E.3 Quantum physics: photons and matter waves

IB Physics HL· 35 min read

1. Photons: The Particle Nature of Light★★☆☆☆⏱ 10 min

📘 Definition

Photon

E = hf

A discrete quantum of electromagnetic radiation with zero rest mass, that travels at speed $c$ in vacuum. Photons are emitted and absorbed only in whole numbers, with energy proportional to their frequency.

Example:

A visible light photon has energy ~2 eV, while an X-ray photon has energy >100 eV.

The photon model was developed to explain phenomena that the classical wave model of light could not account for, most notably the photoelectric effect. Two key relations describe photon energy:

E = hf = \frac{hc}{\lambda}

📐 Worked Example

Calculate the energy of a 450 nm blue photon, in electron volts.

1
Use the simplified relation for energy in eV:
2
$E = \frac{hc}{\lambda} = \frac{1240 \text{ eV·nm}}{450 \text{ nm}}$
3
Calculate the result: $E \approx 2.76$ eV. To confirm in joules:
4
$E = \frac{(6.63 \times 10^{-34} \text{ Js})(3 \times 10^8 \text{ m/s})}{450 \times 10^{-9} \text{ m}} = 4.42 \times 10^{-19} \text{ J}$
5
Convert to eV by dividing by $1.6 \times 1 0^{- 19} J/eV$ , giving ~2.76 eV, matching the quick calculation.

Exam tip:

Always check if the question asks for energy in joules or electron volts, and convert units correctly.

2. De Broglie Hypothesis and Matter Waves★★★☆☆⏱ 15 min

📘 Definition

De Broglie Wavelength

\lambda = \frac{h}{p}

The wavelength of the matter wave associated with any moving massive particle, where $p$ is the particle's momentum $p = m v$ for non-relativistic speeds.

Example:

An electron accelerated through 100 V has a de Broglie wavelength of ~0.1 nm, comparable to atomic spacing in crystals.

Louis de Broglie proposed that just as light exhibits both wave and particle properties, all massive particles also have wave characteristics. This hypothesis was experimentally confirmed by the Davisson-Germer experiment, which observed diffraction of electrons off a nickel crystal lattice.

📐 Worked Example

Calculate the de Broglie wavelength of an electron moving at $1.0 \times 1 0^{6} m/s$ , where $m_{e} = 9.11 \times 1 0^{- 31} kg$ .

1
First calculate momentum $p$ :
2
$p = m v = (9.11 \times 10^{-31} \text{ kg})(1.0 \times 10^6 \text{ m/s}) = 9.11 \times 10^{-25} \text{ kg·m/s}$
3
Substitute into the de Broglie relation:
4
$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34} \text{ Js}}{9.11 \times 10^{-25} \text{ kg·m/s}} \approx 7.3 \times 10^{-10} \text{ m} = 0.73 \text{ nm}$
5
This wavelength is similar to atomic spacing in solids, explaining why electrons diffract through crystalline lattices, proving their wave nature.

3. Wave-Particle Duality★★★☆☆⏱ 12 min

Wave-particle duality is the core quantum concept that all entities (light and matter) exhibit both wave and particle properties. No single experiment can observe both behaviors at the same time; the observed behavior depends on the type of measurement. The table below summarizes experimental evidence:

Entity	Evidence for Particle Behavior	Evidence for Wave Behavior
Light	Photoelectric effect, Compton scattering	Interference, diffraction
Matter	Localized detection, momentum transfer	Electron diffraction, neutron interference

📐 Worked Example

Explain why the wave nature of macroscopic objects (e.g. a 0.1 kg ball moving at 10 m/s) is never observed.

1
First calculate the momentum of the ball:
2
$p = mv = (0.1 \text{ kg})(10 \text{ m/s}) = 1 \text{ kg·m/s}$
3
Calculate the de Broglie wavelength:
4
$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34} \text{ Js}}{1 \text{ kg·m/s}} = 6.63 \times 10^{-34} \text{ m}$
5
This wavelength is many orders of magnitude smaller than any possible aperture or gap that could produce observable diffraction effects, so wave behavior cannot be detected for macroscopic objects.

4. Exam Phrasing and Concept Check★★☆☆☆⏱ 8 min

✓ Quick check

Test your understanding of core concepts:

Which property of a photon does not change when it travels from air to glass?
- A) Wavelength
- B) Frequency
- C) Speed
- D) Energy
Reveal answer
B —
Frequency (and hence energy $E = h f$ ) is an intrinsic property of the photon. Speed and wavelength both decrease when light enters glass from air.
What is the de Broglie wavelength of a proton with momentum $2.0 \times 1 0^{- 22} kg \cdotp m/s$ ?
- A) $3.3 \times 1 0^{- 12} m$
- B) $3.3 \times 1 0^{- 9} m$
- C) $1.3 \times 1 0^{- 11} m$
- D) $3.0 \times 1 0^{11} m$
Reveal answer
A —
$λ = h / p = 6.63 \times 1 0^{- 34} /2.0 \times 1 0^{- 22} \approx 3.3 \times 1 0^{- 12} m$

5. Common Pitfalls

Wrong move:

Forgetting to convert wavelength from nanometers to meters for SI unit calculations

Why:

IB problems often give wavelength in nm, leading to answers 9 orders of magnitude too large if you forget conversion

Correct move:

Always multiply wavelength in nm by $1 0^{- 9}$ to get meters when calculating energy in joules

Wrong move:

Using mass instead of momentum in the de Broglie relation

Why:

Wavelength depends on momentum, not just mass; two particles of the same mass with different speeds have different wavelengths

Correct move:

Always calculate momentum $p = m v$ first before substituting into $λ = h / p$

Wrong move:

Claiming light is sometimes a wave and sometimes a particle

Why:

Wave-particle duality means light always has both properties, only one is observed in a given experiment

Correct move:

State that all quantum entities inherently exhibit both wave and particle properties, with one behavior manifested depending on the measurement

Wrong move:

Using $E = h c / λ$ for matter waves to calculate kinetic energy

Why:

This relation only applies to massless photons, not to massive particles with kinetic energy

Correct move:

Only use $λ = h / p$ for matter; calculate kinetic energy from momentum using $K E = p^{2} / (2 m)$ for non-relativistic speeds

Wrong move:

Forgetting that photon energy depends on frequency, not intensity

Why:

Intensity is the number of photons per unit area, not the energy per photon

Correct move:

Remember that each photon's energy depends only on its frequency, so increasing intensity does not change the energy per photon

6. Quick Reference Cheatsheet

Concept	Key Relation	Useful Exam Note
Photon Energy	$E = h f = \frac{h c}{λ}$	$h c = 1240 eV \cdotp nm$ for quick eV calculations
De Broglie Wavelength	$λ = \frac{h}{p} = \frac{h}{m v}$	p = momentum, non-relativistic for $v << c$
Accelerated charged particle	$λ = \frac{h}{2 m e V}$	For particle with charge $e$ accelerated through p.d. $V$
Wave-Particle Duality Evidence		Particle: Photoelectric effect, Compton scattering \| Wave: Interference, diffraction

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · Paper 1
Photon energy calculation question
2024 · Paper 2
De Broglie wavelength problem
2023 · Paper 1
Wave-particle duality concept question

Going deeper

simulationPhoton and electron diffraction simulationInteractive visualization of matter waves

What's Next

This subtopic lays the foundational quantum concepts you need for all further quantum physics topics in IB Physics HL. The photon model is core to understanding the photoelectric effect, which is explored in the next subtopic. Matter wave concepts underpin the Heisenberg uncertainty principle, quantum tunneling, and the quantum model of the atom, all of which are assessed in IB HL exams. Wave-particle duality also explains the working of modern technologies like electron microscopes and lasers, which often come up in exam context questions.