In an experiment, alpha particles of initial kinetic energy 5.9 MeV are directed at stationary nuclei of lead (82207Pb). Show that the distance of closest approach is about 4×10−14m.
2e AND 82e seen
OR
3.2×10−19 «C» AND 1.312×10−17 «C» seen d=5.9×106×e8.99×109×(2e)(82e)=3.998×10−14≈4×10−14<m≫
Must see either clear substitutions or answer to at least 4 s.f. for MP2.