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Nuclear force - class-XII
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The binding energy per nucleon for $\mathrm { U } ^ { 238 }$ about 7.5Mev where as it is about 8.5 Mev for a nucleus having a mass half of Uranium. If $\mathrm { U } ^ { 238 }$ splits into two exact halves the energy released would be
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A
238 Mev
💡 Explanation:
Energy released = (Total binding energy of products) - (Total binding energy of reactants). Reactants: 238 nucleons * 7.5 MeV = 1785 MeV. Products: 2 * (119 nucleons * 8.5 MeV) = 2023 MeV. Energy released = 2023 - 1785 = 238 MeV.