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If the mass defect of 4X9 is 0⋅09 a.m.u, then the binding energy per nucleon is : (a) 8.38 MeV(b) 83⋅8 MeV (c) 931⋅5 MeV (d) 9⋅315 MeV |
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Answer» Correct option (d) 9⋅315 MeV Explanation: Binding energy of nucleus = Mass defect in amu × 931.48 MeV = 0.09 × 931.48 = 83.83 MeV Binding energy per nucleon = binding energy of the nucleus/number of nucleon (mass number) = 83.83/9 = 9.315 MeV |
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