Question

${}^{228}\text{Th}$ emits an alpha particle to reduce to ${}^{224}\text{Ra}$. Calculate the kinetic energy of the alpha particle emitted in the following decay:

${}^{228}\text{Th}{\to }^{224}{\text{Ra}}^{\ast }+\alpha$

${}^{224}{\text{Ra}}^{\ast }{\to }^{224}\text{Ra}+\gamma \left(217\text{keV}\right)$

Answer 1

${}^{228}\text{Th}{\to }^{224}{\text{Ra}}^{\ast }+\alpha$

${}^{224}{\text{Ra}}^{\ast }{\to }^{224}\text{Ra}+v\left(217\text{keV}\right)$

$\left[\begin{array}{ccc}{\text{Mass}}^{228}\text{Th}& \to & 228.028726u\\ {}^{224}\text{Ra}& \to & 224.020196u\\ \alpha {=}_4^2\text{He}& \to & 4.002650u\end{array}\right]$

Now, Mass of

${}^{224}\text{Ra}=224.020196\times 931+0.217\text{MeV}$

= $228.028726\times 931-[208563.0195+4.00260\times 931]$

= $5.30383\text{MeV}=5.34\text{MeV}$