Question

In a Geiger Marsden experiment, calculate the distance of closest approach to the nucleus of $Z=75$, when an $\alpha$-particle of $5MeV$ energy impinges on it before it comes momentarily to rest and reverse its direction.
How will the distance of closest approach be affected when the kinetic energy of the $\alpha$-particle is doubled?

Answer 1

If $r_0$ be the distance of closest approach between the nucleus and the alpha particle, then the kinetic energy of the alpha particle is $E_k=\frac{1}{4\pi {ϵ}_0}\frac{\left(Ze\right)\left(2e\right)}{r_0}$

Thus, $r_0\propto \frac{1}{E_k}$

Hence, when the kinetic energy of alpha particle is doubled, the closet distance $r_0$ will be halved.