Find the kinetic energy of the `alpha` - particle emitted in the decay `^238 Pu rarr ^234 U + alpha`. The atomic masses needed are as following:
`^238 Pu 238.04955 u`
`^234 U 234.04095 u`
`^4 He 4.002603 u`.
Neglect any recoil of the residual nucleus.

Explanation

Using energy conservation,
`m(^238 Pu) c^2 = m(^234 U) - m (^4 He) c^2 + K`
or, `K = [m(^238 Pu) - m (^234 U) - m (^4 He)] c^2`
`= [238.04955 u - 234.04095 u - 4.002603 u] (931 MeV u^(-1)`
`= 5.58 MeV`.