An -particle of energy 5 MeV is scattered through by a fixed uranium nucleus. The distance of the closest approach is of the order of
- (A)
- (B)
- (C)✓ Correct
- (D)
Answer: (C)
Solution
At closest approach the alpha particle's kinetic energy has turned entirely into electrostatic potential energy, . Putting for the alpha particle, for uranium and MeV and solving for gives about m, which is of the order of cm.
JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option