QJEEM18Ph39b6: JEE Main 2018 Physics - Class 12 JEEM18Ph39b6 Solution: An unstable heavy nucleus at rest breaks into two nuclei which…

The two nuclei have velocity in ratio 8 : 27. By conservation of momentum, we have

${m_1}{v_1} = {m_2}{v_2} \Rightarrow {{{v_1}} \over {{v_2}}} = {{{m_2}} \over {{m_1}}} \Rightarrow {{{m_2}} \over {{m_1}}} = {8 \over {27}}$

Now, since $m = \rho {4 \over 3}\pi {r^3}$

Therefore, ${{{m_2}} \over {{m_1}}} = {{\rho {4 \over 3}\pi r_2^3} \over {\rho {4 \over 3}\pi r_1^3}} \Rightarrow {{{m_2}} \over {{m_1}}} = {\left( {{{{r_2}} \over {{r_1}}}} \right)^3} \Rightarrow {\left( {{{{r_2}} \over {{r_1}}}} \right)^3} = {8 \over {27}}$

$\Rightarrow {{{r_2}} \over {{r_1}}} = {2 \over 3}$

Thus, ratio of radii of nuclei ${r_1}:{r_2} = 3:2$ .