How Is Entropy Calculated in Different Macrostates of Einstein Solids?

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SUMMARY

The entropy of the most likely macrostate for two Einstein solids, with NA = 300 and NB = 200, and a total energy of qtotal = 100, is calculated by distributing the energy according to the number of oscillators. The most likely distribution occurs when solid A has 60 units of energy and solid B has 40 units. It is essential to recognize that the number of oscillators affects the distribution of energy, leading to ΩA ≠ ΩB. Therefore, the assumption that both solids have equal microstates is incorrect due to the differing values of NA and NB.

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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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