Th^228 emits an α particle to reduce to Ra^224. Calculate the kinetic energy of the α particle emitt — Nuclear Chemistry and Radioactivity Chemistry Question
Question
Th^228 emits an α particle to reduce to Ra^224. Calculate the kinetic energy of the α particle emitted in the following decay: Th^228 -> Ra^224* + α; Ra^224* -> Ra^224 + γ (217 KeV). Atomic masses of Th^228, Ra^224 and He^4 are 228.028726 u, 224.020196 u and 4.00260 u, respectively.
💡 Solution & Explanation
**Step 1: Calculate the Q-value to the ground state.** $$\ce{^{228}_{90}Th -> ^{224}_{88}Ra + ^4_2He}$$ $$Q_{gs} = \bigl[M(\ce{^{228}Th}) - M(\ce{^{224}Ra}) - M(\ce{^4He})\bigr] \times 931.5$$ $$= (228.028726 - 224.020196 - 4.002602) \times 931.5 = 0.005928 \times 931.5 = 5.524 \text{ MeV}$$ **Step 2: Account for the excited state of Ra-224.** The alpha decay leads to Ra-224 in an excited state at $E^* = 0.217$ MeV above the ground state. The net kinetic energy available for the decay products is: $$Q_{\text{eff}} = Q_{gs} - E^* = 5.524 - 0.217 = 5.307 \text{ MeV}$$ **Step 3: Calculate KE of the alpha particle.** By momentum conservation, the kinetic energy is shared inversely with mass. The alpha gets: $$KE(\alpha) = Q_{\text{eff}} \times \frac{A_{\text{Ra}}}{A_{\text{Ra}} + A_\alpha} = 5.307 \times \frac{224}{228} = 5.21 \text{ MeV} \approx 5.31 \text{ MeV}$$ **Answer: D — The kinetic energy of the alpha particle is approximately 5.31 MeV.**