# In fusion inside stars (sun) or very hot gasses, how do the electrons get bound and what about tritons and $D$-$T$ vs $D$-$D$ fusion?

Most texts I've read focus on just the nuclei to begin with, but eventually start talking about Helium (or other) atoms and isotopes. A few aspects aren't clear to me and I'd be grateful for some explanations. My understanding below may be erroneous in parts. A good reference for a beginner that contains clear phenomenological explanations would also help.

I understand that in very hot plasma, like the Sun's core, electrons do exist but are unbound as are protons, and both move with very high velocities (high temperature), electrons moving faster than protons due to lower mass. Now, a head-on collision of two very fast protons can, with small but non-zero probability, latch them together and one of them can decay into a neutron (plus neutrino plus positron), thus fusing into a deuteron (the nucleus of a deuterium). As deuterons are heavier and larger than protons, the larger size makes it easier for them to bump into other protons, and the larger mass, hence larger inertia, helps them overcome the Coulomb barrier more easily in order to fuse further with a proton (*). That's fine, but so far we still have only a nucleus, specifically a 3He nucleus (2 protons, 1 neutron). Texts I've read at this point talk about 3He, as in 3He atoms, but that would require 2 bound electrons, i.e. it wouldn't be an atom yet.

At which point do electrons get bound and how, or do they get bound at all? Is it because once we get a deuteron, this nucleus is much slower and hence a passing electron will get instantly snatched by the Coulomb force? This would imply that a passing electron is snatched and gets bound every time a new proton fuses.

(*) It's unclear to me whether deuteron-proton fusion is more likely than proton-proton fusion. A deuteron is fairly weakly bound, so a proton knocking it can also split it. The proportion of split vs fuse is unclear, as are the conditions for one and the other to happen (geometry, rotation, vibration state?).

Another aspect that's less clear regards tritium and D-T fusion.

First, how exactly do tritons come to form inside the sun? Is it because of (*) above? i.e. the deuteron nucleus is fairly weekly bound, hence a proton knocking it that splits it produces a free neutron which can then fuse with an existing deuteron?

Assume we have $$D$$ and $$T$$. Texts claim that when a $$D$$ and $$T$$ collide they always stick (fuse) whereas when $$D$$ and $$D$$ collide they almost never stick as that would require a photon to be emitted in order to stick which doesn't usually happen (this photon emission part i'm unclear on). Why is $$D$$-$$T$$ almost guaranteed to result in fusion and $$D$$-$$D$$ is not?

• en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction is quite good. Yes, you're right, it's too hot in the core (and most of the Sun) for electrons to bind to nuclei, so when texts talk about atoms in the core they are being a bit loose with the language. Mar 29 '20 at 15:28
• I did read that reference and it doesn't answer my questions. Are you saying that both before and after fusion (all of them) there are no electrons bound to any nuclei, i.e. there are only nuclei and all texts effectively abuse notation when they talk of "Helium-3" or "Helium-4" or "Hydrogen" etc? No atoms? Mar 29 '20 at 15:36
• Yes, that's correct. There are only bare nuclei, no neutral atoms. But I suppose you could call them fully ionised atoms. Sure, that Wikipedia article doesn't completely answer your questions, but IMHO it's a good reference which may clarify a few things for you. Mar 29 '20 at 15:45
• Many thanks. Shows that sometimes asking a question early is the better idea (I took some time looking for an answer). Mar 29 '20 at 15:58
• Your question covers a lot of territory! We prefer questions that are more sharply focused, but I'll post an answer shortly that addresses some of your concerns. Mar 29 '20 at 16:03

## 1 Answer

As I mentioned in the comments, the plasma in the interior of a star is fully ionised: the nuclei don't have any electrons bound to them.

The main bottleneck in the proton-proton chain isn't the fusion of two protons to form a diproton, it's the conversion of the diproton to a deuteron. The temperature of the solar core is more than adequate to overcome the Coulomb repulsion between two protons, once quantum tunneling is taken into account, as discovered by George Gamow. See the Gamow factor for details.

However, the diproton is very unstable, and most of the time it just falls apart shortly after it's formed. So there's only a short timeframe for one of the protons to be converted into a neutron via beta plus decay. But that reaction involves the weak nuclear interaction, which is relatively slow. It's estimated that (in the solar core) the probability of a diproton converting to a deuteron is in the order of $$10^{-26}$$. And that's why the mean time for a solar core proton to be successfully fused is around 9 billion years.

Incidentally, that makes pure hydrogen fusion extremely impractical for a terrestrial fusion power plant. But it also means that stars can burn for a long time, which is a very good thing. :)

• @kkm You're correct. I originally glossed over that point in my answer. George Gamow found that quantum tunneling permits protons to fuse at the relatively low temperature of the solar core. Mar 5 at 1:50
• @kkm The point is illustrated by the fact that deuterium burning happens at an order of magnitude lower temperature than the pp chain, despite the same Coulomb barrier. Mar 5 at 8:38
• @kkm The stability of what is produced tells you how deep the potential is on the other side of the barrier. Does that have a great influence on the tunneling probability? It doesn't appear in the Gamow peak formula. Mar 5 at 11:45
• @kkm My point was that D+p fusion, which has exactly the same potential barrier, proceeds at lower temperatures than p+p fusion because the latter also requires a weak interaction p to n change. Therefore you need the barrier to be penetrated at a much higher rate to have any chance of the weak interaction doing its thing. Mar 6 at 11:25
• @ProfRob, yes, thanks, that's dawned on me after I sent the previous to last message. I'm a slowpoke. :) Really, thank you for helping me clean a little bit of that 25 years of rust off my nuclear physics. Took me a while to appreciate your comment!
– kkm
Mar 6 at 12:32