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Why does the photoelectric effect show that light comes in photons?

The photon model and the energy of a photon, the photoelectric effect, the work function and threshold frequency, and Einstein's photoelectric equation.

A CCEA A-Level Physics answer on the photon model and the energy of a photon, the photoelectric effect and why it needs photons, the work function and threshold frequency, and Einstein's photoelectric equation with stopping voltage.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The photon model
  3. The photoelectric effect
  4. Work function and Einstein's equation
  5. Worked example: stopping voltage and the Planck constant
  6. Examples in context
  7. Try this

What this dot point is asking

CCEA wants you to describe the photon model and calculate photon energy, describe the photoelectric effect and explain why it supports the photon model, define the work function and threshold frequency, and apply Einstein's photoelectric equation. Expect a calculation using E=hfE = hf or Einstein's equation, plus an explanation of why the wave model fails.

The photon model

The photon model treats light as both wave-like and particle-like. A beam of light is a stream of photons; its intensity is the number of photons arriving per second per unit area, while the energy of each photon depends only on the frequency. This separation of "how many" from "how energetic" is exactly what the photoelectric effect requires. Photon energies are often quoted in electronvolts, where 1 eV=1.6×1019 J1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J} is the energy gained by an electron accelerated through one volt.

The photoelectric effect

The classic demonstration uses a gold-leaf electroscope with a clean zinc plate. When the plate is negatively charged and ultraviolet light shines on it, the leaf falls as electrons are emitted; ordinary visible light, however intense, has no effect because its frequency is below the threshold. Placing a glass sheet (which absorbs ultraviolet) between the lamp and the plate stops the discharge, confirming that it is the high-frequency ultraviolet doing the work.

These observations cannot be explained by the wave model, which predicts that any frequency would eventually free electrons if bright enough, and predicts a delay while energy accumulates. They are direct evidence for photons.

Work function and Einstein's equation

The fastest electrons are those from the surface, which lose only the work function on escape; electrons from deeper in the metal lose more, so they emerge with a spread of energies up to the maximum.

Worked example: stopping voltage and the Planck constant

Examples in context

Example 1. A photocell light meter. In a vacuum photocell, light above the threshold frequency ejects electrons from a coated cathode, producing a current proportional to the light intensity. This converts a light level into an electrical signal, the basis of older camera light meters and automatic street-light switches, and it works only because the cathode metal has a low enough work function for visible light.

Example 2. Solar-cell physics. While a solar cell uses the related photovoltaic effect rather than electron ejection into vacuum, the same quantum idea applies: only photons with energy above a threshold (the band gap) can free a charge carrier. Photons below that energy pass straight through, which is why a cell has a maximum theoretical efficiency, a direct consequence of light arriving as discrete photons of energy hfhf.

Try this

Q1. A metal has a work function of 3.2×1019 J3.2 \times 10^{-19}\ \text{J}. Light of frequency 7.0×1014 Hz7.0 \times 10^{14}\ \text{Hz} shines on it. Find the maximum kinetic energy of the emitted electrons. Take h=6.6×1034 J sh = 6.6 \times 10^{-34}\ \text{J s}. [3 marks]

  • Cue. Ek(max)=hfϕ=(6.6×1034)(7.0×1014)3.2×1019=1.4×1019 JE_{k(\max)} = hf - \phi = (6.6 \times 10^{-34})(7.0 \times 10^{14}) - 3.2 \times 10^{-19} = 1.4 \times 10^{-19}\ \text{J}.

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

  • Cue. Each photon has too little energy to overcome the work function, and one photon interacts with one electron.

Q3. Calculate the energy of a photon of red light of wavelength 6.5×107 m6.5 \times 10^{-7}\ \text{m}. Take h=6.63×1034 J sh = 6.63 \times 10^{-34}\ \text{J s} and c=3.0×108 m s1c = 3.0 \times 10^{8}\ \text{m s}^{-1}. [2 marks]

  • Cue. E=hcλ=(6.63×1034)(3.0×108)6.5×107=3.1×1019 JE = \dfrac{hc}{\lambda} = \dfrac{(6.63 \times 10^{-34})(3.0 \times 10^{8})}{6.5 \times 10^{-7}} = 3.1 \times 10^{-19}\ \text{J}.

Exam-style practice questions

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

CCEA 20186 marksSodium has a work function of 3.7×1019 J3.7 \times 10^{-19}\ \text{J}. Light of wavelength 4.0×107 m4.0 \times 10^{-7}\ \text{m} is shone on a clean sodium surface. Calculate the threshold frequency of sodium and the maximum kinetic energy of the emitted photoelectrons. Take h=6.63×1034 J sh = 6.63 \times 10^{-34}\ \text{J s} and c=3.0×108 m s1c = 3.0 \times 10^{8}\ \text{m s}^{-1}.
Show worked answer →

The threshold frequency comes from the work function, ϕ=hf0\phi = hf_0:

f0=ϕh=3.7×10196.63×1034=5.6×1014 Hzf_0 = \dfrac{\phi}{h} = \dfrac{3.7 \times 10^{-19}}{6.63 \times 10^{-34}} = 5.6 \times 10^{14}\ \text{Hz}.

The energy of an incident photon is

E=hcλ=(6.63×1034)(3.0×108)4.0×107=4.97×1019 JE = \dfrac{hc}{\lambda} = \dfrac{(6.63 \times 10^{-34})(3.0 \times 10^{8})}{4.0 \times 10^{-7}} = 4.97 \times 10^{-19}\ \text{J}.

Einstein's equation gives the maximum kinetic energy:

Ek(max)=hfϕ=4.97×10193.7×1019=1.3×1019 JE_{k(\max)} = hf - \phi = 4.97 \times 10^{-19} - 3.7 \times 10^{-19} = 1.3 \times 10^{-19}\ \text{J}.

Markers reward f0=ϕ/hf_0 = \phi/h, the photon energy from hc/λhc/\lambda, and a positive maximum kinetic energy from Einstein's equation.

CCEA 20204 marksExplain why the photoelectric effect cannot be explained by the wave model of light, and state what the existence of a threshold frequency tells us about how light delivers energy.
Show worked answer →

The wave model predicts that the energy delivered to a surface depends only on the intensity, so light of any frequency, if bright enough or left long enough, should eventually free electrons. It also predicts a time delay while energy builds up.

Experiment shows neither: below a threshold frequency no electrons are emitted however intense the light, and above the threshold emission is instantaneous.

The threshold frequency shows that light delivers energy in discrete packets (photons) of energy hfhf, and that one photon interacts with one electron. Only if a single photon carries at least the work function can an electron escape, so it is the frequency, not the intensity, that decides whether emission occurs.

Markers reward the wave-model prediction, the contradicting observations (threshold and instant emission), and the conclusion that energy arrives in single photons of energy hfhf.

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