Module 2 : Electrostatics
Lecture 8 : Electrostatic Potential

Example 15 :
An $\alpha$- particle with a kinetic energy of 1 MeV is projected towards a stationary nucleus with a charge $75\mid e\mid$. Neglecting the motion of the nucleus, determine the distance of closest approach of the $\alpha$- particle.

Solution :

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Initial energy of $\alpha$-particle = 1 Mev = $ 10^6\times (1.6\times 10^{-19}) = 1.6\times 10^{-12}$eV.
At the distance of closes approach, the velocity (and the kinetic energy) of the $\alpha$- particle is zero. Hence, all its kinetic energy has been converted into potential energy. The potential energy (reference at infinity) at a distance $d$ is

Equating this to the initial kinetic energy, $ d= 2.16\times 10^{-14}$m.

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