Question

The distance of the closest approach of an alpha particle fired at a nucleus with kinetic energy K is $r_0$. The distance of the closest approach when the alpha particle is fired at the same nucleus with kinetic energy 2K will be

• A. $2r_0$
• B. $4r_0$
• C. $\frac{r_0}{2}$
• D. $\frac{r_0}{4}$

Answer 1

The correct option is C

$\frac{r_0}{2}$

The ditance of the closest approach is given by
$r_0=\frac{1}{4\pi {ϵ}_0}.\frac{2Z{e}^2}{K}$
where $K=\frac{1}{2}m{\nu }^2$. Thus $r_0\propto \frac{1}{K}$. When K is doubled, $r_0$ becomes half. Hence the correct choice is (c).