Question

We are given the following atomic masses:
`._92U^(238)=238.05079u,`
`._90Th^(234)=234.04363u`,
`._91Pa^(237)=237.05121, ._1H^1=1.00783`,
`._2He^2=4.00260u`
(a) Calculate the energy released during `alpha` decay of `._92U^(238)`,
(b) Calculate the kinetic energy of emitted `alpha` particles,
(c) show that `._94U^(238)` cannot spontaneously emit a proton.

Answer 1

The equation representing `alpha` decay of `._92U^(238)` is
`._92U^(238)to ._90Th^(234)+._2He^4+Q`,
where Q is the kinetic energy shared by Th & He. Mass defect in the decay process is
`Deltam =m(._92U^(238))-m(._90Th^(234))-m(._2He^4)`
`=238.05081-234.04363-4.00260`
`=0.00458 a.m.u`
`:. Q=0.00458xx931MeV`
`=4.26 MeV`
This is shared between Th &He. As He is much ligther compared to Th, therefore, most of this energy is K.E. of He alone.