K323 / 2027
Electromagnetic spectrum overview

Lesson 1 of 3

Spectrum order and shared properties

  • I can state that electromagnetic waves are transverse and travel at the same speed in vacuum.Syllabus

    Syllabus K323, 11(a). State that electromagnetic waves are transverse and travel at the same speed in vacuum.

  • I can order the main spectrum regions by wavelength and frequency.Syllabus

    Syllabus K323, 11(b). Order the main spectrum regions by wavelength and frequency.

Make a guess. You are not marked.

A radio wave and an X-ray are sent across the same vacuum at the same moment. Which arrives first?

  1. The radio wave, because it has the longer wavelength and so travels further.
  2. The X-ray, because it has more energy.
  3. They arrive together, because both travel at the speed of light.
Show the answer

They arrive together, because both travel at the speed of light.

Every electromagnetic wave travels through a vacuum at the same speed, 3.0 x 108 m/s. The waves differ in frequency and wavelength, not in speed.

Radio waves, visible light and gamma rays are parts of one electromagnetic spectrum. They differ in wavelength and frequency, while sharing the same speed in vacuum.

Wavelength is a distance between equivalent adjacent points of a wave. Frequency is the number of complete cycles per second, in hertz. A wave transfers energy as it propagates.

What all electromagnetic waves share

All electromagnetic waves are transverse: their oscillations are perpendicular to the direction of propagation. The oscillating quantities are electric and magnetic fields, rather than air particles moving up and down.

They can travel through a vacuum, without a material medium. All travel at the same vacuum speed. For the calculations here, use the supplied value c = 3.0 x 108 m/s.

A sinusoidal drawing can represent how a field varies. It is not a wavy path followed by a photon or a row of air particles carrying the radiation.

Order wavelength and frequency in opposite directions

Read the same order in two directions

Moving down this list, frequency increases and wavelength decreases. All electromagnetic waves have the same speed in vacuum.

The seven main regions ordered by frequency and wavelengthFrom top to bottom, the regions are radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays. A downward arrow on the right marks increasing frequency. An upward arrow on the left marks increasing wavelength. Visible light sits between infrared and ultraviolet. This is a schematic category order: equal-height boxes and spaces do not represent equal frequency ranges or wavelength intervals. No numerical band boundaries or hazard boundary are drawn.Radio wavesMicrowavesInfraredVisible lightUltravioletX-raysGamma raysIncreasing wavelengthIncreasing frequencySchematic order, not a numerical scale

Visible light, enlarged

Red is the longer-wavelength visible end and violet the shorter-wavelength endAn enlarged schematic colour strip has red at the left and violet at the right. The labelled frequency arrow points right, towards violet. The labelled wavelength arrow points left, towards red. Text labels identify both ends independently of colour. The strip is not a calibrated frequency or wavelength axis.RedVioletIncreasing frequencyIncreasing wavelength

The separate region names are useful categories. Microwaves are also commonly described as part of the wider radio-frequency range.

From radio waves towards gamma rays, frequency increases and wavelength decreases. The region sizes are schematic, not equal numerical bandwidths. The enlarged visible section places red at the longer-wavelength end and violet at the shorter-wavelength end.

Increasing frequency: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays.

Increasing wavelength: gamma rays, X-rays, ultraviolet, visible light, infrared, microwaves, radio waves.

Within visible light, red has a longer wavelength and lower frequency than violet. Both are electromagnetic waves. The named regions are convenient divisions of a continuous spectrum, not physical gaps; microwaves are also often included within the wider radio-frequency range.

c = fλ   so   λ = c/fFor a vacuum wave, c is the common speed, f is frequency in Hz and λ is wavelength in m. Higher frequency means shorter wavelength at this fixed speed, not faster travel.

Worked comparison

Two frequencies, one vacuum speed

A 100 MHz signal has f = 100 x 106 = 1.0 x 108 Hz. Its wavelength is λ = (3.0 x 108)/(1.0 x 108) = 3.0 m.

A 2.5 GHz signal has f = 2.5 x 109 Hz. Its wavelength is λ = (3.0 x 108)/(2.5 x 109) = 0.12 m.

The second signal repeats more often and has more closely spaced wave cycles. Both still travel at c in vacuum.

Try it step by step

Work it out in steps: Frequency in Hz, then Wavelength = c / f in m.

  1. Problem 1. An FM radio station broadcasts at 150 MHz. Take c = 300000000 m/s (3.0 x 108 m/s). Find the wavelength. (1 MHz = 1000000 Hz.)
  2. Problem 2. A TV transmitter broadcasts at 600 MHz. Take c = 3.0 x 108 m/s. Find the wavelength.
  3. Problem 3. A Wi-Fi router transmits at 2400 MHz. Take c = 3.0 x 108 m/s. Find the wavelength.
Show the answer

Problem 1.

Frequency in Hz: f = 150 x 1000000 = 150000000 Hz, which is 1.5 x 108 Hz

Wavelength = c / f in m: wavelength = 300000000 / 150000000 = 2.0 m (2 sf)

Problem 2.

Frequency in Hz: f = 600 x 1000000 = 600000000 Hz, which is 6.0 x 108 Hz

Wavelength = c / f in m: wavelength = 300000000 / 600000000 = 0.50 m (2 sf)

Problem 3.

Frequency in Hz: f = 2400 x 1000000 = 2400000000 Hz, which is 2.4 x 109 Hz

Wavelength = c / f in m: wavelength = 300000000 / 2400000000 = 0.125 m, which is 0.13 m to 2 sf

Work from wavelength when that is supplied

One nanometre is 10-9 m. For a supplied wavelength of 600 nm, λ = 6.00 x 10-7 m, so f = c/λ = 5.0 x 1014 Hz.

At 1200 nm, the wavelength is twice as large and the frequency is half as large: 2.5 x 1014 Hz. Wavelength and frequency cannot both increase while their product remains the same speed.

Radio reception ends with a different wave

A radio transmitter sends an electromagnetic signal to a receiver. The receiver processes the information and drives a loudspeaker. The loudspeaker then makes sound waves in the surrounding air.

The radio signal can cross a vacuum; the sound from the loudspeaker needs a material medium. In air, sound is longitudinal, while the radio wave is transverse.

Travel-time comparison

Equal paths through vacuum

Two electromagnetic signals are emitted together and each follows a 900 km vacuum path. Convert the distance to 900000 m.

t = distance/speed = 900000/(3.0 x 108) = 0.0030 s = 3.0 ms.

Both arrive together in this ideal comparison, regardless of their different frequencies.

Keep frequency separate from speed. More cycles per second does not mean faster propagation when the vacuum speed is fixed.

Check your understanding

A 100 MHz signal and a 2.5 GHz signal start together and follow equal 900 km paths through vacuum. Which comparison is correct?

  1. The 2.5 GHz signal arrives first because higher frequency means higher speed.
  2. They arrive together and must have the same wavelength because their speeds match.
  3. They arrive together; the 2.5 GHz signal has the shorter wavelength.
Show the answer

They arrive together; the 2.5 GHz signal has the shorter wavelength.

Both travel at the same vacuum speed and take 900000/(3.0 x 108) = 0.0030 s. At fixed speed, wavelength equals speed divided by frequency: the wavelengths are 3.0 m and 0.12 m respectively.

Practise exam questions

Write your answer first, then tick the mark points it makes.

Exam-style question

State three properties shared by all electromagnetic waves. [3]

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Exam-style question

A radio station broadcasts at a frequency of 98.0 MHz. The speed of electromagnetic waves is 3.0 x 108 m/s. Calculate the wavelength of the radio waves. [3]

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