NVAEasyJEE Main 2026 · 23 January, Shift 2Nuclear Fission & Fusion

Physics Question from JEE Main 2026 · 23 January, Shift 2

The average energy released per fission for the nucleus of 92235U^{235}_{92}U is 190MeV190 \, \text{MeV}. When all the atoms of 47g47 \, \text{g} pure 92235U^{235}_{92}U undergo fission process, the energy released is α×1023\alpha \times 10^{23} MeV\text{MeV}. The value of α\alpha is _____. (Avogadro Number =6×1023= 6 \times 10^{23} per mole)

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