A nucleus ruptures into two nuclear parts whi | vedclass

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(B) According to the law of conservation of linear momentum,the total momentum before rupture is zero,so the magnitudes of the momenta of the two part

Vedclass · Accepted Answer

$1 : 2^{1/3}$

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(B) According to the law of conservation of linear momentum,the total momentum before rupture is zero,so the magnitudes of the momenta of the two parts must be equal:$m_1 v_1 = m_2 v_2$$\frac{m_1}{m_2} = \frac{v_2}{v_1}$Given the velocity ratio $\frac{v_1}{v_2} = \frac{2}{1}$,we have $\frac{v_2}{v_1} = \frac{1}{2}$.Since the mass of a nucleus is proportional to its mass number $A$ $(m \propto A)$,we have $\frac{A_1}{A_2} = \frac{m_1}{m_2} = \frac{1}{2}$.The nuclear radius $R$ is related to the mass number $A$ by the formula $R = R_0 A^{1/3}$.Therefore,the ratio of the radii is $\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} = \left(\frac{1}{2}\right)^{1/3} = 1 : 2^{1/3}$.

$A$ 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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