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An α-particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of the closest approach is of the order of

JEE Main 2004 — Physics Electromagnetism

2004mcqhard

An α\alphaα-particle of energy 5MeV5 \mathrm{MeV}5MeV is scattered through 180∘180^{\circ}180∘ by a fixed uranium nucleus. The distance of the closest approach is of the order of

Official previous-year question

Held on 30 Apr 2004 · Verified 6 Jul 2026.

Options

  1. A

    1A˚1 Å1A˚

  2. B

    10−10 cm10^{-10} \mathrm{~cm}10−10 cm

  3. C

    10−12 cm10^{-12} \mathrm{~cm}10−12 cm

  4. D

    10−15 cm10^{-15} \mathrm{~cm}10−15 cm

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