Consider an ideal Bose gas consisting of molecules with internal degrees of
freedom. Assuming that, besides the ground state \(\varepsilon_{0}=0\), only the
first excited state \(\varepsilon_{1}\) of the internal spectrum needs to be
taken into account, determine the condensation temperature of the gas as a
function of \(\varepsilon_{1}\). Show that, for \(\left(\varepsilon_{1} / k
T_{c}^{0}\right) \gg 1\),
$$
\frac{T_{c}}{T_{c}^{0}} \simeq
1-\frac{\frac{2}{3}}{\zeta\left(\frac{3}{2}\right)} e^{-\pi_{1} / k T_{c}^{0}}
$$
while, for \(\left(\varepsilon_{1} / k T_{c}^{0}\right) \ll 1\),
$$
\frac{T_{c}}{T_{c}^{0}} \simeq\left(\frac{1}{2}\right)^{2 /
3}\left[1+\frac{2^{4 / 3}}{3 \zeta\left(\frac{3}{2}\right)}\left(\frac{\pi
\varepsilon_{1}}{k T_{c}^{0}}\right)^{1 / 2}\right]
$$