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SAPhysicsSyllabus dot point

What is a photon, how is its energy related to frequency, and what does this mean for the dual nature of light?

Apply the photon model to calculate photon energy and explain wave-particle duality.

The photon as a quantum of light energy, the relationship E = hf using Planck's constant, the electronvolt, and the idea of wave-particle duality, with worked examples.

Generated by Claude Opus 4.78 min answer

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  1. What this dot point is asking
  2. Quantisation of light energy
  3. Planck's constant
  4. The electronvolt
  5. Wave-particle duality

What this dot point is asking

You need to apply the photon model to calculate photon energy and explain what wave-particle duality means.

Quantisation of light energy

Classical physics treated light energy as continuous. The photon model says light energy comes in discrete quanta - photons. A beam of light is a stream of photons; a brighter beam has more photons per second, but each photon's energy is fixed by its frequency.

Planck's constant

Planck's constant hh is the fundamental link between a photon's wave property (frequency) and its particle property (energy). Its small size (1034\sim 10^{-34} J s) is why quantum effects are invisible in everyday life but dominate at atomic scales.

The electronvolt

Because photon energies are tiny in joules, the electronvolt is often used: 1 eV=1.60×1019 J1 \text{ eV} = 1.60\times10^{-19} \text{ J}. Visible-light photons carry roughly 1.8 to 3.1 eV. This unit makes atomic energy levels and photon energies easy to compare.

Wave-particle duality

Light behaves as a wave in some experiments and as a particle in others:

Crucially, light is not "sometimes a wave and sometimes a particle" in a contradictory way; it is a single entity whose behaviour is captured by quantum theory, with the wave and particle pictures each applying to particular experiments. The same duality applies to matter: electrons, normally thought of as particles, produce interference patterns too (the de Broglie hypothesis).

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 SACE Stage 22 marksAn X-ray tube has a potential difference of 100 kV across the tube. Calculate the maximum frequency of the X-rays that are produced by this X-ray tube.
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The maximum X-ray photon energy occurs when all of an electron's kinetic energy is converted into a single photon: E = e V = h f max.

Rearrange for the maximum frequency: f max = e V / h.

Substitute e = 1.60 x 10^-19 C, V = 100 kV = 1.00 x 10^5 V, h = 6.63 x 10^-34 J s.

f max = (1.60 x 10^-19)(1.00 x 10^5) / (6.63 x 10^-34) = (1.60 x 10^-14) / (6.63 x 10^-34) = 2.41 x 10^19 Hz.

1 mark for using e V = h f max, 1 mark for the answer of about 2.41 x 10^19 Hz.

2024 SACE Stage 24 marksWhen X-rays of wavelength 1.24 x 10^-10 m pass through aluminium foil they produce an interference pattern identical to that produced by electrons passing through the same foil. Determine the speed of the electrons that produced the identical interference pattern.
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Identical interference patterns mean the electrons have the same de Broglie wavelength as the X-rays, wavelength = 1.24 x 10^-10 m. This is wave-particle duality applied to matter.

The de Broglie relationship is wavelength = h / p, so the electron momentum is p = h / wavelength.

p = (6.63 x 10^-34) / (1.24 x 10^-10) = 5.35 x 10^-24 kg m s-1.

Then v = p / m, using the electron mass 9.11 x 10^-31 kg: v = (5.35 x 10^-24) / (9.11 x 10^-31) = 5.87 x 10^6 m s-1.

1 mark for matching de Broglie wavelengths, 1 mark for p = h / wavelength, 1 mark for the momentum, 1 mark for the speed of about 5.87 x 10^6 m s-1.

2025 SACE Stage 23 marksAn electron gun emits electrons with a de Broglie wavelength of 1.53 x 10^-10 m. Determine the speed of the electrons.
Show worked answer →

Apply the de Broglie relationship wavelength = h / p, so the momentum is p = h / wavelength.

p = (6.63 x 10^-34) / (1.53 x 10^-10) = 4.33 x 10^-24 kg m s-1.

Then v = p / m, with electron mass m = 9.11 x 10^-31 kg.

v = (4.33 x 10^-24) / (9.11 x 10^-31) = 4.76 x 10^6 m s-1.

1 mark for p = h / wavelength, 1 mark for the momentum, 1 mark for the speed of about 4.76 x 10^6 m s-1.