Photon momentum is the momentum carried by a massless photon, with magnitude p = E/c = h/λ = hf/c and a direction along the photon's motion; in AP Physics 2 (Topic 15.1), it's the particle-like property that lets light transfer momentum in collisions and connects directly to de Broglie's λ = h/p.
Photon momentum is one of the strangest and most exam-tested ideas in Unit 15. A photon has zero mass, yet it still carries momentum. You can't use p = mv here (mass is zero, and photons always move at c, so classical mechanics just breaks). Instead, the momentum of a photon comes from its energy: p = E/c. Since photon energy is E = hf, you can rewrite this as p = hf/c, and since c = fλ, you get the most useful form on the exam, p = h/λ.
That last version is the punchline. Momentum and wavelength are tied together by Planck's constant, and that relationship runs both directions. Light (a wave) carries particle-like momentum, and de Broglie flipped the same equation around to say particles (like electrons) have wave-like wavelengths. Photon momentum is a vector, so it has direction along the photon's path, which is why photons can push on things. When a photon hits an electron or gets absorbed by a surface, momentum is transferred and conservation of momentum still applies, just like in any collision from Unit 4 of Physics 1.
Photon momentum lives in Topic 15.1, Quantum Theory and Wave-Particle Duality, and supports learning objective 15.1.A, which asks you to describe the properties and behavior of an object that exhibits both particle-like and wave-like behavior. The CED is explicit that a photon is a massless, electrically neutral particle, and momentum is exactly the property that makes the 'particle' half of that sentence real. A photon with momentum can collide, scatter, and exert pressure, which is behavior classical wave theory can't explain. Photon momentum is also the bridge to the rest of Unit 15. The same equation, written as λ = h/p, gives matter waves, so if you understand photon momentum you've already understood half of wave-particle duality.
Keep studying AP® Physics 2 Unit 15
λ = h/p, the de Broglie wavelength (Unit 15)
This is the same equation as photon momentum, just read in the other direction. For photons, p = h/λ says waves carry particle momentum. De Broglie flipped it to λ = h/p, saying particles with momentum have a wavelength. One equation, two halves of duality.
Photon energy, E = hf (Unit 15)
Energy and momentum of a photon are locked together by p = E/c. Double the frequency and you double both the energy and the momentum. Exam questions love making you convert between the two.
Photon model (Unit 15)
Momentum is evidence that the photon model is more than a bookkeeping trick. A thing that carries momentum and transfers it in collisions behaves like a particle, which is exactly why the photoelectric effect needed photons instead of classical waves.
Quantization (Unit 15)
Because light comes in discrete photons, momentum is delivered in discrete chunks of h/λ rather than as a smooth continuous push. Quantization is the reason a single photon-electron interaction is an all-or-nothing momentum exchange.
Photon momentum shows up almost entirely as multiple-choice ratio reasoning. A typical stem gives you two light sources with frequencies f and 2f (or specific values like 6.0×10¹⁴ Hz and 9.0×10¹⁴ Hz) and asks how the photon momenta compare. The move is always the same. Since p = hf/c, momentum is directly proportional to frequency, so the ratio of momenta equals the ratio of frequencies (2:1 or 1.5:1 in those examples). Triple the frequency, triple the momentum. The other classic stem compares an electron and a photon with the same wavelength. Because p = h/λ applies to both, same wavelength means same momentum, even though their energies are very different. No released FRQ has used the phrase verbatim, but photon momentum supports the kind of qualitative wave-particle duality reasoning Topic 15.1 questions reward, so be ready to explain in words why a massless particle can still push on matter.
The instinct that 'no mass means no momentum' comes from p = mv, but that formula only works for objects with mass moving below c. Photons are massless and always travel at c, so their momentum comes from energy instead, p = E/c = h/λ. If a question gives you a photon, never reach for p = mv. Reach for h/λ or E/c.
A photon has zero mass but still carries momentum, with magnitude p = E/c = hf/c = h/λ, pointing in its direction of travel.
Photon momentum is directly proportional to frequency, so doubling the frequency doubles the momentum, and tripling it triples the momentum.
Photon momentum is inversely proportional to wavelength, so shorter-wavelength light (like X-rays) carries more momentum per photon than longer-wavelength light (like radio waves).
An electron and a photon with the same wavelength have the same momentum, because p = h/λ applies to both; this is the heart of de Broglie's matter-wave idea.
Never use p = mv for a photon; that classical formula fails for massless particles moving at c.
Momentum transfer by photons (in absorption or scattering) is particle-like behavior, which is exactly the evidence learning objective 15.1.A asks you to describe.
Photon momentum is the momentum carried by a photon, calculated as p = E/c = hf/c = h/λ. It's tested in Topic 15.1 as evidence of light's particle-like behavior under learning objective 15.1.A.
Yes, photons really do carry momentum despite being massless. Classical p = mv only applies to massive objects below light speed; a photon's momentum comes from its energy instead, p = E/c, which is why light can exert pressure and scatter off electrons.
They're related but not the same quantity. Energy is E = hf (a scalar in joules), momentum is p = h/λ (a vector in kg·m/s), and they're linked by p = E/c, so anything that changes one changes the other by the same factor.
It doubles. Since p = hf/c, momentum is directly proportional to frequency. This exact ratio reasoning, comparing photons at frequencies f and 2f, is a classic AP Physics 2 multiple-choice setup.
Yes. The relation p = h/λ holds for both photons and matter waves, so equal wavelengths mean equal momenta. Their energies differ, though, because the electron has mass and the photon's energy is E = pc.
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