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 $N_A = 6 \times 10^{23}$)

Question

The average energy released per fission for the nucleus of $^{235_{92}U$ is 190 MeV. When all the atoms of 47 g pure $^{235}_{92}U$ undergo fission process, the energy released is $\alpha \times 10^{23}$ MeV. The value of $\alpha$ is ___. (Avogadro Number $= 6 \times 10^{23}$ per mole)}

Answer 1

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

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

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

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 \(N_A = 6 \times 10^{23}\))

Correct Answer: 6

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