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← JEE Main 2022 · 26 Jul · Shift 2
PhysicsInteger+4 / -1Question 54 of 90

JEE Main 2022 · 26 Jul · Shift 2, Physics question 54

Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below :

12X+12X=24Y{}^2_1X+{}^2_1X={}^4_2Y

The binding energies per nucleon for 12X{}^2_1X and 24Y{}^4_2Y are 1.1 MeV and 7.6 MeV respectively. The energy released in this process is ______ MeV.

Answer and worked solution

Official answer: 26

Binding energies. 4×7.6−2×(2×1.1)=30.4−4.4=264\times7.6-2\times(2\times1.1)=30.4-4.4=26 MeV.

Answer: 26