An alpha particles of energy 5 MeV is scattered through 180o by a fixed uranium nucleus. The distanc

Question

An alpha particles of energy 5 MeV is scattered through 180o by a fixed uranium nucleus. The distance of the closest approach is of the order of

Accepted Answer

From conservation of mechanical energy
decrease in kinetic energy = increase in potential energy
or

= 5 MeV = 5 × 1.6 × 10^-13 J
∴

= 5.3 × 10^-14m
= 5.3 × 10^-12m
i.e., r_min is of the order of 10^-12 cm
∴ Correct option is (c).

Question