A nucleus disintegrates into two nuclear fragments which move apart, having their velocities in the ratio 1 : 3. The ratio of their nuclear sizes, respectively, will be

$\sqrt[3]{3}:1$

The correct answer is Option (1) → $\sqrt[3]{3}:1$

Let the masses of two fragments be $m_1$ and $m_2$, and their velocities $v_1$ and $v_2$.

Given: $v_1 : v_2 = 1 : 3$

By conservation of momentum:

$m_1 v_1 = m_2 v_2 \;\;\;\; \Rightarrow \;\;\; \frac{m_1}{m_2} = \frac{v_2}{v_1} = \frac{3}{1} = 3$

So, $m_1 : m_2 = 3 : 1$

Since nuclear mass $\propto$ nuclear volume $\propto R^3$,

$\frac{R_1}{R_2} = \left(\frac{m_1}{m_2}\right)^{\frac{1}{3}} = \left(\frac{3}{1}\right)^{\frac{1}{3}} = \sqrt[3]{3}$

Hence, the ratio of nuclear sizes is $R_1 : R_2 = \sqrt[3]{3} : 1$