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The electromagnetic spectrum: WACE Year 12 Physics Unit 4

Syllabus dot point

“Describe the electromagnetic spectrum and relate frequency, wavelength and the speed of light”

8 min read

Quick answer

All electromagnetic waves are oscillating electric and magnetic fields that travel through a vacuum at c=3.0×108 m s−1c=3.0\times10^8\ \text{m s}^{-1}, differing only in frequency and wavelength, linked by c=fλc=f\lambda. The spectrum runs from low-frequency radio waves through microwaves, infrared, visible light, ultraviolet and X-rays to high-frequency gamma rays. Higher frequency means shorter wavelength and, by E=hfE=hf, more energetic photons.

What this dot point is asking

WACE wants you to describe the electromagnetic spectrum as a single family of waves, apply the wave equation, and order the regions by frequency and wavelength. The unifying idea is that visible light is just one narrow band of a much larger continuous spectrum.

One family of waves

Every electromagnetic wave is a self-propagating disturbance of electric and magnetic fields oscillating at right angles to each other and to the direction of travel. They need no medium and all travel at the same speed cc in a vacuum. What distinguishes radio from gamma rays is only their frequency and wavelength, not their fundamental nature.

The wave equation

For any electromagnetic wave,

c=fλ,c=f\lambda,

so frequency and wavelength are inversely related at fixed speed. A higher frequency means a shorter wavelength. This single relationship lets you convert between the two anywhere in the spectrum.

Ordering the regions

From lowest frequency (longest wavelength) to highest frequency (shortest wavelength) the regions are radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays. Visible light spans roughly 400 nm400\ \text{nm} (violet) to 700 nm700\ \text{nm} (red). Each region is produced and detected differently: radio waves by oscillating circuits, infrared by warm objects, X-rays by rapidly decelerating electrons, and gamma rays by nuclear processes.

Photon energy across the spectrum

Treating light as photons, each carries energy

E=hf=hcλ.E=hf=\frac{hc}{\lambda}.

So the same ordering by frequency is an ordering by photon energy: radio photons are very low energy, while X-ray and gamma photons are highly energetic and ionising. This is why gamma rays and X-rays are penetrating and biologically damaging, while radio and microwave photons are not ionising.

Frequency and photon energy of an X-ray

From wavelength to frequency to energy

An X-ray has a wavelength of 1.0×10−10 m1.0\times10^{-10}\ \text{m}. Find its frequency and the energy of one photon (h=6.63×10−34 J sh=6.63\times10^{-34}\ \text{J s}).

Frequency from c=fλc=f\lambda:

f=cλ=3.0×1081.0×10−10=3.0×1018 Hz.f=\frac{c}{\lambda}=\frac{3.0\times10^8}{1.0\times10^{-10}}=3.0\times10^{18}\ \text{Hz}.

Photon energy:

E=hf=(6.63×10−34)(3.0×1018)=2.0×10−15 J.E=hf=(6.63\times10^{-34})(3.0\times10^{18})=2.0\times10^{-15}\ \text{J}.

That is about 12000 eV12000\ \text{eV}, far more than a visible photon (a few eV), which is why X-rays are ionising.

Production and uses of each region

WACE expects familiarity with how each region is produced and used, since this connects the abstract spectrum to technology. Radio waves are produced by oscillating currents in aerials and used for broadcasting and communication. Microwaves are produced by devices such as magnetrons and used for cooking, radar and mobile or satellite communication. Infrared is emitted by warm objects and used in thermal imaging and remote controls. Visible light, the narrow band our eyes detect, is produced by hot objects and electronic transitions. Ultraviolet comes from very hot sources such as the Sun and is used in sterilisation. X-rays are produced when fast electrons decelerate sharply in a metal target and are used in medical imaging. Gamma rays come from nuclear transitions and are used in cancer treatment and sterilising equipment. The general trend is that the higher-energy regions are produced by more energetic processes (decelerating electrons, nuclear reactions) and have penetrating, ionising uses.

Linking the relationships

Use c=fλc=f\lambda to move between frequency and wavelength, and E=hfE=hf to find photon energy. Keep wavelengths in metres and watch the powers of ten, since the spectrum spans more than fifteen orders of magnitude in frequency.

Thinking different EM waves travel at different speeds

A common error is assuming gamma rays travel faster than radio waves because they are more energetic. In a vacuum all electromagnetic waves travel at exactly c=3.0×108 m s−1c=3.0\times10^8\ \text{m s}^{-1}. They differ in frequency and wavelength, and therefore in photon energy, but not in speed. Speed only changes when they enter a medium.

Exam-style questions

Questions in the style of SCSA 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.

WACE 2022-style6 marks
A radio station broadcasts at a frequency of 98.5 MHz98.5\ \text{MHz}. (a) Calculate the wavelength of the radio waves. (b) Calculate the energy of one photon at this frequency. (c) Compare this photon energy with that of a visible photon of wavelength 550 nm550\ \text{nm} and comment on why radio waves are non-ionising.
Show worked answer →

A 6 mark calculation rewards the wavelength, both photon energies and a comparison.

(a) Wavelength
From c=fλc=f\lambda: λ=cf=3.0×10898.5×106=3.05 m\lambda=\dfrac{c}{f}=\dfrac{3.0\times10^{8}}{98.5\times10^{6}}=3.05\ \text{m}.
(b) Radio photon energy
E=hf=(6.63×10−34)(98.5×106)=6.5×10−26 JE=hf=(6.63\times10^{-34})(98.5\times10^{6})=6.5\times10^{-26}\ \text{J}.
(c) Visible photon and comparison
Visible: E=hcλ=(6.63×10−34)(3.0×108)550×10−9=3.6×10−19 JE=\dfrac{hc}{\lambda}=\dfrac{(6.63\times10^{-34})(3.0\times10^{8})}{550\times10^{-9}}=3.6\times10^{-19}\ \text{J}. The visible photon carries about 5.6×1065.6\times10^{6} times more energy. Radio photons are far too low in energy to remove electrons from atoms, so they are non-ionising.

Markers reward λ=c/f=3.05 m\lambda=c/f=3.05\ \text{m}, E=hfE=hf for the radio photon, the visible photon energy and the conclusion that low photon energy means non-ionising.

WACE 2021-style5 marks
List the regions of the electromagnetic spectrum in order of increasing frequency, and explain how photon energy and ionising ability change across this order.
Show worked answer →

A 5 mark answer needs the correct order and the energy or ionisation trend.

Order. In order of increasing frequency: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. This is also the order of decreasing wavelength, since c=fλc=f\lambda.

Photon energy and ionisation. Photon energy is E=hfE=hf, so it increases in exactly the same order: radio photons have the least energy and gamma photons the most. Higher-energy photons (ultraviolet, X-rays, gamma rays) carry enough energy to remove electrons from atoms, so they are ionising and biologically damaging, while the lower-energy regions (radio to visible, broadly) are non-ionising.

Markers reward the correct seven-region order, E=hfE=hf giving increasing photon energy, and the link between higher photon energy and ionising ability.

Practise this

Sources & how we know this

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

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