JEE Main 2023 · 12 April · morning · Numerical answer · +4 / −0
A common example of alpha decay is ₉₂²³⁸ U → ₉₀²³⁴ Th+ ₂ He⁴+Q
Given :
₉₂²³⁸ U=238.05060 u,
₉₀²³⁴ Th=234.04360 u,
₂⁴ He=4.00260 u and
1 u=931.5 MeV/c²
The energy released (Q) during the alpha decay of ₉₂²³⁸ U is __________ MeV
Answer and solution
Answer 4
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