# Why is D-T fusion easier than T-T?

At a very basic level, fusion can occur when the electrostatic repulsion between nuclei is overcome and they approach close enough for the strong force to come into play. Thus, the easiest reactions would occur in nuclei with the lowest electrical charge relative to their nucleon count.

So that being the case, why is the cross section for T-T fusion so much smaller than D-T fusion? The electric charge is the same, but the nucleon count is 6 instead of 5.

What am I missing?

• Write out the reactions for each. What are the energetic between He5 and He6 nuclei? How do those nuclei decay? Apr 28, 2017 at 16:21

The title is a bit sloppy (easier), but it is a very good question when you mention cross section.

First - it is more convenient to speak not about cross section, but so called astrophysical S-factor. The cross section goes down like hell ($\exp$) with decreasing the energy. This S-factor removes problems of Coulomb barrier and a natural energetic dependence and one can draw almost flat graphs (sometimes).

$$S(E) = \sigma(E) \exp(-2\pi\eta) E^{-1}$$

See - the d-t reaction is 100x more probable (at low energies) than $t-t$ or $d-d$, as you claimed.

In $^3H - ^{3}H$ scattering, the system feels itself as being $^6He$ in some not well defined state and it has tendency to either just scatter or to "mutate" into some $^6He$ existing state. Energetically, it is possible just to go to (3) lower energy states and waste the remaining energy arbitrarily. Look.

The same happens to $^3H - d$ system, that thinks it is actual $^5He$, but the situation is a little bit different:

The system appears just around the existing level (resonance) 16.84 MeV and this enhances tremendously the time the system stays there. And this way also the probability of a transfer to another state or of irradiation of a neutron. Look at the S-factor:

It peaks at about 50 keV by two ordes of magnitude, when compared to $d-d$:

• So I'm trying to parse this. It seems the "trick" here is that there is an energy level in He5 that just happens to be lower than in He6? I'm not sure I fully understand resonances in this context. Apr 28, 2017 at 18:55
• It happens that it is close enough. The "resonance" - every energy level has some width, it is never "discrete" value unless it lives forever (Heisenberg relations). If the level is already unbound, it is usually quite wide, it could be even MeVs, this one is 74.5 keV. At any small kinetic energy, the particles feel, they are inside. How much they feel at different energies - look at the bump in S-factor picture. Apr 28, 2017 at 19:45
• Ok I think I get this. This also helps explain other texts I've read about resonances that didn't really explain what they meant. So in this case when the D-T gets close enough together it sort of thinks its a single He5 instead of two separate things, as long as the energy is thus. But in T-T the energy level where they see themselves as a single thing is much higher? And that's possible BECAUSE its unstable and thus has a wide resonance. Apr 28, 2017 at 20:01

The T-T cross section is in effect the normal cross section - it is very similar to the D-D cross section - and it is the D-T cross section that is unusually high.

And the D-T cross section is high because there is a resonance of the ${}^5$He nucleus at around the kinetic energies of the D and T nuclei. That is, the D and T fuse to form a (very unstable) ${}^5$He nucleus that immediately falls apart into ${}^4$He and a neutron, and the transient ${}^5$He nucleus has an broad excited state at around the thermal energies in the fusion reactor. The existence of this resonance increases the probability of fusion.

It's the same sort of phenomenon that gives the unusually high cross section for the triple alpha process that forms carbon.

Later:

I see Jaromrax has given all the details of the energy levels involved and I encourage you to have a look at (and upvote :-) his answer.