Question

A common example of alpha decay is 
 

Answer 1

Correct Answer: 4

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.