Question

The energy released per fission of nucleus of ${ }^{240} X$ is $200 \,MeV$ The energy released if all the atoms in $120\, g$ of pure ${ }^{240} X$ undergo fission is _______$\times 10^{25} MeV$. (Given $N _{ A }=6 \times 10^{23}$ )

Answer 1

The correct answer is 6.
No. of mole =240120​=21​
No. of molecules −21​×NA​
Energy released =21​×6×1023×200
=6×1025MeV