Question

The energy released per fission of the 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 _____ \(10^{25} \, \text{MeV}\). (Given \(N_A = 6 \times 10^{23}\))

Answer 1

Correct Answer: 6

The number of moles of \(^{240}\text{X}\) in \(120 \, \text{g}\) is:

\[ \text{No. of moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{120}{240} = 0.5 \, \text{moles} \]

The number of nuclei in \(0.5\) moles is:

\[ \text{No. of nuclei} = 0.5 \cdot N_A = 0.5 \cdot 6 \times 10^{23} = 3 \times 10^{23} \]

The total energy released is:

\[ E = (\text{No. of nuclei}) \cdot (\text{Energy per fission}) \]

Substitute:

\[ E = (3 \times 10^{23}) \cdot (200) = 6 \times 10^{25} \, \text{MeV} \]

Thus, the total energy released is \(6 \times 10^{25} \, \text{MeV}\).

Answer 2

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