Question

The distance of the closest approach of an alpha-particle fired towards a nucleus with momentum $p$ is$r.$ If the momentum of the alpha-particle is $2p$, the corresponding distance of the closest approach is

• A. $4r$
• B. $2r$
• C. $\frac{r}{2}$
• D. $\frac{r}{4}$

Answer 1

The correct option is D

$\frac{r}{4}$

Find the distance of the closest approach

All of the kinetic energy of the alpha particle is converted to the potential energy of the system

$PE=KE$

$\frac{p^2}{2m}=\frac{1}{4\mathrm{\pi }{\in }_{\mathrm{o}}}\times \frac{q_1{q}_2}{r}$

Initially,

$\frac{p^2}{2m}=\frac{2eq}{4\mathrm{\pi }{\in }_{\mathrm{o}}r}.....\left(1\right)$

Now,

$\frac{{\left(2p\right)}^2}{2m}=\frac{2eq}{4\mathrm{\pi }{\in }_{\mathrm{o}}{\mathrm{r}}_1}....\left(2\right)$

Dividing $\left(1\right)$ by $\left(2\right)$

$$\begin{gathered} \Rightarrow \frac{1}{4}=\frac{r_1}{r} \\ \Rightarrow r_1=\frac{r}{4} \end{gathered}$$

Hence, the correct option is$D$.