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Rydberg Equation Calculator

Wavelength, frequency and energy of hydrogen spectral lines

Updated · Free, no signup

1 = Lyman, 2 = Balmer, 3 = Paschen, 4 = Brackett, 5 = Pfund.

Must be larger than n₁.

1 for H, 2 for He⁺, 3 for Li²⁺ (one-electron species only).

Wavelength λ

656.47 nm

Wavenumber

15,233 cm⁻¹

Frequency

4.5667 × 10¹⁴ Hz

Photon energy

1.8887 eV

Photon energy

3.0260 × 10⁻¹⁹ J

Series and region

Balmer series · Visible (red)
  • Emission when the electron falls from n = 3 to n = 2; absorbing a 656.47 nm photon drives the reverse jump.

Wavelengths in this series

Lines of the same series

Transitionλ (nm)Wavenumber (cm⁻¹)Energy (eV)Region
3 → 2656.4715,2331.8887Visible
4 → 2486.27420,564.552.5497Visible
5 → 2434.17323,032.292.8556Visible
6 → 2410.29424,372.83.0218Visible
7 → 2397.12425,181.083.1221Visible
8 → 2389.01925,705.683.1871Visible
9 → 2383.65126,065.353.2317Visible
10 → 2379.90126,322.623.2636Ultraviolet
∞ → 2 (limit)364.70527,419.43.3996Ultraviolet

About the Rydberg Equation Calculator

This Rydberg equation calculator gives the wavelength of light emitted or absorbed when the electron in a hydrogen atom (or a hydrogen-like ion such as He⁺ or Li²⁺) jumps between two energy levels. Enter the lower level n₁ and the upper level n₂, and it returns the wavelength in nanometres, the wavenumber, the frequency, the photon energy in electronvolts and joules, the series name and the part of the spectrum the line falls in.

It is designed for chemistry and physics students working on atomic structure and the Bohr model, and for anyone identifying lines in a hydrogen emission spectrum. The table lists the first lines of the same series plus the series limit, so you can see the lines crowd together as n₂ grows.

By default it uses the Rydberg constant for hydrogen corrected for the finite mass of the proton (1.09678 × 10⁷ m⁻¹), which reproduces measured hydrogen lines closely; you can switch to R∞ (1.09737 × 10⁷ m⁻¹) to match textbook answers or for heavy nuclei. Wavelengths are vacuum values — lines measured in air are about 0.03% shorter (H-alpha is 656.47 nm in vacuum, 656.28 nm in air).

With the default inputs, the wavelength λ is 656.47 nm. Change any value above to recalculate instantly.

How to use the rydberg equation calculator

  1. 1Enter the lower level n₁ — it sets the series (2 for the visible Balmer lines).
  2. 2Enter the upper level n₂, which must be larger.
  3. 3Leave Z = 1 for hydrogen, or enter the nuclear charge of a one-electron ion.
  4. 4Choose R_H for real hydrogen lines or R∞ to match a textbook.
  5. 5Read the wavelength, energy and region, and compare with the rest of the series in the table.

Formula and method

1/λ = R·Z²·(1/n₁² − 1/n₂²) E = hc/λ ν = c/λ

The Rydberg formula gives the reciprocal wavelength (wavenumber) of a line as the Rydberg constant times Z² times the difference between the inverse squares of the two principal quantum numbers. It follows from the Bohr energy levels Eₙ = −hcR·Z²/n², so the photon carries away exactly the energy difference between the two levels.

R∞ = 10,973,731.568157 m⁻¹ is the CODATA 2022 value for an infinitely heavy nucleus. For hydrogen the electron and proton orbit their common centre of mass, so R_H = R∞ ÷ (1 + mₑ/mₚ) ≈ 10,967,758 m⁻¹. Energy and frequency use the exact SI values h = 6.62607015 × 10⁻³⁴ J·s and c = 299,792,458 m/s.

λ
Wavelength of the emitted or absorbed photon (vacuum)
R
Rydberg constant (R_H or R∞), m⁻¹
Z
Nuclear charge (atomic number) of the one-electron species
n₁, n₂
Lower and upper principal quantum numbers
h, c
Planck constant and speed of light

Worked examples

H-alpha (Balmer 3 → 2)

1/λ = 1.0967758 × 10⁷ × (1/4 − 1/9) = 1.5233 × 10⁶ m⁻¹, so λ = 656.47 nm — the red line that gives hydrogen nebulae their colour. Each photon carries about 1.89 eV.

Lyman-alpha (2 → 1)

1/λ = R_H × (1 − 1/4) = 8.2258 × 10⁶ m⁻¹, giving λ ≈ 121.57 nm in the vacuum ultraviolet. This 10.2 eV line is the strongest in the hydrogen spectrum.

He⁺ 4 → 3 line with R∞

For He⁺, Z² = 4: 1/λ = 1.0973732 × 10⁷ × 4 × (1/9 − 1/16) = 2.1338 × 10⁶ m⁻¹, so λ ≈ 468.7 nm — the famous blue He II line seen in hot stars.

Frequently asked questions

What is the Rydberg equation?+

It is 1/λ = R·Z²(1/n₁² − 1/n₂²), which predicts the wavelengths of the spectral lines of hydrogen and other one-electron atoms. R is the Rydberg constant, about 1.097 × 10⁷ m⁻¹.

Which hydrogen lines are visible?+

The Balmer series (transitions down to n = 2) contains the visible lines: 656 nm red (3→2), 486 nm cyan (4→2), 434 nm blue-violet (5→2) and 410 nm violet (6→2). Higher Balmer lines lie in the near ultraviolet.

What value of the Rydberg constant should I use?+

Textbooks often use 1.097 × 10⁷ m⁻¹. For accurate hydrogen wavelengths use R_H = 1.09678 × 10⁷ m⁻¹, which corrects for the proton’s finite mass; the difference is about 0.05%.

Does the Rydberg formula work for helium or other atoms?+

Only for one-electron species such as H, He⁺, Li²⁺ and Be³⁺, using the nuclear charge Z. Atoms with two or more electrons have screening and electron–electron repulsion, so their spectra need quantum-defect or full quantum calculations.

What is the series limit?+

The series limit is the shortest wavelength in a series, reached as n₂ approaches infinity. For the Lyman series it is 91.18 nm (13.6 eV), the energy needed to ionise hydrogen from its ground state.

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