I've been reading up on nuclear physics and I have this dumb idea that I can't seem to poke holes in.

Imagine a linear particle accelerator accelerating tritium nuclei in to a water tank containing D$_{2}$O. At some rate the tritium should collide with deuterium causing fusion.

The helium containing about 3.6 MeV should dissipate its energy quickly into the water. The fast neutron containing about 14 MeV escapes the water tank into a surrounding blanket of $^{238}$U where it may cause fast fission releasing an additional ∼200 MeV + slow neutrons.

These slow neutrons may be absorbed by $^{238}$U to make $^{239}$Pu or escapes into a secondary blanket of lithium where it may make more tritium.

It is to my understanding it takes about 100–200 keV to initiate the fusion reaction. This means that the net energy potential is enormous even at modest reaction rates.

What am I missing? Why wouldn't this approach work?

  • $\begingroup$ The peak of the D-T reaction cross section is at a bit over 100keV. $\endgroup$
    – Jon Custer
    Sep 12, 2021 at 21:48
  • $\begingroup$ Ok I'll update the post. Since D is kept at relatively low temperature in the form of water, would that require the T in the particle beam to have 200keV of kinetic energy? $\endgroup$ Sep 12, 2021 at 21:54
  • $\begingroup$ The standard way would be to accelerate the deuterium to keep from making your accelerator radioactive… $\endgroup$
    – Jon Custer
    Sep 12, 2021 at 23:57

2 Answers 2


It "works" to produce some fusion, but it isn't a net energy producer because the cross-section for the reaction so small.

You won't be able to get modest reaction rates. Instead almost all will miss and you'll only get a fraction of the energy back.


  • $\begingroup$ The fusor is using magnetically confined plasma which is different beast compared to my suggestion, a tritium beam through heavy water. Also the fission component is there to balance the neutron and energy economy, something your answer is not covering. $\endgroup$ Sep 12, 2021 at 21:30
  • $\begingroup$ This is the case today. There is no theoretical reason that you cannot get energy from this reaction. But in practice no one has come close to breakeven with direct acceleration due to the low reaction rates. If you can design something that avoids the energy waste, that would be great. "Why won't it work"? Today the answer is "Because recapturing the energy from the high population of unreacted accelerated particles is inefficient." $\endgroup$
    – BowlOfRed
    Sep 12, 2021 at 23:05
  • $\begingroup$ "Because recapturing the energy from the high population of unreacted accelerated particles is inefficient." - That is what the fission is for, accept the loss and instead be compensated with higher yield from the few reactions that do occur. What I can't find is the probability of fast tritium finding a deuterium in heavy water. Heavy water has concentrations of deuterium that would only be a dream in any plasma. If that probability happens to be anywhere above 0.1% we're at least break even. $\endgroup$ Sep 12, 2021 at 23:24

Essentially, the problem is that the tritium particles in the beam lose their energy very quickly through collisions with particles in the target. While some will undergo fusion before their energy is dissipated, the cross section of the reaction is so small that the number that don't will be several orders of magnitude more. Your scheme for using resulting neutrons in further fission reactions would not be sufficient to produce a net energy gain.

Fusion research tends to focus on confinement of thermal plasmas* rather than accelerators for this reason.

  • $\begingroup$ That's what I don't get. If we look at the fusion equation per unit volume it goes something like this "Rate of fusion = Particle density * Their relative velocity * The cross section". The particle density is conservatively 100,000x that of any plasma so even if the volume of the beam is tiny we shouldn't be off by more than a few orders of magnitude. If that is true then the secondary fission reaction could potentially bridge that gap. $\endgroup$ Sep 13, 2021 at 6:44

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