An α-particle of 5 MeV energy strikes with a nucleus of uranium at stationary at an scattering angle of 180o. The nearest distance upto which α-particle reaches the nucleus will be of the order of 
(a) 1 Å               (b) 10-10 cm 
(c) 10-12 cm      (d) 10-15 cm

(c) At closest distance of approach
Kinetic energy = Potential energy

5×106×1.6×10-19=14πε0×ze2er

For uranium z = 92, so r = 5.3×10-12 cm

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