Energy Levels and Line Spectra

Electrons in an isolated atom can only have certain discrete energies, called energy levels (negative values, with zero meaning the electron is free). When an electron drops from level $E_1$ to a lower level $E_2$ , it emits a photon with $hf=E_1-E_2$ . Because only certain differences exist, only certain wavelengths appear: a line spectrum. An absorption spectrum shows dark lines where the atoms absorb exactly those photon energies from a continuous spectrum.

Cambridge 9702 Paper 4Edexcel YPH11 WPH12AQA 7408 Paper 1≈ 35 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Explain that isolated atoms have discrete electron energy levels.
Use $hf=E_1-E_2$ to find photon energies and wavelengths.
Describe how emission and absorption line spectra form.
Count possible transitions and define ionisation energy.

Pause and think

Street lamps with sodium are a distinctive orange, and fireworks with copper glow green. Why does each element give its own colours?

Section 2 of 14

Core idea

Start hereBarcodes of the elements

Pass light from a neon sign through a prism and you see a few sharp coloured lines, not a rainbow. Every element has its own pattern, like a barcode. Electrons can only sit at certain energies, and each line is a jump between two of them.

The big idea

Discrete energy levels; transitions emit or absorb one photon with $hf=E_1-E_2$ . Hence line spectra.

1. Levels and transitions

Hydrogen energy levels and transitions

Energies are negative: zero means the electron has escaped. Falling to the

-3.4

eV level gives visible light (the Balmer series).

hf=\frac{hc}\lambda=E_{\text{upper}}-E_{\text{lower}}

$E$: level energies (negative; zero = ionised).

Worked exampleHydrogen 3 → 2

Levels −1.51 eV and −3.40 eV.

Goal: Photon wavelength.
Strategy: Energy gap in J, then $\lambda=hc/E$ .

$\Delta E=3.40-1.51=1.89\text{ eV}=3.02\times10^{-19}\text{ J}$
Gap.
$\lambda=\frac{1.99\times10^{-25}}{3.02\times10^{-19}}=658\text{ nm}$
Red.

Answer

658

nm (the red H-alpha line).

✓ CheckVisible range 400–700 nm ✓.

Worked exampleLines from n = 4

Possible emission lines as electrons cascade down from $n=4$ .

Goal: Count.
Strategy: Count pairs of levels.

$4\to3,\ 4\to2,\ 4\to1,\ 3\to2,\ 3\to1,\ 2\to1$
All downward pairs.
$\binom42=6$
Choose 2 of 4 levels.

Answer6 lines.

✓ CheckFormula

\frac{n(n-1)}2

✓.

2. Emission and absorption

Emission spectrum (bright lines) and absorption spectrum (dark lines)

Same wavelengths in both: 410, 434, 486, 656 nm.

Emission: hot gas; bright lines.
Absorption: white light through cool gas; dark lines at the same wavelengths.
Ionisation energy of hydrogen: 13.6 eV.

Common misconceptionPartial absorption A photon must match a gap exactly (unless it ionises): 12.0 eV can't be absorbed by ground-state hydrogen.

Summary

$hf=\Delta E$ .
Line spectra.

Excitation: Electron raised to a higher level.

HistoryHelium found in the Sun first. In

1868

, during a solar eclipse, astronomers saw a yellow line in the Sun's spectrum at

587.5

nm that matched no known element. Norman Lockyer named it helium, after Helios, the Greek Sun god. It wasn't found on Earth until

1895

Fun factNeon signs. Only pure neon glows red-orange. Other "neon" sign colours come from different gases (argon blue, helium pink) or phosphor coatings. Each gas's electrons, excited by the high voltage, drop back down through their own energy levels.

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