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Q.The mass of a 37Li^{7}_{3}\text{Li} nucleus is 0.042 u less than the sum of the masses of all its nucleons. The average binding energy per nucleon of 37Li^{7}_{3}\text{Li} nucleus is nearly :

(a) 23 MeV
(b) 46 MeV
(c) 5.6 MeV
(d) 3.9 MeV
Puducherry TnboardTamil Nadu HSC (DGE) Board 2026MCQ· 1mImportance★★★★★
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Converting the given mass defect to energy and dividing by the 7 nucleons of 37Li^7_3\text{Li} gives a binding energy per nucleon of about 5.6 MeV.

Working

Total binding energy: BE=Δm×931.5 MeV/uBE = \Delta m \times 931.5\ \text{MeV/u} (using E=mc2E=mc^2 with mass in atomic mass units).

Given Δm=0.042\Delta m = 0.042 u:

BE=0.042×931.5≈39.12 MeVBE = 0.042 \times 931.5 \approx 39.12\ \text{MeV}

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