Question

A potential difference of $V$ volts is applied on an $\alpha$-particle at rest in order to get sufficient velocity so that when it is projected from infinity towards $Au$ foil, it can reach closest upto $x$ meters from $Au$ nucleus. If instead of an $\alpha$-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 upto which it can reach is $y$ meters. The ratio $x:y$ is _______.

• A. 2 : 1
• B. 3 : 2
• C. 1 : 1
• D. 2 : 3

Answer 1

The correct option is C 1 : 1

At closest distance of approach : $qV=\frac{K{q}_1{q}_2}{r}$

Therefore,

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

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

From (i) and (ii) equation:

$x:y=1:1$