In a scattering experiment,an alpha particle initially with 8.00 MeV of kinetic energy can approach the nucleus of an unknown element to within 20.0 fm.If the only force between the alpha particle and the nucleus of the unknown element is the electric force,then how close would it approach if its initial kinetic energy was 4.00 MeV instead?
A)10.0 fm
B)20.0 fm
C)40.0 fm
D)80.0 fm

Question

Quizplus · Accepted Answer

The closest approach distance is inversely proportional to the kinetic energy of the alpha particle. Therefore, if the initial kinetic energy is decreased by a factor of 2 (from 8.00 MeV to 4.00 MeV), the closest approach distance will increase by a factor of 2. Therefore, the closest approach distance with 4.00 MeV initial kinetic energy would be 2 x 20.0 fm = 40.0 fm. The answer is C.

In a Scattering Experiment,an Alpha Particle Initially with 8