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Two deuterium nuclei fuse to form a helium-4 nucleus. Given atomic…
Question
Two deuterium nuclei fuse to form a helium-4 nucleus. Given atomic masses m(2H) = 2.014102 u and m(4He) = 4.002602 u, and 1 u = 931.5 MeV, find the energy released (in MeV) in this reaction.
✓ Verified answer: 23.8483checked by our engine — not a guess
Step-by-step solution
Mass defect: dm = 2*m(2H) - m(4He) = 2*2.014102 - 4.002602 = 4.028204 - 4.002602 = 0.025602 u.
Energy released = dm*931.5 = 0.025602*931.5 = 23.8483 MeV.
Final answer23.8483
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