Alpha particle deviation from gold foil (Rutherford Scattering)

[simms_mj]

1. Homework Statement
The alpha particles used by Rutherford had a kinetic energy of 7.7 MeV and, for a head-on collision would get to within a distance of 3×10-14m of the centre of the gold nucleus. However the actual radius of a gold nucleus is approximately 7×10-15m.
2. Homework Equations

(a)Without the use of any other data, including physical constants, calculate the energy at which Alpha particles would start to show a deviation from pure Rutherford scattering off gold.
(b)As the thickness of the gold foil is increased, with the Alpha particle energy remaining at
7.7MeV, a deviation from the thin foil experiment is also observed. Explain

 

[olgranpappy]

attempt at solution?

for part (a) you could try and figure out how much kinetic energy an alpha particle would need to get all the way to the edge of the nucleus in a head-on collision. For higher energies than this the alpha would penetrate the nucleus and, presumably, the scattered distrubtion of alpha particle would deviate from the rutherford case.

 

[simms_mj]

Sorry I should have put down my attempt. Ya I was going to try that but in the question it said I can't use any information other than the stuff provided.

 

[olgranpappy]

yeah... so...

you don't need anything other than what was given. I think you can just use the relation between kinetic and potential energy to show that the distance of closest approach is inversely proportional to the kinetic energy. Use the given information to determine the proportionality constant.