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(C) Let the heavy nucleus break into two nuclei of masses $m_1$ and $m_2$ moving with velocities $V_1$ and $V_2$ respectively.According to the law of

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$3 : 2$

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(C) Let the heavy nucleus break into two nuclei of masses $m_1$ and $m_2$ moving with velocities $V_1$ and $V_2$ respectively.According to the law of conservation of linear momentum,the initial momentum is zero,so $m_1 V_1 = m_2 V_2$.Given the ratio of velocities $\frac{V_1}{V_2} = \frac{8}{27}$,we have $\frac{m_1}{m_2} = \frac{V_2}{V_1} = \frac{27}{8}$.Assuming constant nuclear density $ho$,the mass of a nucleus is $m = ho 	imes \frac{4}{3} \pi R^3$.Therefore,$\frac{m_1}{m_2} = \frac{R_1^3}{R_2^3} = \frac{27}{8}$.Taking the cube root on both sides,$\frac{R_1}{R_2} = \left( \frac{27}{8} ight)^{1/3} = \frac{3}{2}$.Thus,the ratio of the radii is $3:2$.

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

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