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

Answer 1

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