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

To find the total energy released if all the atoms in \(120 \, \text{g}\) of \(^{240}\text{X}\) undergo fission, follow these steps:

1. First, calculate the number of moles of \(^{240}\text{X}\). The molar mass of \(^{240}\text{X}\) is \(240 \, \text{g/mol}\). Therefore, \(\text{moles of } ^{240}\text{X} = \frac{120 \, \text{g}}{240 \, \text{g/mol}} = 0.5 \, \text{mol}\).
2. Next, use Avogadro's number \(N_A = 6 \times 10^{23} \, \text{atoms/mol}\) to find the total number of atoms in \(0.5 \, \text{mol}\): \(\text{Number of atoms} = 0.5 \times 6 \times 10^{23} = 3 \times 10^{23}\).
3. Each fission of a \(^{240}\text{X}\) nucleus releases \(200 \, \text{MeV}\). Thus, the total energy released is \(3 \times 10^{23} \, \text{atoms} \times 200 \, \text{MeV/atom} = 6 \times 10^{25} \, \text{MeV}\).
4. The solution is within the provided range, confirming our calculations.

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