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How does the photoelectric effect show that light behaves as particles?

The photon model of light, the photoelectric effect, the work function and threshold frequency, and the photoelectric equation for the kinetic energy of emitted electrons.

An SQA Higher Physics answer on wave-particle duality and the photoelectric effect, covering the photon model of light, the work function and threshold frequency, and the photoelectric equation for the kinetic energy of emitted electrons.

Generated by Claude Opus 4.811 min answer

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

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  1. What this key area is asking
  2. The photon model
  3. The photoelectric effect
  4. The photoelectric equation
  5. Examples in context
  6. Try this

What this key area is asking

The SQA wants you to describe the photon model of light, explain the photoelectric effect including the work function and threshold frequency, and apply the photoelectric equation Ek=hf−WE_k = hf - W to find the maximum kinetic energy of emitted electrons.

The photon model

This particle picture sits alongside the wave picture (interference and diffraction): light shows wave-particle duality, behaving as a wave in some experiments and as particles in others. The photoelectric effect is the key evidence for the particle behaviour, because the wave model cannot explain it.

The photoelectric effect

When light of high enough frequency shines on a metal surface, electrons are ejected. These are called photoelectrons. The crucial observations are that emission depends on the frequency, not the intensity, of the light, and that it happens instantly once the frequency is high enough.

This one-photon-one-electron rule is why intensity cannot substitute for frequency. The wave model wrongly predicted that any frequency would eventually free electrons if the light were bright enough, which experiment contradicts.

The photoelectric equation

If the photon energy exceeds the work function, the surplus appears as the kinetic energy of the ejected electron.

This is just conservation of energy: photon energy in equals energy to escape (the work function) plus kinetic energy of the electron. If hf<Whf < W there is no emission; if hf=Whf = W electrons are freed with zero kinetic energy at the threshold; and the larger the photon energy above the work function, the faster the ejected electrons.

Examples in context

Solar cells and light meters rely on light ejecting or freeing charge carriers, an application of the photon-energy idea. Photomultiplier tubes in detectors use the photoelectric effect to turn faint light into a measurable electrical pulse, used in medical imaging and particle physics. Automatic doors and night-lights use photoelectric sensors that respond when light above the threshold frequency strikes a surface. The photoelectric effect was the work for which Einstein won the Nobel Prize, because it provided the decisive evidence that light energy is quantised into photons.

Try this

Q1. Calculate the energy of a photon of frequency 5.0×1014 Hz5.0 \times 10^{14}\ \text{Hz}. Take h=6.63×10−34 J sh = 6.63 \times 10^{-34}\ \text{J s}. [2 marks]

  • Cue. E=hf=6.63×10−34×5.0×1014=3.3×10−19 JE = hf = 6.63 \times 10^{-34} \times 5.0 \times 10^{14} = 3.3 \times 10^{-19}\ \text{J}.

Q2. State what is meant by the work function of a metal. [1 mark]

  • Cue. The minimum energy needed to free an electron from the metal surface.

Q3. A metal has a work function of 3.0×10−19 J3.0 \times 10^{-19}\ \text{J}. A photon of energy 5.0×10−19 J5.0 \times 10^{-19}\ \text{J} strikes it. Calculate the maximum kinetic energy of the ejected electron. [2 marks]

  • Cue. Ek=hf−W=5.0×10−19−3.0×10−19=2.0×10−19 JE_k = hf - W = 5.0 \times 10^{-19} - 3.0 \times 10^{-19} = 2.0 \times 10^{-19}\ \text{J}.

Exam-style practice questions

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

SQA Higher 20184 marksA metal surface has a work function of 3.2 times ten to the power minus nineteen J. Light of frequency 7.0 times ten to the power fourteen Hz is shone on it. Calculate the maximum kinetic energy of the emitted photoelectrons. Take h as 6.63 times ten to the power minus thirty-four J s.
Show worked answer →

Use the photoelectric equation.

Relationship: Ek=hf−WE_k = hf - W, where WW is the work function.

Substitution: Ek=(6.63×10−34×7.0×1014)−3.2×10−19E_k = (6.63 \times 10^{-34} \times 7.0 \times 10^{14}) - 3.2 \times 10^{-19}.

Photon energy: hf=4.64×10−19hf = 4.64 \times 10^{-19} J. So Ek=4.64×10−19−3.2×10−19E_k = 4.64 \times 10^{-19} - 3.2 \times 10^{-19}.

Answer: Ek=1.4×10−19E_k = 1.4 \times 10^{-19} J.

Markers reward calculating the photon energy with hfhf, subtracting the work function, and a positive kinetic energy with unit.

SQA Higher 20214 marksExplain what is meant by the threshold frequency of a metal, and explain why no electrons are emitted from a metal surface when light below the threshold frequency is used, no matter how intense the light is.
Show worked answer →

The threshold frequency is the minimum frequency of light needed to release electrons from a metal surface; it corresponds to a photon energy equal to the work function, hf0=Whf_0 = W.

In the photon model, each electron absorbs a single photon. If the photon energy hfhf is less than the work function, that single photon cannot supply enough energy to free the electron, so no emission occurs. Increasing the intensity only increases the number of photons per second, not the energy of each photon, so below the threshold frequency still no electrons are emitted.

Markers reward defining threshold frequency via hf0=Whf_0 = W, the one-photon-one-electron idea, and explaining that intensity adds more photons but not more energy per photon.

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