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An alpha nucleus of energy 1/2 mv² bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to
Options
(a) 1 / Ze
(b) v²
(c) 1 / m
(d) 1 / v⁴
Correct Answer:
1 / m
Explanation:
Kinetic energy of alpha nucleus is equal to electrostatic potential energy of the system of the alpha particle and the heavy nucleus. That is,
1/2 mv² = 1/4π?￢ﾂﾀ . qαZe / r₀
where r₀ is the distance of closest approach
r₀ = 2/4π?￢ﾂﾀ . qαZe / mv²
r₀ ∞ Ze ∞ qα ∞ 1/m ∞ 1/v²
Hence option (c) is correct.
Related Questions: - The kirchoff’s first law (∑i=0) and second law (∑iR=∑E), where the symbols
- Two bodies of masses m₁ and m₂ are initially at rest at infinite distance apart
- If (range)² is 48 times (maximum height)², then angle of projection is
- Light with an energy flux of 25×10⁴ Wm⁻² falls on a perfectly reflecting surface
- The number of significant figures in 0.06900
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The kirchoff’s first law (∑i=0) and second law (∑iR=∑E), where the symbols
- Two bodies of masses m₁ and m₂ are initially at rest at infinite distance apart
- If (range)² is 48 times (maximum height)², then angle of projection is
- Light with an energy flux of 25×10⁴ Wm⁻² falls on a perfectly reflecting surface
- The number of significant figures in 0.06900
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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