Question

In a nuclear reaction of $\alpha$ -decay, the daughter nucleus ${}_{\text{Z}}^{\text{A}}\text{X}$ is moving with some unknown kinetic energy while $\alpha$ -particle is moving with kinetic energy $\text{E}$ . The total energy released if parent nucleus was initially at rest will be

• A. $\text{E}\left(1+\frac{4}{\text{A}}\right)$
• B. $\frac{E}{4}\left(A\right)$
• C. $\frac{E}{4}\left(A-4\right)$
• D. $E\left(A+4\right)$

Answer 1

Answer: (A)

Momentum of $A$ is given by, $p_A=\sqrt{2A{E}_A}$ , where $A\&E_A$ represents atomic mass number and kinetic energy of $A$ .
By linear momentum conservation, we have
$\Rightarrow \sqrt{2A{E}_A}=\sqrt{2\left(4\right)E}$
$\Rightarrow E_A=\frac{4}{A}E$
So Energy released is given by law of conservation of energy,
$E_R=E+\frac{4}{A}E=E\left(1+\frac{4}{A}\right)$