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86–87, 96WavesStretch

Wave–Particle Duality

If light — a wave — can behave like particles, can particles behave like waves? Louis de Broglie\textbf{Louis de Broglie} proposed that they can, with a wavelength set by their momentum. Electron diffraction confirmed it, and the same idea explains why atoms emit light only at particular wavelengths.

40 min Topic 5 · Waves and the Particle Nature of Light

What you'll be able to do

  • Use the de Broglie equation λ=h/p\lambda = h/p
  • Explain electron diffraction as evidence for the wave nature of particles
  • Describe atomic energy levels and calculate photon energies from transitions
  • Explain emission and absorption line spectra
1

The de Broglie wavelength

De Broglie proposed that any particle with momentum pp has an associated wavelength λ=h/p=h/mv\lambda = h/p = h/mv.

For everyday objects the wavelength is absurdly small — a moving tennis ball has λ∼10−34\lambda \sim 10^{-34} m — so wave effects are undetectable. For electrons, with tiny mass, the wavelength can be comparable to the spacing of atoms in a crystal.

For an electron accelerated from rest through a potential difference VV, its kinetic energy is eVeV, so its momentum is 2meV\sqrt{2meV}, which gives the wavelength directly.

λ=hp=hmv\lambda = \dfrac{h}{p} = \dfrac{h}{mv}
De Broglie wavelength.
p=2meVp = \sqrt{2meV}
Momentum of an electron accelerated through VV.
Example 1Find the de Broglie wavelength of an electron travelling at 4.0×106 m s−14.0\times10^{6}\ \mathrm{m\,s^{-1}}. Take me=9.11×10−31m_{e} = 9.11\times10^{-31} kg.
1p=9.11×10−31×4.0×106=3.644×10−24 kg m s−1p = 9.11\times10^{-31} \times 4.0\times10^{6} = 3.644\times10^{-24}\ \mathrm{kg\,m\,s^{-1}}.
2λ=6.63×10−343.644×10−24\lambda = \dfrac{6.63\times10^{-34}}{3.644\times10^{-24}}.
Answer1.8×10−10 m1.8\times10^{-10}\ \mathrm{m} — similar to atomic spacing.

Tip — Only particles with a wavelength comparable to a gap or obstacle show noticeable diffraction. That is why electrons, not footballs, are used.

2

Electron diffraction

Diffraction is a wave phenomenon: particles travelling in straight lines could not spread into a pattern of rings. When a beam of electrons passes through a thin layer of polycrystalline graphite, it produces concentric rings on a screen — a diffraction pattern.

The atomic spacing in graphite is about 10−1010^{-10} m, comparable to the electrons’ de Broglie wavelength, which is exactly the condition for significant diffraction.

Increasing the accelerating voltage increases the electrons’ speed and momentum, shortening the wavelength, so the rings get smaller\textbf{smaller} — just as the de Broglie equation predicts.

3

Energy levels in atoms

Electrons bound in an atom can occupy only certain discrete energy levels\textbf{energy levels}. These are negative, measured relative to zero for an electron that has just escaped; the lowest is the ground state\textbf{ground state}.

An electron moves up a level by absorbing exactly the energy difference, from a photon or a collision. When it drops back, it emits a photon whose energy equals the difference between the levels.

Because only specific differences are possible, only specific photon energies — and therefore specific wavelengths — can be emitted or absorbed.

hf=E1−E2hf = E_{1} - E_{2}
Photon energy equals the level difference.
λ=hcE1−E2\lambda = \dfrac{hc}{E_{1}-E_{2}}
Wavelength of the emitted photon.
Example 1An electron falls from −1.51-1.51 eV to −3.40-3.40 eV in hydrogen. Find the wavelength of the emitted photon.
1Energy difference: 3.40−1.51=1.893.40 - 1.51 = 1.89 eV =3.024×10−19= 3.024\times10^{-19} J.
2λ=(6.63×10−34)(3.00×108)3.024×10−19\lambda = \dfrac{(6.63\times10^{-34})(3.00\times10^{8})}{3.024\times10^{-19}}.
Answer6.58×10−7 m6.58\times10^{-7}\ \mathrm{m} — red, the hydrogen-alpha line.

Tip — Use the difference between levels, not either level value on its own.

4

Line spectra

A hot gas produces an emission line spectrum\textbf{emission line spectrum}: bright lines at specific wavelengths on a dark background, one line for each possible downward transition.

Passing white light through a cooler gas gives an absorption line spectrum\textbf{absorption line spectrum}: a continuous spectrum with dark lines where the gas has absorbed photons of exactly the right energies to excite its electrons.

The dark absorption lines of an element sit at exactly the same wavelengths as its bright emission lines, because the same energy differences are involved. Each element’s set of lines is unique, which is how the composition of distant stars is identified.

Equation recap

λ=h/p\lambda = h/p
De Broglie wavelength.
p=2meVp = \sqrt{2meV}
Accelerated electron.
hf=E1−E2hf = E_{1}-E_{2}
Transition photon.
λ=hc/ΔE\lambda = hc/\Delta E
Emitted wavelength.

Common mistakes to avoid

Saying faster electrons give larger diffraction rings.
Faster means more momentum, shorter wavelength and smaller rings.
Using a single energy level value as the photon energy.
The photon energy is the difference between two levels.
Forgetting to convert eV to J before using hc/λhc/\lambda.
Multiply by 1.60×10−191.60\times10^{-19} first.
Claiming absorption lines are at different wavelengths from emission lines.
They coincide, since the same energy differences are involved.

Key takeaways

  • Particles have a de Broglie wavelength λ=h/p\lambda = h/p.
  • Electron diffraction through graphite shows electrons behave as waves.
  • Higher accelerating voltage gives shorter wavelength and smaller rings.
  • Bound electrons occupy discrete energy levels; transitions emit or absorb hf=ΔEhf = \Delta E.
  • Emission and absorption line spectra are fingerprints of an element’s energy levels.

Test yourself

Answer a few questions, then review anything you missed.

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