A nucleus of mass number 220 decays by $\alpha$ decay. The energy released in the reaction is $5 \mathrm{MeV}$. The kinetic energy of an $\alpha$-particle is

• A

  $\frac{27}{11} \mathrm{MeV}$
• B

  $\frac{27}{11} \mathrm{MeV}$
• Correct answer
  C

  $\frac{54}{11} \mathrm{MeV}$
• D

  $\frac{55}{54} \mathrm{MeV}$

Solution

Let the reaction be represented as ${ }_{\mathrm{Z}}^{220} \mathrm{X} \longrightarrow \underset{\mathrm{Z}-2}{216} \begin{gathered}4 \\ \mathrm{Y}+\mathrm{H} \text { e } \\ 2\end{gathered}$ $\because$ energy released in the reaction is $5 \mathrm{MeV}$ $\Rightarrow \frac{1}{2} m_Y v_Y^2+\frac{1}{2} m_\alpha v_\alpha^2=5 \mathrm{MeV}$
Also using conservation of linear momentum. $\begin{aligned} & m_Y v_Y=-m_\alpha v_\alpha \Rightarrow v_Y=\frac{-m_\alpha}{m_Y} v_\alpha \\ & v_Y=\frac{-4}{216} v_\alpha=\frac{-1}{54} v_\alpha\end{aligned}$ Putting in eqn. (i) $\begin{aligned} & \frac{1}{2}(216)\left(\frac{v_\alpha}{54}\right)^2+\frac{1}{2} 4\left(v_\alpha\right)^2=5 \mathrm{MeV} \\ & \frac{1}{2}(216)\left(\frac{v_\alpha}{54}\right)^2+\mathrm{K}_{\mathrm{E}} \mathrm{E}_\alpha=5 \mathrm{MeV} \\ & \Rightarrow \frac{1}{2} \times 4 \frac{v_\alpha^2}{54}+\mathrm{K} \cdot \mathrm{E}_{\cdot \alpha}=5 \mathrm{MeV} \\ & \Rightarrow \frac{1}{54}\left(\mathrm{~K}_{\cdot} \mathrm{E}_\alpha\right)+\mathrm{K}_{\cdot} \mathrm{E}_{\cdot \alpha}=5 \mathrm{MeV} \\ & \text { K.E. } \alpha=\frac{5 \times 54}{55}=\frac{54}{11} \mathrm{MeV}\end{aligned}$

Hence, the correct answer is option: 'C'.