When 241 Am decays by emitting an alpha particle of kinetic energy $5.5 \mathrm{MeV}$, what is the recoil energy of the daughter nucleus?

Question

Quizplus · Accepted Answer

The recoil energy of the daughter nucleus can be calculated using conservation of momentum. The momentum of the alpha particle and the daughter nucleus must be equal and opposite. Since kinetic energy $KE = \frac{p^2}{2m}$, where $p$ is momentum and $m$ is mass, and given that the mass of an alpha particle is roughly 4 u (atomic mass units) and the mass of the resulting nucleus from $^{241}$Am decay is roughly 237 u, we can set up a ratio of their kinetic energies based on their masses (since $p^2$ is proportional to $m \times KE$). The kinetic energy of the alpha particle is 5.5 MeV, so the recoil energy of the daughter nucleus is $\frac{4}{237} \times 5.5 \mathrm{MeV} \approx 0.093 \mathrm{MeV}$.

When 241 Am Decays by Emitting an Alpha Particle of Kinetic