Quantum physics questions become much easier when you identify the physical model before choosing an equation. This guide shows how to recognise the main A-Level question types, set out calculations, read graphs and write explanations that make your physics visible to the marker. Check your own exam board specification and formula sheet first, because notation, optional content, graphs and practical expectations are not identical across UK boards.
Match the wording to the quantum model
Before calculating, underline what is given, circle what must be found, then name the situation. A single photon with frequency or wavelength suggests E = hf or E = hc/λ. Electrons emitted from a metal suggest hf = φ + Ek(max), where φ is the work function, the minimum energy needed to release an electron from the metal surface. A stopping potential points to Ek(max) = eVs. At threshold, Ek(max) = 0, so φ = hf₀ and f₀ = φ/h. Moving electrons and diffraction point to λ = h/p = h/mv. Transition arrows or spectral lines mean that photon energy equals an energy-level difference. E = hf is often an intermediate result, not necessarily the final answer. For example, if the question asks for photoelectron kinetic energy, you must still account for the work function.
Calculate photon energy and photoelectron energy in the right order
Use a layout a marker can follow: write the equation, rearrange if needed, convert units, substitute values, calculate, give a unit, then state what the result represents. Keep full figures until the final answer, then round to suitable significant figures. If wavelength is supplied in nm, convert it to m before using E = hc/λ. The result is the energy of one photon, not the energy of a beam.
If light of a stated frequency shines on a metal with a stated work function, first calculate the photon energy with E = hf. Keep that photon-energy result as an intermediate step if the question asks for photoelectron kinetic energy and stopping potential.
Worked example: light of wavelength 400 nm falls on a metal with a work function of 2.3 eV. Photon energy E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (400 × 10⁻⁹) = 4.97 × 10⁻¹⁹ J. Convert the work function: 2.3 × 1.60 × 10⁻¹⁹ = 3.68 × 10⁻¹⁹ J. Then Ek(max) = 4.97 × 10⁻¹⁹ − 3.68 × 10⁻¹⁹ = 1.29 × 10⁻¹⁹ J, and the stopping potential is Vs = Ek(max) ÷ e = 1.29 × 10⁻¹⁹ ÷ 1.60 × 10⁻¹⁹ = 0.81 V. Finish with a sentence: the fastest photoelectrons have 1.29 × 10⁻¹⁹ J of kinetic energy, so a stopping potential of 0.81 V is needed to stop them.
The work function must be in the same unit before subtraction. Convert it when necessary, then use hf = φ + Ek(max) to find the maximum photoelectron kinetic energy. If a stopping potential is required, use Ek(max) = eVs. State that this is the stopping potential for the fastest photoelectrons. Above threshold frequency, increasing frequency increases photon energy and therefore the maximum kinetic energy and stopping potential, whereas increasing intensity increases the number of emitted photoelectrons, not their maximum kinetic energy.
Extract physics from threshold and stopping-potential graphs
Read both axis labels and units before recalling a relationship. For an Ek(max)-against-frequency graph, rearrange the photoelectric equation as Ek(max) = hf − φ. With SI axes, the gradient is h, with unit J Hz⁻¹, and the y-intercept, in J, is −φ. For a stopping-potential-against-frequency graph, divide by e: Vs = (h/e)f − φ/e. Its gradient is h/e, with unit V Hz⁻¹, its y-intercept is −φ/e in V, and its frequency-axis intercept is the threshold frequency in Hz. Deriving the line is safer than memorising intercepts, because axes or units may differ.
If an extrapolated line crosses the frequency axis at 5.2 × 10¹⁴ Hz, write that 5.2 × 10¹⁴ Hz is the threshold frequency for that metal. Below it, photons do not have enough energy to overcome the work function, so photoelectrons are not emitted. For an Ek(max)-against-frequency graph, the gradient gives h and the work function is the negative of the y-intercept.
Link momentum to electron diffraction
For an electron with a stated mass and speed, use two stages. Calculate momentum first: p = mv, in kg m s⁻¹. Then calculate λ = h/p and give the wavelength in m in standard form. Do not substitute mass and speed into λ = h/p without showing where p came from if the question asks for momentum or full working.
Worked example: an electron of mass 9.11 × 10⁻³¹ kg moves at 2.0 × 10⁶ m s⁻¹. Momentum p = mv = 9.11 × 10⁻³¹ × 2.0 × 10⁶ = 1.82 × 10⁻²⁴ kg m s⁻¹. Wavelength λ = h/p = 6.63 × 10⁻³⁴ ÷ 1.82 × 10⁻²⁴ = 3.6 × 10⁻¹⁰ m. This is similar to the spacing between atoms in a crystal, which is why these electrons diffract noticeably.
For an explanation, link each idea: greater momentum means a smaller de Broglie wavelength, which changes the diffraction pattern. Lower momentum gives a larger wavelength and greater spreading. State the direction requested by the question. Beam intensity concerns how many electrons arrive; it does not set the de Broglie wavelength of an individual electron.
Explain atomic energy levels and spectral lines
For a transition calculation, find the stated energy difference, check whether it is in J or eV, then use hf = ΔE. Convert between units when necessary, then calculate the frequency of the emitted or absorbed photon. If wavelength is needed, use c = fλ after finding f. Keep energy units consistent with the equation. If a calculated frequency is negative, check the energy-difference convention, the direction of the transition and the units.
Worked example: an electron falls from an energy level at −1.51 eV to one at −3.40 eV. ΔE = 1.89 eV = 1.89 × 1.60 × 10⁻¹⁹ = 3.02 × 10⁻¹⁹ J. The emitted photon has f = ΔE/h = 3.02 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ = 4.56 × 10¹⁴ Hz and λ = c/f = 3.00 × 10⁸ ÷ 4.56 × 10¹⁴ = 6.58 × 10⁻⁷ m, which is 658 nm, red light.
A strong written explanation follows a chain. Atomic energy levels are discrete, meaning electrons can occupy particular permitted energies. An electron absorbs a photon to move up a level, or emits a photon when it drops down. Conservation of energy means hf equals the difference between the two levels. Different transitions have different energy differences, so they produce or absorb particular frequencies and wavelengths. The observed emission or absorption spectrum therefore contains separate lines rather than a continuous range.
Write answers that expose every available mark
Keep full figures through a multi-stage calculation and round only the final answer. Use “maximum kinetic energy” rather than simply “kinetic energy”, because emitted photoelectrons can have a range of kinetic energies and only the fastest have Ek(max).
Match the command word and give the requested level of detail. For a longer response, use claim, relevant principle or equation, consequence, then a conclusion tied to the observation.
Turn marked errors into targeted quantum practice
After marking one question against an official exam-board mark scheme, record the exact missing step, such as “used eV in a joule equation” or “called the intercept h”. If you use an older paper or mark scheme, check that it matches your current specification. Use A-Level revision resources and SimpleStudy’s A-Level topic notes and quizzes for the exact quantum subtopic, then attempt one similar question under timed conditions. Add either the new error or the successful method to the same record.
Before submitting the next answer, check:
- I identified the correct model, not just the first equation I recognised.
- All quantities use compatible units, especially J and eV.
- Working shows conversions, substitutions and sensible significant figures, and I have not reversed a graph gradient.
- The final value has a unit and a sentence explaining its physical meaning.
The aim is to make each choice visible: photon energy, energy balance, stopping potential, matter wave or energy-level transition. Practise the model decision first, then the arithmetic, and use each marked answer to target one precise improvement.
