Rydberg Formula — Physica

Equivalent forms

1/λ = R_H Z² (1/n₁² - 1/n₂²)

One formula predicts every spectral line of hydrogen — and inspired Bohr to quantize the atom.

Symbol	Name	SI unit	Dimension	Range
$λ$	Wavelength of emitted photonoutput Wavelength of light from the n₂ → n₁ transition	m	L	1e-8 – 0.000005
$n_1$	Lower quantum number Principal quantum number of final state	—	1	1 – 10
$n_2$	Upper quantum number Principal quantum number of initial 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	Natural
$λ$	m	cm	1/MeV
$R_H$	1/m	1/cm	MeV

Where it holds

Strictly valid for hydrogen-like (single-electron) atoms. For higher Z, replace R_H with

R_\infty Z^{2}

. Fine-structure and Lamb shifts require relativistic QED corrections.

Dimensional analysis

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

R_H \times

(dimensionless

) \to [L^{-1}]

Both sides are inverse length.

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Discovery

Johannes Rydberg · 1888

Rydberg, generalizing Balmer's 1885 formula, found a single empirical equation describing all hydrogen spectral lines. Bohr later derived it from his quantum model in 1913, revealing R_H = m_e e⁴/(8 ε_0² h³ c).

Try this

Why does hydrogen glow red at exactly 656 nm?

Compute the wavelength of the H-α line (n=3 → n=2) of hydrogen.

Research status: stable

Real-world applications

Astronomical spectroscopy — measuring redshifts and stellar compositions
Plasma diagnostics in fusion research
Calibrating wavelength standards via hydrogen lamps
Tests of QED through precision hydrogen spectroscopy

Common misconceptions

R_H is slightly different from $R_\infty due$ to the reduced mass correction for the finite-mass proton.
The formula predicts only the gross structure; fine and hyperfine structure require relativistic and QED corrections.
It does not directly apply to multi-electron atoms without modification.

Experimental verification

Modern hydrogen spectroscopy measures

R_\infty to 12

significant figures:

10973731.568160(21) m^{-1}

— one of the most precisely known constants in physics.

Derivation

From Bohr's model:

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

.

Photon emitted in transition has energy

\Delta E = E_{n_2} - E_{n_1}

(with E_{n_2} > E_{n_1} when going up).

1/\lambda = \Delta E/(hc)

gives the Rydberg formula.

R_H = m_e e^{4} / (8 \varepsilon _0^{2} h^{3} c)

.

Limiting cases

n_2 \to \infty

⟶

1/\lambda \to R_H/n_1^{2}

Series limit — wavelength becomes shortest in that series (e.g., Lyman limit at 91 nm).

n_1 = 1

⟶ Lyman series (UV)Transitions to ground state emit ultraviolet light.

n_1 = 2

⟶ Balmer series (visible)Transitions to

n=2

produce the famous

H-\alpha

,

H-\beta

,

H-\gamma

visible lines.

What if…

What if R_H were halved?

Every spectral line would shift to twice its current wavelength; visible Balmer lines would move to infrared.

What if we applied it to

He^{+}

?

$Z=2$ → multiply R_H by 4; wavelengths shrink by factor 4.

1

H-α line wavelength

Given ·

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

Find ·

\lambda

Steps

Step 1: $(1/n_1^{2} - 1/n_2^{2}) = (1/4 - 1/9) = (9 - 4)/36 = 5/36 \approx 0.1389$ .
Step 2: $1/\lambda = 10973731.568 \times 0.1389 = 1.524 \times 10^{6} m^{-1}$ .
Step 3: $\lambda = 1/1.524e_{6} = 6.563 \times 10^{-7} m = 656.3\,\mathrm{nm} (red H-\alpha$ line).

Result ·

1/\lambda = 1.097e_{7} \times (1/4 - 1/9) = 1.097e_{7} \times 0.1389 = 1.524e_{6} m^{-1}

;

\lambda = 656.3\,\mathrm{nm}

Bohr Radius→Photoelectric Effect Equation→