Explain why a dim source of light with a high enough frequency can release electrons from a metal surface, but a very bright source of light with a lower frequency cannot.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
In the photoelectric effect, the emission of electrons from a metal surface depends on the properties of the light used.
Model answer (5 marks)
Light is quantised – it consists of photons.
The energy of each photon is given by E=hf, where h is Planck’s constant and f is the frequency of the light. The intensity (brightness) of the light only tells how many photons reach the surface per unit time; it does not change the energy of each photon.
For an electron to be emitted from a metal, a photon must supply at least the work function (ϕ) of that metal. If hf>ϕ, the electron can be released; if hf<ϕ, no electron can be ejected, no matter how many photons arrive.
Thus a dim source of high‑frequency light contains photons whose energy exceeds the work function, so electrons are emitted. A bright source of low‑frequency light contains photons whose energy is below the work function, so no electrons are emitted, even though the intensity is high.
The energy of each photon is given by E=hf, where h is Planck’s constant and f is the frequency of the light. The intensity (brightness) of the light only tells how many photons reach the surface per unit time; it does not change the energy of each photon.
For an electron to be emitted from a metal, a photon must supply at least the work function (ϕ) of that metal. If hf>ϕ, the electron can be released; if hf<ϕ, no electron can be ejected, no matter how many photons arrive.
Thus a dim source of high‑frequency light contains photons whose energy exceeds the work function, so electrons are emitted. A bright source of low‑frequency light contains photons whose energy is below the work function, so no electrons are emitted, even though the intensity is high.
Examiner tips
- Mention photons and E=hf; link frequency to energy, not intensity; state work function condition; give example of high‑freq dim vs low‑freq bright.
Common mistakes
- Confusing intensity with photon energy; writing that more photons mean more energy per electron; neglecting the work function requirement.
Mark scheme (5 marks)
- Light is made up of photons (discrete packets/quanta of energy)
- The energy of a photon depends on its frequency (not intensity)
- A single photon must have enough energy to overcome the work function of the metal in order to release an electron
- The high-frequency (dim) source has photons with enough energy to exceed the work function, so electrons are released
- The low-frequency (bright) source has photons with insufficient energy to overcome the work function, so no electrons are released regardless of intensity
Key terms in this question
Related
- All AQA A-Level Physics (7408) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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