HSC Physics — Module 8
de Broglie Wavelength — Flashcards & Quiz
Louis de Broglie proposed in 1924 that every particle with momentum has an associated wavelength, λ = h/mv, extending wave–particle duality from light to matter. In the NSW Physics syllabus this sits in Year 12 Module 8 From the Universe to the Atom, under the Quantum Mechanical Nature of the Atom, where you investigate de Broglie's matter waves and the experimental evidence for the formula. You should be able to calculate wavelengths for electrons and other particles, explain why everyday objects show no wave behaviour, and describe electron diffraction evidence such as the Davisson–Germer experiment.
Key Points
- Every particle with momentum has an associated wavelength λ = h/mv (that is, λ = h/p), where h is Planck's constant, 6.626 × 10⁻³⁴ J s.
- For macroscopic objects (a cricket ball, a person) the wavelength is around 10⁻³⁴ m, far too small to observe wave behaviour.
- For electrons at typical laboratory speeds the wavelength is around 10⁻¹⁰ m, similar to atomic spacing in crystals, so electrons can be diffracted.
- Davisson and Germer (1927) observed electron diffraction from a nickel crystal, confirming the wave nature of matter.
- De Broglie explained Bohr's quantised orbits as standing waves: a whole number of wavelengths fits the circumference, 2πr = nλ.
- Matter waves led to Schrödinger's wave model of the atom, which replaced fixed orbits with orbitals.
Common Mistakes to Avoid
- Applying λ = h/mv to macroscopic objects without noting that the wavelength is negligibly small.
- Confusing momentum (p = mv) with kinetic energy (½mv²).
- Forgetting to convert electron energies from eV to joules before using p = √(2mE_k).
- Forgetting that de Broglie's hypothesis extended wave–particle duality from light to matter.
- Describing electron diffraction without explaining why only a wave produces a diffraction pattern.
Exam Strategy
Module 8 is Year 12 content and can be examined in the HSC Physics examination. Matter-wave questions often ask you to calculate a wavelength or explain the evidence for the hypothesis. Method: (1) find the momentum, p = mv, or from kinetic energy, p = √(2mE_k), converting eV to joules, (2) apply λ = h/mv, (3) compare the result with atomic spacing (about 10⁻¹⁰ m) to judge whether diffraction is observable, (4) for evidence, describe the electron diffraction pattern and explain why it shows wave behaviour.
Sample Flashcards
Q1: State de Broglie's hypothesis.
In 1924 Louis de Broglie proposed that if light, long understood as a wave, can behave like particles (photons), then particles of matter should also have wave properties. Any particle with momentum has an associated wavelength λ = h/mv, where h is Planck's constant (6.626 × 10⁻³⁴ J s), m is the particle's mass and v is its velocity. These waves are called matter waves.
Q2: Calculate the de Broglie wavelength of an electron moving at 2.0 × 10⁶ m s⁻¹.
λ = h/mv = (6.626 × 10⁻³⁴) ÷ (9.109 × 10⁻³¹ × 2.0 × 10⁶) = 6.626 × 10⁻³⁴ ÷ 1.82 × 10⁻²⁴ ≈ 3.6 × 10⁻¹⁰ m. This is similar to the spacing between atoms in a crystal (about 10⁻¹⁰ m), so electrons at this speed can be diffracted by a crystal lattice, revealing their wave nature.
Q3: Why does a moving cricket ball show no wave behaviour?
A 0.16 kg cricket ball moving at 40 m s⁻¹ has momentum 6.4 kg m s⁻¹, so λ = 6.626 × 10⁻³⁴ ÷ 6.4 ≈ 1.0 × 10⁻³⁴ m. This is vastly smaller than any gap or obstacle, even an atomic nucleus (about 10⁻¹⁵ m), so diffraction and interference cannot be observed. Wave behaviour is only noticeable for particles with very small momentum, such as electrons.
Q4: What experimental evidence supports de Broglie's matter waves?
In 1927 Davisson and Germer fired a beam of electrons at a nickel crystal and found the scattered electrons were strongest at particular angles, forming a diffraction pattern. Diffraction is a wave property, and the wavelength calculated from the pattern matched de Broglie's λ = h/mv. G. P. Thomson independently produced electron diffraction rings by passing electrons through thin metal foil.
Q5: How did de Broglie explain Bohr's quantised electron orbits?
Bohr had simply postulated that electrons occupy only certain allowed orbits, with angular momentum mvr = nh/2π. De Broglie explained this by treating the electron as a standing wave around the nucleus: an orbit is stable only if a whole number of wavelengths fits its circumference, 2πr = nλ. Substituting λ = h/mv gives mvr = nh/2π, so quantisation follows from the electron's wave nature.
Revision Tip
de Broglie calculations are short and formulaic — drill Revizi flashcards with 5-6 different particles (electron, proton, neutron, alpha particle).
Related Concepts
Last updated: 9 October 2026 · 6 sample flashcards