A nucleus of mass number A originally at rest, emits alpha particle with speed v, the recoil speed of the daughter nucleus is:

Question

A nucleus of mass number A originally at rest, emits alpha particle with speed v, the recoil speed of the daughter nucleus is: (A)  (4v/A-4)  (B)  (4v/A+4)  (C)  (v/A-4)  (D)  (v/A+4)

Tardigrade Admin · Accepted Answer

From law of conservation of momentum, we have momentum before collision = momentum after collision The daughter nucleus has mass A-4 ∴ mα vα+mN vN=0 ⇒ 4 v+(A-4) vN=0 vN=(-4 v/A-4)

Q. A nucleus of mass number $A$ originally at rest, emits alpha particle with speed $v$ , the recoil speed of the daughter nucleus is:

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

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

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

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

From law of conservation of momentum,
we have momentum before collision $=$ momentum after collision
The daughter nucleus has mass $A - 4$
$∴ m_{α} v_{α} + m_{N} v_{N} = 0$
$\Rightarrow 4 v + (A - 4) v_{N} = 0$
$v_{N} = \frac{- 4 v}{A - 4}$

Q. A nucleus of mass number A originally at rest, emits alpha particle with speed v, the recoil speed of the daughter nucleus is:

A−44v​

A+44v​

A−4v​

A+4v​