The mass of a 37Li nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of 37Li nucleus is nearly

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Correct answer: (B) 5.6 MeV. For 37Li nucleus, Mass defect, ΔM=0.042 u ∵ 1 u= 931.5 MeV c^(-2) &there4; Δ M = 0.042×931.5 MeV c^(-2) =39.1 MeV c^(-2) Binding energy, E_b=ΔMc^2 = 3 9 · 1 MeV/(c^2) c^2 =39.1 MeV Binding energy per nucleon,⁡E⁡_bn=(E⁡_b)/(A⁡)=(39·1 MeV)/7 ≈5·6 MeV

The mass of a $_{3}^{7} Li$ nucleus is $0.042 u$ less than the sum of the masses of all its nucleons. The binding energy per nucleon of $_{3}^{7} Li$ nucleus is nearly

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