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Photoelectric Effect

Key idea: Understand the photoelectric effect, including threshold frequency, intensity vs kinetic energy, and stopping potential with exam-ready explanations (A Level Physics).

  • GCE A-Level H2 Physics 2027
On this page

Learning objectives

  • Use photon energy and momentum and analyse the photoelectric effect.
Syllabus focus

Use the existence of a threshold frequency as evidence that light transfers energy in photons. The work-function and stopping-potential treatment later on this page is useful legacy support, but keep the evidence statement separate from that calculation model.

1. Definitions (Must Know)

A. Photoelectric effect

The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation is incident on it.

B. Photoelectron

A photoelectron is an electron that has been emitted from the metal during the photoelectric effect.

C. Threshold frequency, f₀

The threshold frequency, f₀, is the minimum frequency of incident radiation needed to just eject photoelectrons from a given metal.

If f < f₀, no photoelectrons are emitted (no matter the intensity).

D. Work function, Φ

The work function, Φ, is the minimum energy needed to remove an electron from the metal surface.

2. Key Ideas (What Earns Marks)

Evidence for photon and matter-wave behaviourTwo evidence chains connect threshold-frequency photoemission to photons and electron diffraction with localised detections to matter-wave behaviour.Light: particulate evidenceThreshold frequencyno emission when f < f₀, however intensePhoton modelone quantum has energy E = hfphoton momentum p = E/c = h/λElectrons: wave evidenceDiffraction and interferencepatterns build from one detection at a timeMatter-wave modelde Broglie wavelength λ = h/pdetection remains localised
Scroll diagram horizontally to read all labels.
No single classical model explains every observation: threshold-frequency photoemission reveals photon behaviour, while diffraction and single-particle interference reveal wave behaviour.
  • Existence of a threshold frequency supports the photon model (energy comes in packets).
  • Increasing intensity increases the rate of emission (number of electrons per second), but does not increase the maximum kinetic energy.
  • Increasing frequency increases the maximum kinetic energy (once above threshold).
  • Emission is effectively instantaneous (no detectable delay) once f ≥ f₀.
  • Link to energy equation (next lesson):
    • Kₘₐₓ = hf-Φ
    • stopping potential: Kₘₐₓ = eVₛ
Two high-frequency statements

“Intensity changes the number of photons per second.”
“Frequency changes the energy per photon.”

Exam pitfall: swapping effects of intensity and frequency

Above threshold, intensity changes emission rate while frequency changes maximum kinetic energy. Keep those two statements separate in explanation questions.

3. Detailed Explanations

One photon ejecting one electron from a metalA photon of frequency f and energy h f reaches a metal surface and transfers energy to one electron. The electron escapes with maximum kinetic energy h f minus work function phi when h f is at least phi.metal surfacephoton: E = hfemitted electronhf = Φ + KEₘₐₓone photon transfers energy to one electron
In the photon model, one photon transfers energy hf to one electron; emission requires hf ≥ Φ.

A. Why threshold frequency exists (photon model)

Photon energy is: E = hf

To eject an electron, one photon must supply at least the work function Φ.

So emission requires: hf ≥ Φ ⇒ f ≥ f₀ = Φ/h

If f < f₀, each photon has too little energy, so no electrons can be emitted even if the light is very intense.

B. Why intensity affects rate, not maximum kinetic energy

At fixed frequency:

  • increasing intensity increases the number of photons per second hitting the surface,
  • so more electrons are emitted per second (higher current),
  • but each photon still has the same energy hf, so Kₘₐₓ is unchanged.

C. Why emission is immediate

Energy is transferred in a single photon–electron interaction.

If a photon has enough energy to overcome Φ, the electron can be emitted without needing to “wait and accumulate energy”.

4. Common Mistakes

  • Saying “photoelectrons are special electrons” (they are ordinary electrons that have been emitted).
  • Using intensity to explain threshold frequency (threshold depends on photon energy, i.e. frequency).
  • Forgetting the condition “f < f₀ gives no emission”.

5. Exam Tips

  • If the question asks for evidence for photons, lead with “threshold frequency”.
  • If a graph is mentioned, remember:
    • stopping potential Vₛ tells you Kₘₐₓ via Kₘₐₓ = eVₛ,
    • Vₛ increases linearly with f above threshold.

6. Worked Examples

Modelled example 1

Photon energy

Core

Problem

Find the energy of a photon of frequency f = 6.0 × 10¹⁴ Hz. Take h = 6.63 × 10⁻³⁴ J s.
Study the worked solution

  1. Choose the photon relation

    Method

    E = hf.

    Reason

    Frequency fixes the energy carried by each photon.

    Working

    E = (6.63 × 10⁻³⁴)(6.0 × 10¹⁴)

  2. Evaluate

    Method

    E = 4.0 × 10⁻¹⁹ J to two significant figures.

    Reason

    Multiplying J s by s⁻¹ gives joules.

    Working

    E = 3.978 × 10⁻¹⁹ J ≈ 4.0 × 10⁻¹⁹ J

Guided practice 2

Threshold frequency

About 4 min

Problem

A metal has work function Φ = 3.2 × 10⁻¹⁹ J. Find its threshold frequency f₀. Take h = 6.63 × 10⁻³⁴ J s.

Try this before viewing the solution

Unit: Hz

Hints

Hint 1: write the threshold condition
Set hf₀ = Φ and isolate f₀.
View solution step by step

  1. Use the threshold condition

    Method

    hf₀ = Φ.

    Reason

    At threshold the emitted electron has zero maximum kinetic energy.

    Working

    f₀ = Φ/h

  2. Calculate

    Method

    f₀ = 4.8 × 10¹⁴ Hz.

    Reason

    The quotient of energy and J s has unit s⁻¹.

    Working

    f₀ = (3.2 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) = 4.8 × 10¹⁴ Hz

Common misconception 3

Work function from threshold frequency

Find and correct the mistake

Learner claim

A metal has threshold frequency f₀ = 5.5 × 10¹⁴ Hz. A learner writes Φ = f₀/h. Diagnose the rearrangement and find Φ. Take h = 6.63 × 10⁻³⁴ J s.

Try this before viewing the solution

Unit: J

View solution step by step

  1. Restore the physical relation

    Method

    At threshold, the photon energy equals the work function: hf₀ = Φ.

    Reason

    f₀/h has the wrong dimensions for energy.

    Working

    [h f₀] = (J s)(s⁻¹) = J

  2. Evaluate

    Method

    Φ = 3.65 × 10⁻¹⁹ J.

    Reason

    Work function is the threshold photon energy.

    Working

    Φ = (6.63 × 10⁻³⁴)(5.5 × 10¹⁴) = 3.65 × 10⁻¹⁹ J

Examiner practice 4

Maximum kinetic energy and stopping potential

4 marks

Legacy-support examination method

Light of frequency f = 9.0 × 10¹⁴ Hz shines on a metal with Φ = 3.2 × 10⁻¹⁹ J. Find Kₘₐₓ and the stopping potential Vₛ. Take h = 6.63 × 10⁻³⁴ J s and e = 1.60 × 10⁻¹⁹ C. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Use Einstein’s equation

    1 mark

    Method

    Kₘₐₓ = hf-Φ.

    Reason

    One photon supplies the removal energy and the remaining energy becomes maximum electron kinetic energy.

    Working

    Kₘₐₓ = (6.63 × 10⁻³⁴)(9.0 × 10¹⁴)-3.2 × 10⁻¹⁹

  2. Find maximum kinetic energy

    1 mark

    Method

    Kₘₐₓ = 2.77 × 10⁻¹⁹ J.

    Reason

    The incident frequency exceeds the threshold frequency.

    Working

    Kₘₐₓ = 2.77 × 10⁻¹⁹ J

  3. Link to stopping potential

    1 mark

    Method

    eVₛ = Kₘₐₓ.

    Reason

    At the stopping potential even the fastest photoelectrons are just prevented from reaching the collector.

    Working

    Vₛ = Kₘₐₓ/e

  4. Evaluate stopping potential

    1 mark

    Method

    Vₛ = 1.7 V to two significant figures.

    Reason

    Energy per unit charge has unit volts.

    Working

    Vₛ = (2.77 × 10⁻¹⁹)/(1.60 × 10⁻¹⁹) = 1.73 V ≈ 1.7 V

Challenge 5

Photon rate and photoelectric current (idealised)

Minimal support

Independent transfer

Monochromatic light of power P = 2.0 mW has photon frequency f = 6.0 × 10¹⁴ Hz. Assuming one emitted electron per photon, estimate the photoelectric current. Take h = 6.63 × 10⁻³⁴ J s and e = 1.60 × 10⁻¹⁹ C.

Try this before viewing the solution

Hints

Hint 1: build a rate chain
Find energy per photon, divide power by that energy, then convert electrons per second to charge per second.
View solution step by step

  1. Find photon energy

    Method

    E = 4.0 × 10⁻¹⁹ J.

    Reason

    Each photon has energy hf.

    Working

    E = (6.63 × 10⁻³⁴)(6.0 × 10¹⁴) = 4.0 × 10⁻¹⁹ J

  2. Find photon rate

    Method

    N = 5.0 × 10¹⁵ s⁻¹.

    Reason

    Power is energy transferred per second, so divide it by energy per photon.

    Working

    N = (2.0 × 10⁻³)/(4.0 × 10⁻¹⁹) = 5.0 × 10¹⁵ s⁻¹

  3. Convert rate to current

    Method

    I = 0.80 mA.

    Reason

    Under the stated idealisation, each photon contributes one electron of charge e.

    Working

    I = eN = (1.60 × 10⁻¹⁹)(5.0 × 10¹⁵) = 8.0 × 10⁻⁴ A

7. Mind Stretchers

Mind stretcher 1: Intensity vs frequency testExtension

A metal emits photoelectrons for violet light but not for red light. Which change is guaranteed to make it emit for red light: increasing red intensity, or increasing frequency? Explain.

Show Answer

Increasing intensity at the same red frequency does not increase photon energy, so it cannot overcome the threshold if f < f₀.

Increasing frequency increases photon energy and can exceed the threshold, so increasing frequency is the guaranteed fix.

Mind stretcher 2: Changing the metalExtension

Light of fixed frequency shines on two different metals. Metal X has a larger work function than metal Y. Compare their threshold frequencies and the stopping potentials (if both emit).

Show Answer

Threshold frequency: f₀ = Φ/h, so the metal with larger Φ has a larger f₀.

If both emit at the given frequency, then Kₘₐₓ = hf-Φ is smaller for the larger-Φ metal, so its stopping potential Vₛ = Kₘₐₓ/e is also smaller.

8. Optional (Enrichment)

A. A quick video demo

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027