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An alpha particle with a kinetic energy of 6.00 MeV makes a head-on collision with a...

An alpha particle with a kinetic energy of 6.00 MeV makes a head-on collision with a gold nucleus at rest.

What is the distance of closest approach of the two particles? (Assume that the gold nucleus remains stationary and that it may be treated as a point charge. A gold nucleus has 79 protons, and an alpha particle is a helium nucleus consisting of two protons and two neutrons. The mass of an alpha particle is 6.64424 x 10-27 kg. 1MeV = 1.60 x 10-13 J. )

d=______ m

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Answer #1

The alpha particle's  kinetic energy, after head on collision with gold nucleus converts to potential energy of system.

Kinetic energy of alpha particle

Potential energy of system

Charge on alpha particle

Charge on gold nucleus

Conserving the energy of the system of particles,

Distance of closest approach of the two particles is

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