Difference between a diproton and H2

I was reading about helium recently and found that there is a theoretical isotope of helium known as a diproton. Apparently a diproton is a helium atom without any neutrons, meaning it is two protons and two electrons. The article said that a diproton is not possible because the strong force, that would hold the protons toghether, is not strong enough. When I picture the diproton in my head it looks very much like two atoms of hydrogen toghether, which I know exists as H2. What I want to know is what is the difference between the theoretical diproton and H2?

The difference would be how close they are, among other things.

A diproton would be two protons in a single nucleus - presumably with two electrons orbiting them further away, as in any other helium atom.

H2 is two separate atoms with two nuclei, each of which contains only a single proton. Again, there are two electrons, which are shared by both nuclei, but the protons aren’t any closer to each other than they are to the electrons.

Of course, uncertainty principle makes all of this approximate, but I think it’s a good simplification to begin with. Does it help?

On a basic level, the difference would be the distance between the protons. An H[sub]2[/sub] molecule has protons that are about 7 x 10[sup]-11[/sup] metres long, while a diproton (during the brief instant it would hypothetically exist) would have protons about 10[sup]-15[/sup] metres apart.

Another way to look at it is that a hydrogen molecule only exists because of the presence of the electrons around the hydrogen molecule; it’s the forces between the protons and the electrons that hold the molecule together, and if you were to remove the electrons from a hydrogen molecule it would fly apart. On the other hand, a hypothetical diproton doesn’t really care whether or not there are electrons surrounding it; it’s equally unstable either way.

H[sub]2[/sub] is a molecule with two atoms - it has two separate (single proton) nuclei, and the electrons move in a molecular bonding orbital defined by the interaction of those two atoms. H[sub]2[/sub] will participate in chemical reactions, by the breaking of the molecular bond or by electron loss (H[sup]+[/sup]).

Diproton will have a single nuclei, and the two electrons will form a full shell - it will be monatomic, and unreactive. Like He, only lighter. And it can’t exist due to the proton charge being higher than the strong nuclear force


Actually, I don’t think that the charge of the protons is what makes it unstable. IIRC, it’s due to a subtlety of the strong nuclear force itself, and how it interacts with various sorts of symmetry. For the same reason, two neutrons together wouldn’t work, either, but one proton and one neutron (a deuterium nucleus) does.

Charge can be relevant for larger nuclei, but that’s because the nucleons are, on average, further apart, and the electric force doesn’t fall off as rapidly with distance as the strong force does.

If it were possible to make a superatom of hydrogen, the boundary might be a little fuzzier–but still pretty clear.

A superatom is a (typically supercooled) cluster of atoms where some of the electrons orbit the entire cluster instead of individual nuclei. (For hydrogen, “some of the electrons” would have to mean “all of them”, there being only 1.) This causes the whole cluster to act in some ways (chemical bonding, paramagnetism, etc.) to act like a single, different atom.

Of course that’s not the same as it actually being a different atom. An Al7 cluster acts like a germanium atom, but it’s not a germanium atom, nor is it an isotope of proactinium (the element with 91 protons). The nuclei are at something on the order of molecular distances from each other, not nuclear distances.

Plus, as far as I know, hydrogen won’t cluster this way, and (as others have explained) there’s no He2 isotope to compare it to anyway.

I believe:

The strong nuclear force is stronger between nucleons with the same spin than between nucleons with opposite spins. The Pauli exclusion principle prevents two protons (or two neutrons) with the same spin from being in the same “ground” state in a nucleus so a diproton or dineutron has to have opposite spin protons or neutrons in it (if it’s in its ground state) and these are not bound as strongly as a neutron and a proton in an nucleus (heavy hydrogen) in which the spins are aligned.

I do not have a cite.

This is correct. There was some related discussion in this thread, but your post is a succinct statement of the effect.

Regarding the Coulombic repulsion in p-p itself: one can get a sense of the magnitude of its importance by comparing the binding energies of [sup]3[/sup]H and [sup]3[/sup]He, which differ only by the conversion of a neutron in [sup]3[/sup]H into a proton. (Importantly, though, the nuclear structures match – one n and two p vs. one p and two n – which keeps spin-dependent issues from entering.)

Anyway, the binding energy of [sup]3[/sup]H is only 9% lower (736 keV) than that of [sup]3[/sup]He (data from here). So, the proton repulsion is indeed more of a perturbation in these low-nucleon-number situations.

We can take this further. Consider the binding energy of deuterium, [sup]2[/sup]H, from the nuclide table above: 2225 keV. If the only relevant unstabilizing force in p-p were electrostatic repulsion, we could determine if p-p were bound simply by noting that:

(1) The protons in [sup]3[/sup]He won’t be a whole lot farther apart than the would be in p-p. Geometrically, their centers-of-charge could be at most twice as far apart, but that would only be if [sup]3[/sup]He was arranged in a line – p-n-p – which it is not. Even in this extreme case, electrostatic repulsion would only cause twice the loss in binding energy going from n-p to p-p compared with what is seen going from [sup]3[/sup]H to [sup]3[/sup]He: 2*736 keV = 1472 keV. In reality, the energy cost likely differs by only a modest amount, so I’ll just move up to the next round number: 800 keV.

(2) Since n-p has a binding energy of 2225 keV, and since (2225 keV - 800 keV) > 0, then p-p should be bound if n-p is. But p-p isn’t bound, so this must be due to something other than electrostatic repulsion.

And while we’re at it, note that He-3 is completely stable, while H-3 has a halflife of a couple decades (which is admittedly a very long time, in particle terms). So even with the extra electrostatic repulsion, the helium actually turns out to be more stable.

Now that really is weird and I’d never thought about it. It’s been too long since I took physics, but I thought the strong force treated neutrons and protons identically. Presumably H-3 has an up and down neutron and either spin proton, while He-3 has an up and down proton and either spin neutron. These should be bound identically by the strong force, I’d have thought, though that’s clearly wrong.

They are treated identically by the strong force, or tritium wouldn’t last anywhere remotely near that long. Tritium decay is a weak-force process.

To be pedantic, there is a difference between the quarks based on mass in the strong force, but the masses of the up and down quarks are both so much smaller than the binding energies that they’re almost irrelevant. It does become somewhat relevant for strange hadrons, though, and very relevant for charmed, truthful, and beautiful ones.

Isn’t that mass difference the reason the neutron decays into a proton, and not vice-versa? And presumably why [sup]3[/sup]He is stable and [sup]3[/sup]H not?

I should have been clearer: The mass difference is relevant for both the strong and weak forces, but the strong force difference isn’t relevant here. The mass difference is of course relevant for the simple reason that nothing can decay into something heavier than it.