Q 12-13-016NEETJEE MainMedium
The mass defect of $^4_2$He is $0.0304$ u. Its binding energy per nucleon is about ($1$ u $= 931$ MeV)
Answer: (C) $7.1$ MeV
Concept. Binding energy $=$ mass defect $\times 931$ MeV/u; divide by the number of nucleons.
Step 1. Binding energy. $0.0304 \times 931 \approx 28.3$ MeV.
Step 2. Per nucleon. $\dfrac{28.3}{4} \approx 7.1$ MeV.
Answer. $7.1$ MeV (Option C).
Solution by Sreeraj P, M.Sc Physics