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


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. $\endgroup$ – John Rennie Nov 26 '13 at 16:07
  • $\begingroup$ @fibonatic ${}^2H$ doesn't form, it immediately beta decays to deuterium. $\endgroup$ – Brandon Enright Nov 26 '13 at 16:08
  • $\begingroup$ @JohnRennie oops I misread ${}^2He$ and typed it in wrong too. $\endgroup$ – Brandon Enright Nov 26 '13 at 16:12

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.