Q 12-13-139JEE MainJEE Main 2025 (2 Apr, Shift 2)Easy
Energy released when two deuterons ($_1\text{H}^2$) fuse to form a helium nucleus ($_2\text{He}^4$) is: (Given: binding energy per nucleon of $_1\text{H}^2 = 1.1\ \text{MeV}$ and binding energy per nucleon of $_2\text{He}^4 = 7.0\ \text{MeV}$)
Answer: (C) $23.6\ \text{MeV}$
Energy released = total binding energy of products − total binding energy of reactants.
Two deuterons: $2\times(2\times1.1) = 4.4\ \text{MeV}$. Helium: $4\times7.0 = 28.0\ \text{MeV}$.
$$Q = 28.0 - 4.4 = 23.6\ \text{MeV}$$
Solution by Sreeraj P, M.Sc Physics