Q32 / 199 · NTA Abhyas (NEET) 2020 · Physics · Nuclear Physics

A nucleus disintegrates into two nuclear parts which have their velocities in the ratio $2 : 1$ . The ratio of their nuclear size will be

A
$2^{\frac{1}{3}} : 1$
B
$1 : 3^{\frac{1}{2}}$
C
$3^{\frac{1}{2}} : 1$
Correct answer
D
$1 : 2^{\frac{1}{3}}$

Solution

According to the law of conservation of momentum, we get

m_{1} v_{1} = m_{2} v_{2}

Or

(\frac{m_{1}}{m_{2}}) = (\frac{v_{2}}{v_{1}})

But

m = \frac{4}{3} π r^{3} ρ o r m \propto r^{3}

\Rightarrow \frac{m_{1}}{m_{2}} = \frac{r_{1}^{2}}{r_{2}^{3}}

∴ \frac{r_{1}^{3}}{r_{2}^{3}} = \frac{v_{2}}{v_{1}} = \frac{1}{2}

\Rightarrow \frac{r_{1}}{r_{2}} = {(\frac{1}{2})}^{\frac{1}{3}}

∴ r_{1} : r_{2} = 1 : {(2)}^{\frac{1}{3}}

Hence, the correct answer is option: 'D'.

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