Wave nature of matter
De Broglie’s bold idea
Light has a dual nature. In interference and diffraction it behaves as a wave. In the photoelectric effect it behaves as a stream of particles, photons.
In 1924, the French physicist Louis de Broglie asked a daring question. Nature loves symmetry. If waves can behave like particles, can particles, like electrons, behave like waves?
He proposed that every moving material particle has a wave associated with it, called a matter wave. Its wavelength, the de Broglie wavelength, is λ equals h over p, where p equals m v is the particle’s momentum.
This is the same relation that holds for a photon. A photon’s momentum is h ν over c, which is h over λ. So λ equals h over p for light and for matter alike.
The relation joins a wave property, the wavelength λ, to a particle property, the momentum p. The bigger the momentum, the shorter the wave. A heavier or faster particle has a smaller wavelength.
Let us try it on an electron moving at 5.4 × 10⁶ metre per second. Its momentum is 9.11 × 10⁻³¹ times 5.4 × 10⁶, which is 4.92 × 10⁻²⁴ kilogram metre per second.
So λ equals 6.63 × 10⁻³⁴ divided by 4.92 × 10⁻²⁴: 1.35 × 10⁻¹⁰ metre, or 0.135 nanometre. That is about the spacing between atoms in a crystal, and in the range of X-ray wavelengths.
Now a cricket ball of mass 150 grams, moving at 30 metre per second. Its momentum is 0.15 times 30, which is 4.5 kilogram metre per second. λ equals 6.63 × 10⁻³⁴ divided by 4.5: 1.47 × 10⁻³⁴ metre.
That is far, far smaller than even a nucleus. No experiment could ever detect such a wave, so the ball behaves just as a particle. For the electron, the wave is big enough to show up, and electrons really can be diffracted by crystals.
Electrons accelerated through a potential
In the laboratory, we get fast electrons by accelerating them from rest through a potential difference V. Each electron gains kinetic energy K equals e V.
Kinetic energy and momentum are related: K equals p squared over 2m. So p equals √(2 m K). The de Broglie wavelength is λ equals h over √(2 m K).
For an electron accelerated through V volts, K equals e V, so λ equals h over √(2 m e V).
Put in the numbers: h over √(2 m e) for an electron works out to 1.227 × 10⁻⁹. So λ equals 1.227 over √V nanometres, with V in volts.
For 100 volts, √V is 10, and λ is 0.123 nanometre, about the size of an atom. For 10 000 volts, λ is ten times smaller, 0.0123 nanometre. Four times the voltage halves the wavelength.
The formula λ equals h over √(2 m K) also compares different particles with the same kinetic energy. The heavier the particle, the shorter its wavelength.
Take a proton and an electron with the same kinetic energy. The proton is about 1836 times heavier, so its wavelength is √1836, about 43 times shorter.
Wave effects show up only when the wavelength is comparable to the size of what the particle meets. A 100 volt electron meets atoms spaced about 0.1 nanometre apart in a crystal, and diffracts, just like X-rays.
This was confirmed in 1927, when electron beams were diffracted by crystals. Matter, like radiation, has a dual nature: it behaves as particles and as waves.
Try it: matter waves
- Electron through 100 V: check that λ = 1.227/√V nm gives 0.123 nm. Now try 400 V: λ halves.
- Choose “Set the speed” and set 5.4 × 10⁶ m/s: λ ≈ 0.135 nm.
- Switch to a proton or an α particle accelerated through the same V. Which has the shortest wavelength, and why?
- Try the cricket ball and the dust grain. Where do their wavelengths land on the ruler, compared with a nucleus?
Same kinetic energy
An electron, a proton and an α particle all have the same kinetic energy. Which has the shortest de Broglie wavelength?
- The electron
- The proton
- The α particle
- All three have the same wavelength
Answer
C. λ = h/p = h/√(2mK). With the same K, λ ∝ 1/√m, so the heaviest particle has the shortest wavelength. The α particle (about 4 times the proton’s mass) wins; the electron, the lightest, has the longest wavelength. Same kinetic energy does not mean the same momentum.
Identify the particle
A particle moves three times as fast as an electron. The ratio of its de Broglie wavelength to that of the electron is 1.813 × 10⁻⁴. What is its mass? (mₑ = 9.11 × 10⁻³¹ kg)
- 5.02 × 10⁻²⁷ kg
- 1.675 × 10⁻²⁷ kg: a neutron
- 6.64 × 10⁻²⁷ kg: an α particle
- 1.65 × 10⁻³⁴ kg
Answer
B. λ = h/(mv), so λ/λₑ = mₑvₑ/(mv) = mₑ/(3m). Then m = mₑ/(3 × 1.813 × 10⁻⁴) = 9.11 × 10⁻³¹ / 5.439 × 10⁻⁴ = 1.675 × 10⁻²⁷ kg, the mass of a neutron (a proton’s mass is almost the same). Leaving out the factor 3 gives 5.02 × 10⁻²⁷ kg; 1.65 × 10⁻³⁴ kg comes from multiplying by the ratio instead of dividing.
Summary
- de Broglie wavelength:
λ = h/p = h/(mv) - In terms of kinetic energy:
λ = h / √(2mK) - Electron accelerated through V:
λ = h / √(2meV) = 1.227/√V nm
- De Broglie (1924): every moving particle has a matter wave with λ = h/p, the same relation as for a photon.
- λ is tiny for everyday objects (a 150 g ball at 30 m/s: 1.47 × 10⁻³⁴ m), so they show no wave effects.
- For electrons, λ is comparable to atomic spacings (≈ 0.1 nm at about 100 V), so electrons can be diffracted by crystals.
- For the same kinetic energy, λ ∝ 1/√m: heavier particles have shorter wavelengths.
- Radiation and matter both have a dual nature: wave-like and particle-like.