Question

A heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio 3 : 1. The ratio of radii of the fragments is

• A. $1 : 3^{1/3}$
• B. $3^{1/3} : 4$
• C. $4 : 1$
• D. $2 : 1$

Answer 1

Answer: (A)

$1 : 3^{1/3}$

Answer 2

As the heavy nucleus at rest breaks, therefore according to law of conservation of momentum, we get $m_{1}\upsilon_{1} + m_{2}\upsilon _{2} = 0$ or $\frac{\upsilon _{1}}{\upsilon _{2}} = \frac{m_{2}}{m_{1}} = \frac{3}{1} \quad.... \left(i\right)$ As nuclear density is same, $\therefore\quad\frac{m_{2}}{m_{1}} = \frac{\rho \frac{4}{3}\pi R^{3}_{1}}{\rho \frac{4}{3}\pi R^{3}_{2}} = \frac{R^{3}_{1}}{R^{3}_{2}}$ or $\frac{R^{3}_{1}}{R^{3}_{2}} = \frac{m_{1}}{m_{2}} = \frac{1}{3}\quad$ (Using $\left(i\right)$) $\therefore \quad R_{1} : R_{2} = 1 : 3^{1/3}$