Question
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
-
1 Å
-
10−10 cm
-
10−12 cm
-
10−15 cm
Solution
C.
10−12 cm
At closest approach, all the kinetic energy of the α-particle will converted into the potential energy of the system, K.E. = P.E.
