A heavy nucleus at rest breaks into two fragm | vedclass

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(A) By the law of conservation of momentum,for a nucleus at rest,the magnitudes of momenta of the two fragments must be equal:$m_1 v_1 = m_2 v_2$$\fra

Vedclass · Accepted Answer

$1 : 2$

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(A) By the law of conservation of momentum,for a nucleus at rest,the magnitudes of momenta of the two fragments must be equal:$m_1 v_1 = m_2 v_2$$\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 $r = r_0 A^{1/3}$,where $r_0$ is a constant.Therefore,the ratio of the radii is:$\frac{r_1}{r_2} = \left( \frac{A_1}{A_2} \right)^{1/3} = \left( \frac{1}{8} \right)^{1/3} = \frac{1}{2}$.Thus,the ratio of the radii is $1 : 2$.

$A$ heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio $8 : 1$. The ratio of radii of the fragments is

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