De Broglie's Hypothesis Revision Notes | A Level AQA Physics
De Broglie's hypothesis is that any particle with momentum p has an associated wavelength λ = h/p, where h is the Planck constant. 13 questions answered.
De Broglie's Hypothesis is part of Wave-particle duality (A-level only) in A Level AQA Physics.
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What this lesson covers
De Broglie's hypothesis is that any particle with momentum p has an associated wavelength λ = h/p, where h is the Planck constant. 13 questions answered.
State de Broglie's hypothesis and explain why it extended wave-particle duality to matter
Apply the de Broglie relation $p = h/\lambda$ to calculate the wavelength of a moving particle
Derive and apply the expression $\lambda = h/\sqrt{2meV}$ for electrons accelerated through a potential difference
Evaluate the significance of de Broglie's hypothesis as a turning point in physics
Key ideas to keep in view
Use this lesson to stay inside Wave-particle duality (A-level only) while connecting the detail back to the wider Turning points in physics (A-level only) topic.
de Broglie wavelength: The wavelength associated with a moving particle, given by $\lambda = h/p$, where $h$ is the Planck constant and $p$ is the particle's momentum. Used when calculating the wave properties of matter.
Wave-particle duality: The property by which all matter and radiation exhibit both wave-like and particle-like behaviour depending on the experimental context. Central to quantum mechanics.
Momentum (p): The product of a particle's mass and velocity, $p = mv$, measured in kg m/s. Appears in the de Broglie relation as the quantity that determines the associated wavelength.
Accelerating potential difference (V): The voltage through which a charged particle is accelerated, transferring electrical potential energy into kinetic energy. Used in the derived expression $\lambda = h/\sqrt{2meV}$ for electrons.
Revision notes
Read the core explanations from De Broglie's Hypothesis before testing yourself from memory.
1. The Question That Changed Physics
By the early 1920s, physicists had already accepted something strange: light, which everyone thought was a wave, also behaved as a stream of particles called photons. Einstein had shown this through the photoelectric effect. But a young French physicist named Louis de Broglie asked an even bolder question — if waves can behave like particles, can particles behave like waves?
This was not a minor tweak to existing theory. It was a proposal that every piece of matter — electrons, protons, even a football — has a wavelength associated with its motion. The idea sounds extraordinary, and in 1924 it was. What makes it a genuine turning point in physics is that it turned out to be experimentally correct.
2. De Broglie's Hypothesis and the Core Equation
De Broglie's reasoning was elegant. For a photon, Einstein had shown that its energy is $E = hf$ and its momentum is $p = h/\lambda$. De Broglie proposed that the same relationship between momentum and wavelength should apply to any particle with momentum, not just photons. This is the hypothesis: matter has an associated wavelength determined by its momentum.
Since momentum is $p = mv$ for a non-relativistic particle, the relation can also be written as $\lambda = h/(mv)$. Notice what this tells you: a larger mass or a higher speed both increase the momentum, which decreases the wavelength. This is why we never notice wave behaviour in everyday objects — their wavelengths are unimaginably small.
To feel the scale of this, consider a tennis ball of mass 0.057 kg moving at 50 m/s. Its momentum is $p = 0.057 \times 50 = 2.85$ kg m/s, giving a de Broglie wavelength of roughly $2.3 \times 10^{-34}$ m. That is about $10^{19}$ times smaller than a proton. No experiment could ever detect such a wavelength, which is why tennis balls do not diffract through doorways.
3. Why Electrons Are the Key Test Case
Electrons have a very small mass ($9.11 \times 10^{-31}$ kg), so even at modest speeds their de Broglie wavelengths fall in the range of $10^{-10}$ m — comparable to the spacing between atoms in a crystal lattice. That match in scale is crucial: diffraction only occurs when the wavelength is similar to the gap size. Electrons, unlike tennis balls, are in the right regime for their wave nature to be detectable.
In 1927, Davisson and Germer fired electrons at a nickel crystal and observed a diffraction pattern — a result that could only be explained if the electrons were behaving as waves with exactly the wavelength predicted by de Broglie's relation. This was direct experimental confirmation of the hypothesis and one of the most significant results in twentieth-century physics.
4. Electrons Accelerated Through a Potential Difference
In many experimental setups — including electron diffraction tubes and electron microscopes — electrons are accelerated from rest through a potential difference $V$. The AQA specification requires you to derive and apply a specific expression for the de Broglie wavelength in this situation, so it is worth understanding where it comes from, not just memorising the result.
When a charge $e$ moves through a potential difference $V$, it gains kinetic energy equal to the work done on it by the electric field. Setting the kinetic energy equal to $eV$ gives the starting point for the derivation.
Worked example: Deriving $\lambda = h / \sqrt{2meV}$
An electron (charge $e$, mass $m$) is accelerated from rest through a potential difference $V$. Derive an expression for its de Broglie wavelength.
$eV = \frac{1}{2}mv^2$ — The electron gains kinetic energy equal to the work done by the electric field. Starting from rest means all the energy goes into kinetic energy. [M1]
$2meV = m^2v^2 = (mv)^2$ — Multiply both sides by $2m$. The right-hand side becomes $(mv)^2$, which is momentum squared. This is the key algebraic step.
$p = mv = \sqrt{2meV}$ — Taking the positive square root gives the magnitude of the electron's momentum after acceleration. [M1]
$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$ — Replace $p$ in $\lambda = h/p$ with the expression from Step 3. This is the required result. [A1]
Answer: $$\lambda = \frac{h}{\sqrt{2meV}}$$
Notice the structure of this result: increasing $V$ increases the electron's kinetic energy and therefore its momentum, which decreases $\lambda$. Higher accelerating voltages produce electrons with shorter wavelengths — a fact exploited in electron microscopes, where higher voltages allow finer resolution.
5. Worked Calculations
Exam questions on this topic typically ask you to calculate a wavelength given either a speed or an accelerating voltage. The two worked examples below cover both cases. Work through each step before revealing the answer.
Worked example: Calculating de Broglie wavelength from speed
An electron travels at $2.0 \times 10^6$ m/s. Calculate its de Broglie wavelength. ($h = 6.63 \times 10^{-34}$ J s, $m_e = 9.11 \times 10^{-31}$ kg) [3 marks]
$\lambda = \frac{h}{mv}$ — Since $p = mv$ for a non-relativistic particle, substitute directly. [M1]
$\lambda = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 2.0 \times 10^6}$ — All values are in SI units, so no conversion is needed. [M1]
$9.11 \times 10^{-31} \times 2.0 \times 10^6 = 18.22 \times 10^{-25} = 1.822 \times 10^{-24}$ kg m/s — Multiply the coefficient: $9.11 \times 2.0 = 18.22$. Add the exponents: $-31 + 6 = -25$. Then rewrite in standard form.
Worked example: Calculating de Broglie wavelength from accelerating voltage
An electron is accelerated from rest through a potential difference of 1500 V. Calculate its de Broglie wavelength. ($h = 6.63 \times 10^{-34}$ J s, $m_e = 9.11 \times 10^{-31}$ kg, $e = 1.60 \times 10^{-19}$ C) [3 marks]
$\lambda = \frac{h}{\sqrt{2meV}}$ — Use the derived expression for an electron accelerated from rest. [M1]
$2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19} \times 1500$ — Substitute all values before taking the square root. Keep the full expression to avoid rounding errors.
$2 \times 9.11 = 18.22$
$18.22 \times 1.60 = 29.152$
$29.152 \times 1500 = 43728$
$\text{Exponent: } 10^{-31} \times 10^{-19} = 10^{-50}$
$2meV = 4.3728 \times 10^{4} \times 10^{-50} = 4.3728 \times 10^{-46}$ J kg — Work through the coefficients and exponents separately to avoid errors. The unit J kg is equivalent to kg² m² s⁻², so the square root will give kg m/s (momentum units).
$\sqrt{4.3728 \times 10^{-46}} = \sqrt{4.3728} \times 10^{-23} \approx 2.091 \times 10^{-23}$ kg m/s — $\sqrt{4.3728} \approx 2.091$. For the exponent: $\sqrt{10^{-46}} = 10^{-23}$. [M1]
Before de Broglie, wave-particle duality was a property attributed only to electromagnetic radiation. The photoelectric effect had shown that light could behave as particles, but matter was still thought of as purely particle-like. De Broglie's hypothesis unified these ideas: wave-particle duality is a universal feature of nature, not a quirk of light.
The experimental confirmation by Davisson and Germer, and independently by G.P. Thomson, provided the empirical foundation for quantum mechanics. Schrödinger's wave equation — the central equation of quantum mechanics — is built on the idea that particles have associated wavefunctions. De Broglie's hypothesis was the conceptual bridge that made that framework possible.
From a practical standpoint, the hypothesis underpins electron microscopy. Visible light has wavelengths of roughly $400$–$700$ nm, which limits the resolution of optical microscopes. Electrons accelerated through tens of kilovolts have wavelengths thousands of times shorter, allowing structures at the atomic scale to be imaged — something impossible with any optical instrument.
7. Exam Technique and Common Mistakes
Questions on de Broglie's hypothesis appear in the Turning Points section of AQA A Level Physics and tend to reward precise use of the derivation, correct substitution of constants, and clear evaluative language. The following guidance addresses the patterns that most frequently cost marks.
For extended-response questions asking you to evaluate de Broglie's hypothesis as a turning point, structure your answer around three elements: the prior state of knowledge (duality was only attributed to radiation), the new theoretical claim (all matter has an associated wavelength), and the experimental evidence (electron diffraction confirming the predicted wavelengths). A conclusion that explicitly weighs the theoretical and practical impact will access the highest mark band.
De Broglie's hypothesis — that $p = h/\lambda$ applies to all matter — and the derived expression $\lambda = h/\sqrt{2meV}$ for accelerated electrons together cover both specification requirements for this topic. Mastering the derivation, not just the formula, is what separates a confident answer from a memorised one.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course.
De Broglie wavelength is calculated with λ = h/(mv). An electron moving at 2.0 × 10⁶ m/s has λ = 6.63 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 2.0 × 10⁶) ≈ 3.6 × 10⁻¹⁰ m. A larger mass or a higher speed gives a shorter wavelength.
What is de Broglie's hypothesis?
De Broglie's hypothesis is that any particle with momentum p has an associated wavelength λ = h/p, where h is the Planck constant. It extends wave-particle duality from radiation to all matter, and since p = mv, the wavelength can also be written λ = h/(mv).
What was the Davisson-Germer experiment?
The Davisson-Germer experiment, in 1927, fired electrons at a nickel crystal and observed a diffraction pattern. It could only be explained if the electrons behaved as waves with exactly the wavelength predicted by de Broglie's relation, giving direct experimental confirmation of his hypothesis.
How do you derive λ = h/√(2meV)?
The expression λ = h/√(2meV) is derived from eV = ½mv² for an electron accelerated from rest through a potential difference V. Multiplying both sides by 2m gives 2meV = (mv)², so momentum p = √(2meV), and substituting into λ = h/p gives the result.
What evidence supports de Broglie's hypothesis?
Electron diffraction is the key evidence for de Broglie's hypothesis. In 1927 Davisson and Germer fired electrons at a nickel crystal and observed a diffraction pattern with exactly the wavelength de Broglie's relation predicted, which could only be explained if the electrons were behaving as waves.
What is the de Broglie wavelength of a tennis ball?
A tennis ball of mass 0.057 kg moving at 50 m/s has a momentum of 2.85 kg m/s and a de Broglie wavelength of roughly 2.3 × 10⁻³⁴ m. That is about 10¹⁹ times smaller than a proton, far too small to detect, which is why tennis balls do not diffract through doorways.
What is the de Broglie wavelength of an electron moving at 2.0 × 10⁶ m/s?
An electron moving at 2.0 × 10⁶ m/s has a de Broglie wavelength of about 3.6 × 10⁻¹⁰ m. Using λ = h ÷ mv, the momentum is 9.11 × 10⁻³¹ × 2.0 × 10⁶ = 1.822 × 10⁻²⁴ kg m/s, and 6.63 × 10⁻³⁴ divided by that gives 3.64 × 10⁻¹⁰ m.
What is the de Broglie wavelength of an electron accelerated through 1500 V?
An electron accelerated from rest through 1500 V has a de Broglie wavelength of about 3.2 × 10⁻¹¹ m. Using λ = h ÷ √(2meV), 2meV = 4.3728 × 10⁻⁴⁶, its square root gives a momentum of 2.091 × 10⁻²³ kg m/s, and dividing h by that gives 3.17 × 10⁻¹¹ m.
When did de Broglie propose that matter has a wavelength?
Louis de Broglie proposed in 1924 that any particle with momentum has an associated wavelength, λ = h ÷ p. He asked whether, since waves can behave like particles, particles could behave like waves; the idea was confirmed by electron diffraction in 1927.
Why was de Broglie's hypothesis a turning point in physics?
De Broglie's hypothesis was a turning point because it showed wave-particle duality is a universal feature of nature, not just of electromagnetic radiation. Confirmed by Davisson and Germer and by G.P. Thomson, it gave quantum mechanics its empirical foundation and was the conceptual bridge to Schrödinger's wave equation.
How does mass affect de Broglie wavelength?
A larger mass gives a shorter de Broglie wavelength, because λ = h ÷ mv and a larger mass or higher speed both increase the momentum. Electrons, with a very small mass of 9.11 × 10⁻³¹ kg, have wavelengths around 10⁻¹⁰ m, comparable to atomic spacing, while everyday objects have unimaginably small wavelengths.
How is de Broglie's hypothesis used in electron microscopes?
De Broglie's hypothesis underpins electron microscopy: increasing the accelerating voltage increases an electron's momentum and so decreases its wavelength. Electrons accelerated through tens of kilovolts have wavelengths thousands of times shorter than visible light, at 400–700 nm, so atomic-scale structures can be imaged.
Why must electrons have a wavelength similar to atomic spacing to diffract?
Electrons must have a wavelength similar to atomic spacing to diffract because diffraction only occurs when the wavelength is similar to the gap size. Even at modest speeds, electrons have de Broglie wavelengths around 10⁻¹⁰ m, comparable to the spacing between atoms in a crystal lattice, so their wave nature can be detected.