Electromagnetic waves and the electromagnetic spectrum: VCE Physics Unit 4

What this dot point is asking

VCAA wants you to describe electromagnetic waves at a conceptual level (transverse oscillation of electric and magnetic fields, speed $c$), apply the universal wave equation $c = f \lambda$, identify the regions of the EM spectrum in order, and state representative applications of each.

What electromagnetic waves are

An electromagnetic (EM) wave is a transverse wave consisting of:

• An oscillating electric field $\vec{E}$.
• An oscillating magnetic field $\vec{B}$.
• Both perpendicular to the direction of propagation.
• Both perpendicular to each other.

The fields oscillate in phase. As the wave propagates, $\vec{E}$ and $\vec{B}$ at each point oscillate sinusoidally. A changing electric field generates a magnetic field (Maxwell's displacement-current term), and a changing magnetic field generates an electric field (Faraday's law). These two effects sustain each other, allowing the wave to propagate without a medium.

Key universal properties

All EM waves share the following:

1. Speed $c$ in vacuum. $c = 2.998 \times 10^8$ m s$^{-1} \approx 3.0 \times 10^8$ m s$^{-1}$. Independent of frequency, wavelength, intensity, or source motion.
2. Transverse. $\vec{E}$ and $\vec{B}$ both perpendicular to propagation direction. Hence polarisation is possible (covered in the polarisation dot point).
3. No medium required. Unlike sound, water, or seismic waves, EM waves propagate through vacuum. This is why light from the Sun reaches Earth across empty space.
4. Wave behaviour. All EM waves obey reflection, refraction (with frequency-dependent refractive index, hence dispersion), diffraction, interference, and polarisation, subject to the relative size of the wavelength.
5. Universal wave equation. $c = f \lambda$ holds for all EM waves.

In a medium, the speed reduces to $v = c / n$ where $n$ is the refractive index. The frequency stays the same; the wavelength is $\lambda_{\text{medium}} = \lambda_0 / n$ where $\lambda_0$ is the vacuum wavelength.

Cross-link: see the wavelength-frequency calculator for conversions across regions.

The electromagnetic spectrum

The EM spectrum is the full range of EM waves classified by frequency / wavelength. There are no sharp boundaries; the named regions are conventional.

The boundaries between regions are conventions; "X-ray" and "gamma ray" overlap, distinguished historically by source (X-ray = electron deceleration, gamma = nuclear decay).

Visible light sub-bands

Within visible light, the standard rainbow order (long wavelength to short) is:

• Red (around 700 to 620 nm)
• Orange (620 to 590)
• Yellow (590 to 570)
• Green (570 to 495)
• Blue (495 to 450)
• Violet (450 to 400)

Visible light spans less than one octave (a factor of about 1.7), the smallest band of any major EM region.

Applications by region

Radio

AM (530 kHz to 1700 kHz), FM (87.5 MHz to 108 MHz), TV broadcast (VHF / UHF up to about 800 MHz). Long-range communication uses long wavelengths because they diffract around obstacles and reflect off the ionosphere.

Microwave

Mobile phones (around 0.7 to 2.7 GHz), Wi-Fi (2.4 GHz and 5 GHz), Bluetooth (2.4 GHz), radar (microwave + sub-microwave), satellite (1 to 30 GHz). Microwave ovens (2.45 GHz) heat water through dielectric absorption.

Infrared

Thermal imaging cameras detect body heat. Remote controls use near-IR LEDs around 940 nm. Optical fibres operate at IR wavelengths (typically 1310 nm and 1550 nm) where silica is most transparent.

Visible

Direct human vision. Photography, microscopy, plant photosynthesis, photovoltaic cells.

Ultraviolet

Sterilisation (UV-C around 254 nm destroys bacterial DNA). Fluorescence (UV light absorbed and re-emitted at visible wavelengths). Sunburn (UV-B). Vitamin D production in skin (UV-B).

X-ray

Medical radiography (X-ray photons penetrate soft tissue but are absorbed by bone). CT scans (computed tomography). Crystallography (X-ray wavelengths comparable to atomic spacings, so diffraction reveals crystal structures).

Gamma

Cancer radiotherapy (gamma photons damage cancer cell DNA). Sterilisation of medical equipment and some foods. Gamma-ray astronomy (gamma sources include pulsars, supernovae, active galactic nuclei).

Energy and biological effect

Photon energy increases with frequency: $E = h f$. The higher-frequency regions (UV, X-ray, gamma) have photon energies sufficient to ionise atoms, which is why they are biologically dangerous and require shielding.

• Photons below about 10 eV (visible, IR, microwave, radio and most UV): non-ionising. Cannot strip electrons from atoms. Damage, if any, is via heating (microwave) or eye / skin burns (UV-A, very high intensity).
• Photons above about 10 eV (extreme UV, X-ray, gamma): ionising. Strip electrons from biological molecules, damaging DNA. Significant cancer risk above modest doses.

This is why X-ray operators wear lead aprons, but Wi-Fi exposure (microwave, $\sim 10^{-5}$ eV per photon) does not cause ionisation regardless of intensity.

Worked conversions

FM radio

Frequency 100 MHz. Wavelength $\lambda = c / f = 3 \times 10^8 / 10^8 = 3$ m. Radio region.

Green light

Wavelength 550 nm. Frequency $f = c / \lambda = 3 \times 10^8 / (550 \times 10^{-9}) \approx 5.45 \times 10^{14}$ Hz. Photon energy $E = h f \approx 2.26$ eV.

Medical X-ray

Photon energy 30 keV. Frequency $f = E / h = 30 \times 10^3 \times 1.6 \times 10^{-19} / 6.626 \times 10^{-34} \approx 7.25 \times 10^{18}$ Hz. Wavelength $\lambda = c / f \approx 4.1 \times 10^{-11}$ m $= 41$ pm.

Examples in context

Example 1. Square Kilometre Array radio observations from Murchison. SKA-Low at Murchison observes radio waves in the $50$ to $350$ MHz range. At $150$ MHz, wavelength is $\lambda = c/f = 3 \times 10^8/1.5 \times 10^8 = 2.0$ m. Photon energy is $E = hf = 6.63 \times 10^{-34} \times 1.5 \times 10^8 = 9.9 \times 10^{-26}$ J or $6.2 \times 10^{-7}$ eV. Each tiny photon contributes very little energy, so radio astronomy requires huge collecting areas; SKA-Low's $\sim 130\,000$ dipole antennas spread over $65$ km of desert capture neutral-hydrogen signals from the cosmic dawn era $13$ billion years ago.

Example 2. Australian Synchrotron X-ray beamline at Clayton. The Australian Synchrotron's X-ray imaging beamline produces photons at $10$ keV with wavelength $\lambda = hc/E = 1.24 \times 10^{-10}$ m = $0.124$ nm, well within the X-ray band. These photons have frequency $f = E/h = 1.6 \times 10^{-15}/6.63 \times 10^{-34} = 2.4 \times 10^{18}$ Hz. The same beam can be tuned across the $2$-$30$ keV range for different applications (protein crystallography, lung imaging, materials analysis). Both ends propagate at $c = 3 \times 10^8$ m s$^{-1}$ in vacuum, with $E = hf$ photon energy scaling inversely with wavelength.

Try this

Q1. State three regions of the electromagnetic spectrum in order of decreasing wavelength, with one application each. [3 marks]

• Cue. Radio (broadcasting), microwave (cooking, telecoms), infrared (thermal imaging) or visible-UV-X-ray-gamma at the short end.

Q2. A radar transmits at $10$ GHz. Calculate (a) the wavelength, and (b) the energy per photon in joules and in eV. [4 marks]

• Cue. (a) $\lambda = 3 \times 10^8/10^{10} = 0.03$ m. (b) $E = hf = 6.63 \times 10^{-24}$ J = $4.14 \times 10^{-5}$ eV.

Q3. Refer to Synchrotron X-rays at $10$ keV. (a) Calculate the wavelength. (b) Determine the frequency. (c) Outline one application that requires this energy range. [2+2+2 marks]

• Cue. (a) $0.124$ nm. (b) $2.4 \times 10^{18}$ Hz. (c) Protein crystallography requires wavelengths comparable to atomic spacing.