Question

If in $\alpha$ decay process, $Q-\text{value}$ is $5\text{MeV}$ and $K.E$ of $\alpha$ particle is $4.8\text{MeV}$ then, find out the mass number of parent nuclei.

• A. $100$
• B. $50$
• C. $20$
• D. $10$

Answer 1

The correct option is A $100$

$\alpha -\text{decay}$ is represented by,

${}_Z{X}^A\to {}_{Z-2}{Y}^{A-4}+\alpha +Q$

Where, $A$ is the mass number of the parent nucleus.

We know that,

$Q=K.E_Y+K.E_{\alpha }$

$K.E_Y=Q-K.E_{\alpha }$

$\Rightarrow K.E_Y=5\text{MeV}-4.8\text{MeV}$

$\Rightarrow K.E_Y=0.2\text{MeV}$

By conservation of momentum,

$\Rightarrow \frac{K.E_Y}{K.E_{\alpha }}=\frac{m_{\alpha }}{m_Y}$

We know that, the mass of the nucleus is proportional to the mass number.

$\Rightarrow \frac{K.E_Y}{K.E_{\alpha }}=\frac{4}{A-4}$

$\Rightarrow \frac{0.2}{4.8}=\frac{4}{A-4}$

$\Rightarrow 0.2A-0.8=19.2$

$\Rightarrow A=100$

Hence, option $\left(A\right)$ is correct.

Key concept: By conservation of momentum, $0=P_Y+P_{\alpha }$ $\Rightarrow \left|P_Y\right|=\left|P_{\alpha }\right|$ We know that, $K.E=\frac{P^2}{2m}$ $\Rightarrow \frac{K.E_Y}{K.E_{\alpha }}=\frac{m_{\alpha }}{m_Y}$ If $A$ is the mass number of parent nuclei, then $K{E}_{\propto }=Q\left[\frac{A-4}{A}\right]$