A common example of alpha decay is
A common example of alpha decay is
Correct Answer: 4
Solution and Explanation
Energy released \( Q \) is given by the equation: \[ Q = (\Delta m)_{\text{amu}} \times 931.5 \, \text{MeV}. \] Here, the mass defect \( \Delta m \) is: \[ \Delta m = m_u - m_{\text{Th}} - m_{\text{He}} = 238.05060 \, u - 234.04360 \, u - 4.00260 \, u = 0.0044 \, u. \] Thus, \[ Q = 0.0044 \times 931.5 \, \text{MeV} = 4.0986 \, \text{MeV}. \] Therefore, the energy released during the alpha decay of \( ^{238}_{92} U \) is 4.0986 MeV.
Top Questions on work, energy and power
- Given below are two statements:
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Statement II: Any object moving from one location to another location can follow infinite number of paths. Therefore, the amount of work done by the object changes with the path it follows for a conservative force.
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