De Broglie's Hypothesis Revision Notes | A Level AQA Physics

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.

1. $eV = \frac{1}{2}mv^2$ — 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]
2. $2meV = m^2v^2 = (mv)^2$ — Multiply both sides by $2m$. The right-hand side becomes $(mv)^2$, which is momentum squared. This is the key algebraic step.
3. $p = mv = \sqrt{2meV}$ — Taking the positive square root gives the magnitude of the electron's momentum after acceleration. [M1]
4. $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$ — 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]

1. $\lambda = \frac{h}{mv}$ — Since $p = mv$ for a non-relativistic particle, substitute directly. [M1]
2. $\lambda = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 2.0 \times 10^6}$ — All values are in SI units, so no conversion is needed. [M1]
3. $9.11 \times 10^{-31} \times 2.0 \times 10^6 = 18.22 \times 10^{-25} = 1.822 \times 10^{-24}$ kg m/s — Multiply the coefficient: $9.11 \times 2.0 = 18.22$. Add the exponents: $-31 + 6 = -25$. Then rewrite in standard form.
4. $\lambda = \frac{6.63 \times 10^{-34}}{1.822 \times 10^{-24}} = \frac{6.63}{1.822} \times 10^{-34-(-24)} = 3.64 \times 10^{-10}$ m — $6.63 \div 1.822 \approx 3.64$. Exponent: $-34 - (-24) = -34 + 24 = -10$. [A1]

Answer: $\lambda \approx 3.6 \times 10^{-10}$ m (to 2 significant figures)

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]

1. $\lambda = \frac{h}{\sqrt{2meV}}$ — Use the derived expression for an electron accelerated from rest. [M1]
2. $2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19} \times 1500$ — Substitute all values before taking the square root. Keep the full expression to avoid rounding errors.
3. $2 \times 9.11 = 18.22$ $18.22 \times 1.60 = 29.152$ $29.152 \times 1500 = 43728$ $\text{Exponent: } 10^{-31} \times 10^{-19} = 10^{-50}$ $2meV = 4.3728 \times 10^{4} \times 10^{-50} = 4.3728 \times 10^{-46}$ 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).
4. $\sqrt{4.3728 \times 10^{-46}} = \sqrt{4.3728} \times 10^{-23} \approx 2.091 \times 10^{-23}$ kg m/s — $\sqrt{4.3728} \approx 2.091$. For the exponent: $\sqrt{10^{-46}} = 10^{-23}$. [M1]
5. $\lambda = \frac{6.63 \times 10^{-34}}{2.091 \times 10^{-23}} = \frac{6.63}{2.091} \times 10^{-34-(-23)} = 3.17 \times 10^{-11}$ m — $6.63 \div 2.091 \approx 3.17$. Exponent: $-34 + 23 = -11$. [A1]

Answer: $\lambda \approx 3.2 \times 10^{-11}$ m (to 2 significant figures)

6. Why This Is a Turning Point in Physics

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.