Physics · Nuclear Physics
A common example of alpha decay is: 23892U → 23490Th + 42He + Q Given atomic masses: 23892U = 238.05079 u23490Th = 234.04363 u42He = 4.00260 u and 1 u = 931.5 MeV/c2. The energy released (Q) during this alpha decay is approximately:
Verified Solution4 OptionsUpdated: October 2026👁️1 View
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- 12.25 MeV
- 23.15 MeV
- 35.50 MeV
- 44.25 MeV✓
Correct answer
4.25 MeV
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