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A nucleus of mass number $A$, originally at rest, emits an $\alpha$-particle with speed $v$. The daughter nucleus recoils with a speed
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Verified Answer
The correct answer is:
$\frac{4 v}{A-4}$
${ }^A X \quad \rightarrow{ }^{A-4} Y+\alpha^4$ (parent nucleus) (daughter nucleus)
As the parent nucleus is at rest, then according to conservation of momentum.
$$
\begin{aligned}
& \Rightarrow \quad 0=m_Y v_Y+m_\alpha v_\alpha \\
& \Rightarrow \quad 0=(A-4) v_Y+4 v
\end{aligned}
$$
$$
\Rightarrow \quad v_Y=\frac{4 v}{A-4} \text {, the recoil speed of daughter nucleus. }
$$
As the parent nucleus is at rest, then according to conservation of momentum.
$$
\begin{aligned}
& \Rightarrow \quad 0=m_Y v_Y+m_\alpha v_\alpha \\
& \Rightarrow \quad 0=(A-4) v_Y+4 v
\end{aligned}
$$
$$
\Rightarrow \quad v_Y=\frac{4 v}{A-4} \text {, the recoil speed of daughter nucleus. }
$$
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