Wave-Particle Duality: de Broglie

By 1923 the photon theory was established. Compton's 1923 experiment on X-ray scattering from electrons confirmed that photons carry momentum 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).

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The de Broglie Hypothesis

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

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.

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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.

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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):

p = mv = (0.060)(30) = 1.8 kg m s⁻¹
λ = h/p = (6.63 × 10⁻³⁴)/1.8 ≈ 3.7 × 10⁻³⁴ m

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.

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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 × 10⁻¹⁹)(200) = 3.20 × 10⁻¹⁷ J

Or directly: 200 eV.

(b) Speed. Use KE = (1/2)mv²:

v = √(2 KE/m)
  = √(2 × 3.20 × 10⁻¹⁷ / 9.11 × 10⁻³¹)
  = √(7.03 × 10¹³)
  = 8.38 × 10⁶ m s⁻¹

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

(c) Momentum:

p = mv = (9.11 × 10⁻³¹)(8.38 × 10⁶) = 7.63 × 10⁻²⁴ kg m s⁻¹

(d) de Broglie wavelength:

λ = h/p = (6.63 × 10⁻³⁴)/(7.63 × 10⁻²⁴) = 8.69 × 10⁻¹¹ m ≈ 87 pm

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.

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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:

p = √(2 m eV)
λ = h/p = h/√(2 m eV)

Plugging in the constants for an electron:

λ (in metres) ≈ √(1.50/V) × 10⁻⁹
            ≈ 1.226 × 10⁻⁹ / √V

where V is in volts. Equivalently:

λ (in nm) ≈ 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.

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Worked Example 3: Neutron Diffraction