Home Physics Atomic and Nuclear Physics JEE Main 2026 - ( Nuclear Physics ) The average energy released per fission for …
Physics Atomic and Nuclear Physics JEE Main 2026 - ( Nuclear Physics ) MCQ (Single Correct)

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)

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The number of atoms in 47 g is determined using Avogadro's number.

Molar mass of is , so the number of moles is .
The total number of atoms is .
Since each fission releases 190 MeV , the total energy released is:
.

Calculating the numerator: .
Therefore: , giving .

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