An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus…
Physics · JEE Advanced · NTA Exams — Electrostatics
An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of closest approach is of the order of
- 1 Å
- 10–10 cm
- 10–12 cm
- 10–15 cm
Answer
(C) 10 –12 cm
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