Wave-particle duality
Introduced| English | Meaning |
|---|---|
| diffraction/dɪˈfrækʃn/ | The spreading or redistribution of waves as they pass an edge or aperture. |
| wave-particle duality/weɪv ˈpɑːtɪkl djuːˈælɪti/ | Radiation and matter both show wave behaviour (interference, diffraction) and particle behaviour (the photoelectric effect). |
| de Broglie wavelength/də ˈbrəʊli ˈweɪvleŋθ/ | The wavelength associated with a particle's momentum, equal to the Planck constant divided by that momentum. |
| interference/ˌɪntəˈfɪərəns/ | The superposition of waves, producing a combined displacement that depends on their relative phase. |
| diffraction grating/dɪˈfrækʃn ˈɡreɪtɪŋ/ | An optical component with many equally spaced slits or grooves that produces interference maxima at particular angles. |
| electron diffraction/ɪˈlektrɒn dɪˈfrækʃn/ | The spreading and interference of electron waves when they encounter structures such as a crystal lattice. |
| Planck constant/plæŋk ˈkɒnstənt/ | The constant relating a photon's energy to its frequency and setting the scale of quantum effects. |
| polycrystalline/ˌpɒlɪˈkrɪstəlaɪn/ | Made of many small crystals with different orientations. |
Fire electrons at graphite and you get rings
- Electrons are particles. Everyone agrees: they have mass, they carry charge, they arrive one at a time.
- Fire a beam of them at a thin sheet of graphite and the screen shows concentric bright rings.
- Rings are a diffraction 衍射 pattern, and only waves diffract. Yet these are electrons.
- This lesson is that contradiction and its resolution: wave-particle duality 波粒二象性 and the de Broglie wavelength 德布罗意波长 $\lambda = h/p$.
The two-sided evidence
- Electromagnetic radiation, and matter, can show both wave properties such as interference and diffraction and particle properties such as the photoelectric effect. That is the two-mark answer.
- Asked for one piece of evidence each, for radiation: particle side, the photoelectric effect (a threshold frequency, and immediate emission even when dim); wave side, interference 干涉 or diffraction (Young's slits, a diffraction grating 衍射光栅).
- For matter, the wave evidence is electron diffraction 电子衍射. Name the phenomenon, not just the word "wave".
Light shows its wave nature in interference, and its particle nature in:
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The photoelectric effect needs photons (particles); interference and diffraction need waves.
Match each nature to the evidence for it.
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Name the phenomenon, not the category. "It behaves like a wave" is a restatement of the question, not evidence.
The de Broglie hypothesis
- If a wave can act like particles, perhaps particles can act like waves. Every moving particle has a wavelength:
- The de Broglie wavelength is the wavelength associated with a moving particle, where $h$ is the Planck constant 普朗克常量 and $p$ the momentum of the particle. Name both symbols when asked.
- It is the same $\lambda = h/p$ that a photon obeys. One relation, two kinds of object.
The de Broglie wavelength
λ = h/p
A particle's wavelength is inversely proportional to its momentum — faster, heavier particles have shorter waves.
The de Broglie wavelength of a particle is:
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$\lambda = \dfrac{h}{p}$ — wavelength is Planck's constant over momentum.
Why you never diffract in a doorway
- $h$ is tiny, so an everyday object has an absurdly small wavelength. A $0.10\ \text{kg}$ ball at $10\ \text{m/s}$ has
- That is far smaller than any slit or lattice spacing, and diffraction only shows when the gap is comparable to the wavelength.
- An electron is different: at $4.9\times10^{7}\ \text{m/s}$ its momentum is $4.46\times10^{-23}\ \text{kg m/s}$, giving $\lambda = 1.5\times10^{-11}\ \text{m}$, about $0.015\ \text{nm}$. That is the size of the gaps between atoms in a crystal, which is exactly why a crystal can diffract it.
A 0.10 kg ball moves at 10 m/s. What is its de Broglie wavelength, in units of 1e-34 m?
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p = 1.0 kg m/s, so lambda = 6.63e-34 m. That is far smaller than any slit or lattice spacing, which is why everyday objects never diffract.
Electron diffraction, described
- The four-mark description runs in four moves, and each is a mark.
- One. Electrons from a heated filament are accelerated through a high p.d. into a beam.
- Two. The beam passes through a thin polycrystalline 多晶的 graphite film, whose atomic spacing of about $10^{-10}\ \text{m}$ acts as a diffraction grating.
- Three. On a fluorescent screen they produce a bright central spot surrounded by concentric rings.
- Four. Rings are a diffraction pattern, diffraction is a wave property, so the electrons behave as waves, and the ring radii match $\lambda = h/p$.
Rings, not spots
Electron diffraction shows that electrons can behave as waves.
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Only waves diffract, yet a beam of electrons makes a diffraction pattern — so electrons have a wave nature.
Speeding up the electrons makes the diffraction rings:
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Faster electrons → shorter $\lambda$ → less diffraction → the rings close in.
Put the description of the electron-diffraction experiment in order.
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Each move is a mark in the four-marker. The last one is the conclusion, and it is the one most often left out.
Worked example: accelerating an electron
- An electron is accelerated from rest through $2500\ \text{V}$. Find its de Broglie wavelength.
- Kinetic energy gained: $E_K = eV$, and $p = \sqrt{2mE_K} = \sqrt{2m_e eV}$.
- So $\lambda = \dfrac{h}{\sqrt{2m_e eV}} = \dfrac{6.63\times10^{-34}}{\sqrt{2(9.11\times10^{-31})(1.6\times10^{-19})(2500)}} = 2.5\times10^{-11}\ \text{m}$.
- That is close to the atomic spacing in a crystal, which is why the electrons diffract off graphite at all.
- Learn the chain $qV = \tfrac{1}{2}mv^2 \Rightarrow p = \sqrt{2mqV} \Rightarrow \lambda = h/\sqrt{2mqV}$. It is asked for directly.
An electron is accelerated from rest through 2500 V. What is its de Broglie wavelength, in units of 1e-11 m?
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p = sqrt(2 m e V) = 2.7e-23 kg m/s, so lambda = h/p = 2.5e-11 m, close to the atomic spacing in a crystal. Use momentum, never speed, in h/p.
What changing the speed does
- Faster electrons have more momentum, so a shorter wavelength, so less diffraction: the rings move closer together.
- Slower electrons spread the rings apart.
- Because $\lambda \propto 1/\sqrt{V}$, halving the wavelength needs four times the accelerating p.d.
- This is the standard follow-up question, and the sign of the effect is what it tests. More volts, smaller rings.
A faster particle (more momentum) has a ____ de Broglie wavelength.
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$\lambda = \dfrac{h}{p}$, so larger $p$ gives a smaller $\lambda$.
The accelerating p.d. in an electron-diffraction tube is increased. Which happen? Select all that apply.
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More volts, more momentum, shorter wavelength, less diffraction, smaller rings. Because lambda goes as 1/sqrt(V), halving the wavelength takes four times the p.d.
Marks that slip away
- Name the phenomenon as evidence, not the category: "electron diffraction", not "it behaves like a wave".
- Use momentum, not speed, in $\lambda = h/p$. For an accelerated electron go through $p = \sqrt{2mqV}$.
- Sketch the pattern as rings, never as spots or as a straight-line fringe pattern.
- A higher accelerating voltage gives smaller rings. Getting this backwards is the commonest error here.
- Everyday objects have wavelengths around $10^{-34}\ \text{m}$, which is why duality is invisible outside the laboratory.
You've got it
- wave-particle duality: radiation and matter both show wave behaviour (interference, diffraction) and particle behaviour (the photoelectric effect)
- the de Broglie wavelength of a moving particle is $\lambda = h/p$, with $h$ the Planck constant and $p$ the momentum
- electron diffraction through polycrystalline graphite gives concentric rings, which only a wave could produce
- for an electron accelerated through $V$, $\lambda = h/\sqrt{2m_e eV}$, so a larger p.d. gives a shorter wavelength and smaller rings