Rydberg Equation Calculator
Enter the initial and final principal quantum numbers of a hydrogen electron transition and get the wavelength, frequency and photon energy of the spectral line, plus its series (Lyman, Balmer, Paschen…).
Find Hydrogen Line Wavelength from Quantum Numbers
Rydberg Equation Formula
Where:
- \(\lambda\) — wavelength of the emitted or absorbed photon (m)
- \(R_H = 1.0967758 \times 10^7\ \text{m}^{-1}\) — Rydberg constant for hydrogen (corrected for the reduced mass of the atom)
- \(n_i\) — initial principal quantum number of the electron
- \(n_f\) — final principal quantum number of the electron
The formula gives the inverse wavelength (wavenumber) of the photon produced when an electron in a hydrogen atom jumps between two energy levels. The companion relations \(\nu = c/\lambda\) and \(E = h\nu = hc/\lambda\) convert the wavelength into frequency and photon energy.
Rydberg Equation – Calculation Example
Example: an electron in a hydrogen atom falls from level \(n_i = 3\) to level \(n_f = 2\) (the transition that produces the famous red Hα line).
\[ \frac{1}{\lambda} = 1.0967758 \times 10^7 \times \left| \frac{1}{2^2} - \frac{1}{3^2} \right| = 1.0967758 \times 10^7 \times \frac{5}{36} = 1.5233 \times 10^6\ \text{m}^{-1} \]
\[ \lambda = \frac{1}{1.5233 \times 10^6} \approx 6.5647 \times 10^{-7}\ \text{m} = 656.47\ \text{nm} \]
\[ \nu = \frac{c}{\lambda} = \frac{2.9979 \times 10^8}{6.5647 \times 10^{-7}} \approx 4.568 \times 10^{14}\ \text{Hz} \]
Result: \(\lambda \approx 656.47\) nm, \(\nu \approx 456.81\) THz, photon energy \(E \approx 1.89\) eV \((3.03 \times 10^{-19}\) J) — the red Hα line of the Balmer series.
How to Use the Rydberg Equation Calculator
- Enter the Initial Level (nᵢ) — the principal quantum number the electron starts from. For an emission line this is the higher level (e.g. 3).
- Enter the Final Level (n_f) — the level the electron ends on. For an emission line this is the lower level (e.g. 2).
- Both numbers must be whole numbers from 1 to 100, and they must differ — identical levels produce no photon.
- Read the results: the Wavelength in nanometres, the Frequency in terahertz, and the Photon Energy in both eV and joules, together with the transition type, spectral series and wavelength region.
If \(n_i > n_f\) the calculator describes an emission line (the electron falls and releases a photon); if \(n_i < n_f\) it describes an absorption line (the electron is lifted up by absorbing a photon of exactly the same wavelength).
What Is the Rydberg Equation?
The Rydberg equation, formulated by Johannes Rydberg in 1888, predicts the wavelengths of all the spectral lines of atomic hydrogen. It generalised Balmer's empirical formula for the visible lines into a single expression covering the entire spectrum. In 1913 Niels Bohr showed why it works: the electron in a hydrogen atom can occupy only the discrete levels \(n = 1, 2, 3, \ldots\), and a photon is emitted or absorbed when the electron jumps between them, carrying exactly the energy difference of the two levels.
Because each level has a fixed energy, every pair of levels corresponds to one specific wavelength. This is why hydrogen's spectrum is not a continuous rainbow but a set of sharp lines — and why those lines act as a fingerprint of hydrogen in laboratory analysis and astronomy alike.
The lines group into series named after their lower level: the Lyman series (to \(n = 1\), ultraviolet), the Balmer series (to \(n = 2\), the visible lines Hα 656 nm, Hβ 486 nm, Hγ 434 nm and Hδ 410 nm), the Paschen (to \(n = 3\)), Brackett (to \(n = 4\)) and Pfund (to \(n = 5\)) series, all in the infrared. As the upper level grows, the lines crowd together and converge to the series limit, beyond which the atom is ionised. The calculator identifies the series automatically and recognises several famous lines such as Hα and Lyman-alpha.
The photon energy is worth interpreting physically: \(E \approx 1.89\) eV for Hα corresponds to the gap between the \(n = 2\) and \(n = 3\) levels, and 13.6 eV — the series limit of the Lyman series — is the ionisation energy of hydrogen from the ground state.
Assumptions and Limitations
- The formula applies to the hydrogen atom only (\(Z = 1\)) in the gas phase. For hydrogen-like ions such as He⁺ or Li²⁺, multiply the right-hand side by \(Z^2\).
- Wavelengths are vacuum values computed with the reduced-mass constant \(R_H = 1.0967758 \times 10^7\ \text{m}^{-1}\). In air the lines shift by roughly 0.1–0.3 nm (Hα appears at 656.28 nm instead of 656.47 nm).
- Fine structure, the Lamb shift and isotope effects (deuterium, tritium) are neglected; these cause sub-nanometre splittings that matter only in high-resolution spectroscopy.
- The model treats the atom as a single electron around a single nucleus — it does not apply to multi-electron atoms, molecules or solids.
If you are analysing other types of spectra or energy conversions, you may also find our Bragg's Law calculator useful for diffraction-based wavelength analysis, or the Standard Cell Potential calculator for electrochemical energy calculations. For the kinetic behaviour of gas-phase atoms, see the Molecular Speed of a Gas calculator.
Frequently Asked Questions
How do I calculate a hydrogen wavelength by hand with the Rydberg formula?
Take the reciprocals of the squared quantum numbers, subtract them, and multiply by the Rydberg constant: \(1/\lambda = R_H \cdot |1/n_f^2 - 1/n_i^2|\). Then invert the result to get the wavelength in metres. For example, for \(n_i = 4, n_f = 2\): \(1/\lambda = 1.0967758\times10^7 \times 3/16 = 2.056\times10^6\ \text{m}^{-1}\), so \(\lambda \approx 486\) nm — the cyan Hβ line.
Does it matter which level I enter first — initial or final?
No, the wavelength, frequency and photon energy are identical either way, because the formula uses an absolute value. Only the interpretation changes: entering the higher level first describes emission (the electron falls and releases a photon), while entering the lower level first describes absorption of a photon with the same wavelength. The one case to avoid is entering the same number twice — equal levels produce no photon at all.
Why does the calculator refuse my input when both quantum numbers are the same?
When \(n_i = n_f\) the term \(1/n_f^2 - 1/n_i^2\) equals zero, so the wavenumber is zero and the wavelength would be infinite — physically, no transition and no photon occur because the electron's energy does not change. Choose two different whole numbers, each at least 1. Also note that n must be an integer: values like 1.5 are not valid energy levels.
What range of wavelengths can hydrogen lines have?
The shortest possible line is the Lyman limit at about 91.2 nm (far ultraviolet), which corresponds to the full 13.6 eV ionisation energy from the ground state. Lines with higher upper levels get longer wavelengths and crowd together as they approach each series limit. Transitions ending on n = 2 fall between 410 nm and 656 nm — the only hydrogen lines visible to the human eye; everything to n = 1 is ultraviolet and everything to n ≥ 3 is infrared.
Can I use this calculator for He⁺, Li²⁺ or other hydrogen-like ions?
Not directly, but the fix is simple. For a single-electron ion with nuclear charge Z, multiply the wavenumber by \(Z^2\): \(1/\lambda = R_H Z^2 |1/n_f^2 - 1/n_i^2|\). Take the wavelength from this calculator, then divide it by \(Z^2\). For He⁺ (Z = 2) every line is four times shorter than the corresponding hydrogen line. The formula does not work for neutral helium or any multi-electron atom.
Why does my measured Hα wavelength differ slightly from the calculated 656.47 nm?
Two common reasons. First, the calculator returns vacuum wavelengths; in air the line shifts by roughly 0.1–0.3 nm (Hα is seen at 656.28 nm in standard air). Second, high-resolution instruments resolve fine structure, the Lamb shift and isotope splittings, which this simple model ignores. In astronomy, motion of the gas also shifts lines via the Doppler effect — a redshifted or blueshifted Hα line is normal and is actually used to measure velocities.
What is the difference between R_H and the Rydberg constant R∞ I see in textbooks?
\(R_\infty = 1.0973732\times10^7\ \text{m}^{-1}\) assumes an infinitely heavy nucleus, while \(R_H = 1.0967758\times10^7\ \text{m}^{-1}\) is corrected for the finite reduced mass of the electron–proton system in real hydrogen. The difference is about 0.05%, which shifts Hα by roughly a third of a nanometre. This calculator uses \(R_H\), so its results match observed hydrogen lines more closely; use \(R_\infty\) only when the nucleus is treated as fixed.