Atomic spectra

Hold a prism up to sunlight and you get the familiar rainbow — every colour blending smoothly into the next, with no colour missing. Now put a low-pressure hydrogen gas discharge tube in front of that same prism instead. The rainbow vanishes. In its place: a handful of sharp, bright, isolated lines of colour on an otherwise black background — a red line here, a blue-green line there, a violet line further along — and nothing at all in between. Nature, it turns out, does not let a free hydrogen atom glow in just any colour it pleases. It glows in a fixed, reproducible set of colours, as if each element were issued its own barcode at the factory, and no two elements share a barcode. This single observation — first catalogued patiently through the nineteenth century, decades before anyone could explain it — turned out to be one of the most consequential clues in the history of physics. A classical, Rutherford-style atom, with an electron freely spiralling in toward the nucleus while radiating energy, has no reason to prefer some colours over others; it should bleed out a continuous smear of light as it dies. Real atoms instead emit and absorb at only a few sharp, discrete wavelengths, always the same for a given element. That single stubborn fact told physicists something the classical picture flatly refused to explain: an electron bound inside an atom cannot have just any energy — it is confined to a fixed ladder of allowed energies, and light appears or disappears only when the electron jumps a rung. This article builds that reasoning up carefully: how continuous and line spectra differ, how emission and absorption spectra are two faces of the same coin, how the discreteness of spectral lines forced physicists toward the idea of quantised atomic energy levels, and how a jump between two such levels fixes the energy — and hence the colour — of the photon involved.

Key takeaways

A continuous spectrum (hot solid, liquid, or dense gas) is an unbroken band of every wavelength; a line spectrum (rarefied excited gas) shows only a few sharp, discrete wavelengths.
Emission spectrum: an excited gas glows with bright lines against a dark background, at wavelengths unique to that element.
Absorption spectrum: white light through a cooler sample of the same gas shows the continuous spectrum with dark lines missing at exactly those same wavelengths.
Each element's line pattern is a fingerprint — reproducible, unique, and usable to identify elements even in starlight millions of light-years away.
Classical physics fails here: a continuously accelerating, spiralling electron (Rutherford's picture) should radiate a smooth spread of frequencies, not sharp lines.
The resolution is quantisation: an atom's electron is restricted to discrete allowed energies $E_n$ ; a photon is emitted or absorbed only when the electron jumps between two such levels, so only certain photon energies — and hence only certain sharp wavelengths — are possible.
Key idea: sharp, discrete atomic spectral lines exist because an atom's electron energy is quantised into fixed levels, so a photon of energy $E_i - E_f$ appears only when the electron jumps between two of those levels, never for a continuous range of energies.

The core insight

Atomic spectra are sharp and discrete, not continuous, because an electron bound in an atom can only occupy a fixed ladder of allowed energies $E_n$ , so light is emitted or absorbed only in fixed packets equal to the energy gap between two such levels.

Where it fits in the syllabus

This topic sits inside the CBSE Class 12 Physics chapter "Atoms" (Unit IX, Atoms and Nuclei), right at the hinge between Rutherford's nuclear model and Bohr's model of the hydrogen atom. The board syllabus treats hydrogen line spectra qualitatively only — you are expected to know that discrete series such as the Lyman and Balmer series exist and to explain, in words, why quantised energy levels are needed to account for them, but you are not expected to compute exact wavelengths from the Rydberg formula, which is explicitly kept out of this syllabus's numerical scope. Wherever this article does use numbers, it uses only the orbit-energy formula $E_n = -13.6/n^2$ eV, which is squarely inside the prescribed Bohr-model scope, to find the energy of a transition — never to grind out a precise wavelength.

The physical picture

When a substance is heated to incandescence — the filament of a bulb, molten metal, the interior of the Sun — it emits light of every wavelength smoothly blended together. Pass this light through a prism and you see an unbroken rainbow: a continuous spectrum. Contrast this with a rarefied gas of free atoms, excited by heat or an electric discharge (as in a neon sign or a hydrogen discharge tube). Such a gas emits light only at certain sharp wavelengths, seen as bright separated lines against a dark background — a line spectrum, specifically the gas's emission spectrum. Run the experiment the other way: send white light (which already contains every wavelength) through a sample of the same gas, kept cooler than the light source, and look at what comes out the far side with a spectrometer. The continuous rainbow reappears, but now with a few sharp dark lines missing — exactly the wavelengths the gas would otherwise emit. This is the gas's absorption spectrum. The coincidence is not an accident: an atom absorbs light most efficiently at precisely the wavelengths at which it can also emit, because both processes involve the same underlying transitions between the same fixed set of allowed energies.

E_{\text{photon}} = h\nu = E_i - E_f, \qquad E_i > E_f. \qquad (1)

where: $h = 6.63\times10^{-34}\ \text{J s}$ is Planck's constant, $\nu$ is the frequency of the photon involved, and $E_i$ , $E_f$ are the (quantised) energies of the atom's electron before and after the transition. The physical reading of equation (1) is the entire reason line spectra are sharp: because $E_i$ and $E_f$ can only take values from a fixed discrete list — never anything in between — the photon energy $E_i - E_f$ , and therefore its frequency and wavelength, can also only take values from a correspondingly fixed, discrete list. There is no continuum of allowed jumps, so there is no continuum of allowed colours.

Figure

Energy level ladder for hydrogen with three downward arrows of different lengths landing on the lowest level, next to a dark background showing three separated bright spectral lines at different positions matching the three arrow lengths

Derivation AWhy atomic energy quantisation forces spectral lines to be sharp

The goal of this derivation is to show, starting only from the idea that a bound electron's energy is restricted to a discrete list of values $E_1, E_2, E_3, \ldots$ , that the light an atom can emit or absorb must itself be confined to a discrete list of frequencies rather than spread continuously — which is exactly what a line spectrum is, and exactly what a purely classical atom could never produce.

Step 1. Start from energy conservation for a single emission event. When an electron inside an atom drops from a higher allowed energy $E_i$ to a lower allowed energy $E_f$ , the energy it loses does not vanish — it must appear somewhere. Experiment shows it appears as a single photon of electromagnetic radiation, carrying away exactly the energy the electron lost: $E_{\text{photon}} = E_i - E_f$ . The same logic runs in reverse for absorption: an atom can absorb a photon only if that photon supplies exactly enough energy to lift the electron from one allowed level to another.

Step 2. Bring in the quantisation postulate. The Bohr model asserts that a bound electron cannot possess an arbitrary energy — only the members of a fixed, discrete list $E_1, E_2, E_3, \ldots$ , indexed by a whole number $n$ , are allowed at all. Whatever the exact list happens to be for a given atom, the crucial feature is that it is a list, with gaps, not a continuous range. This is the postulate itself, quoted from the Bohr model rather than derived here — the derivation of its origin belongs to the Bohr-model article that follows this one.

Step 3. Combine the two. Because $E_i$ and $E_f$ can only be chosen from that same fixed, discrete list, the difference $E_i - E_f$ can also only take a fixed, discrete set of values — one for every possible pair of levels the electron could jump between. Using $E = h\nu$ , each allowed energy difference corresponds to exactly one allowed frequency $\nu = (E_i - E_f)/h$ , and hence one allowed wavelength $\lambda = c/\nu$ . A continuous range of electron energies would have given a continuous range of possible jumps and therefore a continuous spread of colours; a discrete ladder of energies gives a discrete, finite set of possible jumps, and therefore a discrete set of sharp spectral lines.

Step 4. Read off the conclusion. Every distinct pair of levels $(n_i, n_f)$ with $n_i > n_f$ gives one spectral line. Grouping all jumps that share the same final level gives a series: all jumps ending on $n_f = 1$ form the Lyman series (in the ultraviolet), all jumps ending on $n_f = 2$ form the Balmer series (partly in the visible), and all jumps ending on $n_f = 3$ form the Paschen series (in the infrared) — the syllabus asks you to know that these series exist and roughly where they lie, not to compute their exact wavelengths.

Worked examples

Board level - identifying emission and absorption spectra of hydrogen

A student places a low-pressure hydrogen discharge tube in front of a spectrometer, then separately shines white light through cool hydrogen gas onto the same spectrometer. Describe what she observes in each case and explain how the two observations are related.

Step 1. For the glowing discharge tube, the gas atoms are excited and their electrons fall back down through the allowed energy levels, each jump emitting one photon of a fixed energy. The spectrometer therefore shows a set of individual bright, sharp coloured lines against an otherwise dark background — hydrogen's emission line spectrum.

Step 2. For white light passed through cool hydrogen gas, the incoming light already contains every wavelength. Ground-state hydrogen atoms can only absorb photons whose energy exactly matches the gap between two of their allowed levels, so only those particular wavelengths get removed from the beam. The spectrometer shows the continuous rainbow with a few sharp dark lines missing — hydrogen's absorption line spectrum.

Step 3. Since both processes involve the same fixed ladder of allowed energies $E_n = -13.6/n^2$ eV, the wavelengths that hydrogen can absorb are exactly the wavelengths it can emit; the dark lines in the absorption spectrum sit at precisely the same positions as the bright lines in the emission spectrum.

Answer: the discharge tube shows bright discrete lines (emission spectrum); the white light through cool gas shows the continuous spectrum with dark lines missing at the same wavelengths (absorption spectrum) — both arise from the same set of allowed transitions between hydrogen's quantised energy levels.

Competitive level - comparing the energy of two hydrogen transitions

Using $E_n = -13.6/n^2$ eV for the allowed energies of the hydrogen atom, find the photon energy released when the electron falls (a) from $n = 2$ to $n = 1$ and (b) from $n = 3$ to $n = 2$ . Which transition releases the more energetic photon, and roughly which part of the electromagnetic spectrum does each fall in?

Step 1. Compute the levels needed: $E_1 = -13.6/1^2 = -13.6\ \text{eV}$ , $E_2 = -13.6/2^2 = -3.4\ \text{eV}$ , and $E_3 = -13.6/3^2 = -1.51\ \text{eV}$ .

Step 2. For $n = 2 \to n = 1$ : $E_{\text{photon}} = E_2 - E_1 = -3.4 - (-13.6) = 10.2\ \text{eV}$ . This transition ends on $n_f = 1$ , so it belongs to the Lyman series — a large energy gap that, qualitatively, is known to lie in the ultraviolet.

Step 3. For $n = 3 \to n = 2$ : $E_{\text{photon}} = E_3 - E_2 = -1.51 - (-3.4) = 1.89\ \text{eV}$ . This transition ends on $n_f = 2$ , so it belongs to the Balmer series — a much smaller gap, qualitatively known to lie in the visible part of the spectrum (this particular line is the red Balmer-alpha line).

Answer: the $n = 2 \to 1$ transition releases the far more energetic photon, $10.2$ eV versus $1.89$ eV, because it drops all the way to the ground state; the Lyman-series photon is ultraviolet, while the Balmer-series photon is visible light — consistent with the fact that transitions landing on lower $n_f$ always release more energy than ones landing on higher $n_f$ . Timing note: under ninety seconds. The whole calculation is two subtractions once the three $E_n$ values are written down; do not be tempted to push further into an exact wavelength — the syllabus wants the energy comparison and the qualitative spectral region (UV/visible/IR), not a Rydberg-formula numeric wavelength.

Common traps

Confusing a continuous spectrum (produced by hot, dense matter such as a filament) with a line spectrum (produced by a rarefied, excited gas of free atoms) — the physical source and the resulting pattern are completely different.
Treating emission and absorption spectra of the same element as unrelated observations, when in fact the dark absorption lines sit at exactly the same wavelengths as the bright emission lines, because both come from the same set of allowed transitions.
Believing the Rutherford model explained atomic spectra — it explained nothing about them; the sharp-line nature of spectra is precisely the puzzle the Rutherford model could not solve, and it is what motivated Bohr's quantisation postulate.
Reaching for the Rydberg formula and grinding out exact wavelengths in nanometres — this syllabus keeps hydrogen line spectra at the qualitative level (series exist, and roughly where they sit), and only asks for numeric transition energies using $E_n = -13.6/n^2$ eV, not numeric wavelengths from $1/\lambda = R(1/n_1^2 - 1/n_2^2)$ .
Forgetting that a bigger drop in $n$ (especially landing on $n_f = 1$ ) always means a larger photon energy and hence a shorter wavelength — students sometimes assume, wrongly, that "further apart in $n$ " and "more energy" are unrelated.

Common trapA continuous-looking classical atom cannot produce sharp lines, no matter how you tune it

It is tempting to think that the Rutherford model just needed a small tweak — a faster orbit, a heavier nucleus — to explain line spectra. It cannot, in principle, no matter what you tune. In that model, a spiralling, continuously accelerating electron radiates continuously as it loses energy and falls inward, so its orbital radius, its orbital frequency, and hence the frequency of light it emits are all free to take any value as the spiral proceeds. That directly predicts a smooth, continuous spread of emitted wavelengths — precisely a continuous spectrum, not a line spectrum. The only way to forbid the electron from radiating away a continuous range of energies is to forbid it from having a continuous range of energies in the first place — which is exactly the quantisation postulate Bohr introduced. Sharp spectral lines are not weak evidence for quantised energy levels; they are direct, unavoidable evidence that classical continuous orbits cannot be the whole story.

You should be able to…

Distinguish a continuous spectrum from a line spectrum, and state what kind of source produces each.
Distinguish an emission spectrum from an absorption spectrum, and explain why their line positions coincide for a given element.
Explain, in words, why a classical (Rutherford) atom with a continuously spiralling electron cannot produce sharp spectral lines.
State the relation $E_{\text{photon}} = h\nu = E_i - E_f$ and use it to compare the energy of two given hydrogen transitions using $E_n = -13.6/n^2$ eV.
Name the Lyman, Balmer and Paschen series by their common final level and state, qualitatively, which region of the spectrum each occupies.
Explain why line spectra count as experimental evidence for quantised atomic energy levels.

Ask the tutor: want to see how astronomers actually use absorption lines in starlight to figure out what a star is made of, or how far away it is moving? Ask the AI tutor to connect atomic spectra to real observational astronomy.

Why can't the electron just have any energy it wants inside the atom?

That is exactly the puzzle Bohr's model was built to solve, and this article only shows you the consequence of quantisation, not why it happens at a deeper level — that deeper justification (in terms of standing electron waves) is usually met later, if at all, in school physics. For your board exam, it is enough to state the postulate as given: a bound electron in an atom occupies one of a fixed list of allowed energies $E_n$ , and treat the sharp spectral lines as the experimental evidence that convinced physicists this postulate must be true, since nothing in classical physics forces energy to be discrete.

If absorption and emission lines match, why do stars show absorption lines rather than emission lines when we look at them?

A star's hot, dense interior produces a continuous spectrum of light — much like the filament example in this article. That light then has to pass outward through the star's own cooler outer atmosphere, a much less dense gas of atoms, before it reaches us. Those cooler outer-atmosphere atoms absorb light at their characteristic wavelengths on the way out, carving dark absorption lines into the otherwise continuous spectrum. That is why the light astronomers receive from most stars, including the Sun, shows an absorption spectrum, and it is how astronomers identify which elements are present in a star without ever visiting it.

Do I need to memorise the Lyman, Balmer and Paschen series and their exact wavelengths for my board exam?

No — CBSE keeps hydrogen line spectra at a qualitative level for this chapter. You should be able to say that these series exist, that they correspond to sets of transitions landing on a common final level ( $n_f = 1$ for Lyman, $n_f = 2$ for Balmer, $n_f = 3$ for Paschen), and roughly which part of the spectrum each occupies (ultraviolet, visible, infrared respectively). You are not expected to compute or memorise exact numeric wavelengths using the Rydberg formula for this syllabus.

Is the photon energy formula

E = h\nu

the same idea as the photoelectric effect I studied earlier?

Yes, it is the very same relation, $E = h\nu$ , that you met when studying the photoelectric effect and the dual nature of radiation — Planck and Einstein's idea that light comes in discrete packets called photons, each carrying an energy proportional to its frequency. Here you are simply applying that same packet picture to light emitted or absorbed by an atom, where the packet's energy is fixed by the gap between two of the atom's allowed energy levels rather than by anything to do with a metal surface.