Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 33C / c / Solution

Let $x=\mathcal{E}/(k_{B}T)>0$ and $r=e^{-x}$. The one-particle internal partition function is $z=(1-r^q)/(1-r)$, and independence gives $Z=z^N$. The mean level index isThus the full internal entropy isThis describes the specified internal degrees of freedom, with no translational or mixing contribution added. For a positive level spacing, $S\to 0$ as $T\to 0$, since the unique lowest internal level becomes certain, while $S\to N{k}_B\log q$ as $T\to \infty$, since all $q^N$ internal configurations become equally likely. The formula includes $q=1$, where the entropy is identically zero.