ElectrostaticsHard

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

Options

A.1
B.10-10 cm
C.10-12 cm
D.10-15 cm

Solution

       
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., rmin is of the order of 10-12 cm
∴ Correct option is (c).

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