STANDARD Physics Scattering MCQ Question
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 A
10⁻¹⁰ cm
10⁻⁴ m
10⁻¹⁵ cm
Correct Answer
Detailed Explanation
The distance of closest approach is calculated using the formula for scattering, which involves the energy of the alpha particle and the charge of the nucleus. For a 5 MeV alpha particle, this distance is approximately 10⁻¹⁰ cm.
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