The average energy released per fission for the nucleus of ²³⁵₉₂U is 190 MeV. When all the atoms of 47 g pure ²³⁵₉₂U undergo fission process, the energy released is α×10²³ MeV. The value of α is _ _ _ _. (Avogadro Number =6×10²³ per mole)

Atoms and Nuclei: JEE Main 2026 Physics question

JEE Main 2026 · 23 Jan, Evening shift · Q49

JEE Main 202623 Jan, Evening shiftPhysicsMediumAtoms and Nuclei183× since 2002

The average energy released per fission for the nucleus of

^{235}_{92}\mathrm{U}

is 190 MeV. When all the atoms of 47 g pure

^{235}_{92}\mathrm{U}

undergo fission process, the energy released is

\alpha\times10^{23}

MeV. The value of

\alpha

is _______. (Avogadro Number

=6\times10^{23}

per mole)

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2026 Atoms and Nuclei Question with Solution (23 Jan Evening shift, Q49) | JEE Arc