My book says that ordinary dihydrogen contains 75% ortho and 25% para forms of hydrogen, while at significantly lower temperatures (like 20K) ortho and para hydrogens are 0.18% and 99.82% respectively. It also says that pure para form can be obtained at very low temperatures.

So what is the relation between the nuclear spins and temperature, and also why do the nuclear spin changes with temperature variance?


The effect is due to the symmetry properties of the rotational energy levels and those of the nuclear spin. The change with temperature is, as usual, traced back to the Boltzmann distribution.

In homonuclear diatomic molecules such as $\ce{H2}$ it is necessary to consider the spin of the overall wavefunction to exchange of nuclei. If this is $\Psi = \psi_t\psi_v\psi_R\psi_n$. The translational wavefunction $\psi_t$ depends on the coordinates of centre of mass, the vibration only on separation of nuclei. Thus we need only consider the product of rotational and nuclear wavefunctions, $\Psi = \psi_R\psi_n$. As there are identical nuclei the degeneracy is $g=2s_n+1$ with nuclear spin $s_n$ ($=1/2$) for hydrogen atoms.

As the product wavefunction $\Psi$ is neither odd nor even, it is necessary to ‘symmetrise’ by making the symmetric form in terms of states $\psi_1$ and $\psi_2$ for particle $(1)$ and $(2)$, as these also satisfy Schrodinger’s equation.


and asymmetric form


The number of symmetric functions is $g_n(g_n+1)/2$ and asymmetric $g_n(g_n-1)/2$ making a total of $g_n^2$.

The rotational wavefunctions are symmetric for even J rotational states and asymmteric for odd rotational quantum numbers J. As the product $\psi_R\psi_n$ must be asymmetrical for a molecule with an odd mass number for even J only asymmetrical nuclear states are accessible and and vice versa.

The partition function for odd mass number is $$Z= \frac{g_n(g_n-1)}{2}\sum _{J=0,2,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT)) +\\ \frac{g_n(g_n+1)}{2}\sum_{J=1,3,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT)) $$

and for $\ce{H2}$ the partition function is $$Z= 1\sum_{J=0,2,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT)) + 3\sum_{J=1,3,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT)) $$

From the equilibrium distribution equation the ratio of number of ortho- to para- molecules is therefore

$$\frac{N_{ortho}}{N_{para}} = \frac{ 3\sum_{J=1,3,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT)) } { \sum_{J=0,2,..} (2J+1)\exp(-B_JJ(J+1)/(k_BT))} $$

As the temperature changes so does this ratio; as the temperature tends to infinity the ratio tends to a limit of 3 and to zero as the temperature tends to zero.

For any spin degeneracy the ratio in the limit of very high temperature is $$\frac{N_{ortho}}{N_{para}}=\frac{g_n+1}{g_n-1}$$ and for $\ce{D2}$ with spin of $1$ and $g_n=3 $ the ratio is $2$.

The figure shows the ratio for hydrogen.



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