The problem is about an alpha particle with an initial kinetic energy K = 5MeV, which hits a thin metal foil with Z=50 and gets scattered 180 degrees afterwards. The question is: what must have been the alpha particle’s distance of closest approach to the scattering nucleus?

They way I approached the problem was by thinking that the alpha particle could have gotten its energy after been accelerated through a voltage difference of 2.5x10^6 Volts (2e x 2.5x10^6 = 5MeV). If the alpha particle is to bounce back after hitting the thin metal foil then it is necessary that as the alpha particle gets closer to the nucleus it "sees" or "feels" at least a similar voltage through which it was accelerated in order to balance its initial kinetic energy. In this case is 5MeV. The voltage the alpha particle’s sees will depend on how close it gets to the nucleus of one particular atom in the thin metal foil and of course depends on the electrical charge of that nucleus.

The electrical potential created by the nucleus (Z=50) is:

V = K x q / d

V = (9x10^9) x (50 x 1.6x10^-19) / d

and it has to be equal to

(2.5x10^6) = (9x10^9) x (50 x 1.6x10^-19) / d

solving for "d" I got 2.88x10^-14 meters. The book says 2.9x10^-14 meters.

The questions I have are:

1. Is the alpha particle bouncing back just entirely because of electric field repulsion or there is also a real physical collision (contact) between the alpha particle and the nucleus?

2. The problem was asking for "the distance of closest approach between the alpha particle and the nucleus to have it scattering 180 degrees". I was thinking why is this distance the closest distance? I thought the reason why this distance is the “closest” is because the conservation of energy will not allow the alpha particle to get any closer than 2.9x10^-14 meters from the nucleus because beyond that point the electrical potential energy will be equivalent to more than 5MeV and this could not be. Is this correct?

3. Assuming the alpha particle does not lose energy on other collisions, an alpha particle with kinetic energy of 5MeV will always reach 2.9x10^-14 meters from this nucleus (Z=50) and will not get any closer. Is this statement correct?

4. If the scattering angle were not 180 degrees but, for example, 165 degrees, how would have this changed the answer of the problem? My thoughts are that this change in scattering angle will definitely affect momentum calculations but not necessary kinetic energy of the alpha particle after the collision. I suppose the velocity of the scattered alpha particle could have had the same magnitude in both cases, 180 degrees and 165 degrees. The same velocity after the collision means same kinetic energy after the collision and that would mean that in both cases the particles have reached the same distance from the scattering nucleus, in this case 2.9x10^-14 meters. The scattering deviation angle depends on the distance between the alpha particle's straight trajectory and a parallel line to it passing through the nucleus center. Again my question is whether these thoughts are correct or not.

From what I have read, Rutherford worked out a formula for the expected distribution in scattering angles using classical dynamics. The answer that Rutherford found showed dependence on the charge to mass ratio of the alpha particle and on the charge of the nucleus, Ze. Where can I find a paper/summary/analysis of how Rutherford arrived to this conclusion?