An \alpha-particle of energy 5 {\MeV} is scattered through 180^{\circ} by a fixed uranium nucleus. The distance of closest approach is of the order of [AIEEE 2004]

Atoms and Nuclei Part 1

Section: Physics

Question : 87 of 100

Marks: +1, -0

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

[AIEEE 2004]

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