An α -particle of energy 5   MeV is scattered through 180 ° by a fixed uranium nucleus. The distance of the closest approach is of the order of

Step-by-step solution

Let, at the distance of closest approach r 0 from the nucleus, the α -particle stops a moment and gets scattered by the angle 180 ° . Now charge on an α -particle q 1 = + 2 e and charge on a scattering nucleus q 2 = + Z e where Z = 92 atomic number of uranium. The initial kinetic energy of the α -particle gets converted into electrostatic potential energy, K α = U . K α = 1 2 m v 2 = k q 1 q 2 r 0 . . . . . ( 1 ) , ( k = 9 × 10 9   N   m 2   C - 2 ) Given- K α