Rydberg Formula and Spectral Series Calculator

Q: Does the formula work for helium?

For He+, the singly ionised atom with one electron left, yes — with every wavenumber multiplied by Z² = 4. For neutral helium, no, and not approximately either. With two electrons each one screens the nucleus from the other, the energy no longer depends on n alone, and the spectrum has a structure this formula cannot produce. The same goes for every element past hydrogen in its neutral state. Rydberg’s own original formula did include an empirical correction term for heavier elements, which is a different and more complicated object from the clean two-term formula on this page.

Give any two levels and get the line: 1/λ = RZ²(1/n₁² − 1/n₂²), its wavenumber, its photon energy, which series it belongs to and whether your eye could see it — with a switch between the Rₕ that matches real hydrogen, the CODATA R∞ and the rounded 1.097×10⁷ your textbook prints.

Last updated 11 Nov 2026

The line emitted when an electron drops from n₂ to n₁

Two levels and Z → λ, wavenumber, photon energy, series and visibility

Lower level n₁The level the electron falls TO, and the one that names the series: 1 is Lyman, 2 Balmer, 3 Paschen, 4 Brackett, 5 Pfund, 6 Humphreys. Every Balmer line without exception is a drop to n = 2, which is why they all lie in one small band of the spectrum. Some books write this level as n and the upper one as m; others reverse the two, so read which is which before substituting.

Upper level n₂The level it falls FROM, and it must be above n₁ or there is no emitted line to compute. For the series limit, put a very large number here: 100,000 is indistinguishable from infinity to twelve figures, and the series-limit row below gives the exact value directly.

Nuclear charge ZZ = 1 is hydrogen. Z = 2 is He⁺, Z = 3 is Li²⁺ — any ion stripped down to one electron, where the same formula works with every wavenumber multiplied by Z². It does NOT work for neutral helium or anything else with two or more electrons, because those electrons screen each other and the simple 1/n² ladder breaks. Read the note under the table before using Z above 1.

Which Rydberg constantR∞ is the measured constant and assumes the nucleus is infinitely heavy. A real proton is not, so the electron and proton both orbit their common centre of mass and the effective constant for hydrogen is R∞/(1 + m_e/m_p) = 10,967,758.34 m⁻¹, which is the default here because it is the one that reproduces measured hydrogen wavelengths. The difference is 0.054 per cent — small, but larger than the precision of every number on this page.

Left: hydrogen’s levels at their true spacing — they crowd towards the ionisation limit because E goes as −1/n² — with your own transition drawn as a red arrow between the right two. Below: the visible spectrum from 380 to 750 nm, with your line marked at its true wavelength and coloured only if it actually falls inside that window. Push it into the ultraviolet or the infrared and the marker pins to the end of the bar and says so.

656.4696nmExample

The red hydrogen line: an electron dropping from n₂ = 3 to n₁ = 2 in hydrogen, with Rₕ

One formula, and the three decisions you have to make before you use it

1/λ = R Z² (1/n₁² − 1/n₂²), n₂ > n₁ · R_H = R_∞/(1 + m_e/m_p) · series limit: 1/λ = R Z²/n₁² · E = hc/λ = 1239.842/λ_nm eV

1/λ: the wavenumber, and what the formula is really about. It is additive in a way wavelengths are not: the 3 → 1 wavenumber is exactly the sum of the 3 → 2 and 2 → 1 ones. That is the Ritz combination principle, known before any theory explained it, and it is why spectroscopists tabulate cm⁻¹
n₁: the LOWER level, which fixes the series and nothing else. Lyman 1, Balmer 2, Paschen 3, Brackett 4, Pfund 5. The series limit — the short-wavelength end — depends only on n₁, and so does the ionisation energy from that level
Z²: for a one-electron ion. He⁺ has every wavenumber four times hydrogen’s, Li²⁺ nine times. It is not valid for a many-electron atom: neutral helium does not follow this at all, because its two electrons screen the nucleus from each other
R∞: 10,973,731.568 157(12) m⁻¹ from CODATA 2022 — one of the most precisely measured constants in physics, known to one part in 10¹². It is defined for an infinitely heavy nucleus, which no real atom has
Rₕ: R∞ divided by (1 + m_e/m_p) = 1.000 544 617, giving 10,967,758.34 m⁻¹. The correction is the reduced mass: the proton recoils, so the electron is effectively a little lighter. It shifts every hydrogen wavelength by 0.054 per cent, 0.36 nm on the red line
vacuum: the formula gives the wavelength in vacuum. Light slows in air, so the air wavelength is shorter by the refractive index, roughly a factor of 1.0003 in the visible. That is why the red line is quoted as 656.47 nm in vacuum and 656.28 nm in air, and the two are the same line

Worked example

The red hydrogen line: an electron dropping from n₂ = 3 to n₁ = 2 in hydrogen, with Rₕ
WHICH SERIES. The lower level is n₁ = 2, so this is a Balmer line — the first one, usually written Hα. The series is named by the level the electron lands on and by nothing else.
THE BRACKET. 1/n₁² − 1/n₂² = 1/4 − 1/9 = 9/36 − 4/36 = 5/36 = 0.138 888 89. Keep it as a fraction for as long as you can; rounding it to 0.139 costs you the fourth figure of the answer.
THE WAVENUMBER. 1/λ = R_H × 5/36 = 10,967,758.34 × 0.138 888 89 = 1,523,299.77 m⁻¹, which spectroscopists write as 15,232.998 cm⁻¹. Z = 1 for hydrogen so the Z² factor does nothing here.
THE WAVELENGTH. λ = 1/1,523,299.77 = 6.5647×10⁻⁷ m = 656.4696 nm. That is in the red, at 74.7 per cent of the way across the 380–750 nm visible window, and it is the colour of a hydrogen discharge tube and of every H-alpha filter ever sold to an astronomer.
THE PHOTON. E = 1239.842/656.4696 = 1.888 651 eV, a frequency of 456.67 THz. Check it against the levels: −13.6/9 − (−13.6/4) = 1.889 eV. The two routes agree, as they must, because the Rydberg formula is the energy-level difference divided by hc.
WHICH CONSTANT, AND WHAT IT COSTS. With R∞ instead of R_H the answer is 656.1123 nm, and with the rounded 1.097×10⁷ it is 656.3355 nm. Those differ in the fourth figure. The measured vacuum wavelength of Hα is close to the R_H value, because real hydrogen has a proton of finite mass — so if a question gives you R∞ and expects four figures, it is quietly asking for a slightly wrong answer.
AND THE AIR TRAP. Many tables print 656.28 nm for this line. That is the same line measured in air, where light travels slower and the wavelength is shorter by the refractive index. The formula gives vacuum wavelengths, this page prints vacuum wavelengths, and a 0.03 per cent discrepancy against a printed table is usually this and not an arithmetic error.

The six series of hydrogen: which one you are in, and where it ends

Computed on this page’s own expressions with Rₕ and Z = 1. The answer to “which series is visible” is Balmer and only Balmer, and even then only part of it: seven of its lines fall inside the 380 nm cut this page uses, and the eye gives up after about the fourth. Notice also that the series overlap in wavelength — the Balmer limit at 364.71 nm is shorter than the first Paschen line at 1875.63 nm but longer than the Lyman limit at 91.18 nm — so you cannot name a series from a wavelength alone. You need the lower level.
n₁	Series	First line (n₁+1 → n₁)	Series limit	Spectral region	Visible to the eye?
1	Lyman	121.5684 nm	91.1763 nm	far ultraviolet	no — none of it
2	Balmer	656.4696 nm	364.7053 nm	visible, with an ultraviolet tail	yes — four to the eye, seven inside this page’s 380 nm cut
3	Paschen	1875.6274 nm	820.5870 nm	near infrared	no
4	Brackett	4052.2815 nm	1458.8213 nm	infrared	no
5	Pfund	7459.8819 nm	2279.4084 nm	infrared	no
6	Humphreys	12371.9272 nm	3282.3480 nm	far infrared	no

The Balmer series line by line, and where each one sits in the visible window

The lines crowd together towards the limit: the first four span 246 nm and the next infinitely many fit into the 45 nm below them, with n₂ = 10 already outside the 380 nm cut. That crowding is the 1/n² in the formula made visible, and it is the opposite of the particle in a box where the levels spread out as you climb. Where the eye stops is genuinely not a sharp number — the CIE definition cited below says so explicitly, giving the violet limit as somewhere between 360 and 400 nm — so the 410.29 nm line is called faintly visible here and the 397.12 nm one is not, and a different reasonable convention would move that boundary. This page implements 380 to 750 nm throughout and says so.
Transition	λ (vacuum)	Wavenumber	Photon energy	Colour	Visible?
3 → 2 (Hα)	656.4696 nm	15,232.998 cm⁻¹	1.888651 eV	red	yes
4 → 2 (Hβ)	486.2738 nm	20,564.547 cm⁻¹	2.549679 eV	blue-green	yes
5 → 2 (Hγ)	434.1730 nm	23,032.293 cm⁻¹	2.855640 eV	violet	yes
6 → 2 (Hδ)	410.2935 nm	24,372.796 cm⁻¹	3.021842 eV	deep violet	yes, faintly
7 → 2	397.1236 nm	25,181.078 cm⁻¹	3.122056 eV	extreme violet	inside the 380 nm cut, but not to the eye
8 → 2	389.0190 nm	25,705.684 cm⁻¹	3.187099 eV	extreme violet	inside the 380 nm cut, but not to the eye
9 → 2	383.6511 nm	26,065.352 cm⁻¹	3.231692 eV	extreme violet	the last one inside the 380 nm cut
10 → 2	379.9014 nm	26,322.620 cm⁻¹	3.263589 eV	ultraviolet	no — just outside the cut
∞ → 2 (limit)	364.7053 nm	27,419.396 cm⁻¹	3.399572 eV	ultraviolet	no

An empirical formula that was right for 28 years before anyone knew why

The formula came before the physics. In 1885 Balmer noticed that the four visible hydrogen lines fitted a simple arithmetic pattern involving small whole numbers. Rydberg generalised it to other series and other elements a few years later, in the wavenumber form this page uses. Nobody could say why integers should appear in the colours of a gas. Bohr supplied the reason in 1913 by quantising the atom’s angular momentum, and the Rydberg constant then dropped out of h, m_e, e and c. That is the order of events, and it matters for how you should think about the formula: it is a FIT to measurements that a later theory explained, not a consequence of a theory that was then confirmed.

Why spectroscopists work in wavenumbers. Because 1/λ adds and λ does not. The wavenumber of the 3 → 1 line is exactly that of 3 → 2 plus that of 2 → 1, since both are differences of the same 1/n² terms: on this page, 97,491.19 = 15,232.998 + 82,258.19 cm⁻¹. Wavelengths obey no such rule. Ritz spotted this before Bohr, and it is the plainest sign that each line is a difference of two fixed energy levels — which is what the formula says once you multiply it by hc. One reciprocal centimetre is 29.979 GHz or 1.2398×10⁻⁴ eV; watch the factor of 100 between m⁻¹ and cm⁻¹.

Where this goes wrong. Four places, and the first two are the same mistake twice. Swapping n₁ and n₂ gives a negative wavenumber and a negative wavelength; the lower level always goes first. Forgetting to invert at the end leaves you with a wavenumber labelled as a wavelength, and since the numbers are of order 10⁶ rather than 10⁻⁷ that error is at least obvious. The third is mixing units: R in m⁻¹ gives λ in metres, R in cm⁻¹ gives λ in centimetres, and the two differ by 100. The fourth is applying it to a many-electron atom; sodium’s yellow doublet does not come from this formula and cannot be made to.

Hydrogen-like ions, and the trap in them. Multiplying by Z² works for any ion with exactly one electron left. It is worth doing for a famous near-coincidence: the He⁺ line from n = 4 to n = 2 has exactly four times the bracket of hydrogen’s 2 → 1 line, so with the same R the two would be identical. They are not, quite — 121.5023 nm against 121.5684 nm — and the 0.07 nm between them is the reduced mass, because a helium nucleus is four times heavier than a proton. Those two lines were confused in stellar spectra for years, and their small difference is proof that the nucleus moves. Put Z = 2, n₁ = 2, n₂ = 4 with R∞ to see it.

How this differs from the Bohr-model page here. The Bohr model energy levels page starts from the energy levels and works down to the line, and is where the 13.6 eV ionisation energy and the orbital radii live. This page starts from the empirical formula, works in wavenumbers, adds Z² for hydrogen-like ions and judges where the line falls in the spectrum. Both agree on every wavelength: it is the same arithmetic written from the two ends.

Frequently asked questions

Which hydrogen lines can I actually see?

Four, and they are all Balmer lines — drops to n = 2. Hα at 656.47 nm is red and much the brightest; Hβ at 486.27 nm is blue-green; Hγ at 434.17 nm is violet; Hδ at 410.29 nm is a deep violet most people struggle with. Everything from n₂ = 7 upwards sits where the eye’s response has all but gone: this page’s 380 nm cut still calls n₂ = 7, 8 and 9 visible, at 397.12, 389.02 and 383.65 nm, and they are real lines you will not see. The series then piles up on its limit at 364.71 nm. No Lyman line is visible — the whole series is ultraviolet — and no Paschen line is either, because they are all infrared. So the answer to “which series is visible” is: part of exactly one of them.

Why does my textbook give 656.28 nm when this page gives 656.47 nm?

Two separate reasons, and it is worth knowing which is which. The big one is air: 656.28 nm is the wavelength measured in air, where light travels slower, and 656.47 nm is the vacuum wavelength the formula gives. The ratio is the refractive index of air, roughly 1.0003 in the visible, so the air value is about 0.03 per cent shorter. The exact dispersion of air was not established for this build, so this page does not convert. The smaller reason is which Rydberg constant you used: R∞ instead of R_H moves the answer by 0.054 per cent, to 656.1123 nm. Both effects are in the fourth figure and both are real.

What does n₂ = ∞ mean?

The electron arrives from outside the atom with no kinetic energy of its own and is captured straight into level n₁, or in reverse, a photon just barely strips it out of that level. The 1/n₂² term goes to zero and the wavenumber becomes RZ²/n₁², which is the short-wavelength end of the series, called the series limit. For n₁ = 1 that is 91.18 nm and for n₁ = 2 it is 364.71 nm, which is why the Balmer series fades out just short of the violet rather than continuing. The series-limit row on this page gives it exactly, so you do not have to type a large number for n₂ — though 100,000 is close enough for twelve figures if you want to watch it converge.

Does the formula work for helium?

For He⁺, the singly ionised atom with one electron left, yes — with every wavenumber multiplied by Z² = 4. For neutral helium, no, and not approximately either. With two electrons each one screens the nucleus from the other, the energy no longer depends on n alone, and the spectrum has a structure this formula cannot produce. The same goes for every element past hydrogen in its neutral state. Rydberg’s own original formula did include an empirical correction term for heavier elements, which is a different and more complicated object from the clean two-term formula on this page.

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References

CODATA Internationally Recommended 2022 Values of the Fundamental Physical Constants, NIST Standard Reference Database 121. Read on 11 November 2026 from the machine-readable table at physics.nist.gov/cuu/Constants/Table/allascii.txt. Every constant on this page comes from that file. It marks h = 6.626 070 15×10⁻³⁴ J Hz⁻¹, c = 299 792 458 m/s and e = 1.602 176 634×10⁻¹⁹ C as (exact), because the 2019 SI defines the kilogram, the metre and the ampere through them; the masses are measured, the electron being 9.109 383 7139(28)×10⁻³¹ kg.
CIE e-ILV term 17-21-003, visible radiation, International Commission on Illumination; read 11 November 2026 at cie.co.at/eilvterm/17-21-003. Quoted verbatim: Note 1 states that “there are no precise limits for the spectral range of visible radiation” because they “depend upon the amount of radiant flux reaching the retina and the responsivity of the observer”, with “the lower limit … generally taken between 360 nm and 400 nm and the upper limit between 760 nm and 830 nm”. This page implements 380–750 nm, inside both ranges. The band labels are therefore a convention, not a measurement, and a table that calls the 410.29 nm line invisible is not wrong.
The International System of Units (SI), 9th edition, Bureau International des Poids et Mesures, 2019 (the copy read on 11 November 2026 is marked V4.01, June 2026). §2.2 fixes the numerical values of h, c and e, which is why those three carry no uncertainty here while every mass does. A school question that says “take h = 6.63×10⁻³⁴ J s” is now rounding an exact number rather than quoting a measured one.
R_H on this page is not quoted from a source: it is derived here as R_∞/(1 + m_e/m_p) from the two CODATA 2022 entries above, R_∞ = 10 973 731.568 157(12) m⁻¹ and the electron-proton mass ratio 5.446 170 214 889(94)×10⁻⁴, giving 10 967 758.34 m⁻¹. CODATA does not publish a constant called “the Rydberg constant for hydrogen”, so the derivation is shown rather than a citation given.
J. J. Balmer (1885) and J. R. Rydberg (1888–1890) are credited above for the empirical formula and its generalisation. Neither original paper was read for this build and no number here comes from either; the dates are the conventional ones. Treat the history as orientation and the arithmetic as sourced to CODATA.
NCERT, Physics, Textbook for Class XII, Part II — the syllabus most Indian readers of this page are working to. This could not be read for this build: the chapter PDFs at ncert.nic.in covering these topics refused automated retrieval, and every other copy a search returns is an unattributed mirror. Nothing here is attributed to NCERT, and where this page and a school table differ the difference is stated against sources that could be read — a weaker claim than “the textbook is wrong”, and deliberately the weaker one.