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JEE Main 2023 · 24 Jan · Shift 2 · numerical answer

JEE Main 2023 Physics question on Atoms and Nuclei

The energy released per fission of nucleus of 240X^{240}\text{X} is 200 MeV. The energy released if all the atoms in 120g of pure 240X^{240}\text{X} undergo fission is __________ ×1025\times 10^{25} MeV.
(Given NA=6×1023N_A = 6 \times 10^{23})
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Answer: 6

Solution

Number of atoms in 120g of ²⁴⁰X = (120/240) × Nₐ = 0.5 × 6×10²³ = 3×10²³.

Total energy = 3×10²³ × 200 MeV = 6×10²⁵ MeV = 6 × 10²⁵ MeV.

So the answer is 6.

[JEE Main 2023 · 24 Jan · Shift 2]

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