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. 4:09
• C. 3:02
• D. 2:03

Answer 1

Answer: (C)

3:02

Answer 2

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 $r_{1}: r_{2}=3: 2$.