The big idea: If light (a wave) can act like a particle, then matter (particles) can act like a wave. De Broglie said every moving particle has a wavelength set by its momentum p — the bigger the momentum, the shorter the wavelength.
- de Broglie wavelength (m)
- Planck constant, 6.63×10⁻³⁴ J s
- momentum of the particle (kg m s⁻¹)
Worked example — wavelength of an electron
An electron (mass 9.11 × 10⁻³¹ kg) moves at 2.0 × 10⁶ m s⁻¹. Find its de Broglie wavelength.
Solution
- First the momentum, then the given formula:
- Substitute into λ = h/p:
- Work it out — keep the unit:
Final answer
λ = 3.6 × 10⁻¹⁰ m — about the size of an atom, so it can diffract off crystals.
Electrons make diffraction patterns: Fire a beam of electrons at a thin crystal and you get a diffraction pattern of rings — exactly what waves do when they pass through gaps about the size of their wavelength.
The atomic spacing in the crystal (~10⁻¹⁰ m) matches an electron's de Broglie wavelength, so the effect shows up. This is direct proof that particles have a wave nature.
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Big, everyday objects have huge momentum, so their de Broglie wavelength is far too small to ever notice. Wave behaviour only shows up for tiny particles like electrons.
Worked example — a cricket ball
A 0.16 kg cricket ball is bowled at 40 m s⁻¹. Find its de Broglie wavelength, and comment.
Solution
- Momentum, then λ = h/p:
- Substitute:
- Compare: this is far smaller than any gap or atom, so no diffraction is ever observed.
Final answer
λ ≈ 1 × 10⁻³⁴ m — unimaginably small, so a cricket ball never behaves like a wave.
You can't have it both ways: Because particles behave as waves, you cannot know both a particle's position and its momentum exactly at the same time. The more precisely you pin down one, the less precisely you can know the other.
- uncertainty in position (m)
- uncertainty in momentum (kg m s⁻¹)
- Planck constant, 6.63×10⁻³⁴ J s
Worked example — minimum uncertainty
An electron's position is known to within Δx = 1.0 × 10⁻¹⁰ m. Find the minimum uncertainty in its momentum.
Solution
- At the minimum, the ≥ becomes =. Rearrange the given relation for Δp:
- Substitute:
- Work it out:
Final answer
Δp ≥ 5.3 × 10⁻²⁵ kg m s⁻¹.
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Where it shows up: Matter waves are HL only (E.2):
- Paper 1A — a quick λ = h/p, or 'why do electrons diffract but cricket balls don't?'. - Paper 2 — find a de Broglie wavelength (often after finding the momentum first), or a minimum uncertainty from Δx·Δp ≥ h/4π.
Three easy marks: (1) Find the momentum p = mv first, then λ = h/p. (2) A bigger momentum → a shorter wavelength. (3) Diffraction needs a gap about the size of the wavelength.
IB-style question — wavelength of a proton
A proton has a momentum of 3.0 × 10⁻²¹ kg m s⁻¹. Determine its de Broglie wavelength.
Solution
- Write the given formula first:
- Substitute:
- Work it out:
Final answer
λ = 2.2 × 10⁻¹³ m.