The distance of closest approach of an alpha particle to a nucleus when the alpha particle moves towards the nucleus with linear momentum $\mathrm{P}$ is $\mathrm{d}$. The distance of closest approach of alpha particle to nucleus, if the linear momentum of the alpha particle is $1.5 \mathrm{P}$

• A

  $\frac{2 \mathrm{~d}}{3}$
• B

  $\frac{3 \mathrm{~d}}{2}$
• Correct answer
  C

  $\frac{4 d}{9}$
• D

  $\frac{9 \mathrm{~d}}{4}$

Solution

At sufficient distance, electric potential energy is zero. So, $\mathrm{U}_1=0$
at (1) $\mathrm{K}_1=\frac{1}{2} \mathrm{mv}^2=\frac{\mathrm{P}^2}{2 \mathrm{~m}}$ at (2) $\mathrm{K}_2=0$ $\mathrm{U}_2=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}_\alpha \mathrm{q}_{\mathrm{n}}}{\mathrm{d}_{\mathrm{c}}}$ By energy conservation $\begin{aligned} & \mathrm{U}_1+\mathrm{K}_1=\mathrm{U}_2+\mathrm{K}_2 \\ & 0+\frac{\mathrm{p}^2}{2 \mathrm{~m}}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}_\alpha \mathrm{q}_{\mathrm{n}}}{\mathrm{d}_{\mathrm{c}}}+0\end{aligned}$ Distance of closest approach, $d_c \propto \frac{1}{p^2}$ $\begin{aligned} & \frac{\mathrm{d}_2}{\mathrm{~d}_1}=\frac{\mathrm{p}_1^2}{\mathrm{p}_2^2} \\ & =\mathrm{d} \frac{\mathrm{p}^2}{(1.5 \mathrm{p})^2}=\frac{4 \mathrm{~d}}{9}\end{aligned}$

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