Q 12-13-045JEE MainJEE Main 2024 (5 Apr, Shift 1)Easy
If three helium nuclei combine to form a carbon nucleus, then the energy released in this reaction is ______ $\times10^{-2}\ \text{MeV}$. (Given $1\,u = 931\ \text{MeV}/c^2$, atomic mass of helium $= 4.002603\,u$)
Numerical value type. Enter your answer.
Answer: 727
The atomic mass of ${}^{12}\text{C}$ is exactly $12\,u$.
$$\Delta m = 3\times4.002603 - 12 = 0.007809\,u$$
$$E = 0.007809\times931 = 7.27\ \text{MeV} = 727\times10^{-2}\ \text{MeV}$$
Solution by Sreeraj P, M.Sc Physics