Question

A potential difference of V volts is applied on a $\alpha$-particle at rest in order to get sufficient velocity so that when it is projected from infinity towards gold foil, it can reach closest up to $x$ meters from Au nucleus. If instead of a $\alpha$-particle, a proton is used in the above experiment and is accelerated from rest by the same potential difference, then the closest distance from the Au nucleus up to which it can reach is $y$ meters. Then, the ratio $x:y$ is:

Answer 1

At closest distance of approach :

$qV=\frac{K{q}_1{q}_2}{r}$

Now,

$2eV=\frac{K\left(2e\right)\left(79e\right)}{x}$ (for $\alpha$-particle) ....... (i)

$eV=\frac{K\left(e\right)\left(79e\right)}{y}$ (for proton) ....... (ii).

From (i) and (ii), $x=y,x:y=1:1$