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The photon model and Planck's constant (SACE Stage 2 Physics)

Syllabus dot point

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

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Quick answer

A photon is a discrete packet (quantum) of light energy. Its energy is set entirely by frequency: E=hf=hcλE = hf = \dfrac{hc}{\lambda}, where h=6.63×10−34h = 6.63\times10^{-34} J s is Planck's constant. Light shows wave behaviour (interference, diffraction) and particle behaviour (the photoelectric effect) - wave-particle duality - and neither model alone is complete.

Jump to a section
  1. What this dot point is asking
  2. Quantisation of light energy
  3. Planck's constant
  4. The electronvolt
  5. Wave-particle duality
  6. How SACE assesses this
  7. Exam-style questions

What this dot point is asking

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

Photon energy of red light

Find the energy of a photon of red light, wavelength 700 nm

Use h=6.63×10−34h = 6.63\times10^{-34} J s and c=3.00×108c = 3.00\times10^8 m s−1^{-1}.

Step 1: Convert the wavelength to metres and apply E=hc/λE = hc/\lambda

The wavelength must be in metres before substituting. Converting 700 nm gives 700×10−9 m=7.00×10−7 m700\times10^{-9}\ \text{m} = 7.00\times10^{-7}\ \text{m}. The photon energy formula E=hc/λE = hc/\lambda then gives the energy directly in joules.

E=hcλ=(6.63×10−34)(3.00×108)700×10−9=1.99×10−257.00×10−7=2.84×10−19 J.E = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3.00\times10^8)}{700\times10^{-9}} = \frac{1.99\times10^{-25}}{7.00\times10^{-7}} = 2.84\times10^{-19}\ \text{J}.

Step 2: Convert to electronvolts

Because photon energies are tiny in joules, electronvolts are more convenient for comparison with atomic energy scales. Dividing by the conversion factor 1.60×10−19 J eV−11.60\times10^{-19}\ \text{J eV}^{-1} gives the energy in eV.

E=2.84×10−191.60×10−19=1.8 eV.E = \frac{2.84\times10^{-19}}{1.60\times10^{-19}} = 1.8\ \text{eV}.

Final answer: E=2.84×10−19 J=1.8 eVE = 2.84\times10^{-19}\ \text{J} = 1.8\ \text{eV}.

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.

Photon energy

E=hf=hcλE = hf = \frac{hc}{\lambda}

EE is the photon energy (J), ff the frequency (Hz), λ\lambda the wavelength (m), c=3.00×108 m s−1c = 3.00\times10^8\ \text{m s}^{-1}, and h=6.63×10−34 J sh = 6.63\times10^{-34}\ \text{J s} Planck's constant. Higher frequency (shorter wavelength) means more energetic photons, which is why ultraviolet is more harmful than visible light.

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 (∼10−34\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=1.60×10−19 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:

Wave-particle duality

Light exhibits both wave and particle properties depending on how it is observed. Interference and diffraction (the double-slit experiment) reveal its wave nature; the photoelectric effect reveals its particle nature. Neither model alone fully describes light - both are needed.

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 relationship λ=hp\lambda = \dfrac{h}{p} gives the wavelength of a moving particle of momentum pp, and is the basis of the electron-diffraction question above.

How SACE assesses this

SACE Stage 2 questions on the photon model take three main forms. The first is a direct photon-energy calculation using E=hfE = hf or E=hcλE = \dfrac{hc}{\lambda}, sometimes converting the result to electronvolts. The second uses energy conservation in an X-ray tube, where an accelerated electron's energy becomes a single photon, eV=hfmaxeV = hf_\text{max}. The third applies de Broglie's relationship to matter: given a particle's wavelength (or one matched to an X-ray wavelength), find its momentum from p=hλp = \dfrac{h}{\lambda} and then its speed from v=pmv = \dfrac{p}{m}. Across all three, convert any wavelength to metres before substituting and keep frequency in hertz. A short conceptual part may ask you to state what wave-particle duality means, where the expected answer names interference or diffraction as the wave evidence and the photoelectric effect as the particle evidence.

Common errors
Thinking brighter light means more energetic photons
Brightness (intensity) means more photons per second; each photon's energy still depends only on frequency.
Forgetting to convert wavelength to metres
A wavelength in nanometres must become metres before using E=hc/λE = hc/\lambda.
Treating duality as a contradiction
Light is one thing; the wave and particle models each describe different observed behaviours. It is not "switching" between two substances.
Using E=hfE = hf with the wrong constant
h=6.63×10−34h = 6.63\times10^{-34} J s. Do not confuse it with other constants, and keep frequency in hertz.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2023-style2 marks
An X-ray tube has a potential difference of 100 kV100\ \text{kV} across it. Calculate the maximum frequency of the X-rays produced. Use e=1.60×10−19 Ce = 1.60\times10^{-19}\ \text{C} and h=6.63×10−34 J sh = 6.63\times10^{-34}\ \text{J s}.
Show worked answer →

The maximum photon energy occurs when all of an electron's kinetic energy becomes a single photon: eV=hfmaxeV = hf_\text{max}.

Rearrange: fmax=eVhf_\text{max} = \dfrac{eV}{h}, with V=100 kV=1.00×105 VV = 100\ \text{kV} = 1.00\times10^5\ \text{V}.

fmax=(1.60×10−19)(1.00×105)6.63×10−34=1.60×10−146.63×10−34=2.41×1019 Hz.f_\text{max} = \dfrac{(1.60\times10^{-19})(1.00\times10^5)}{6.63\times10^{-34}} = \dfrac{1.60\times10^{-14}}{6.63\times10^{-34}} = 2.41\times10^{19}\ \text{Hz}.

1 mark for using eV=hfmaxeV = hf_\text{max}, 1 mark for the answer of about 2.41×1019 Hz2.41\times10^{19}\ \text{Hz}.

SACE 2024-style4 marks
X-rays of wavelength 1.24×10−10 m1.24\times10^{-10}\ \text{m} passing through aluminium foil produce an interference pattern identical to that of electrons passing through the same foil. Determine the speed of those electrons. Use h=6.63×10−34 J sh = 6.63\times10^{-34}\ \text{J s} and me=9.11×10−31 kgm_e = 9.11\times10^{-31}\ \text{kg}.
Show worked answer →

Identical interference patterns mean the electrons have the same de Broglie wavelength as the X-rays, λ=1.24×10−10 m\lambda = 1.24\times10^{-10}\ \text{m}. This is wave-particle duality applied to matter.

The de Broglie relationship is λ=hp\lambda = \dfrac{h}{p}, so p=hλp = \dfrac{h}{\lambda}:

p=6.63×10−341.24×10−10=5.35×10−24 kg m s−1.p = \dfrac{6.63\times10^{-34}}{1.24\times10^{-10}} = 5.35\times10^{-24}\ \text{kg m s}^{-1}.

Then v=pmv = \dfrac{p}{m}:
v=5.35×10−249.11×10−31=5.87×106 m s−1.v = \dfrac{5.35\times10^{-24}}{9.11\times10^{-31}} = 5.87\times10^6\ \text{m s}^{-1}.

1 mark for matching de Broglie wavelengths, 1 mark for p=h/λp = h/\lambda, 1 mark for the momentum, 1 mark for the speed of about 5.87×106 m s−15.87\times10^6\ \text{m s}^{-1}.

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Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

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