Matter Wavelength Calculations

In 1924 Louis de Broglie proposed that every 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. For a non-relativistic particle of mass $m$ moving with speed $v$ , the momentum is $p = mv$ , so

$\lambda = \frac{h}{mv}.$

This is not a metaphor. In 1927 Davisson and Germer scattered electrons off a nickel crystal and observed diffraction peaks at exactly the angles predicted by treating the electrons as waves with the de Broglie wavelength — direct experimental confirmation.

Finding the momentum from kinetic energy

In the lab, electrons are rarely characterised by their speed directly. Instead they are accelerated through a potential difference $V$ . Starting from rest, the work done on the electron equals its kinetic energy:

$eV = \frac{p^2}{2m_e} \implies p = \sqrt{2 m_e e V},$

where $e = 1.602 \times 10^{-19}\ \text{C}$ is the elementary charge and $m_e = 9.109 \times 10^{-31}\ \text{kg}$ is the electron mass. Substituting into de Broglie's formula:

$\lambda = \frac{h}{\sqrt{2 m_e e V}}.$

A handy approximate form (derived by inserting the numerical values) is

$\lambda \approx \frac{1.226\,\text{nm}}{\sqrt{V/\text{V}}},$

so a $100\ \text{V}$ electron has $\lambda \approx 0.123\ \text{nm}$ , on the scale of atomic spacings — which is exactly why electron diffraction is a powerful probe of crystal structure.

Atoms and molecules

The same formula applies to atoms. Because atomic masses are thousands of times larger than $m_e$ , thermal atoms at room temperature have wavelengths in the sub-nanometre range. For a gas of atoms at temperature $T$ , a natural momentum scale is set by the equipartition theorem: $\frac{3}{2}k_B T = \frac{p^2}{2m}$ gives $p = \sqrt{3 m k_B T}$ , and

$\lambda_{\text{thermal}} = \frac{h}{\sqrt{3 m k_B T}}.$

For helium ( $m \approx 4 \times 1.661 \times 10^{-27}\ \text{kg}$ ) at $300\ \text{K}$ this works out to about $0.073\ \text{nm}$ — still measurable, and the basis of atom-interferometry experiments.

Try it

This is a numerical exercise — return a number. In the original Davisson–Germer experiment the electrons were accelerated through $54\ \text{V}$ . Compute the de Broglie wavelength and return it in picometres ( $1\ \text{m} = 10^{12}\ \text{pm}$ ).

Run your code to see the quantum state.