Rydberg Formula — Physica

Equivalent forms

\nu = R_H c \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)

E_\gamma = 13.6\,\text{eV}\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)

Three decades before Schrödinger, two integers and one constant predicted every visible hydrogen line.

Symbol	Name	SI unit	Dimension	Range
$lambda$	Photon wavelengthoutput Wavelength of the emitted or absorbed photon	m	L	1e-8 – 0.01
$R_H$	Rydberg constant Inverse-length Rydberg constant for hydrogen	1/m	L^-1	10973731.568 – 10973731.568
$n_1$	Lower level Quantum number of the lower state	1	1	1 – 6
$n_2$	Upper level Quantum number of the upper state (n_2 > n_1)	1	1	2 – 20

Unit systems

Symbol	SI	CGS	Imperial
$lambda$	m	cm	ft
$R_H$	1/m	1/cm	1/ft
$n_1$	1	1	1
$n_2$	1	1	1

Where it holds

Hydrogen-like one-electron systems. Use R_inf with reduced mass correction for muonic atoms, positronium, helium+. Breaks down for multi-electron atoms where screening matters.

Dimensional analysis

1/\lambda \to [L^-1]

R_H * (dimensionless

) \to [L^-1]

Both sides are inverse length.

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Discovery

Johannes Rydberg · 1888

Rydberg distilled Balmer's empirical formula for hydrogen visible lines into a universal relation valid for all series. Bohr later derived it from his atomic model in 1913, vindicating quantum theory.

Try this

Why does hydrogen emit only specific colors of light?

Heat a hydrogen lamp and you see four lines in the visible spectrum — red, blue-green, blue, violet. Why these exact wavelengths, and how did Rydberg predict them 30 years before quantum mechanics existed?

Research status: stable

Real-world applications

Astronomy — H-alpha imaging of nebulae and solar prominences
21 cm line cosmology (hyperfine, but spectroscopy methodology)
Atomic clocks and frequency metrology
Plasma diagnostics in tokamaks and stellar atmospheres

Common misconceptions

The formula applies to one-electron systems only; helium needs explicit electron-electron repulsion
Series labels (Lyman, Balmer, Paschen) are conventions tied to n_1, not physical regions
Rydberg's constant R_H differs slightly from R_inf because the proton is not infinitely heavy

Experimental verification

Balmer (1885) fit the four visible H lines; Lyman (1906) extended to UV; Paschen (1908) to IR. Modern frequency-comb spectroscopy verifies R_inf to 13 significant figures — the most precisely measured fundamental constant.

Derivation

From Bohr's model

E_n = -13.6\,\mathrm{eV} / n^2

.

A transition

n_2 \to n_1

emits

E_\gamma = |E_n_{2} - E_n_{1}| = 13.6\,\mathrm{eV} * (1/n_1^2 - 1/n_2^2)

.

Photon energy

E_\gamma = h*c/\lambda

, so

1/\lambda = (13.6\,\mathrm{eV} / (h*c)) * (1/n_1^2 - 1/n_2^2) = R_H * (1/n_1^2 - 1/n_2^2)

, where

R_H = 13.6\,\mathrm{eV} / (h*c) \approx 1.097e_{7} / m

.

Limiting cases

n_1 = 1

,

n_2 = 2

⟶

\lambda \approx 121.6\,\mathrm{nm}

(Lyman-alpha, UV)Strongest hydrogen emission line in the universe; dominates interstellar UV background.

n_1 = 2

,

n_2 = 3

⟶

\lambda \approx 656.3\,\mathrm{nm} (H-\alpha

, red)Defines the color of nebulae and the chromosphere of the Sun.

n_2 \to

infinity⟶

1/\lambda = R_H/n_1^2

(series limit)Photon at this wavelength just ionizes from level n_1; corresponds to binding energy 13.6/n_1^2 eV.

What if…

What if R_H were twice as large?

All hydrogen lines would shift to half their wavelength — visible lines would become UV.

What if we apply the formula to deuterium?

R_D > $R_H by \sim 0.027$ % from reduced-mass correction — gives the isotope shift Urey used to discover deuterium.

What if you replace the electron with a muon?

Muonic hydrogen: R scales by $m_\mu /m_e \approx 207$ — all lines move to X-ray; basis of the proton-radius puzzle.

1

H-alpha wavelength (n_1=2, n_2=3)

Given ·

n 1:: 2
n 2:: 3
R H:: 10973731.568

Find · lambda

Steps

$1/n_1^2 - 1/n_2^2 = 0.25 - 0.1111 = 0.1389$
$1/\lambda = 10973731.568 * 0.1389 = 1.524e_{6} /m$
$\lambda = 6.563e-7\,\mathrm{m} = 656.3\,\mathrm{nm}$ — red H-alpha line

Result ·

1/\lambda = 1.097e_{7} * (1/4 - 1/9) = 1.097e_{7} * 0.1389 \approx 1.524e_{6} /m \to \lambda \approx 656.3\,\mathrm{nm}

2

Lyman series limit (ionization from ground)

Given ·

n 1:: 1
n 2:: 9999999
R H:: 10973731.568

Find · lambda_min

Steps

$1/\lambda \approx R_H * 1 = 1.097e_{7} /m$
$\lambda \approx 9.118e-8\,\mathrm{m} = 91.18\,\mathrm{nm}$
$E_\gamma = h*c/\lambda \approx 2.18e-18\,\mathrm{J} = 13.6\,\mathrm{eV}$ — hydrogen ionization energy

Result ·

1/\lambda \approx R_H = 1.097e_{7} /m \to \lambda \approx 91.18\,\mathrm{nm}

(corresponds to 13.6 eV)

bohr hydrogen energy levelsPhoton Energy (Planck Relation)→Time-Independent Schrödinger Equation→