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

The decay is represented by the nuclear reaction: `._94Pu^(238) to ._92U^(234)+._2He^4(alpha "particle")+Q`
Mass defect, `Deltam=mass of Pu^(238)-("mass of" U^(234)+"mass of" He^4)`
`=238.04954-(234.04096+4.002603)`
`=238.04954-238.043563=0.005977` a.m.u.
Q value of the reaction `=0.005977xx931MeV`
`=5.564MeV`.
This energy is shared by `._92U^(234)` and `alpha`- particle. As `._92U^(234)` is very heavy compared to `alpha` particle, therefore, this energy is carried mostly by the alpha particle.
`:.` K.E. of alpha particle `=5.564MeV`