Question

An $\alpha -$particle and a proton are both moving with the same speed. They approach a gold foil. If the distance of the closest approach for proton is $y,$ what will be its value for the $\alpha -$particle?

• A. $2y$
• B. $4y$

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 the particle.

$\Rightarrow v^2=\frac{2k{q}_{n}q}{mr}$

As speed are equal,

$\frac{2k{q}_{n}q}{mr}=$ constant

So,

${\frac{q_{n}q}{mr}}_{\alpha -\text{particle}}={\frac{q_{n}q}{mr}}_{\text{proton}}$

$\Rightarrow \frac{q_n\times 2e}{4m\times r}=\frac{q_n\times e}{m\times y}$

$\Rightarrow r=\frac{y}{2}$

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