Nuclear spin coupling in molecular (protium) hydrogen

[hkyriazi]

I'm not sure where this post belongs--here, or nuclear chemistry, quantum mechanics, NMR spectroscopy, etc. Moderator--please feel free to move it to a better location.
I'm wondering if a container of liquid hydrogen subjected to a strong magnetic field would have both nuclei of each atom pointed in the same direction, or whether there's some sort of "spin coupling" of those (protium--let's ignore deuterium and tritium) nuclei that would prevent such an alignment.
Also, I'm wondering how tight would be any amount of alignment if the liquid were kept at close to absolute zero (how small a precession angle of the nuclear spins?), and what percentage of molecules would be oriented in this way.

 

[snorkack]

I'm not sure where this post belongs--here, or nuclear chemistry, quantum mechanics, NMR spectroscopy, etc. Moderator--please feel free to move it to a better location.
I'm wondering if a container of liquid hydrogen subjected to a strong magnetic field would have both nuclei of each atom pointed in the same direction, or whether there's some sort of "spin coupling" of those (protium--let's ignore deuterium and tritium) nuclei that would prevent such an alignment.

There is a spin coupling that would resist this.
Two protium nuclei in the same molecule, same spin and not having an orbital motion relative to each other cannot coexist because they are indistinguishable fermions.
You can get around this if the molecule has an odd orbital momentum - at least 1. But this needs energy.
At high temperatures, orthohydrogen is the prevalent form of diprotium, in the ratio of 3:1. Because there are 3 different states for hydrogen having spin 1 - spin projections +1, 0 and -1, while spin 0 has only the lone state of spin projection 0.
At low temperatures, the lower energy of parahydrogen (no orbital motion) prevails.
The changeover to prevalence of parahydrogen is around 100 K - at equilibrium. The equilibrium is not quick to establish.
Now suppose that hydrogen is in a strong magnetic field.Then the 3 spin projections of orthohydrogen are no longer equal energy. Like, +1 gets lower and -1 gets higher. There is still the energy of orbital movement. But at some strength of external magnetic field, the +1 spin projection state should lie below the state of spin 0. How strong field is this?

Also, I'm wondering how tight would be any amount of alignment if the liquid were kept at close to absolute zero (how small a precession angle of the nuclear spins?), and what percentage of molecules would be oriented in this way.

Dihydrogen freezes at low temperatures. The freezing point is affected a lot by isotopic composition and a bit by molecular spin. It should be affected a bit if the spins are aligned with external field, but I do not see why dihydrogen should stay liquid at absolute zero like helium does.

 

[hkyriazi]

Thank you. I didn't know there was such a thing as solid hydrogen! Interesting! I wonder, then, if we placed the liquid, at, say, 15 deg. Kelvin (slightly above its melting temperature), and subjected it to the strongest magnetic field we can make (a Japanese team was reported to have generated a field of 1,200 Tesla in Sept. of 2018), could we get the molecules all oriented in the same direction, and then freeze them into place in that orientation? Almost certainly not, as the crystal packing would have its own orientation requirements. I'm assuming that the molecular/electron orbital coupling of the two protium atoms itself has already "frozen" their nuclei (protons) in some set mutual orientation, and that the molecules (in the liquid state) might be oriented with respect to the magnetic field.

Regarding magnetic field strength, it seems the very strong fields are over much too small a volume for what I have in mind. Would fields of typical MRI and NMR machine strength be capable of orienting dihydrogen?

I also didn't know about ortho- (spins aligned) and parahydrogen (spins anti-parallel). It seems that as the temp is lowered to ~15 deg Kelvin, the percentage of ortho should fall to zero. I wonder if the strong magnetic field could render all of it ortho, along with orienting all of the molecules?

 

[snorkack]

Nuclear magneton is 5,05*10^-27 J/T.
The dipole moment of a proton is 2,8 magnetons, and orthohydrogen 5,6 magnetons - making 2,8*10^-26 J/T
The energy of orthohydrogen over parahydrogen is quoted as 1,06*10³ J/mol. With Avogadro number being 6,02*10²³, this would make the energy of orthohydrogen about 1,75*10^-21 J above parahydrogen.
Which would suggest that magnetic field should favour the aligned state of orthohydrogen over parahydrogen from field strengths a bit over 60 000 T.
Nonlinear effects might happen at these magnitudes. In any case, can anyone check my sources for gross errors of magnitude?

 

[hkyriazi]

...Which would suggest that magnetic field should favour the aligned state of orthohydrogen over parahydrogen from field strengths a bit over 60 000 T.

Yikes! If that means what I think it means--that if your calculation is correct, it'd require a field about 15 thousand times the strength of those in MRI machines to get close to 100% orthohydrogen, perhaps the easier route would be to use liquid H2 at its highest convenient temperature--perhaps 77K (liquid N2 temp)--which would, for a while at least, consist of 75% orthohydrogen (the percentage present at RT).

Any thoughts on the overall magnetic moment of the orthohydrogen molecule, and the ability of currently available field strengths to align these molecules?

 

[snorkack]

Yikes! If that means what I think it means--that if your calculation is correct, it'd require a field about 15 thousand times the strength of those in MRI machines to get close to 100% orthohydrogen, perhaps the easier route would be to use liquid H2 at its highest convenient temperature--perhaps 77K (liquid N2 temp)

Diprotium critical temperature is 33 K.

--which would, for a while at least, consist of 75% orthohydrogen (the percentage present at RT).

Any thoughts on the overall magnetic moment of the orthohydrogen molecule, and the ability of currently available field strengths to align these molecules?

If they hold water then use the numbers above.
No, there would not be much orientation requirements of crystal.
Both parahydrogen and orthohydrogen are bosons (spin 0 or spin 1 - integer either way). Parahydrogen is 1 type of indistinguishable bosons, 1 state (spin 0). Ordinary hydrogen in 2 types, 4 states (spin 0, and spin 1 states projection 1, 0 and -1). Melting points do differ... parahydrogen melts at 13,88 K, ordinary hydrogen at 13,93 K.
So, if the spins were aligned with external field, 25 % spin 0, 75 % spin 1 projection +1, it would behave much like ordinary solid hydrogen (except that slow spontaneous spin-flip would be releasing heat).
At 1200 T, I get energy 3,35*10^-23 J. Which is about 20 J/mol.
Apply R (8,314 J/K*mol), and below about 2,4 K, the magnetic field of 1200 T should favour spin alignment. How fast do orthohydrogen molecules align with external field, compared to decay to parahydrogen?