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6.4.4 de Broglie and Matter Waves

By the end of this lesson you can:

  • State de Broglie's hypothesis
  • Calculate the wavelength of a particle
  • Describe evidence for matter waves
Find your course: NSW Year 11NSW Year 12: Module 8, IQ3QLD Unit 4: Quantum theoryVIC Unit 4: Light, matter and relativityAP Physics 2: Unit 15IB: E.2 Quantum physics
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Learn the idea: a short explanation, then a check.

In the previous lesson, we explored the Bohr model of the atom. While it represented a major advancement over Rutherford's planetary model—which itself improved upon J.J. Thomson’s Plum Pudding Model—it still suffered from key limitations.

Chief among these was Bohr’s inability to explain why electrons were restricted to discrete, quantized energy levels and stable orbits. Louis de Broglie addressed this fundamental question by introducing a revolutionary concept: electrons don't just act as particles—they also behave as standing waves.

De Broglie's Matter Wave Hypothesis

To explain why electron orbits are quantized, Louis de Broglie hypothesized in 1924 that matter—like light—exhibits dual wave-particle properties. He proposed that any moving particle with momentum (p = m v) has an associated de Broglie wavelength (λ):

λ = h⁄p = h⁄m v

When applied to an electron orbiting a nucleus at radius r, de Broglie suggested that the electron acts as a standing circular wave.

For a standing wave to sustain itself without destructive interference, the circumference of the electron's orbit must equal an integer number (n) of whole wavelengths:

2πr = n λ   (where n = 1, 2, 3, …)

Substituting λ = h⁄m v directly yields Bohr's angular momentum quantization condition:

m v r = n h⁄2π = nħ

This provided the physical foundation for Bohr's model: electrons do not radiate energy in stable orbits because they exist as stationary standing waves rather than accelerating point charges.

Watch the video to see how this standing wave condition dictates stable atomic orbitals.

Demonstration
We can model de Broglie standing wave model with a simple apparatus.
​In this video I examine a useful demonstration to help understand the wavelike nature of the electron.
📝Practise
Worked examples first, then problems to try, then worksheets.
Sample Problem
We are now ready to try a sample problem
Below is a sample problem with a video that explain how to solve it. It is suggested you try the problem beforehand, as this actually aids understanding, even if you are unsure if you are correct.
Picture

Practice Problems: Matter Waves and the de Broglie Wavelength

The de Broglie wavelength of a particle is λ = h / p = h / (mv), where Planck’s constant h = 6.626 × 10-34 J s.

1. Determine the de Broglie wavelength of the matter wave associated with a cricket ball of mass 0.175 kg and velocity 23.6 m s-1. Use the answer to this question to explain why we do not observe the matter waves associated with macroscopic objects.

Given: m = 0.175 kg, v = 23.6 m s-1

Formula: λ = h / p = h / (mv)

Calculation: λ = (6.626 × 10-34) / (0.175 × 23.6) = (6.626 × 10-34) / 4.13 = 1.60 × 10-34 m

Explanation: This wavelength is incredibly small. It is about 1019 times smaller than an atomic nucleus (about 10-15 m). Wave effects such as diffraction and interference only show up when the wavelength is similar to the size of the gaps or objects the wave meets. Nothing is small enough to diffract a wave of this length, so the wave nature of the cricket ball is far too small to be observed. Macroscopic objects behave as particles.

Answer: λ = 1.6 × 10-34 m, far too small to observe
2. Calculate the de Broglie wavelength of an electron travelling at 106 m s-1. (Mass of an electron is 9.11 × 10-31 kg.) Use the answer to this question to explain why we would expect to observe the effects of the matter waves associated with electrons. Give one example of these effects.

Given: m = 9.11 × 10-31 kg, v = 1.0 × 106 m s-1

Formula: λ = h / (mv)

Calculation: λ = (6.626 × 10-34) / (9.11 × 10-31 × 1.0 × 106) = (6.626 × 10-34) / (9.11 × 10-25) = 7.27 × 10-10 m

Explanation: This wavelength is of the order of the size of an atom and the spacing between atoms in a crystal (about 10-10 m). The mass of the electron is so small that its wavelength is big enough for the gaps between atoms to diffract it, so wave effects can be observed.

Example: Electron diffraction. When a beam of electrons is directed at a thin crystal or a thin film of graphite, the electrons produce a diffraction pattern (rings or spots) like the pattern for waves, as in the Davisson-Germer experiment. Electron microscopes also use the wave nature of electrons.

Answer: λ = 7.3 × 10-10 m, similar to atomic spacing, so electrons can be diffracted
3. A proton is travelling at a speed of 1.5 × 107 m s-1. Determine the de Broglie wavelength of the proton, given its mass is 1.67 × 10-27 kg.

Given: m = 1.67 × 10-27 kg, v = 1.5 × 107 m s-1

Formula: λ = h / (mv)

Calculation: λ = (6.626 × 10-34) / (1.67 × 10-27 × 1.5 × 107) = (6.626 × 10-34) / (2.505 × 10-20) = 2.65 × 10-14 m

Answer: λ = 2.6 × 10-14 m
4. Calculate the momentum of a neutron if it has a de Broglie wavelength of 1.59 × 10-13 m.

Given: λ = 1.59 × 10-13 m

Formula: λ = h / p, so p = h / λ

Calculation: p = (6.626 × 10-34) / (1.59 × 10-13) = 4.17 × 10-21 kg m s-1 (N s)

Answer: p = 4.17 × 10-21 N s
5. Determine the speed of the neutron in question 4, given that the mass of a neutron is 1.67 × 10-27 kg.

Given: p = 4.17 × 10-21 N s, m = 1.67 × 10-27 kg

Formula: p = mv, so v = p / m

Calculation: v = (4.17 × 10-21) / (1.67 × 10-27) = 2.50 × 106 m s-1

This is only about 0.8% of the speed of light, so there is no need to use relativity.

Answer: v = 2.50 × 106 m s-1
6. An electron volt (eV) is an energy unit equivalent to the work done when an electron is moved through a potential difference of 1 volt. If an electron has a kinetic energy of 100 eV, what is its associated de Broglie wavelength? (Charge on an electron is 1.6 × 10-19 C.)

Given: K = 100 eV, q = 1.6 × 10-19 C, m = 9.11 × 10-31 kg

Step 1 (Convert to joules): 1 eV = 1.6 × 10-19 J, so K = 100 × 1.6 × 10-19 = 1.6 × 10-17 J

Step 2 (Momentum): K = p2 / (2m), so p = √(2mK) = √(2 × 9.11 × 10-31 × 1.6 × 10-17) = √(2.92 × 10-47) = 5.40 × 10-24 N s

Step 3 (Wavelength): λ = h / p = (6.626 × 10-34) / (5.40 × 10-24) = 1.23 × 10-10 m

(The electron’s speed is about 5.9 × 106 m s-1, about 2% of the speed of light, so the non-relativistic formula is fine.)

Answer: λ = 1.2 × 10-10 m
🕹️Explore
Test your predictions in a simulation.
Interactive
This animation looks that the standing waves that De Broglie hypothesised for the Bohr atom. By Tom Walsh


​The experiment that demonstrated that electrons act like waved

Experimental Proof: The Davisson–Germer Experiment

At the time of its publication in 1924, Louis de Broglie’s matter wave hypothesis was purely theoretical and lacked empirical evidence.

That changed in 1927 when American physicists Clinton Davisson and Lester Germer experimentally confirmed the wave properties of electrons. While firing a beam of slow-moving electrons at a crystalline nickel target, they observed a distinct pattern of constructive and destructive interference—a phenomenon unique to wave behavior.

By applying Bragg’s Law for crystal diffraction (n λ = 2 d sin θ) to the observed scattering angles, they calculated an experimental electron wavelength that matched de Broglie’s theoretical prediction (λ = h⁄p) with remarkable precision.

Watch the video to explore how the Davisson–Germer experiment provided the definitive proof that electrons behave as waves.