The energy released per fission of nucleus of ²⁴⁰ X is 200 MeV. The energy released if all the atoms in 120g of pure ²⁴⁰ X undergo fission is ____________ × 10 ²⁵ MeV. (Given NA = 6 × 10²³) Numerical answer type. Work it out, then open the answer.

Answer: 6. Number of moles = (120)/(240) = 0.5, so number of nuclei N = 0.5 × 6×10²³ = 3×10²³. Total energy E = N × 200 MeV = 3×10²³× 200 = 6×10²⁵ MeV. So the answer is 6.

The energy released per fission of nucleus of ²⁴⁰ X is 200 MeV. The… | JEE Main 2023 (24 Jan, Shift 2)

Q17JEE Main 2023 · 24 JAN · SHIFT 2Easy

Nuclear Fission & Fusion

The energy released per fission of nucleus of ${}^{240}$ X is 200 MeV. The energy released if all the atoms in 120g of pure ${}^{240}$ X undergo fission is ____________ $\times$ 10 ${}^{25}$ MeV.

(Given $\text{N}_{\text{A}} = 6 \times 10^{23}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 6

Number of moles $= \dfrac{120}{240} = 0.5$ , so number of nuclei $N = 0.5 \times 6\times10^{23} = 3\times10^{23}$ .
Total energy $E = N \times 200\text{ MeV} = 3\times10^{23}\times 200 = 6\times10^{25}\text{ MeV}$ .
So the answer is $6$ .

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