Ratio of Ortho and Para Hydrogen This question is actually 29.5 of Blundell and Blundell.
The question says ortho-hydrogen has spin $S=1$ (so degeneracy of 3) and total angular momentum $J(J+1)\hbar^2$ where $J$ is odd (degeneracy of $2J+1$).  And then para-hydrogen has even $J$ and $S=0$ (no degeneracy from spin). Show that the ratio of ortho to parahydrogen is:
$$
f=3\frac{\sum_{J=1,3..}(2J+1)e^{-J(J+1)\hbar^2/2Ik_BT}}{\sum_{J=0,2..}(2J+1)e^{-J(J+1)\hbar^2/2Ik_BT}}
$$ 
We have Bose-Einstein distribution function:
$$
F(E)=\frac{\text{degeneracy}}{e^{\beta(E-\mu)}-1}
$$
where $E=J(J+1)\hbar^2/2Ik_BT$  Since the translation part of the energy is the same for both we can ignore it and just include rotation energy.
so surely, 
$$
f=\frac{F_{ortho}}{F_{para}}=\frac{3\sum_{J=1,3,...}\frac{\text{2J+1}}{e^{\beta(E-\mu)}-1}}{\sum_{J=0,2,...}\frac{\text{2J+1}}{e^{\beta(E-\mu)}-1}}
$$  
With the relevant expression for $E$ above.  The answer comes easily if I assume that the $-1$ is negligible, however I cannot justify why this is the case.
 A: The Bose-Einstein and Fermi-Dirac distribution functions,
$$ \left<n\right> = 
\frac1{\pm1 + \exp \beta(E-\mu)},
$$
tell you about the mean occupation number of any particular state.
But you're not interested in, say, how many orthohydrogens are in the $J=5$ state; you're interested in how many hydrogen molecules will be orthohydrogen.  To find this you have to go back to the partition function.  The probabilities that a given molecule is either orthohydrogen or parahydrogen must obey
\begin{align}
P_\text{ortho} &\propto \sum_\text{ortho states}  d\ e^{-\beta E}
\\
P_\text{para} &\propto \sum_\text{para states}  d \ e^{-\beta E}
,
\end{align}
where $d$ is the degeneracy of each state and $e^{-\beta E}$ is the usual Boltzmann factor.
If those are the only two options, the partition function $Z$ is their sum --- but it has to be the same partition function in any case, whether you include other physics (like chemical reactions that dissociate or remove the hydrogen molecules) or not.
Note that the Bose-Einstein and Fermi-Dirac distributions are derived from the partition function, rather than the other way around.  If you're working from Blunden and Blunden, the derivation is in that same chapter.
For physical hydrogen at a temperature of 3 kelvin, the thermal wavelength of the molecules is
\begin{align}
\lambda_\text{thermal} &= \frac h{p_\text{thermal}} 
\approx \frac{hc}{\sqrt{2 mc^2 \frac 32 k_B T}}
% \\ &= \frac{1240\rm\,eV\,nm}{\sqrt{3\cdot 1\rm\,GeV \cdot 0.25\rm\,meV}}
\approx 1.5\rm\,pm
\end{align}
This is lots smaller than the intermolecular spacing in liquid or solid hydrogen, so there's not really a reason to expect the difference between classical and quantum statistics to be very important.
