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Q.

A radioactive nuclei $_{z} X^{A}$ emits an $α ({He}^{+ +})$ particle. Speed of $α$ particle is v then the relative velocity of separation between them.

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a

$\frac{Av}{A - 4}$

b

$\frac{(A - 4) v}{4}$

c

$\frac{(A + 4) v}{4}$

d

$2 v$

answer is A.

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Detailed Solution

$z^{X} ⟶_{z - 2} X^{A - 4} +_{2} {He}^{4}$

$\bar{P_{i}} = \bar{P_{f}}$

$0 = (A - 4) {\vec{v}}_{1} + 4 \vec{v}$

${\bar{v}}_{1} = - \frac{4 \bar{v}}{A - 4}$

So relative velocity of separation

$= v + V_{1}$

$= v + \frac{4 v}{A - 4} = \frac{Av}{A - 4}$

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