Question

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 :

• A. 8 : 27
• B. 4 : 9
• C. 3 : 2
• D. 2 : 3

Answer 1

Answer: (C)

The two nuclei have velocity in ratio $8:27$. By conservation of momentum, we have
$m_1{v}_1=m_2{v}_2$
$\Rightarrow \frac{v_1}{v_2}=\frac{m_2}{m_1}$
$\Rightarrow \frac{m_2}{m_1}=\frac{8}{27}$
Now, since $m=\rho \frac{4}{3}\pi r^3$
Therefore $\frac{m_2}{m_1}=\frac{\rho \frac{4}{3}\pi r_2^3}{\rho \frac{4}{3}\pi r_1^3}$
$\Rightarrow \frac{m_2}{m_1}={\left(\frac{r_2}{r_1}\right)}^3$
$\Rightarrow {\left(\frac{r_2}{r_1}\right)}^3=\frac{8}{27}$
$\Rightarrow \frac{r_2}{r_1}=\frac{2}{3}$
Thus, ratio of radii of nuclei $r_1:r_2=3:2$.