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 ${\displaystyle \frac{d}{2}}$. What is the speed of projection of the alpha particle in this case?

विकल्प

• ${\displaystyle \frac{V}{2}}$

• ${\displaystyle \sqrt{2}V}$

• 2 V

• 4 V

Answer 1

${\displaystyle \sqrt{2}V}$

Explanation:

The distance of closest approach

${\displaystyle d=\frac{\text{const}}{V_1^2}}$   ...(1)

${\displaystyle \frac{d}{2}=\frac{\text{const}}{V_2^2}}$   ...(2)

From equations (1) and (2),

2 = ${\displaystyle \frac{V_2^2}{V_1^2}\Rightarrow V_2=\sqrt{2}{V}_1}$

${\displaystyle \therefore V_2=\sqrt{2}V}$  ....Given, (V₁ = V)