Bose Einstein Statistics Questions and Answers

Answer: B

The specific heat obeys \(C_{V} \propto T^{2}\).

In two dimensions the number of photon modes with frequency below \(\nu\) grows as the area of a disc in k-space, \(\propto \nu^{2}\), so the density of states is \(g_{2d}(\nu) \propto \nu\). Substituting \(x = h\nu/k_{B}T\):

\[ \begin{aligned} E &\propto \int_{0}^{\infty} \frac{\nu \cdot h\nu}{e^{h\nu/k_{B}T} - 1}\,d\nu \\ &\propto T^{3}\int_{0}^{\infty} \frac{x^{2}}{e^{x} - 1}\,dx \propto T^{3} \\ C_{V} &= \frac{dE}{dT} \propto T^{2} \end{aligned} \]

\(T^{3}\) is the three-dimensional result (from \(E \propto T^{4}\), the Stefan-Boltzmann law), and \(T^{4}\) confuses the 3D energy with its specific heat. \(C_{V} \propto T\) would be the one-dimensional photon gas.

Answer: A

The partition function is \( Z = e^{-4\beta\epsilon} + e^{-3\beta\epsilon} + e^{-2\beta\epsilon} \).

For indistinguishable bosons a state is fixed by how many particles are in each level, and any number may share a level. With two particles and two levels there are three states:

• both in \( \epsilon \): total energy \( 2\epsilon \);
• one in each level: \( 3\epsilon \), counted once, since the particles cannot be told apart;
• both in \( 2\epsilon \): \( 4\epsilon \).

Summing \( e^{-\beta E} \) over these three states gives the keyed result. The other options keep a single Boltzmann factor, as if the system had only one state.