An α - particle of energy 5 MeV is scattered through 180 ° by a fixed Uranium nucleus. The distance of the closest approach is (Atomic number of uranium = 92 )

Question

An α - particle of energy 5 MeV is scattered through 180 ° by a fixed Uranium nucleus. The distance of the closest approach is (Atomic number of uranium = 92 ), 5 . 3 × 1 0 - 1 2 m, 5 . 3 × 1 0 - 1 3 m, 5 . 3 × 1 0 - 1 4 m, 5 . 3 × 1 0 - 1 5 m

MARKS App · Accepted Answer

According to the law of conservation of energy, The kinetic energy of α - particle = Potential energy of alpha α - particle at a distance of closest approach. ⇒ 1 2 m v 2 = 1 4 π ε 0 q 1 q 2 r ⇒ 5 MeV = 9 × 1 0 9 2 e 9 2 e r r = 5 . 3 × 1 0 - 1 4 m

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