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Suppose we have a single tritium nucleus ( $^3_1H$ so two neutrons and one proton). Tritium is a highly unstable nucleic configuration and hence decays into $^3_2He$ with a half life of roughly 12 years. $\,^4_2He$ (the alpha particle) on the other hand is a highly stable configuration and does not decay at all. Thus, if we should precisely fire a solitary proton at the tritium nucleus we should expect a fusion reaction and a release of energy (in the form of photons or kinetic energy of the resulting $^4_2He$ nucleus ?) as the nucleus attains the more favorable and stable configuration of $^4_2He$.

This reaction should look something of the form $$p^++\,^3_1H\rightarrow \,\,^4_2He+x_0MeV$$ where $x_0MeV\neq 0$ is some positive (presumably large) amount of energy. My question can simply be stated as follows: is this reaction actually possible and if so, why is it not used or referred to in texts since it appears to be the simplest form of a fusion reaction imaginable? Also, if this reaction is possible, is energy released in the form of photons or kinetic energy or perhaps both? The reason I ask is that I want to be able to use this example to frame my thoughts on the release of energy in fusion reactions and this seems like the perfect example due to its simplicity. Any help on this would be most appreciated!

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  • $\begingroup$ You should try to calculate $x_0$ and include it in your question. $\endgroup$
    – PM 2Ring
    May 27 at 7:12
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Your formula has a critical problem with it- the "energy" term on the right-hand side. If you look at formulae for fusion, they tend to have multiple particles on the right-hand side. This is because, in the center of momentum frame, if there is only one particle on the right-hand side, that particle has no momentum and thus no kinetic energy. So this energy would have to be released in some other way.

Imagine that the proton and triton came together to form an excited form of helium-4, $\rm ^4He^*$. This particle could decay into stable helium-4 by emitting a photon, or it could decay into a proton and a triton via the strong force. Since the electromagnetic interaction is much weaker than the strong force, you would expect the latter interaction to happen much more often, which keeps the interaction $\rm p+^{3}H\to ^{4}He+\gamma$ from occurring readily.

An interaction like $\rm ^2H+^3H\to^4He+n+K.E.$ is overall much simpler because the kinetic energy can be shared between the two particles so you don't need to worry about how the energy comes out.

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  • $\begingroup$ (1/2) Thanks for the great answer! I think you've pretty much covered all my issues. Before I accept your answer, is my understanding correct (which can be summarized as follows): When a proton is fired at and initially merges with a tritium, the resulting $^4He$ must necessarily be in an excited state because before the collision, the system (proton + tritium) had a non zero $E_K$ in the center of momentum frame. After the collision, we have only an $^4He$ which trivially has zero $E_K$ in the center of momentum RF. So the kinetic energy has to have gone into exciting the $^4He$... $\endgroup$ May 27 at 8:07
  • $\begingroup$ (2/2). So immediately after the collision, we have an excited $^4He^*$ nuclei. Now because the strong force is much stronger than the EM force, it is more favorable for the $^4He^*$ to immediately decay back into a proton and a tritium rather than emitting a gamma ray and so the fusion described in my original post does not readily occur? $\endgroup$ May 27 at 8:09
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    $\begingroup$ @SalahTheGoat Yes, basically. The particles can also emit a photon without actually forming an excited state (and such an excited state may not in fact exist), but that would also be rare for the same reason that the electromagnetic force is much weaker than the strong force. It's basically the same reason we can't fuse two protons, just with the weak force substituted for the electromagnetic force. Generally speaking you drastically slow down a strong interaction if you start adding electromagnetic interactions to it. $\endgroup$
    – Chris
    May 27 at 8:21

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