Nuclear Physics and RadioactivityHard
Question
An α - particle of energy 5 MeV is scattered through 180o by a fixed uranium nucleus. The distance of the closest approach is of the order of
Options
A.1 
B.10-10 cm
C.10-12 cm
D.10-15 cm
Solution
At closest approach, all the kinetic energy of the α-particle will converted into the potential energy of the system, K.E. = P.E.
5 MeV =
5 × 106 × e = 9 × 109
r =
∴ r = 5.3 × 10-14 m = 5.3 × 10-12 cm.
5 MeV =
5 × 106 × e = 9 × 109
r =
∴ r = 5.3 × 10-14 m = 5.3 × 10-12 cm.
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