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An α-particle of energy 5 MeV is scattered through 180⁰ by a fixed uranium nucleus. The distance of the closest approach is of the order of
Options
(a) 1Å
(b) 10⁻¹⁰ cm
(c) 10⁻¹² cm
(d) 10⁻¹⁵ cm
Correct Answer:
10⁻¹² cm
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
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- The kinetic energy of α-particle emitted in the α-dacay of ₈₈Ra²²⁶ is
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Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two radioactive materials X₁ and X₂ have decay constants 5λ and λ respectively
- The kinetic energy of α-particle emitted in the α-dacay of ₈₈Ra²²⁶ is
- A short magnet of magnetic moment M, is placed on a straight line. The ratio
- Point masses m₁ and m₂ are placed at the opposite ends of a rigid rod of length L,
- A body of mass 2 kg is moved by a force F=3x where x is the distance covered.
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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