Rydberg Formula — Physica

Equivalent forms

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

R_H \approx 1.0973731568 \times 10^7\,\text{m}^{-1}

Two integers reproduce hundreds of spectral lines across infrared, visible, and UV.

Symbol	Name	SI unit	Dimension	Range
$\lambda$	Wavelengthoutput Photon wavelength of emitted/absorbed line	m	L	1e-8 – 0.00001
$n_1$	Lower level Principal quantum number of lower state	—	1	1 – 10
$n_2$	Upper level Principal quantum number of upper state	—	1	2 – 20
$R_H$	Rydberg constant Rydberg constant for hydrogen	1/m	L⁻¹	10973731.568 – 10973731.568

Unit systems

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

Where it holds

Exact for hydrogen and hydrogenic ions

(use R_Z = Z^{2}R_H

for ions with one electron). Multi-electron atoms have similar but quantum-defect-corrected spectra. Relativistic + spin-orbit fine structure adds small splittings

\sim \alpha ^{2} \cdot R_H

.

Dimensional analysis

$[1/\lambda ] = L^{-1} = [R_H] \times$ (dimensionless $) \checkmark$

Discovery

Johannes Rydberg · 1888

Rydberg distilled Balmer's empirical formula for visible hydrogen lines into a universal form using integer quantum numbers — 25 years before anyone knew why atoms had quantized energy levels. Bohr's 1913 model finally explained it.

Try this

Why does hydrogen glow in such precise, predictable colors?

When excited hydrogen gas emits light, it does so at specific wavelengths organized into series (Lyman, Balmer, Paschen). One formula predicts every line from two integers. How?

Research status: stable

Real-world applications

Stellar spectroscopy — Balmer lines determine star temperatures and compositions.
Hubble's law — redshift of Lyman- $\alpha$ calibrates cosmic distances.
Plasma diagnostics — $H\alpha$ emission monitors fusion plasmas (ITER, tokamaks).
Lasers — H-alpha at 656 nm and other hydrogenic transitions seed many laser systems.

Common misconceptions

The Rydberg formula applies only to hydrogen — actually it generalizes to any hydrogenic system $(He^{+}$ , $Li^{2+}$ , Rydberg states in alkalis).
Each spectral line is infinitely sharp — natural linewidth and Doppler broadening give finite widths.
n_1 < n_2 is for absorption; emission swaps — both interpretations give the same formula since the energy difference is symmetric.

Experimental verification

Balmer series visible lines

(H\alpha 656\,\mathrm{nm}

,

H\beta 486\,\mathrm{nm}

, ...) match to

10^{-5}

. Modern frequency-comb spectroscopy of the 1S–2S transition pins R_H to 12 significant digits — among the most precisely measured constants in physics.

Derivation

Bohr's energy levels:

E_n = -R_H hc / n^{2}

.

Photon energy from transition:

\Delta E = hc/\lambda = R_H hc (1/n_{1}^{2} - 1/n_{2}^{2})

.

Divide by hc to get

1/\lambda

form.

Limiting cases

n_{2} \to \infty

⟶

1/\lambda \to R_H / n_{1}^{2}

Series limit (ionization edge). For Lyman: 91.2 nm; for Balmer: 364.6 nm.

n_{2} = n_{1} + 1

⟶ Longest wavelength of the seriesSmallest energy jump — first line of each series (e.g.,

H\alpha at 656\,\mathrm{nm}

for

n=2\to 3)

.

n_{1} = 1

(Lyman)⟶ UV emissionAll transitions to ground state are in deep UV — invisible to the eye, drives interstellar ionization.

What if…

What about

He^{+}

(single-electron helium)?

Use R_He+ $= 4\cdot R_H$ (since $Z=2 \to Z^{2}=4)$ . All wavelengths shrink by $4\times$ ; the Pickering series in stars matched this prediction in 1912.

What if R_H were 10% larger?

All hydrogen wavelengths shift 10% bluer — H-alpha would be 596 nm (yellow-green). Stellar spectra would look completely different.

1

H-alpha line (Balmer, n=2→3)

Given ·

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

Find · \lambda

Steps

$1/\lambda = 1.097e_{7} \times (1/4 - 1/9) = 1.097e_{7} \times 5/36$
$1/\lambda \approx 1.524e_{6} m^{-1}$
$\lambda \approx 6.563 \times 10^{-7} m = 656.3\,\mathrm{nm}$ — the iconic red of glowing hydrogen.

Result ·

\lambda \approx 656.3\,\mathrm{nm}

(deep red)

2

Lyman-alpha (n=1→2)

Given ·

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

Find · \lambda

Steps

$1/\lambda = 1.097e_{7} \times (1 - 1/4) = 1.097e_{7} \times 3/4$
$1/\lambda \approx 8.23e_{6} m^{-1}$
$\lambda \approx 121.6\,\mathrm{nm}$ — dominant UV line in astrophysics; used to detect distant galaxies.

Result ·

\lambda \approx 121.6\,\mathrm{nm} (UV)

Bohr Energy Levels (Hydrogen)→Bohr Radius→Planck Radiation Law→