If an `alpha`-particle is incident directly towards a nucleus then it will stop when the distance between them is the distance of closest approach. Which of these is correct?

Options

A. Its kinetic energy is converted into released radiation
B. When its kinetic energy becomes zero then it converts into electrostatic potential energy
C. At distance of closest approach the potential energy of the `alpha`-particle becomes zero
D. A `beta`-particle will show the same behaviour as an `alpha`-particle

Answer

B

Explanation

When an `alpha`-particle is incident towards the nucleus with a velocity then repulsions arising due to identical charges on both operate to stop the `alpha`-particle at a particular distance. At this point, the kinetic energy of the `alpha`-particle becomes zero but as per law of energy conservation it converts into electrostatic potential energy.