Question

A radioactive nuclei $\_{z}^{}X_{}^{A}$ emits an $\alpha \left(\left(He\right)^{+ +}\right)$ particle. Speed of $\alpha $ particle is $v$ then the relative velocity of sepration between them.

• A. $\frac{A v}{A - 4}$
• B. $\frac{\left(\right. A - 4 \left.\right) v}{4}$
• C. $\frac{\left(\right. A + 4 \left.\right) v}{4}$
• D. $2v$

Answer 1

Answer: (A)

$z^{X} \rightarrow \_{z - 2}^{}X_{}^{A - 4}+\_{2}^{}He_{}^{4}$
$\bar{P_{i}}=\bar{P_{f}}$
$0=\left(\right.A-4\left.\right)\left(\overset{ \rightarrow }{v}\right)_{1}+4\overset{ \rightarrow }{v}$
$\bar{v}_{1}=-\frac{4 \bar{v}}{A - 4}$
So relative velocity of separation
$=v+V_{1}$
$=v+\frac{4 v}{A - 4}=\frac{A v}{A - 4}$