An alpha particle with a kinetic energy of 13.0 mev makes a head-on collision with a gold nucleus at rest. part a: 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. the atomic number of gold is 79, and an alpha particle is a helium nucleus consisting of two protons and two neutrons.)

8.75 fm

Explanation:

We use this expression

Which is derived from conservation of energy. E is the kinetic energy. d is the distance to find, Ze and 2e are the charges of the gold nucleus and alpha particle respectively. is the coulomb constant which is 9*10^9. Converting the energy to Joules

So, solving for distance (having all units in international system, distance ends up in meters)

In fermi:

the inner core.

good luck: )

klicker

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.