Rydberg Formula Calculator
Written by Thierno Sadou Diallo, formula verified per our methodology • Last checked on 9/9/2026
The Rydberg formula gives 1/λ = R_H × (1/n1² − 1/n2²), where R_H is the Rydberg constant for hydrogen. For the transition n1=2 to n2=3 (the Hα line of the Balmer series), the emitted wavelength is about 656.5 nm — hydrogen's famous red line.
Explanation
When an electron in the hydrogen atom drops from a higher energy level (n2) to a lower one (n1), it emits a photon whose wavelength is precisely determined by the energy difference between these two levels — this is exactly what the Rydberg formula describes, one of the earliest empirical equations in atomic physics, discovered even before quantum mechanics explained its theoretical origin (the Bohr model, then full quantum mechanics, later confirmed this formula from first principles). Transitions down to level n1=1 form the Lyman series, in the ultraviolet; those down to n1=2 form the Balmer series, the only one with several lines falling in the visible range — it's this Balmer series that allowed hydrogen's spectrum to be observed directly as early as the 19th century, well before the full theoretical formulation of atomic structure. The Hα line (n1=2, n2=3), at about 656.5 nm, is the most intense and characteristic line of this series: it gives its characteristic red color to many emission nebulae observed in astronomy, where ionized hydrogen recombines by emitting precisely this wavelength. This formula uses the Rydberg constant specific to hydrogen (R_H), slightly lower than the "infinite" Rydberg constant (R∞), which assumes an infinitely heavy nucleus relative to the electron: the correction, proportional to the ratio of the proton and electron masses, remains small (less than 0.1%) but measurable, and becomes more significant for lighter hydrogen-like atoms such as positronium — the same fundamental quantum effect as the one demonstrated by the photoelectric effect at the very start of the 20th century.
Example: the Hα line of the Balmer series
Inputs
Final level: n1 = 2. Initial level: n2 = 3.
Calculation
1/λ = 1.09677584×10⁷ × (1/2² − 1/3²) = 1.09677584×10⁷ × (0.25 − 0.1111) ≈ 1,523,891 m⁻¹. λ = 1 ÷ 1,523,891 ≈ 6.565×10⁻⁷ m, or about 656.5 nm.
Result
This transition emits a photon of about 656.5 nm, in the visible red — the Hα line, the most intense of hydrogen's Balmer series.
Frequently asked questions
Why is only the Balmer series visible to the naked eye?
Because transitions down to n1=1 (Lyman series) release too much energy, corresponding to wavelengths in the ultraviolet, invisible to the human eye; transitions down to n1=3 and beyond (Paschen, Brackett series...) instead release too little energy, falling into the infrared. Only transitions down to n1=2 produce photons whose energy corresponds precisely to the wavelength range visible to the human eye (about 380 to 700 nm), which explains why the Balmer series was historically the first to be observed.
Why use R_H rather than R∞ (the "infinite" Rydberg constant)?
R∞ assumes a nucleus infinitely heavier than the electron, an approximation valid for very heavy atoms but imprecise for hydrogen, whose proton is "only" 1836 times more massive than the electron. In practice, the electron and proton both orbit their common center of mass, not a perfectly stationary nucleus: the reduced-mass correction (R_H = R∞ × mp/(mp+me)) accounts for this shared motion and gives slightly more accurate wavelengths, consistent with actual spectroscopic measurements.
Does this formula apply to atoms other than hydrogen?
A generalized version applies to so-called "hydrogen-like" ions (a single electron, like He⁺ or Li²⁺), by multiplying the right-hand term by the square of the atomic number Z. For a neutral atom with several electrons, however, electron-electron interactions considerably complicate the spectrum, which no longer follows this simple single-parameter formula — the precise energy of each emitted photon is then derived directly from its wavelength via our photon energy calculator.