27
$\begingroup$

In the hearts of stars, hydrogen atoms fuse together to make helium. After the hydrogen in the core is depleted, the star changes state and conditions at the heart of the star make it possible for helium atoms to fuse together.

There are parts of a star where hydrogen and helium are in contact which makes me wonder why there isn't any fusion going on between the two.

Can hydrogen and helium fuse together? If so, under what conditions? If not, why not?

$\endgroup$
6

2 Answers 2

34
$\begingroup$

Hydrogen and helium can briefly bind together to make lithium-5, but this is an extremely unstable nuclide which falls apart instantly (with a half-life of ${\sim}4\times 10^{-22}\:\rm s$) and which actively requires energy to make (i.e. it is an endothermic process, as opposed to how we normally think of nuclear fusion).

The reason for this is that helium-4 is a particularly stable system, and it has a huge binding energy $-$ much bigger than anything immediately higher up in size. In lithium-5, you have three protons, which you can think of as two of them paired up and one lone guy in a nuclear shell of its own at much higher energy. This energy is so high that it's simpler for the extra proton to just peel off and go away to become a separate hydrogen nucleus.

To make stable lithium, you need more neutrons to stabilize the nuclide, so only lithium-6 and lithium-7 are stable.


This raises the question of whether it might be possible to combine suitable isotopes to make those, for which the only candidates are \begin{align} \rm ^2H+{}^4He\to{}^6Li, \\ \rm ^3H+{}^4He\to{}^7Li, \\ \rm ^3H+{}^3He\to{}^6Li. \end{align} From these:

  • The first reaction does happen, and e.g. this paper calls it "radiative capture of deuterium on alpha particles". But it is extremely unlikely and it only produced trace amounts of lithium-6 (w.r.t. lithium-7 production) in Big-Bang nucleosynthesis. (And, in addition, deuterium is not stable in stellar cores.)
  • The second one does happen and it does produce energy. However, it is unlikely in stellar nucleosynthesis since it requires tritium, which is unstable.
  • The third reaction can also happen (studied e.g. in this paper) but again it is extremely unlikely, and it requires tritium, which is unstable.

For what it's worth, these reactions are exothermic, releasing 1.5, 2.4 and 934 MeV of energy, respectively, so they are allowed to happen on their own without needing to supply initial energy to the reactants for them to fuse.

In other words, the higher isotopes of hydrogen do have an open channel of fusing with helium to produce lithium. However, these channels are so suppressed, due to the details of how likely the reactions are to happen, that they are negligible in stellar nucleosynthesis.


And, finally, there's an even bigger problem, known as lithium burning: if you just release a nucleus of lithium (either the -6 or -7 isotopes) into a stellar core, the star will just tend to eat it raw:

  • Lithium-7 can fuse with hydrogen to make beryllium-8, which promptly breaks in half to give two helium-4 nuclei. Again, this is a consequence of the extreme stability of alpha particles compared to any of its neighbours in the table of nuclides.
  • Lithium-6 can fuse with hydrogen to make beryllium-7, which decays via electron capture to lithium-7. The resulting lithium-7 will then end up catching another proton, as above.

The net result of this mechanism is that developed stars have less lithium than the primordial soup they started out with.

$\endgroup$
2
  • 1
    $\begingroup$ In other words, the most relevant hydrogen-helium reaction amounts to $\mathrm{{}^3H + {}^1H + {}^4He \to {}^4He + {}^4 He}$, where the Helium is both a product and catalyst, instead of being consumed. $\endgroup$ Commented Aug 25, 2021 at 12:58
  • $\begingroup$ @leftaround That's a valid way to understand the reactions in this post. But it's not the most relevant hydrogen-helium reaction - there are other processes which are much more common, starting with the p-p chain. The reactions in this post are just those with naive fusion between one hydrogen and one helium (which must produce lithium). $\endgroup$ Commented Aug 25, 2021 at 13:05
4
$\begingroup$

One problem is the conservation of baryon number, that is the number of nucleons before and after reaction needs to be the same. They shouldn't change protons into neutrons or vice versa (unless there's no other way), because it would require weak interaction - and if there's a way to react without relying on weak interaction, the nuclei will usually choose this way. Another is the conservation of energy, and momentum: for both of them to be conserved, there has to be another particle besides the new nucleus created in the reaction so that both energy and momentum be conserved.

The fusion of hydrogen-1 and helium-4 would need to have 5 nucleons, but no stable nucleus of that size 5 exists. Helium-5 has the halflife of $7×10^{−22}$ s and lithium-5 has the halflife of $3.7×10^{−22}$ s. That's too short for them to take part in another reaction and possibly create a heavier, stable nucleus before it decays. That's why, when hydrogen-2 reacts with helium-3 they only create helium-4 and a proton.

If hydrogen-2 and helium-4 were to fuse into lithium-6, in this reaction no nucleons would need to change from proton to neutron or vice versa, and with no such change, there would be no side products (like electrons and neutrinos) that are necessary to balance the energy and momentum. It is possible for it to happen with the emmission of a photon, but this means involving electromagnetic interaction in the reraction, which is much weaker than the strong interaction, and that makes the reaction much less likely to occur (and because of that less effective in producing energy). The fusion of hydrogen-3 and helium-3 to lithium-6, or hydrogen-3 and helium-4 to lithium-7, has the same problem.

It seems it could be possible a reaction in which hydrogen-3 and helium-4 create lithium-6 and neutron, as it doesn't violate any rules. However, this reaction uses up energy instead of producing it (the product weighs more than the reagents) and as such, it can't fuel the burning of a star, especially since tritium-3 is a rare isotope.

To sum up: the fusion reaction between helium and hydrogen may

  1. happen without the creation of new elements, or
  2. are not an effective energy source.
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.