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 i $s _______.$
(Avogadro Number = 6 × 10²³ per mole)

Question

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 i $s _______.$

(Avogadro Number = 6 × 10²³ per mole)

G1 Physics · Accepted Answer

Molar mass of U-235 is 235 g/mol. Therefore, 47 g = 47/235 mol = 1/5 mol. Number of U-235 nuclei: N = (1/5)(6×10²³) = 1.2×10²³. Energy released per fission is 190 MeV. Hence the total energy is E = N×190 = (1.2×10²³)(190) = 228×10²³ MeV. Therefore, α = 228.

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 i s _______. (Avogadro Number = 6 × 10²³ per mole)