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What does the photoelectric effect tell us about light?

The photon model and E=hfE = hf, the photoelectric effect and the photoelectric equation, threshold frequency and work function, and the electronvolt and atomic energy levels.

A focused answer to the Edexcel 9PH0 quantum content, covering the photon model and E=hfE = hf, the photoelectric effect and its equation, threshold frequency and work function, the electronvolt, and atomic energy levels.

Generated by Claude Opus 4.812 min answer

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

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  1. What this dot point is asking
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What this dot point is asking

Edexcel wants you to use the photon model and E=hfE = hf, explain the photoelectric effect and apply the photoelectric equation, define threshold frequency and work function, use the electronvolt, and relate photon emission and absorption to atomic energy levels.

The answer

The photon model

The photon model says light energy is delivered in indivisible lumps, not continuously as the wave model assumes. The intensity of a beam sets how many photons arrive per second; the frequency sets the energy of each one.

The photoelectric effect

These facts defeated the wave model, which predicted that any frequency would eventually free electrons given enough intensity, and that brighter light would give faster electrons. The photon model explains them: one photon transfers all its energy to one electron in a single interaction, so only a photon above the threshold energy can free an electron, and it does so at once.

The photoelectric equation

A graph of Ek,max⁡E_{k,\max} against ff is a straight line of gradient hh with xx-intercept f0f_0 and yy-intercept −ϕ-\phi, which is how the Planck constant is measured experimentally.

The electronvolt and energy levels

Because the levels are discrete, atoms emit and absorb only specific photon energies, producing the line spectra used to identify elements. The largest jump (to ionisation) sets the ionisation energy.

Examples in context

Solar cells rely on the photoelectric (photovoltaic) effect to convert light into electricity. Digital camera sensors and light meters count photoelectrons. The line spectra produced by electron transitions let astronomers identify elements in distant stars and detect their motion by Doppler shift. Fluorescent lights and LEDs emit photons of specific energies from electron transitions. Einstein's explanation of the photoelectric effect, not relativity, earned him the Nobel Prize.

Try this

Q1. State what is meant by the work function of a metal. [1 mark]

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

Q2. Find the energy of a photon of frequency 5.0×10145.0 \times 10^{14} Hz. Take h=6.63×10−34h = 6.63 \times 10^{-34} J s. [2 marks]

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

Q3. Explain why no electrons are emitted below the threshold frequency, however bright the light. [2 marks]

  • Cue. Each photon gives all its energy to one electron; below the threshold a single photon has less than the work function, so no electron can be freed regardless of how many arrive.

Exam-style practice questions

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

Edexcel 20184 marksA metal has a work function of 2.02.0 eV. Light of frequency 8.0×10148.0 \times 10^{14} Hz strikes it. Calculate the maximum kinetic energy of the emitted photoelectrons. Take h=6.63×10−34h = 6.63 \times 10^{-34} J s and 11 eV =1.6×10−19= 1.6 \times 10^{-19} J.
Show worked answer →

Photon energy: E=hf=6.63×10−34×8.0×1014=5.30×10−19E = hf = 6.63 \times 10^{-34} \times 8.0 \times 10^{14} = 5.30 \times 10^{-19} J.

Work function in joules: ϕ=2.0×1.6×10−19=3.2×10−19\phi = 2.0 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} J.

Photoelectric equation: Ek,max⁡=hf−ϕ=5.30×10−19−3.2×10−19=2.1×10−19E_{k,\max} = hf - \phi = 5.30 \times 10^{-19} - 3.2 \times 10^{-19} = 2.1 \times 10^{-19} J (about 1.31.3 eV).

Markers reward E=hfE = hf, converting the work function to joules, and Ek,max⁡=hf−ϕE_{k,\max} = hf - \phi.

Edexcel 20215 marksExplain why the photoelectric effect cannot be explained by the wave model of light, and determine the threshold frequency of a metal with work function 3.03.0 eV. Take h=6.63×10−34h = 6.63 \times 10^{-34} J s and 11 eV =1.6×10−19= 1.6 \times 10^{-19} J.
Show worked answer →

The wave model predicts that any frequency, given enough intensity or time, should eventually free electrons, and that brighter light should give faster electrons. Experiment shows neither: below a threshold frequency no electrons are emitted however bright the light, and emission is instantaneous above it. This is explained by the photon model: one photon gives all its energy hfhf to one electron, so only photons above the threshold energy can free an electron.

Threshold frequency: emission just begins when hf0=ϕhf_0 = \phi, so f0=ϕh=3.0×1.6×10−196.63×10−34=7.2×1014f_0 = \frac{\phi}{h} = \frac{3.0 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} = 7.2 \times 10^{14} Hz.

Markers reward the failure of the wave model (threshold and instant emission), the one-photon-one-electron idea, and f0=ϕh=7.2×1014f_0 = \frac{\phi}{h} = 7.2 \times 10^{14} Hz.

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