If a nucleus D at rest breaks up into two fra | vedclass

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(A) According to the law of conservation of linear momentum,since the initial nucleus is at rest,the magnitudes of the momenta of the two fragments mu

Vedclass · Accepted Answer

$1 : 2$

Vedclass · Answer

(A) According to the law of conservation of linear momentum,since the initial nucleus is at rest,the magnitudes of the momenta of the two fragments must be equal:$m_1 v_1 = m_2 v_2$Given the ratio of velocities is $v_1 / v_2 = 8 / 1$,we have:$m_1 / m_2 = v_2 / v_1 = 1 / 8$Since the mass of a nucleus is proportional to its mass number $A$ (i.e.,$m \propto A$),the ratio of mass numbers is $A_1 / A_2 = m_1 / m_2 = 1 / 8$.The radius of a nucleus is given by $r = R_0 A^{1/3}$,so the ratio of the radii is:$r_1 / r_2 = (A_1 / A_2)^{1/3}$$r_1 / r_2 = (1 / 8)^{1/3} = 1 / 2$Thus,the ratio of the radii is $1 : 2$.

If a nucleus $D$ at rest breaks up into two fragments $A$ and $C$ which fly off with velocities in the ratio $8 : 1$,then the ratio of the radii of the fragments is

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