Calculate photoelectric maximum kinetic energy
Apply Einstein's photoelectric equation to connect photon energy, work function and the fastest emitted electrons.
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 does the excess photon energy hf − Φ become for the fastest emitted photoelectrons?
What you will learn
- Use hf = Φ + ½mvmax²
- Calculate maximum photoelectron kinetic energy
- Identify the threshold case and check energy units
Photon energy is shared between release and electron motion
For photoelectric emission, the incoming photon supplies energy hf. A minimum amount Φ releases the electron from the metal. Any excess becomes the maximum kinetic energy of the emitted photoelectron: hf = Φ + ½m vₘₐₓ².
Therefore Kmax = ½m vₘₐₓ² = hf − Φ. At threshold, hf = Φ and Kmax = 0. Above threshold, Kmax is positive. The equation describes the maximum kinetic energy; emitted electrons can have smaller kinetic energies if they lose energy inside the metal before escaping.
Keep the photon and work-function units consistent
If h is in J s, calculate hf in joules. Convert a work function in eV to joules before subtracting, or convert both energies to eV and use a consistent photon-energy conversion. The difference must be positive for emission.
If electron mass and speed are required, use m = 9.11 × 10⁻³¹ kg and calculate vₘₐₓ = √(2Kmax/m) with Kmax in joules. Check that Kmax is not greater than the incident photon energy, because some photon energy is used to overcome the work function.
Pause and explain
At the threshold frequency, what is Kmax?
Worked example
Light of frequency 8.0 × 10¹⁴ Hz falls on a metal with work function 2.0 eV. Find the maximum kinetic energy of the emitted electrons in eV and joules. Use h = 6.63 × 10⁻³⁴ J s.
Show the worked solution
- Calculate hf in joules, then convert to eV.
- Subtract the 2.0 eV work function.
- Convert the excess energy back to joules if required.
Answer hf = 5.304 × 10⁻¹⁹ J ≈ 3.315 eV; Kmax ≈ 1.315 eV ≈ 2.10 × 10⁻¹⁹ J.
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Test this topic →From guided practice to a new situation
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 photon has energy 4.0 eV and the metal work function is 1.5 eV. Find Kmax in eV and joules.
Give me a hint
Subtract energies in eV first, then multiply by 1.60 × 10⁻¹⁹ J/eV.
Compare my reasoning
- Kmax = 4.0 − 1.5 = 2.5 eV.
- Kmax = 2.5 × 1.60 × 10⁻¹⁹ J.
- Kmax = 4.0 × 10⁻¹⁹ J.
Kmax = 2.5 eV = 4.0 × 10⁻¹⁹ J.
Look for these in your work
- Work function subtracted
- Energy conversion correct
- Both units stated
A metal has Φ = 3.0 × 10⁻¹⁹ J and is illuminated at f = 7.0 × 10¹⁴ Hz. Calculate Kmax.
Give me a hint
Use Kmax = hf − Φ in joules.
Compare my reasoning
- hf = (6.63 × 10⁻³⁴)(7.0 × 10¹⁴) = 4.641 × 10⁻¹⁹ J.
- Kmax = 4.641 × 10⁻¹⁹ − 3.0 × 10⁻¹⁹ J.
- Kmax = 1.641 × 10⁻¹⁹ J.
Kmax ≈ 1.64 × 10⁻¹⁹ J.
Look for these in your work
- Photon energy calculated
- Work function subtracted
- Positive kinetic energy reported
Two metals are illuminated by the same frequency above both thresholds. Metal A has work function 1.8 eV and metal B has work function 2.4 eV. Compare their maximum photoelectron kinetic energies.
Give me a hint
The photon energy is common; subtract each metal's work function.
Compare my reasoning
- For each metal, Kmax = hf − Φ.
- The photon energy hf is the same for both.
- Metal A's smaller work function gives it 0.6 eV more maximum kinetic energy.
Metal A's maximum kinetic energy is 0.6 eV greater than metal B's.
Look for these in your work
- Same photon energy recognized
- Separate work functions applied
- Kinetic-energy difference compared
Common mistakes
- Adding Φ to hf instead of subtracting for Kmax
- Using the average kinetic energy instead of maximum kinetic energy
- Subtracting eV from joules without conversion
- Predicting emission when hf < Φ
What can you explain now?
What does the excess photon energy hf − Φ become for the fastest emitted photoelectrons?
Compare with the explanation
Their maximum kinetic energy
The work function releases the electron; excess energy becomes kinetic energy.
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Make it stick
Why is Kmax the difference between photon energy and work function?
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