Suppose we have the technology to create high enough temperatures and pressures inside a confined space to fuse together deuterium and tritium, and create a Pure Fusion Bomb.

How would the explosion of this Pure Fusion Bomb differ from the explosion of a standard Nuclear Bomb that uses fissile materials?

I am assuming, ton for ton, the Pure Fusion Bomb would be more destructive. I am also assuming that a Pure Fusion Bomb would also produce a lot more neutrons than a normal Nuke.

Are my assumptions correct? I gathered them after reading this Wikipedia page on Pure Fusion Weapons.

Would both bombs produce a similar output of radiation? And would they both produce an electro magnetic pulse?

Just to clarify, I am not talking about a Fission - Fusion Bomb.

  • 1
    $\begingroup$ Just to be clear, the comparison is between pure fusion and fusion with a fission detonator? You're not trying to compare against pure fission? $\endgroup$
    – user10851
    Commented Jan 16, 2015 at 19:52
  • $\begingroup$ There is a lot of material on the Internet on these bombs. $\endgroup$
    – Sofia
    Commented Jan 16, 2015 at 19:57
  • $\begingroup$ @ChrisWhite I am trying to compare against pure fission yes. $\endgroup$
    – Jimmery
    Commented Jan 16, 2015 at 20:14
  • $\begingroup$ @Sofia I did search but I couldn't find anything that answered these specific questions. Has my Google skills failed me? $\endgroup$
    – Jimmery
    Commented Jan 16, 2015 at 20:16
  • 3
    $\begingroup$ @Jimmery What's confusing is your wording makes fusion sound hypothetical. In fact most nuclear weapons today are fusion based. $\endgroup$
    – user10851
    Commented Jan 16, 2015 at 20:17

1 Answer 1


Your technology supposition is a huge one. Real fusion explosives must use small conventional and fission explosions to create the confinement and temperature needed to get a rapid fusion explosion. But, assuming that -- You can calculate what is known as the Q of each reaction: $$Q=m(reactants)-m(products)$$ so for your fusion reaction in the strongest (only?) channel we have$$Q_{fusion}=m(^3H)+m(^2H)-m(^4He)-m(^1n)$$ where the $m$'s are nuclear masses. Because all the missing electron masses will cancel, you can use atomic masses which are in tables here and here. A typical fission reaction would be $$Q_{fission}=m(^{235}U)+m(^1n)-m(^{137}Cs)-m(^{96}Rb)-3m(^1n)$$

If you calculate the Qs for these reactions and compute the Q/m(reactants) for each you will see that the fusion event releases over 4$\times$ the energy per reactant mass.

In this scenario, the fusion event produces a helium nucleus and a neutron with no direct radioactive by-products. The fission event produces 2-3 neutrons, depending on the exact products (a bi-modal distribution and not a single channel), but your numbers will show that for equal energy output, the fusion event produces about 3$\times$ the neutrons.

The fission explosion would produce a wide variety of radioactive by-products along with a tremendous spectrum of x-rays and gamma rays. The neutron production in the fusion device could produce radioactive products via neutron activation reactions with the bomb casing and the atmosphere. The actual energy release may produce some high energy gamma rays due to excitation of atmospheric oxygen and nitrogen.

Another thing to keep in mind is that a very small fraction of the reactant ingredients in these devices actually react, so there will be leftover tritium in the fusion device and leftover uranium (or plutonium) in the fission device.

  • $\begingroup$ Great answer, I am just confused about the phrasing: "In this scenario, the fusion event produces a helium nucleus and a neutron with no direct radioactive by-products..." Isn't a free neutron a radioactive byproduct? Or am I misunderstanding something? [random curiosity mostly, not really a criticism] $\endgroup$ Commented Sep 14, 2022 at 16:22

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