A fusion reaction that might be used in a power reactor involves two deuterium:

Question

A fusion reaction that might be used in a power reactor involves two deuterium:

2 1H + 2 1H --> 3 1H + 1 1H

Calculate the energy released in this reaction.

(masses are: 2 1H: 2.014102u, 3 1H:3.016049u, 1 H:1.007825u)

*I know forumula is E=mc^2, my question is that do I:

subtracte all the masses and multiply C^2, OR do I add the two deutrium masses together and sub tract from the other two masses, OR do I add the two detrium and the other two masses and then subtract those from each other? I dont know the correct way to do this. PLEASE HELP !!!! Thanks!

Explanation / Answer

[(2.014102u)2 -3.016049u-1.007825u]c^2 = 4.33x10^-3 x u x c^2 = 4.33x10^-3 x 931.5 MeV = 4.033 MeV

rest mass energy of nucleon = u x c^2 = 931.5 MeV

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