Q 12-13-068JEE MainJEE Main 2023 (24 Jan, Shift 2)Easy
The energy released per fission of nucleus of $^{240}\text{X}$ is $200\ \text{MeV}$. The energy released if all the atoms in $120\ \text{g}$ of pure $^{240}\text{X}$ undergo fission is ______ $\times10^{25}\ \text{MeV}$. (Given $N_A=6\times10^{23}$)
Numerical value type. Enter your answer.
Answer: 6
Number of atoms $=\dfrac{120}{240}\times6\times10^{23}=3\times10^{23}$.
Energy $=3\times10^{23}\times200=6\times10^{25}\ \text{MeV}$.
Solution by Sreeraj P, M.Sc Physics