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

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