The Bohr Model and Atomic Line Spectra

In the Bohr model of the hydrogen atom, the electron moves in circular orbits with quantized angular momentum L = nħ and energies E_n = −13.6 eV ÷ n²; the atom emits or absorbs a photon of energy hf = E_i − E_f when the electron jumps between levels, which explains the discrete line spectrum of hydrogen. In brief: the orbit radii are r_n = 0.0529 nm × n² (the Bohr radius is a₀ = 0.0529 nm). The wavelengths of the lines follow 1 ÷ λ = R(1 ÷ n_f² − 1 ÷ n_i²), with R = 1.097 × 10⁷ m⁻¹. The series are named Lyman (to n = 1, ultraviolet), Balmer (to n = 2, visible) and Paschen (to n = 3, infrared). The model works for hydrogen-like ions but not for atoms with several electrons, and was replaced by quantum mechanics.

Heated gases emit light only at certain wavelengths, and each element has its own set of lines, like a fingerprint. Classical physics could not explain why. This guide describes line spectra, the postulates of the Bohr model, the energy levels and transitions, compares the series in a table, works through numerical examples and corrects common mistakes.

What are atomic line spectra?

A hot, low-pressure gas emits an emission spectrum consisting of bright lines at specific wavelengths on a dark background. When white light passes through a cool gas, the gas absorbs the same wavelengths, leaving dark lines in a continuous spectrum (absorption spectrum), as seen in the dark Fraunhofer lines in sunlight. Each element has a unique set of lines, which allows identification of the chemical composition of stars. Johann Balmer found in 1885 that the visible lines of hydrogen (656.3 nm red, 486.1 nm blue-green, 434.0 nm violet, 410.2 nm violet) fit a simple formula, and Johannes Rydberg generalized it in 1888. Classical physics predicted that an orbiting electron would radiate continuously and spiral into the nucleus in about 10⁻¹¹ s, and could not explain the discrete lines.

What are the postulates of the Bohr model?

Niels Bohr proposed in 1913 that:

1. The electron moves in certain circular orbits (stationary states) around the nucleus without radiating energy, in spite of its acceleration.
2. The allowed orbits are those in which the angular momentum is an integer multiple of ħ = h ÷ 2π: mvr = nħ (n = 1, 2, 3, …). (De Broglie later explained this as the condition for a standing electron wave: 2πr = nλ.)
3. The atom emits or absorbs radiation only when the electron jumps between two orbits, and the photon has the energy of the difference: hf = E_i − E_f.

Combining the quantization with Coulomb’s law and Newton’s second law gives the radius of the nth orbit r_n = n²a₀, with the Bohr radius a₀ = 0.0529 nm, and the speed v_n = 2.19 × 10⁶ m/s ÷ n (about 1/137 of the speed of light for n = 1).

What are the energy levels?

The total energy of the electron in the nth orbit is E_n = −13.6 eV ÷ n² (for hydrogen; for an ion with nuclear charge Z, E_n = −13.6 Z² ÷ n² eV). The ground state (n = 1) has −13.6 eV; the first excited state (n = 2), −3.40 eV; n = 3, −1.51 eV; and E = 0 at n = ∞ (the electron free). The ionization energy of hydrogen is therefore 13.6 eV. The negative sign means that the electron is bound. An atom in an excited state returns to a lower state within about 10⁻⁸ s, emitting a photon. The wavelength: λ = hc ÷ ΔE = 1,240 eV·nm ÷ ΔE. In absorption, the photon must have exactly the energy of a transition (or more than the ionization energy, in which case any excess becomes kinetic energy of the freed electron). The model explains the spectral series of hydrogen: transitions to n = 1 (Lyman), n = 2 (Balmer), n = 3 (Paschen), n = 4 (Brackett).

The hydrogen series compared

Worked example: hydrogen lines

1. Balmer alpha (n = 3 → 2). ΔE = 13.6 × (1/4 − 1/9) = 13.6 × 0.1389 = 1.89 eV. λ = 1,240 ÷ 1.89 = 656 nm (red). Using the Rydberg formula: 1 ÷ λ = 1.097 × 10⁷ × (1/4 − 1/9) = 1.524 × 10⁶ m⁻¹, giving λ = 656 nm ✓.
2. Balmer beta (4 → 2). ΔE = 13.6 × (1/4 − 1/16) = 2.55 eV; λ = 1,240 ÷ 2.55 = 486 nm (blue-green).
3. Lyman alpha (2 → 1). ΔE = 13.6 × (1 − 1/4) = 10.2 eV; λ = 1,240 ÷ 10.2 = 122 nm (ultraviolet).
4. Paschen alpha (4 → 3). ΔE = 13.6 × (1/9 − 1/16) = 0.661 eV; λ = 1,240 ÷ 0.661 = 1,876 nm (infrared).
5. Ionization. A photon of 14.0 eV ionizes a ground-state hydrogen atom, with the electron emerging with kinetic energy 14.0 − 13.6 = 0.4 eV. The longest wavelength that ionizes: 1,240 ÷ 13.6 = 91.2 nm.
6. Orbit radius and speed. For n = 2: r = 4 × 0.0529 = 0.212 nm, v = 2.19 × 10⁶ ÷ 2 = 1.1 × 10⁶ m/s. In He⁺ (Z = 2), the ground state energy is −13.6 × 4 = −54.4 eV.

How was the model developed?

Gustav Kirchhoff and Robert Bunsen established spectroscopy in 1859, showing that each element has a characteristic spectrum. Rutherford’s nuclear atom (1911) posed the stability problem, and Bohr published his model in 1913, which explained the Balmer formula and the Rydberg constant from fundamental constants. The Franck-Hertz experiment (1914) confirmed the discrete energy levels by electron collisions with mercury atoms. Bohr received the Nobel Prize in Physics in 1922. The model was extended by Arnold Sommerfeld (elliptical orbits, 1916) and replaced by the quantum mechanics of Heisenberg and Schrödinger in 1925 to 1926, which describes electrons in orbitals (probability distributions) rather than orbits.

Which mistakes are common?

Mistake 1: Thinking that electrons orbit like planets

In quantum mechanics, electrons are described by orbitals, not by well-defined orbits. The Bohr model is a useful simplification.

Mistake 2: Mixing up emission and absorption

Emission: the electron falls to a lower level and a photon is emitted. Absorption: a photon of exactly the right energy lifts the electron to a higher level.

Mistake 3: Forgetting the negative sign of the energy

E_n is negative for bound states. The photon energy is the difference of two levels, E_i − E_f > 0.

Mistake 4: Using the formula for atoms with several electrons

The simple 13.6 ÷ n² formula applies to one-electron systems (H, He⁺, Li²⁺, with the Z² factor).