Physics 9702 · syllabus 22
Cambridge A Level Physics 9702 — Quantum physics topical past papers
Light and matter as both wave and particle: photon energy and momentum, the photoelectric effect, de Broglie wavelengths and atomic energy levels.
- Questions
- 145
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 35
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What Quantum physics covers
Electromagnetic radiation is emitted and absorbed in packets called photons, each carrying energy E = hf and momentum p = E/c. At this scale the electronvolt is a convenient unit of energy: it is the energy an electron gains in moving through a p.d. of one volt, 1.60 × 10⁻¹⁹ J.
The photoelectric effect is the evidence. Radiation falling on a metal releases electrons only if each photon's energy exceeds the work function, so there is a threshold frequency below which no electrons appear however intense the radiation. Above it, the maximum kinetic energy of the photoelectrons depends on the frequency, not the intensity, while the number emitted per second, and so the photoelectric current, is proportional to the intensity. The equation hf = Φ + ½mvₘₐₓ² sums this up.
Interference and diffraction show light behaving as a wave; the photoelectric effect shows it behaving as particles. Matter has the same dual nature: electrons diffracted by thin graphite show that moving particles have a wavelength λ = h/p. Electrons in isolated atoms occupy discrete energy levels, and a transition between levels emits or absorbs a photon of energy hf = E₁ − E₂, which gives line spectra. This is A Level content, examined on Paper 4.
In the syllabus
The headings this topic has in the Cambridge International 2025–2027 syllabus, as the board prints them.
- 22 Quantum physics
- 22.1 Energy and momentum of a photon
- 22.2 Photoelectric effect
- 22.3 Wave-particle duality
- 22.4 Energy levels in atoms and line spectra
What you should be able to do
- Use E = hf = hc/λ and p = E/c, and convert between joules and electronvolts
- Explain the photoelectric effect in terms of photons and the work function, including why a wave model fails
- Use hf = Φ + ½mvₘₐₓ² to find a threshold frequency, threshold wavelength or maximum kinetic energy
- Explain why the maximum kinetic energy is independent of intensity while the photoelectric current is proportional to it
- Describe electron diffraction as evidence for the wave nature of particles, and use λ = h/p
- Explain emission and absorption line spectra, and calculate photon energies from an energy-level diagram
Key ideas
- Photon energy E = hf = hc/λ, with h = 6.63 × 10⁻³⁴ J s; photon momentum p = E/c
- 1 eV = 1.60 × 10⁻¹⁹ J
- Photoelectric equation hf = Φ + ½mvₘₐₓ²; threshold frequency f₀ = Φ/h
- de Broglie wavelength λ = h/p = h/mv
- Line spectra: hf = E₁ − E₂ for a transition between levels of energy E₁ and E₂
How Quantum physics has been examined
From 2022 to 2026, Quantum physics was on Paper 4 · A Level Structured Questions in 11 of 11 exam series, worth about 10% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 1 · Multiple Choice | 4 | 1% | 2 of 11 |
| Paper 4 · A Level Structured Questions | 31 | 10% | 11 of 11 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 10.
Filed in Momentum under: Quantum Physics.
A worked example
Ultraviolet radiation of wavelength 250 nm falls on a metal surface with a work function of 3.0 eV. Find the maximum kinetic energy of the emitted electrons, in eV, and the threshold wavelength of the metal.
- Photon energy: E = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 250 × 10⁻⁹ = 7.96 × 10⁻¹⁹ J
- Work function in joules: Φ = 3.0 × 1.60 × 10⁻¹⁹ = 4.80 × 10⁻¹⁹ J
- Maximum kinetic energy = 7.96 × 10⁻¹⁹ − 4.80 × 10⁻¹⁹ = 3.16 × 10⁻¹⁹ J = 1.97 eV
- Threshold wavelength: λ₀ = hc/Φ = 1.989 × 10⁻²⁵ / 4.80 × 10⁻¹⁹ = 4.14 × 10⁻⁷ m
Answer: Maximum kinetic energy 2.0 eV; threshold wavelength 414 nm
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Saying brighter light gives photoelectrons more energy — intensity sets the number of photons per second and so the current; each electron's maximum energy depends only on the frequency
- Subtracting a work function in eV from a photon energy in joules — convert one so the units match
- Treating ½mvₘₐₓ² as the energy of every emitted electron — the work function is the minimum energy needed to escape, so only the fastest electrons have the maximum kinetic energy
- Using E = hc/λ to relate an electron's kinetic energy to its wavelength — that formula is for photons; for a particle with mass use λ = h/p, with p = √(2mEₖ)
- Counting only the transitions down to the ground state when listing emission lines — any downward jump between two levels gives a line
Exam technique
- When explaining the photoelectric effect, say that one photon interacts with one electron and describe the energy transfer; this is the step a wave model cannot explain
- Set out energy-level calculations as ΔE in eV, then joules, then f or λ, showing each conversion
- For 'explain how electron diffraction shows…', name the observation, a pattern of rings like that from light, and state that diffraction is a wave property
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.