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Find the energy released during the following fission reaction. - Physics (Theory)

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प्रश्न

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
संख्यात्मक

उत्तर

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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Nuclear Reactions
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2024-2025 (April) Specimen Paper
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