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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