An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of $8:…
An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of $8: 27$. The ratio of the radii of the nuclei (assumed to be spherical ) is:
$8: 27$
$2: 3$
$3: 2$
$4: 9$
Solution
Let heavy nucleus breaks into two nuclei of mass $m_1$ and $m_2$ and move away with velocities $V_1$ and $V_2$ respectively.
According to question, $\frac{V_1}{V_2}=\frac{8}{27}$ $m_1 V_1=m_2 V_2$ (Law of momentum conservation)
$
\begin{aligned}
&\Rightarrow \frac{m_1}{m_2}=\frac{V_2}{V_1}=\frac{27}{8} \\
&\frac{\rho \times \frac{4}{3} \pi R_1^3}{\rho \times \frac{4}{3} \pi R_2^3} \quad\left(\because \text { density } \rho=\frac{\text { mass }}{\text { volume }}\right)
\end{aligned}
$
$
\begin{aligned}
&\Rightarrow\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)=\left(\frac{27}{8}\right)^{\frac{1}{3}}=\left(\frac{3}{2}\right)^{3 \times \frac{1}{3}} \\
&\therefore \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{3}{2}
\end{aligned}
$