Question

The ‘distance of closest approach’ of an alpha-particle is \(d\) when it moves with a velocity \(v\) head-on towards the target nucleus. If the velocity of alpha particle is halved, the new ‘distance of closest approach’ will be –

• A. \(\dfrac{d}{2}\)
• B. \(2d\)
• C. \(\dfrac{d}{4}\)
• D. \(4d\)

Hint:

For a head-on collision of an alpha particle with a nucleus, \[ d\propto \frac{1}{v^2}. \] If the velocity becomes \(n\) times smaller, the distance of closest approach becomes \(n^2\) times larger.

Answer 1

The Correct Option is C