A nucleus ruptures into two nuclear parts whi | vedclass

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(B) According to the law of conservation of linear momentum,the initial momentum of the nucleus is zero,so the final momenta of the two parts must be

Vedclass · Accepted Answer

$1 : 2^{1/3}$

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(B) According to the law of conservation of linear momentum,the initial momentum of the nucleus is zero,so the final momenta of the two parts must be equal and opposite: $m_1 v_1 = m_2 v_2$.Given the velocity ratio $\frac{v_1}{v_2} = \frac{2}{1}$,we have $\frac{m_2}{m_1} = \frac{v_1}{v_2} = \frac{2}{1}$.Since the density $\rho$ of nuclear matter is constant,the mass $m$ is proportional to the volume $V$,where $V = \frac{4}{3} \pi r^3$. Thus,$\frac{m_2}{m_1} = \frac{r_2^3}{r_1^3}$.Equating the ratios: $\frac{r_2^3}{r_1^3} = \frac{2}{1}$.Taking the cube root on both sides: $\frac{r_2}{r_1} = 2^{1/3}$.Therefore,the ratio of their radii $r_1 : r_2 = 1 : 2^{1/3}$.

$A$ nucleus ruptures into two nuclear parts which have their velocity ratio equal to $2 : 1$. What will be the ratio of their nuclear size (nuclear radius)?

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