I just finished reading this research about antimatter induced fusion and thermonuclear reactions. And one conclusion I could make is that very little mass of antimatter (in range of micrograms) is needed to initiate a fusion reaction in lithium-deuteride fuel.

Also, in page 14 in this PDF file, there is a theoretical design of a 1-kiloton antimatter induced fusion bomb.

Now, I actually have 2 questions. Firstly, 100 grams of lithium-deuteride is used in this theoretical 1 kt design. But according to this, 100 grams of lithium-deuteride should yield 6.4 kilotons not only 1 kiloton, so is there any explanation of this ?

Secondly, since I seem to find only low yield designs of antimatter induced/catalyzed fusion bombs, a doubt about the feasibility of larger yield came to my mind. So, if a single kiloton fusion reaction is feasible with a certain amount of antimatter, then should we consider fusing more fuel with even more antimatter feasible too ?

Note : I completely understand the difficulty of making, handling and storing antimatter, and I am not saying this thing is going to be made any time soon. I am just curious about the physics part behind it.

  • $\begingroup$ Also how can pure deuterium fusion happen ? shouldn't be there tritium to fuse with the deuterium to make He ? $\endgroup$ Aug 24, 2013 at 16:57
  • $\begingroup$ Just guessing here, but could it be that part of the difference between those 6.4kt and that 1kt is that a conventional fusion weapon has an influx of neutrons from the primary, boosting the reaction? Or that it is because the secondary gets compressed via a Teller-Ulam device rather than being torn apart from an internal matter-antimatter explosion? $\endgroup$
    – Hennes
    Aug 24, 2013 at 21:21
  • $\begingroup$ @Hennes, The first guess might be correct. But not the second since in the hypothetical bomb, explosive lenses are used to compress the fuel and collapse it into the antihydrogen pellet. I noticed something else, this research says pellets of Li-DT is used, which I don't really understand what this is. is it lithium, deuterium and tritium ? $\endgroup$ Aug 24, 2013 at 21:45
  • $\begingroup$ Yes. Lithium, Deuterium and Tritium. Lithium is used in traditional fusion devices to generate tritium. (H3 is expensive. H3 has a 12.3 year half life and H3 decays into He-3, which likes to absorb neutrons). For these reasons H3 is best made fresh. This can be done by bombarding lithium with neutrons. Li-6 seems to be the preferred choice for this. Li-6 + neutron -> H3 + He-4 + 4.78 MeV. The H3 can then undergo fusion with other H3 atoms). Li-7 can be used in the same way if you have higher energy neutrons and it releases an extra neutron. $\endgroup$
    – Hennes
    Aug 24, 2013 at 22:02
  • $\begingroup$ @Hennes, that's great, thanks for the info. But IIRC, Tritium is considered very rare and only a few kilograms of it is made each year, so it shouldn't be considered available I think, so Li-DT is just nonsense. If I am not mistaken, Li-6 can produce Tritium by being bombarded by a neutron with any energy, so I was thinking if we could compensate the neutrons from the fission reaction by an external neutron source to produce "fresh" Tritium. The D-T reaction also produces an extra neutron, so it should look like a fission chain reaction. What do you think ? $\endgroup$ Aug 24, 2013 at 22:08

1 Answer 1


Aside: What's with this obsession with multimegaton explosions? End aside.

There are no inherent limits on the size of fusion explosion. Moreover, with the use of staged weapon design, antimatter amount needed to initiate the arbitrary powerful explosion would essentially remain the same.

Ordinary fusion explosion (Teller-Ulam design) is initiated by the x-rays released by fission primary that compress the secondary fusion fuel that is simultaneously heated with fissile sparkplug.

If we try to eliminate the fission, while already having the antimatter initiated small scale fusion device, the obvious way would be to use this device as primary to compress and heat the secondary fusion fuel dose.

Note, that since we also want to eliminate uranium sparkplug we need to provide mechanisms to not only compress but also heat the deuterium-tritium of the secondary. And the energy needed for such heating needed to be extracted from the primary explosion. This means that the hohlraum needs to be somewhat redesigned.

This secondary explosion could then be used to initiate the explosion of the third fusion stage.

In Tsar bomba ~1 megaton fission primary is enough to initiate 56 megaton fusion secondary. So using the value 50 for increase in yield between stages and using as primary the device mentioned in the question we can estimate the following yield for three stages: primary ~ 1 kiloton yield, secondary ~ 50 kiloton yield, tertiary ~ 2.5 megaton.


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