Question

A radioactive nucleus of mass number $A$, initially at rest, emits an $\alpha$ - particle with a speed $v$. The recoil speed of the daughter nucleus will be

• A. $\frac{2v}{A + 4}$
• B. $\frac{4v}{A + 4}$
• C. $\frac{4v}{A - 4}$
• D. $\frac{2v}{A - 4}$

Answer 1

Answer: (C)

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

Answer 2

$\alpha$-particle is equivalent to helium nucleus.
The emission of an $\alpha$-particle from the atom of an element reduces its atomic number by $2$ , and mass number by $4$ .
Hence, the radioactive emission is as follows :
${ }_{Z} X ^{A} \xrightarrow{\alpha \text {-particle }}{ }_{Z-2} Y ^{A-4}+{ }_{Z} He ^{4}$ ( $\alpha$-particle)
Also form law of conservation of momentum,
$m \times 0 =m_{y} v_{y}+m_{\alpha} v_{\alpha}$
$=(A-4) v_{y}+4 v$
$\Rightarrow v_{y} =-\frac{4 v}{A-4}$