A radioactive nuclei zXA emits an α ((He)+ +) particle. Speed of α particle is v then the relative velocity of sepration between them.

Question

A radioactive nuclei zXA emits an α ((He)+ +) particle. Speed of α particle is v then the relative velocity of sepration between them. (A) (A v/A - 4) (B) ((. A - 4 .) v/4) (C) ((. A + 4 .) v/4) (D) 2v

Tardigrade Admin · Accepted Answer

zX arrow z - 2XA - 4+ 2He4 barPi= barPf 0=(.A-4.)( overset arrow v)1+4 overset arrow v barv1=-(4 barv/A - 4) So relative velocity of separation =v+V1 =v+(4 v/A - 4)=(A v/A - 4)

Q. A radioactive nuclei $_z X_{A}$ emits an $α ((He)^{++})$ particle. Speed of $α$ particle is $v$ then the relative velocity of sepration between them.

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

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

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

$2 v$

$z^{X} \to_z - 2 X_{A - 4} +_2 H e_{4}$
$\overset{ˉ}{P_{i}} = \overset{ˉ}{P_{f}}$
$0 = (A - 4) (v \to)_{1} + 4 v \to$
$\overset{v}{ˉ}_{1} = - \frac{4 v ˉ}{A - 4}$
So relative velocity of separation
$= v + V_{1}$
$= v + \frac{4 v}{A - 4} = \frac{A v}{A - 4}$

Q. A radioactive nuclei _zXA​ emits an α((He)++) particle. Speed of α particle is v then the relative velocity of sepration between them.

A−4Av​

4(A−4)v​

4(A+4)v​

2v