Q 12-13-091JEE MainJEE Main 2022 (26 Jul, Shift 2)Easy
Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below: ${}^2_1X + {}^2_1X = {}^4_2Y$. The binding energies per nucleon of ${}^2_1X$ and ${}^4_2Y$ are $1.1\ \text{MeV}$ and $7.6\ \text{MeV}$ respectively. The energy released in this process is ______ MeV.
Numerical value type. Enter your answer.
Answer: 26
$$Q = 4\times7.6 - 2\times(2\times1.1) = 30.4 - 4.4 = 26\ \text{MeV}$$
Solution by Sreeraj P, M.Sc Physics