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A common example of alpha decay is ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He + Q Given: ²³⁸₉₂U = 238.05060 u ²³⁴₉₀Th = 234.04360 u ⁴₂He = 4.00260 u and 1u = 931.5 MeV/c² The energy released (Q) during the alpha decay of ²³⁸₉₂U is ____ MeV.
- 4
- 0.4
- 40
- 400
Correct answer: 4
Solution
The energy released during alpha decay can be calculated using the mass defect, which is the difference in mass between the original nucleus and the sum of the masses of the decay products. In this case, the mass difference corresponds to a release of approximately 4 MeV, making option A the correct choice.
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