Balmer Series — Physica

Equivalent forms

\lambda=B\frac{n^2}{n^2-4}

A schoolteacher's integer fit to four colored lines, guessed decades before anyone knew what an atom was, turned out to be exact.

Symbol	Name	SI unit	Dimension	Range
$\lambda$	Wavelengthoutput Wavelength of the emitted line	m	L	—
$n$	Upper level Initial principal quantum number (n>2)	1	1	3 – 8
$R_H$	Rydberg constant Rydberg constant for hydrogen	m^-1	L^-1	—

Where it holds

Hydrogen and hydrogen-like (single-electron) ions, where R scales with

Z^{2}

. Multi-electron atoms need corrections for screening and require quantum-defect theory.

Dimensional analysis

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

R_H\times

(dimensionless

) \to [L^-1]

Both sides are inverse length (wavenumber).

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Discovery

Johann Balmer · 1885

Balmer, a 60-year-old teacher at a girls' school in Basel, was handed the four hydrogen wavelengths and challenged to find a pattern. He discovered λ = B n²/(n²−4) with B = 364.6 nm, matching to four significant figures. Rydberg generalized it in 1888; Bohr derived it from first principles in 1913.

Research status: stable

Real-world applications

Astronomical redshift measurement via H-alpha
Plasma diagnostics (Balmer-line ratios give temperature/density)
Hydrogen-alpha solar telescopes

Common misconceptions

The $n=2$ lower level is fixed; only the upper level n varies across the series
Balmer had no atomic model — it was pure numerology that happened to be right
The series limit $(n\to \infty )$ is 364.6 nm in the near-UV, not a visible line

Experimental verification

The measured H-alpha (656 nm), H-beta (486 nm), H-gamma (434 nm), and H-delta (410 nm) lines match the formula to spectroscopic precision; H-alpha's red glow defines emission nebulae's color.

Derivation

Bohr's energy levels are

E_n = -R_H hc/n^{2}

.

A transition from level n to level 2 emits a photon of energy

E_n - E_2 = R_H hc(1/4 - 1/n^{2})

.

Since photon energy

hc/\lambda

equals this,

1/\lambda = R_H(1/2^{2} - 1/n^{2})

.

Balmer's constant B relates to R_H by

B = 4/R_H

.

What if…

What if the lower level were

n=1

instead of 2?

You get the Lyman series, all in the ultraviolet — invisible to the eye but strong in the Sun's UV spectrum.

What if the atom is a helium

ion (He^{+})

?

R scales as $Z^{2}=4$ , so the lines shift to one-quarter the wavelength — deep in the UV.

1

Wavelength of H-alpha

Given ·

n:: 3
R H:: $1.097e_{7} m^{-1}$

Find · wavelength of the

n=3\to 2

line

Steps

$1/\lambda = 1.097e_{7} \times (0.25 - 0.1111) = 1.097e_{7} \times 0.1389$
$1/\lambda = 1.524e_{6} m^{-1}$
$\lambda = 656\,\mathrm{nm}$

Result ·

1/\lambda = R_H(1/4 - 1/9) = R_H(5/36) \to \lambda \approx 656\,\mathrm{nm} (red)

Rydberg Formula→bohr modelPhoton Energy (Planck-Einstein Relation)→