Discussion
A system of three indistinguishable particles has a total energy of \(4\varepsilon\). There are four single particle energy states with energies 0, \(\varepsilon\), \(2\varepsilon\) and \(3\varepsilon\). The number of microstates accessible to the system will be :
Answer: D
On the key, there are 3 accessible microstates.
For indistinguishable particles a microstate is fixed by how many particles occupy each level, not by which particle is where. With no limit on the number per level (Bose counting), we need three energies from {0, 1, 2, 3} in units of \(\varepsilon\) adding to 4:
\[ \begin{aligned} 4 &= 3 + 1 + 0 \\ 4 &= 2 + 2 + 0 \\ 4 &= 2 + 1 + 1 \end{aligned} \]
(4 + 0 + 0 is impossible because the highest level is \(3\varepsilon\).) That gives 3 microstates. One microstate would be the answer for fermions, which forbid the doubly occupied levels in the second and third sets. Four would come from counting an arrangement twice; distinguishable particles would give many more, since each set could be permuted.
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