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OxfordAQA International GCSE Physics: The electromagnetic spectrum: uses, hazards, infrared and X-rays (9203)

Study guide to the electromagnetic spectrum, infrared emission, black-body radiation, uses, hazards and X-rays for OxfordAQA International GCSE Physics.

Subject
Physics
Level
IGCSE
Topic
The electromagnetic spectrum: uses, hazards, infrared and X-rays
Updated

Aligned to OxfordAQA IGCSE Physics (9203), For exams May/June 2018 onwards. Official specification .

Syllabus page (what it covers and how it is assessed): OxfordAQA IGCSE Physics.

Syllabus points this page covers

9203

  • 3.3.2 The electromagnetic spectrum

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This study guide covers section 3.3.2, The electromagnetic spectrum (statements a to m), of the OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards. The qualification is untiered. Some statements in 3.3.2 are for the full Physics award only (marked P in the specification): the extra detail on emission in statement d, and statements e and f on black-body radiation. They are included below. The wave equation used in the calculations comes from section 3.3.1.

Use it with the revision notes and the practice questions. Course pages: the OxfordAQA IGCSE Physics hub and the printable checklist. To find weak spots first, try the free 10-minute diagnostics.

What this unit covers

Statement What you must be able to do
3.3.2 a Say that EM waves are transverse and transfer energy from a source to an absorber
3.3.2 b Describe the spectrum as continuous, with one speed in a vacuum; give the order by energy, frequency and wavelength; know the wavelength range
3.3.2 c Link visible light to what the eye detects, and colour to wavelength
3.3.2 d Explain infrared emission and absorption, and the effect of the surface (emission detail is Physics only)
3.3.2 e, f Describe how emission changes as an object heats up; define black-body radiation (Physics only)
3.3.2 g, h Give communication uses and the listed applications
3.3.2 i Describe hazards (heating, ionisation) and simple protection
3.3.2 j to m State X-ray properties, medical uses and the precautions for ionising radiation

3.3.2 a: what an electromagnetic wave is

Electromagnetic (EM) waves are transverse: the vibrations (of electric and magnetic fields) are perpendicular to the way the wave moves. They transfer energy from a source to an absorber. A lamp is a source and your skin is an absorber. The wave does not carry matter from one to the other.

Unlike sound, EM waves need no medium. That is why sunlight reaches Earth across space.

3.3.2 b: the spectrum

The EM waves form a continuous spectrum: there are no gaps, and the named groups merge into each other. All of them travel at the same speed through a vacuum. In air the speed is almost the same, so calculations here use c = 3.00 × 10⁸ m/s.

The order, from low energy to high energy, is:

Group Frequency Wavelength Energy
Radio waves lowest longest lowest
Microwaves
Infrared
Visible light (red to violet)
Ultraviolet
X-rays
Gamma rays highest shortest highest

Read down the table for increasing frequency and energy, and decreasing wavelength. The wavelengths run from about 10⁻¹⁵ m to 10⁴ m and beyond. The boundaries between groups are not sharp and the ranges overlap.

Because the speed is the same, the wave equation links frequency and wavelength for every group:

wave speed (m/s) = frequency (Hz) × wavelength (m), so λ = c / f and f = c / λ

Worked example 1: a mobile phone signal

A phone network uses microwaves of frequency 1.8 GHz. Calculate the wavelength.

  1. Convert: 1.8 GHz = 1.8 × 10⁹ Hz.
  2. λ = c / f = (3.00 × 10⁸) / (1.8 × 10⁹) = 0.1667 m.
  3. Answer: 0.17 m (2 significant figures).

Worked example 2: green laser light

A laser pointer emits green light of wavelength 532 nm. Calculate the frequency.

  1. Convert: 532 nm = 532 × 10⁻⁹ m = 5.32 × 10⁻⁷ m.
  2. f = c / λ = (3.00 × 10⁸) / (5.32 × 10⁻⁷) = 5.639 × 10¹⁴ Hz.
  3. Answer: 5.64 × 10¹⁴ Hz.

A signal travels at 3.00 × 10⁸ m/s from a ground station to a satellite 3.6 × 10⁷ m away. Find the time for the signal to reach the satellite and for a reply to come back.

  1. Time = distance / speed = (3.6 × 10⁷) / (3.00 × 10⁸) = 0.12 s.
  2. The reply covers the same distance, so the round trip takes 2 × 0.12 = 0.24 s.

Worked example 4: the ends of the range

Calculate the frequency of EM waves of wavelength 10⁴ m and of wavelength 10⁻¹⁵ m.

  1. For 10⁴ m: f = (3.00 × 10⁸) / 10⁴ = 3.0 × 10⁴ Hz.
  2. For 10⁻¹⁵ m: f = (3.00 × 10⁸) / 10⁻¹⁵ = 3.0 × 10²³ Hz.

The shortest wavelength has the highest frequency, and so the highest energy.

3.3.2 c: visible light

Visible light is the part of the spectrum detected by our eyes. We see different wavelengths as different colours. Red light has the longest wavelength we can see and violet the shortest, roughly 700 nm down to 400 nm. Light with a mixture of wavelengths looks white.

3.3.2 d: infrared radiation

All objects emit and absorb infrared radiation, whatever their temperature. For the Physics award, add the reason: objects emit infrared because of the motion of their particles, and the amount and frequency of the radiation depend on the temperature and surface of the object.

The hotter an object is, the more infrared radiation it radiates in a given time. A hot drink radiates more than the same drink once it has cooled to room temperature. An object also absorbs infrared from its surroundings. If it emits more than it absorbs, it cools. If it absorbs more than it emits, it warms up.

The surface matters:

Surface Absorbs infrared Emits infrared Reflects infrared
Dark, matt well well poorly
Light, shiny poorly poorly well

Worked example 5: interpreting an experiment

Two identical metal cans each hold 200 g of water at 20 °C. One has a matt black surface and the other a shiny silver surface. They stand the same distance from a radiant heater. After ten minutes the black can’s water is at 38 °C and the silver can’s water is at 26 °C.

  1. Rise for the black can = 38 − 20 = 18 °C. Rise for the silver can = 26 − 20 = 6 °C.
  2. The black can’s rise is 18 / 6 = 3.0 times larger.
  3. Conclusion: the matt black surface absorbed more infrared radiation than the shiny surface, so it is the better absorber.

In a second test both cans are filled with water at 85 °C and left to cool. After ten minutes the black can is at 61 °C and the silver can at 73 °C. The black can fell by 24 °C and the silver can by 12 °C, so the black can lost heat twice as fast. This shows that dark, matt surfaces are also better emitters.

Controlling variables. Use the same mass and starting temperature of water, cans of the same size and material, and the same distance from the heater. Only the surface should change.

3.3.2 e and f: black-body radiation (Physics only)

As an object heats up, it radiates more and more infrared, and it also radiates at higher frequencies. A cool electric heater element gives out infrared that you feel but cannot see. As it gets hotter it begins to glow dull red. A very hot filament glows brighter and looks whiter, because the radiation now includes a wider range of visible frequencies.

Black-body radiation is the range of electromagnetic radiation emitted by an object at a particular temperature. Two objects at different temperatures emit different ranges, with the hotter one giving out more radiation overall and reaching higher frequencies. A good answer states both points: more radiation, and higher frequencies.

3.3.2 g and h: communication and applications

Radio waves, microwaves, infrared and visible light can all be used for communication. The applications listed in the specification are:

Wave Application
Radio waves Television and radio systems, including Bluetooth
Microwaves Mobile phones and satellite television systems
Infrared TV remote controls, night vision devices, heating
Visible light Photography, fibre optic communications
Ultraviolet Security marking
X-rays Medical imaging
Gamma rays Sterilising surgical instruments, killing harmful bacteria in food

To choose a wave for a job, match the property to the use. Gamma rays sterilise instruments because they are ionising and kill bacteria. Night vision devices detect the infrared that warm bodies emit.

3.3.2 i: hazards and protection

Too much exposure of the human body to EM waves can be harmful. Low-energy waves have a heating effect. Higher-energy waves have enough energy to cause ionisation, which means removing an electron from an atom or molecule.

Wave Hazard Simple protection
Microwaves Heating of body tissue Keep the oven door shut and sealed; do not use a damaged oven
Infrared Skin burns Keep away from hot objects; wear protective clothing and gloves
Ultraviolet Skin cancer and blindness Wear sunscreen, cover skin and use eye protection that blocks UV
X-rays High doses kill cells Limit exposure time; shield with lead; stay behind a screen
Gamma rays Genetic mutations Shield with thick lead or concrete; keep time short and distance large

Ionising waves can damage the molecules inside cells. Damage to a cell’s DNA can cause a mutation.

3.3.2 j to l: X-rays

X-rays are part of the EM spectrum. They have a very short wavelength, high energy and cause ionisation.

Properties of X-rays:

  • they affect a photographic film in the same way as light;
  • they are absorbed strongly by metal and bone;
  • they are transmitted by healthy tissue.

Medical diagnosis

A patient is placed between the X-ray source and a detector or photographic film. Soft tissue transmits most of the X-rays, so the film is exposed and turns dark. Bone absorbs them, so little reaches the film and the bone appears as a pale shadow. This is how X-rays show bone fractures and dental problems. In computed tomography (CT) scanning, X-rays are passed through the body from many angles and a computer builds up detailed images of the inside.

Treatment

X-rays are also used to treat some conditions. The radiation is aimed at a tumour, and the high dose kills cancer cells. Because it also damages healthy cells, the beam is directed carefully and the dose is controlled.

Worked example 6: comparing X-rays with visible light

An X-ray has wavelength 0.020 nm. Green light has wavelength 550 nm. Find the frequency of the X-ray and compare the wavelengths.

  1. 0.020 nm = 2.0 × 10⁻¹¹ m.
  2. f = c / λ = (3.00 × 10⁸) / (2.0 × 10⁻¹¹) = 1.5 × 10¹⁹ Hz.
  3. Wavelength ratio = 550 / 0.020 = 27 500, so the light’s wavelength is about 2.8 × 10⁴ times longer.
  4. In a vacuum both have the speed 3.00 × 10⁸ m/s. The X-ray has the higher frequency and higher energy.

3.3.2 m: working safely with ionising radiation

High-energy ionising radiation can be dangerous. People who work with it need precautions to monitor and minimise their exposure. Examples:

  • reduce the time spent near the source;
  • increase the distance from the source;
  • use shielding such as lead aprons, lead-lined walls and screens;
  • wear a monitoring badge, for example a film badge, so that the exposure can be measured and checked;
  • limit how often patients are scanned, so that each scan has a clear benefit.

Common errors

  • Writing that EM waves need a medium, or that different groups travel at different speeds in a vacuum.
  • Reversing the order, for example putting microwaves above infrared in frequency.
  • Saying that only black objects emit infrared. All objects emit it. Dark, matt surfaces are the better emitters.
  • Saying that shiny surfaces “reflect heat”. Say that they reflect infrared radiation.
  • Writing that ultraviolet or X-rays “cause radiation”. Say that they cause ionisation.
  • Forgetting to convert nm, MHz or GHz to m and Hz before using the wave equation.
  • Saying that X-rays “pass through bone”. Bone absorbs them strongly; soft tissue transmits them.

Next steps

Condense this with the revision notes, then test yourself with the practice questions (both linked at the top). Tick off progress on the checklist.

Official syllabus

OxfordAQA International GCSE Physics (9203), for exams May/June 2018 onwards, published by OxfordAQA, section 3.3.2 The electromagnetic spectrum.

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