§ 44. Bohr Atom Model, de Broglie Standing Waves & Spectral Series (Grade 9–11)
Quantization of angular momentum Lₙ = nℏ, de Broglie standing waves 2πrₙ = nλ_dB, and Rydberg spectral transitions. An allowed Bohr orbit accommodates an exact integer number of electron de Broglie wavelengths, forming a non-radiating standing wave.
Theory Summary & Scientific Mechanism
In classical electrodynamics, an orbiting electron undergoing centripetal acceleration aₙ = v²/r would radiate away its energy and spiral into the nucleus within sim 10⁻¹¹ s. In 1913, Niels Bohr resolved this instability by postulating that electrons occupy discrete stationary orbits with quantized angular momentum Lₙ = mₑ vₙ rₙ = nℏ (n = 1, 2, 3, …) that do not radiate.
In 1924, Louis de Broglie explained Bohr’s quantization rule geometrically: a stationary orbit is stable if and only if its circumference 2π rₙ accommodates an integer number n of electron de Broglie wavelengths λ_B = (h)/(mₑ vₙ), forming a closed standing matter wave 2π rₙ = nλ_B! Allowed hydrogen radii scale as rₙ = n² a₀ (a₀ = 0.0529 nm) and energy levels follow Eₙ = -(13.6 eV)/(n²).
A photon is emitted or absorbed only during a quantum jump nᵢ → n_f: hν = |Eₙᵢ - Eₙ_f|, yielding the Rydberg spectral formula (1)/(λ) = R_H ≤ft|(1)/(n_f²) - (1)/(nᵢ²) | (R_H = 1.097× 10⁷ m⁻¹) across the Lyman (n_f = 1, UV), Balmer (n_f = 2, visible), and Paschen (n_f = 3, IR) series.
Everyday Analogy
An electron in an atom is like a person on a staircase: it can stand on steps n = 1, 2, 3…, never in mid-air between them! Why? Because the electron is a matter wave that must wrap around the orbit in a whole number of wavelengths 2π rₙ = nλ_B. Jumping down a step emits a photon of light!
Interactive Question & Derivation
Why, according to de Broglie's hypothesis, can an electron occupy only discrete Bohr orbits of radii rₙ = n² a₀ with angular momentum Lₙ = nℏ?
- A: Because the circumference must contain an integer number n of de Broglie wavelengths (2π rₙ = n (h)/(pₙ)); otherwise the matter wave destructively interferes with itself
- B: Because the Coulomb force acts only at integer distances from the nucleus
- C: Because the electron speed is identical on every orbit
Study Hint: How many full de Broglie wavelengths λ_B must fit around the circumference 2π rₙ of the n-th stationary Bohr orbit?
Proof & Explanation: Substituting the de Broglie wavelength λ_dB = (h)/(mₑ vₙ) into the circular standing-wave condition 2π rₙ = nλ_dB directly yields Bohr's quantization rule mₑ vₙ rₙ = n(h)/(2π) = nℏ. Jumping from nᵢ to n_f emits a photon of energy hν = Eₙᵢ - Eₙ_f.
Key Equations & 3D Simulation Parameters
2π rₙ = nλ_dB, Eₙ = -(13.6 eV)/(n²), Δ E = 13.6≤ft((1)/(n_f²) - (1)/(nᵢ²) ) eV
- Initial quantum level nᵢ: 2 … 4 (default: 3)
- Final quantum level n_f: 1 … 3 (default: 2)
- de Broglie standing-wave amplitude: 0.1 … 0.45 rel. (default: 0.25 rel.)