 Solution

Set the ground electronic energy to zero and write the excitation energy as $\Delta =hc(120\mathrm{c}{\mathrm{m}}^{-1})$, or the characteristic temperature $\theta =\Delta /k_{\mathrm{B}}=172.65$ K. Counting each doubly degenerate manifold once gives the single-molecule electronic partition functionFor $N$ independent molecules the electronic factor is $q_{\mathrm{el}}^N$. Hence the electronic internal energy isThe equal degeneracies cancel from the excited-state probability. At low temperature the ground manifold dominates and $U_{\mathrm{el}}\to 0$; at high temperature both manifolds have equal populations, giving $U_{\mathrm{el}}\to N\Delta /2$. These are electronic contributions only, excluding translation, rotation and vibration.