A nucleus splits into two parts. If the ratio | vedclass

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(A) 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 b

Vedclass · Accepted Answer

$1:2$

Vedclass · Answer

(A) 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 in magnitude: $m_1 v_1 = m_2 v_2$.Given the ratio of velocities $\frac{v_1}{v_2} = \frac{8}{1}$, we have $\frac{m_1}{m_2} = \frac{v_2}{v_1} = \frac{1}{8}$.Since the mass of a nucleus is proportional to its mass number $A$ $(m \propto A)$, we have $\frac{A_1}{A_2} = \frac{m_1}{m_2} = \frac{1}{8}$.The radius $r$ of a nucleus is related to its mass number $A$ by the relation $r = R_0 A^{1/3}$, which implies $r \propto A^{1/3}$.Therefore, the ratio of the radii is $\frac{r_1}{r_2} = \left( \frac{A_1}{A_2} \right)^{1/3}$.Substituting the values: $\frac{r_1}{r_2} = \left( \frac{1}{8} \right)^{1/3} = \frac{1}{2}$.Thus, the ratio of the radii is $1:2$.

$A$ nucleus splits into two parts. If the ratio of their velocities is $8 : 1$, what is the ratio of their radii?

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