The Electromagnetic Spectrum

All electromagnetic (EM) waves are transverse waves consisting of oscillating electric and magnetic fields perpendicular to each other and to the direction of travel. They all travel at the speed of light in a vacuum (c = 3.00 × 10⁸ m s⁻¹) and require no medium for propagation. What distinguishes one type of EM wave from another is its wavelength and frequency — and these determine its properties, uses, and hazards.

Properties Common to All EM Waves

Every electromagnetic wave shares the following characteristics:

Transverse — the oscillations of the electric and magnetic fields are perpendicular to the direction of energy transfer
Travels at c = 3.00 × 10⁸ m s⁻¹ in a vacuum — all EM waves travel at the same speed in free space
Does not require a medium — EM waves can travel through a vacuum (unlike sound)
Obeys the wave equation: c = fλ — since c is constant, a longer wavelength means a lower frequency and vice versa
Can be reflected, refracted, diffracted, and can interfere — demonstrating wave behaviour
Transfers energy — the energy carried by an EM wave is related to its frequency (E = hf)
Can be polarised — since they are transverse waves

The Regions of the Spectrum

The electromagnetic spectrum is divided into seven main regions, listed from longest wavelength (lowest frequency) to shortest wavelength (highest frequency):

Region	Wavelength range	Frequency range	Photon energy range
Radio waves	> 0.1 m	< 3 GHz	< 1.2 × 10⁻⁵ eV
Microwaves	1 mm – 0.1 m	3 GHz – 300 GHz	1.2 × 10⁻⁵ – 1.2 × 10⁻³ eV
Infrared (IR)	700 nm – 1 mm	300 GHz – 430 THz	1.2 × 10⁻³ – 1.77 eV
Visible light	400 nm – 700 nm	430 THz – 750 THz	1.77 – 3.10 eV
Ultraviolet (UV)	10 nm – 400 nm	750 THz – 30 PHz	3.10 – 124 eV
X-rays	0.01 nm – 10 nm	30 PHz – 30 EHz	124 eV – 124 keV
Gamma rays (γ)	< 0.01 nm	> 30 EHz	> 124 keV

The boundaries between regions are not sharp — they overlap. The classification depends on how the radiation is produced, not just its wavelength.

Visible light occupies a tiny fraction of the spectrum, from about 400 nm (violet) to 700 nm (red). Our eyes are sensitive only to this narrow band.

graph LR
    A["Radio\n> 0.1 m"] --> B["Microwaves\n1 mm – 0.1 m"]
    B --> C["Infrared\n700 nm – 1 mm"]
    C --> D["Visible\n400 – 700 nm"]
    D --> E["Ultraviolet\n10 – 400 nm"]
    E --> F["X-rays\n0.01 – 10 nm"]
    F --> G["Gamma\n< 0.01 nm"]
    style A fill:#ff9999
    style B fill:#ffcc99
    style C fill:#ffdd99
    style D fill:#99ff99
    style E fill:#99ccff
    style F fill:#cc99ff
    style G fill:#ff99ff

Production and Detection

Each region of the EM spectrum has characteristic methods of production and detection.

Region	Produced by	Detected by	Typical source
Radio	Oscillating charges in an antenna	Aerial/antenna	Radio transmitter
Microwaves	Magnetron, klystron	Microwave receiver	Microwave oven, radar
Infrared	Hot objects, thermal emission	Thermometer, IR camera, skin (warmth)	Heater, human body
Visible	Very hot objects, electronic transitions	Eyes, photographic film, CCD	Sun, light bulb, LED, laser
Ultraviolet	Very hot objects (e.g., Sun), gas discharge	Fluorescent materials, UV-sensitive film	Sun, UV lamp, mercury lamp
X-rays	Decelerating high-energy electrons hitting a metal target	Photographic film, fluorescent screen, Geiger tube	X-ray tube
Gamma	Nuclear decay, matter-antimatter annihilation	Geiger-Müller tube, scintillation counter	Radioactive nuclei

Uses of Electromagnetic Radiation

Region	Key uses
Radio	Broadcasting (AM, FM, DAB), communication, radio astronomy
Microwaves	Cooking, satellite communication, mobile phones, radar, Wi-Fi
Infrared	Heating, thermal imaging, TV remote controls, night-vision, fibre optic communication
Visible	Vision, photography, illumination, optical fibres, laser surgery
Ultraviolet	Sterilisation, fluorescent lamps, security marking, detecting forged banknotes
X-rays	Medical imaging, airport security, crystallography
Gamma	Sterilisation of medical equipment and food, cancer treatment (radiotherapy), PET scans

Hazards of Electromagnetic Radiation

Higher-frequency (shorter-wavelength) EM radiation carries more energy per photon and is generally more hazardous to living tissue.

Region	Hazard	Mechanism
Radio / Microwaves	Low risk at normal intensities; intense microwaves can cause heating of tissue	Internal heating of water in cells
Infrared	Skin burns, eye damage	Heating of surface tissue
Visible (intense)	Eye damage	Retinal damage from intense sources (lasers, direct sunlight)
Ultraviolet	Sunburn, skin cancer, eye damage (cataracts)	Ionisation and DNA damage in skin cells
X-rays	Cell damage, cancer, genetic mutations	Ionisation — enough energy to remove electrons from atoms
Gamma	Severe cell damage, cancer, radiation sickness	Ionisation — very high energy, deeply penetrating

UV, X-rays, and gamma rays are all classified as ionising radiation because their photons have enough energy to knock electrons out of atoms. This can damage DNA and lead to mutations and cancer.

Worked Calculations

Worked example 1: Calculate the frequency of green light with a wavelength of 530 nm.

f = c/λ = 3.00 × 10⁸ / (530 × 10⁻⁹) = 5.66 × 10¹⁴ Hz

Worked example 2: A microwave oven operates at 2.45 GHz. Calculate the wavelength of the microwaves.

λ = c/f = 3.00 × 10⁸ / (2.45 × 10⁹) = 0.122 m ≈ 12.2 cm

This is why microwave ovens have a cooking cavity that is at least ~30 cm across — it needs to be large compared to the wavelength.

Worked example 3: The photon energy of a particular UV photon is 6.0 eV. Calculate its wavelength. (1 eV = 1.60 × 10⁻¹⁹ J, h = 6.63 × 10⁻³⁴ J s)

First convert to joules: E = 6.0 × 1.60 × 10⁻¹⁹ = 9.60 × 10⁻¹⁹ J

λ = hc/E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (9.60 × 10⁻¹⁹) = 1.989 × 10⁻²⁵ / 9.60 × 10⁻¹⁹ = 2.07 × 10⁻⁷ m = 207 nm

This is in the UV-C range — germicidal UV used for sterilisation.

Worked example 4: Compare the photon energies of radio waves (f = 100 MHz) and gamma rays (f = 3.0 × 10²⁰ Hz).

E_radio = hf = 6.63 × 10⁻³⁴ × 100 × 10⁶ = 6.63 × 10⁻²⁶ J = 4.1 × 10⁻⁷ eV

E_gamma = hf = 6.63 × 10⁻³⁴ × 3.0 × 10²⁰ = 1.99 × 10⁻¹³ J = 1.24 × 10⁶ eV = 1.24 MeV

The gamma ray photon has about 3 × 10¹² (3 trillion) times more energy than the radio photon. This enormous difference in photon energy explains why gamma rays can ionise atoms and damage cells, while radio waves cannot.

The Visible Spectrum

Within the visible range, different wavelengths correspond to different colours:

Colour	Wavelength / nm	Frequency / THz
Red	620 – 700	430 – 484
Orange	590 – 620	484 – 508
Yellow	570 – 590	508 – 526
Green	495 – 570	526 – 606
Blue	450 – 495	606 – 668
Violet	380 – 450	668 – 789

Real-world application: Spectroscopy uses the EM spectrum to identify elements. When an element is heated or excited in a gas discharge tube, it emits light at specific wavelengths (its emission spectrum). Each element has a unique spectral "fingerprint." Astronomers use this to determine the composition, temperature, and velocity of distant stars and galaxies. The redshift of spectral lines tells us how fast a galaxy is receding.

Polarisation

Polarisation is the restriction of oscillations of a transverse wave to a single plane. Only transverse waves can be polarised — this is an important piece of evidence that light is a transverse wave.

Unpolarised light has oscillations in all planes perpendicular to the direction of travel. A polarising filter (Polaroid) transmits only the component oscillating in one plane, producing plane-polarised light.

If a second polariser (the "analyser") is placed after the first:

When the transmission axes are parallel, maximum intensity is transmitted
When the axes are perpendicular (crossed polarisers), no light passes through — the intensity is zero
At intermediate angles, the transmitted intensity follows Malus's law: I = I₀ cos²θ

Applications of polarisation:

Polaroid sunglasses reduce glare from reflected light (which is partially polarised horizontally)
LCD screens use two polarising filters and liquid crystals that rotate the plane of polarisation
Stress analysis in engineering — stressed transparent materials produce coloured patterns when viewed between crossed polarisers

Common exam mistake: Students sometimes claim that "polarisation proves light is a wave." Polarisation proves that light is a transverse wave specifically. Longitudinal waves (like sound) cannot be polarised. Interference and diffraction prove wave behaviour in general; polarisation distinguishes transverse from longitudinal.

A-Level Deep Dive: The Electromagnetic Spectrum

Spec mapping

Edexcel 9PH0 specification Topic 5 — Waves and the Particle Nature of Light, in the sub-strands relating to the electromagnetic spectrum, requires students to recall that all electromagnetic waves are transverse, travel at $c = 3.00 \times 10^8\ \text{m s}^{-1}$ in vacuum, and span a continuous spectrum from radio through microwaves, infrared, visible, ultraviolet, X-rays and gamma rays in order of increasing frequency (refer to the official Pearson Edexcel specification document for exact wording). The spec also requires use of $c = f\lambda$ and the photon energy relation $E = hf$ (equivalently $E = hc/\lambda$ ). The ideas are load-bearing across the Topic 5 photoelectric effect sub-strand, Topic 5 atomic line spectra, Topic 7 (electric and magnetic fields) for radio-frequency radiation, and Topic 12 (turning points / nuclear) for X-ray and gamma production. The Edexcel data-and-formulae booklet lists $c = f\lambda$ , $E = hf$ and $h = 6.63 \times 10^{-34}\ \text{J s}$ , but does not list the regional wavelength boundaries — those approximate ranges (radio $> 0.1\ \text{m}$ , visible $400$ – $700\ \text{nm}$ , X-rays $0.01$ – $10\ \text{nm}$ , etc.) must be memorised.

Worked example with full mark scheme

Question (8 marks):

A radio station transmits at $f = 95.8\ \text{MHz}$ . A medical X-ray tube produces X-rays of wavelength $\lambda = 0.050\ \text{nm}$ . (Take $c = 3.00 \times 10^8\ \text{m s}^{-1}$ and $h = 6.63 \times 10^{-34}\ \text{J s}$ .)

(a) Calculate the wavelength of the radio waves and identify the region of the electromagnetic spectrum to which they belong. (2)

(b) Calculate the photon energy of the X-ray, expressing your answer in both joules and electronvolts. (3)

(c) Calculate the ratio of the X-ray photon energy to the radio photon energy, and use this ratio to comment on the difference in biological hazard between the two types of radiation. (3)

Solution with mark scheme:

(a) Step 1 — apply the wave equation.

$\lambda = \frac{c}{f} = \frac{3.00 \times 10^8}{95.8 \times 10^6} = 3.13\ \text{m}$

M1 — correct rearrangement of $c = f\lambda$ with the frequency converted from MHz to Hz ( $95.8\ \text{MHz} = 95.8 \times 10^6\ \text{Hz}$ ). Leaving the frequency in MHz produces $\lambda = 3.13 \times 10^6\ \text{m}$ , off by a factor of $10^6$ .

A1 — value $\lambda = 3.13\ \text{m}$ identified as a radio wave (any wavelength greater than approximately $0.1\ \text{m}$ is in the radio region).

(b) Step 1 — apply the photon energy relation.

$E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{0.050 \times 10^{-9}}$

M1 — correct selection of $E = hc/\lambda$ with the wavelength converted from nm to m ( $0.050\ \text{nm} = 5.0 \times 10^{-11}\ \text{m}$ ).

Step 2 — evaluate in joules.

$E = \frac{1.989 \times 10^{-25}}{5.0 \times 10^{-11}} = 3.98 \times 10^{-15}\ \text{J}$

A1 — value to 2 or 3 sf, $E \approx 4.0 \times 10^{-15}\ \text{J}$ .

Step 3 — convert to eV.

$E = \frac{3.98 \times 10^{-15}}{1.60 \times 10^{-19}} = 2.49 \times 10^4\ \text{eV} \approx 25\ \text{keV}$

A1 — value in eV (or keV) with correct unit. A photon of $\sim 25\ \text{keV}$ is a typical diagnostic medical X-ray.

$E_\text{radio} = hf = 6.63 \times 10^{-34} \times 95.8 \times 10^6 = 6.35 \times 10^{-26}\ \text{J}$

M1 — application of $E = hf$ to the radio frequency.

Step 2 — form the ratio.

$\frac{E_\text{X-ray}}{E_\text{radio}} = \frac{3.98 \times 10^{-15}}{6.35 \times 10^{-26}} \approx 6.3 \times 10^{10}$

A1 — ratio of order $10^{10}$ to $10^{11}$ .

Step 3 — link to biological hazard.

B1 — explanation that the X-ray photon carries roughly $10^{10}$ times more energy per photon and exceeds the ionisation energy of typical atoms ( $\sim$ a few eV), so it can knock electrons out of atoms in DNA and damage cells; the radio photon, at $\sim 4 \times 10^{-7}\ \text{eV}$ , is far below any ionisation threshold and cannot damage tissue chemically — at most it causes mild thermal heating at very high intensities.

Total: 8 marks (M3 A4 B1).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A school physics laboratory uses a low-pressure sodium discharge lamp. The lamp emits a strong yellow line at a wavelength of $589\ \text{nm}$ . A separate fluorescent tube emits ultraviolet radiation at a wavelength of $254\ \text{nm}$ , which is absorbed by the tube's phosphor coating and re-emitted as visible light.

(a) Calculate the frequency and the photon energy (in joules) of the sodium yellow line. (3)

(b) State, with a numerical justification, whether the $254\ \text{nm}$ ultraviolet photon has sufficient energy to ionise a hydrogen atom (ionisation energy of hydrogen $= 13.6\ \text{eV}$ ). (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1) — frequency $f = c/\lambda = 3.00 \times 10^8 / (589 \times 10^{-9}) = 5.09 \times 10^{14}\ \text{Hz}$ .
M1 (AO1) — substitution into $E = hf$ .
A1 (AO1) — $E = 6.63 \times 10^{-34} \times 5.09 \times 10^{14} = 3.38 \times 10^{-19}\ \text{J}$ .

(b)

M1 (AO1) — UV photon energy $E = hc/\lambda = (6.63 \times 10^{-34} \times 3.00 \times 10^8)/(254 \times 10^{-9}) = 7.83 \times 10^{-19}\ \text{J}$ .
M1 (AO2) — conversion to eV: $E = 7.83 \times 10^{-19} / 1.60 \times 10^{-19} = 4.89\ \text{eV}$ .
A1 (AO3) — comparison: $4.89\ \text{eV} < 13.6\ \text{eV}$ , therefore the UV photon cannot ionise a hydrogen atom in its ground state; an explicit conclusion sentence is required for the AO3 mark.

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. This question typifies the Edexcel pattern of asking candidates to combine $c = f\lambda$ and $E = hf$ in part (a), then in part (b) to compare a calculated photon energy against a stated threshold and reach a justified conclusion.

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