Wave–particle duality and de Broglie wavelength

Key ideas

Wave–particle duality is the idea that light and matter both show wave-like behaviour (diffraction, interference) and particle-like behaviour (discrete, localised energy packets), depending on the experiment used to observe them.
- Light behaves as a wave in interference and diffraction experiments (Level 2/3 wave optics), but as a particle (photons) in the photoelectric effect.
- This duality is not limited to light — matter shows it too.
De Broglie's hypothesis: every moving particle with momentum $p$ has an associated wavelength:

$\lambda = \frac{h}{p}$

Each variable, with sub-bullets:
- $\lambda$ — the de Broglie wavelength of the particle (m)
- $h$ — the Planck constant, $6.63\times10^{-34}$ J s
- $p$ — the particle's momentum, $p = mv$ for a non-relativistic particle (kg m s⁻¹)
Because $h$ is so small, everyday objects (a thrown ball, a car) have an immeasurably tiny de Broglie wavelength — their wave nature is completely undetectable. Only very light, fast particles (electrons, protons) have a wavelength large enough to observe wave effects.

Evidence: electron diffraction

Electron diffraction is the key experimental evidence for de Broglie's hypothesis.
- A beam of electrons is fired at a thin crystal (or diffraction grating with spacing comparable to $\lambda$ ).
- Diffraction and interference patterns — bright and dark rings or fringes — are observed on a detector behind the crystal.
- Diffraction and interference are wave phenomena: only waves can produce this pattern of reinforcement and cancellation. Their appearance with a beam of electrons — which are unambiguously particles, with mass and charge — is direct evidence that matter has wave-like properties, exactly as de Broglie predicted.
- The spacing of the diffraction pattern matches the wavelength calculated from $\lambda = h/p$ using the electrons' known momentum, confirming the formula quantitatively as well as qualitatively.

Worked ExampleDe Broglie wavelength of an electron

An electron ( $m = 9.11\times10^{-31}$ kg) is accelerated to a speed of $v = 4.00\times10^6$ m s⁻¹. Find its de Broglie wavelength.

Step 1 — Find the electron's momentum

$p = mv = (9.11\times10^{-31})(4.00\times10^6) = 3.644\times10^{-24} \text{ kg m s}^{-1}$

Step 2 — Apply the de Broglie relationship

$\lambda = \frac{h}{p} = \frac{6.63\times10^{-34}}{3.644\times10^{-24}} = 1.819...\times10^{-10} \text{ m} \approx 1.82\times10^{-10} \text{ m}$

This is comparable to the spacing between atoms in a crystal ( $\sim10^{-10}$ m) — which is exactly why electron beams diffract from crystals, the same way X-rays do.

Tips

Say which behaviour is being tested by which experiment. "Light behaves as a wave in double-slit interference, and as a particle in the photoelectric effect" is the kind of explicit statement Excellence answers need — don't just assert duality without naming the evidence for each side.
For electron diffraction, the logical chain examiners want is: diffraction patterns require waves → electrons (particles) produce diffraction patterns → therefore electrons have wave properties. State all three links, not just the conclusion.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

An electron has momentum $p = 5.00\times10^{-24}$ kg m s⁻¹.

Calculate its de Broglie wavelength.

Merit

A proton ( $m = 1.67\times10^{-27}$ kg) is accelerated from rest through a potential difference of $500$ V, gaining kinetic energy $E_k = eV = 500$ eV $= 8.01\times10^{-17}$ J.

Calculate the proton's speed and its de Broglie wavelength.

Excellence

Explain how electron diffraction provides evidence for wave–particle duality, and explain why this effect is never observed for everyday macroscopic objects such as a thrown ball.

Wave–particle duality and de Broglie wavelength

Covers AS91525

Key ideas

Wave–particle duality is the idea that light and matter both show wave-like behaviour (diffraction, interference) and particle-like behaviour (discrete, localised energy packets), depending on the experiment used to observe them.
- Light behaves as a wave in interference and diffraction experiments (Level 2/3 wave optics), but as a particle (photons) in the photoelectric effect.
- This duality is not limited to light — matter shows it too.
De Broglie's hypothesis: every moving particle with momentum $p$ has an associated wavelength:

$\lambda = \frac{h}{p}$

Each variable, with sub-bullets:
- $\lambda$ — the de Broglie wavelength of the particle (m)
- $h$ — the Planck constant, $6.63\times10^{-34}$ J s
- $p$ — the particle's momentum, $p = mv$ for a non-relativistic particle (kg m s⁻¹)
Because $h$ is so small, everyday objects (a thrown ball, a car) have an immeasurably tiny de Broglie wavelength — their wave nature is completely undetectable. Only very light, fast particles (electrons, protons) have a wavelength large enough to observe wave effects.

Evidence: electron diffraction

Electron diffraction is the key experimental evidence for de Broglie's hypothesis.
- A beam of electrons is fired at a thin crystal (or diffraction grating with spacing comparable to $\lambda$ ).
- Diffraction and interference patterns — bright and dark rings or fringes — are observed on a detector behind the crystal.
- Diffraction and interference are wave phenomena: only waves can produce this pattern of reinforcement and cancellation. Their appearance with a beam of electrons — which are unambiguously particles, with mass and charge — is direct evidence that matter has wave-like properties, exactly as de Broglie predicted.
- The spacing of the diffraction pattern matches the wavelength calculated from $\lambda = h/p$ using the electrons' known momentum, confirming the formula quantitatively as well as qualitatively.

Worked ExampleDe Broglie wavelength of an electron

An electron ( $m = 9.11\times10^{-31}$ kg) is accelerated to a speed of $v = 4.00\times10^6$ m s⁻¹. Find its de Broglie wavelength.

Step 1 — Find the electron's momentum

$p = mv = (9.11\times10^{-31})(4.00\times10^6) = 3.644\times10^{-24} \text{ kg m s}^{-1}$

Step 2 — Apply the de Broglie relationship

$\lambda = \frac{h}{p} = \frac{6.63\times10^{-34}}{3.644\times10^{-24}} = 1.819...\times10^{-10} \text{ m} \approx 1.82\times10^{-10} \text{ m}$

This is comparable to the spacing between atoms in a crystal ( $\sim10^{-10}$ m) — which is exactly why electron beams diffract from crystals, the same way X-rays do.

Tips

Say which behaviour is being tested by which experiment. "Light behaves as a wave in double-slit interference, and as a particle in the photoelectric effect" is the kind of explicit statement Excellence answers need — don't just assert duality without naming the evidence for each side.
For electron diffraction, the logical chain examiners want is: diffraction patterns require waves → electrons (particles) produce diffraction patterns → therefore electrons have wave properties. State all three links, not just the conclusion.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

An electron has momentum $p = 5.00\times10^{-24}$ kg m s⁻¹.

Calculate its de Broglie wavelength.

Merit

A proton ( $m = 1.67\times10^{-27}$ kg) is accelerated from rest through a potential difference of $500$ V, gaining kinetic energy $E_k = eV = 500$ eV $= 8.01\times10^{-17}$ J.

Calculate the proton's speed and its de Broglie wavelength.

Excellence

Explain how electron diffraction provides evidence for wave–particle duality, and explain why this effect is never observed for everyday macroscopic objects such as a thrown ball.