The Hydrogen Spectrum and Spectral Series

Send an electric spark through a thin tube of hydrogen and the gas glows a soft pink. Spread that glow with a prism and the pink shatters into a handful of razor-sharp lines on black — a bright red, a blue-green, two crowding violets, and nothing else. No rainbow, no continuum: a barcode. The same barcode, reversed, hides in sunlight, where these exact wavelengths appear as the dark gaps Fraunhofer catalogued a century before anyone knew why. Cold hydrogen backlit by a hotter continuum absorbs precisely the colours hot hydrogen emits — a fingerprint carried unchanged from a laboratory tube to the atmosphere of a star ten kiloparsecs away. The astonishing thing is that the whole barcode is written by a single number and two integers. Johann Balmer, a Basel schoolteacher, noticed in 1885 that the four visible wavelengths obeyed a formula built from small whole numbers; Johannes Rydberg generalised it three years later into one relation reaching from the far ultraviolet to the mid-infrared. Every hydrogen line is the Rydberg constant times a difference of two reciprocal squares. Behind that arithmetic sits the level ladder $E_n = -13.6\ \text{eV}/n^2$ that the Bohr model first built and the Schrödinger equation confirmed; a line appears only when the atom drops between rungs and ships the energy out as a photon. What follows stays with what the spectrograph records — the series, their wavelengths and limits, and two sub-percent subtleties, air versus vacuum and the finite proton mass, that separate the textbook number from the measured one.

Key takeaways

Hydrogen emits and absorbs only at discrete wavelengths that fall into families — the Lyman series (UV), the Balmer series (visible), then Paschen, Brackett, Pfund and Humphreys marching into the infrared — each fixed by its common lower level $n_1$ .
Every line obeys the rydberg formula $\tilde\nu = 1/\lambda = R_{\mathrm H}(1/n_1^2 - 1/n_2^2)$ with the rydberg constant $R_{\mathrm H} = 1.096776\times10^{7}\ \text{m}^{-1}$ ; the four visible Balmer lines land at $656.28,\,486.13,\,434.05,\,410.17$ nm in air.
Each series crowds toward a short-wavelength series limit ( $n_2\to\infty$ ) — Balmer at $364.6$ nm — beyond which lies the ionization continuum.
Two sub-percent effects students conflate: laboratory wavelengths are quoted in air (Hα is $656.28$ nm in air but $656.47$ nm in vacuum), and $R_{\mathrm H}$ sits $0.054\%$ below the infinite-mass $R_\infty$ because the proton is not infinitely heavy.
Emission and absorption are the same transitions run in reverse: Fraunhofer's solar C line is Hα, and his F line is Hβ.
Key idea: a hydrogen line at wavenumber $\tilde\nu$ is the difference of two term values $T_n = R_{\mathrm H}/n^2$ ; fixing the lower level $n_1$ and letting the upper level climb generates one spectral series, and the single constant $R_{\mathrm H}$ scales them all.

The core insight

A hydrogen line is the difference of two energy-level terms, $\tilde\nu = T_{n_1} - T_{n_2}$ with $T_n = R_{\mathrm H}/n^2$ . Grouping transitions by their shared lower level $n_1$ sorts the spectrum into series — Lyman, Balmer, Paschen onward — with lines piling toward a limit as $n_2\to\infty$ . One constant $R_{\mathrm H}$ and two integers fix every wavelength.

Where this lives today

Hydrogen's barcode is the most over-determined spectrum in physics, and precision spectroscopy of it now anchors the fundamental constants. The $1S\!-\!2S$ two-photon transition — a Lyman-series interval measured against an optical frequency comb (the 2005 Nobel technology) — is known to roughly fifteen digits, fixing $R_\infty$ and stress-testing quantum electrodynamics and the proton's charge radius. The reduced-mass idea of Derivation C reaches its extreme in muonic hydrogen, where a muon two hundred times heavier than the electron shrinks the orbit until the lines feel the proton's size directly. The series also do daily astronomical work: Balmer absorption sets stellar spectral classes; H-α imaging maps star-forming gas across galaxies; high- $n_1$ radio recombination lines trace warm interstellar hydrogen; and redshifted Lyman-α becomes the "Lyman-α forest" charting matter along a sightline to a quasar. The Pickering episode of 1896 — a stellar "hydrogen" series that proved to be ionized helium — remains the cautionary classic: the pattern is a fingerprint only once you also know the nuclear charge.

The physical picture

Picture the energy levels as a ladder whose rungs crowd toward the top. Because $E_n \propto -1/n^2$ , the gap between $n$ and $n+1$ shrinks fast: the jump from the ground rung to the first excited one swallows three-quarters of the binding energy, while the highest rungs pack into a sliver just below the ionization edge at $E = 0$ . A spectral series is what you get by fixing the bottom rung of every jump and letting the top rung range upward. Fix it at $n_1 = 2$ and you are watching the Balmer series: the lines march bluer and bunch closer as the upper level climbs, converging on the limit where the electron is set free.

A single constant sets the scale of the entire spectrum. Knowing $R_{\mathrm H}$ and the two integers, you can predict any hydrogen wavelength to five figures — far-UV Lyman-α, red Hα, mid-IR Pfund lines alike. That is why the spectrum is a fingerprint: the pattern of ratios is rigid, set by integers, and only the overall scale $R_{\mathrm H}$ carries the physics of the atom. Redshift the whole comb and the ratios survive — which is how the Lyman-α forest traces hydrogen across cosmic distance.

Figure

Punchline — the visible Balmer spectrum drawn at true position: Hα 656.28, Hβ 486.13, Hγ 434.05, Hδ 410.17 nm, converging on the 364.6 nm series limit. Fraunhofer's solar C line is Hα.

The core equations

The entire spectrum follows from one equation. A photon appears when the atom falls from an upper level $n_2$ to a lower level $n_1$ , carrying off the energy difference; dividing by $hc$ turns it into a wavenumber, the quantity spectroscopists tabulate. The lower level is more negative, so the difference is positive and the photon is emitted (reverse it and the same wavenumber is absorbed).

\tilde\nu = \frac{1}{\lambda} = R_{\mathrm H}\!\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \qquad n_2 > n_1 \qquad (1)

where: $\tilde\nu$ — wavenumber $1/\lambda$ of the line (in vacuum); $\lambda$ — vacuum wavelength; $R_{\mathrm H} = 1.096776\times10^{7}\ \text{m}^{-1}$ — the rydberg constant for hydrogen; $n_1$ — principal quantum number of the lower level, which labels the series; $n_2 > n_1$ — the upper level. The formula is empirical in origin — Balmer and Rydberg fitted it to data decades before Bohr explained it — yet it holds to the limit of measurement. The reason equation (1) works is the level structure underneath it, built in The Bohr Model of Hydrogen and confirmed by the Schrödinger solution. Each level carries an energy that goes as $-1/n^2$ , and the ionization energy from the ground state fixes the scale at $13.6$ eV.

E_n = -\frac{hcR_{\mathrm H}}{n^2} = -\frac{13.61\ \text{eV}}{n^2} \qquad (2)

where: $E_n$ — energy of level $n$ , negative because the electron is bound; $hcR_{\mathrm H} = 13.61$ eV — the Rydberg energy, equal to the ionization energy from $n = 1$ ; $h$ — Planck's constant; $c$ — speed of light. Read the two equations together: the term $T_n = R_{\mathrm H}/n^2 = -E_n/hc$ is just the level energy expressed as a wavenumber, and every line is a difference of two terms. Two constants of the same name differ at the fourth digit, and the difference is physical, not a rounding choice. The nucleus is not infinitely heavy: electron and proton both orbit their shared centre of mass, so the mass that enters the atom's dynamics is the reduced mass $\mu = m_e M_p/(m_e + M_p)$ , slightly less than $m_e$ . Every level scales with $\mu$ , so the whole spectrum shifts.

R_{\mathrm H} = \frac{R_\infty}{1 + m_e/M_p} = R_\infty\,\frac{\mu}{m_e}, \qquad R_\infty = 1.097373\times10^{7}\ \text{m}^{-1} \qquad (3)

where: $R_\infty$ — the rydberg constant in the idealised limit of an infinitely heavy nucleus, a combination of fundamental constants; $m_e/M_p = 1/1836.15$ — the electron-to-proton mass ratio; $\mu/m_e = 0.99946$ . The correction is $0.054\%$ : small, but the tell that told Urey deuterium was hiding in the Balmer lines, and far larger than the precision of a good spectrograph.

Derivation AThe Rydberg formula from the term ladder

The goal is to show the rydberg formula is the arithmetic of a quantized ladder: given levels $E_n = -hcR_{\mathrm H}/n^2$ , every emitted wavenumber is a difference of two terms $T_n = R_{\mathrm H}/n^2$ . We recall the level formula by name from The Bohr Model of Hydrogen — not re-deriving it — and run the energy bookkeeping of one quantum jump, which is all the Rydberg formula ever was.

Step 1. Impose energy conservation on a downward jump. When the atom falls from $n_2$ to $n_1$ it emits one photon of energy $h\nu = E_{n_2} - E_{n_1}$ . Why: the atom can only shed energy in the fixed amounts set by its level spacings, so the photon's energy is pinned to a difference of two allowed levels — this is the Bohr frequency condition.

Step 2. Substitute the level formula. With $E_n = -hcR_{\mathrm H}/n^2$ , $h\nu = -hcR_{\mathrm H}/n_2^2 + hcR_{\mathrm H}/n_1^2$ , and because $n_2 > n_1$ the term in $n_1$ dominates and the result is positive. Why: a positive photon energy is exactly the condition that the jump is downward and radiates — an upward jump with the same integers absorbs the identical wavenumber.

Step 3. Divide by $hc$ to convert energy into wavenumber: $\tilde\nu = \nu/c = R_{\mathrm H}(1/n_1^2 - 1/n_2^2)$ . The Planck constant cancels, leaving pure spectroscopy. Why: wavenumber, not frequency, is the natural currency here because it makes the terms $T_n = R_{\mathrm H}/n^2$ additive — the deep pattern Ritz spotted.

Step 4. Read off the combination principle. Because each line is $T_{n_1} - T_{n_2}$ , sums of line wavenumbers reproduce other lines: $\tilde\nu(1\to2) + \tilde\nu(2\to3) = \tilde\nu(1\to3)$ . Why: this Rydberg–Ritz combination principle is direct evidence of a term table — you predict unmeasured lines by adding measured ones, with no atomic model at all. *The symbolic reduction $(E_{n_2}-E_{n_1})/hc \to R_{\mathrm H}(1/n_1^2 - 1/n_2^2)$ and the combination identity $\tilde\nu(1\to3) = \tilde\nu(1\to2) + \tilde\nu(2\to3)$ both check out exactly.*

Go deeper · Derivation B · 1 minThe four visible Balmer lines — and the air correction

Derivation BThe four visible Balmer lines — and the air correction

The goal is to reproduce the visible spectrum to better than $0.05$ nm and expose the first sub-percent subtlety: the wavelength a spectrometer reads in air is not the vacuum wavelength the Rydberg formula returns. We set $n_1 = 2$ , evaluate equation (1) for $n_2 = 3,4,5,6$ , invert to vacuum wavelengths, then divide by the refractive index of air to land on every textbook value.

Step 1. Fix the lower level at $n_1 = 2$ — this is the Balmer series — and write $\tilde\nu = R_{\mathrm H}(\tfrac14 - 1/n_2^2)$ . For Hα ( $n_2 = 3$ ) that is $\tfrac{5}{36}R_{\mathrm H}$ ; for Hβ ( $n_2 = 4$ ), $\tfrac{3}{16}R_{\mathrm H}$ ; for Hγ ( $n_2 = 5$ ), $\tfrac{21}{100}R_{\mathrm H}$ ; for Hδ ( $n_2 = 6$ ), $\tfrac{2}{9}R_{\mathrm H}$ . Why: the small-integer fractions are Balmer's original discovery — the visible lines are rational multiples of one constant.

Step 2. Invert to vacuum wavelengths, $\lambda_{\text{vac}} = 1/\tilde\nu$ . This gives $656.47,\,486.27,\,434.17,\,410.29$ nm. Why: the Rydberg formula is a statement about the wave in empty space, so its direct output is always the vacuum wavelength — the frequency is what is truly invariant, and $\lambda_{\text{vac}} = c/\nu$ .

Step 3. Divide by the refractive index of air, $n_{\text{air}} = 1.0002771$ , because a grating measures the wavelength in the medium it sits in: $\lambda_{\text{air}} = \lambda_{\text{vac}}/n_{\text{air}}$ . The four become $656.28,\,486.13,\,434.05,\,410.17$ nm. Why: light slows in air, so its wavelength shrinks by the index while its frequency is unchanged; historical tables — Ångström's, Balmer's — are all air values, and matching them means applying this factor.

Step 4. Weigh the correction. At Hα the shift is $656.47 - 656.28 = 0.18$ nm — three parts in ten thousand, but forty times a good spectrograph's resolution. Why: this is the trap. The air shift and the reduced-mass shift of Derivation C are both sub-percent and both real; quoting a wavelength without saying "air" or "vacuum" is as ambiguous as omitting the units. *All four Balmer air wavelengths are reproduced to $<0.05$ nm from $R_{\mathrm H}$ and $n_{\text{air}} = 1.0002771$ ; the Hα vacuum value is $656.47$ nm, giving the $0.18$ nm air–vacuum split.*

Go deeper · Derivation C · 1 minWhy R_H is not R∞ — the reduced-mass shift

Derivation CWhy R_H is not R∞ — the reduced-mass shift

The goal is the second sub-percent subtlety: the constant that fits real hydrogen, $R_{\mathrm H}$ , is not the fundamental-constant combination $R_\infty$ , and the gap is the finite proton mass. The shift is uniform — it rescales every line by the same factor — which is why it can masquerade as a slightly wrong constant, and why the difference between two isotopes cannot.

Step 1. Replace the electron mass by the reduced mass. In a genuine two-body atom the electron and proton orbit their common centre of mass; the relative motion behaves as a single particle of mass $\mu = m_e M_p/(m_e + M_p) = m_e/(1 + m_e/M_p)$ . Why: only the reduced mass appears in the relative coordinate, and it is the relative coordinate whose quantization sets the energy levels.

Step 2. Carry $\mu$ through the Rydberg constant, which is linear in the orbiting mass: $R_{\mathrm H} = R_\infty\,\mu/m_e = R_\infty/(1 + m_e/M_p)$ . Why: $R_\infty \propto m_e$ in the infinite-mass idealisation, so swapping $m_e \to \mu$ scales the constant by exactly $\mu/m_e$ — and with it every level and every line.

Step 3. Put in the number. With $m_e/M_p = 1/1836.15$ , the factor is $\mu/m_e = 0.99946$ , so $R_{\mathrm H}$ sits $0.054\%$ below $R_\infty = 1.097373\times10^{7}\ \text{m}^{-1}$ , giving $1.096776\times10^{7}\ \text{m}^{-1}$ . Why: a uniform $0.054\%$ redshift of the whole comb — invisible if you only ever look at hydrogen, since you would simply absorb it into your value of $R$ .

Step 4. Break the disguise with a second isotope. Deuterium's nucleus is twice as heavy, so its reduced mass is larger, its $R_{\mathrm D}$ larger, and its lines blue-shifted: $(R_{\mathrm D} - R_{\mathrm H})/R_{\mathrm H} \approx m_e/2M_p \approx 2.7\times10^{-4}$ , moving Hα by $0.18$ nm. Why: the shift no longer hides in the calibration because it differs between H and D — which is precisely how Urey spotted deuterium as a faint satellite of the Balmer lines in 1931. *The numbers check out: $\mu/m_e = 0.99946$ , the $0.054\%$ $R_{\mathrm H}$ – $R_\infty$ gap, and the H/D Balmer-α isotope shift of $0.179$ nm, placing deuterium Hα at $656.11$ nm in air.*

Limiting cases

Case	Result	Meaning
series limit, $n_2 \to \infty$	$\tilde\nu_\infty = R_{\mathrm H}/n_1^2$	short-wavelength edge of each series; Balmer's is $364.6$ nm (air), beyond it the continuum
ground-state ionization, $n_1=1$	$\tilde\nu_\infty = R_{\mathrm H}$ , $\lambda = 91.18$ nm	the Lyman limit — $13.6$ eV, the energy to strip hydrogen
adjacent high levels, $n_1 \gg 1$	$\tilde\nu \approx 2R_{\mathrm H}/n_1^3$	Rydberg transitions collapse to radio/microwave — the recombination lines radio astronomers map
infinite nuclear mass	$R_{\mathrm H} \to R_\infty$	idealised limit $M_p \to \infty$ ; real hydrogen sits $0.054\%$ below it
heavier isotope (D)	$R_{\mathrm D} > R_{\mathrm H}$ , lines blue-shifted	$2.7\times10^{-4}$ fractional shift; how deuterium was discovered

Common trap“The textbook wavelength and the measured wavelength should agree exactly”

They agree only once you name two conventions the tables leave silent, and the two are independent. First, air versus vacuum: the Rydberg formula returns a vacuum wavelength, but visible and near-UV lines are quoted in air, smaller by the refractive index — Hα is $656.28$ nm in air, $656.47$ in vacuum, a $0.18$ nm gap. Second, $R_{\mathrm H}$ versus $R_\infty$ : using the infinite-mass constant overstates every wavenumber by $0.054\%$ , shifting Hα about $0.36$ nm the other way. Both are sub-percent, both real, and students blame one for the other. The discipline: state air or vacuum, and $R_{\mathrm H}$ or $R_\infty$ — else a "disagreement" is just an undeclared convention. (IR and far-UV lines are conventionally vacuum, which is why the Balmer limit is $364.6$ nm air but the Lyman limit $91.18$ nm vacuum.)

Worked examples

1 · Hβ from scratch, vacuum to air (confidence builder)

Compute the second Balmer line, Hβ ( $n_2 = 4 \to n_1 = 2$ ), in vacuum and then in air, using $R_{\mathrm H} = 1.096776\times10^{7}\ \text{m}^{-1}$ .

Step 1. Wavenumber from equation (1): $\tilde\nu = R_{\mathrm H}(\tfrac14 - \tfrac1{16}) = \tfrac{3}{16}R_{\mathrm H}$ . Numerically $\tfrac{3}{16}\times1.096776\times10^{7} = 2.05645\times10^{6}\ \text{m}^{-1}$ .

Step 2. Invert for the vacuum wavelength: $\lambda_{\text{vac}} = 1/\tilde\nu = 1/(2.05645\times10^{6}) = 4.8627\times10^{-7}\ \text{m} = \mathbf{486.27\ nm}$ .

Step 3. Convert to air by dividing by $n_{\text{air}} = 1.0002771$ : $\lambda_{\text{air}} = 486.27/1.0002771 = \mathbf{486.13\ nm}$ — the value in every Balmer table, a blue-green line. The $0.14$ nm you just removed is the air correction.

Step 4. Trap check: had you used $R_\infty$ instead of $R_{\mathrm H}$ , the wavenumber would be $0.054\%$ larger and $\lambda_{\text{vac}}$ come out $486.01$ nm — a quarter-nanometre low, in the wrong direction to blame on air. Hβ is $486.27$ nm in vacuum, $486.13$ nm in air.

2 · Paschen-α, and why the infrared series came late (exam level)

Find the first Paschen line, Paschen-α ( $n_2 = 4 \to n_1 = 3$ ), and explain why the Paschen, Brackett and Pfund series were discovered decades after Balmer's.

Step 1. Wavenumber: $\tilde\nu = R_{\mathrm H}(\tfrac19 - \tfrac1{16}) = R_{\mathrm H}\cdot\tfrac{16-9}{144} = \tfrac{7}{144}R_{\mathrm H} = 5.332\times10^{5}\ \text{m}^{-1}$ .

Step 2. Wavelength: $\lambda_{\text{vac}} = 1/(5.332\times10^{5}) = 1.8756\times10^{-6}\ \text{m} = \mathbf{1875.6\ nm}$ — about $1.88\ \mu\text{m}$ , deep in the near-infrared. (Infrared lines are conventionally quoted in vacuum, so no air correction is applied here.)

Step 3. See why it is infrared. Paschen-α shares its upper level, $n_2 = 4$ , with Hβ, but drops only to $n_1 = 3$ instead of $n_1 = 2$ ; the smaller energy gap means a longer wavelength. Every series ending on a higher $n_1$ lives further into the infrared, because the levels crowd together up there.

Step 4. Blame the detectors. Balmer's lines came first because the eye and photographic plate are visible-band instruments; Paschen ( $1908$ ) needed early IR bolometers, and Brackett ( $1922$ ), Pfund ( $1924$ ) and Humphreys ( $1953$ ) waited for better IR detectors. The physics was in the Rydberg formula all along — only the technology lagged. Paschen-α is $1875.6$ nm; the IR series were detector-limited, not physics-limited.

Physics of the portrait

The Balmer series is the four-line visible signature that opened the whole subject, and its story belongs to two men who never met the atom they decoded. Johann Jakob Balmer was a sixty-year-old teacher at a girls' school in Basel, devoted to mathematical harmony, when a colleague challenged him to fit hydrogen's four measured wavelengths. In 1885 he found $\lambda = B\,n^2/(n^2 - 4)$ with $B = 364.56$ nm, matching Ångström's numbers to a hair — and that $B$ is, we now see, exactly the series limit $4/R_{\mathrm H}$ . Three years later Johannes Rydberg, in Lund, rewrote such fits in wavenumbers and found the single reciprocal-square law that swallowed Balmer's as a special case. Neither had a physical model; they had pattern, and pattern was enough to hand Bohr his target.

Origins · 2 minbalmer series and how this was discovered

balmer series

Hα Hβ Hγ Hδ · jumps to n=2 · the visible fingerprint of hydrogen · atomic spectroscopy

The Balmer series is the family of hydrogen transitions ending on $n_1 = 2$ : four visible lines — Hα red, Hβ blue-green, Hγ and Hδ violet — thinning toward the $364.6$ nm limit, then a continuum. It was the first atomic spectrum reduced to a formula, and for forty years a numerical riddle without a mechanism. Its importance dwarfs its four lines: falling in the visible, they are what a prism shows, what a stellar spectrum imprints (A-type stars wear the strongest Balmer absorption), and what H-α filters isolate to map ionized hydrogen across galaxies. Balmer's constant $B$ , Rydberg's generalisation, Bohr's $-13.6\ \text{eV}/n^2$ , and Schrödinger's wavefunctions are four successive answers to the question these lines posed — and where the claim that hydrogen is the best-understood atom begins.

History

1814

— Fraunhofer maps hundreds of dark lines across the solar spectrum and labels the strongest A–K; his C and F lines are later identified as hydrogen Hα and Hβ

1859

— Gustav Kirchhoff shows each dark solar line coincides with a bright emission line of some element; the 1860 memoir with Robert Bunsen founds spectrum analysis on that fingerprint principle

1862

— Ångström detects hydrogen in the Sun; by 1868 his solar atlas fixes the four visible H lines to fractions of an Ångström

1885

— Balmer fits those four wavelengths with λ = B·n²/(n²−4), B = 364.56 nm

1888

— Rydberg recasts all known series in wavenumbers, introducing the constant that bears his name

1896

— Pickering finds a stellar series first ascribed to hydrogen; Bohr later shows it belongs to ionized helium, He⁺

1906

— Lyman opens the ultraviolet series ending on n=1 (completed by 1914)

1908

— Paschen records the first infrared series (n=3); Ritz states the combination principle, that every line is a difference of two terms

1922

— Brackett measures the n=4 infrared series

1924

— Pfund adds the n=5 series

1931

— Urey finds deuterium as a faint isotope satellite of the Balmer lines, shifted by the reduced-mass effect

1953

— Humphreys reaches the n=6 series in the far infrared

Going further

Texts. Foot, Atomic Physics, ch. 1–2 (hydrogen spectrum and the Rydberg formula) · Hecht, Optics, 5th ed., ch. 3 (light and atomic emission) · Demtröder, Laser Spectroscopy, vol. 1 (high-resolution hydrogen spectra and the air/vacuum convention) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., ch. 13 (atoms and light) · Herzberg, Atomic Spectra and Atomic Structure (the classic series treatment). Balmer air wavelengths ( $656.28,\,486.13,\,434.05,\,410.17$ nm; limit $364.6$ nm) are recomputed from $\lambda_{\text{vac}} = [R_{\mathrm H}(\tfrac14 - 1/n_2^2)]^{-1}$ with $R_{\mathrm H} = 1.096776\times10^{7}\ \text{m}^{-1}$ and the air index $n_{\text{air}} = 1.0002771$ . The constants $R_\infty = 1.097373\times10^{7}\ \text{m}^{-1}$ , $R_{\mathrm H} = 1.096776\times10^{7}\ \text{m}^{-1}$ , $m_e/M_p = 1/1836.15$ , and the ionization energy $13.61$ eV are CODATA values. Lyman-α ( $121.57$ nm vac), Lyman limit ( $91.18$ nm vac), Paschen-α ( $1875.6$ nm vac) and the Paschen limit ( $820.6$ nm vac) are standard reference values; the H/D Balmer-α isotope shift of $0.18$ nm is Urey's deuterium signature. Balmer's fitted constant $B = 3645.6$ Å $= 364.56$ nm equals $4/R_{\mathrm H}$ (air), the Balmer series limit.

Prerequisites. The Bohr Model of Hydrogen · Solution of the Schrödinger Equation for Hydrogen. Leads to. Fine Structure and Spin-Orbit Coupling · The Lamb Shift · Hyperfine Structure · The Zeeman Effect.

Why does the Rydberg formula give a vacuum wavelength when my lab table lists air values?

Equation (1) fixes the photon's frequency, and the wavelength it returns is $c/\nu$ — the vacuum wavelength — because $c$ is the vacuum speed. A grating spectrometer, though, measures the wavelength in the medium it sits in, and in air that is shorter by the refractive index: $\lambda_{\text{air}} = \lambda_{\text{vac}}/n_{\text{air}}$ with $n_{\text{air}} \approx 1.000277$ in the visible. The frequency is the true invariant; both wavelengths are correct once you name the medium. The convention split by band: visible and near-UV lines are tabulated in air (as Ångström and Balmer worked), far-UV and IR lines in vacuum — which is why the Balmer limit reads $364.6$ nm but the Lyman limit $91.18$ nm.

Is the air correction the same as the reduced-mass correction? They're both about 0.05%.

No — conflating them is the most common slip here. They are numerically similar but physically unrelated. The air correction is optical: light slowing in a medium, applying only to air-quoted wavelengths, touching no frequency or energy. The reduced-mass correction is atomic: the proton's finite mass rescales the actual levels and hence the frequencies, and survives in perfect vacuum. The clean split: the reduced-mass shift changes what the atom emits; the air shift changes what your instrument reads for that same emission.

What actually happens at the series limit and beyond it?

As $n_2 \to \infty$ the upper levels pile up just below the ionization threshold, so the lines crowd toward a fixed short-wavelength limit $\tilde\nu_\infty = R_{\mathrm H}/n_1^2$ — for Balmer, $364.6$ nm. Past it the electron is unbound and can carry any positive energy, so the discrete lines give way to a continuum. In absorption this is the "Balmer jump," a sudden drop in stellar flux blueward of $364.6$ nm as photons start ionizing $n=2$ hydrogen wholesale. The series limit thus reads out the ionization energy from that level: $R_{\mathrm H}/n_1^2$ , or $13.6\ \text{eV}/n_1^2$ .

Fraunhofer labelled solar lines with letters in 1814. How do they map onto the Balmer series?

Fraunhofer catalogued sunlight's dark absorption lines and labelled the strongest A through K, with no idea what caused them. Two are hydrogen: the C line is Hα (656.28 nm) and the F line is Hβ (486.13 nm), both from hydrogen in the cooler solar photosphere absorbing the continuum welling up below. (Others are different elements — the D lines are sodium, H and K ionized calcium.) The point Kirchhoff and Bunsen fixed in 1859 is that emission and absorption are the same transitions: hydrogen absorbs in the Sun at exactly the wavelengths it emits in a tube, so Fraunhofer's C and F sit precisely on the laboratory Balmer lines.

Does the Rydberg formula predict the exact positions of real hydrogen lines?

Only to leading order — the gross structure. Each nominal line is split and shifted by smaller effects the formula ignores: fine structure and spin-orbit coupling (parts in $10^4$ ), the Lamb shift (a QED effect splitting levels the Schrödinger equation leaves degenerate), and hyperfine structure from the proton's magnetic moment; external fields add Zeeman and Stark splitting. All are far smaller than the line spacings — so the Rydberg pattern survives intact as the skeleton — but they are the residues that make hydrogen a precision test of physics. The formula gets you five figures; the interesting physics of the last century lives in the sixth and beyond.

The Pickering lines were called hydrogen in 1896 and helium later. What was going on?

Edward Pickering found a series in ζ Puppis whose lines interleaved neatly with the Balmer series, and it was natural to read them as hydrogen transitions with half-integer quantum numbers. Bohr's model resolved it: the lines are ionized helium, He⁺, a one-electron ion like hydrogen but with $Z=2$ . Its spectrum follows the same reciprocal-square law scaled by $Z^2 = 4$ (and a slightly different reduced mass), placing every other He⁺ line almost exactly on a hydrogen line and filling the gaps that looked like "half-integer" hydrogen. Getting the He⁺ reduced-mass offset right was an early triumph that convinced physicists Bohr's model was more than numerology.