Question

A nucleus $\text{X}$ initially at rest, undergoes alpha decay according to the equation,

${}_Z{\text{X}}^{232}\to {}_{90}{\text{Y}}^A+\alpha$

What fraction of the total energy released in the decay will be the kinetic energy of the alpha particle.

• A. $\frac{1}{2}$
• B. $\frac{90}{92}$
• C. $\sqrt{\frac{228}{232}}$
• D. $\frac{228}{232}$

Answer 1

The correct option is D $\frac{228}{232}$

The given equation can be represented as,

${}_Z{\text{X}}^{232}\to {}_{90}{\text{Y}}^{228}+{}_2{\text{He}}^4(\alpha -\text{particle})$

As nucleus $\text{X}$ is initially at rest,

According to the conservation of momentum,

$0=P_{\text{Y}}+P_{\alpha }$

$\therefore$ The momentum of nucleus $\text{Y}$ and $\alpha$ particle will be same.

We know that, $K.E=\frac{P^2}{2m}$

$\Rightarrow K.E\propto \frac{1}{m}$

$\Rightarrow \frac{K.E_{\alpha }}{K.E_{\text{Y}}}=\frac{m_{\text{Y}}}{m_{\alpha }}$

Here, $m\propto A$

$\Rightarrow \frac{K.E_{\alpha }}{K.E_{\text{Y}}}=\frac{228}{4}$

$\Rightarrow \frac{K.E_{\alpha }}{K.E_{Total}}=\frac{K.E_{\alpha }}{K.E_{\alpha }+K.E_{\text{Y}}}$

$\Rightarrow \frac{K.E_{\alpha }}{K.E_{Total}}=\frac{228}{232}$

Hence, option $\left(\text{D}\right)$ is correct.

Why this Question? This question gives and idea of how to apply law of conservation of linear momentum in the nuclear emission.