Question

The distance of 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?

(A) V /2 

(B) \(\sqrt{2}\ V\) 

(C) 2 V 

(D) 4 V

Answer 1

Correct option is : (B) \(\sqrt{2}\ V\)  

The distance of closest approach

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

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

From equations (1) and (2)

\(2 = \frac{V_2^2}{V_1^2}\Rightarrow V_2 = \sqrt{2}\ V_1\) 

\(\therefore\ V_2 = \sqrt{2}\ V\)   Given, \((V_1 = V)\)