A scientist is investigating the photoelectric effect using a clean zinc metal surface. When ultraviolet light is directed at the zinc surface, electrons are emitted immediately. The scientist then reduces the intensity of the ultraviolet light to a very low level. Explain why electrons are still emitted immediately, even at very low intensity, and why a beam of visible light of much greater intensity produces no electron emission at all from the same surface.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Model answer (5 marks)
Light consists of photons, each carrying energy (E=h
u). The energy of a photon increases with its frequency. An electron in the metal can absorb only one photon; if that photon’s energy exceeds the metal’s work function, the electron is ejected immediately. Ultraviolet light has a high frequency, so its photons have enough energy to overcome zinc’s work function, so electrons are emitted even when the intensity is very low – the rate of emission is proportional to the photon flux, but each photon causes an instant ejection. Visible light has a lower frequency; its photons do not have sufficient energy to exceed the work function, so no electrons are emitted regardless of how many photons are present.
u). The energy of a photon increases with its frequency. An electron in the metal can absorb only one photon; if that photon’s energy exceeds the metal’s work function, the electron is ejected immediately. Ultraviolet light has a high frequency, so its photons have enough energy to overcome zinc’s work function, so electrons are emitted even when the intensity is very low – the rate of emission is proportional to the photon flux, but each photon causes an instant ejection. Visible light has a lower frequency; its photons do not have sufficient energy to exceed the work function, so no electrons are emitted regardless of how many photons are present.
Examiner tips
- Use the term ‘photon’ and ‘work function’ explicitly; link photon energy to frequency; explain single‑photon absorption; mention intensity affects count rate but not emission instantness; contrast UV and visible photon energies.
Common mistakes
- Confusing intensity with photon energy; saying electrons need many photons; forgetting that work function is a threshold; not using correct terminology such as ‘work function’ or ‘photon’.
Mark scheme (5 marks)
- Light (electromagnetic radiation) consists of discrete packets of energy called photons
- The energy of a photon depends on its frequency (higher frequency = higher energy photon)
- Each electron interacts with a single photon, so if that one photon has enough energy to overcome the work function, the electron is emitted immediately
- Ultraviolet photons have sufficient energy to exceed the work function of zinc, so electrons are released
- Visible light photons have lower frequency than ultraviolet, so each photon has insufficient energy to overcome the work function, and no electrons are emitted regardless of intensity
Key terms in this question
intensity · photoelectric effect
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