Find the kinetic energy of emitted a-particles in the following nuclear reaction:
`""_(94)Pu^(238) to ""_(92)U^(234) + ""_(2)He^(4)`
`m(Pu^(238)) = 238.04954` amu
`m(U^(234)) = 234.04096` amu
`m(""_(2)He^(4)) = 4.002603` amu

Explanation

To find the kinetic energy of the emitted alpha particles in the nuclear reaction:

\[ _{94}^{238}\text{Pu} \rightarrow _{92}^{234}\text{U} + _{2}^{4}\text{He} \]

we will follow these steps:

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