Rydberg Equation Calculator
Enter the atomic number (Z) and the two principal quantum numbers to calculate the wavelength, wavenumber, frequency, and photon energy of the emitted or absorbed photon. Works for hydrogen (Z = 1) and any hydrogen-like ion (He+, Li2+, Be3+, and others). The spectral series (Lyman, Balmer, Paschen, and so on) is identified automatically, and the step-by-step panel walks through every calculation with your actual numbers.
Formula
Worked example
For hydrogen (Z=1) transitioning from n2=3 to n1=1 (Lyman series): 1/lambda = 1.0974e7 x 1 x (1/1 - 1/9) = 1.0974e7 x 0.8889 = 9.754e6 m^-1, so lambda = 102.6 nm (UV). Photon energy = 12.09 eV.
What is the Rydberg equation?
The Rydberg equation describes the wavelengths of light emitted or absorbed when an electron in a hydrogen atom (or a hydrogen-like ion) jumps between two discrete energy levels. It was proposed empirically by Johannes Rydberg in 1888 to unify earlier formulas by Balmer, Lyman, and Paschen into a single expression. The formula is 1/lambda = R*Z^2*(1/n1^2 - 1/n2^2), where R is the Rydberg constant (about 1.0974 * 10^7 m^-1), Z is the atomic number, n1 is the lower principal quantum number, and n2 is the upper one. When the electron falls to a lower level (n2 > n1), the atom emits a photon; when it absorbs a photon and jumps upward, the same wavelength is absorbed.
Hydrogen spectral series and the visible spectrum
For hydrogen (Z = 1), the spectral lines group into named series depending on which shell the electron lands in. The Lyman series (n1 = 1) produces UV photons. The Balmer series (n1 = 2) is the only one with lines visible to the naked eye: H-alpha at 656.3 nm (red), H-beta at 486.1 nm (cyan), H-gamma at 434.0 nm (violet), and H-delta at 410.2 nm (deep violet). The Paschen, Brackett, Pfund, and Humphreys series fall in the infrared. These series are the fingerprint of hydrogen and appear in stellar atmospheres, nebulae, and laboratory discharge tubes around the world.
Hydrogen-like ions (Z > 1)
The Rydberg equation also applies to any ion that has been stripped down to a single electron, called a hydrogen-like or hydrogenic ion. Examples include He+ (Z = 2), Li2+ (Z = 3), and Be3+ (Z = 4). Because the Z^2 factor multiplies the wavenumber, all spectral lines shift to shorter wavelengths and higher energies compared to hydrogen. A He+ transition between the same quantum numbers as a hydrogen Balmer line produces a photon with four times the energy, landing deep in the UV. This scaling is exact under the Bohr model and remains a good approximation for low-Z ions.
From wavelength to energy and frequency
Once the wavelength is known, frequency follows from f = c/lambda (c = 2.998 * 10^8 m/s) and photon energy from E = h*f (h = 6.626 * 10^-34 J*s). Spectroscopists often prefer the wavenumber (reciprocal wavelength in cm^-1) because it is directly proportional to energy and avoids handling very small or very large numbers. The Rydberg constant expressed in wavenumber units is R_inf = 1.09737 * 10^5 cm^-1. Energies are most conveniently quoted in electron-volts: the ionization energy of hydrogen from n=1 is exactly 13.6 eV, the series limit of the Lyman series.
Hydrogen emission spectral series
| Series | n1 (lower) | Spectral region | Series limit (nm) | Longest line (nm) |
|---|---|---|---|---|
| Lyman | 1 | Ultraviolet | 91.18 | 121.57 |
| Balmer | 2 | Visible / near-UV | 364.60 | 656.33 |
| Paschen | 3 | Near infrared | 820.36 | 1875.10 |
| Brackett | 4 | Mid infrared | 1458.03 | 4051.28 |
| Pfund | 5 | Mid infrared | 2278.17 | 7458.63 |
| Humphreys | 6 | Far infrared | 3280.56 | 12368.55 |
Each series is named for its discoverer. The series limit is the shortest wavelength in that series (n2 approaches infinity).
Frequently asked questions
What is the Rydberg constant?
The Rydberg constant (R_infinity) is a fundamental physical constant equal to approximately 1.09737 * 10^7 m^-1 (or 10973731.57 m^-1). It represents the wavenumber of the longest-wavelength photon that can ionize a hydrogen atom from its ground state. It appears in the Rydberg formula and can be derived from first principles using the electron mass, electron charge, Planck constant, and the speed of light.
Why does the formula use n1 and n2 with n2 > n1?
The principal quantum numbers n1 and n2 label the two energy shells involved in the transition. The convention used here is that n1 is the lower (final) state and n2 is the upper (initial) state, so the electron drops from n2 to n1 and releases a photon. If n2 were less than n1, the difference inside the parentheses would be negative, giving a negative wavenumber, which is physically meaningless for emission. Absorption works in reverse (n1 is the initial lower state, n2 the final higher state) but the formula gives the same wavelength because the magnitude is identical.
Does this calculator work for helium or other elements?
It works for hydrogen-like ions only, meaning atoms or ions that have exactly one electron: H (Z=1), He+ (Z=2), Li2+ (Z=3), Be3+ (Z=4), and so on. Multi-electron atoms like neutral helium or lithium have inter-electron repulsion that the simple Rydberg formula does not account for. For those, you need quantum mechanical calculations beyond the Bohr model.
What is the series limit and what happens beyond it?
The series limit is the shortest wavelength in a spectral series, corresponding to the transition from n2 = infinity down to n1. At this limit, 1/n2^2 approaches zero and the wavenumber equals R*Z^2/n1^2. Beyond this limit (shorter wavelengths), the electron has enough energy to escape the atom entirely, which is ionization. For hydrogen the Lyman series limit is about 91.18 nm, meaning any UV photon shorter than that can ionize hydrogen from the ground state.
What is the difference between emission and absorption spectra?
In an emission spectrum, excited atoms release photons as electrons fall to lower levels, producing bright lines on a dark background. In an absorption spectrum, cool gas in front of a hot continuum source absorbs photons at exactly the same wavelengths, producing dark lines. The Rydberg equation gives identical wavelengths for both processes because the energy gap between two levels is the same regardless of which direction the electron moves.
What units are used for the wavenumber output?
The wavenumber output is in inverse centimetres (cm^-1), the standard spectroscopic unit. It equals the number of wavelengths that fit into one centimetre. Multiply by 100 to convert to m^-1, or divide wavenumber in cm^-1 by 8065.54 to get photon energy in eV.