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CHEM 121 Studioby Learn4Less · UBC CHEM 121

3.1 · Quantum fundamentals

Waves and wave–particle duality

Light behaves as a wave and as particles; so does matter.

By the end you should be able to:

  • Describe waves and the evidence for wave–particle duality
  • Calculate de Broglie wavelengths

Key idea

Describing a wave

  • Wavelength λ\lambda: the distance between successive crests (m or nm).
  • Frequency ν\nu: the number of crests passing a point each second (Hz = s⁻¹).
  • Amplitude: the height of the wave; the intensity (brightness) is proportional to amplitude squared.
  • Node: a point where a standing wave has zero amplitude at all times. More nodes means a shorter wavelength.

All electromagnetic waves travel at the same speed cc in a vacuum, so wavelength and frequency are inversely proportional.

Formula

Speed, wavelength and frequency

c=λν,c=2.998×108 m s−1c = \lambda\nu, \qquad c = 2.998 \times 10^{8}\ \mathrm{m\,s^{-1}}

Longer wavelength means lower frequency. Red light at 650 nm has

ν=2.998×108 m s−1650×10−9 m=4.61×1014 Hz\nu = \frac{2.998\times10^{8}\ \mathrm{m\,s^{-1}}}{650\times10^{-9}\ \mathrm{m}} = 4.61\times10^{14}\ \mathrm{Hz}

Key idea

Evidence for wave–particle duality

ObservationWhat it shows
Diffraction and two-slit interference of lightlight is a wave: overlapping waves add (constructive) or cancel (destructive)
Photoelectric effectlight comes as particles (photons), each with E=hνE = h\nu
Line spectra of atomsthe energy of atoms is quantized
Electron diffraction by crystals (Davisson and Germer, 1927)matter is a wave, with λ=h/mv\lambda = h/mv

Neither picture alone explains everything. Light and electrons show wave behaviour as they travel and interfere, and particle behaviour when they are emitted, absorbed or detected.

Formula

The de Broglie wavelength

λ=hmv=hp=h2m KE\lambda = \frac{h}{mv} = \frac{h}{p} = \frac{h}{\sqrt{2m\,\mathrm{KE}}}

Use mm in kg and vv in m/s with h=6.626×10−34h = 6.626\times10^{-34} J s (1 J = 1 kg m² s⁻²), and λ\lambda comes out in metres. Wave behaviour such as diffraction is noticeable only when λ\lambda is comparable to the size of the slit or obstacle; for electrons that is the spacing of atoms in a crystal (about 0.1 nm).

An electron with KE = 100 eV = 1.602×10−171.602\times10^{-17} J has λ=6.626×10−342(9.109×10−31)(1.602×10−17)=1.23×10−10\lambda = \frac{6.626\times10^{-34}}{\sqrt{2(9.109\times10^{-31})(1.602\times10^{-17})}} = 1.23\times10^{-10} m.

Common mistake

Wave and de Broglie traps

Wrong: finding an electron's wavelength from E=hc/λE = hc/\lambda or c=λνc = \lambda\nu. Right: those equations are for photons only; for a particle with mass use λ=h/mv\lambda = h/mv.

Wrong: putting the mass in grams. Right: hh contains kg, so a 145 g baseball is 0.145 kg.

Wrong: "brighter light has a higher frequency." Right: intensity depends on amplitude (the number of photons); colour depends on frequency.

Worked example

Worked example: why baseballs do not diffract

An electron moving at 1.00×1061.00\times10^{6} m/s:

λ=6.626×10−34(9.109×10−31)(1.00×106)=7.27×10−10 m=0.727 nm\lambda = \frac{6.626\times10^{-34}}{(9.109\times10^{-31})(1.00\times10^{6})} = 7.27\times10^{-10}\ \mathrm{m} = 0.727\ \mathrm{nm}

This is comparable to the spacing of atoms, so a crystal diffracts electrons.

A baseball (0.145 kg) thrown at 40.0 m/s:

λ=6.626×10−34(0.145)(40.0)=1.14×10−34 m\lambda = \frac{6.626\times10^{-34}}{(0.145)(40.0)} = 1.14\times10^{-34}\ \mathrm{m}

This is far smaller than even a proton (about 10−1510^{-15} m), so nothing exists that could diffract it.

Answer: 0.727 nm for the electron and 1.14×10−341.14\times10^{-34} m for the baseball; the wave nature of everyday objects is unobservable.

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