JEE Main 2026 — Nuclear Physics Question with Solution

The nuclear reaction is given by:

$2X^3+Y^4\to Z^{10}$

The total binding energy of the reactants is the sum of the binding energies of the two nuclei of mass number $3$ and the nucleus of mass number $4$.

Binding energy of reactants $=2\times (3\times 5.6)+(4\times 7.4)$

$=33.6+29.6=63.2$ MeV

The total binding energy of the product nucleus of mass number $10$ is:

Binding energy of product $=10\times 6.1=61.0$ MeV

The change in mass energy $\Delta M{c}^2$ is equal to the difference in the total binding energies of the reactants and the product.

$\Delta M{c}^2=|63.2-61.0|=2.2$ MeV

Answer: $2.2$