A ^ 228 Th nucleus emits an alpha particle, reducing to ^ 224 Ra ....

The mass defect for the decay to the ground state is given by:
$\Delta m = m(^{228}\text{Th}) - m(^{224}\text{Ra}) - m(^{4}_{2}\text{He})$
$\Delta m = 228.028726 - 224.020196 - 4.00260 = 0.005930 \text{ u}$

The total energy released (Q-value) in the decay to the ground state is:
$Q = \Delta m \times 931 \text{ MeV/u}$
$Q = 0.005930 \times 931 \approx 5.5208 \text{ MeV}$

Since the decay leaves the ^{224}\text{Ra}$nucleus in an excited state which subsequently emits a$217 \text{ keV} = 0.217 \text{ MeV}$gamma ray, the energy available for the kinetic energy of the alpha particle and the recoiling nucleus is:<br>$Q' = Q - E_{\gamma}$<br>$Q' = 5.5208 \text{ MeV} - 0.217 \text{ MeV} = 5.3038 \text{ MeV}$<br><br>By conservation of linear momentum, the kinetic energy of the emitted alpha particle is shared inversely with their masses:<br>$K_{\alpha} = \dfrac{A - 4}{A} Q'$<br>where$A = 228$is the mass number of the parent nucleus.<br><br>$K_{\alpha} = \dfrac{224}{228} \times 5.3038 \text{ MeV}$<br>$K_{\alpha} = \dfrac{56}{57} \times 5.3038 \text{ MeV} \approx 5.211 \text{ MeV}