Q72 / 75 · JEE Main 2026 (23 January Shift 2) · Physics
The average energy released per fission for the nucleus of ${ }_{92}^{235} \mathrm{U}$ is 190 MeV. When all the atoms of 47 g pure ${ }_{92}^{235} \mathrm{U}$ undergo fission process, the energy released is $\alpha \times 10^{23} \mathrm{MeV}$. The value of $\alpha$ is $\_\_\_\_$.
(Avogadro Number $=6 \times 10^{23}$ per mole)
(Avogadro Number $=6 \times 10^{23}$ per mole)
Correct answer: 228
Solution
The number of $^{235}\text{U}$ atoms in 47 g is determined using Avogadro's number.
Molar mass of $^{235}\text{U}$ is 235 g/mol, so the number of moles is $n = 47/235$ mol.
The total number of atoms is $N = (47/235) \times 6 \times 10^{23}$.
Since each fission releases 190 MeV, the total energy released is:
$E_{total} = N \times 190 = \frac{47 \times 6 \times 10^{23}}{235} \times 190 = \frac{47 \times 6 \times 190}{235} \times 10^{23}$ MeV.
Calculating the numerator: $47 \times 6 \times 190 = 282 \times 190 = 53580$.
Therefore: $E_{total} = \frac{53580}{235} \times 10^{23} = 228 \times 10^{23}$ MeV, giving $\alpha = 228$.
Molar mass of $^{235}\text{U}$ is 235 g/mol, so the number of moles is $n = 47/235$ mol.
The total number of atoms is $N = (47/235) \times 6 \times 10^{23}$.
Since each fission releases 190 MeV, the total energy released is:
$E_{total} = N \times 190 = \frac{47 \times 6 \times 10^{23}}{235} \times 190 = \frac{47 \times 6 \times 190}{235} \times 10^{23}$ MeV.
Calculating the numerator: $47 \times 6 \times 190 = 282 \times 190 = 53580$.
Therefore: $E_{total} = \frac{53580}{235} \times 10^{23} = 228 \times 10^{23}$ MeV, giving $\alpha = 228$.
Hence, the correct answer is 228.