Scattering in central force fields (introductory)
Central forces and rigid bodies · Physics
Study notes
Worked example - Rutherford backscattering. Alpha particle (Z1 = 2) with E = 5 MeV approaches a gold nucleus (Z2 = 79) head-on (b = 0). Step 1: distance of closest approach: all kinetic energy becomes potential, E = k/d, k = Z1 Z2 e^2/(4 pi epsilon_0) = 2 x 79 x 1.44 eV.nm = 227.5 eV.nm = 2.275 x 10^-7 eV.m. Step 2: d = k/E = 2.275 x 10^-7/5 x 10^6 = 4.55 x 10^-14 m = 45.5 fm. Gold's nuclear radius ~ 7 fm, so the alpha turns around well outside the nucleus - pure Coulomb scattering, no nuclear forces involved. This is why Rutherford could probe nuclear size: increasing E until deviations appear measures the radius.