Question 49

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

Correct Answer: 228

Moles of U-235 = 47/235 = 0.2. Atoms = 0.2 × 6×10²³ = 1.2×10²³. Energy = 1.2×10²³ × 190 = 228×10²³ MeV. So α = 228.

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