Hydrogen Spectral Lines & the Rydberg Formula

Pick a series and a transition to see where hydrogen puts its lines, and how many of them your eye can actually reach.

Worked example — Hydrogen Spectral Lines & the Rydberg Formula

Wavelength (vacuum)	656.47 nm
Photon energy	1.889 eV
Wavelength in air	656.29 nm
Band	Visible
Series limit (vacuum)	364.71 nm
Limit energy	3.400 eV
Lines in this window (all series)	7

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Hydrogen Spectral Lines & the Rydberg Formula

Hydrogen's electron can sit only at certain energies, and a jump between two of them emits or absorbs one photon carrying exactly the difference: E = hc/λ. A series is every jump that ends on the same lower level — Lyman on n = 1, Balmer on n = 2, Paschen on n = 3 — which is why the lines come in families that crowd together towards a limit instead of spreading evenly.

What each symbol means

n₁: the lower level, the one that names the series
n₂: the upper level the electron falls from
R∞: the Rydberg constant, the one measured number in the whole formula

Assumes: A single electron and a nucleus at rest, which is why the same formula needs a correction for deuterium and a different constant for helium. The page prints a vacuum wavelength and an air wavelength separately because air slows light by about a part in 3,700: a line catalogued from the ground and one catalogued from a satellite are the same line with different numbers.

Two integers and one constant give you the whole spectrum 🖖

The four bright lines a school spectroscope pulls out of hydrogen are one electron falling to n₁ = 2 from n₂ = 3, 4, 5 and 6. In vacuum they sit at 656.47, 486.27, 434.17 and 410.29 nm: red, cyan, blue, violet. Rydberg's constant sets the scale and the two integers pick the line. Nothing else goes in. A ground-based spectrograph reads 656.29 nm for the first of them, because air's refractive index shortens the wavelength, and the Wavelength in air row prints that value too.

Press n₁ = 1 and touch nothing else. One integer different, and the line you were looking at moves from 656.47 nm to 102.57 nm: same atom, same formula, nothing left in the visible. The Band row reads Ultraviolet for every n₂ you can pick. Bring n₂ down to 2 and you have Lyman-alpha at 121.57 nm, the longest line this series has and still 258 nm short of violet. The Wavelength in air row stays empty throughout: molecular oxygen absorbs the whole Lyman series, so there is no air wavelength to quote and those lines are measured from above the atmosphere.

How many lines you can see is a fact about your eye 🖖

Five lines, at a window edge of 390 nm. Drag that edge to 380 and the count row says 7. Push it to 370, raise "show lines up to n₂" to 16, and you get 14. All Balmer: nothing else in hydrogen lands in that window, because Lyman finishes at 91.18 nm and no Paschen line is shorter than 820.59 nm.

The lines are stacking against the Balmer limit at 364.71 nm, and there is no last one. Between 365 and 750 nm hydrogen puts 68 lines, needing n₂ up to 70; the tool draws at most 40, so its count stops at 38. The rest are still there, packed into the final nanometre.

The series limit is an ionisation energy 🖖

With Balmer selected, the Limit energy row reads 3.400 eV. Press n₁ = 1 and it reads 13.598 eV, which is hydrogen's ionisation energy from the ground state: the work of taking the electron from n = 1 to unbound. The series limit is that same quantity, read as a photon.

The factor between them is four, and it is exact rather than measured. The limit energy is hcR_H/n₁², so moving from n₁ = 2 to n₁ = 1 multiplies it by 2². The row prints three decimals, so dividing what you see gives 3.9994; the exactness lives in the formula, not in the rounding.

Past the limit there are no lines. Watch the amber dashed edge on the strip while you raise n₂: bars crowd towards it from the long-wavelength side and never cross. A photon carrying more than 3.400 eV can still be absorbed by an atom sitting at n = 2, but it ionises instead of promoting, and the surplus becomes the free electron's kinetic energy, which takes any value at all. The picket fence turns into a smooth band there.

Problem solved in full

One line, and the photon that goes one step further 7 steps
A laboratory spectrum of hydrogen shows a line at 486.13 nm, measured in air. Identify the transition. Then decide whether a 3.00 eV photon could ionise a hydrogen atom already sitting in that same lower level, and say what a 4.00 eV photon would do instead.

Open this problem in the tool
1. The Rydberg formula has two integers in it and you already know one of them. A visible hydrogen line means the electron landed on n₁ = 2, so the only unknown is where it fell from.
  
  $\frac{1}{λ} = R_{H} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}), n_{1} = 2$
2. Before inverting anything, deal with the medium. That 486.13 nm was measured through air, and air slows light, so the line arrives shorter than it left. The formula is written for vacuum. Multiply by the refractive index, 1.0002763, and you get 486.266 nm. Skip this and the arithmetic below returns 4.0018, which is not an integer and not a transition.
  
  $λ = n_{air} λ^{'} = 1.0002763 \times 486.13 = 486.266 nm$
3. Now rearrange. 1/λ = R_H(1/4 − 1/n₂²) becomes 1/n₂² = 0.25 − 1/(λR_H), and the vacuum wavelength hands back n₂ = 4.0000989.
  
  $\frac{1}{n _{2}^{2}} = \frac{1}{4} - \frac{1}{λ R _{H}} = 0.25 - \frac{1}{486.266 \times 1 0 ^{- 9} \times 1.09677576 \times 1 0 ^{7}}$
4. It is 4, and the tail is worth a moment because it is not what it looks like. It is not the Rydberg model running out: put the tool’s own unrounded air value, 486.138 nm, through the same three lines and n₂ comes back 4.0000000. The 0.0000989 is the two-decimal rounding of the 486.13 you were handed. The model does have a limit of its own, near one part in 10⁵, from the fine structure and the Lamb shift it knows nothing about. You cannot see it at this precision.
  
  $n_{2} = 4.0000989 \Rightarrow n_{2} = 4 (λ^{'} = 486.138 \Rightarrow n_{2} = 4.0000000)$
5. Now the energy. E = hc/λ on the vacuum wavelength gives 2.550 eV, and the Photon energy row agrees. That is one photon of this line: what the atom emits dropping from n = 4 to n = 2, or absorbs going the other way. It is not what it costs to remove the electron.
  
  $E = \frac{h c}{λ} = 2.550 eV$
6. Removing it is the series limit. Let n₂ run to infinity, 1/n₂² vanishes, and what is left is hcR_H/n₁², which for n₁ = 2 is 3.400 eV. The Limit energy row prints it. Every Balmer line carries less than that, and they crowd up towards it without ever reaching it.
  
  $E_{\infty} (n_{1} = 2) = \frac{h c R _{H}}{2 ^{2}} = 3.400 eV$
7. So the 3.00 eV photon fails, and it fails by 0.400 eV. There is no state at that energy for the electron to land in, and not enough energy to free it either. The atom stays where it is and the photon goes past.
  
  $3.00 eV < 3.400 eV$
Answer

The line is n = 4 → 2, the second line of the Balmer series, H-beta at 486.27 nm in vacuum and 486.14 nm in air, carrying 2.550 eV. A 3.00 eV photon cannot ionise from n = 2, because that costs 3.400 eV. A 4.00 eV photon can, and the 0.600 eV left over becomes the freed electron’s kinetic energy — which is where the line spectrum ends and a continuum begins, since that surplus can take any value at all.

$E_{kin} = 4.00 - 3.400 = 0.600 eV$

By the 42Math team · Reviewed against published sources. Corrections welcome.