# How would the explosion from a Pure Fusion Bomb differ from the explosion from a Fission Nuclear Bomb?

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.

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.

• 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? – user10851 Jan 16 '15 at 19:52
• There is a lot of material on the Internet on these bombs. – Sofia Jan 16 '15 at 19:57
• @ChrisWhite I am trying to compare against pure fission yes. – Jimmery Jan 16 '15 at 20:14
• @Sofia I did search but I couldn't find anything that answered these specific questions. Has my Google skills failed me? – Jimmery Jan 16 '15 at 20:16
• @Jimmery What's confusing is your wording makes fusion sound hypothetical. In fact most nuclear weapons today are fusion based. – user10851 Jan 16 '15 at 20:17

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.