F. Revolutions in Modern Physics: Quantum Mechanics and Special Relativity · Quantum MechanicsMatter Waves: The de Broglie Wavelength
F2.2 F3.2 F2.1 F1.1About 70 minutes
Learning goals- I can state de Broglie's hypothesis and calculate the wavelength of a moving particle with λ = h/p = h/(mv).
- I can find the de Broglie wavelength of an electron accelerated through a potential difference.
- I can explain why wave properties of matter are noticeable for electrons but not for everyday objects.
- I can explain how de Broglie's idea used symmetry to propose a new model of matter.
Words to know
matter wave
The wave associated with any moving particle; also called a de Broglie wave.
de Broglie wavelength
λ = h/p = h/(mv), the wavelength of a particle with momentum p.
momentum (p)
For a particle, p = mv; for a photon, p = h/λ.
accelerating potential difference
The voltage that speeds up a charged particle; an electron gains kinetic energy eV.
symmetry (in physics)
The idea that a pattern seen in one part of nature may also hold in a matching part, used as a guide to new theories.
If waves can be particles...
By the early 1920s, physicists had accepted that light, long thought to be a wave, also behaves like a stream of particles with momentum p = h/λ. In his 1924 doctoral thesis, the French physicist Louis de Broglie turned the idea around. Nature loves symmetry, he argued: if waves can act like particles, perhaps particles can act like waves.
He proposed that every moving particle has a wavelength, given by the same relationship as for photons, solved for λ:
λ = h/p = h/(mv)
At the time there was no experimental evidence at all. Einstein supported the idea, and within three years experiments with electrons confirmed it (next lesson). De Broglie received the 1929 Nobel Prize in Physics. His hypothesis is a famous example of a theory driven by an idea of symmetry rather than by a puzzling measurement.
Worked exampleExample 1: A baseball and an electron
Find the de Broglie wavelength of (a) a 0.145 kg baseball moving at 40 m/s and (b) an electron (9.11 × 10⁻³¹ kg) moving at 1.0 × 10⁶ m/s.
- (a) λ = h/(mv) = 6.63 × 10⁻³⁴ ÷ (0.145 × 40) = 6.63 × 10⁻³⁴ ÷ 5.8 = 1.1 × 10⁻³⁴ m.
- (b) λ = 6.63 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 1.0 × 10⁶) = 6.63 × 10⁻³⁴ ÷ 9.11 × 10⁻²⁵ = 7.3 × 10⁻¹⁰ m.
(a) 1.1 × 10⁻³⁴ m, about 10¹⁹ times smaller than a proton. Nothing can diffract it. (b) 7.3 × 10⁻¹⁰ m = 0.73 nm, a few times the spacing between atoms: an electron's wave properties can be detected.
Why don't we notice matter waves?
Waves only show their wave nature (diffraction and interference) when they meet openings or obstacles comparable in size to their wavelength. You saw this in The Wave Nature of Light: diffraction is strong when the slit width is close to λ.
Planck's constant is tiny, so for anything with an everyday mass, h/(mv) is far smaller than any opening that could exist. A baseball's wavelength is 10¹⁹ times smaller than a proton. But an electron's mass is so small that its wavelength can be comparable to the spacing of atoms in a crystal (about 0.1 nm to 0.3 nm). A crystal is a natural diffraction grating for electrons.
Electrons accelerated through a voltage
In experiments, electrons are usually sped up from rest by a potential difference V (as in a cathode-ray tube or electron microscope). Each electron gains kinetic energy:
eV = ½mv² = p²/(2m), so p = √(2meV)
and the de Broglie wavelength is
λ = h / √(2meV)
For electrons this simplifies to a handy shortcut: λ (in nm) ≈ 1.226 ÷ √V, with V in volts. (This assumes the electron is much slower than light, which is accurate below about 10 kV; above that, relativity is needed for precise values.)
Worked exampleExample 2: An electron through 100 V
An electron starts from rest and is accelerated through 100 V. Find its kinetic energy, momentum and de Broglie wavelength.
- E(k) = eV = (1.60 × 10⁻¹⁹ C)(100 V) = 1.60 × 10⁻¹⁷ J.
- p = √(2mE(k)) = √(2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁷) = √(2.92 × 10⁻⁴⁷) = 5.40 × 10⁻²⁴ kg·m/s.
- λ = h/p = 6.63 × 10⁻³⁴ ÷ 5.40 × 10⁻²⁴ = 1.23 × 10⁻¹⁰ m.
- Shortcut check: 1.226 ÷ √100 = 0.123 nm. ✓
λ = 0.123 nm, about the spacing of atoms in a crystal: just right for electron diffraction.
de Broglie wavelength of an electron against accelerating voltage (λ ≈ 1.226/√V nm). Between roughly 15 V and 150 V the wavelength matches the spacing of atoms in crystals.Worked exampleExample 3: What speed gives a chosen wavelength?
What speed and kinetic energy (in eV) must an electron have for λ = 0.10 nm? Compare with the energy of an X-ray photon of the same wavelength.
- v = h/(mλ) = 6.63 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 1.0 × 10⁻¹⁰) = 7.3 × 10⁶ m/s.
- E(k) = ½mv² = ½(9.11 × 10⁻³¹)(7.28 × 10⁶)² = 2.41 × 10⁻¹⁷ J = 151 eV.
- Photon with λ = 0.10 nm: E = 1240 ÷ 0.10 = 12 400 eV.
v ≈ 7.3 × 10⁶ m/s, E(k) ≈ 150 eV. An X-ray photon of the same wavelength carries about 80 times more energy. This is one reason electron beams can probe surfaces gently, and why electron microscopes get short wavelengths at modest voltages.
Same formula, two directions
For a photon, p = h/λ tells you its momentum from its wavelength. For a particle, λ = h/p tells you its wavelength from its momentum. It is the same relationship. That is the symmetry de Broglie saw. Be careful, though: for a particle, E(k) = ½mv² = p²/(2m), whereas for a photon, E = pc. Do not use E = hc/λ for an electron.
Just for funYour de Broglie wavelength while walking down the hallway is about 6 × 10⁻³⁶ m. So if you are ever late for class, you cannot claim you diffracted around the door frame.
Hands-on activityWhy size matters for diffraction
You need: a sink or bathtub with a few centimetres of still water, a small plastic container or your fingertip, and two objects to make a gap (two books standing in the water work; or use your hands).
- Make ripples by tapping the water steadily with your fingertip at one end.
- Make a narrow gap (about the size of the ripple spacing, 1 cm to 2 cm) and watch the ripples spread out in a half-circle on the other side.
- Make the gap wide (10 cm or more). The ripples now go straight through with little spreading.
- Stand a large object, like a mug, in the path of the ripples: do the ripples bend around it?
What to look for: waves spread (diffract) noticeably only when the gap or obstacle is about as small as their wavelength. A baseball's matter wave is 10¹⁹ times smaller than a proton, so there is no gap small enough; an electron's wavelength (about 0.1 nm) matches the gaps between atoms. Wipe up spills so no one slips.
Your turn
1Find the de Broglie wavelength of a 1.0 g marble rolling at 0.50 m/s.
λ = 6.63 × 10⁻³⁴ ÷ (1.0 × 10⁻³ × 0.50) = 1.3 × 10⁻³⁰ m.
2Find the wavelength of an electron moving at 2.2 × 10⁶ m/s.
λ = 6.63 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 2.2 × 10⁶) = 3.3 × 10⁻¹⁰ m (0.33 nm).
3Find the wavelength of an electron accelerated from rest through 54 V.
λ ≈ 1.226 ÷ √54 = 0.167 nm (1.67 × 10⁻¹⁰ m).
4A neutron (1.675 × 10⁻²⁷ kg) moves at 2.2 × 10³ m/s. Find its wavelength. Could neutrons be diffracted by crystals?
λ = 6.63 × 10⁻³⁴ ÷ (1.675 × 10⁻²⁷ × 2.2 × 10³) = 1.8 × 10⁻¹⁰ m (0.18 nm). Yes: comparable to atomic spacing. (Neutron diffraction is used to study materials.)
5If an electron's speed doubles, what happens to its de Broglie wavelength?
λ = h/(mv), so λ is halved.
6If the accelerating voltage on an electron is multiplied by 4, what happens to its wavelength?
λ ∝ 1/√V, so λ becomes 1/2 as long.
7An electron and a proton have the same speed. Which has the longer wavelength, and by about how much? (mₚ ≈ 1836 mₑ)
The electron: λ ∝ 1/m at the same speed, so its wavelength is about 1836 times longer.
More practice: warm-up
1Write de Broglie's equation.
2Who proposed that matter has wave properties, and in what year?
3What is the kinetic energy of an electron accelerated through 200 V, in eV?
4Do heavier particles at the same speed have longer or shorter wavelengths?
5Why can't we see a baseball diffract?
Its wavelength (about 10⁻³⁴ m) is far too small compared with any opening or obstacle.
More practice: on your own
1An electron has a de Broglie wavelength of 0.20 nm. Find its speed and its kinetic energy in eV.
v = h/(mλ) = 6.63 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 2.0 × 10⁻¹⁰) = 3.6 × 10⁶ m/s. E(k) = p²/(2m) = (3.32 × 10⁻²⁴)² ÷ (2 × 9.11 × 10⁻³¹) = 6.0 × 10⁻¹⁸ J ≈ 38 eV.
2Through what potential difference must an electron be accelerated to have λ = 0.050 nm?
λ = 1.226/√V → √V = 1.226 ÷ 0.050 = 24.5, so V ≈ 600 V (601 V).
3A proton (1.67 × 10⁻²⁷ kg) moves at 1.0 × 10⁵ m/s. Find its wavelength.
λ = 6.63 × 10⁻³⁴ ÷ (1.67 × 10⁻²⁷ × 1.0 × 10⁵) = 4.0 × 10⁻¹² m.
4A helium atom (6.64 × 10⁻²⁷ kg) in a gas moves at about 1300 m/s. Find its wavelength.
λ = 6.63 × 10⁻³⁴ ÷ (6.64 × 10⁻²⁷ × 1300) = 7.7 × 10⁻¹¹ m (0.077 nm). Whole atoms have wave properties too; atom interferometers use this.
5Explain de Broglie's reasoning from symmetry in two or three sentences.
Light, a wave, had been shown to behave like particles, with momentum p = h/λ. De Broglie reasoned that nature should be symmetric, so particles like electrons should also behave like waves, with the same link between wavelength and momentum, λ = h/p. He proposed this with no direct evidence; experiments confirmed it a few years later.
Challenge
1Show that for a particle with kinetic energy E(k), λ = h/√(2mE(k)), and use it to find λ for a 1.0 keV electron.
E(k) = ½mv² = p²/(2m), so p = √(2mE(k)) and λ = h/p = h/√(2mE(k)). For 1.0 keV: E(k) = 1.60 × 10⁻¹⁶ J; p = √(2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁶) = 1.71 × 10⁻²³ kg·m/s; λ = 3.9 × 10⁻¹¹ m (0.039 nm).
2An electron and a photon both have wavelength 0.50 nm. Find (a) the photon's energy and (b) the electron's kinetic energy, both in eV. Which carries more momentum?
(a) E = 1240 ÷ 0.50 = 2480 eV. (b) p = h/λ = 1.33 × 10⁻²⁴ kg·m/s; E(k) = p²/(2m) = (1.33 × 10⁻²⁴)² ÷ (1.82 × 10⁻³⁰) = 9.7 × 10⁻¹⁹ J = 6.0 eV. They have the same momentum (both h/λ).
3Through what voltage must a proton be accelerated to have the same wavelength as an electron accelerated through 100 V? Explain.
λ = h/√(2meV). Same λ needs the same product mV. So V(proton) = 100 × mₑ/mₚ = 100 ÷ 1836 ≈ 0.054 V. The proton is so much heavier that it needs far less energy to reach the same momentum.
Quiz
1. The de Broglie wavelength of a particle is
2. An electron is accelerated from rest through 100 V. Its de Broglie wavelength is about
3. Why are matter waves not observed for everyday objects?
4. If the speed of an electron is tripled (still much less than c), its de Broglie wavelength
5. De Broglie proposed matter waves mainly because of
6. An electron and a proton move at the same speed. Compared with the proton, the electron has
Big idea (F2.2, F3.2): every moving particle has a wavelength λ = h/(mv). For electrons it is about the size of atoms, so electrons can diffract; for everyday objects it is unimaginably small.
How to help: the water-ripple activity builds intuition: waves spread only when the gap is about their own size. Ask your teen to explain why an electron can diffract through a crystal but a baseball cannot get through a doorway as a wave.
Common difficulties:
- Using E = hc/λ for electrons. That formula is only for photons; for particles use λ = h/(mv) or h/√(2mE(k)).
- Forgetting to square-root the voltage: λ ∝ 1/√V, not 1/V.
- Unit errors with grams (convert to kg) and eV (convert to J for SI calculations).