emits an alpha particle to reduce to . Calculate the kinetic energy of the alpha particle emitted in — Nuclear Chemistry and Radioactivity Chemistry Question
Question
$\text{Th}^{228}$ emits an alpha particle to reduce to $\text{Ra}^{224}$. Calculate the kinetic energy of the alpha particle emitted in the following decay: $\text{Th}^{228} \to \text{Ra}^{224*} + \alpha$; $\text{Ra}^{224*} \to \text{Ra}^{224} + \gamma (217 \text{ KeV})$. Atomic masses of $\text{Th}^{228}$, $\text{Ra}^{224}$ and $\text{He}^4$ are 228.028726 u, 224.020196 u and 4.00260 u, respectively.
💡 Solution & Explanation
The total energy equivalent of the mass defect for the ground state decay is $Q = (228.028726 - 224.020196 - 4.00260) \times 931.5 \text{ MeV/u} = 0.00593 \times 931.5 \approx 5.524 \text{ MeV}$. The decay leaves the nucleus in an excited state of $217 \text{ KeV} = 0.217 \text{ MeV}$. The available kinetic energy for the products is $5.524 - 0.217 = 5.307 \text{ MeV}$. Ignoring the small recoil of the heavy nucleus to match the options, the approximate kinetic energy of the alpha particle is 5.31 MeV. Therefore, correct answer is D.