Electron diffraction and de Broglie wavelength

What you'll learn

Why diffraction is strong evidence that electrons can behave like waves.
How to use the de Broglie equation $λ=hp\lambda = \frac{h}{p}$ .
How accelerating voltage affects an electron’s wavelength.
How electron diffraction patterns are produced and interpreted.

Starting point: what diffraction tells you

You already know that diffraction is the spreading of waves when they pass through a gap or around an obstacle. Diffraction effects are most noticeable when the wavelength is similar in size to the gap, obstacle, or spacing between scattering centres.

For visible light, diffraction by a grating or slit shows that light has wave behaviour. The surprising discovery in this topic is that electrons, which we often treat as particles, can also diffract.

Definition

Diffraction

Diffraction is the spreading or bending of waves when they pass through an aperture or around an obstacle. A clear diffraction pattern is evidence of wave behaviour.

Key Idea

The big idea

If electrons produce a diffraction pattern, they must have a wave-like property: a wavelength.

Wave-particle duality

In earlier work on the photoelectric effect, you saw that light sometimes behaves like particles called photons. Here, the direction of the surprise is reversed: electrons are usually introduced as particles, but they can show wave behaviour.

Definition

Wave-particle duality

Wave-particle duality is the idea that objects such as light and electrons can show both wave-like and particle-like behaviour, depending on the experiment.

Electrons are still detected as individual particles. For example, an electron hits a fluorescent screen at one point. But the overall pattern built up by many electrons can be a diffraction pattern, which is a wave effect.

Common Mistake

Thinking the electron becomes a wave instead of a particle

An electron is not “just a wave” or “just a particle”. In this topic, you use a particle property, momentum, to calculate a wave property, wavelength.

de Broglie wavelength

Louis de Broglie proposed that any moving particle has an associated wavelength. This is called the de Broglie wavelength.

Definition

de Broglie wavelength

The de Broglie wavelength of a particle is the wavelength associated with its motion:

λ=hp\lambda = \frac{h}{p}

where $λ\lambda$ is wavelength in metres, $h$ is the Planck constant, and $p$ is momentum in kilogram metres per second.

The Planck constant is:

\times 10^{-34}\ \text{J s}

Since momentum is $p = m v$ for a non-relativistic particle, you can also write:

λ=hmv\lambda = \frac{h}{mv}

This means a particle with greater momentum has a smaller de Broglie wavelength.

Key Idea

Momentum and wavelength are inversely related

For matter waves, increasing momentum decreases wavelength: $λ∝1p\lambda \propto \frac{1}{p}$ .

Example

Finding the de Broglie wavelength of an electron

An electron travels at $2.0 \times 10^6\ \text{m s}^{-1}$ . Calculate its de Broglie wavelength. Use $kgm_e = 9.11 \times 10^{-31}\ \text{kg}$ .

Calculate the electron’s momentum using $p = m v$ :
$\left(9.11 \times 10^{-31}\right)\left(2.0 \times 10^6\right)$ $\times 10^{-24}\ \text{kg m s}^{-1}$
Substitute into $λ=hp\lambda = \frac{h}{p}$ :
$λ=6.63×10−341.82×10−24\lambda = \frac{6.63 \times 10^{-34}}{1.82 \times 10^{-24}}$
Calculate the wavelength:
$m\lambda = 3.64 \times 10^{-10}\ \text{m}$
So the electron’s de Broglie wavelength is about $3.6 \times 10^{-10}\ \text{m}$ .

Why electron diffraction needs tiny spacings

Electron wavelengths in typical A-Level calculations are often around $m10^{-10}\ \text{m}$ . This is about the same size as the spacing between atoms in a crystal.

That is why crystals are useful for electron diffraction: their atoms act like a very closely spaced diffraction structure.

Tip

Useful scale comparison

Atomic spacings in crystals are around $m10^{-10}\ \text{m}$ , so electrons with wavelengths of this order can diffract strongly from crystal planes.

Electron diffraction apparatus

A common demonstration uses an electron diffraction tube. Electrons are emitted by a heated cathode, accelerated through a potential difference, pass through a thin graphite foil, and hit a fluorescent screen.

The graphite is polycrystalline, meaning it contains many tiny crystal regions at different orientations. This produces circular diffraction rings rather than just a few spots.

$Schematic of an electron diffraction tube showing electrons accelerated through a potential difference, passing through graphite foil, and forming diffraction rings on a fluorescent screen$

Definition

Fluorescent screen

A fluorescent screen emits visible light when struck by energetic electrons, allowing the electron impact positions to be seen.

How the electron gains kinetic energy

An electron accelerated through a potential difference $V$ gains energy from the electric field.

The energy gained by a charge moving through a potential difference is:

E = Q V

For an electron, the magnitude of the charge is the elementary charge:

\times 10^{-19}\ \text{C}

So the electron gains kinetic energy:

E_k = eV

If the electron starts from rest and is not moving relativistically, then:

\frac{1}{2}m_ev^2

This allows you to connect the accelerating voltage to the electron speed, momentum, and wavelength.

Common Mistake

Non-relativistic assumption

At A-Level, you usually use $\frac{1}{2}m_ev^2$ for electron diffraction tube calculations. At very high voltages, relativistic effects become important and this simple equation is no longer accurate.

Combining voltage with de Broglie wavelength

You often want the electron wavelength directly from the accelerating voltage.

Start with:

\frac{1}{2}m_ev^2

Rearrange to find speed:

\sqrt{\frac{2eV}{m_e}}

Momentum is:

p = m_ev

So:

\sqrt{2m_eeV}

Substitute into the de Broglie equation:

λ=h2meeV\lambda = \frac{h}{\sqrt{2m_eeV}}

Key Idea

Voltage controls wavelength

A larger accelerating voltage gives electrons more kinetic energy and momentum, so their de Broglie wavelength becomes smaller.

Example

Calculating wavelength from accelerating voltage

Electrons are accelerated through a potential difference of $2.50 \times 10^3\ \text{V}$ . Calculate their de Broglie wavelength.

Use $\times 10^{-34}\ \text{J s}$ , $kgm_e = 9.11 \times 10^{-31}\ \text{kg}$ , and $\times 10^{-19}\ \text{C}$ .

Use the combined equation:
$λ=h2meeV\lambda = \frac{h}{\sqrt{2m_eeV}}$
Substitute the values, keeping the voltage in volts:
$λ=6.63×10−342(9.11×10−31)(1.60×10−19)(2.50×103)\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2\left(9.11 \times 10^{-31}\right)\left(1.60 \times 10^{-19}\right)\left(2.50 \times 10^3\right)}}$
Evaluate the denominator as the electron momentum:
$s−1\sqrt{2m_eeV} = 2.70 \times 10^{-23}\ \text{kg m s}^{-1}$
Calculate the wavelength:
$λ=6.63×10−342.70×10−23\lambda = \frac{6.63 \times 10^{-34}}{2.70 \times 10^{-23}}$ $m\lambda = 2.46 \times 10^{-11}\ \text{m}$
The electron wavelength is $2.46 \times 10^{-11}\ \text{m}$ .

Common Mistake

Forgetting that voltage is not energy

The energy gained is $e V$ , not just $V$ . A volt is a joule per coulomb, so you must multiply by the electron charge to get energy in joules.

Why rings form on the screen

When the electron beam passes through graphite, electrons scatter from regularly spaced atomic planes. For certain angles, waves scattered from different planes arrive in phase and interfere constructively.

Definition

Constructive interference

Constructive interference occurs when waves meet in phase, producing a larger resultant amplitude.

In a polycrystalline graphite foil, there are many tiny crystals pointing in many directions. For each allowed diffraction angle, many crystal orientations contribute, forming a cone of diffracted electrons. Where this cone hits the screen, you see a ring.

A smaller electron wavelength gives a smaller diffraction angle, so the rings move closer to the central spot.

Key Idea

What the rings prove

The diffraction rings show that electrons have wave-like behaviour, because diffraction and interference are wave phenomena.

Linking ring size to accelerating voltage

You are usually expected to know the qualitative relationship:

increasing $V$ increases electron kinetic energy
increasing kinetic energy increases momentum $p$
increasing $p$ decreases wavelength $λ\lambda$
decreasing $λ\lambda$ decreases the diffraction angle
the diffraction rings become smaller

So:

\uparrow \Rightarrow p \uparrow \Rightarrow \lambda \downarrow \Rightarrow \text{ring radius decreases}

Example

Predicting the effect of increasing voltage

In an electron diffraction tube, the accelerating potential difference is increased. Explain what happens to the diffraction rings.

Increasing the potential difference increases the kinetic energy gained by the electrons because $E_k = eV$ .
Greater kinetic energy means the electrons have greater momentum, since $p = m v$ for these non-relativistic speeds.
From $λ=hp\lambda = \frac{h}{p}$ , greater momentum gives a smaller de Broglie wavelength.
A smaller wavelength produces a smaller diffraction angle for the same graphite lattice spacing, so the rings on the screen become closer to the central spot.

Evidence for matter waves

Electron diffraction is important because it provides experimental evidence for de Broglie’s hypothesis. Electrons have measurable momentum, but they also produce interference patterns.

This connects neatly with the bigger theme of the parent section: light and matter cannot always be described using only a simple particle model or only a simple wave model.

Tip

How to phrase the conclusion

A strong exam answer says: “The diffraction pattern shows that electrons have wave-like behaviour, and the measured wavelength agrees with $λ=hp\lambda = \frac{h}{p}$ .”

Quick equation summary

de Broglie wavelength:
$λ=hp\lambda = \frac{h}{p}$
momentum of a non-relativistic particle:
$p = m v$
kinetic energy gained by an electron accelerated through a potential difference:
$E_k = eV$
non-relativistic kinetic energy:
$Ek=12mev2E_k = \frac{1}{2}m_ev^2$
electron wavelength from accelerating voltage:
$λ=h2meeV\lambda = \frac{h}{\sqrt{2m_eeV}}$

Exam technique

In the exam

If you see an accelerating voltage, start with $E_k = eV$ and then connect it to momentum or wavelength.
If asked what happens when voltage increases, follow the chain: $V$ increases, $p$ increases, $λ\lambda$ decreases, ring radius decreases.
If asked why electron diffraction is significant, link the observed diffraction pattern directly to wave behaviour and de Broglie wavelength.

Self review

Check yourself

Why does a diffraction pattern provide evidence that electrons have wave-like behaviour?
An electron’s momentum doubles. What happens to its de Broglie wavelength?
Why is graphite suitable for producing electron diffraction patterns?

Electron diffraction and de Broglie wavelength

What you'll learn

Starting point: what diffraction tells you

Wave-particle duality

de Broglie wavelength

Why electron diffraction needs tiny spacings

Electron diffraction apparatus

How the electron gains kinetic energy

Combining voltage with de Broglie wavelength

Why rings form on the screen

Linking ring size to accelerating voltage

Evidence for matter waves

Quick equation summary

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Electron diffraction and de Broglie wavelength Revision Guide