JEE Main 2026 — Jan 23, Shift 2 — Physics solution

JEE Main2026Jan 23, Shift 2Physics

The average energy released per fission for the nucleus of $_{92}^{235} U$ is $190$ MeV. When all the atoms of $47$ g pure $_{92}^{235} U$ undergo fission process, the energy released is $α \times 1 0^{23}$ MeV. The value of $α$ is ______.

(Avogadro Number $= 6 \times 1 0^{23}$ per mole)

Your answer

Solution

Number of moles in 47 g of $^{235} U$ : $n = \frac{47}{235} = \frac{1}{5}$ .

Total atoms: $N = \frac{1}{5} \cdot 6 \times 1 0^{23} = 1.2 \times 1 0^{23}$ .

Total energy: $E = N \cdot 190 = 1.2 \times 1 0^{23} \cdot 190 = 228 \times 1 0^{23}$ MeV.

$∴ α = 228$ .

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