Question

A stationary $P{b}^{200}$ nucleus emits an $\alpha -$particle with a $K.E.$, $K=5.77MeV$. Find the recoil velocity of daughter nucleus. What fraction of the total energy liberated in this decay is accounted for by the daughter nucleus?

Answer 1

The momentum of the $\alpha -$particle is given by,
$p_{\alpha }=\sqrt{2m_{\alpha k}}$ . . .($1$)
Let the recoiled momentum of the daughter nucleus be $p_d=m_d{v}_d$ where $m_d$ and $v_d$ are the mass and velocity of daughter nucleus respectively. Using the principle of conservation of momentum we get,
$P_d=P_{\alpha }=\sqrt{2m_{\alpha k}}$
$\Rightarrow v_d=\frac{\sqrt{{2m}_{\alpha }k}}{m_d}$ . . .($2$)
$\Rightarrow v_d=\frac{1}{196}\sqrt{\frac{2\times 4\times k}{m_p}}$
$=\frac{2}{196}\sqrt{\frac{2k}{m_p}}$ where $m_p$ is the mass of the proton.
$\Rightarrow v_d=3.39\times {10}^{5}m/s$
Let the $K.E.$ of the daughter nucleus be $K$, then,
$\frac{k}{k}=\frac{m_{\alpha }}{m_d}$ as the momenta are same.
$\therefore \frac{k}{k+k}=\frac{m_{\alpha }}{m_{\alpha }{+}_d}$
But $K+K$ is the total energy, $K_t$(say).
$\therefore \frac{k}{k_t}=\frac{m_{\alpha }}{m_{\alpha }+m_d}$
$\Rightarrow k=\frac{m_{\alpha }}{m_{\alpha }+m_d}{k}_1=\frac{4}{196+4}{k}_t$
$\Rightarrow k=0.02k_t$
$\Rightarrow \frac{k}{k_t}=0.02$.