SACE Board SACE 2023-style practice: An X-ray tube has a potential difference of 100\ kV across it. Calculate the maximum frequency of the X-rays produced. Use e = 1.6010^-19\ C and h = 6.6310^-34\ J s.

The maximum photon energy occurs when all of an electron's kinetic energy becomes a single photon: eV = hf_max. Rearrange: f_max = eVh, with V = 100\ kV = 1.0010^5\ V. $f_max = (1.6010^-19)(1.0010^5)6.6310^-34 = 1.6010^-146.6310^-34 = 2.4110^19\ Hz.$ 1 mark for using eV = hf_max, 1 mark for the answer of about 2.4110^19\ Hz.

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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 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Quantisation of light energy
Planck's constant
The electronvolt
Wave-particle duality
How SACE assesses this

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 $h = 6.63\times10^{-34}$ J s and $c = 3.00\times10^8$ m s $^{-1}$ .

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

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

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 $1.60\times10^{-19}\ \text{J eV}^{-1}$ gives the energy in eV.

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

Final answer: $E = 2.84\times10^{-19}\ \text{J} = 1.8\ \text{eV}$ .

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.

Planck's constant

Planck's constant $h$ is the fundamental link between a photon's wave property (frequency) and its particle property (energy). Its small size ( $\sim 10^{-34}$ 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: $1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$ . 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 relationship $\lambda = \dfrac{h}{p}$ gives the wavelength of a moving particle of momentum $p$ , 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 $E = hf$ or $E = \dfrac{hc}{\lambda}$ , 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, $eV = hf_\text{max}$ . 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 $p = \dfrac{h}{\lambda}$ and then its speed from $v = \dfrac{p}{m}$ . 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.

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.

SACE 20232 marksAn X-ray tube has a potential difference of

100\ \text{kV}

across it. Calculate the maximum frequency of the X-rays produced. Use

e = 1.60\times10^{-19}\ \text{C}

and

h = 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: $eV = hf_\text{max}$ .

Rearrange: $f_\text{max} = \dfrac{eV}{h}$ , with $V = 100\ \text{kV} = 1.00\times10^5\ \text{V}$ .

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 = hf_\text{max}

, 1 mark for the answer of about

2.41\times10^{19}\ \text{Hz}

SACE 20244 marksX-rays of wavelength

1.24\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\times10^{-34}\ \text{J s}

and

m_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, $\lambda = 1.24\times10^{-10}\ \text{m}$ . This is wave-particle duality applied to matter.

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

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

Then

v = \dfrac{p}{m}

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/\lambda

, 1 mark for the momentum, 1 mark for the speed of about

5.87\times10^6\ \text{m s}^{-1}