Explain work function and threshold frequency
Define the work function and use threshold frequency and wavelength to predict photoelectric emission.
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 22. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use wave frequency and wavelength, energy, momentum, SI prefixes, electronvolts and scientific notation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the work function of a metal surface?
What you will learn
- Describe photoelectron emission from an illuminated metal
- Define threshold frequency and threshold wavelength
- Relate work function to the minimum photon energy
One photon can release an electron if it supplies enough energy
When electromagnetic radiation illuminates a metal, an electron may be emitted from the surface. A photon transfers its energy to one electron. The minimum energy needed to release an electron from that metal surface is the work function Φ.
For emission, a photon must supply at least the work function energy: hf ≥ Φ. If the photon energy is below Φ, that photon cannot release an electron in the basic photoelectric model. Work function depends on the metal surface, not on the intensity of the incident radiation.
Threshold values mark the emission boundary
The threshold frequency f₀ is the minimum radiation frequency that can cause photoelectric emission from a given surface. At the threshold, hf₀ = Φ. Frequencies below f₀ do not emit photoelectrons, however intense the light is in this model.
The corresponding threshold wavelength λ₀ is the maximum wavelength that can cause emission. Since c = fλ, λ₀ = c/f₀ = hc/Φ. Shorter wavelengths have higher frequency and photon energy. Convert Φ to joules when using h in J s and c in m s⁻¹.
Pause and explain
What happens if incident photon frequency is below the threshold frequency?
Worked example
A metal has work function 2.0 eV. Find its threshold frequency and threshold wavelength. Use h = 6.63 × 10⁻³⁴ J s and c = 3.00 × 10⁸ m s⁻¹.
Show the worked solution
- Convert Φ = 2.0 eV to 3.2 × 10⁻¹⁹ J.
- Use f₀ = Φ/h.
- Use λ₀ = c/f₀.
Answer f₀ ≈ 4.82 × 10¹⁴ Hz and λ₀ ≈ 6.22 × 10⁻⁷ m (622 nm).
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A metal's work function is 3.2 × 10⁻¹⁹ J. Calculate its threshold frequency.
Give me a hint
At threshold, hf₀ = Φ.
Compare my reasoning
- f₀ = Φ/h.
- f₀ = 3.2 × 10⁻¹⁹/(6.63 × 10⁻³⁴).
- f₀ ≈ 4.83 × 10¹⁴ Hz.
f₀ ≈ 4.8 × 10¹⁴ Hz.
Look for these in your work
- Threshold equation selected
- Values substituted in SI
- Frequency and unit stated
Radiation of wavelength 700 nm illuminates a surface whose threshold wavelength is 500 nm. Will photoelectrons be emitted? Explain.
Give me a hint
Compare wavelengths through frequency: a shorter wavelength means a higher frequency.
Compare my reasoning
- The incident wavelength is longer than the threshold wavelength.
- Its frequency is therefore below the threshold frequency.
- Each photon has less energy than the work function, so no emission occurs in the basic model.
No. 700 nm is longer than 500 nm, so the photon frequency and energy are below threshold.
Look for these in your work
- Wavelength comparison made
- Frequency relation used
- Emission prediction justified
A student doubles the intensity of light but keeps its frequency below a metal's threshold. Predict the emission and explain why.
Give me a hint
Apply the threshold condition to each photon, not to the total beam energy.
Compare my reasoning
- Each photon still has energy hf below Φ.
- Greater intensity means more photons arrive per second.
- No individual photon can supply the missing work-function energy, so photoelectrons are still not emitted.
There is still no photoelectric emission in the basic model because each photon remains below the work-function energy.
Look for these in your work
- Per-photon energy considered
- Intensity interpreted as photon rate
- No-emission result explained
Common mistakes
- Assuming any frequency emits if intensity is high enough
- Calling threshold wavelength the minimum wavelength
- Using work function in eV with Planck's constant in J s
- Confusing threshold frequency with the frequency of emitted electrons
What can you explain now?
What is the work function of a metal surface?
Compare with the explanation
Minimum energy needed to release an electron from its surface
A photon must supply at least this energy for emission.
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Make it stick
How do the work function and threshold frequency set the boundary for photoelectric emission?
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