# E.3 Quantum physics: photons and matter waves

> IB Physics HL · Theme E: Nuclear and quantum physics
> Source: https://www.owlsprep.com/study/ib-physics-hl-u5-e-3-quantum-physics-photons/

This subtopic introduces the core quantum concepts of the particle nature of light as photons, the wave nature of matter, and wave-particle duality. You will learn key relations and apply them to common IB exam problems.

**Prerequisites:** [Basic wave properties of electromagnetic radiation](https://www.owlsprep.com/study/ib-physics-hl-wave-behaviour-fundamentals/); [Classical momentum and kinetic energy](https://www.owlsprep.com/study/ib-physics-hl-mechanics-momentum/)

## Learning objectives

- Describe the particle nature of electromagnetic radiation as discrete photons
- Apply the de Broglie relation to calculate matter wavelengths for moving particles
- Summarize experimental evidence for wave-particle duality for both light and matter
- Solve common IB exam problems involving photon energy and matter waves

## Photons: The Particle Nature of Light

**Photon** — 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.

*Notation:* E = hf

*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:

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

> **tip**
>
> For IB exams, remember that $hc \approx 1240 \text{ eV·nm}$, which lets you quickly calculate photon energy in electron volts without unit conversions for wavelength given in nm.

**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 $1.6 \times 10^{-19} \text{ J/eV}$, 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.

## De Broglie Hypothesis and Matter Waves

**De Broglie Wavelength** — The wavelength of the matter wave associated with any moving massive particle, where $p$ is the particle's momentum $p=mv$ for non-relativistic speeds.

*Notation:* \lambda = \frac{h}{p}

*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.

> **tip**
>
> For a charged particle accelerated through potential difference $V$, kinetic energy $KE = eV = \frac{p^2}{2m}$, so momentum $p = \sqrt{2meV}$. This is a common starting point for IB problems.

**Worked example:** Calculate the de Broglie wavelength of an electron moving at $1.0 \times 10^6 \text{ m/s}$, where $m_e = 9.11 \times 10^{-31} \text{ kg}$.

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.

## Wave-Particle Duality

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.

## Exam Phrasing and Concept Check

**Exam command terms**

- **Explain wave-particle duality** — You must state that all quantum entities have both wave and particle properties, and give one piece of evidence for each for light and matter for full marks. *(A 5 mark question expects 2 pieces of evidence for each behavior.)*

- **Calculate the de Broglie wavelength** — Always start from $\lambda = h/p$, show calculation of momentum first, and give final answer with correct units.

- **Show that** — You must show all intermediate calculation steps to reach the approximate value given in the question.

**Check your understanding**

Test your understanding of core concepts:

1. Which property of a photon does not change when it travels from air to glass?

   - A) Wavelength
   - B) Frequency
   - C) Speed
   - D) Energy

   *Why:* Frequency (and hence energy $E=hf$) is an intrinsic property of the photon. Speed and wavelength both decrease when light enters glass from air.

2. What is the de Broglie wavelength of a proton with momentum $2.0 \times 10^{-22} \text{ kg·m/s}$?

   - A) $3.3 \times 10^{-12} \text{ m}$
   - B) $3.3 \times 10^{-9} \text{ m}$
   - C) $1.3 \times 10^{-11} \text{ m}$
   - D) $3.0 \times 10^{11} \text{ m}$

   *Why:* $\lambda = h/p = 6.63 \times 10^{-34} / 2.0 \times 10^{-22} \approx 3.3 \times 10^{-12} \text{ m}$

## Common pitfalls

- **Wrong:** Forgetting to convert wavelength from nanometers to meters for SI unit calculations
  - Why it fails: IB problems often give wavelength in nm, leading to answers 9 orders of magnitude too large if you forget conversion
  - Correct: Always multiply wavelength in nm by $10^{-9}$ to get meters when calculating energy in joules
- **Wrong:** Using mass instead of momentum in the de Broglie relation
  - Why it fails: Wavelength depends on momentum, not just mass; two particles of the same mass with different speeds have different wavelengths
  - Correct: Always calculate momentum $p=mv$ first before substituting into $\lambda = h/p$
- **Wrong:** Claiming light is sometimes a wave and sometimes a particle
  - Why it fails: Wave-particle duality means light always has both properties, only one is observed in a given experiment
  - Correct: State that all quantum entities inherently exhibit both wave and particle properties, with one behavior manifested depending on the measurement
- **Wrong:** Using $E = hc/\lambda$ for matter waves to calculate kinetic energy
  - Why it fails: This relation only applies to massless photons, not to massive particles with kinetic energy
  - Correct: Only use $\lambda = h/p$ for matter; calculate kinetic energy from momentum using $KE = p^2/(2m)$ for non-relativistic speeds
- **Wrong:** Forgetting that photon energy depends on frequency, not intensity
  - Why it fails: Intensity is the number of photons per unit area, not the energy per photon
  - Correct: Remember that each photon's energy depends only on its frequency, so increasing intensity does not change the energy per photon

## Cheatsheet

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

## 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.

- [E.4 The nuclear atom](https://www.owlsprep.com/study/ib-physics-hl-u5-e-4-the-nuclear-atom/)
- [E.6 Fission and fusion (AHL)](https://www.owlsprep.com/study/ib-physics-hl-u5-e-6-fission-and-fusion/)

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