Checkpoint: Matter Waves

This checkpoint consolidates the core quantitative skills from module 6: applying the de Broglie relation, tracing how momentum follows from energy, and connecting the result to observable wave properties.

The de Broglie relation

In 1924, de Broglie proposed that any particle with momentum $p$ has an associated wavelength

$\lambda = \frac{h}{p},$

where $h = 6.626 \times 10^{-34}\ \text{J\,s}$ is Planck's constant. This relation is the matter-wave counterpart of Einstein's $E = h\nu$ for photons; it was confirmed two years later by Davisson and Germer, who observed electron diffraction from a nickel crystal with a lattice spacing of about $2.15\ \text{Å}$ , finding exactly the wavelength this formula predicted.

Momentum from kinetic energy

When an electron (charge $e$ , mass $m$ ) is accelerated through a potential difference $V$ , it gains kinetic energy

$KE = eV.$

For a non-relativistic electron, $KE = p^2 / (2m)$ , so

$p = \sqrt{2m \cdot KE} = \sqrt{2meV}.$

Substituting into the de Broglie relation gives

$\lambda = \frac{h}{\sqrt{2meV}}.$

For $V = 150\ \text{V}$ , the numbers work out as follows. First, the kinetic energy in joules:

$KE = 150 \times 1.602 \times 10^{-19}\ \text{J} = 2.403 \times 10^{-17}\ \text{J}.$

The momentum is then

= \sqrt{4.377 \times 10^{-47}}\ \text{kg\,m/s} \approx 6.616 \times 10^{-24}\ \text{kg\,m/s}.$$ The de Broglie wavelength follows immediately: $$\lambda = \frac{6.626 \times 10^{-34}}{6.616 \times 10^{-24}}\ \text{m} \approx 1.00 \times 10^{-10}\ \text{m} = 1.00\ \text{Å}.$$ This is comfortably within the range of typical atomic spacings ($1$–$3\ \text{Å}$), which is precisely why low-energy electron beams can be diffracted by crystalline solids — a fact the Davisson–Germer experiment exploited directly. ## Packet picture A real electron beam is not a single plane wave of perfectly defined momentum; it is a **wave packet** formed by superposing plane waves over a spread of wave-numbers $\Delta k$ around the central value $k_0 = 2\pi/\lambda$. Two speeds characterize the packet: - The **phase velocity** $v_p = \omega/k = \hbar k/(2m)$, which equals half the classical particle speed and is not directly observable. - The **group velocity** $v_g = d\omega/dk = \hbar k/m = p/m$, which matches the classical particle speed and determines when the electron arrives at a detector. A narrower packet (smaller $\Delta x$) requires a broader range of momenta ($\Delta p \geq \hbar / (2\Delta x)$), a relationship that will become the Heisenberg uncertainty principle in the next module. <Callout type="tip"> The $150\ \text{V}$ electron is a classic benchmark: its wavelength of roughly $1\ \text{Å}$ is just right for probing crystal lattices. At $10\ \text{keV}$ (used in electron microscopes) the wavelength drops to about $0.12\ \text{Å}$, allowing resolution of individual atoms. </Callout> ## Try it This is a numerical exercise — your code should `return` a number, not a circuit. Compute the de Broglie wavelength (in metres) of an electron accelerated through $150\ \text{V}$. Use the formula $\lambda = h / \sqrt{2meV}$ with $h = 6.626 \times 10^{-34}\ \text{J\,s}$, $m = 9.109 \times 10^{-31}\ \text{kg}$, $e = 1.602 \times 10^{-19}\ \text{C}$, and $V = 150\ \text{V}$. <RunnableExample />