De Broglie Wavelength and Diffraction Revision Notes | A Level AQA Physics
The de Broglie wavelength is λ = h / mv, where h is the Planck constant, m the mass and v the speed of the particle. Free AQA notes and 15 questions answered.
Updated 12 Sept 2026 · Revise Right Team
De Broglie Wavelength and Diffraction is part of Electromagnetic radiation and quantum phenomena in A Level AQA Physics.
7 min
Reading time
15
Lessons in this topic
A Level AQA
Pathway
Reviewed by Revise Right
Content review
What this lesson covers
The de Broglie wavelength is λ = h / mv, where h is the Planck constant, m the mass and v the speed of the particle. Free AQA notes and 15 questions answered.
State the de Broglie wavelength equation and define each symbol with correct units
Calculate the de Broglie wavelength of a particle given its mass and velocity
Explain how and why the degree of diffraction changes when the momentum of a particle is changed
Analyse the wave-particle duality concept and evaluate the evidence that particles exhibit wave behaviour
Key ideas to keep in view
Use this lesson to stay inside Electromagnetic radiation and quantum phenomena while connecting the detail back to the wider Particles and radiation topic.
de Broglie wavelength: The wavelength associated with a moving particle, given by $\lambda = h / mv$, where $h$ is the Planck constant, $m$ is the mass, and $v$ is the speed. Measured in metres (m).
wave-particle duality: The principle that particles such as electrons can exhibit both particle-like and wave-like behaviour depending on the experimental context.
momentum: The product of a particle's mass and velocity, $p = mv$. Measured in kg m s$^{-1}$. In the de Broglie equation, momentum appears in the denominator, so larger momentum means shorter wavelength.
diffraction: The spreading of a wave as it passes through a gap or around an obstacle. Significant diffraction occurs when the wavelength is comparable in size to the gap width.
Revision notes
Read the core explanations from De Broglie Wavelength and Diffraction before testing yourself from memory.
1. The Puzzle: Can Particles Behave Like Waves?
You already know that light behaves like a wave — it diffracts, it interferes, and it has a wavelength. You also know that light behaves like a particle — photons carry discrete packets of energy. That dual nature is strange enough. But here is the deeper question: if light can behave like a particle, can particles behave like waves?
In 1924, Louis de Broglie proposed exactly that. He argued that any moving particle — an electron, a proton, even a football — has a wavelength associated with its motion. This idea, called wave-particle duality, is one of the most profound results in modern physics, and it is directly testable through diffraction experiments.
The key evidence comes from electron diffraction. When a beam of electrons is fired at a thin crystalline material, the electrons spread out and form a pattern of rings on a detector — exactly the kind of diffraction pattern you would expect from waves passing through a regular array of gaps. Particles, behaving as waves. That is the phenomenon this lesson unpacks.
2. The de Broglie Wavelength Equation
De Broglie's insight was to connect the wave description to a quantity you already know from mechanics: momentum. Recall that momentum $p = mv$, where $m$ is mass in kg and $v$ is speed in m s$^{-1}$. De Broglie proposed that the wavelength $\lambda$ of a particle is inversely proportional to its momentum.
Notice the structure of the equation: $h$ is a constant, so the wavelength depends entirely on the momentum $mv$ in the denominator. Increase the momentum and the wavelength gets shorter. Decrease the momentum and the wavelength gets longer. That inverse relationship is the engine behind everything else in this topic.
One thing worth noting immediately: the AQA specification does not require you to know the details of any particular method of particle diffraction. You do not need to describe the crystal structure of graphite or explain how electron guns work. What you do need is the equation, the reasoning behind it, and the ability to connect momentum to diffraction behaviour.
3. Calculating the de Broglie Wavelength
The equation is straightforward to apply, but the numbers involved are very small — electron masses are of the order $10^{-31}$ kg, so the wavelengths come out in the range of $10^{-10}$ m, comparable to the spacing between atoms in a crystal. That is precisely why electron diffraction works: the wavelength matches the gap size.
Worked example: Calculating the de Broglie Wavelength of an Electron
An electron travels at $2.0 \times 10^{7}$ m s$^{-1}$. Calculate its de Broglie wavelength. (Mass of electron $m_e = 9.11 \times 10^{-31}$ kg, Planck constant $h = 6.63 \times 10^{-34}$ J s.)
$\lambda = \dfrac{h}{mv}$ — [M1] Selecting the correct equation is the first method mark.
$h = 6.63 \times 10^{-34}$ J s, $m = 9.11 \times 10^{-31}$ kg, $v = 2.0 \times 10^{7}$ m s$^{-1}$ — List values with units before substituting — this prevents unit errors.
$mv = 9.11 \times 10^{-31} \times 2.0 \times 10^{7} = 1.822 \times 10^{-23}$ kg m s$^{-1}$ — [M1] Working out $mv$ separately keeps the arithmetic clean. Check: $9.11 \times 2.0 = 18.22$, and $10^{-31} \times 10^{7} = 10^{-24}$, so $18.22 \times 10^{-24} = 1.822 \times 10^{-23}$.
$\lambda \approx 3.6 \times 10^{-11}$ m — [A1] The answer is given to 2 significant figures, matching the data. The unit is metres. This wavelength is of the same order as atomic spacings — which is why electrons diffract through crystal lattices.
Answer: $\lambda \approx 3.6 \times 10^{-11}$ m
Compare that wavelength — around $3.6 \times 10^{-11}$ m — to the wavelength of visible light, which is roughly $4 \times 10^{-7}$ m. The electron's de Broglie wavelength is about ten thousand times shorter. That is why electrons diffract through crystal lattices but not through the slits used in light experiments: the gap size must be comparable to the wavelength for significant diffraction to occur.
4. Momentum, Wavelength, and Diffraction
Now that you can calculate a de Broglie wavelength, the next step is to connect it to observable behaviour — specifically, diffraction. You already know from wave physics that diffraction is most pronounced when the wavelength of a wave is comparable to the size of the gap it passes through. The same principle applies to matter waves.
Here is the chain of reasoning the specification requires you to be able to articulate. Start with the de Broglie equation: $\lambda = h / mv$. The Planck constant $h$ never changes. So if you increase the momentum $mv$ of a particle, the wavelength $\lambda$ decreases. A shorter wavelength means less diffraction through a gap of fixed size. Conversely, reducing the momentum increases the wavelength, which increases the amount of diffraction.
In an electron diffraction experiment, you can change the momentum of the electrons by changing the accelerating voltage. Higher voltage gives the electrons more kinetic energy, which means greater speed and therefore greater momentum. The result: shorter de Broglie wavelength, narrower diffraction rings on the detector. Lower voltage produces slower electrons, longer wavelength, and wider rings. This is directly observable and provides compelling evidence for wave-particle duality.
It is worth being precise about what 'amount of diffraction' means here. When physicists say diffraction increases, they mean the waves spread out more as they pass through the gap. For a fixed gap width, a longer wavelength produces a wider central maximum and more spreading. A shorter wavelength produces a narrower pattern with less spreading. The ratio of wavelength to gap size is what governs this — and the de Broglie equation tells you exactly how to control that ratio by adjusting momentum.
Higher momentum $\Rightarrow$ shorter $\lambda$ $\Rightarrow$ less diffraction (narrower pattern)
Lower momentum $\Rightarrow$ longer $\lambda$ $\Rightarrow$ more diffraction (wider pattern)
Maximum diffraction occurs when $\lambda$ is approximately equal to the gap width
The Planck constant $h$ is fixed, so momentum is the only variable you can change
5. Why Everyday Objects Do Not Diffract
A natural question arises: if all moving particles have a de Broglie wavelength, why do we never see a cricket ball diffracting around a doorframe? The answer lies in the numbers. A cricket ball has a mass of roughly 0.16 kg and might travel at 40 m s$^{-1}$. Its momentum is therefore about 6.4 kg m s$^{-1}$. Dividing the Planck constant by this gives a wavelength of around $10^{-34}$ m — many orders of magnitude smaller than an atomic nucleus.
No physical gap is anywhere near that small, so the cricket ball never encounters a gap comparable to its wavelength. Diffraction effects are completely undetectable. Wave-particle duality is a real phenomenon for all matter, but it only becomes experimentally significant for particles with very small mass — primarily electrons, protons, and neutrons — where the momentum is small enough to give a wavelength in the range of atomic or nuclear dimensions.
6. Exam Technique and Common Mistakes
Calculation questions on de Broglie wavelength are reliable marks if you follow a consistent method. The most common errors are powers-of-ten mistakes when computing $mv$, and forgetting to state units on the final answer. Examiners award a method mark for writing the correct equation, a second method mark for a correct substitution, and an accuracy mark for the correct numerical answer with units — so even if your arithmetic slips, you can still earn two out of three marks by showing clear method.
'Explain' questions on diffraction and momentum require a clear logical chain. The mark scheme rewards three linked ideas: state how momentum changes, use $\lambda = h / mv$ to explain the effect on wavelength, and then connect the changed wavelength to the changed amount of diffraction. Missing any link in that chain typically costs a mark. Write it as a sequence: momentum up $\rightarrow$ wavelength down $\rightarrow$ less diffraction.
Bringing this together: the de Broglie wavelength equation $\lambda = h / mv$ quantifies the wave nature of any moving particle. The wavelength is inversely proportional to momentum, so controlling momentum — for example by changing accelerating voltage in an electron beam — directly controls the degree of diffraction observed. This is the core reasoning chain the AQA specification requires you to construct fluently, both in calculation and in extended written explanation.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-12.
The de Broglie wavelength equation is λ = h / mv, where h is the Planck constant (6.63 × 10⁻³⁴ J s), m is the particle's mass in kg and v is its speed in m s⁻¹. The wavelength is measured in metres and is inversely proportional to the particle's momentum mv.
How do you calculate the de Broglie wavelength of an electron?
To calculate the de Broglie wavelength of an electron, work out its momentum mv, then divide the Planck constant by it. An electron of mass 9.11 × 10⁻³¹ kg moving at 2.0 × 10⁷ m s⁻¹ has momentum 1.822 × 10⁻²³ kg m s⁻¹, giving λ = 6.63 × 10⁻³⁴ ÷ 1.822 × 10⁻²³ ≈ 3.6 × 10⁻¹¹ m.
What are the units of de Broglie wavelength?
The de Broglie wavelength is measured in metres. In λ = h / mv, h is the Planck constant, 6.63 × 10⁻³⁴ J s, m is the particle's mass in kg and v is its speed in m s⁻¹, so the momentum mv has units of kg m s⁻¹.
How does momentum affect diffraction?
Increasing a particle's momentum reduces the amount of diffraction. By λ = h / mv, higher momentum gives a shorter de Broglie wavelength, which spreads out less through a gap of fixed size, giving a narrower pattern. Lower momentum gives a longer wavelength and more diffraction, with maximum diffraction when λ is approximately equal to the gap width.
Why don't everyday objects diffract?
Everyday objects do not diffract because their de Broglie wavelength is far too small. A cricket ball of about 0.16 kg moving at 40 m s⁻¹ has momentum of about 6.4 kg m s⁻¹, giving a wavelength around 10⁻³⁴ m, many orders of magnitude smaller than an atomic nucleus, so no gap is ever comparable to it.
How does the de Broglie wavelength of an electron compare to visible light?
An electron's de Broglie wavelength is far shorter than visible light: an electron at 2.0 × 10⁷ m s⁻¹ has a wavelength around 3.6 × 10⁻¹¹ m, while visible light is roughly 4 × 10⁻⁷ m, about ten thousand times longer. That is why electrons diffract through crystal lattices but not through the slits used in light experiments.
Is de Broglie wavelength inversely proportional to momentum?
The de Broglie wavelength is inversely proportional to a particle's momentum, because the Planck constant h in λ = h / mv never changes. Increasing the momentum makes the wavelength shorter and decreasing the momentum makes it longer, so momentum is the only variable that controls the wavelength.
How does accelerating voltage affect the de Broglie wavelength of electrons?
Increasing the accelerating voltage gives a shorter de Broglie wavelength of electrons. Higher voltage gives the electrons more kinetic energy, so greater speed and greater momentum, and by λ = h / mv the wavelength falls, giving narrower diffraction rings. Lower voltage produces slower electrons, a longer wavelength and wider rings.
What is the de Broglie wavelength of a cricket ball?
A cricket ball has a de Broglie wavelength of around 10⁻³⁴ m. With a mass of roughly 0.16 kg moving at 40 m s⁻¹, its momentum is about 6.4 kg m s⁻¹, and dividing the Planck constant by this gives a wavelength far smaller than an atomic nucleus, so no diffraction is detectable.
Which particles show noticeable wave behaviour?
Wave behaviour is only experimentally significant for particles with very small mass, mainly electrons, protons and neutrons. Their momentum is small enough to give a de Broglie wavelength in the range of atomic or nuclear dimensions, so they can meet gaps comparable to their wavelength and diffract.
When does maximum diffraction occur for a particle?
Maximum diffraction occurs when the de Broglie wavelength is approximately equal to the gap width. For a fixed gap, a longer wavelength gives a wider central maximum and more spreading, while a shorter wavelength gives a narrower pattern; the ratio of wavelength to gap size governs the effect.
What are the common mistakes in de Broglie wavelength calculations?
The most common errors in de Broglie wavelength calculations are powers-of-ten mistakes when computing the momentum mv and forgetting to state units on the final answer. Working out mv separately, listing values with units before substituting, and giving the wavelength in metres avoids both.
Who proposed that particles have a wavelength?
Louis de Broglie proposed in 1924 that any moving particle, such as an electron, a proton or even a football, has a wavelength associated with its motion. This idea, wave-particle duality, is directly testable through diffraction experiments, and electron diffraction provides the key evidence.
What is the typical de Broglie wavelength of an electron?
A typical electron de Broglie wavelength is in the range of 10⁻¹⁰ m, because electron masses are of the order 10⁻³¹ kg. That is comparable to the spacing between atoms in a crystal, which is precisely why electrons diffract through crystal lattices.
How do you find the speed of a particle from its de Broglie wavelength?
The speed of a particle comes from changing λ = h / mv into v = h / mλ, the Planck constant divided by the mass times the de Broglie wavelength. Mass must be in kg and wavelength in metres for the speed to come out in m s⁻¹.