Bohr quantization for electron orbits If m, v, and r are the electron mass, orbital speed, and orbit radius, and h is Planck’s constant, what is the Bohr quantization condition for the allowed circular orbits in the hydrogen atom?

Difficulty: Easy

Correct Answer: m v r = n h / (2π)

Explanation:


Introduction / Context:
Bohr’s model introduces the postulate that the orbital angular momentum of an electron in a hydrogen-like atom is quantized in integer multiples of ħ = h / (2π). Although superseded by full quantum mechanics, this condition still yields correct energy levels for hydrogen and is foundational in the historical development of atomic theory.


Given Data / Assumptions:

  • Electron moves in a circular orbit under Coulomb attraction.
  • Non-relativistic speeds; single electron atom.
  • Quantization postulate applies to angular momentum.


Concept / Approach:

Bohr’s postulate: L = m v r = n ħ = n h / (2π), where n = 1, 2, 3, …. Only those orbits with angular momentum equal to an integer multiple of ħ are allowed. This leads to quantized radii r_n and energy levels E_n that match observed spectral lines via the Rydberg formula.


Step-by-Step Solution:

Angular momentum L = m v r.Quantization: L = n ħ = n h / (2π).Therefore allowed condition: m v r = n h / (2π).


Verification / Alternative check:

Using m v r = n h / (2π) and Coulomb force balance m v^2 / r = k e^2 / r^2 gives r_n ∝ n^2 and E_n ∝ −1/n^2, in agreement with hydrogen spectra.


Why Other Options Are Wrong:

(b) misses the 2π factor; (c) uses wrong mechanical quantity; (d) inverts the variables and is dimensionally inconsistent.


Common Pitfalls:

Confusing angular momentum quantization with de Broglie standing wave condition; they are equivalent but require careful use of 2π.


Final Answer:

m v r = n h / (2π)

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