Question

The distance of the closest approach of an alpha particle is d when it moves with a speed V towards a nucleus.

Another alpha particle is projected with higher energy, such that the new distance of the closest approach is `d/2`. What is the speed of projection of the alpha particle in this case?

पर्याय

• `V/2`

• `sqrt2V`

• 2 V

• 4 V

Answer 1

`bb(sqrt2V)`

Explanation:

The distance of closest approach

`d = "const"/V_1^2`   ...(1)

`d/2 = "const"/V_2^2`   ...(2)

From equations (1) and (2),

2 = `(V_2^2)/(V_1^2) ⇒ V_2 = sqrt2 V_1`

`therefore V_2 = sqrt2 V`  ....Given, (V₁ = V)