Question

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

• A. $2r$
• B. $4r$
• C. $r/2$
• D. $r/4$

Answer 1

Answer: (D)

$r/4$

$\frac{1}{2}m{v}^2=\frac{q_1{q}_2}{4\pi {ϵ}_{0r}}$

Initially, $\frac{p^2}{2m}=\frac{2eq}{4\pi {ϵ}_{0}r}$ ----(1)

Now, $\frac{(2p{)}^2}{2m}=\frac{2eq}{4\pi {ϵ}_0{r}^1}$ ----(II)

Equation $\left(I\right)÷\left(ii\right)$

$\frac{1}{4}=\frac{r^1}{r}$

$r^1=\frac{r}{4}$