Question

An $\alpha -$particle and a proton, moving with same kinetic energy approach a gold foil. If the distance of closest approach for proton is $x$, what will be its value for the $\alpha -$particle?

• A. $\frac{x}{2}$
• B. $\frac{x}{4}$

Answer 1

From law of conservation of mechanical energy,

$\frac{1}{2}m{v}^2=\frac{k{q}_{n}q}{r};q$ is charge of particle.

As kinetic energies are equal,

$\frac{k{q}_{n}q}{r}=$ constant

So,

${\frac{q_{n}q}{r}}_{(\alpha -\text{particle})}={\frac{q_{n}q}{r}}_{\left(\text{proton}\right)}$


$\Rightarrow \frac{q_n\times 2e}{r}=\frac{q_n\times e}{x}$

$\Rightarrow r=2x$

Hence, $\left(C\right)$ is the correct answer.