I would expect two deuterium nuclei to fuse straight into a single helium-4 nucleus, because that's by far the most stable way to arrange 2 protons and 2 neutrons. But instead, any two fusing deuterons have a 50-50 chance of producing either a tritium nucleus and a neutron or a helium-3 nucleus and a proton. Why is this?
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2 Answers
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The deuterium fusion reaction is extremely exothermic. It releases about a million times more energy than a typical chemical reaction, and that energy has to go somewhere. If we had two deuterium nuclei fusing to form a helium-4 nucleus there is nowhere for the energy to go and the helium nucleus would just split up again.
So the newly formed helium nucleus has to get rid of all that energy, and there are three ways to do this:
the helium nucleus could release the energy as a gamma ray and form ${}^4\mathrm{He}$ directly
the helium nucleus could release a proton to form ${}^3\mathrm{H}$. Then the energy is carried away as the kinetic energy of the proton and the ${}^3\mathrm{H}$ nucleus.
the helium nucleus could release a neutron to form ${}^3\mathrm{He}$. Then the energy is carried away as the kinetic energy of the neutron and the ${}^3\mathrm{He}$ nucleus.
But these three branches have very different probabilities. About 55% of the time reaction (3) occurs and we end up with helium-3. About 45% of the time reaction (2) occurs and we end up with tritium. Reaction (1) happens only about 0.0001% of the time so it's very rare for the fusion to form helium-4 in one step.
Now the next question is why emitting a photon is so much less probable than emitting a proton or neutron, and as Chris commented below we can answer this in a handwaving way. Creating a photon involves an electromagnetic interaction, while ejecting a proton or neutron requires only a strong force interaction. The EM force is much, much weaker than the strong force so in general any process involving the EM force is much slower than interactions involving the strong force.
answered Aug 25, 2022 at 5:11
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5$\begingroup$ Emitting a photon is much less likely than emitting a proton or neutron because the electromagnetic force is much, much weaker than the residual strong force that dominates nuclear reactions. It's a similar reason to why $\rm pp\to {}^{2}H\nu e^{+}$ is so slow in the sun- that involves the even weaker weak force and so is greatly suppressed. $\endgroup$– Chris ♦Commented Aug 25, 2022 at 6:03
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$\begingroup$ I think that EM interaction is pretty much involved in either of the possible outcomes. The transition of the excitet He to He + gamma is probably forbidden and is outcompeted by the other processes. $\endgroup$– fraxinusCommented Aug 25, 2022 at 11:30
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$\begingroup$ I might be being dumb, but isn't phase space worth a mention here too -- Fermi's golden rule tells us that the matrix element is obviously very important, but don't you have more degrees of freedom in final states with two massive particles? $\endgroup$– LandakCommented Aug 25, 2022 at 15:30
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2$\begingroup$ I think it might be interesting if someone asked a new question specifically about why photon production is so improbable compared to p or n emission in this reaction. $\endgroup$ Commented Aug 25, 2022 at 16:01
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1$\begingroup$ @nickblack The excess energy cannot go into the KE of the ⁴He nucleus as that would not conserve momentum. $\endgroup$ Commented Oct 30, 2023 at 5:18
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Actually, fusion of two deuterons can yield helium-4, although more often it yields some other results.
