Bohr's Quantization Condition — de Broglie Waves Explained | Engineering Physics notes

Bohr’s Quantization Condition — how de Broglie waves explain allowed atomic orbits

By Mohan Dangi Gold Medalist

Contents 1. Quick overview 2. Derivation: from de Broglie to Bohr 3. Numeric example: 1st Bohr orbit (hydrogen) 4. Physical interpretation (stationary waves) 5. Significance & implications 6. FAQ & common confusions

1. Quick overview

The de Broglie hypothesis assigns a wavelength λ = h/p to any particle of momentum p. When applied to electrons orbiting an atomic nucleus, this wave nature explains why only certain orbits (with specific radii) are allowed: the electron's matter-wave must form a stationary standing wave around the orbit. This requirement leads directly to Bohr's quantization of angular momentum: L = m v r = n h / (2π).

2. Derivation: from de Broglie to Bohr

Start with de Broglie's relation for wavelength:

λ = h / p = h / (m v)

For an electron moving in a circular orbit of radius r, the circumference must accommodate an integral number n of wavelengths to form a stationary (standing) wave:

2π r = n λ

Substitute λ = h / (m v):

2π r = n (h / (m v))

Rearrange to get:

m v r = n h / (2π)

But the angular momentum of the electron is L = m v r, therefore:

L = n h / (2π) = n ħ where ħ = h / (2π). This is exactly Bohr's quantization rule.

Key idea: The quantization of angular momentum is equivalent to insisting that the electron's de Broglie wave fits an integer number of wavelengths around the orbit.

3. Numeric example: 1st Bohr orbit (hydrogen)

Bohr's model gives the radius of the first orbit (n = 1) as:

r₁ = 5.3 × 10⁻¹¹ m

The circumference is:

C = 2π r₁ ≈ 2 × 3.1416 × 5.3 × 10⁻¹¹ m ≈ 3.33 × 10⁻¹⁰ m

Compute de Broglie wavelength for the electron in the orbit using λ = h / (m v). Using Bohr's expression for v in the first orbit (non-relativistic):

v₁ = e² / (2 ε₀ h) × (constants...) (it's simpler to observe that the calculated λ numerically equals the circumference for n=1)

Direct substitution (using standard constants) yields

λ ≈ 3.33 × 10⁻¹⁰ m ≈ circumference

So 2π r₁ = λ, which corresponds to n = 1. For higher n, the circumference must be n times the wavelength.

4. Physical interpretation (stationary waves)

Because a stationary wave has nodes and antinodes fixed in space, it does not transport energy away from the orbit — matching Bohr's postulate that electrons in stationary orbits do not radiate. The standing matter-wave picture thus gives an intuitive explanation for both the stability of specific orbits and the discrete energy levels in atoms.

5. Significance & implications

• Bohr's rule, re-interpreted via de Broglie, links classical orbital motion to quantum wave properties.
• It explains why only discrete electron energies exist in atoms — only certain standing waves (and hence energies) are allowed.
• While Bohr's model has limitations (works best for hydrogen-like atoms), the standing-wave idea influenced later quantum mechanics and the Schrödinger equation.

6. FAQ & common confusions

Q: Is Bohr's model fully correct?

A: No — Bohr's model is semi-classical and provides exact results only for hydrogen-like atoms. Quantum mechanics (Schrödinger wave mechanics) gives a more complete and general description: discrete energy levels emerge naturally from boundary conditions on the wavefunction.

Q: Why integers only?

A: Standing waves around a closed loop require the phase to match after each circuit. This forces the circumference to equal an integer number of wavelengths; fractional wavelengths would not match phase after a full turn.

Q: Does this mean electrons are literally little waves circling the nucleus?

A: The modern view is that the electron is described by a probability wave (wavefunction). The standing-wave picture is a helpful semiclassical visualization; full quantum mechanics describes orbitals with shapes that can be stationary but not necessarily simple circular waves.

Written By Mohan Dangi (Gold medalist)