The Rydberg Equation with Worked Examples

In 1888, the Swedish physicist Johannes Rydberg found a single formula that predicted the wavelength of every line in the spectrum of hydrogen. He found it by studying the data, decades before anyone understood why it worked. Bohr’s model explained it in 1913, and quantum mechanics confirmed it. Today, the Rydberg equation is the quickest way to calculate the wavelengths in hydrogen’s spectrum. This article explains the equation and works through the problems you’re likely to meet.

The equation

1/λ = R_H × (1/n₁² − 1/n₂²)

• λ = wavelength of the emitted (or absorbed) light, in metres
• R_H = 1.097 × 10⁷ m⁻¹, the Rydberg constant for hydrogen
• n₁ = the lower energy level
• n₂ = the higher energy level (n₂ > n₁)

The quantity 1/λ is called the wavenumber (often written ν̃). Spectroscopists commonly use it in cm⁻¹, in which case R_H = 109,678 cm⁻¹ (often rounded to 109,700 cm⁻¹).

For one-electron ions

For species with one electron and nuclear charge Z (He⁺, Li²⁺, Be³⁺…):

1/λ = R × Z² × (1/n₁² − 1/n₂²)

Connection to energy

The Rydberg equation is equivalent to the Bohr energy formula. Multiplying by hc gives the photon energy:

E = hc/λ = hcR_H × (1/n₁² − 1/n₂²), and hcR_H = 2.18 × 10⁻¹⁸ J (13.6 eV).

See Bohr model calculations.

Which n₁ for which series?

Series	n₁	Region
Lyman	1	ultraviolet
Balmer	2	visible
Paschen	3	infrared
Brackett	4	infrared
Pfund	5	far infrared

See the hydrogen emission spectrum.

Worked example 1: the first Balmer line (H-alpha)

n₁ = 2, n₂ = 3

1/λ = 1.097 × 10⁷ × (1/4 − 1/9) = 1.097 × 10⁷ × (0.25000 − 0.11111) = 1.097 × 10⁷ × 0.13889 = 1.5236 × 10⁶ m⁻¹

λ = 1 ÷ 1.5236 × 10⁶ = 6.563 × 10⁻⁷ m = 656.3 nm (red)

Worked example 2: the second Balmer line (H-beta)

n₁ = 2, n₂ = 4

1/λ = 1.097 × 10⁷ × (1/4 − 1/16) = 1.097 × 10⁷ × 0.1875 = 2.0569 × 10⁶ m⁻¹

λ = 486.2 nm (blue-green)

Worked example 3: the first Lyman line

n₁ = 1, n₂ = 2

1/λ = 1.097 × 10⁷ × (1 − 1/4) = 1.097 × 10⁷ × 0.75 = 8.2275 × 10⁶ m⁻¹

λ = 121.5 nm (ultraviolet)

Worked example 4: the first Paschen line

n₁ = 3, n₂ = 4

1/λ = 1.097 × 10⁷ × (1/9 − 1/16) = 1.097 × 10⁷ × 0.048611 = 5.3326 × 10⁵ m⁻¹

λ = 1.875 × 10⁻⁶ m = 1,875 nm (infrared)

Notice the pattern: series ending on lower levels (Lyman) have much shorter wavelengths than those ending on higher levels (Paschen).

Worked example 5: series limits

The series limit (convergence limit) is the shortest wavelength in a series, when n₂ = ∞ and 1/n₂² = 0.

Balmer limit: 1/λ = 1.097 × 10⁷ × (1/4 − 0) = 2.7425 × 10⁶ m⁻¹ → λ = 364.6 nm (near UV)

Lyman limit: 1/λ = 1.097 × 10⁷ × 1 → λ = 91.16 nm

The Lyman limit corresponds to ionisation of a ground-state hydrogen atom:

E = hc/λ = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) ÷ (9.116 × 10⁻⁸) = 2.179 × 10⁻¹⁸ J

× 6.022 × 10²³ = 1,312 kJ/mol, the first ionisation energy of hydrogen. See ionization energy trend.

Worked example 6: working backwards to find n₂

A line in the Balmer series has a wavelength of 434.0 nm. Which transition produced it?

1/λ = 1 ÷ (434.0 × 10⁻⁹) = 2.3041 × 10⁶ m⁻¹

2.3041 × 10⁶ = 1.097 × 10⁷ × (1/4 − 1/n₂²)

1/4 − 1/n₂² = 0.21004

1/n₂² = 0.25 − 0.21004 = 0.03996

n₂² = 25.03 → n₂ = 5

The line is the n = 5 → n = 2 transition (H-gamma).

Because n must be a whole number, small rounding differences are fine; always round to the nearest integer.

Worked example 7: absorption

Which wavelength must a ground-state hydrogen atom absorb to jump to n = 3?

1/λ = 1.097 × 10⁷ × (1 − 1/9) = 1.097 × 10⁷ × 0.8889 = 9.751 × 10⁶ m⁻¹

λ = 102.6 nm

The same formula works for absorption; the wavelength is identical to that emitted in the reverse transition. See emission vs absorption spectra.

Worked example 8: He⁺

Calculate the wavelength of the n = 4 → n = 3 transition in He⁺ (Z = 2).

1/λ = 1.097 × 10⁷ × 2² × (1/9 − 1/16) = 1.097 × 10⁷ × 4 × 0.048611 = 2.1330 × 10⁶ m⁻¹

λ = 468.8 nm (blue)

This line was important historically. It was seen in stellar spectra and at first attributed to hydrogen, until Bohr showed it came from ionised helium, one of the early confirmations of his model.

Why the Rydberg equation works

Rydberg found his formula empirically, simply by looking for a pattern in the measured wavelengths. The explanation came from Bohr’s model: the energy of level n in hydrogen is proportional to −1/n², so the energy of a photon emitted between two levels is proportional to (1/n₁² − 1/n₂²). Dividing energy by hc converts it to 1/λ, and the constant of proportionality becomes the Rydberg constant. Bohr was able to calculate R_H from fundamental constants (the electron’s mass and charge, Planck’s constant and the speed of light), and his value matched Rydberg’s experimental one almost exactly. That agreement was one of the strongest early pieces of evidence for quantum theory.

Today, the Rydberg constant is one of the most precisely measured constants in physics, known to about twelve significant figures. The value for hydrogen, R_H, is very slightly smaller than the “infinite-mass” Rydberg constant R∞ = 1.0974 × 10⁷ m⁻¹, because the hydrogen nucleus isn’t infinitely heavy and moves slightly as the electron moves. For school and university calculations, 1.097 × 10⁷ m⁻¹ is precise enough.

A quick check on any answer

• Balmer lines must fall between 364.6 nm (the series limit) and 656.3 nm (the first line).
• Lyman lines must fall between 91.2 nm and 121.5 nm.
• Paschen lines must fall between 820.4 nm and 1,875 nm.

If your answer for a given series lies outside its range, check your values of n₁ and n₂.

Converting between units

Quantity	Relationship
Wavenumber (m⁻¹) → wavelength (m)	λ = 1 ÷ ν̃
Wavelength (m) → frequency (Hz)	f = c ÷ λ
Wavelength (m) → energy per photon (J)	E = hc ÷ λ
Energy per photon → per mole	× 6.022 × 10²³
m⁻¹ → cm⁻¹	÷ 100

For practice with these conversions, see wavelength, frequency and photon energy calculations.

Common mistakes

1. Forgetting to take the reciprocal. The equation gives 1/λ; you must invert to get λ.
2. Swapping n₁ and n₂. n₁ is always the lower level; otherwise you get a negative answer.
3. Mixing units. R in m⁻¹ gives λ in metres; convert to nm by multiplying by 10⁹.
4. Using the wrong n₁ for a series. Balmer is always n₁ = 2.
5. Applying it to multi-electron atoms. It works only for hydrogen and one-electron ions.

Practice questions

1. Calculate the wavelength of the n = 6 → n = 2 transition in hydrogen.
2. Calculate the wavelength of the Paschen series limit.
3. A hydrogen line has λ = 97.25 nm. Identify the transition.

Answers:

1. 1/λ = 1.097 × 10⁷ × (1/4 − 1/36) = 2.4378 × 10⁶ → 410.2 nm (H-delta, violet)
2. 1/λ = 1.097 × 10⁷ × (1/9) = 1.2189 × 10⁶ → 820.4 nm
3. 1/λ = 1.0283 × 10⁷; ÷ R = 0.9374, which fits the Lyman series: 1 − 1/n₂² = 0.9374 → n₂² = 16 → n = 4 → n = 1

Key takeaways

• 1/λ = R_H (1/n₁² − 1/n₂²), with R_H = 1.097 × 10⁷ m⁻¹ and n₁ the lower level.
• n₁ = 1, 2 and 3 give the Lyman, Balmer and Paschen series.
• Setting n₂ = ∞ gives the series limit; the Lyman limit gives hydrogen’s ionisation energy.
• For one-electron ions, multiply R by Z².
• Always take the reciprocal, check units and round n to a whole number when working backwards.