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 the alpha particle is halved, the new distance of closest approach will be ______.

पर्याय

• `d/2`

• 2d

• `d/4`

• 4d

Answer 1

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 the alpha particle is halved, the new distance of closest approach will be 4d.

Explanation:

The distance of closest approach (d) is determined by equating the initial kinetic energy (K.E.) of the alpha particle to the electrostatic potential energy (U) at distance d:

`1/2 m v^2 = 1/(4 pi epsilon_0) (q_1 q_2)/d`

From this formula, we can see that the distance of closest approach is inversely proportional to the square of the velocity (v):

d ∝ `1/v^2`

If the initial velocity v is halved (becomes v/2), we can find the new distance d' by substituting the new velocity into the proportionality:

d' ∝ `1/((v/2)^2)`

d' ∝ `1/(V^2/4)`

d' ∝ `4 * 1/v^2`

Since the original distance d was proportional to `1/v^2`, the new distance is:

d' = 4d