Wave-Particle Duality: de Broglie

Spec mapping: OCR H556 Module 4.5 — Quantum physics (wave-particle duality of matter; de Broglie wavelength $\lambda = h/p$λ=h/p for a moving particle; the use of $p = \sqrt{2 m_e e V}$p=2me​eV​ to compute the wavelength of an electron accelerated through a potential difference $V$V; the order-of-magnitude argument showing why classical objects do not exhibit observable wave behaviour). Refer to the official OCR H556 specification document for exact wording.

By 1923 the photon theory was established. Compton's 1923 experiment on X-ray scattering from electrons confirmed that photons carry momentum $p = h/\lambda$p=h/λ, just as particles do. Light, the archetypal wave, was now also known to behave as a particle in its interactions with matter.

That same year, a young French physicist named Louis de Broglie — who was preparing his doctoral thesis at the University of Paris — asked a bold and beautiful question: if waves can behave like particles, can particles behave like waves?

His answer, published in 1924 and awarded the Nobel Prize in 1929, is now known as the de Broglie hypothesis. It is one of the cornerstones of quantum physics and a central topic in Module 4.5 of the OCR A-Level Physics A specification (H556).

---

The de Broglie Hypothesis

Every moving particle has an associated wave, whose wavelength is given by

$$\lambda = h/p = h/(mv)$$λ=h/p=h/(mv)

Here:

• λ is the de Broglie wavelength in metres
• h is Planck's constant, 6.63 × 10⁻³⁴ J s
• p = mv is the linear momentum of the particle in kg m s⁻¹
• m is the mass of the particle in kg
• v is its speed in m s⁻¹

The formula is a direct generalisation of the photon momentum relation. For a photon, p = h/λ (as Compton confirmed). De Broglie's insight was to invert this: if any particle has momentum p, then there must be an associated wavelength λ = h/p.

This hypothesis was initially speculative. It had no experimental support in 1924. But within three years it had been dramatically confirmed by electron diffraction experiments (Lesson 7), and it has since become one of the most thoroughly tested predictions in physics.

---

What Does the Wavelength Mean?

The de Broglie wavelength is the characteristic length scale of quantum behaviour for a particle. When the particle encounters structures of size comparable to λ, it diffracts and interferes like a wave. When it encounters structures much larger than λ, it behaves like a classical particle travelling in a straight line.

The wave is not a wave of the particle's displacement, or of the medium through which it moves (the particle may be moving through vacuum). It is — on one interpretation — a wave of probability: the intensity of the wave at each point tells you how likely the particle is to be found there. This is the Born interpretation, developed in 1926, and you will meet it in university quantum mechanics. At A-Level, the wave is simply "the wave associated with the particle", and you can use it to compute wavelengths and diffraction patterns.

---

Why Is Quantum Behaviour Not Obvious in Everyday Life?

Because Planck's constant is tiny, the de Broglie wavelength of any macroscopic object is absurdly small. Consider a tennis ball of mass 60 g travelling at 30 m s⁻¹ (a fast serve):

$$\begin{aligned} p = mv = (0.060)(30) = 1.8 kg m s^{-1} \\ \lambda = h/p = (6.63 \times 10^{-34})/1.8 \approx 3.7 \times 10^{-34} m \end{aligned}$$p=mv=(0.060)(30)=1.8kgms−1λ=h/p=(6.63×10−34)/1.8≈3.7×10−34m​

This is about 10⁻¹⁹ times the diameter of a proton. There is no conceivable experiment that could detect the wave nature of a tennis ball — you would need diffraction slits smaller than a subnuclear particle, which is obviously impossible.

Now compare with an electron accelerated through a potential difference of 100 V. We shall derive the speed more carefully in a moment, but the de Broglie wavelength works out to about 1.2 × 10⁻¹⁰ m — comparable to the spacing of atoms in a crystal. This is a wavelength we can easily work with: pass the electron through a crystal and it will diffract.

The moral: quantum effects are hidden at everyday scales because h is small, and they become visible only when we deal with very small masses and low speeds — i.e. when p is small enough that h/p is comparable to the dimensions of the apparatus.

---

Worked Example 1: Electron Accelerated Through 200 V

An electron is accelerated from rest through a potential difference of 200 V. Calculate (a) its final kinetic energy, (b) its final speed, (c) its final momentum, and (d) its de Broglie wavelength.

Solution.

(a) Kinetic energy gained from the potential difference:

$$KE = eV = (1.60 \times 10^{-19})(200) = 3.20 \times 10^{-17} J$$KE=eV=(1.60×10−19)(200)=3.20×10−17J

Or directly: 200 eV.

(b) Speed. Use KE = (1/2)mv²:

$$\begin{aligned} v = \sqrt{2 KE/m} \\ = \sqrt{2 \times 3.20 \times 10^{-17} / 9.11 \times 10^{-31}} \\ = \sqrt{7.03 \times 10^{13}} \\ = 8.38 \times 10^{6} m s^{-1} \end{aligned}$$v=2KE/m​=2×3.20×10−17/9.11×10−31​=7.03×1013​=8.38×106ms−1​

About 2.8% of the speed of light — borderline for non-relativistic treatment, but still comfortable.

(c) Momentum:

$$p = mv = (9.11 \times 10^{-31})(8.38 \times 10^{6}) = 7.63 \times 10^{-24} kg m s^{-1}$$p=mv=(9.11×10−31)(8.38×106)=7.63×10−24kgms−1

(d) de Broglie wavelength:

$$\lambda = h/p = (6.63 \times 10^{-34})/(7.63 \times 10^{-24}) = 8.69 \times 10^{-11} m \approx 87 pm$$λ=h/p=(6.63×10−34)/(7.63×10−24)=8.69×10−11m≈87pm

Answer: λ ≈ 87 pm, or about 0.087 nm. This is comparable to the spacing of atoms in a crystal — which is exactly why electron diffraction is possible using crystals as gratings.

---

Worked Example 2: A Useful Shortcut for Electrons

For an electron accelerated through a p.d. V, you can derive a closed-form expression that avoids the intermediate speed calculation entirely. The kinetic energy is KE = eV, and KE = p²/(2m) (which follows from KE = (1/2)mv² and p = mv). So:

$$\begin{aligned} p = \sqrt{2 m eV} \\ \lambda = h/p = h/ \sqrt{2 m eV} \end{aligned}$$p=2meV​λ=h/p=h/2meV​​

Plugging in the constants for an electron:

$$\begin{aligned} \lambda\,(\text{in metres}) &\approx \sqrt{1.50/V} \times 10^{-9} \\ &\approx 1.226 \times 10^{-9} / \sqrt{V} \end{aligned}$$λ(in metres)​≈1.50/V​×10−9≈1.226×10−9/V​​

where V is in volts. Equivalently:

$$\lambda (in nm) \approx 1.226/ \sqrt{V}$$λ(innm)≈1.226/V​

For V = 200 V: λ ≈ 1.226/√200 ≈ 0.0867 nm — agreeing exactly with our step-by-step calculation. This shortcut is well worth knowing; it lets you compute electron de Broglie wavelengths in seconds without any intermediate steps.

Exam Tip: If you only need an electron's de Broglie wavelength, and you know the accelerating potential, use the shortcut λ_nm ≈ 1.226/√V. This speeds up multi-part questions substantially.

---

Worked Example 3: Neutron Diffraction

A neutron has mass m_n = 1.67 × 10⁻²⁷ kg and travels at 2.0 × 10³ m s⁻¹ (a "thermal neutron", in the speed range typical of reactor neutrons moderated to room temperature). Calculate its de Broglie wavelength.

Solution.

$$\begin{aligned} p = mv = (1.67 \times 10^{-27})(2.0 \times 10^{3}) = 3.34 \times 10^{-24} kg m s^{-1} \\ \lambda = h/p = (6.63 \times 10^{-34})/(3.34 \times 10^{-24}) = 1.98 \times 10^{-10} m \approx 0.2 nm \end{aligned}$$p=mv=(1.67×10−27)(2.0×103)=3.34×10−24kgms−1λ=h/p=(6.63×10−34)/(3.34×10−24)=1.98×10−10m≈0.2nm​

Answer: λ ≈ 0.2 nm. Again, this is of order crystal-lattice spacing. Thermal neutrons can therefore be diffracted from crystals, and neutron diffraction is in fact a standard technique in condensed-matter and materials physics — complementary to X-ray diffraction because neutrons interact with nuclei (not electrons) and so reveal different structural information.

---

Worked Example 4: A Buckyball

Can a large molecule behave as a wave? In 1999, researchers at the University of Vienna demonstrated diffraction of C₆₀ ("buckyball") molecules through a grating. Each C₆₀ has mass 1.2 × 10⁻²⁴ kg and, in their experiment, moved at about 220 m s⁻¹.

$$\begin{aligned} p = (1.2 \times 10^{-24})(220) \approx 2.64 \times 10^{-22} kg m s^{-1} \\ \lambda = h/p = (6.63 \times 10^{-34})/(2.64 \times 10^{-22}) \approx 2.5 \times 10^{-12} m \end{aligned}$$p=(1.2×10−24)(220)≈2.64×10−22kgms−1λ=h/p=(6.63×10−34)/(2.64×10−22)≈2.5×10−12m​

About 2.5 picometres — smaller than the atoms in the molecule itself! Yet diffraction patterns were observed through a grating with 100 nm spacing, by averaging over many molecules. This is an astonishing demonstration of quantum behaviour in a 60-atom molecule, and it continues to be pushed to ever larger molecules (as of the mid-2020s, molecules with thousands of atoms have shown diffraction).

---

The Double-Slit Experiment with Particles

Perhaps the most famous demonstration of wave-particle duality is the double-slit experiment, originally performed with light by Young in 1801. Since the 1960s it has been done with electrons, neutrons, atoms, and large molecules.

The result is always the same: when particles are sent one at a time through a pair of slits, each particle lands at a single, definite point on the detector (particle behaviour). But the statistical distribution of many particles builds up an interference pattern with characteristic fringe spacing (wave behaviour):

$$\text{fringe spacing} = \frac{\lambda D}{a}$$fringe spacing=aλD​

where a is the slit separation, D is the distance from slit to screen, and λ is the de Broglie wavelength.

The interference pattern appears even when particles are sent one at a time — meaning each particle somehow "interferes with itself". The particle goes through both slits. The wave-like spread of probability governs where each individual particle can land, while the total number of particles landing at each point builds the statistical pattern.

flowchart LR
    S["Single particle source"] --> T["Two slits"]
    T --> W["Wave of probability passes through both"]
    W --> D["Particle detected at single point"]
    D --> P["Many particles build<br/>interference pattern"]

This is the most striking demonstration of wave-particle duality. A single electron behaves as a wave as it propagates, and as a particle as it is detected. The wave and the particle are two aspects of the same entity.

---

The Momentum–Wavelength Relation in Context

System	Typical mass	Typical speed	Typical wavelength
Visible-light photon	0 (rest)	c	5 × 10⁻⁷ m
Electron at 100 V	9.1 × 10⁻³¹ kg	6 × 10⁶ m s⁻¹	1.2 × 10⁻¹⁰ m
Thermal neutron	1.7 × 10⁻²⁷ kg	2 × 10³ m s⁻¹	2 × 10⁻¹⁰ m
Helium atom (room T)	6.7 × 10⁻²⁷ kg	1.4 × 10³ m s⁻¹	7 × 10⁻¹¹ m
Buckyball (220 m/s)	1.2 × 10⁻²⁴ kg	220 m s⁻¹	2.5 × 10⁻¹² m
Tennis ball (30 m/s)	0.06 kg	30 m s⁻¹	3.7 × 10⁻³⁴ m

Notice how the de Broglie wavelength shrinks rapidly as mass increases — which is why we never see tennis balls diffract.

---

De Broglie Wavelength Geometry — Visual Reference

The de Broglie wave is associated with the particle and has wavelength inversely proportional to momentum. The following SVG sketches the comparative wavelengths for a slow electron and a fast electron (same mass, different speeds), showing how higher momentum produces a tighter wave.

The slow electron's wave covers a few wavelengths over the visible region; the fast electron's wave covers many. This is the operational meaning of $\lambda = h/p$λ=h/p: as momentum increases, wavelength decreases proportionally, and the particle "looks more like a point" at the scale of any fixed apparatus.

---

Synoptic Links

• Photon momentum (Module 4.5, lesson 1) — $p = h/\lambda$p=h/λ for a photon is the same relation, run backwards. De Broglie noted that if a wave (photon) has momentum $h/\lambda$h/λ, then by symmetry a particle of momentum $p$p should have an associated wavelength $h/p$h/p. The matter wave is the converse of the photon-as-particle.
• Capacitor / charged particle acceleration (Module 6.1) — the work-energy theorem $W = qV$W=qV gives an electron accelerated through $V$V a kinetic energy $eV$eV. Combining with $KE = p^2/(2m_e)$KE=p2/(2me​) yields $p = \sqrt{2 m_e e V}$p=2me​eV​ and hence $\lambda = h/\sqrt{2 m_e e V}$λ=h/2me​eV​. This is the standard A-Level closed form.
• Diffraction grating (Module 4.4) — $d \sin\theta = n\lambda$dsinθ=nλ holds for matter waves through a crystal grating exactly as for light through a ruled grating; the next lesson on electron diffraction is the experimental embodiment.
• Standing waves in a string (Module 4.4) — bound matter waves form standing-wave patterns confined by boundary conditions, just as transverse string waves do. This is the conceptual link from de Broglie (lesson 6) to discrete energy levels (lesson 9).

---

Specimen question modelled on the OCR H556 paper format

Question (10 marks): An electron is accelerated from rest through a potential difference of $V = 150$V=150 V in an electron-diffraction apparatus.

(a) Show that the momentum of the electron after acceleration is given by $p = \sqrt{2 m_e e V}$p=2me​eV​, stating clearly the assumptions made. [3]

(b) Calculate (i) the momentum and (ii) the de Broglie wavelength of the electron. Take $m_e = 9.11 \times 10^{-31}$me​=9.11×10−31 kg, $e = 1.60 \times 10^{-19}$e=1.60×10−19 C, $h = 6.63 \times 10^{-34}$h=6.63×10−34 J s. [4]

(c) A tennis ball of mass $60$60 g is served at $30$30 m s$^{-1}$−1. Calculate its de Broglie wavelength and use the result to explain, with reference to a typical diffraction grating, why no wave behaviour is ever observed for the ball. [3]

AO breakdown