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