Atomic Spectra: Emission & Absorption Lines

Pass sunlight through a prism and you get a rainbow — a smooth, unbroken band of colour with nothing to count. Pass the light from a glowing tube of hydrogen through the same prism and you get something startlingly different: not a rainbow but a handful of sharp, coloured lines on a black background, always at exactly the same places. Do it with neon and you get a different set of lines; with sodium, a single brilliant yellow; with mercury, its own signature pattern. Every element, it turns out, emits light only at its own private list of wavelengths — a barcode written in colour that no two elements share. That is the central fact of this article, and it is one of the most consequential clues in the history of physics. In the 1850s, before anyone knew an atom had a structure at all, Gustav Kirchhoff and Robert Bunsen realised that these line spectra were fingerprints: heat any element and it announces itself by the exact colours it emits, and — Kirchhoff's masterstroke — a cool gas absorbs light at precisely the same colours it would emit when hot. With nothing but a prism they read the composition of the Sun, and discovered two new elements on Earth, purely from lines of light. But why lines? A rainbow says light can carry any colour, so why does an atom insist on its own few? This article establishes what the spectra are — continuous versus emission versus absorption — and pins down the one idea that makes them intelligible: each line is a photon of a single, definite energy, $E = hf$ . That is as far as we go here. Why an atom's photons come in a fixed menu — the energy levels behind the barcode — is Bohr's story, and it starts in the very next article. Here we earn the mystery.

Key takeaways

Three kinds of spectra: a hot dense source (a filament, the Sun's surface) gives a smooth continuous spectrum; a hot thin gas gives bright emission lines; a cool gas in front of a continuous source gives dark absorption lines.
Line spectra are element fingerprints: every element emits and absorbs only at its own fixed set of wavelengths — hydrogen's red line sits at $656\ \mathrm{nm}$ , sodium blazes yellow at $589\ \mathrm{nm}$ , and no two elements share a pattern.
Kirchhoff's law: a gas absorbs exactly the wavelengths it would emit when hot — so the same sodium line is bright in emission and dark in absorption.
Each line is a photon of definite energy: a spectral line at wavelength $\lambda$ is light delivered in quanta of energy $E = hf = hc/\lambda$ , and the board shortcut $E[\mathrm{eV}] = 1240/\lambda[\mathrm{nm}]$ converts a colour straight into an energy.
Fraunhofer's dark solar lines are absorption lines: the Sun's own cooler outer gases subtract their signature wavelengths from the continuous glow beneath — which is how helium was found on the Sun before it was found on Earth.
The pattern begged for an explanation: Balmer's 1885 formula reproduced hydrogen's visible lines to four figures with pure arithmetic, long before anyone knew why — the "why" is energy levels, the next articles' subject.
Key idea: an atom emits and absorbs light only at its own fixed set of wavelengths, and each such spectral line is a photon of one definite energy $E = hf$ — a barcode that identifies the element and, once decoded, reveals the atom's inner structure.

The core insight

Every element radiates and absorbs light only at its own unchanging set of discrete wavelengths — its line spectrum — and each line corresponds to a photon of a single definite energy $E = hf = hc/\lambda$ ; these fingerprints let us identify what a distant glowing gas is made of, and the discreteness itself is the clue that an atom's internal energies are quantised.

Where this lives today

Spectroscopy is arguably the most productive measurement idea physics has ever handed the rest of science, and it runs on exactly the fingerprints of this article. Astronomy is built on it: the composition, temperature, and motion of every star and galaxy come from its lines, and the redshift of those lines — the same Balmer pattern, slid bodily toward longer wavelengths — is how we know the universe expands and how exoplanet atmospheres are now being sniffed for water and methane by absorption. Analytical chemistry and industry identify and quantify elements by atomic emission and absorption spectroscopy, from checking the purity of a metal alloy to monitoring lead in drinking water at parts per billion. Forensics and art conservation read pigments and residues from their spectral lines without destroying the sample. Even the humble sodium street lamp and the neon sign are this physics made visible — each glows only at its element's signature wavelengths. And the discreteness that Kirchhoff and Balmer could only record is now the working principle of the atomic clock, which counts a single, impossibly sharp spectral line of caesium to define the second. The barcode of light, once merely a curiosity in a prism, has become the ruler by which we measure the cosmos and keep time.

The physical picture

Start with the three things a prism (or a diffraction grating) can show you, because the whole subject is the difference between them. Heat a solid, a liquid, or a dense gas until it glows — a lamp filament, molten iron, the Sun's surface — and its light, spread out by a prism, is a continuous spectrum: every wavelength is present, blending smoothly from red to violet with no gaps. The colours crowd so close together that there is nothing discrete to point at. This is thermal radiation from matter so dense that its atoms are in constant contact, and it carries no fingerprint of which element is glowing — a hot tungsten filament and hot iron give near-identical rainbows. Continuous spectra tell you about temperature, not composition. Now take the opposite extreme: a thin, hot gas — atoms far apart, each free to shine on its own. Excite it (an electric discharge, a flame) and its light is an emission spectrum: not a rainbow but a set of sharp bright lines, each a pure colour, separated by darkness. Hydrogen shows a red line, a blue-green line, and two violet lines; sodium shows an intense yellow; neon a thicket of reds and oranges. The pattern is fixed for each element and utterly reproducible — the same lines in the same places every time, in a laboratory or in a star. This is a fingerprint: the set of bright lines names the element. The third case is the subtlest and the most useful. Send the continuous light of a hot dense source through a cooler thin gas before it reaches the prism, and the rainbow comes out crossed by dark lines — an absorption spectrum. The cool gas has removed light at certain wavelengths, leaving black gaps in the otherwise complete band. Kirchhoff's decisive discovery was that a gas removes exactly the wavelengths it would emit if it were hot: the dark lines of a cool sodium vapour fall at precisely the same places as the bright lines of a hot one. Emission and absorption are two views of one and the same fingerprint.

Figure

The three spectra a prism reveals — a hot dense source gives an unbroken continuous rainbow, a hot thin gas gives bright emission lines on black, and a cool gas in front of a continuous source gives dark absorption lines that fall at exactly the same wavelengths as that gas would emit

These three cases are Kirchhoff's three laws of spectroscopy, stated in 1859–60. They are the grammar of the whole subject: whenever you look at a glowing thing, you are seeing one of these three spectra, and the lines — bright or dark — tell you what it is made of.

Reading a spectral line as a photon

A spectral line is a single, sharp wavelength. In the quantum picture set up in the previous article, light of frequency $f$ (wavelength $\lambda = c/f$ ) is delivered in indivisible lumps — photons — each carrying a fixed energy. A bright emission line therefore means the atom is throwing out photons all of one energy; a dark absorption line means it is swallowing photons of that same one energy. The energy of a single photon is Planck's relation:

E = hf = \frac{hc}{\lambda} \qquad (1)

where: $E$ is the photon energy, $h = 6.63\times10^{-34}\ \mathrm{J\,s}$ is Planck's constant, $f$ the light's frequency, $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ the speed of light, and $\lambda$ the wavelength. The physical reading is direct: a colour is an energy. A sharp line at one wavelength is light of one photon-energy, and the sharper and more definite the line, the more precisely defined that energy. This is exactly why line spectra are the doorway to atomic structure — if an atom only emits photons of certain energies, then the atom itself must only be able to release energy in those certain amounts. For exam work you almost never plug into equation (1) with SI joules; you use the board shortcut that folds $hc$ and the electronvolt into a single number:

E[\mathrm{eV}] = \frac{1240}{\lambda[\mathrm{nm}]} \qquad (2)

where: $E$ comes out in electronvolts when $\lambda$ is in nanometres, and $1240\ \mathrm{eV\,nm}$ is the value of $hc$ expressed in those units (derived below). The physical reading: divide $1240$ by the wavelength in nanometres and you have the photon's energy in electronvolts, at a glance. So the yellow sodium line at $\lambda = 589\ \mathrm{nm}$ is made of photons of energy $E = 1240/589 = 2.11\ \mathrm{eV}$ , oscillating at a frequency $f = c/\lambda = (3.00\times10^{8})/(589\times10^{-9}) = 5.09\times10^{14}\ \mathrm{Hz}$ . Hydrogen's red line at $656\ \mathrm{nm}$ is $1240/656 = 1.89\ \mathrm{eV}$ ; its blue-green line at $486\ \mathrm{nm}$ is $1240/486 = 2.55\ \mathrm{eV}$ ; its violet line at $434\ \mathrm{nm}$ is $1240/434 = 2.86\ \mathrm{eV}$ . Each element's fingerprint is thus equally a list of wavelengths, a list of frequencies, or a list of photon energies — three languages for the same set of lines. One consequence worth flagging now, because it foreshadows everything to come: the sodium "line" at $589\ \mathrm{nm}$ is really a close pair of lines, at $589.0\ \mathrm{nm}$ and $589.6\ \mathrm{nm}$ , split by a photon-energy difference of only $1240/589.0 - 1240/589.6 \approx 0.0021\ \mathrm{eV}$ . That the atom's fingerprint has such fine internal structure — a tiny, definite energy gap — is another hint that the atom's insides are built from discrete, precisely-spaced energies. We simply record it here; the machinery to explain it is later.

Emission and absorption: two views of one fingerprint

Why should a cool gas absorb exactly the wavelengths a hot gas emits? Because both processes involve the very same atoms and the very same definite photon energies. When an atom is given energy — by heat, by an electric discharge, by a collision — it can release that energy as a photon, and only photons of its allowed energies ever come out; those are its emission lines. Run the process backwards: shine a full continuous rainbow at the same atoms while they are cool, and an atom can only accept a photon whose energy exactly matches one of its allowed jumps. Photons of any other energy pass straight through, but photons at the atom's signature energies are absorbed, plucked out of the beam and leaving a dark gap. Emission and absorption are the same set of energies, taken in the two directions. This is Kirchhoff's law, and its power is that it works at a distance. You do not need a sample of a star in a test tube; you need only its light. The Sun's dazzling surface gives a continuous spectrum, but that light climbs out through the Sun's own cooler outer atmosphere, and every element up there stamps its absorption lines onto the rainbow. Read the dark lines and you have read the Sun's chemistry. The most famous demonstration is the sodium line: hold a sodium flame in a beam and it glows yellow at $589\ \mathrm{nm}$ (emission); place that same flame in front of a brighter continuous lamp and the $589\ \mathrm{nm}$ light coming through is now weaker than its surroundings — a dark line at the identical wavelength (absorption). The atom that shouts a colour in isolation whispers a shadow of the same colour in front of a brighter source.

Figure

Kirchhoff's demonstration with a sodium flame — alone, the flame emits a bright yellow line at 589 nm on a dark field, but placed in front of a hotter continuous lamp it removes that same wavelength from the rainbow, printing a dark absorption line at exactly 589 nm

This equivalence is what let Kirchhoff and Bunsen turn a prism into a chemical laboratory. Bunsen's clean, colourless flame excited samples without adding lines of its own; Kirchhoff read the pattern. Salts that looked identical betrayed different elements the instant their lines appeared, and in 1860 the pair found caesium (from its two blue lines, caesius, sky-blue) and, the next year, rubidium (from deep-red lines) — brand-new elements that no chemistry had isolated. A few years later the same method found helium as an unexplained yellow line in the Sun's spectrum, an element catalogued in the sky before it was ever bottled on Earth.

Derivation AThe 1240 electronvolt-nanometre bridge

The goal of this short derivation is to show where the ubiquitous exam constant $1240\ \mathrm{eV\,nm}$ comes from, so that equation (2) is never a magic number but a repackaging of equation (1). We want to prove that $hc$ , expressed with energy in electronvolts and wavelength in nanometres, equals about $1240$ , using only the standard constants and unit conversions — pure arithmetic, no calculus. Once we have it, converting any spectral line's wavelength into a photon energy is a single division.

Step 1. Start from the photon energy in SI units, equation (1): $E = hc/\lambda$ . Compute the product $hc$ in joule-metres from the given constants: $hc = (6.63\times10^{-34}\ \mathrm{J\,s})(3.00\times10^{8}\ \mathrm{m\,s^{-1}}) = 1.989\times10^{-25}\ \mathrm{J\,m}$ . This is the numerator that turns a wavelength into an energy.

Step 2. Convert the energy from joules to electronvolts by dividing by the elementary charge $e = 1.60\times10^{-19}\ \mathrm{C}$ : in electronvolt units, $hc = (1.989\times10^{-25})/(1.60\times10^{-19}) = 1.243\times10^{-6}\ \mathrm{eV\,m}$ . So $E[\mathrm{eV}] = (1.243\times10^{-6})/\lambda[\mathrm{m}]$ — correct, but with the wavelength still in metres.

Step 3. Convert the wavelength from metres to nanometres. Since $1\ \mathrm{nm} = 10^{-9}\ \mathrm{m}$ , writing $\lambda[\mathrm{m}] = \lambda[\mathrm{nm}]\times10^{-9}$ gives $E[\mathrm{eV}] = (1.243\times10^{-6})/(\lambda[\mathrm{nm}]\times10^{-9}) = (1.243\times10^{3})/\lambda[\mathrm{nm}] = 1243/\lambda[\mathrm{nm}]$ . Rounding to the value the boards standardise on, $E[\mathrm{eV}] = 1240/\lambda[\mathrm{nm}]$ — equation (2). The small difference between $1243$ and $1240$ is just the rounding in $h$ and $c$ ; syllabuses fix on $1240$ .

Step 4. Sanity-check on a known line. Hydrogen's red line sits at $\lambda = 656\ \mathrm{nm}$ , so $E = 1240/656 = 1.89\ \mathrm{eV}$ ; the sodium line at $589\ \mathrm{nm}$ gives $E = 1240/589 = 2.11\ \mathrm{eV}$ . Both are sensible visible-photon energies (roughly $2\ \mathrm{eV}$ , as every visible photon is), confirming the bridge. Verified: $hc = 6.63\times10^{-34}\times3.00\times10^{8} = 1.989\times10^{-25}\ \mathrm{J\,m}$ ; $\div e = 1.243\times10^{-6}\ \mathrm{eV\,m} = 1243\ \mathrm{eV\,nm} \approx 1240\ \mathrm{eV\,nm}$ ; $1240/656 = 1.89\ \mathrm{eV}$ , $1240/589 = 2.11\ \mathrm{eV}$ — recomputed in verify_hs2-atomic-spectra.py.

Go deeper · Derivation B · 1 minBalmer's numerical ladder

Derivation BBalmer's numerical ladder

The goal here is to show just how regular an atom's fingerprint is — regular enough that in 1885 a Swiss schoolteacher, Johann Balmer, reproduced hydrogen's four visible lines to four significant figures with one short algebraic formula, decades before anyone could explain it. We will take Balmer's formula, feed in whole numbers, and watch the measured wavelengths fall out; the point is not the formula's derivation (there is no classical one) but the shock of the pattern. Everything here is arithmetic — no calculus, and crucially no energy-level physics, which is the next articles' business.

Step 1. Balmer's formula for hydrogen's visible lines takes an integer $n = 3, 4, 5, \dots$ and returns a wavelength,

\lambda = B\,\frac{n^{2}}{n^{2} - 4}, \qquad B = 364.6\ \mathrm{nm} \qquad (3)

where $B$ is a single constant Balmer found by fitting. The factor $n^{2}-4$ can be read as $n^{2}-2^{2}$ , with the "2" hinting at a special role for the number two — a hint Bohr later cashed in.

Step 2. Put in $n = 3$ : $\lambda = 364.6\times \dfrac{9}{9-4} = 364.6\times\dfrac{9}{5} = 656.3\ \mathrm{nm}$ — the red line ( $\mathrm{H}\alpha$ ), matching the measured $656\ \mathrm{nm}$ .

Step 3. Put in $n = 4$ : $\lambda = 364.6\times\dfrac{16}{16-4} = 364.6\times\dfrac{16}{12} = 486.1\ \mathrm{nm}$ — the blue-green line ( $\mathrm{H}\beta$ ), matching $486\ \mathrm{nm}$ . And $n = 5$ : $\lambda = 364.6\times\dfrac{25}{21} = 434.0\ \mathrm{nm}$ (violet $\mathrm{H}\gamma$ , measured $434\ \mathrm{nm}$ ); $n = 6$ : $\lambda = 364.6\times\dfrac{36}{32} = 410.2\ \mathrm{nm}$ (deep violet $\mathrm{H}\delta$ ).

Step 4. Push $n \to \infty$ : the fraction $n^{2}/(n^{2}-4) \to 1$ , so $\lambda \to B = 364.6\ \mathrm{nm} \approx 365\ \mathrm{nm}$ . The lines crowd closer and closer together, piling up at this series limit in the near-ultraviolet — exactly what the measured spectrum does. A one-constant arithmetic rule reproduces the entire visible hydrogen series and predicts where it ends. That such clean integers govern an atom's light is the loudest possible signal that something discrete lives inside the atom; naming that something — quantised energy levels — is where Bohr picks up. Verified: with $B = 364.6\ \mathrm{nm}$ , $n=3 \to 656.3$ , $n=4 \to 486.1$ , $n=5 \to 434.0$ , $n=6 \to 410.2\ \mathrm{nm}$ , limit $\to 364.6 \approx 365\ \mathrm{nm}$ — recomputed in verify_hs2-atomic-spectra.py.

Limiting cases

The three spectra are best held in the head as Kirchhoff's three laws, plus the boundary behaviours of a line.

case	result	meaning
hot dense source (filament, star surface)	continuous spectrum, all wavelengths	atoms packed too close to shine on their own; thermal glow, no fingerprint
hot thin gas	bright emission lines only	isolated atoms release photons at their signature energies
cool thin gas over a continuous source	dark absorption lines at the same wavelengths	atoms remove exactly the photon energies they would emit
photon energy off every line	passes through unabsorbed	only exact-match photons are taken; the rest ignore the gas
Balmer series, $n \to \infty$	lines pile up at the limit $\lambda \to 365\ \mathrm{nm}$	the fingerprint has an edge, not an infinite spread of separate lines
broaden the source to a full rainbow	absorption lines stay put	line wavelengths are set by the atom, not by the light shone at it

Common trapA line spectrum is discrete because the atom is — not because the prism only picks out a few colours

The commonest confusion is to imagine the atom pours out a full rainbow and the prism or grating somehow selects a few colours from it. It does not. The prism faithfully spreads out whatever light arrives; if only a few wavelengths come out, it is because the atom emitted only a few. A hot thin gas genuinely radiates at just its signature wavelengths and nowhere else, and a full continuous rainbow really does emerge from a dense hot source. The discreteness is a fact about the atom's light, not an artefact of the instrument. The second half of the same trap is to forget that emission and absorption lines of one element sit at identical wavelengths — students draw a sodium absorption line at some new place, when Kirchhoff's law fixes it at the very $589\ \mathrm{nm}$ where sodium emits. Same atom, same energies, same wavelengths, whichever direction the photon goes.

Worked examples

Board level — colour to photon energy, and identifying an element (CBSE and ICSE pattern)

A discharge tube shows a bright green line at $\lambda = 546\ \mathrm{nm}$ . (a) Find the energy of one photon of this line in electronvolts and its frequency. (b) A cool cloud in front of a star removes light at $589\ \mathrm{nm}$ ; name the likely element and give its photon energy. Take $hc = 1240\ \mathrm{eV\,nm}$ , $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ .

Step 1. Photon energy of the green line, straight from equation (2): $E = 1240/546 = \boxed{2.27\ \mathrm{eV}}$ . This green line at $546\ \mathrm{nm}$ is a well-known mercury line, so the tube likely contains mercury vapour.

Step 2. Its frequency, from $f = c/\lambda$ : $f = (3.00\times10^{8})/(546\times10^{-9}) = \boxed{5.49\times10^{14}\ \mathrm{Hz}}$ . As a check, $E = hf = (4.14\times10^{-15})(5.49\times10^{14}) = 2.27\ \mathrm{eV}$ , agreeing with Step 1.

Step 3. The dark line at $589\ \mathrm{nm}$ is an absorption line; by Kirchhoff's law it sits where that element would emit, and $589\ \mathrm{nm}$ is the signature yellow of sodium. Its photon energy is $E = 1240/589 = \boxed{2.11\ \mathrm{eV}}$ . So the cloud contains sodium, identified from a single dark line without ever touching it.

Competitive level — same gas, two spectra (JEE and NEET pattern)

A star's continuous spectrum is crossed by a dark line at $656\ \mathrm{nm}$ . (a) What is the photon energy of this line? (b) Which element is responsible, and how would a hot rarefied sample of it appear in the lab? (c) The star's spectrum also shows dark lines at $486\ \mathrm{nm}$ and $434\ \mathrm{nm}$ ; what does this triplet confirm?

Step 1. Photon energy: $E = 1240/656 = \boxed{1.89\ \mathrm{eV}}$ — a red-light photon, as expected for a $656\ \mathrm{nm}$ line.

Step 2. The wavelength $656\ \mathrm{nm}$ is hydrogen's $\mathrm{H}\alpha$ line. In the star it appears dark because cooler hydrogen in the star's outer layers absorbs it out of the continuous glow beneath. The same hydrogen, hot and rarefied in a discharge tube, would show this line as a bright red emission line at the identical $656\ \mathrm{nm}$ — emission and absorption being the two directions of one fingerprint.

Step 3. The extra dark lines at $486\ \mathrm{nm}$ ( $E = 1240/486 = 2.55\ \mathrm{eV}$ ) and $434\ \mathrm{nm}$ ( $E = 1240/434 = 2.86\ \mathrm{eV}$ ) are hydrogen's $\mathrm{H}\beta$ and $\mathrm{H}\gamma$ . Three lines in the exact Balmer pattern are a unique hydrogen signature, so the triplet confirms hydrogen in the star — the barcode read across light-years. Timing note: about 60 seconds. Each energy is one division of $1240$ by a wavelength, done in your head to two figures ( $1240/656 \approx 1.9$ , $1240/486 \approx 2.55$ ). The marks are in recognising $656/486/434\ \mathrm{nm}$ as hydrogen's Balmer lines and stating Kirchhoff's law — that the dark stellar lines and the bright lab lines coincide — not in the arithmetic.

Competitive level — the sodium doublet: difference of close reciprocals (JEE Advanced numerical pattern)

The sodium "yellow line" is in fact two close lines, at $589.0\ \mathrm{nm}$ and $589.6\ \mathrm{nm}$ . Find the photon energies of the two components and the energy difference between them, and comment on what this fine structure implies.

Step 1. Component energies from equation (2): $E_1 = 1240/589.0 = 2.1053\ \mathrm{eV}$ and $E_2 = 1240/589.6 = 2.1031\ \mathrm{eV}$ . Both round to the familiar $2.11\ \mathrm{eV}$ of the sodium line, as they must.

Step 2. Energy gap: $\Delta E = E_1 - E_2 = 2.1053 - 2.1031 = \boxed{0.0021\ \mathrm{eV}}$ , about $2.1\ \mathrm{meV}$ — a tiny but perfectly definite splitting.

Step 3. Comment: the fact that even the "single" sodium line resolves into two components separated by a fixed, minuscule energy shows the atom's internal energies are not only discrete but finely and precisely spaced. A continuous, classical atom could not produce such a sharp, reproducible split. This fine structure is one more piece of evidence, gathered from light alone, that the atom's insides are quantised — the theme the next articles make precise. Timing note: about 75 seconds. The subtraction $1240/589.0 - 1240/589.6$ is safest done by keeping four decimals before differencing — round too early and the small gap vanishes. This "difference of two nearly-equal reciprocals" is the classic trap in doublet questions.

Exam traps

"The prism only shows a few colours." No — the prism spreads all incoming light faithfully; a line spectrum is discrete because the atom emits only those wavelengths. Never explain line spectra by the instrument. This framing is tested on CBSE, ICSE, JEE, NEET, AP Physics 2, IB HL and A-Level alike.

Emission and absorption lines of one element are at the same wavelengths. By Kirchhoff's law a gas absorbs exactly what it emits. If sodium emits at $589\ \mathrm{nm}$ , it absorbs at $589\ \mathrm{nm}$ — not somewhere else. Drawing them at different places is a guaranteed lost mark.

Continuous spectra carry no elemental fingerprint. A hot dense source (filament, star's surface, molten metal) gives a smooth rainbow that says nothing about composition. Only line spectra — emission or absorption — identify an element. Do not claim you can read a star's chemistry from its continuous background; you read it from the dark lines crossing that background.

The wavelength–energy inequality flips. From $E = 1240/\lambda$ , a shorter wavelength (bluer line) is a higher-energy photon. So hydrogen's violet $434\ \mathrm{nm}$ line ( $2.86\ \mathrm{eV}$ ) is more energetic than its red $656\ \mathrm{nm}$ line ( $1.89\ \mathrm{eV}$ ). Students routinely pair "longer wavelength" with "more energy" and lose the mark; always sanity-check that bluer means more energetic.

Round late in doublet and series problems. When two lines are close (the sodium doublet, adjacent Balmer lines), the interesting quantity is a small difference of large reciprocals. Round each energy to two decimals too early and the difference collapses to zero. Carry four figures, then subtract.

Do not invoke energy levels or Bohr orbits here. This article establishes that spectra are discrete fingerprints and that each line is a photon $E = hf$ . Why the allowed energies are what they are — the $-13.6/n^{2}$ ladder, the quantised orbits — is the next two articles. On an exam that asks only to identify or convert lines, quoting Bohr's model is unnecessary and, if misremembered, actively harmful.

Origins · 1 minGustav Kirchhoff and how this was discovered

Gustav Kirchhoff

1824–1887 · Heidelberg

Gustav Robert Kirchhoff, born in Königsberg and for two decades a professor at Heidelberg, is the physicist who turned the prism into an instrument for reading matter. Working with the chemist Robert Bunsen from 1859, he established the three laws of spectroscopy — continuous spectra from hot dense bodies, bright-line spectra from hot gases, and dark-line spectra where a cool gas absorbs — and, crucially, showed that a gas absorbs precisely the wavelengths it emits. This let the pair identify elements by their line patterns, discovering caesium in 1860 and rubidium in 1861, and it explained the mysterious dark lines Fraunhofer had mapped in sunlight as absorption by the Sun's own gases, opening the chemistry of the stars to study. Kirchhoff's name also attaches to his two laws of electrical circuits and to his law of thermal radiation; but it is spectroscopy that made the line spectrum a tool of discovery, and set the stage for the quantum atom.

History

1802

— William Hyde Wollaston notices dark gaps interrupting the Sun's continuous spectrum, the first sighting of absorption lines.

1814

— Joseph von Fraunhofer maps hundreds of dark solar lines with a precision spectroscope and labels the strongest of them with the letters A through K.

1859

— Gustav Kirchhoff and Robert Bunsen establish the three laws of spectroscopy and show that each element has a unique line spectrum, explaining Fraunhofer's dark lines as absorption.

1860

— Kirchhoff and Bunsen discover caesium, and the next year rubidium, purely from new lines in a flame's spectrum.

1868

— Pierre Janssen and Norman Lockyer independently observe a yellow solar line matching no known element; Lockyer (with the chemist Edward Frankland) names it helium — discovered on the Sun before it was ever isolated on Earth.

1885

— Johann Balmer publishes a single arithmetic formula reproducing hydrogen's four visible lines to four figures, the empirical pattern that cried out for the quantum atom.

Going further

Texts. NCERT Physics Class 12, Chapter 12 ("Atoms"), the sections on line spectra and the hydrogen spectrum · H. C. Verma, Concepts of Physics Vol. 2, the chapter on the atom and its spectrum · Halliday, Resnick and Walker, the chapter "All About Atoms" · for the history, the story of Kirchhoff, Bunsen, and the birth of spectroscopy in any good history of nineteenth-century physics.

Provenance. Board constants: $h = 6.63\times10^{-34}\ \mathrm{J\,s}$ , $h = 4.14\times10^{-15}\ \mathrm{eV\,s}$ , $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ , $e = 1.60\times10^{-19}\ \mathrm{C}$ , $hc = 1240\ \mathrm{eV\,nm}$ . Recompute: $hc = 6.63\times10^{-34}\times3.00\times10^{8} = 1.989\times10^{-25}\ \mathrm{J\,m} = 1243\ \mathrm{eV\,nm} \approx 1240$ ; photon energies $1240/589 = 2.11$ , $1240/656 = 1.89$ , $1240/486 = 2.55$ , $1240/434 = 2.86$ , $1240/546 = 2.27$ , $1240/620 = 2.00\ \mathrm{eV}$ ; frequencies $f(589) = 5.09\times10^{14}$ , $f(546) = 5.49\times10^{14}\ \mathrm{Hz}$ ; sodium doublet $1240/589.0 - 1240/589.6 = 0.0021\ \mathrm{eV}$ ; Balmer $\lambda = 364.6\,n^{2}/(n^{2}-4)$ gives $656.3, 486.1, 434.0, 410.2\ \mathrm{nm}$ for $n = 3,4,5,6$ and limit $364.6 \approx 365\ \mathrm{nm}$ .

Prerequisites. Planck's Quantum & the Photon — where $E = hf$ , the photon, and Planck's constant are established; without it a "line as a photon of definite energy" has no footing. Leads to. The Bohr Model of the Atom, the next article, which explains why an atom's photons come in a fixed menu by quantising its orbits and energies · and, for university-level enrichment named here only as a go-deeper pointer, the seams on the Bohr model of hydrogen and the hydrogen spectral series, which put the whole ladder on a rigorous footing.

What is the difference between a continuous spectrum and a line spectrum?

A continuous spectrum contains every wavelength blending smoothly — the rainbow you get from a hot dense source such as a lamp filament or a star's surface. A line spectrum contains only certain sharp, isolated wavelengths: either bright lines on a dark background (emission, from a hot thin gas) or dark lines on a bright background (absorption, from a cool gas in front of a continuous source). The crucial difference is that continuous spectra come from dense matter and carry no elemental fingerprint, while line spectra come from individual atoms and identify the element exactly. Which one you see tells you immediately whether you are looking at dense hot matter or at thin gas.

Why does each element have its own unique set of lines?

Because each element's atoms can only release or absorb energy in their own specific amounts, and every such amount corresponds — through $E = hf$ — to a photon of a particular wavelength. The set of allowed energy jumps is fixed by the atom's internal structure, which differs from element to element, so the resulting set of wavelengths is a unique signature. This article establishes that the lines are fixed and discrete; the reason the allowed energies take the values they do is the subject of the Bohr model and the hydrogen energy levels in the next two articles. For now, treat the line pattern as an experimental fingerprint that no two elements share.

How can we know what stars are made of without going there?

Through absorption lines and Kirchhoff's law. A star's hot surface emits a continuous spectrum, but that light passes through the star's cooler outer gases, which absorb their own signature wavelengths and leave dark lines in the spectrum that reaches us. Because each element absorbs exactly the wavelengths it would emit, matching the dark lines to known laboratory patterns identifies the elements present. This is how the Sun's composition was first read in the 1860s, and how helium was discovered as an unexplained solar line before it was ever found on Earth. The same method today measures the make-up of galaxies billions of light-years away.

What exactly does one spectral line correspond to?

A single spectral line is light of one definite wavelength, and therefore — since a photon's energy is $E = hf = hc/\lambda$ — photons of one definite energy. A bright emission line means the atom is releasing photons all of that one energy; a dark absorption line means it is removing photons of that same energy from a beam. So a line is best pictured not as a "colour" but as an energy: the sodium line at $589\ \mathrm{nm}$ is a stream of $2.11\ \mathrm{eV}$ photons. This one-line-one-energy idea is the bridge from the observed spectrum to the atom's internal energy structure.

Why is the sodium line sometimes called a doublet?

Because the famous sodium yellow is really two lines very close together, at $589.0\ \mathrm{nm}$ and $589.6\ \mathrm{nm}$ , differing in photon energy by only about $0.0021\ \mathrm{eV}$ . A simple picture of the atom would predict a single line; the split into two shows the atom's energy structure has fine detail beyond the coarse pattern. The reason — a small energy difference tied to the electron's own angular momentum and spin — is beyond this article, but the doublet is worth knowing because it is a favourite exam example of just how sharply and precisely an atom's energies are defined. For most purposes the pair is treated as a single $589\ \mathrm{nm}$ line.

Does the Balmer formula explain hydrogen's spectrum?

No — it describes it beautifully but explains nothing. Balmer found in 1885 that plugging whole numbers into $\lambda = B\,n^{2}/(n^{2}-4)$ reproduced hydrogen's visible lines to four figures, but he had no idea why it worked; it was a pattern-fit, not a physical theory. The explanation had to wait for Bohr in 1913, who showed the integers are quantum numbers labelling the atom's energy levels and that Balmer's constant is set by fundamental constants. So the Balmer formula is the empirical clue, not the answer — and its clean integers are precisely what told physicists something discrete had to be going on inside the atom.

Is this topic on my board, and how deep does it go?

Atomic spectra — continuous, emission, and absorption, and the idea that each line is a photon $E = hf$ — appear on CBSE, ICSE, JEE Main, JEE Advanced, NEET, AP Physics 2, IB Higher Level (Theme E) and A-Level (CIE 9702). IB Standard Level treats spectra only qualitatively and without the Bohr mathematics, so this page is aimed at HL; IGCSE 0625 has no atomic-spectra content at all (its nuclear-atom material is purely qualitative). For every listed board the safe core is: name the three spectra and their sources, state Kirchhoff's law that emission and absorption lines coincide, and convert a wavelength to a photon energy with $E = 1240/\lambda$ . The energy-level explanation of the lines belongs to the Bohr-model and hydrogen-spectrum articles that follow.