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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 ofa)1 Åb)10–10 cmc)10–12cmd)10–15 cmCorrect answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 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 ofa)1 Åb)10–10 cmc)10–12cmd)10–15 cmCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 ofa)1 Åb)10–10 cmc)10–12cmd)10–15 cmCorrect answer is option 'C'. Can you explain this answer?.

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