Wave-particle duality
Quantum Physics - OCR A-Level Physics
- Wave-particle dualityThe concept that all matter and radiation exhibit both wave-like and particle-like properties. Particles have a de Broglie wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m).The wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). associated with a moving particle, demonstrating wave-particle dualityThe concept that all matter and radiation exhibit both wave-like and particle-like properties. Particles have a de Broglie wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m).; photons exhibit particle behaviour in the photoelectric effectThe emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). is incident on it...; photons exhibit particle behaviour in the photoelectric effectThe emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). is incident on it..: all matter and radiation exhibit both wave and particle properties.
- Light behaves as a wave in diffraction and interference experiments, and as a particle in the photoelectric effectThe emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). is incident on it..
- Electrons behave as particles in thermionic emission and as waves in electron diffractionThe wave-like scattering of electrons when passed through a crystal lattice, providing evidence for wave-particle duality. experiments.
- De Broglie proposed that all moving particles have an associated wavelength, known as the de Broglie wavelengthThe wavelength associated with a moving particle, demonstrating wave-particle dualityThe concept that all matter and radiation exhibit both wave-like and particle-like properties. Particles have a de Broglie wavelength; photons exhibit particle behaviour in the photoelectric effect...
- The de Broglie wavelengthThe wavelength associated with a moving particle, demonstrating wave-particle dualityThe concept that all matter and radiation exhibit both wave-like and particle-like properties. Particles have a de Broglie wavelength; photons exhibit particle behaviour in the photoelectric effect.. is significant only for particles with very small mass (electrons, neutrons). Macroscopic objects have negligibly small wavelengths.
$$\begin{aligned}
\lambda &= \frac{h}{p} \\
&= \frac{h}{mv}
\end{aligned}$$
- Faster particles have shorter de Broglie wavelengths (λ decreases as v increases).
- Heavier particles have shorter de Broglie wavelengths (λ decreases as m increases).
- In electron diffractionThe wave-like scattering of electrons when passed through a crystal lattice, providing evidence for wave-particle duality., electrons accelerated through a thin polycrystalline graphite target produce a pattern of concentric rings, confirming their wave nature.
- Increasing the accelerating voltageThe energyThe capacity to do work. Measured in joules (J). transferred per unit chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). between two points. Measured in volts (V). Informal term for potential difference. increases electron speed, decreasing λ, so the diffraction rings become smaller and closer together.
Worked Example [3 marks]
Calculate the de Broglie wavelengthThe wavelength associated with a moving particle, demonstrating wave-particle duality. of an electron travelling at 2.0 × 10⁶ m s⁻¹. (m_e = 9.11 × 10⁻³¹ kg)
Show Solution
1
Use λ = h/(mv) = (6.63 × 10⁻³⁴) / (9.11 × 10⁻³¹ × 2.0 × 10⁶)
[1]2
$\lambda = (6.63 \times 10⁻^{3}⁴) / (1.822 \times 10⁻^{2}⁴) = 3.64 \times 10⁻¹⁰\;\text{m}$
[1]3
This is comparable to atomic spacing (~10⁻¹⁰ m), explaining why electron diffraction is observable.
[1]Answer
$\lambda = 3.6 \times 10⁻¹⁰\;\text{m} (0.36 nm)$
Common Mistake
MEDIUM
Students often: Stating that only electrons exhibit wave-particle duality.
Instead: All particles exhibit wave-particle duality, including protons, neutrons, and even atoms. However, the de Broglie wavelength is only detectable for very light particles because $\lambda = h/(mv) gives negligibly small wavelengths$ for macroscopic masses.
Instead: All particles exhibit wave-particle duality, including protons, neutrons, and even atoms. However, the de Broglie wavelength is only detectable for very light particles because $\lambda = h/(mv) gives negligibly small wavelengths$ for macroscopic masses.
Examiner Tips and Tricks
- Electron diffraction does not prove electrons ARE waves.
- It proves electrons EXHIBIT wave-like behaviour.
- They are still detected as individual particles on the screen.
- This distinction matters for full marks.