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United StatesPhysics 2Syllabus dot point

What experiments prove that light delivers its energy and momentum as particles?

Topic 15.5 The Photoelectric Effect and Compton Scattering: apply the photoelectric equation and describe Compton scattering as evidence of the photon.

A focused answer to AP Physics 2 Topics 15.5 and 15.6, covering the photoelectric effect, the work function and threshold frequency, the photoelectric equation Kmax = hf - phi, why the effect proves the photon model, and Compton scattering as evidence that photons carry momentum, with full worked examples.

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  1. What this topic is asking
  2. The photoelectric effect
  3. The photoelectric equation
  4. Compton scattering: photons carry momentum
  5. Try this

What this topic is asking

The College Board (Topics 15.5 and 15.6) want you to apply the photoelectric equation, Kmax=hfϕK_{max} = hf - \phi, explain why the photoelectric effect proves the photon model, and describe Compton scattering as evidence that photons carry momentum.

The photoelectric effect

The photoelectric effect is the experiment that forced physics to accept photons. Classically, a bright enough light of any frequency should eventually shake electrons loose, but experiment shows the opposite: below a threshold frequency, no electrons are emitted however bright the light, while above it electrons are emitted instantly even for dim light. This makes sense only if light arrives as photons, each carrying energy hfhf: an electron is freed only if the single photon that hits it has enough energy (hfϕhf \geq \phi).

The photoelectric equation

The equation is energy conservation for a single photon-electron interaction: the photon's energy hfhf pays the "exit fee" ϕ\phi to free the electron, and the leftover becomes the electron's kinetic energy. Two predictions distinguish it sharply from the wave model. First, there is a hard threshold: below f0f_0, nothing happens at any intensity. Second, intensity changes count, not energy: brighter light means more photons, so more electrons, but each electron's energy still depends only on hfhf. These are the most-tested points.

Compton scattering: photons carry momentum

The photoelectric effect shows photons carry energy; Compton scattering shows they also carry momentum. When X-ray photons scatter off electrons, the scattered light comes out with a longer wavelength (lower energy), and the electron recoils, exactly as in a particle collision where momentum and energy are conserved. The photon behaves like a particle with momentum p=hλp = \dfrac{h}{\lambda}, transferring some to the electron. This is decisive: only a particle can collide and share momentum like this; a spread-out wave cannot. The strategic role of these topics is that they provide the experimental backbone of the photon model introduced in Topic 15.1: the photoelectric effect proves light delivers energy in quanta, and Compton scattering proves photons carry momentum. Together with interference (which shows the wave side), they complete the case for wave-particle duality and set up the quantised atomic energy levels of Topic 15.2.

Try this

Q1. State what happens to the number of ejected electrons and their maximum kinetic energy when the light's intensity is increased (frequency fixed). [2 points]

  • Cue. More electrons are ejected, but their maximum kinetic energy is unchanged.

Q2. State what Compton scattering shows about photons. [1 point]

  • Cue. That photons carry momentum (they collide with electrons like particles).

Exam-style practice questions

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

AP 2024 (style)7 marksSection II (long FRQ). A metal has a work function of 3.0×10193.0 \times 10^{-19} J. Light of frequency 7.0×10147.0 \times 10^{14} Hz shines on it. Take h=6.63×1034h = 6.63 \times 10^{-34} J s. (a) Calculate the energy of each incident photon. (b) Calculate the maximum kinetic energy of the ejected electrons. (c) Explain why increasing the light's intensity (but not its frequency) does not change the maximum kinetic energy of the electrons.
Show worked answer →

A 7-point FRQ on the photoelectric effect.

(a) Photon energy (2 points): E=hf=(6.63×1034)(7.0×1014)=4.64×1019E = hf = (6.63 \times 10^{-34})(7.0 \times 10^{14}) = 4.64 \times 10^{-19} J.
(b) Maximum kinetic energy (3 points): Kmax=hfϕ=4.64×10193.0×1019=1.64×1019K_{max} = hf - \phi = 4.64 \times 10^{-19} - 3.0 \times 10^{-19} = 1.64 \times 10^{-19} J.
(c) Intensity (2 points): each electron is ejected by a single photon, so the kinetic energy depends on the energy per photon (hfhf), not on how many photons arrive. Higher intensity means more photons (more electrons), but the same energy per photon, so KmaxK_{max} is unchanged.

Markers reward the photon energy, the photoelectric equation, and the one-photon-per-electron explanation of why intensity does not change KmaxK_{max}.

AP 2023 (style)1 marksSection I (multiple choice). In the photoelectric effect, no electrons are emitted when the light's frequency is below a certain threshold, no matter how bright the light. What does this show? (A) light is purely a wave (B) the energy of light is delivered in photons of energy hf (C) electrons have no mass (D) the metal is transparent. Justify your reasoning.
Show worked answer →

A 1-point MCQ on the photoelectric effect. The answer is (B).

A threshold frequency, below which no electrons are emitted however bright the light, can only be explained if light delivers energy in discrete photons of energy hfhf: below the threshold no single photon has enough energy to free an electron. A wave model would predict emission at any frequency given enough intensity. The trap is (A): the threshold is exactly what the wave model cannot explain.

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