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An α-particle of 7 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

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a

1 Å

b

10-10 cm

c

10-12 cm

d

10-15 cm

answer is C.

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Detailed Solution

At closest distance of approach Kinetic energy = Potential energy ⇒7×106×1.6×10−19=14πε0×(ze)(2e)rFor uranium z= 92, so r=5.3×10−12cm

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