How this instrument works
The Rydberg equation gives the wavelength of light emitted or absorbed when a hydrogen electron jumps between two quantized energy levels: 1/λ = R(1/n1² − 1/n2²). Here n1 and n2 are the principal quantum numbers of the lower and upper levels, positive integers with n2 greater than n1, and R is the Rydberg constant, close to 1.097×10⁷ per metre. Johann Balmer found the visible-light pattern by trial in 1885; Johannes Rydberg generalized it to every hydrogen series three years later, before anyone knew why it worked.
The shape comes straight from the Bohr model. Each level n has energy E_n = −13.6 eV ⁄ n², so a transition releases a photon whose energy is the difference between two such terms. Photon energy is hc ⁄ λ, and rearranging E_n2 − E_n1 = hc ⁄ λ produces exactly the inverse-square difference in the formula — the constant R is just 13.6 eV repackaged into hc's units. Because both terms are squared and inverted, closely spaced high levels crowd the emitted wavelengths together near a series limit, while the lowest transitions are spread far apart.
The formula is exact only for hydrogen and hydrogen-like ions with a single electron; multiply R by Z² and it works for He⁺ or Li²⁺ too. Add a second electron and the neat 1/n² energy ladder breaks down, because electron-electron repulsion makes each level depend on more than just n. Which series a line belongs to is set entirely by n1: n1 = 1 gives the ultraviolet Lyman series, n1 = 2 gives the visible Balmer series, and n1 = 3 gives the infrared Paschen series.
- Enter the lower energy level, n₁ — the orbit the electron ends up on after emitting a photon, or starts from before absorbing one.
- Enter the upper energy level, n₂ — it must be a larger integer than n₁, or the equation returns a wavelength with no physical meaning.
- Read Emitted wavelength off the result field; switch its unit menu to see the same figure in nanometres or ångströms.
- To scan a whole series, hold n₁ fixed and step n₂ upward — 2, 3, 4 — and watch the wavelengths converge toward the series limit.
Worked example — the Balmer H-alpha line
Set n₁ = 2 and n₂ = 3 — an electron dropping from the third hydrogen orbit to the second, the transition Balmer's formula was originally built to fit. Compute the bracket first: 1/n1² − 1/n2² = 1/4 − 1/9 = 5/36. Multiply by the Rydberg constant: 10,973,731.568160 × 5/36 = 1,524,129.38 m⁻¹. Invert that to get the wavelength: λ = 6.56112276419×10⁻⁷ m, or 656.112 nm.
That figure is the Balmer-alpha, or Hα, line — deep red, the strongest visible hydrogen emission and the reason star-forming nebulae like the Orion Nebula glow crimson in astrophotographs. Astronomers use its known rest wavelength, 656.1 nm, as a fixed reference: measuring how far a galaxy's Hα line has drifted toward red gives its recession velocity directly.
Questions
What do n1 and n2 actually represent?
They are principal quantum numbers — integers labeling which orbit, or energy shell, the electron occupies before and after the transition. n₁ is always the lower, smaller-radius orbit; n₂ is the higher one. Only whole numbers 1, 2, 3 and so on are physically valid; the equation will still compute a result for fractional entries, but no such orbit exists in a real hydrogen atom.
Why does n2 have to be larger than n1?
Because the bracket 1/n1² − 1/n2² must stay positive for the wavelength to come out positive. Physically, n₂ greater than n₁ means the electron starts higher and finishes lower, releasing a photon — emission. Swap the two and you are describing absorption instead: the atom takes in a photon of that same wavelength to push the electron from n₁ up to n₂. The arithmetic is identical either way; only the direction of the photon changes.
Is R here the same as R_H, the hydrogen-specific Rydberg constant?
Not quite. This instrument uses R∞ = 10,973,731.568160 m⁻¹, the Rydberg constant for an infinitely heavy nucleus. The true hydrogen value, R_H, is smaller by the electron-to-proton mass ratio correction, about 10,967,758 m⁻¹, because the proton actually recoils slightly. The difference shows up only in the fifth significant figure of the wavelength, well below what most spectroscopy problems need.
Does the formula work for atoms besides hydrogen?
Only for hydrogen-like ions — a single electron orbiting a bare nucleus of charge Z, such as He⁺ or Li²⁺ — and only after multiplying R by Z². Neutral helium, carbon, or any atom with more than one electron does not follow this pattern, because electron-electron repulsion shifts each energy level away from the simple −13.6 eV ⁄ n² rule the formula assumes.
What are the Lyman, Balmer, and Paschen series?
They are the families of lines sharing the same lower level. n₁ = 1 gives the Lyman series, all in the ultraviolet; n₁ = 2 gives the Balmer series, the only one with lines in visible light, including the 656.1 nm Hα line; n₁ = 3 gives the Paschen series, in the infrared. Each series converges to a shortest-wavelength limit as n₂ approaches infinity.
Why does the wavelength stop changing much as n2 gets large?
Because 1/n2² shrinks toward zero as n₂ grows, so each extra step contributes less than the last. By n₂ = 10 the bracket is already close to its limiting value of 1/n1², and by n₂ = 20 the wavelength has all but stopped moving — that limiting value, 1/(R·n1²), is called the series limit and marks where the atom is about to ionize rather than emit a line.