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Question
Find the energy released during the following fission reaction.
\[\ce{_92^235U + _0^1n-> _92^236U->_36^90Kr + _56^143Ba + 3 _0^1n}\]
| Mass(u) | |
| 235U | 235.0439 |
| 90Kr | 89.9195 |
| 143Ba | 142.9206 |
| 1n | 1.0087 |
Solution
Mass of reactants = 235.0439 + 1.0087 = 236.0526u
Mass of products = 89.9195 + 142.9206 + 3(1.0087) = 235.8662u
change in mass = ∆m = 236.0526 − 235.8662 = 0.1864u
Q = Δm × 931.5 MeV
= 0.1864 × 931.5 MeV
= 173.63
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