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)

$\alpha$-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 :
${}_Z{X}^A\stackrel{\alpha \text{-particle }}{\to }{}_{Z-2}{Y}^{A-4}+{}_{Z}H{e}^4$ ( $\alpha$-particle)
Also form law of conservation of momentum,
$m\times 0=m_y{v}_y+m_{\alpha }{v}_{\alpha }$
$=(A-4)v_y+4v$
$\Rightarrow v_y=-\frac{4v}{A-4}$