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The mass of a \(^7_3Li\) nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of \(^7_3Li\) nucleus is nearly(a) 46 MeV (b) 5.6 MeV (c) 3.9 MeV (d) 23 MeV |
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Answer» (b) 5.6 MeV If w = 1 u, C = 3 x 108 ms-1 then, E = 931 MeV 1 u = 931 Mev Binding energy = 0. 042 x 931 = 39. 10 MeV ∴ B.E 39.10 Binding energy per nucleon = \(\frac{B.E}{A}\) = \(\frac{39.10}{7}\) 5.58 = 5.6 MeV |
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