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What does the photoelectric effect tell us about the nature of light?

The photon model and E = hf, the photoelectric effect, Einstein's photoelectric equation, the work function and threshold frequency.

A focused answer to WJEC A-Level Physics Unit 2 photons, covering the photon model and E = hf, the photoelectric effect, Einstein's photoelectric equation, the work function and threshold frequency, and why the wave model fails.

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

WJEC wants you to use the photon model and E=hfE = hf, describe the photoelectric effect, apply Einstein's photoelectric equation, and define the work function and threshold frequency, explaining why the wave model cannot account for the observations. This is the topic where classical physics breaks down and the quantum picture takes over, and the examiners reward a clear contrast between what the wave model predicts and what experiment shows.

The answer

The photon model

The photoelectric effect

The photoelectric effect is the emission of electrons from a metal surface when light shines on it. Key observations: emission is instant, occurs only above a threshold frequency, the maximum electron kinetic energy depends on frequency, not intensity, and brighter light above the threshold simply ejects more electrons per second.

Einstein's equation

The photon energy hfhf equals the work function ϕ\phi (the minimum energy to free an electron from the surface) plus the maximum kinetic energy of the emitted electron. The threshold frequency f0f_0 is the lowest frequency that causes emission, where hf0=ϕhf_0 = \phi.

Examples in context

Example 1. A solar cell. A photovoltaic cell relies on photons each freeing one charge carrier in a semiconductor. Only photons with energy above the band gap (the semiconductor's equivalent of a work function) generate current, which is why silicon cells waste the low-energy infrared part of sunlight. This single-photon-single-electron rule is the same physics as the photoelectric effect.

Example 2. Light meters in cameras. A photodiode produces a current proportional to the number of photons striking it per second. Because intensity sets the photon rate, the current measures brightness, while the threshold behaviour ensures only light above a certain energy registers. The camera converts this current into the exposure setting.

Try this

Q1. A metal has a work function of 3.2×10−19 J3.2\times10^{-19}\,\text{J}. Light of frequency 7.0×1014 Hz7.0\times10^{14}\,\text{Hz} shines on it. Find the maximum kinetic energy of the emitted electrons. Take h=6.63×10−34 J sh = 6.63\times10^{-34}\,\text{J s}. [3 marks]

  • Cue. hf=6.63×10−34×7.0×1014=4.64×10−19 Jhf = 6.63\times10^{-34}\times7.0\times10^{14} = 4.64\times10^{-19}\,\text{J}; Ek(max)=4.64×10−19−3.2×10−19=1.4×10−19 JE_{k(max)} = 4.64\times10^{-19} - 3.2\times10^{-19} = 1.4\times10^{-19}\,\text{J}.

Q2. Explain why no electrons are emitted below the threshold frequency, however bright the light. [2 marks]

  • Cue. Each photon has too little energy (hf<ϕhf < \phi); intensity only adds more such photons, none of which can free an electron.

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC 20206 marksSodium has a work function of 3.6×10−19 J3.6 \times 10^{-19}\,\text{J}. Light of wavelength 4.5×10−7 m4.5 \times 10^{-7}\,\text{m} illuminates a clean sodium surface. Calculate the threshold frequency, show that emission occurs, and find the maximum kinetic energy of the emitted electrons. Take h=6.63×10−34 J sh = 6.63 \times 10^{-34}\,\text{J s} and c=3.0×108 m s−1c = 3.0 \times 10^{8}\,\text{m s}^{-1}.
Show worked answer →

Threshold frequency from ϕ=hf0\phi = hf_0:

f0=ϕh=3.6×10−196.63×10−34=5.43×1014 Hzf_0 = \dfrac{\phi}{h} = \dfrac{3.6 \times 10^{-19}}{6.63 \times 10^{-34}} = 5.43 \times 10^{14}\,\text{Hz}.

Frequency of the incident light: f=cλ=3.0×1084.5×10−7=6.67×1014 Hzf = \dfrac{c}{\lambda} = \dfrac{3.0 \times 10^{8}}{4.5 \times 10^{-7}} = 6.67 \times 10^{14}\,\text{Hz}.

Since f>f0f > f_0, emission occurs.

Maximum kinetic energy from Einstein's equation Ek(max)=hf−ϕE_{k(\text{max})} = hf - \phi:

hf=6.63×10−34×6.67×1014=4.42×10−19 Jhf = 6.63 \times 10^{-34} \times 6.67 \times 10^{14} = 4.42 \times 10^{-19}\,\text{J}, so Ek(max)=4.42×10−19−3.6×10−19=8.2×10−20 JE_{k(\text{max})} = 4.42 \times 10^{-19} - 3.6 \times 10^{-19} = 8.2 \times 10^{-20}\,\text{J}.

Markers reward the threshold frequency, the comparison showing emission, and the maximum kinetic energy from Einstein's equation.

WJEC 20183 marksExplain why the photoelectric effect provides evidence for the particle nature of light, referring to the existence of a threshold frequency.
Show worked answer →

A wave delivers energy continuously and spread across the surface, so any frequency of light, if bright enough or left long enough, should eventually free an electron. Experiment shows this does not happen: below a threshold frequency no electrons are emitted, however intense the light.

The photon model explains this. Light arrives in quanta of energy hfhf, and one photon gives all its energy to one electron. If hfhf is less than the work function ϕ\phi, no single photon can free an electron, so there is a threshold frequency f0=ϕ/hf_0 = \phi/h. The existence of this threshold is direct evidence that light behaves as particles. Markers reward contrasting the wave prediction with the photon explanation tied to the threshold.

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