A LevelPhysicsAQARevision NotesTurning Points in PhysicsWave-Particle DualityElectron Diffraction

Electron Diffraction (AQA A Level Physics): Revision Note

Exam code: 7408

Written by: Dan Mitchell-Garnett

Reviewed by: Tim

Updated on 19 October 2026

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Electron diffraction

Do electrons diffract?

To investigate whether electrons did exhibit wave-like properties, an electron diffraction tube was produced
The electrons were accelerated in an electron gun with a high potential difference, such as 5000 V, and were then directed through a thin film of graphite
- The gaps between the carbon atoms were sufficiently small to cause diffraction, using the predicted de Broglie wavelength of electrons
- Diffraction is a property of waves when passing through a small gap - if the electrons were seen to diffract then this was proof of their wave-like properties
The electrons were indeed seen to diffract from the gaps between carbon atoms and produce a circular pattern on a fluorescent screen made from phosphor

A cathode at +6 volts emits electrons, which are then accelerated by a +5 kilovolt anode. The accelerated electrons are incident on graphite, and their paths diffract by an angle theta and form concentric rings on a phosphor screen. — A low voltage causes thermionic emission of electrons at the cathode, then these are accelerated towards anode directed through the graphite. The electrons diffract through the gaps between atoms in the graphite, forming a circular diffraction pattern made visible by the phosphor screen

The effect of potential difference on the diffraction pattern

Recall that de Broglie's hypothesis predicted that, if more work was done on the electrons, their wavelength would be shorter
- Recall from the diffraction grating subtopic of the Waves topic, that a shorter wavelength leads to a smaller diffraction angle $θ$ for a given gap width
Therefore, if de Broglie was right, increasing the potential difference would cause the diffraction rings to move closer to the centre of the phosphor screen

An electron gun directs electrons through a thin graphite film in a vacuum tube. Electrons show wave properties by producing diffraction rings which are wider at low voltage and closer together at high voltage. — The left of this diagram shows the equipment from another angle. The right of this diagram shows that increasing the voltage caused the diffraction rings to move closer to the centre of the screen, following de Broglie's equation

This was the first experimental proof of electrons exhibiting wave-like properties
- The wavelength of electrons was significantly smaller than high energy forms of electromagnetic radiation, such as X-rays

Examiner Tips and Tricks

This topic builds on knowledge acquired from studying diffraction gratings. As is becoming a common theme with Turning Points, go back and make sure your understanding of diffraction is solid before getting stuck in with this topic.

Estimate of anode voltage

Changing the voltage across the anode (it is the voltage that causes the electrons to accelerate), scientists can manipulate the wavelength of the electron
A microscope's resolving power (its ability to differentiate between two nearby points on an object) depends on the wavelength of the radiation being used
- A shorter wavelength means the microscope can be used to see finer details in an object
- Electrons had wavelengths much shorter than X-rays
The size of an atom is roughly 10⁻¹⁰ m
- By using de Broglie's equation for an accelerated electron, researchers could find the accelerating voltage needed to make the electron's wavelength the same size as an atom

Worked Example

An atom of carbon has a diameter of 1.5 × 10⁻¹⁰ m. Determine the anode potential difference needed in an electron diffraction tube to produce electrons with a wavelength equal to the diameter of a carbon atom.

[3]

Answer:

Step 1: Write down the equation for wavelength of accelerated electrons from the data and formulae booklet

The wavelength of an accelerated electron is:

$λ = \frac{h}{\sqrt{2 m e V}}$

Step 2: List the known quantities

Mass of an electron, $m = 9.11 \times 10^{- 31} kg$
Magnitude of charge on an electron, $e = 1.60 \times 10^{- 19} C$
Planck's constant, $h = 6.63 \times 10^{- 34} J s$
Required wavelength of the electron, $λ = 1.5 \times 10^{- 10} m$

Step 3: Rearrange the wavelength equation to make anode potential difference the subject

Separate the $V$ term from the square root:

$λ = \frac{h}{\sqrt{2 m e} \sqrt{V}}$

$\sqrt{V} = \frac{h}{\sqrt{2 m e} \times λ}$

Square the whole equation:

$V = \frac{h^{2}}{2 m e λ^{2}}$ [1 mark]

Step 4: Substitute the known quantities

$V = \frac{{(6.63 \times 10^{- 34})}^{2}}{2 \times (9.11 \times 10^{- 31}) \times (1.60 \times 10^{- 19}) \times {(1.5 \times 10^{- 10})}^{2}}$ [1 mark]

$V = 67.0 V$ [1 mark]

Examiner Tips and Tricks

In a high pressure situation like an exam, it's easy to make mistakes on your calculator when substituting values in standard form. Use brackets carefully and retype your equation to ensure you get the same value when checking over your answers.

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Electron Diffraction (AQA A Level Physics): Revision Note

Electron diffraction

Do electrons diffract?

The effect of potential difference on the diffraction pattern

Examiner Tips and Tricks

Estimate of anode voltage

Worked Example

Examiner Tips and Tricks

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Dan Mitchell-Garnett

Reviewer: Tim