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

Question

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 (A) 46 MeV (B) 5.6 MeV (C) 3.9 MeV (D) 23 MeV

Tardigrade Admin · Accepted Answer

For 37Li nucleus, Mass defect, Δ M=0.042u âµ 1u=931.5MeVc- 2 â´ Δ M=0.042× 931.5MeVc- 2 =39.1MeVc- 2 Binding energy, Eb=Δ Mc2 =(3 9 ⋅ 1 MeV c- 2)c2 =39.1MeV Binding energy per nucleon Eb n=(Eb/A)=(39 ⋅ 1 MeV/7)≈5⋅ 6MeV

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

46 MeV

5.6 MeV

3.9 MeV

23 MeV

For 37​Li nucleus, Mass defect, ΔM=0.042u ∵ 1u=931.5MeVc−2 ∴ ΔM=0.042×931.5MeVc−2 =39.1MeVc−2 Binding energy, Eb​=ΔMc2 =(39⋅1MeVc−2)c2 =39.1MeV Binding energy per nucleon Ebn​=AEb​​=739⋅1MeV​≈5⋅6MeV

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