Question

The distance of the closest approach of an alpha particle fired at a nucleus with kinetic energy $K_1$ is $r_o$. The distance of the closest approach when the α – particle is fired at the same nucleus with kinetic energy $2K_1$ will be.

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

Answer 1

The correct option is A $\frac{r_o}{2}$

At a distance of closest approach, kinetic energy of α particle = Potential energy of system.
$\therefore \frac{K{q}_1{q}_2}{r}=\frac{1}{2}m{v}^2$
Initially kinetic energy,
$K_1=\frac{K{q}_1{q}_2}{r_0}$ .....(i)
When $K_2=2K_1$
$2K_1=\frac{K{q}_1{q}_2}{r^{\prime }}$ ....(ii)
From the above equation we get,
$r^{\prime }=\frac{r_o}{2}$