An unstable heavy nucleus at rest breaks into | vedclass

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Question

Let heavy nucleus breaks into two nuclei of mass $m_{1}$ and $m_{2}$ and move away with velocities ${V}_{1}^1$ and $V_{2}^2$ respectively.

Accordin

Vedclass · Accepted Answer

$3 : 2$

Vedclass · Answer

Let heavy nucleus breaks into two nuclei of mass $m_{1}$ and $m_{2}$ and move away with velocities ${V}_{1}^1$ and $V_{2}^2$ respectively.

According to question, $\frac{V_{1}}{V_{2}}=\frac{8}{27}$ $m_{1} V_{1}=m_{2} V_{2}$ (Law of momentum conservation)

$\Rightarrow$ $\frac{{{m_1}}}{{{m_2}}} = \frac{{{V_2}}}{{{V_1}}} = \frac{{27}}{8}$

$ho  	imes \frac{4}{3}\pi R_1^3$     $\left( {\because {	ext{ density }}ho  = \frac{{{	ext{ mass }}}}{{{	ext{ volume }}}}} ight)$

$ho  	imes \frac{4}{3}\pi R_2^3$  

$ \Rightarrow \left( {\frac{{{R_1}}}{{{R_2}}}} ight) = $ ${\left( {\frac{{27}}{8}} ight)^{\frac{1}{3}}} = {\left( {\frac{3}{2}} ight)^{3 	imes \frac{1}{3}}}$

$	herefore \frac{R_{1}}{R_{2}}=\frac{3}{2}$

An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of $8:27$. The ratio of the radii of the nuclei (assumed to be spherical ) is

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