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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: - In which of the processes, does the internal energy of the system remain constant?
- A black body at 227⁰C radiates heat at the rate of 7cals/cm²s. At a temperature
- If R is the radius of the earth and g is the acceleration due to gravity
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In which of the processes, does the internal energy of the system remain constant?
- A black body at 227⁰C radiates heat at the rate of 7cals/cm²s. At a temperature
- If R is the radius of the earth and g is the acceleration due to gravity
- In nuclear fusion, two nuclei come together to form a large nucleus.
- The period of oscillation of a mass M suspended from a spring of negligible mass is T
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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