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Numerical · Q25

Q.In a fission reaction of 92235U^{235}_{92}\text{U} induced by a slow neutron, the total mass defect is found to be 0.215 u0.215\ \text{u}. Calculate the energy released per fission event, in MeV and in joules.

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Using E=Δm c2E=\Delta m\,c^2 with the mass defect expressed directly in atomic mass units:

E=0.215×931.5≈200.3 MeVE = 0.215\times931.5 \approx 200.3\ \text{MeV}

Converting to joules using 1 MeV=1.602×10−13 J1\ \text{MeV}=1.602\times10^{-13}\ \text{J}:

E=200.3×1.602×10−13≈3.21×10−11 JE = 200.3\times1.602\times10^{-13} \approx 3.21\times10^{-11}\ \text{J} …

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