Photoelectric evidence for photons
Key idea: Photoelectric emission needs a photon with enough energy to overcome the work function. Above threshold, higher frequency raises maximum electron energy; greater intensity raises emission rate.
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Guided lessons · coming soon
Guided lesson
Guided steps
- Connect the escape energy to the threshold frequency.
- Separate photon energy from photon arrival rate.
- Decide which beam frees electrons.
- Decide whether dim light causes a delay.
- Compare the wave prediction with the observations.
- Predict emission before changing the explorer controls.
- Test both frequencies at two intensities.
- Explain why the current changes in only one trial.
- Compare the electron energies and beam powers.
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You stopped at:
ContinueShine 250 nm ultraviolet light on a clean zinc plate and electrons fly off it. Shine a far brighter red light on it and nothing happens at all. The wave theory of light cannot explain this, but it follows at once if light arrives in packets of energy, called photons.
The threshold frequency
Light can release electrons from a metal surface: this is the photoelectric effect. The work function Φ of the metal is the minimum energy needed to remove an electron from its surface.
Each photon of light of frequency f carries energy hf, and one absorbed photon gives all its energy to one electron in the single-photon model. An electron can only escape if that energy is at least the work function. The minimum frequency for escape is the threshold frequency f₀, given by
Below the threshold, no electrons are emitted however intense the light, because every photon has too little energy. Above it, the extra photon energy, hf-Φ, can become the kinetic energy of the escaping electron.
Intensity and frequency
For light of one frequency, a higher intensity means more photons arriving per second; each photon still has the same energy hf. Above the threshold, more photons per second free more electrons per second. A higher frequency instead gives every photon more energy. At fixed frequency, intensity controls the emission rate. Frequency sets the maximum kinetic energy; electrons leaving from deeper in the metal can lose energy before escaping.
Worked example 1
Intensity or frequency?
Problem
Light above the threshold frequency shines on a metal. Predict what happens to the emitted electrons when the intensity is doubled at the same frequency, and when the frequency is increased.
Show full solution
Doubling the intensity
Method
Decide whether the number of photons or the energy of each photon changes.Reason
At a fixed frequency each photon carries hf; the intensity sets how many arrive per second.Working
Answer
Twice as many photons arrive per second, so about twice as many electrons are emitted per second. Each electron gains no more energy than before.
Increasing the frequency
Working
Answer
Each photon now has more energy, so the emitted electrons can have more kinetic energy.
Quick check
Only the frequency changes what a single photon can do. The intensity changes how many photons there are.
Guided practice 2
Which beam frees electrons?
Problem
A metal has a threshold frequency of 5.00 × 10¹⁴ Hz. It is lit in turn by light of 4.00 × 10¹⁴ Hz and 6.00 × 10¹⁴ Hz. Which beam frees electrons? Would doubling the intensity of the other beam make it free electrons?
Decide
Hints
Hint 1
one photon, one electron
Compare each frequency with f₀. Doubling the intensity doubles the number of photons, not the energy of each.
Show full solution
Compare with the threshold
Working
Answer
Only the 6.00 × 10¹⁴ Hz beam is above the threshold. Doubling the intensity of the 4.00 × 10¹⁴ Hz beam doubles the number of photons, but each still has too little energy, so no electrons are emitted.
Spot the mistake 3
Should dim light cause a delay?
Learner claim
A student says that with very dim light above the threshold frequency, electrons should be emitted only after a delay, while each electron gathers enough energy from the wave. What is actually observed?
Try this before viewing the solution
Show solution step by step
Observation
Method
Compare the claim with experiment.Reason
A model that disagrees with measurement has to be changed.Working
Answer
Emission starts with no measurable delay, even in very dim light.
Photon explanation
Working
Answer
One photon gives its whole energy hf to one electron at once, so the electron does not need to gather energy. Dim light means fewer photons per second, so fewer electrons are emitted per second, but each is emitted as soon as a photon arrives.
Common mistake
The delay belongs to the wave picture, and experiment rules it out.
Evidence for photons
In the wave theory, light delivers energy continuously, spread over the surface. An electron could then gather energy gradually, so light of any frequency should free electrons if it is intense enough, and dim light should free them only after a delay while the energy builds up. Neither is observed:
- there is a sharp threshold frequency, below which even very intense light frees no electrons;
- above the threshold, emission starts at once, even in very dim light.
Both follow if the energy arrives in photons, each given whole to one electron. The existence of a threshold frequency is therefore evidence for the particulate nature of electromagnetic radiation. Interference and diffraction of light, on the other hand, are evidence for its wave nature. Light shows both kinds of behaviour.
Check your understanding 1 Before you try the explorer
The explorer gives sodium a work function of 2.36 eV, corresponding to f₀ = 5.71 × 10¹⁴ Hz. At zero collector p.d., predict the emission when the intensity rises from 30% to 60% at 5.0 × 10¹⁴ Hz, then at 7.0 × 10¹⁴ Hz.
Show answer
At 5.0 × 10¹⁴ Hz each photon has too little energy, so neither intensity frees electrons. At 7.0 × 10¹⁴ Hz both intensities free electrons; raising intensity from 30% to 60% doubles their rate but leaves their maximum kinetic energy unchanged.
Try your prediction
Select sodium and set collector p.d. to zero. Set frequency to 5.0, then 7.0, in the displayed units of 10¹⁴ Hz. At each frequency, compare intensities of 30% and 60%. Watch the electrons and the current reading.
Sodium lit by light of frequency 7.0 × 10¹⁴ Hz at 60% intensity, collector at 0.00 V. Electrons are emitted; the current is 20.73 nA.
- Photon energy, hf
- 2.89 eV
- Work function, Φ
- 2.36 eV
- Wavelength, λ
- 428 nm
- Current, I
- 20.73 nA
- Last transition
- —
- Photon energy, ΔE
- — eV
- Wavelength, λ = hc / ΔE
- — nm
- Series
- —
Try this
0 of 4 doneShine very bright light below the threshold frequency, then dim light above it. (not done yet)
Bright light below the threshold frees no electrons; dim light above it does at once. One photon gives one electron its energy hf.
Find the stopping potential at four frequencies and record each one. (not done yet)
KEmax = hf − Φ: the gradient of your line is the Planck constant (× 10⁻³³ here) and the y-intercept is −Φ.
Make a hydrogen atom emit its red line (656 nm). (not done yet)
n = 3 → 2: ΔE = 1.89 eV, so λ = hc / ΔE = 656 nm. All transitions down to n = 2 form the Balmer series.
In the absorption spectrum, make a dark line where you saw a bright one. (not done yet)
An atom absorbs only the photons it could emit: the same energy gap gives the same wavelength.
You’ve tried every task.
Your readings
| # | f / 1014 Hz | Vs / V | KEmax / 10−19 J | Remove |
|---|---|---|---|---|
| No readings yet. Set up a measurement, then record it. | ||||
Check your understanding 2 Explain the two observations
Why does increasing intensity change the current in one trial but not the other? Include what stays the same about each photon.
Show answer
At a fixed frequency, each photon keeps the same energy hf as intensity changes. At 5.0, that energy is below sodium’s work function, so there are no emitted electrons and no current. At 7.0, photons can free electrons; more photons per second give more electrons per second and a larger current. Check that your explanation uses energy per photon for escape, and photon arrival rate for current.
Try it yourself 4
Equal photon arrival rates
Problem
Two sources illuminate the same metal with equal photon arrival rates. Source 1 has frequency 1.2f₀ and Source 2 has 1.8f₀. Which gives the larger maximum photoelectron energy, and which supplies the greater light power? Give the ratios.
Decide
Show the steps Optional working
Hints
Hint 1
per photon, then per second
Use Φ = hf₀ and Kₘₐₓ = hf-Φ. Power is photon arrival rate multiplied by photon energy.
Show solution step by step
Photon model
Reason
Both frequencies exceed the threshold, but the surplus energies differ.Working
Answer
Source 1 gives Kₘₐₓ = 0.2hf₀ and Source 2 gives 0.8hf₀. Source 2’s maximum photoelectron energy is four times as large.
Beam power
Working
Answer
At equal photon arrival rates, P₂/P₁ = f₂/f₁ = 1.8/1.2 = 1.5. Source 2 supplies 1.5 times the power. This ratio differs from the maximum-electron-energy ratio because the same work function must be supplied for each emitted electron.
Common mistakes
- Saying brighter light gives electrons more energy; it gives more electrons per second.
- Expecting intense light below the threshold frequency to emit electrons eventually.
- Thinking every photon above the threshold frees an electron; many do not.
- Concluding that light is only particles; interference and diffraction show it also behaves as a wave.
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Syllabus and review details
- GCE A-Level H2 Physics 2027 · 2027
Content Overview, PDF pages 9–10; Subject Content, PDF pages 11–30
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