A nucleus zAX emits an -particle with velocity v- The recoil speed of the daughter nucleus is-

A $^{238} U$ nucleus decays by emitting an alpha particle of speed $^{'} v^{'} m s^{- 1}$ . The recoil speed of the residual nucleus (in $m s^{- 1}$ ) is:

Q. A nucleus of mass number '

A

' originally at rest emits an

α

particle with a speed '

v

' the recoil speed of the daughter nucleus is

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A nucleus AzX emits an α-particle with velocity v. The recoil speed of the daughter nucleus is:

The correct option is B 4vA−4Let the daughter nucleus by Y.AZX→A−4Z−2Y+(42α)+++2β−Let recoil velocity of Y be −vr.Using conservation of momentum:pi=pf0=(A−4)vr−4vvr=4vA−4

A nucleus $_{z}^{A} X$ emits an $α$ -particle with velocity $v$ . The recoil speed of the daughter nucleus is:

The correct option is B $\frac{4 v}{A - 4}$
Let the daughter nucleus by $Y$ .
$_{Z}^{A} X \to_{Z - 2}^{A - 4} Y + (_{2}^{4} α)^{+ +} + 2 β^{-}$

Let recoil velocity of $Y$ be $- v_{r}$ .

Using conservation of momentum:
$p_{i} = p_{f}$
$0 = (A - 4) v_{r} - 4 v$
$v_{r} = \frac{4 v}{A - 4}$