JEE Main 2026 — Nuclear Physics Question with Solution
JEE Main 2026 (23 January Shift 2)
Question
The average energy released per fission for the nucleus of is 190 MeV. When all the atoms of 47 g pure undergo fission process, the energy released is . The value of is .
(Avogadro Number per mole)
(Avogadro Number per mole)
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Show full solutionCorrect answer: 228
Correct answer
228
Step-by-step explanation
The number of atoms in 47 g is determined using Avogadro's number.
Molar mass of is 235 g/mol, so the number of moles is mol.
The total number of atoms is .
Since each fission releases 190 MeV, the total energy released is:
MeV.
Calculating the numerator: .
Therefore: MeV, giving .
Molar mass of is 235 g/mol, so the number of moles is mol.
The total number of atoms is .
Since each fission releases 190 MeV, the total energy released is:
MeV.
Calculating the numerator: .
Therefore: MeV, giving .
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