Question

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

Answer 1

Answer: 6

No. of mole $=\frac{120}{240}=\frac{1}{2}$
No. of molecules $-\frac{1}{2}\times N_A$
Energy released $=\frac{1}{2}\times 6\times 10^{23}\times 200$
$=6\times 10^{25}MeV$