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In a fission reaction U 92236X  117+Y  117+n+n, the binding energy per nucleon of X and Y is 8.5 MeV where as of U  236 is 7.6 MeV. The total energy liberated will be about

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By Expert Faculty of Sri Chaitanya
a
200 KeV
b
2 MKeV
c
200 MeV
d
2000 MeV

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detailed solution

Correct option is C

∆E=8.5 x 234 - 7.6 x 236=195.4 MeV                                                   =200 MeV

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The mass defect for the nucleus of helium is 0.0303 amu. What is the binding energy per nucleon for helium in MeV (approx)

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