The de Broglie Wavelength with Worked Examples

In 1924, Louis de Broglie proposed that everything that moves has a wavelength. For a cricket ball, it’s unimaginably tiny; for an electron in an atom, it’s about the size of the atom itself. That single idea explains electron diffraction, electron microscopes and the quantised energy levels of atoms. This article gives the equation and works through the calculations you’re likely to meet.

For the background, see wave-particle duality.

The equation

λ = h ÷ p = h ÷ (mv)

• λ = de Broglie wavelength (m)
• h = Planck’s constant = 6.626 × 10⁻³⁴ J s
• p = momentum = mass × velocity (kg m s⁻¹)
• m = mass (kg)
• v = velocity (m s⁻¹)

Useful masses:

Particle	Mass (kg)
electron	9.109 × 10⁻³¹
proton	1.673 × 10⁻²⁷
neutron	1.675 × 10⁻²⁷
1 atomic mass unit (u)	1.661 × 10⁻²⁷

Worked example 1: an electron

Calculate the de Broglie wavelength of an electron travelling at 2.00 × 10⁶ m/s.

p = 9.109 × 10⁻³¹ × 2.00 × 10⁶ = 1.822 × 10⁻²⁴ kg m/s

λ = 6.626 × 10⁻³⁴ ÷ 1.822 × 10⁻²⁴ = 3.64 × 10⁻¹⁰ m = 0.364 nm (364 pm)

That’s similar to the size of an atom and the spacing between atoms in a crystal. So electrons at this speed will diffract from crystals, exactly what Davisson and Germer observed in 1927.

Worked example 2: a tennis ball

A 58 g tennis ball is served at 50 m/s. Calculate its de Broglie wavelength.

m = 0.058 kg

λ = 6.626 × 10⁻³⁴ ÷ (0.058 × 50) = 6.626 × 10⁻³⁴ ÷ 2.9 = 2.3 × 10⁻³⁴ m

That’s about 10¹⁹ times smaller than a proton. No gap, crystal or experiment could ever reveal wave behaviour at this scale, which is why we never see tennis balls diffract. Quantum effects matter only when the wavelength is comparable to the size of the things the object interacts with.

Worked example 3: comparing particles at the same speed

Compare the wavelengths of an electron and a proton, both moving at 1.00 × 10⁵ m/s.

• Electron: λ = 6.626 × 10⁻³⁴ ÷ (9.109 × 10⁻³¹ × 1.00 × 10⁵) = 7.27 × 10⁻⁹ m (7.27 nm)
• Proton: λ = 6.626 × 10⁻³⁴ ÷ (1.673 × 10⁻²⁷ × 1.00 × 10⁵) = 3.96 × 10⁻¹² m (3.96 pm)

The electron’s wavelength is about 1,836 times longer, the ratio of the two masses. Lighter particles show wave behaviour much more readily.

Worked example 4: an electron accelerated through a voltage

In electron microscopes and diffraction experiments, electrons are accelerated through a potential difference V. They gain kinetic energy:

KE = eV (e = 1.602 × 10⁻¹⁹ C)

Since KE = p²/2m, the momentum is p = √(2m × eV), and:

λ = h ÷ √(2meV)

Calculate the wavelength of electrons accelerated through 150 V.

KE = 1.602 × 10⁻¹⁹ × 150 = 2.403 × 10⁻¹⁷ J

p = √(2 × 9.109 × 10⁻³¹ × 2.403 × 10⁻¹⁷) = √(4.378 × 10⁻⁴⁷) = 6.617 × 10⁻²⁴ kg m/s

λ = 6.626 × 10⁻³⁴ ÷ 6.617 × 10⁻²⁴ = 1.00 × 10⁻¹⁰ m = 0.100 nm (100 pm)

A handy approximation for electrons (non-relativistic): λ (nm) ≈ 1.226 ÷ √V. For 150 V: 1.226 ÷ 12.25 = 0.100 nm ✓.

Worked example 5: electron microscope

A transmission electron microscope accelerates electrons through 100,000 V. Estimate their wavelength (ignoring relativity).

λ ≈ 1.226 ÷ √100,000 = 1.226 ÷ 316.2 = 0.00388 nm (3.88 pm)

That’s about 100,000 times shorter than visible light (400–700 nm), which is why electron microscopes can resolve far finer detail than light microscopes. (At these energies, relativistic effects reduce the true wavelength slightly, to about 3.7 pm.)

Worked example 6: finding speed from wavelength

What speed must an electron have for its de Broglie wavelength to be 0.500 nm?

v = h ÷ (mλ) = 6.626 × 10⁻³⁴ ÷ (9.109 × 10⁻³¹ × 5.00 × 10⁻¹⁰) = 6.626 × 10⁻³⁴ ÷ 4.555 × 10⁻⁴⁰ = 1.45 × 10⁶ m/s

Worked example 7: thermal neutrons

Neutrons from a nuclear reactor, slowed to room-temperature energies (“thermal” neutrons), travel at about 2,200 m/s.

λ = 6.626 × 10⁻³⁴ ÷ (1.675 × 10⁻²⁷ × 2,200) = 6.626 × 10⁻³⁴ ÷ 3.685 × 10⁻²⁴ = 1.80 × 10⁻¹⁰ m (0.180 nm)

This is ideal for neutron diffraction, which reveals crystal structures and is especially good at locating light atoms such as hydrogen.

Worked example 8: an atom

Calculate the de Broglie wavelength of a helium atom (mass 4.00 u) moving at 1,300 m/s (a typical speed at room temperature).

m = 4.00 × 1.661 × 10⁻²⁷ = 6.644 × 10⁻²⁷ kg

λ = 6.626 × 10⁻³⁴ ÷ (6.644 × 10⁻²⁷ × 1,300) = 7.67 × 10⁻¹¹ m (77 pm)

Even whole atoms can diffract, and atom interferometers use this to make extremely precise measurements of gravity and fundamental constants.

De Broglie and Bohr’s orbits

De Broglie’s idea gave a beautiful explanation of Bohr’s mysterious rule for allowed orbits. If the electron is a wave travelling around the nucleus, the orbit is only stable if a whole number of wavelengths fits exactly around the circumference:

nλ = 2πr (n = 1, 2, 3…)

Otherwise, the wave would interfere with itself destructively and cancel out. Substituting λ = h/mv gives mvr = nh/2π, which is exactly Bohr’s quantisation condition. Allowed energy levels are simply the standing-wave patterns that fit. See the Bohr model and Bohr model calculations.

When is wave behaviour noticeable?

Wave effects become important when the de Broglie wavelength is comparable to or larger than the relevant length scale:

Object	Typical λ	Compared with	Wave effects?
electron in an atom	about 0.3 nm	atom size (about 0.1–0.3 nm)	yes, essential
100 V electron	about 0.12 nm	atomic spacing in crystals	yes, diffraction
thermal neutron	about 0.18 nm	atomic spacing	yes, diffraction
dust particle	about 10⁻²⁰ m	anything measurable	no
person walking	about 10⁻³⁶ m	anything	no

A three-step routine

1. Identify the particle and write its mass in kilograms.
2. Find the momentum, either directly (p = mv) or from kinetic energy (p = √(2m × KE)); for charged particles accelerated through a voltage, KE = charge × V.
3. Divide Planck’s constant by the momentum and convert the answer to a sensible unit (nm or pm), then compare it with atomic sizes to judge whether wave effects will matter.

Common mistakes

1. Using grams instead of kilograms. Always convert mass to kg.
2. Forgetting to take the square root in λ = h/√(2meV).
3. Using the photon equation. For photons, λ = hc/E; for particles with mass, use λ = h/mv. They’re not interchangeable.
4. Mixing up electron and proton masses.
5. Expecting large objects to diffract. Their wavelengths are far too small.

Practice questions

1. Calculate the wavelength of an electron moving at 5.00 × 10⁵ m/s.
2. Calculate the wavelength of electrons accelerated through 54 V (Davisson and Germer’s famous setting).
3. What speed gives a proton a wavelength of 1.00 × 10⁻¹¹ m?

Answers:

1. λ = 6.626 × 10⁻³⁴ ÷ (9.109 × 10⁻³¹ × 5.00 × 10⁵) = 1.45 × 10⁻⁹ m (1.45 nm)
2. λ ≈ 1.226 ÷ √54 = 0.167 nm, comparable to the spacing between atoms in their nickel crystal, which is why diffraction was observed.
3. v = 6.626 × 10⁻³⁴ ÷ (1.673 × 10⁻²⁷ × 1.00 × 10⁻¹¹) = 3.96 × 10⁴ m/s

Key takeaways

• Every moving particle has a de Broglie wavelength λ = h/mv.
• Electrons at typical speeds have wavelengths similar to atomic spacings, so they diffract from crystals.
• For electrons accelerated through V volts, λ = h/√(2meV) ≈ 1.226/√V nm.
• Everyday objects have wavelengths far too small to show wave behaviour.
• Fitting whole wavelengths around an orbit explains why only certain energy levels are allowed.