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?
(correct answer + 1, wrong answer - 0.25)

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

Answer 1

The correct option is 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,
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
$\Rightarrow 0=-(A-4)v^{\prime }+4v$
$\Rightarrow v^{\prime }=\frac{4v}{A-4}$