9 min read · about 1 h 30 min with practice3 quick checks≈2% of the testCore: Core: tested on most papers
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Electrons in atoms have only certain energies, jumps between them produce line spectra, and moving particles have a wavelength. Exam questions typically combine an energy-level calculation, an explanation (fluorescent tube or absorption spectra) and a de Broglie calculation in one structured question worth 8–12 marks.
By the end you’ll be able to
Explain excitation, de-excitation and ionisation, and calculate photon energies from energy-level diagrams
Distinguish continuous, emission and absorption spectra and explain the operation of a fluorescent tube
Use λ = h/p = h/mv for particles and calculate the de Broglie wavelength of accelerated electrons
Describe electron diffraction as evidence of wave behaviour and relate diffraction to atomic spacing
Apply the phasor/sum-over-paths model to explain reflection, refraction and diffraction of photons (OCR B)
What the exam asks
Excitation, de-excitation and ionisation, by photon absorption or electron collision.
Energy-level diagrams: photon energy, frequency and wavelength for a transition; the number of lines; the ionisation energy.
Line spectra: emission and absorption, and why they prove levels are discrete.
The fluorescent tube: a favourite AQA explanation worth 4–6 marks.
Wave–particle duality: λ=h/p, electron diffraction, and how diffraction changes with momentum.
OCR B: phasors (sum over paths). Eduqas and WJEC: lasers.
vii.Check your understanding
3 questions on energy levels, spectra and wave–particle duality. Every option is explained once you answer.
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PromptCard 1 of 3
Why are atomic energy levels given negative values?
=
E1−
E2
λ=h/mv
OCR A
λ=h/p
Edexcel 9PH0 and IAL
CCEA
Cambridge 9702
recall
λ=h/p
hf=E1−E2
OCR B
Eduqas and WJEC
Core ideas
Energy levels
An electron in an isolated atom can have only certain discrete energies, called energy levels. They are drawn as horizontal lines with negative values. Zero is the level at which the electron is just free (the ionisation level). The lowest level is the ground state.
Excitation: an electron moves up to a higher level.
De-excitation: an electron drops to a lower level and emits one photon whose energy equals the difference between the levels:
hf=λhc=E1−E2
Ionisation: the electron is removed completely. From the ground state, the ionisation energy equals the size of the ground-state energy (13.6 eV for hydrogen).
Excitation by a photon
Excitation by a colliding electron
Energy needed
ExactlyΔE: the photon is absorbed completely or not at all
At leastΔE: the electron keeps the rest as kinetic energy
Hydrogen, ground state
A 10.2 eV photon excites to n=2; an 11.0 eV photon passes straight through
An 11.0 eV electron can excite to n=2 and leaves with 0.8 eV
The exception: a photon with more than the ionisation energy can ionise the atom, and the surplus becomes the kinetic energy of the freed electron.
An electron excited to level n can return to the ground state by different routes, so the number of possible lines is 2n(n−1). From n=4 there are 6 lines.
Line spectra
Continuous spectrum: every wavelength, emitted by a hot dense source such as a filament.
Emission line spectrum: bright lines on a dark background, from a hot gas at low pressure. Each line is one transition.
Absorption line spectrum: dark lines on a continuous spectrum. It forms when white light passes through a cooler gas. The atoms absorb photons whose energies exactly match their transitions, then re-emit them in all directions, so fewer photons of those wavelengths travel on to the observer.
Only certain wavelengths appear, so only certain energy differences exist: this is the evidence for discrete levels. Each element’s lines are a fingerprint, used to find the composition of stars and measure red shift. A cool gas shows fewer absorption lines than emission lines, because almost all its atoms are in the ground state and only transitions starting from n=1 can absorb.
The fluorescent tube (AQA)
A high p.d. across the tube accelerates free electrons. They collide with mercury atoms and excite them (and ionise some, which releases more free electrons).
The mercury atoms de-excite and emit ultraviolet photons.
The phosphor coating on the inside of the glass absorbs the UV photons, and its atoms are excited to high levels.
The phosphor atoms de-excite in several smaller steps, emitting lower-energy visible photons.
Wave–particle duality
Light behaves as a wave (diffraction, interference, polarisation) and as a particle (the photoelectric effect).
Electrons behave as particles (they have a definite mass and charge and are deflected by fields) and as waves (they diffract).
De Broglie proposed that every moving particle has a wavelength:
λ=ph=mvh
For an electron accelerated from rest through a p.d. V (non-relativistic), eV=2mp2, so
λ=2meVh
Electron diffraction
A beam of electrons, accelerated through a few kilovolts, passes through a thin film of polycrystalline graphite in an evacuated tube. Concentric rings appear on a fluorescent screen. The rings are diffraction maxima, so the electrons are behaving as waves.
Diffraction is noticeable only when λ is similar to the spacing of the atoms (about 10−10 m). Electrons accelerated through about 150 V have exactly this wavelength.
Increasing V increases the speed and momentum, so λdecreases. There is less diffraction and the rings get smaller. For small angles the ring radius is proportional to λ, and so to 1/V.
A heavier particle at the same speed has a shorter wavelength.
De Broglie’s 1924 hypothesis was accepted only after electron diffraction experiments confirmed it in 1927: AQA’s example of ideas being validated by experiment and peer review.
OCR B: phasors and the sum over paths
A photon or electron “explores” every path from source to detector. Along each path a phasor rotates at frequency f; the number of turns is f× trip time (path length ÷ λ for light).
Add the phasors for all paths tip to tail. The probability of arrival is proportional to the square of the resultant amplitude.
Paths near the least-time path have almost equal trip times, so their phasors line up and add; paths far away cancel. This explains straight-line travel, equal angles of reflection and incidence, and refraction along the least-time path.
A narrow slit removes paths that would otherwise cancel, so the light spreads out (diffraction). Paths differing by λ/2 (or 1.5λ, 2.5λ …) give opposite phasors, which cancel.
Lasers (Eduqas and WJEC)
Stimulated emission: a passing photon of energy exactly ΔE makes an excited electron drop, emitting a second photon with the same frequency, phase, direction and polarisation.
Population inversion: more atoms in the upper (metastable) level than the lower laser level, so stimulated emission beats absorption. It needs pumping; a four-level system reaches it more easily because the lower laser level empties quickly.
Mirrors at each end (one partly transmitting) form a cavity that amplifies the light over many passes.
Worked examples
Exam technique
Energy differences: subtract the two (negative) level values, then convert to joules before using λ=hc/ΔE. The smallest gap gives the longest wavelength; visible light is roughly 1.8–3.1 eV.
Photon or electron? Always ask “exact match, or at least?”. This decides many multiple-choice items.
Fluorescent tube: a four-step chain (collisions excite mercury → UV → phosphor absorbs → cascades to visible) using excite, de-excite, photon, phosphor.
de Broglie with a p.d.: write p=2meV first, then λ=h/p.
“How does the pattern change?”: p.d. → speed and momentum → wavelength → diffraction angle → ring size, one step per mark.
Common mistakes
Quick recap
Levels are discrete and negative; 0 means ionised. A transition emits or absorbs a photon with hf=E1−E2.
A photon must match ΔEexactly. A colliding electron needs at leastΔE and keeps the rest.
Emission spectra are bright lines; absorption spectra are dark lines on a continuum. Both are evidence of discrete levels.
Fluorescent tube: electron collisions excite mercury → UV → phosphor → visible light in smaller steps.
λ=h/p=h/mv. An electron accelerated through V has λ=h/.
Electron diffraction rings show wave behaviour. A higher p.d. gives a shorter λ and smaller rings.
OCR B: probability ∝ (resultant phasor amplitude)², and paths near the least-time path dominate.