Two nuclei of mass number $3$ combine with another nucleus of mass number $4$ to yield a nucleus of mass number $10$. If the binding energy per nucleon for the mass numbers $3, 4$ and $10$ are $5.6$ MeV, $7.4$ MeV and $6.1$ MeV, respectively, then in the process, $\Delta Mc^2 =$ ______ MeV.
Answer: (C) $2.2$
Concept. Compare total binding energies before and after. If the products are less bound, energy must be supplied and the mass increases by $\Delta M = \dfrac{\Delta E}{c^2}$.
Step 1. Before. $2 \times 3 \times 5.6 + 4 \times 7.4 = 33.6 + 29.6 = 63.2$ MeV.
Step 2. After. $10 \times 6.1 = 61.0$ MeV.
Step 3. Difference. The product is less tightly bound by $63.2 - 61.0 = 2.2$ MeV, so $\Delta Mc^2 = 2.2$ MeV.
Answer. $2.2$ (Option C).
Solution by Sreeraj P, M.Sc Physics
