA does not equal ΩB due to differing values of NA and NB. The relationship between energy distribution and the number of oscillators is crucial for determining the entropy of each macrostate. Understanding these calculations is essential for accurately assessing the entropy in different macrostates of Einstein solids.
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This is the problem:
Consider a system of two Einstein solids, with NA = 300, NB = 200, and qtotal = 100. Compute the entropy of the most likely macrostate and of the least likely macro state.

I only have a doubt. Is the most likely macro state when each solid has half the energy (in this case qA=qB=50 units of energy)? Or does NA get a larger amount of energy for having a larger number of oscillators (In which case I assume that ΩA = ΩB)?
 
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The most likely macrostate will be when the energy is uniformly distributed over all the oscillators of each solid.

If there are 500 oscillators in total and 100 units of energy then each oscillator will have on average a fifth of a unit of energy.

In the most likely macrostate, solid A would have 60 units of energy, and solid B would have 40 units of energy.

You can't assume that \Omega_{A}=\Omega_{B} since N_{A}\neq N_{B} and q_{A}\neq q_{B}.

Knowing what q_{A} and q_{B} are for the most likely macrostate, and knowing N_{A} and N_{B}, you can figure out \Omega_{A} and \Omega_{B}
 
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