
Bounded modern-physics survey · Interactive lesson
Atomic spectra and quantised energy levels
Hydrogen's lines were measured to six figures, and fitted by one formula, decades before anyone could say why they exist. This topic reads them the way the spectroscopists did: first as energy differences, then as a ladder of levels anchored by the series limits, and it marks exactly what the formula cannot tell you.
UPIII-10.01 · Lesson 1 of 9 in this unit. Position is not a completion score.Bounded modern-physics survey. This modern-physics material is an introductory survey, not a complete course in relativity, quantum, atomic, nuclear or particle physics. Your university determines the required depth and assessment.
Build the model
A spectrum is a table of energy differences; its series limits turn it into a ladder.
The key idea
An atomic line spectrum is the set of sharp wavelengths an atom emits or absorbs; each line's photon energy hf equals the difference between two of the atom's allowed energy levels, so the spectrum is a map of level differences.
Example. Hydrogen's red Balmer line at 656.47 nm carries hf = 1239.84 eV nm ÷ 656.47 nm = 1.889 eV, exactly the gap between the levels at −1.511 eV (n = 3) and −3.400 eV (n = 2).
The complete physical model and assumptions
A hot, dilute gas does not glow in a continuous rainbow. It emits a handful of sharp wavelengths, and the same gas held cool in front of a continuous source removes some of them. For hydrogen the lines obey one empirical rule. Written in wavenumber σ = 1/λ rather than wavelength, every line is a difference of two terms, σ = RH/n₁² − RH/n₂², with integers n₂ > n₁ and a single fitted constant RH = 1.09678 × 10⁷ m⁻¹. Multiply by hc and the terms become energies: a photon carries away exactly the gap between two levels, hf = E₂ − E₁, and the wavenumbers of two lines that share a level add or subtract to give a third, which is Ritz's combination principle and the real evidence that the levels exist. Lines alone fix only differences, since shifting every level by the same amount moves no wavelength. But each series converges to a limit beyond which absorption turns continuous, and that limit is the energy needed to free the electron from the series' lower level. It pins the ladder: Eₙ = −13.598 eV/n² below the ionisation threshold. There the model stops. The formula is a fit, not a mechanism: it cannot say why the terms go as 1/n², why RH has its value, or how bright any line is, which needs the populations of the upper levels and the rates of the transitions. It fails at a few parts in 10⁵, where fine structure splits the lines, and outright for atoms with more than one electron, whose terms need an adjustable quantum defect.
1Read every line as an energy difference
A spectrometer measures wavelength; the physics lives in energy. A photon of wavelength λ carries E = hf = hc/λ, and with E in eV and λ in nm, hc = 1239.84 eV nm, so hydrogen's red line at 656.47 nm carries 1.8886 eV. The atom that emitted it did not possess 1.8886 eV in any useful sense: it fell by 1.8886 eV from one level to another. A hot, dilute gas gives emission lines because collisions keep lifting atoms into upper levels; a cooler gas in front of a continuous source gives dark absorption lines where photons match an upward jump from a level that is actually occupied. Quote vacuum wavelengths. Air has a refractive index near 1.00028 in the visible, so a line reads about 3 parts in 10⁴ shorter in air: Hα, 656.47 nm in vacuum, is listed near 656.28 nm in air-wavelength tables. That correction is about ten times larger than the fine-structure splitting this unit meets later.
2From Balmer's pattern to Rydberg's formula
In 1885 Balmer found that the four visible hydrogen lines fit λ = B n²/(n² − 4) with n = 3, 4, 5, 6 and B = 364.56 nm. The rule is opaque in wavelength and transparent in wavenumber σ = 1/λ, where it becomes σ = (4/B)(1/2² − 1/n²), a difference of two terms of the same shape. Rydberg generalised it to σ = RH(1/n₁² − 1/n₂²), and σ is the natural variable because it is proportional to energy, E = hcσ. The generalisation made predictions. With n₁ = 1 it puts a series in the ultraviolet from 121.57 nm down to 91.18 nm, which Lyman found from 1906; with n₁ = 3 it puts one in the infrared from 1875.6 nm down to 820.6 nm, which Paschen found in 1908. Each series is the family of lines sharing one lower level. In hydrogen the families occupy separate bands until n₁ = 4, whose Brackett lines, 4052 nm to 1459 nm, overlap the long-wavelength end of the Paschen series.
3The combination principle is the evidence for levels
Ritz saw in 1908 that the content is not the 1/n² but the structure underneath it: every line of an atom has a wavenumber equal to the difference of two entries in one list of terms, T(n). Lines that share a term therefore combine into a third. The cascade 3 → 2 → 1 emits Hα at 15 233 cm⁻¹ and then Lyman α at 82 258 cm⁻¹, and their sum, 97 491 cm⁻¹, is Lyman β at 102.57 nm, observed on its own. Two lines that share an upper level subtract instead: σ(3 → 1) − σ(3 → 2) = σ(2 → 1). Multiply every term by hc and the principle is energy conservation for an atom with a fixed set of levels. It also marks what lines cannot fix. Add the same constant to every term and every difference, and so every line, is unchanged: a spectrum measures the spacing of the ladder, never the height of its floor.
4Series limits pin the ladder to the ionisation threshold
As n₂ → ∞ the lines of a series crowd toward σₗᵢₘ = RH/n₁², and adjacent lines close up as 2RH/(n₂ + ½)³, the derivative of the term. Past the limit absorption does not stop; it turns continuous, because a photon above the limit frees the electron and the surplus becomes kinetic energy, which is not quantised. So the limit photon is the one that just ionises the atom from the lower level, and its energy is that level's binding energy: the Balmer limit at 364.71 nm gives 3.400 eV for n = 2, and the Lyman limit at 91.18 nm gives 13.598 eV for n = 1. Put the zero at the threshold and every level sits at Eₙ = −13.598 eV/n². Electron-impact experiments need the same 13.6 eV to ionise hydrogen, an independent check that the ladder is real. The crowding is the correspondence check: for large n the gap to the next level is about 2/n of the binding energy, so near the threshold the energy becomes effectively continuous, as classical physics expects.
5Absorption needs atoms in the lower level
An absorption line needs atoms waiting in its lower level. In thermal equilibrium the populations follow a Boltzmann factor, Nₙ/N₁ = (gₙ/g₁)e(−ΔE/kT), where ΔE = Eₙ − E₁, k is Boltzmann's constant and gₙ = 2n² counts the states in level n. For n = 2 the gap is 10.199 eV while kT = 0.0259 eV at 300 K, so N₂/N₁ ≈ 2 × 10⁻¹⁷¹: a laboratory cell of atomic hydrogen absorbs only from n = 1, in the Lyman series and the continuum beyond its limit, all of it in the vacuum ultraviolet, and is transparent to every Balmer line. At 9500 K, kT = 0.819 eV and N₂/N₁ = 4e(−12.46) ≈ 1.6 × 10⁻⁵. That fraction is small, but a stellar atmosphere holds so many atoms along the line of sight that Balmer absorption lines are the strongest features in the spectra of stars near that temperature. Hotter still, most of the hydrogen is ionised and the lines weaken again, a competition the Boltzmann factor alone does not describe.
6What the formula cannot tell you
Rydberg's formula places every hydrogen line and says nothing else. It does not explain why the terms fall as 1/n² or why RH = 1.09678 × 10⁷ m⁻¹; the next topic derives both, and any such derivation must also explain why deuterium's constant is larger by 2.7 parts in 10⁴, the 0.18 nm shift of Hα by which heavy hydrogen was discovered in 1932. It is silent on brightness. An emission line radiates power N Γ hf, where N counts the atoms in the upper level and Γ is the rate at which that level decays by that line, and Γ ranges from 6.3 × 10⁸ s⁻¹ for the fast states of n = 2 to almost zero for the metastable 2s state, which survives about 0.12 s. Those rates need wavefunctions. And the fit fails in two places: at a few parts in 10⁵, where fine structure splits each line, and at once for atoms with more than one electron, whose terms fit only as R/(n − δ)² with a quantum defect δ, about 1.35 for sodium's s levels.
Key equations
Read the units, assumptions and limits before substituting.
Photon energy equals the level gap
hf = hc/λ = E₂ − E₁
Turns every measured wavelength into an energy difference: 656.47 nm means a drop of 1.8886 eV between two levels, not the energy of either one.
h = 6.626 × 10⁻³⁴ J s, f in Hz, λ in m, E₂ > E₁ the two level energies in J; hc = 1239.84 eV nmRydberg formula for hydrogen
1/λ = RH (1/n₁² − 1/n₂²), n₂ > n₁
n₁ = 1, 2, 3 give the Lyman (ultraviolet), Balmer (visible and near ultraviolet) and Paschen (infrared) series; one constant places them all to a few parts in 10⁵.
λ the vacuum wavelength in m; n₁, n₂ positive integers; RH = 1.09678 × 10⁷ m⁻¹, fitted to the measured linesTerms and the combination principle
σ(n₂ → n₁) = T(n₁) − T(n₂), with T(n) = RH/n²
Wavenumbers of a cascade add, wavelengths never do: σ(3 → 1) = σ(3 → 2) + σ(2 → 1), 97 491 = 15 233 + 82 258 cm⁻¹.
σ = 1/λ the wavenumber and T the term of a level, both in m⁻¹ (spectroscopists quote cm⁻¹)Series limit gives the level energy
Eₙ = −hcσₗᵢₘ(n) = −hcRH/n² = −13.598 eV/n²
The limit photon just frees the electron from level n, so it fixes absolute energies: the Lyman limit at 91.18 nm gives 13.598 eV for n = 1.
σₗᵢₘ(n) the limit of the series whose lower level is n, in m⁻¹; Eₙ measured from the ionisation threshold, in eVAdjacent lines near the limit
σ(n + 1 → n) = RH (2n + 1)/[n²(n + 1)²] ≈ 2RH/(n + ½)³
Lines crowd as n⁻³ into the continuum at the limit; at n = 124 the adjacent-level line has fallen to 3.408 GHz, a radio wave.
σ in m⁻¹, n the lower level; the approximation is low by a fraction 1/[2(n + ½)²], 3 parts in 10⁵ at n = 124Population of an absorbing level
Nₙ/N₁ = (gₙ/g₁) e(−ΔE/kT), ΔE = Eₙ − E₁
Absorption needs atoms in the lower level: at 300 K, N₂/N₁ ≈ 2 × 10⁻¹⁷¹, so cold hydrogen absorbs only from n = 1, in the Lyman series and the continuum beyond its limit.
N the atoms in a level, gₙ = 2n² its number of states, ΔE in eV, k = 8.617 × 10⁻⁵ eV K⁻¹ (Boltzmann), T in KInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Keep n₁ = 2 and raise n₂: the marked line climbs through the Balmer comb toward 3.400 eV but never passes the dashed limit, which sits as far from hf = 0 as level 2 sits below E = 0. Then set n₁ = 1: each Lyman line lands directly under the level it starts from.
Live results
- PHOTON ENERGY hf
- 1.889 eV
- WAVELENGTH λ (VACUUM)
- 656.5 nm
- BINDING ENERGY OF n₁
- 3.400 eV
- SERIES LIMIT λ
- 364.7 nm
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Lower level n₁ (series): 2; Upper level above it, n₂ − n₁: 1
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
“A gas absorbs at the same wavelengths it emits, so a cold cell of hydrogen in front of a lamp should show dark Balmer lines.” The premise is right and the conclusion is not. A transition between two levels has one wavelength whichever way it runs, but absorption starts from the lower level, so the line appears only if atoms occupy that level. A discharge keeps atoms in every level, so its emission shows every series; a gas at 300 K has essentially all its atoms in n = 1, with N₂/N₁ ≈ 2 × 10⁻¹⁷¹, so it absorbs only the Lyman series and the continuum beyond it, all below 121.6 nm, and is transparent to Hα. Dark Balmer lines are therefore a thermometer as well as a fingerprint: they say the gas is hot enough, near 10⁴ K, to hold a measurable fraction of its atoms in n = 2.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A hydrogen discharge spectrum shows the Balmer lines converging on a limit at 364.71 nm, and its red line Hα, the 3 → 2 transition, at 656.47 nm, both vacuum wavelengths. Using only these two measurements and hc = 1239.84 eV nm, find the energies of levels 2 and 3 measured from the ionisation threshold.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- The series limit is the photon that just frees the electron from the series' lower level, n = 2. Its energy is hc/λ = 1239.84 eV nm ÷ 364.71 nm = 3.3995 eV, so level 2 is bound by 3.3995 eV.
- With the zero of energy at the ionisation threshold, a bound level has negative energy: E₂ = −3.3995 eV.
- Hα is a difference, not a level: hf = 1239.84 eV nm ÷ 656.47 nm = 1.8886 eV = E₃ − E₂.
- So E₃ = E₂ + 1.8886 eV = −3.3995 eV + 1.8886 eV = −1.5109 eV. Check: level 3 must sit closer to the threshold than level 2, and it does.
Answer. E₂ = −3.400 eV and E₃ = −1.511 eV. No formula for the levels was needed: one series limit fixed the floor and one line fixed the gap.
MediumExample 2
A grating spectrometer gives two Balmer lines, Hα (3 → 2) at 656.5 ± 0.1 nm and Hδ (6 → 2) at 410.3 ± 0.1 nm, vacuum-corrected. Assuming only that the lines fit σ = σₗᵢₘ − R/n² with n the upper level, find R and the series limit, then the binding energy of level 2 with its standard uncertainty.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Work in wavenumber, where the model is linear: σ₃ = 1/656.5 nm = 1.523229 × 10⁶ m⁻¹ and σ₆ = 1/410.3 nm = 2.437241 × 10⁶ m⁻¹. An uncertainty in λ becomes u(σ) = u(λ)/λ², giving 232 m⁻¹ and 594 m⁻¹.
- Subtract to remove the limit: σ₆ − σ₃ = R(1/9 − 1/36) = R/12, so R = 12 × 9.14012 × 10⁵ m⁻¹ = 1.09681 × 10⁷ m⁻¹, with u(R) = 12√(232² + 594²) m⁻¹ = 7.7 × 10³ m⁻¹.
- Extrapolate to 1/n² = 0: σₗᵢₘ = σ₆ + R/36 = (4σ₆ − σ₃)/3 = 2.741912 × 10⁶ m⁻¹. Written in the measured inputs, which are independent, u(σₗᵢₘ) = √[(4/3 × 594)² + (1/3 × 232)²] m⁻¹ = 796 m⁻¹, almost all of it from Hδ.
- Binding energy of level 2: EB = hcσₗᵢₘ = 1239.84 eV nm × 2.741912 × 10⁻³ nm⁻¹ = 3.3995 eV, with u = 1239.84 eV nm × 7.96 × 10⁻⁷ nm⁻¹ = 0.0010 eV.
- Check: the limit wavelength 1/σₗᵢₘ = 364.71 ± 0.11 nm, the accepted binding energy 3.3996 eV lies well inside one standard uncertainty, and R matches RH = 1.09678 × 10⁷ m⁻¹ within its 0.07%.
Answer. R = (1.0968 ± 0.0008) × 10⁷ m⁻¹, σₗᵢₘ = (2.7419 ± 0.0008) × 10⁶ m⁻¹ and EB(n = 2) = 3.3995 ± 0.0010 eV. The extrapolation weights Hδ by 4/3, so the line nearest the limit sets the precision.
HardExample 3
A radio telescope aimed at an ionised nebula records a narrow emission line at 3.4078 GHz, read as hydrogen falling between adjacent levels, n + 1 → n. Find n without trial and error, confirm it with the exact Rydberg formula, and compare the photon energy with the binding energy of level n. Take cRH = 3.28805 × 10¹⁵ Hz.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- In frequency the formula reads f = cRH[1/n² − 1/(n + 1)²] = cRH(2n + 1)/[n²(n + 1)²]. With m = n + ½ this is (2cRH/m³)[1 + 1/(2m²) + …], and near m = 125 the correction is 3 parts in 10⁵.
- Invert: m = (2cRH/f)¹⁄³ = (6.57610 × 10¹⁵ Hz ÷ 3.4078 × 10⁹ Hz)¹⁄³ = (1.92972 × 10⁶)¹⁄³ = 124.50, so n = 124. The cruder 2cRH/n³, the bare derivative, gives n ≈ 124.5 and cannot choose between 124 and 125.
- Exact check: 1/124² − 1/125² = 6.503642 × 10⁻⁵ − 6.400000 × 10⁻⁵ = 1.036420 × 10⁻⁶. The first two digits cancel, so each term must be carried to seven figures. Then f = 3.28805 × 10¹⁵ Hz × 1.036420 × 10⁻⁶ = 3.4078 GHz.
- Neighbours: the same formula puts 124 → 123 at 3.4913 GHz and 126 → 125 at 3.3270 GHz, 83 MHz and 81 MHz away, far more than a narrow line's width, so the identification is unambiguous.
- Energies: hf = 4.1357 × 10⁻¹⁵ eV s × 3.4078 × 10⁹ Hz = 1.409 × 10⁻⁵ eV, while level 124 is bound by 13.598 eV/124² = 8.844 × 10⁻⁴ eV. The ratio is 1.59%, close to 2/n = 1.61%: the spacing falls as n⁻³ but the binding only as n⁻², so near the threshold the levels blur into the continuous energies of classical physics.
Answer. n = 124: the line is 125 → 124 at 3.4078 GHz (λ = 8.80 cm). Its photon carries 1.41 × 10⁻⁵ eV, only 1.6% of the level's 8.84 × 10⁻⁴ eV binding energy.
Quick check
Select an answer and explain it before checking.
Review & continue
An atomic line spectrum is the set of sharp wavelengths an atom emits or absorbs; each line's photon energy hf equals the difference between two of the atom's allowed energy levels, so the spectrum is a map of level differences.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics III completion tasks.