Question

²²⁸Th emits an alpha particle to reduce to ²²⁴Ra. Calculate the kinetic energy of the alpha particle emitted in the following decay:
$$\begin{gathered} {}^{228}\mathrm{Th}\to {}^{224}\mathrm{Ra}*+\alpha \\ {}^{224}\mathrm{Ra}*{\to }^{224}\mathrm{Ra}+{\rm Y}(217\mathrm{keV}). \end{gathered}$$
Atomic mass of ²²⁸Th is 228.028726 u, that of ²²⁴Ra is 224.020196 u and that of ${}_2{}^4\mathrm{H}$ is 4.00260 u.

Answer 1

Given:
Atomic mass of ²²⁸Th, m(²²⁸Th) = 228.028726 u
Atomic mass of ²²⁴Ra, m(²²⁴Ra) = 224.020196 u
Atomic mass of ${}_2{}^4\mathrm{H}$, m(${}_2{}^4\mathrm{H}$) = 4.00260 u
Mass of ²²⁴Ra = 224.020196 × 931 + 0.217 MeV = 208563.0195 MeV

Kinetic energy of alpha particle, K = $\left[m\left({}^{228}\mathrm{Th}\right)-\left[m\left({}^{224}\mathrm{Ra}\right)+m\left({}_2{}^4\mathrm{H}\right)\right]\right]$c²
= (228.028726 × 931) − [(208563.0195 + 4.00260 × 931]
= 5.30383 MeV = 5.304 MeV