AQA AQA 2018-style practice: Light of frequency 9.0 × 10^14 Hz is shone on a metal surface of work function 3.4 × 10^-19 J. Calculate the maximum kinetic energy of the emitted photoelectrons. Take h = 6.63 × 10^-34 J s.

Find the photon energy: hf = (6.63 × 10^-34)(9.0 × 10^14) = 5.97 × 10^-19 J. Apply the photoelectric equation E_k(max) = hf - = 5.97 × 10^-19 - 3.4 × 10^-19. E_k(max) = 2.6 × 10^-19 J. Markers reward calculating the photon energy, subtracting the work function, and a positive answer (confirming the frequency is above threshold).

EnglandPhysicsSyllabus dot point

Why does light release electrons from a metal only above a threshold frequency, and how does this prove light is made of photons?

The photoelectric effect, the threshold frequency and work function, the photoelectric equation, and how the effect provides evidence for the particle nature of electromagnetic radiation.

A focused answer to AQA A-Level Physics 3.2.2.1, covering the photoelectric effect, threshold frequency and work function, the photoelectric equation, and why the wave model fails while the photon model succeeds.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The photoelectric effect
Threshold frequency and work function
The photoelectric equation
Why the wave model fails
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What this dot point is asking

AQA specification point 3.2.2.1 wants you to describe the photoelectric effect, define the threshold frequency and work function, use the photoelectric equation, and explain why the observations require the photon model rather than the wave model.

The photoelectric effect

Threshold frequency and work function

Different metals have different work functions, so the same light may eject electrons from one metal but not another. The work function is typically a few electronvolts.

The photoelectric equation

Plotting $E_{k(max)}$ against frequency $f$ gives a straight line of gradient $h$ (the Planck constant) and an x-intercept at the threshold frequency $f_0$ , which is a classic experimental confirmation of the photon model.

Why the wave model fails

The wave model predicts that any frequency of light would eventually free electrons if intense enough, and that brighter light would give faster electrons. Experiment shows neither: below $f_0$ no electrons appear at any intensity, and above $f_0$ emission is instant. The photon model explains this because each electron absorbs one photon of energy $hf$ ; only if $hf \geq \phi$ can it escape. Intensity simply sets the number of photons (and so electrons), not their individual energy.

Finding the maximum kinetic energy of a photoelectron

Light of frequency $f = 8.0 \times 10^{14} \text{ Hz}$ hits a metal of work function $\phi = 3.0 \times 10^{-19} \text{ J}$ . Find the maximum kinetic energy of the photoelectrons. Take $h = 6.63 \times 10^{-34} \text{ J s}$ .

step 1: Find the photon energy

$hf = (6.63 \times 10^{-34})(8.0 \times 10^{14}) = 5.3 \times 10^{-19} \text{ J}$ .

step 2: Compare with the work function

Since $hf = 5.3 \times 10^{-19} \text{ J}$ exceeds $\phi = 3.0 \times 10^{-19} \text{ J}$ , electrons are emitted.

step 3: Apply the photoelectric equation

$E_{k(max)} = hf - \phi = 5.3 \times 10^{-19} - 3.0 \times 10^{-19}$ .

step 4: Evaluate

$E_{k(max)} = 2.3 \times 10^{-19} \text{ J}$ .

Try this

Q1. Define the work function of a metal. [1 mark]

Cue. The minimum energy needed to remove an electron from the surface.

Q2. Explain why no electrons are emitted below the threshold frequency, however intense the light. [3 marks]

Cue. Each electron absorbs one photon of energy $hf$ ; below $f_0$ this is less than the work function, so no electron can escape regardless of intensity.

Q3. State what the gradient of a graph of maximum kinetic energy against frequency represents. [1 mark]

Cue. The Planck constant $h$ .

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20184 marksLight of frequency

9.0 \times 10^{14} \text{ Hz}

is shone on a metal surface of work function

3.4 \times 10^{-19} \text{ J}

. Calculate the maximum kinetic energy of the emitted photoelectrons. Take

h = 6.63 \times 10^{-34} \text{ J s}

Show worked answer →

Find the photon energy: $hf = (6.63 \times 10^{-34})(9.0 \times 10^{14}) = 5.97 \times 10^{-19} \text{ J}$ .

Apply the photoelectric equation $E_{k(max)} = hf - \phi = 5.97 \times 10^{-19} - 3.4 \times 10^{-19}$ .

$E_{k(max)} = 2.6 \times 10^{-19} \text{ J}$ .

Markers reward calculating the photon energy, subtracting the work function, and a positive answer (confirming the frequency is above threshold).

AQA 20214 marksExplain how the photoelectric effect provides evidence for the particle nature of electromagnetic radiation, referring to the threshold frequency and the wave model.

Show worked answer →

The wave model predicts that light of any frequency would eventually release electrons if it were intense enough, and that brighter light would give more energetic electrons. Neither is observed: below the threshold frequency no electrons are emitted at any intensity, and emission is instantaneous above it.

The photon model explains this: each electron absorbs a single photon of energy $hf$ . Only if $hf$ is at least the work function can the electron escape, so there is a threshold frequency. Brighter light means more photons (more electrons per second), not more energetic ones.

This one-to-one, quantised energy transfer is particle-like behaviour, so the photoelectric effect is evidence for the particle nature of light.

Markers reward the failures of the wave model, the one photon per electron mechanism, the threshold condition, and the conclusion about particle behaviour.

Related dot points

Sources & how we know this

AQA A-level Physics (7408) specification — AQA (2017)