Wave-particle duality, electron diffraction as evidence for the wave nature of particles, the de Broglie wavelength, and atomic energy levels and line spectra.

What this dot point is asking

CCEA wants you to explain wave-particle duality, use electron diffraction as evidence for the wave nature of matter, calculate the de Broglie wavelength, and explain how discrete atomic energy levels give rise to line spectra. Calculations usually combine an accelerating voltage with the de Broglie equation, or a level difference with $\Delta E = hf$.

Wave-particle duality and the de Broglie wavelength

Light shows its particle nature in the photoelectric effect (photons of energy $hf$) and its wave nature in interference and diffraction. De Broglie's insight was that the relationship works both ways: a particle of momentum $p$ has a wavelength too.

For an electron accelerated through a voltage $V$, the kinetic energy is $E_k = eV$, the momentum is $p = \sqrt{2mE_k} = \sqrt{2meV}$, and so the wavelength is $\lambda = \dfrac{h}{\sqrt{2meV}}$, falling as the voltage rises.

Electron diffraction

The regular spacing of carbon atoms in graphite acts like the slits of a diffraction grating, but for the electron matter wave rather than for light. The first such patterns were seen by Davisson and Germer, confirming de Broglie's prediction and earning it a Nobel Prize.

Atomic energy levels and line spectra

Electrons in an atom can only occupy discrete energy levels, conventionally given negative values measured from the ionised state at zero. When an electron falls from a higher level $E_2$ to a lower level $E_1$, a photon is emitted with energy

This gives a line emission spectrum of bright lines at specific wavelengths. A line absorption spectrum appears as dark lines where photons of exactly those energies are absorbed from an otherwise continuous spectrum and re-emitted in all directions. The pattern of lines is unique to each element, so spectra act as atomic fingerprints used to identify the composition of stars.

Worked example: energy of an emitted photon

Examples in context

Example 1. The electron microscope. Because fast electrons have a de Broglie wavelength thousands of times shorter than visible light, an electron microscope can resolve far finer detail than an optical one. Increasing the accelerating voltage shortens the wavelength further and sharpens the resolution, a direct application of $\lambda = \dfrac{h}{p}$.

Example 2. Identifying elements in stars. Starlight passing through the cooler outer gas of a star shows dark absorption lines at the exact wavelengths the atoms can absorb. Matching these line positions to laboratory spectra reveals which elements are present, so the discrete energy levels of atoms let astronomers read the composition of objects light-years away.

Try this

Q1. An electron of mass $9.1 \times 10^{-31}\ \text{kg}$ moves at $2.0 \times 10^{6}\ \text{m s}^{-1}$. Find its de Broglie wavelength. Take $h = 6.6 \times 10^{-34}\ \text{J s}$. [2 marks]

• Cue. $\lambda = \dfrac{h}{mv} = \dfrac{6.6 \times 10^{-34}}{(9.1 \times 10^{-31})(2.0 \times 10^{6})} \approx 3.6 \times 10^{-10}\ \text{m}$.

Q2. Explain why atomic line spectra contain only certain wavelengths. [2 marks]

• Cue. Electrons occupy discrete energy levels, so only fixed energy differences, and therefore fixed photon energies, are possible.

Q3. An electron falls between two levels, emitting a photon of energy $4.8 \times 10^{-19}\ \text{J}$. Find the wavelength of the photon. Take $h = 6.63 \times 10^{-34}\ \text{J s}$ and $c = 3.0 \times 10^{8}\ \text{m s}^{-1}$. [2 marks]

• Cue. $\lambda = \dfrac{hc}{\Delta E} = \dfrac{(6.63 \times 10^{-34})(3.0 \times 10^{8})}{4.8 \times 10^{-19}} = 4.1 \times 10^{-7}\ \text{m}$.