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A nuclear fission is represented by the following reaction: `U^(236) = X^(111) +Y^(122) +3n` If the binding energies per nucleon of `X^(111), Y^(122)` and `U^(236)` are `8.6MeV, 8.5 MeV` and `7.6 MeV` respectively, then the energy released in the reaction will be-A. `200 MeV`B. `202 MeV`C. `195 MeV`D. `198 MeV` |
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Answer» Correct Answer - D Energy released `= BE` of products - BE of reactants `= 111 xx 8.6 +122 xx 8.5 - 236 xx 7.6` `= 954.6 + 1037.0 - 173.6 =198 MeV` |
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