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Rydberg Equation Calculator

Result

656.112276nm

Result: 656.112276 nm
How the result moves → nm

An electron falling from level n2 to the lower level n1 emits light whose wavelength the Rydberg equation fixes exactly. The lower level names the series: n1 = 1 is Lyman and ultraviolet, n1 = 2 is Balmer and visible, n1 = 3 is Paschen and infrared. The upper level must be the larger of the two.

Worked examples

How it's calculated

1 ÷ λ = R × (1 ÷ n1² − 1 ÷ n2²)

  1. StepEnter the lower level n1 — where the electron lands.
  2. StepEnter the upper level n2 — where it starts. It must exceed n1.
  3. StepThe bracket times the Rydberg constant gives one over the wavelength.
  4. ResultTake the reciprocal and read the wavelength in nanometres.

Reference table

n1, n2Line and seriesWavelength (nm)
1, 2Lyman α — ultraviolet121.502273
2, 5Balmer Hγ — violet433.936691
2, 4Balmer Hβ — blue-green486.009094
2, 3Balmer Hα — red656.112276
3, 4Paschen α — infrared1874.606504

Questions

How do I use the Rydberg equation?

Work out the bracket 1/n1² − 1/n2², multiply it by the Rydberg constant to get one over the wavelength in inverse metres, then take the reciprocal. Scaling by a billion turns metres into nanometres. For n1 = 2 and n2 = 3 the result is about 656.11 nm.

What are the Lyman, Balmer and Paschen series?

They are families of spectral lines grouped by the level the electron lands on. Transitions ending at n1 = 1 form the Lyman series in the ultraviolet, those ending at n1 = 2 the Balmer series in visible light, and those ending at n1 = 3 the Paschen series in the infrared.

Why must n2 be greater than n1?

The bracket 1/n1² − 1/n2² has to be positive for the equation to return a real, positive wavelength, and that happens only when n2 exceeds n1. Physically the electron starts high and drops low, releasing a photon. Equal or reversed levels emit no line.

What is the value of the Rydberg constant?

This calculator uses R = 10973731.56816 per metre, the CODATA 2018 recommended value. It is a measured quantity, not a defined one: the current NIST figure is 10973731.568157 with an uncertainty of 0.000012, which changes a visible wavelength by far less than a billionth of a nanometre.

Why is my textbook's Hα value 656.28 nm and not 656.11?

This formula treats the nucleus as infinitely heavy, which is a very good but not perfect approximation. Correcting for the finite mass of the proton shifts every hydrogen line by about 0.05 percent, which turns 656.11 nm into the measured 656.28 nm.

Sources and last check

  1. physics.nist.gov

Information, not professional advice.