Atomic Emission Spectra and the Hydrogen Spectrum

An atomic emission spectrum is the set of discrete wavelengths of light emitted by excited atoms of an element, and the hydrogen spectrum, described by the Rydberg equation 1/λ = R_H(1/n₁² − 1/n₂²), shows that electrons in atoms have quantized energy levels. In brief: the red Balmer line of hydrogen at 656.3 nm comes from the transition of an electron from level n = 3 to n = 2, with R_H = 1.097 × 10⁷ m⁻¹.

Every element has a characteristic line spectrum that serves as a fingerprint, and the simplicity of the hydrogen spectrum led to the first quantum model of the atom. This article explains how emission spectra arise, describes the Bohr model and the Rydberg equation, gives a table of the series of hydrogen lines, shows how to calculate wavelengths and energies and explains flame tests and the use of spectra in astronomy.

What is a line spectrum?

A hot solid emits a continuous spectrum, with all wavelengths. When a gas of atoms is excited, for example in an electric discharge or a flame, it emits light only at specific wavelengths, which appear as bright lines on a dark background when the light is dispersed by a prism or grating. In an absorption spectrum, white light passing through the gas has dark lines at the same wavelengths. The line positions are characteristic for each element. This was a surprise for classical physics, which predicted a continuous range of frequencies from an electron orbiting in an atom.

What is the Bohr model?

In 1913, Niels Bohr proposed that the electron in a hydrogen atom can occupy only certain circular orbits with fixed energies, and emits or absorbs radiation only when it jumps between them. The energy levels are:

En = −RHhc / n2 = −2.18 × 10−18 J / n2 = −13.6 eV / n2 (n = 1, 2, 3, …)

The negative sign means that the electron is bound: the energy is zero when the electron is free and at rest (n = ∞). The lowest level (n = 1, the ground state) has an energy of −2.18 × 10⁻¹⁸ J, which means that 2.18 × 10⁻¹⁸ J (1312 kJ/mol, or 13.6 eV) is needed to ionize hydrogen. A transition from a higher level n₂ to a lower level n₁ emits a photon with the energy ΔE = hν = 2.18 × 10⁻¹⁸ (1/n₁² − 1/n₂²) J. The Bohr model works well for hydrogen and one-electron ions such as He⁺ (with E_n multiplied by Z²), but fails for atoms with more electrons and has been replaced by quantum mechanics, which keeps the energy levels.

What is the Rydberg equation?

Johannes Rydberg gave an empirical formula that fits all the lines in the hydrogen spectrum:

1/λ = RH (1/n12 − 1/n22), with n₂ > n₁ and R_H = 1.097 × 10⁷ m⁻¹

The Bohr model gives a theoretical explanation of this formula and of the value of R_H in terms of fundamental constants (m_e, e, h, c, ε₀).

Series of lines in the hydrogen spectrum

The visible lines of the Balmer series are Hα (656.3 nm, red, 3 → 2), Hβ (486.1 nm, blue-green, 4 → 2), Hγ (434.0 nm, violet, 5 → 2) and Hδ (410.2 nm, violet, 6 → 2). The lines come closer together toward the series limit, where n₂ → ∞, which corresponds to ionization from the level n₁.

Worked examples

Example 1: the red Balmer line

For n₂ = 3 → n₁ = 2: 1/λ = 1.097 × 10⁷ × (1/4 − 1/9) = 1.097 × 10⁷ × 0.1389 = 1.524 × 10⁶ m⁻¹. λ = 6.563 × 10⁻⁷ m = 656.3 nm.

Example 2: the first Lyman line

For 2 → 1: 1/λ = 1.097 × 10⁷ × (1 − 1/4) = 8.228 × 10⁶ m⁻¹, so λ = 121.5 nm, in the ultraviolet. The photon energy is E = hc/λ = 1.988 × 10⁻²⁵ / 1.215 × 10⁻⁷ = 1.64 × 10⁻¹⁸ J (10.2 eV).

Example 3: energy and wavelength of the 4 → 2 transition

ΔE = 2.18 × 10⁻¹⁸ × (1/4 − 1/16) = 2.18 × 10⁻¹⁸ × 0.1875 = 4.09 × 10⁻¹⁹ J. λ = hc/ΔE = 1.988 × 10⁻²⁵ / 4.09 × 10⁻¹⁹ = 4.86 × 10⁻⁷ m = 486 nm, the blue-green line. The frequency is 6.17 × 10¹⁴ Hz.

Example 4: the ionization energy of hydrogen

Ionization from n = 1 corresponds to n₂ = ∞: ΔE = 2.18 × 10⁻¹⁸ J per atom. Per mole: 2.18 × 10⁻¹⁸ × 6.022 × 10²³ = 1.31 × 10⁶ J/mol = 1312 kJ/mol, in agreement with the measured first ionization energy of hydrogen. The corresponding wavelength is 91.2 nm.

Example 5: which transition gives a given wavelength?

A line is observed at 434 nm. Try the Balmer series: 1/λ = 2.304 × 10⁶ m⁻¹, so (1/4 − 1/n₂²) = 2.304 × 10⁶ / 1.097 × 10⁷ = 0.2100. Then 1/n₂² = 0.25 − 0.2100 = 0.0400, so n₂ = 5. The line is the 5 → 2 transition.

What are flame tests?

When a salt is heated in a flame, the metal atoms are excited and emit light of characteristic colors. Sodium gives an intense yellow (589 nm), lithium a crimson red (671 nm), potassium a lilac (766 and 770 nm), calcium an orange-red, strontium a deep red, barium a pale green and copper a blue-green. Fireworks use the same principle with metal salts. In atomic emission spectroscopy, the intensity of the lines is used to measure the concentration of metals in water, soil and blood.

How are spectra used in astronomy?

The composition of stars is found from the absorption lines in their light. Helium was discovered in the spectrum of the Sun in 1868 by Pierre Janssen and Norman Lockyer, before it was found on Earth, and its name comes from the Greek word for the Sun, helios. The shift of the lines to longer wavelengths (the redshift) in the light of distant galaxies shows that they are receding from us, and it was the basis for the discovery of the expansion of the universe.

How was the hydrogen spectrum understood?

Joseph von Fraunhofer observed dark lines in the solar spectrum in 1814. Gustav Kirchhoff and Robert Bunsen showed around 1859 that each element has its own set of lines, and used it to discover cesium and rubidium. Johann Balmer found in 1885 a formula for the visible lines of hydrogen, which Johannes Rydberg generalized in 1888. Niels Bohr explained the formula in 1913 with his quantized orbits, and received the Nobel Prize in Physics in 1922.

Which mistakes are common?

• Swapping n₁ and n₂: n₂ is the higher level; swapping gives a negative 1/λ.
• Forgetting the units of R_H: m⁻¹, so λ comes out in meters.
• Thinking that all Balmer lines are visible: the series limit is in the ultraviolet.
• Applying the Bohr model to atoms with many electrons: it works only for one-electron species.
• Confusing emission and absorption: the same wavelengths are involved, but the processes are opposite.