86-87 - Electron Diffraction and de Broglie Wavelength

86-87 - Electron Diffraction and de Broglie Wavelength

Use electron diffraction as evidence for wave behaviour and relate a particle’s wavelength to its momentum.

Electron diffraction as wave evidence

Imagine a narrow beam of electrons travelling through a thin crystalline material. If electrons behaved only like tiny pellets, you might expect a bright patch where the beam hits the screen, perhaps broadened by scattering.

Instead, an electron diffraction experiment produces a structured pattern, often rings or bright regions. That structure is the key evidence: diffraction is a wave behaviour. The regular spacing of atoms in the crystal acts like many tiny scattering centres. Electron waves scattered from different parts of the crystal can reinforce in some directions and cancel in others, producing a pattern rather than a single smeared patch.

The careful wording is important. Electron diffraction does not mean an electron is "just a wave" in the same way a water ripple is a wave. An electron is still detected at particular points on the screen. The wave model explains the pattern built up by many detections.

Misconception contrast:

  • Tempting but wrong: "Diffraction means electrons stop being particles."
  • Better model: electrons are detected as particles, but their probability pattern shows wave behaviour.

For Edexcel, an explain answer needs the link, not just the name of the experiment:

An electron diffraction pattern is observed. Diffraction is a wave property, so the experiment provides evidence for the wave nature of electrons.

Using the de Broglie equation

de Broglie connected the wave and particle descriptions by assigning a wavelength to a moving particle. The wavelength is shorter when the momentum is larger:

de Broglie wavelength

λ=hp \lambda = \frac{h}{p}

λ\lambda: de Broglie wavelength (m); hh: Planck constant, 6.63×10−34 J s6.63 \times 10^{-34}\,\mathrm{J\,s}; pp: momentum of the particle (use p=mvp=mv only at non-relativistic speeds) (kg m s⁻¹).

This equation is supplied in the data booklet, so the exam skill is not memorising it. The skill is choosing it, using powers of ten cleanly, and interpreting whether the wavelength is physically sensible.

Worked example:

An electron in a diffraction experiment has momentum 4.0×10−24kg m s−14.0 \times 10^{-24} \mathrm{kg}\,\mathrm{m}\,\mathrm{s}^{-1}. Calculate its de Broglie wavelength.

Decision point 1: Which equation?
The question gives momentum and asks for wavelength, so use:

λ=hp\lambda= \frac{h}{p}

Decision point 2: What value of hh?
Use the data-booklet value:

h=6.63×10−34 J s\begin{aligned} h=6.63\times10^{-34}\,\mathrm{J\,s} \end{aligned}

Decision point 3: Do the units become metres?
Since 1 J=1 kg m2 s−21\,\mathrm{J} = 1\,\mathrm{kg}\,\mathrm{m}^{2}\,\mathrm{s}^{-2}, then:

J s=kg m2 s−1\begin{aligned} \mathrm{J\,s}=\mathrm{kg\,m^2\,s^{-1}} \end{aligned}

So:

J skg m s−1=m\begin{aligned} \frac{\mathrm{J\,s}}{\mathrm{kg\,m\,s^{-1}}}=\mathrm{m} \end{aligned}

Now substitute:

λ=6.63×10−344.0×10−24=1.66×10−10 m\begin{aligned} \lambda=\frac{6.63\times10^{-34}}{4.0\times10^{-24}}=1.66\times10^{-10}\,\mathrm{m} \end{aligned}

Interpretation: 1.66×10−10 m1.66 \times 10^{-10}\,\mathrm{m} is about 0.17 nm0.17\,\mathrm{nm}, similar in scale to atomic spacing in a crystal. That is why a crystal can act as a useful diffraction object for electrons.

Guided check:

If the same electron beam had twice the momentum, the denominator in h/ph/p would double, so the wavelength would halve. Higher momentum means shorter wavelength. That inverse link is one of the most common places to catch an error before it becomes a wrong conclusion.

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