JEE Main 2004Annual, Paper 1PhysicsMediumAtoms and Nuclei183× since 2002An α\alphaα-particle of energy 555 MeVMeVMeV is scattered through 180∘{180^ \circ }180∘ by a fixed uranium nucleus. The distance of closest approach is of the order ofAOption A. 10−12 cm{10^{ - 12}}\,cm10−12cmBOption B. 10−10 cm{10^{ - 10}}\,cm10−10cmCOption C. 1A1A1ADOption D. 10−15a cm{10^{ - 15a}}\,cm10−15acmCheckSkip · show solution
