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Question:

In a Geiger Marsden experiment, calculate the distance of closest approach to the nucleus of Z=75, when an α-particle of 5 MeV energy impinges on it before it comes momentarily to rest and reverses its direction. How will the distance of closest approach be affected when the kinetic energy of the α-particle is doubled?

Solution:

If r0 be the distance of closest approach between the nucleus and the alpha particle, then the kinetic energy of the alpha particle is Ek = 1/4πε0(Ze)(2e)/r0. Thus, r0 ∝ 1/Ek. Hence, when the kinetic energy of the alpha particle is doubled, the closest distance r0 will be halved.