In an α -particle scattering experiment, the distance of closest...

The distance of closest approach $r_0$ for a head-on collision ($\theta = 180^\circ$) of an $\alpha$-particle with kinetic energy $K$ is given by equating kinetic energy to electrostatic potential energy:

$K = \dfrac{1}{4\pi\epsilon_0} \dfrac{(2e)(Ze)}{r_0} \Rightarrow r_0 = \dfrac{1}{4\pi\epsilon_0} \dfrac{2Ze^2}{K}$

The impact parameter $b$ for an $\alpha$-particle scattered at an angle $\theta$ is given by:

$b = \dfrac{1}{4\pi\epsilon_0} \dfrac{Ze^2 \cot(\theta/2)}{K}$

Comparing the two expressions, we can write $b$ in terms of $r_0$:

$b = \dfrac{r_0}{2} \cot\left(\dfrac{\theta}{2}\right)$

For a scattering angle $\theta = 90^\circ$:

$b = \dfrac{r_0}{2} \cot(45^\circ)$

Since $\cot(45^\circ) = 1$, we get:

$b = \dfrac{r_0}{2}$

Answer: $\dfrac{r_0}{2}$