A student shines ultraviolet light onto a negatively charged metal plate. The plate loses its charge. When the student replaces the ultraviolet light with red light of much greater intensity, the plate no longer loses its charge. Explain these observations in terms of photons and the photoelectric effect.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Model answer (5 marks)
1. Light is made of photons.
2. Each photon carries energy E = hν, which depends on its frequency.
3. UV photons have a high ν, so their E exceeds the metal’s work function and electrons are ejected.
4. Red light has a lower ν, so each photon’s E is below the work function.
5. Because no single red photon can supply enough energy, increasing intensity (more photons) does not help – electrons cannot accumulate energy from several photons.
2. Each photon carries energy E = hν, which depends on its frequency.
3. UV photons have a high ν, so their E exceeds the metal’s work function and electrons are ejected.
4. Red light has a lower ν, so each photon’s E is below the work function.
5. Because no single red photon can supply enough energy, increasing intensity (more photons) does not help – electrons cannot accumulate energy from several photons.
Examiner tips
- Use the formula E = hν to show the link between frequency and photon energy.
- Explain why intensity (number of photons) does not affect the ability to eject electrons.
- Show the step‑by‑step reasoning: UV → enough energy → electrons ejected; red → insufficient energy → no ejection.
Common mistakes
- Confusing intensity with frequency; claiming more photons can raise energy.
- Saying electrons can absorb multiple photons to reach the work function.
- Using the wrong equation for photon energy (e.g., E = hc/λ without recognising λ relates to ν).
Mark scheme (5 marks)
- Light (electromagnetic radiation) is composed of discrete packets of energy called photons
- Each photon has a specific energy that depends on the frequency (not intensity) of the light
- Ultraviolet photons have sufficient energy to exceed the work function of the metal, so electrons are released
- Red light has a lower frequency than ultraviolet, so each red photon has less energy than a UV photon
- Red photons cannot release electrons regardless of intensity because no single photon has enough energy to overcome the work function / electrons cannot accumulate energy from multiple photons
Key terms in this question
photon · photoelectric effect · intensity
Related
- All WJEC A-Level Physics (Wales) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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