[Physics] Why is D-T fusion easier than T-T

atomic-physicsfusion

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?

Best Answer

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.

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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.

t-t g

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

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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:

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

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