I would like to know why scientists try to use deuterium and tritium for fusion and not just the ordinary isotope of Hydrogen ${}^1H$?

  • $\begingroup$ What prior research did you do? $\endgroup$
    – my2cts
    Oct 29 '20 at 12:14

The problem with attempting to fuse two protons is that there is no bound state $^2$He, for the rather obvious reason that there are no neutrons present to hold the two protons together. The fusion of two protons requires one of them to undergo beta plus decay while the two protons are close, and the probability of this is vanishingly small. It happens in the Sun because there are an awful lot of proton collisions in the Sun's core and even the tiny probability of fusion produces a sizable overall reaction rate.

By contrast fusing deuterium and tritium produces $^5$He, which does have a bound state, so this has a relatively large probability. The deuterium and tritium fuse to form $^5$He, and this then decays to $^4$He and a neutron with a half life of about $7 \times 10^{-22}$ seconds.

See the related question: How much faster is the fusion we make on earth compared to the fusion that happens in the sun?

  • $\begingroup$ But why not use ${}^1H$ and ${}^3H$ or just ${}^2H$ to get ${}^4He$ directly? Or ${}^1H$ and ${}^2H$, since ${}^3He$ is also a stable isotope of helium. Or does this combination yields the highest energy ouput? $\endgroup$
    – fibonatic
    Nov 26 '13 at 15:53
  • 2
    $\begingroup$ The D-D reaction rate is slower than D-T. I don't know exactly why this is, but I would guess it's because if you collide two deuterons the resulting $^4$He nucleus is formed with a greater energy than it's dissociation energy and it immediately falls apart again. With the D-T fusion the escaping neutron can carry away the excess energy and allow the $^4$He nucleus to relax. I think the D-D fusion actually produces $^3$He - one of the neutrons undergoes beta decay and the ejected proton carries away the excess energy. $\endgroup$ Nov 26 '13 at 16:07
  • $\begingroup$ @fibonatic ${}^2H$ doesn't form, it immediately beta decays to deuterium. $\endgroup$ Nov 26 '13 at 16:08
  • $\begingroup$ @JohnRennie oops I misread ${}^2He$ and typed it in wrong too. $\endgroup$ Nov 26 '13 at 16:12
  • $\begingroup$ FWIW, at the solar core temperature, the probability that a diproton converts to a deuteron, rather than just falling apart into 2 protons, is $\approx 10^{-26}$. $\endgroup$
    – PM 2Ring
    Oct 29 '20 at 12:25

I agree with @john rennie, but I think it's worth noting that reaction rates for D-T are higher at a lower temperature as compared to D-D:

Higher temperatures present many difficult engineering challenges (even more so than the still high temperatures for D-T reactions).


The choice of fusion fuel is dictated by many factors:

  1. Its abundance or how easy/expensive it is to synthesise
  2. Its characteristics/behaviour in a reactor e.g. how does it move in an electromagnetic field? Is it easy to ionise?
  3. The types of fusion reactions that the fuel enables

For D and T:

  • The fuel is relatively easy to synthesise (not very, but more practical than He3 for example)
  • Using D and T fuel leads to reactions such as DD fusion and DT fusion. The latter releases a lot of energy, and is much more likely at achievable temperatures - hence the total power output is higher. The downside is that most of this power is carried by neutrons - which cannot be confined by electromagnetic fields. To achieve the same power output with other fuels, we would usually need an extremely hot and dense plasma, which would lead to the plasma collapsing due to Bremsstrahlung radiation. Likewise for fuels with other cross sections, one may find the fusion reaction is actually thermally unstable.

If we had much better confinement of the fuel, we would be able to use these alternative fuels at a lower temperature. Improving confinement is what most fusion science research is about :)


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