A nucleus disintegrates into two nuclear part | vedclass

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Question

(A) The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $A$ is the mass number.Given the ratio of radii $\frac{R_1}{R_2} = \frac{1}{2^{1/3}}$,

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$2$

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(A) The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $A$ is the mass number.Given the ratio of radii $\frac{R_1}{R_2} = \frac{1}{2^{1/3}}$,we have $\frac{A_1^{1/3}}{A_2^{1/3}} = \frac{1}{2^{1/3}}$,which implies $\frac{A_1}{A_2} = \frac{1}{2}$.Thus,the ratio of their masses is $\frac{m_1}{m_2} = \frac{A_1}{A_2} = \frac{1}{2}$.According to the law of conservation of linear momentum,$m_1 v_1 = m_2 v_2$ (assuming the initial nucleus was at rest).Therefore,the ratio of their speeds is $\frac{v_1}{v_2} = \frac{m_2}{m_1} = \frac{2}{1}$.Given that the ratio of speeds is $n:1$,we find $n = 2$.

$A$ nucleus disintegrates into two nuclear parts,in such a way that the ratio of their nuclear radii is $1: 2^{1/3}$. Their respective speeds have a ratio of $n: 1$. The value of $n$ is $.....$.

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