De Broglie's hypothesis and electron diffraction

Turning Points in Physics | AQA A-Level Physics

Key Definition

De Broglie wavelength: Every moving particle has an associated wavelength given by $\lambda = h/p = h/(mv)$, where $h$ is Planck's constant and $p$ is the particle's momentum.

De Broglie's hypothesis (1924)

Louis de Broglie proposed that if light (a wave) can behave as particles (photons), then particles like electrons should also exhibit wave-like behaviour.
The wavelength of a particle accelerated from rest through a p.d. $V$ is found by combining $eV = \frac{1}{2}mv^2$ with $\lambda = h/(mv)$:

$$\lambda = \frac{h}{\sqrt{2m_e eV}}$$

For an electron accelerated through 100 V, this gives $\lambda \approx 1.2 \times 10^{-10}$ m, which is comparable to atomic spacings. This is why electrons can be diffracted by crystalline structures.

Electron diffraction: the evidence

In 1927, Davisson and Germer (and independently G.P. Thomson) fired electrons at thin layers of polycrystalline graphiteA form of carbon made up of many tiny crystals oriented randomly. The regular atomic spacing acts as a diffraction grating for electrons. and observed concentric rings on a fluorescent screen beyond.
This ring pattern is identical to what you see when X-rays are diffracted by a crystal. It can only be explained if the electrons behave as waves with a wavelength matching de Broglie's prediction.
Increasing the accelerating voltage increases the electrons' momentum, which decreases their de Broglie wavelength. This causes the diffraction rings to become smaller (less spreading), which is exactly what is observed.

Why everyday objects do not show wave behaviour

For a macroscopic object (say, a tennis ball of mass 0.06 kg moving at 50 m s$^{-1}$), the de Broglie wavelength is roughly $2 \times 10^{-34}$ m.
This is far too small to produce any observable diffraction. Wave behaviour only becomes detectable when $\lambda$ is comparable to the size of obstacles or slits the particle encounters.

Common Mistake

Students sometimes say "increasing the voltage increases the wavelength." It is the opposite. Higher voltage means higher speed, higher momentum, and therefore a shorter wavelength ($\lambda = h/p$). The diffraction rings get smaller, not larger.