Question

A nucleus $X^{220}$ at rest decays by emitting an $\alpha$ particle. If energy of the daughter nucleus is $0.2\text{MeV}$, then the $Q$ value of the reaction is

• A. $10.8\text{MeV}$
• B. $10.9\text{MeV}$
• C. $11\text{MeV}$
• D. $11.1\text{MeV}$

Answer 1

The correct option is C $11\text{MeV}$

The given decay is represented as

$X^{220}\to Y^{216}+\alpha$

As the $X^{220}$ is initially at rest,

According to conservation of momentum,

$0=P_Y+P_{\alpha }$

$\Rightarrow \left|P_Y\right|=\left|P_{\alpha }\right|$

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

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

The mass of the nucleus is proportional to the mass number $\left(A\right)$.

$\Rightarrow \frac{K.E_Y}{K.E_{\alpha }}=\frac{4}{216}=\frac{1}{54}$

$\Rightarrow K.E_{\alpha }=54(K.E_Y)$

Given that, $K.E_Y=0.2\text{MeV}$

$\Rightarrow K.E_{\alpha }=10.8\text{MeV}$

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

$\Rightarrow Q=11\text{MeV}$

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

Alternate solution : $X^{220}\to Y^{216}+\alpha$ $K{E}_Y=\frac{m_{\alpha }}{m_X}Q=0.2$ $\Rightarrow \frac{4u}{220u}Q=0.2$ $\therefore Q=11\text{MeV}$