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

Answer 1

The correct option is A $\frac{r_0}{2}$

From law of conservation of mechanical energy,

$K{E}_i+P{E}_i=K{E}_f+P{E}_f$

$\Rightarrow KE+0=0+\frac{kq\left(2e\right)}{r}$

$\Rightarrow r\propto \frac{1}{KE}$

Therefore, if kinetic energy is doubled, the distance of the closest approach will be halved.

​​​​​​​​​​​​​​Hence, option $\left(A\right)$ is the correct answer.