OCR GCSE Physics A: Electromagnetic waves and the spectrum (J249)

This guide covers statements P5.2a to P5.2f of topic P5.2, The electromagnetic spectrum, in the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026). None of these statements is in bold in the specification, so all of them are assessed in both Foundation and Higher Tier papers. Uses and hazards of the different groups are separate statements and are not taught here. Course pages: OCR GCSE Physics hub and the printable checklist. Companion pages: revision notes and practice questions. To find out which topics need most work, try the free 10-minute diagnostics.

What this guide covers

Statement	What you must be able to do
P5.2a	Recall that electromagnetic waves are transverse and travel through space, all with the same velocity
P5.2b	Explain that they transfer energy from a source to an absorber
P5.2c	Apply the relationships between frequency and wavelength across the spectrum
P5.2d	Describe the main groupings, from long to short wavelength and from low to high frequency
P5.2e	Describe that our eyes detect only a limited range of the spectrum
P5.2f	Recall that light is an electromagnetic wave

The specification names seven groupings: radio, microwave, infrared, visible light, ultraviolet, X-rays and gamma rays. It also warns that learners often treat these as unrelated things instead of one spectrum. The aim of this guide is to show what they share and how they differ.

P5.2a: transverse waves, one speed

Electromagnetic (EM) waves are transverse. The oscillations are at right angles to the direction the wave travels. In an EM wave the thing that oscillates is a pair of linked electric and magnetic fields, not particles of a material. That is why EM waves need no medium and travel through empty space. Sound is different: it is a longitudinal wave in which particles vibrate, so it cannot cross a vacuum.

All EM waves travel at the same speed through a vacuum, whatever their frequency or wavelength. A common value is 3.0 × 10⁸ m/s. If a question does not supply a value, this is the standard figure to use. In air the speed is almost the same, so the difference is ignored at GCSE.

Because the speed is fixed, the relationship speed = distance ÷ time gives the delay for a signal to cross a distance.

Worked example 1. A radio signal travels from Earth to a spacecraft orbiting the Moon, a distance of 3.84 × 10⁸ m. How long does it take?

time = distance / speed
     = 3.84 × 10⁸ / 3.0 × 10⁸
     = 1.28 s

The answer to two significant figures is 1.3 s. A visible light flash and an X-ray pulse sent at the same moment would arrive together.

Worked example 2. Light takes 500 s to travel from the Sun to Earth. Find the distance.

distance = speed × time
         = 3.0 × 10⁸ × 500
         = 1.5 × 10¹¹ m

Written in minutes, 500 s is 8.3 minutes (500 ÷ 60 = 8.33).

P5.2b: energy from source to absorber

A wave carries energy without carrying matter along with it. An EM wave starts at a source, crosses the gap, and is absorbed by a material. Energy is transferred from the source to the absorber. When the energy is absorbed, the absorber changes. Typical changes are a rise in temperature, a current in a circuit or a chemical change.

Examples to use in an explanation:

• Infrared from a hot object. A glowing grill element is the source. Your hand is the absorber. The skin warms up.
• Microwaves in an oven. The oven’s emitter is the source. Water in the food absorbs the energy, so the food warms up.
• Visible light from the Sun. The Sun is the source. A black road surface absorbs the light and warms up.
• Radio waves. A transmitting aerial is the source. A receiving aerial absorbs some of the energy, which is how a signal is detected.

A full explanation names the source, says that energy is transferred by the wave, names the absorber and says what the absorber does with the energy. The energy transferred depends on the wave and on how much of it the absorber takes in. A surface that reflects a wave takes in less energy than one that absorbs it.

P5.2d: the groupings in order

The groups run in a continuous spectrum. Going from radio to gamma rays, wavelength gets shorter and frequency gets higher. The order, long wavelength first, is:

radio, microwave, infrared, visible light, ultraviolet, X-rays, gamma rays

Visible light itself is ordered from red (longest wavelength, lowest frequency) to violet (shortest wavelength, highest frequency).

The sizes below are rounded guide values. Sources give slightly different boundaries, and the groups run into each other, so use the table to get a feel for size and order.

Group	Typical wavelength	Typical frequency
Radio	from about 0.1 m to many km	below about 3 × 10⁹ Hz
Microwave	about 1 mm to 0.1 m	about 3 × 10⁹ to 3 × 10¹¹ Hz
Infrared	about 700 nm to 1 mm	about 3 × 10¹¹ to 4 × 10¹⁴ Hz
Visible	about 400 nm to 700 nm	about 4 × 10¹⁴ to 7.5 × 10¹⁴ Hz
Ultraviolet	about 10 nm to 400 nm	about 7.5 × 10¹⁴ to 3 × 10¹⁶ Hz
X-rays	about 0.01 nm to 10 nm	about 3 × 10¹⁶ to 3 × 10¹⁹ Hz
Gamma rays	shorter than about 0.01 nm	higher than about 3 × 10¹⁹ Hz

Here 1 nm (nanometre) is 1 × 10⁻⁹ m. A memory phrase helps for the order, but write the names in full in an exam.

What the groups share: each is a transverse wave, each moves at one common speed through a vacuum and each transfers energy. What differs: wavelength, frequency, how they interact with matter, and how they are produced and detected.

P5.2c: frequency and wavelength

For any wave, wave speed = frequency × wavelength, or v = f × λ. Here v is in metres per second (m/s), f in hertz (Hz) and λ in metres (m). For EM waves in a vacuum, v is the same constant for every group, so frequency and wavelength are inversely proportional. If the wavelength halves, the frequency doubles. If the wavelength becomes ten times bigger, the frequency becomes ten times smaller.

Rearranged:

f = v / λ        λ = v / f

Unit prefixes you will meet:

Prefix	Meaning
kHz (kilohertz)	× 10³ Hz
MHz (megahertz)	× 10⁶ Hz
GHz (gigahertz)	× 10⁹ Hz
nm (nanometre)	× 10⁻⁹ m
μm (micrometre)	× 10⁻⁶ m

Convert to hertz and metres first, then substitute.

Worked example 3 (radio, find wavelength). A radio station broadcasts at 97.5 MHz. Calculate the wavelength.

f = 97.5 MHz = 97.5 × 10⁶ Hz
λ = v / f = 3.0 × 10⁸ / 97.5 × 10⁶
  = 3.08 m

To two significant figures this is 3.1 m.

Worked example 4 (visible, find frequency). Red light has wavelength 650 nm. Calculate its frequency.

λ = 650 nm = 650 × 10⁻⁹ m = 6.5 × 10⁻⁷ m
f = v / λ = 3.0 × 10⁸ / 6.5 × 10⁻⁷
  = 4.6 × 10¹⁴ Hz

Worked example 5 (X-ray). An X-ray has wavelength 0.050 nm. Find its frequency.

λ = 0.050 × 10⁻⁹ m = 5.0 × 10⁻¹¹ m
f = 3.0 × 10⁸ / 5.0 × 10⁻¹¹
  = 6.0 × 10¹⁸ Hz

Compare examples 3 to 5. From the radio wave to the X-ray the wavelength falls by a factor of about 6 × 10¹⁰, and the frequency rises by the same factor.

Worked example 6 (comparing). Wave A has wavelength 1.5 m. Wave B has wavelength 0.30 m. Both are EM waves in a vacuum. How do their frequencies compare?

Wave B’s wavelength is 1.5 ÷ 0.30 = 5 times smaller, so its frequency is 5 times greater. No calculation of the actual frequencies is needed.

Practical: wavelength of microwaves

The specification suggests an investigation using chocolate or processed cheese in a microwave oven. With the turntable removed, the food melts at spots that are half a wavelength apart, because the waves set up a pattern of strong and weak heating. A student measures the distance across several melted spots and divides to get the spacing.

Worked example 7. The first and fifth melted spots are 23.2 cm apart. Four gaps lie between them. Find the wavelength and the frequency.

spacing = 23.2 / 4 = 5.8 cm = 0.058 m
wavelength = 2 × 0.058 = 0.116 m
f = v / λ = 3.0 × 10⁸ / 0.116 = 2.6 × 10⁹ Hz

The answer is 0.12 m and 2.6 GHz. Spacing is half a wavelength, so forgetting to double it gives half the correct wavelength. Measuring over several gaps and dividing reduces the effect of reading error on a single gap.

P5.2e: what the eye can detect

Our eyes detect only a limited range of the spectrum: the visible group, roughly 400 nm to 700 nm. Waves just outside it, infrared and ultraviolet, are there but do not produce a sensation of sight. This is why a remote control’s infrared signal cannot be seen by eye. It is a feature of the detector, not of the wave. Some other animals respond differently. Bees can detect ultraviolet, for example.

Within the visible range, different wavelengths give different colours: red at the long end, then orange, yellow, green, blue and violet at the short end. Converting the ends of the range to frequency with f = v / λ:

400 nm: f = 3.0 × 10⁸ / 4.0 × 10⁻⁷ = 7.5 × 10¹⁴ Hz
700 nm: f = 3.0 × 10⁸ / 7.0 × 10⁻⁷ = 4.3 × 10¹⁴ Hz

This range is very narrow compared with the whole spectrum, which covers many powers of ten in wavelength.

P5.2f: light is an electromagnetic wave

Visible light is an EM wave. Everything above applies to it:

• It is transverse.
• It travels through a vacuum. Sunlight reaches Earth across empty space.
• Its speed in a vacuum is 3.0 × 10⁸ m/s, equal to that of radio waves or X-rays.
• It transfers energy from source to absorber. Light from a lamp warms the paper it falls on.

Do not treat light as a separate kind of wave from the rest of the spectrum. A common slip is to say that light travels faster than radio waves. They travel at the same speed. They differ in wavelength and frequency.

Common errors

• Writing that sound is an EM wave, or that EM waves need a medium.
• Saying X-rays travel faster than microwaves. In a vacuum they do not.
• Reversing the order: ultraviolet has a shorter wavelength than visible light but a higher frequency.
• Forgetting to convert MHz, GHz, nm or μm before substituting.
• Giving an answer in standard form with the wrong power of ten. Check by asking whether the group fits: a radio wave should not come out at 10⁻⁷ m.
• Describing energy transfer without naming the absorber and what happens to it.

Next steps

Work through the revision notes, then the practice questions. Tick off statements on the checklist and use the free 10-minute diagnostics to find weak areas. You can also return to the OCR GCSE Physics hub.

Official syllabus

OCR GCSE (9-1) Gateway Science Physics A (J249), Version 5.0 (July 2026), first assessment 2018, published by OCR. Topic P5.2, statements P5.2a to P5.2f.