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The ratio of speed of an electron in ground state in Bohr’s first orbit of hydrogen atom to velocity of light in air (c ) is
Options
(a) e²/2ε₀hc
(b) 2e²ε₀/hc
(c) e³/2ε₀hc
(d) 2ε₀hc/e²
Correct Answer:
e²/2ε₀hc
Explanation:
Velocity of electron in nᵗʰ orbit = αcz / n
where α = (e²/4πε₀) [1 / c(h/π)] = e²/2ε₀hc
v / c = αz / n = (e² × 1) / 2ε₀hc × 1 = e²/2ε₀hc
Related Questions: - The energy released in the fission of 1 kg of ₉₂U²³⁵ is (energy per fission = 200 MeV)
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The energy released in the fission of 1 kg of ₉₂U²³⁵ is (energy per fission = 200 MeV)
- Depletion layer contains
- Which logic gate is represented by the following combination of logic gate?
- If h is Planck’s constant, the momentum of a photon of wavelength 0.01 Å is
- A transverse wave is represented by y = a sin (?t – kx)
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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