Answer to Question #153052 in Atomic and Nuclear Physics for Muhammad

Question

Find the energy released in the alpha decay
238
92 U →234
90 T h +
4
2 He.

Answer 1

m_i = 238.050788 u

m_f = 234.043601 u + 4.002603 u = 238.046204 u

Δm = |m_f-m_i| = $4.584 \times 10^{-3} \;u$

$E = Δmc^2 \\ E = 4.584 \times 10^{-3} \times 931.5 = 4.26 \;MeV$

Answer: 4.26 MeV