JEE Main 23 Jan 2026 Shift 2 Paper Solved Paper

Section: Physics

Question : 47 of 75

Marks: +1, -0

The average energy released per fission for the nucleus of ‌92235U is 190 MeV . When all the atoms of 47 g pure ‌92235U undergo fission process, the energy released is α×1023MeV. The value of α is ____ .
(Avogadro Number =6×1023 per mole)

[23 jan 2026 Shift 2]

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