Q.
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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