A stationary $Pb^{200}$ nucleus undergoes alpha decay, emitting an alpha particle with a kinetic ene

Question

A stationary $Pb^{200}$ nucleus undergoes alpha decay, emitting an alpha particle with a kinetic energy $T_{\alpha} = 5.77 	ext{ MeV}$. What fraction of the total energy liberated in this decay ($Q$-value) is accounted for by the recoil energy of the daughter nucleus?

Professor PC Thomas Classes · Answer

By conservation of momentum, $p_{\alpha} = p_d$ (where $d$ is the daughter nucleus).
Kinetic energy $K = p^2 / 2m$, so $K \propto 1/m$.
The recoil energy of the daughter nucleus is $T_d = T_{\alpha} (m_{\alpha} / m_d)$.
The total energy liberated is $Q = T_{\alpha} + T_d = T_{\alpha} (1 + m_{\alpha}/m_d) = T_{\alpha} (m_d + m_{\alpha}) / m_d$.
The fraction is $f = T_d / Q = m_{\alpha} / (m_{\alpha} + m_d)$.
Here, $m_{\alpha} \approx 4 	ext{ u}$ and the parent mass is $200 	ext{ u}$, so $m_{\alpha} + m_d = 200 	ext{ u}$.
$f = 4 / 200 = 1/50 = 2\%$.