Question

A nucleus of mass number $A$ originally at rest emits $\alpha$ particle with speed $v$. What will be recoil speed of the daughter nucleus?

• A. $\frac{v}{A-4}$
• B. $\frac{v}{A+4}$
• C. $\frac{4v}{A-4}$
• D. $\frac{4v}{A+4}$

Answer 1

Answer: (C)

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

When a nucleus emits an $\alpha$ particle its mass number decreases by $4$, hence mass of an $\alpha$ particle is $4$. The mass of daughter nucleus is $(A-4)$.
In the emission of an $\alpha$ particle the momentum remains conserved if $v^{{}^{\prime }}$ is the recoil velocity of daughter nucleus then,
$m_1{u}_1+m_2{u}_2=(m_1-m_2)v$
$A\left(0\right)+4v=(A-4)v^{{}^{\prime }}$
$v^{{}^{\prime }}=\frac{4v}{A-4}$