Electromagnetic waves

Class 10 Physics, unit 20, book pages 281–308

Electromagnetic waves are transverse waves of oscillating electric and magnetic fields that need no medium and travel at $3 \times 1 0^{8} m s^{- 1}$ in a vacuum. This unit sets out the electromagnetic spectrum from radio waves to gamma rays and links energy, frequency and wavelength through $E = h f$ and $c = f λ$ . It then describes the uses and hazards of each region, explains the blue sky and the red sunset by the scattering of light, and ends with the photon picture of light, radiation pressure and solar sails.

Learning outcomes (7)

State the main regions of the electromagnetic spectrum in order of frequency and in order of wavelength.
State that the speed of all electromagnetic waves in a vacuum is $3.0 \times 1 0^{8} m s^{- 1}$ .
State that the speed of electromagnetic waves in air is approximately the same as in a vacuum.
Describe the applications of electromagnetic waves in society.
Describe the damage caused by electromagnetic radiation.
Explain qualitatively how the scattering of light by molecules in the air gives the sky its blue colour during the day and its shades of red at sunset.
State that, theoretically, light can also be considered to be made of massless particles that carry energy and momentum, called photons.

20.1Electromagnetic wavesOften examined

Electromagnetic waves are essential to our lives. Their exploration started with the work of Michael Faraday and James Clerk Maxwell, whose experiments showed the link between electricity and magnetism and gave Physics the theory of electromagnetic waves. Since then these waves have played a vital role in human progress and in technology, from sending wireless messages to studying far-off galaxies.

What is an electromagnetic wave?

Electromagnetic waves: Transverse waves produced by the oscillations of electric and magnetic fields that are perpendicular to each other. They do not need a material medium for their production and propagation.

In an electromagnetic wave the electric field $E$ and the magnetic field $B$ vibrate at right angles to each other, and the direction in which the wave travels is perpendicular to the vibrations of both fields (Figure 20.1). Because no medium is needed, these waves can travel through the vacuum of space.

Figure 20.1: An electromagnetic wave. The electric field (coloured) and the magnetic field (dark) vibrate at right angles to each other and to the direction of propagation. One wavelength is the distance between two successive crests.

They are transverse waves: the fields vibrate at right angles to the direction of travel.
The electric field, the magnetic field and the direction of propagation are mutually perpendicular.
They need no medium, so they can cross the vacuum of space.
Different types differ in frequency, wavelength and energy, and together they form the electromagnetic spectrum.

Speed of electromagnetic waves

In a vacuum all electromagnetic waves travel at the speed of light. The speed of light in a vacuum is a universal constant, denoted by $c$ :

c = 3.0 \times 1 0^{8} m s^{- 1}

In a material medium the speed of light does not stay the same: it slows down because of the interaction of light with the particles of the medium. In air, however, the speed is approximately the same as in a vacuum.

20.1Electromagnetic spectrumOften examined

Electromagnetic waves are grouped into different types according to their frequency, wavelength or energy. Together the groups form the electromagnetic spectrum.

Electromagnetic spectrum: The range of different frequencies and wavelengths of electromagnetic waves. It includes radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays.

Table 20.1: Spectrum of electromagnetic waves
Type of radiation	Wavelength range	Frequency range (Hz)
Radio	$> 1 mm$	$< 3 \times 1 0^{11}$
Microwave	$1 mm$ to $25 μ m$	$3 \times 1 0^{11}$ to $3 \times 1 0^{13}$
Infrared	$25 μ m$ to $2.5 μ m$	$1 0^{13}$ to $4 \times 1 0^{14}$
Visible	$750 nm$ to $400 nm$	$4 \times 1 0^{14}$ to $7.5 \times 1 0^{14}$
Ultraviolet	$400 nm$ to $1 nm$	$1 0^{15}$ to $1 0^{17}$
X-rays	$1 nm$ to $1 pm$	$1 0^{17}$ to $1 0^{20}$
Gamma rays	$< 1 0^{- 12} m$	$1 0^{20}$ to $1 0^{24}$

Figure 20.2: The electromagnetic spectrum, drawn on a scale where each step is a factor of 10. Frequency increases to the left and wavelength to the right. The visible part (coloured) is only a very thin slice of the whole spectrum.

Order of the regions

Going from radio waves to gamma rays, the frequency and the energy increase while the wavelength decreases.

The two ends of the electromagnetic spectrum
Property	Radio waves (one end)	Gamma rays (other end)
Frequency	lowest	highest
Wavelength	longest	shortest
Energy	lowest	highest

Visible light and its colours

The visible light is the portion of the spectrum that the human eye can detect. It is divided into sub-ranges, the colours of light: red, orange, yellow, green, blue and violet. The spectrum of light can be studied with the help of a prism or a diffraction grating.

Mini lab: the spectrum of visible light

Aim: to study the spectrum of visible light. Apparatus: a prism, white paper, and light coming through a window or from a white light source (such as a torch).

Take a prism, which has two triangular faces and three rectangular faces.
Cover the window of the room so that only a small portion lets light into the room. Sunlight is white light, which is not a pure colour but a mixture of many colours. If there is no window light, use a torch.
Place the prism in the path of the light so that one rectangular face faces the incident light.
Observe that the white light splits into its component colours on leaving the prism.

Mini lab: a prism splits white light into its component colours. Red is bent least and violet most.

Conclusion: the white light from the Sun is a mixture of different wavelengths of different colours. Each component deviates by a different amount on passing through the prism, which splits the white light into its constituent colours. The prism bends light because the speed of light differs in air and in glass, and also because the components of white light have different wavelengths.

20.1Energy, frequency and wavelengthOften examined

The relation between the energy $E$ and the frequency $f$ of an electromagnetic wave was found by Max Planck. It is called Planck's equation:

E = h f

(20.1)

Here $h$ is Planck's constant, $h = 6.63 \times 1 0^{- 34} J s$ . Equation (20.1) shows that the energy is directly proportional to the frequency: an increase in frequency means an increase in the energy of the electromagnetic wave.

The speed, wavelength and frequency of a wave are related by

c = f λ or f = \frac{c}{λ}

(20.2)

where $c = 3 \times 1 0^{8} m s^{- 1}$ is the speed of light. Since $c$ is constant in a vacuum, equation (20.2) shows that frequency and wavelength are inversely proportional: an increase in frequency means a decrease in wavelength.

Putting $f = c / λ$ from (20.2) into (20.1) gives the energy in terms of the wavelength:

E = \frac{h c}{λ}

(20.3)

The speed of light $c$ and Planck's constant $h$ are constants, so energy and wavelength are inversely proportional: an increase in energy means a decrease in the wavelength of the electromagnetic waves.

How the three quantities change together
If the frequency increases	Wavelength	Energy
$f$ becomes larger	becomes smaller ( $λ = c / f$ )	becomes larger ( $E = h f$ )
$f$ becomes smaller	becomes larger	becomes smaller

Wavelength against frequency

250

500

750

1000

1250

1500

f (1 0^{14} Hz)

λ (nm)

750 nm

400 nm

λ = c / f

The graph of

λ = c / f

. The wavelength falls as the frequency rises, and the curve never touches either axis. The dashed lines mark the ends of the visible range,

400 nm

and

750 nm

Energy against frequency

f (1 0^{14} Hz)

E (1 0^{- 19} J)

E = h f

The graph of

E = h f

is a straight line through the origin, so the photon energy is directly proportional to the frequency.

Example 20.1: energy ranges of infrared, visible and ultraviolet

Find the energy range for the infrared, visible and ultraviolet for their corresponding maximum and minimum wavelengths.

Given

$h = 6.63 \times 1 0^{- 34} J s$
$c = 3 \times 1 0^{8} m s^{- 1}$
$Infrared = λ_{L} = 250 μ m, λ_{S} = 2.5 μ m$
$Visible = λ_{L} = 750 nm, λ_{S} = 400 nm$
$Ultraviolet = λ_{L} = 400 nm, λ_{S} = 1 nm$

To find

maximum energy $E_{ma x}$ and minimum energy $E_{min}$ for each region

The energy of the radiation is given by (20.3).
$E = \frac{h c}{λ}$
Energy is inversely proportional to wavelength, so the maximum energy goes with the smallest wavelength $λ_{S}$ and the minimum energy with the largest wavelength $λ_{L}$ . Work out the product $h c$ once, because it is the same every time.
$h c = 6.63 \times 1 0^{- 34} \times 3 \times 1 0^{8} = 1.989 \times 1 0^{- 25} J m$
Infrared. Maximum energy, using $λ_{S} = 2.5 \times 1 0^{- 6} m$ .
$E_{ma x} = \frac{1.989 \times 1 0 ^{- 25}}{2.5 \times 1 0 ^{- 6}} = 7.96 \times 1 0^{- 20} J$
Minimum energy, using $λ_{L} = 250 \times 1 0^{- 6} m$ .
$E_{min} = \frac{1.989 \times 1 0 ^{- 25}}{250 \times 1 0 ^{- 6}} = 7.96 \times 1 0^{- 22} J$
Visible. Maximum energy, using $λ_{S} = 400 \times 1 0^{- 9} m$ .
$E_{ma x} = \frac{1.989 \times 1 0 ^{- 25}}{400 \times 1 0 ^{- 9}} = 4.97 \times 1 0^{- 19} J$
Minimum energy, using $λ_{L} = 750 \times 1 0^{- 9} m$ .
$E_{min} = \frac{1.989 \times 1 0 ^{- 25}}{750 \times 1 0 ^{- 9}} = 2.65 \times 1 0^{- 19} J$
Ultraviolet. Maximum energy, using $λ_{S} = 1 \times 1 0^{- 9} m$ .
$E_{ma x} = \frac{1.989 \times 1 0 ^{- 25}}{1 \times 1 0 ^{- 9}} = 1.99 \times 1 0^{- 16} J$
Minimum energy, using $λ_{L} = 400 \times 1 0^{- 9} m$ .
$E_{min} = \frac{1.989 \times 1 0 ^{- 25}}{400 \times 1 0 ^{- 9}} = 4.97 \times 1 0^{- 19} J$

Energy ranges in joules
Region	$E_{min}$ (J)	$E_{ma x}$ (J)
Infrared	$7.96 \times 1 0^{- 22}$	$7.96 \times 1 0^{- 20}$
Visible	$2.65 \times 1 0^{- 19}$	$4.97 \times 1 0^{- 19}$
Ultraviolet	$4.97 \times 1 0^{- 19}$	$1.99 \times 1 0^{- 16}$

IR: 7.96 \times 1 0^{- 22} to 7.96 \times 1 0^{- 20} J; visible: 2.65 \times 1 0^{- 19} to 4.97 \times 1 0^{- 19} J; UV: 4.97 \times 1 0^{- 19} to 1.99 \times 1 0^{- 16} J

Worked example: doubling the frequency of a radio wave

A radio station broadcasts at a frequency of

90 MHz

(

9.0 \times 1 0^{7} Hz

). Find the wavelength and the energy of one photon of this radiation. Then the frequency is doubled to

180 MHz

in the same vacuum. What happens to the speed, the wavelength and the energy?

Given

$f_{1} = 9.0 \times 1 0^{7} Hz$
$f_{2} = 2 f_{1} = 1.8 \times 1 0^{8} Hz$
$c = 3 \times 1 0^{8} m s^{- 1}$
$h = 6.63 \times 1 0^{- 34} J s$

To find

$λ_{1}$ , $E_{1}$ , and the changes at $f_{2}$

Wavelength. Use $c = f λ$ and make $λ$ the subject.
$λ_{1} = \frac{c}{f _{1}} = \frac{3 \times 1 0 ^{8}}{9.0 \times 1 0 ^{7}} = 3.33 m$
Energy. Use Planck's equation $E = h f$ .
$E_{1} = h f_{1} = 6.63 \times 1 0^{- 34} \times 9.0 \times 1 0^{7} = 5.97 \times 1 0^{- 26} J$
Doubled frequency. The medium is the same (a vacuum), so the speed stays $3 \times 1 0^{8} m s^{- 1}$ . Since $λ = c / f$ and $c$ is fixed, doubling $f$ halves $λ$ .
$λ_{2} = \frac{3 \times 1 0 ^{8}}{1.8 \times 1 0 ^{8}} = 1.67 m (half of 3.33 m)$
Since $E = h f$ and $h$ is fixed, doubling $f$ doubles $E$ .
$E_{2} = 6.63 \times 1 0^{- 34} \times 1.8 \times 1 0^{8} = 1.19 \times 1 0^{- 25} J (twice 5.97 \times 1 0^{- 26} J)$
Check using (20.3): $E_{1} = h c / λ_{1}$ .
$\frac{1.989 \times 1 0 ^{- 25}}{3.33} = 5.97 \times 1 0^{- 26} J ✓$

Energy against frequency

f (1 0^{7} Hz)

E (1 0^{- 26} J)

90 MHz

180 MHz

E = h f

The graph of

E = h f

is a straight line through the origin: the energy is directly proportional to the frequency. The two points are the

90 MHz

and

180 MHz

waves; doubling

f

doubles

E

λ_{1} = 3.33 m, E_{1} = 5.97 \times 1 0^{- 26} J; at 180 MHz : c unchanged, λ_{2} = 1.67 m, E_{2} = 1.19 \times 1 0^{- 25} J

Worked example: green light and blue light in electron-volts

Green light has a wavelength of

550 nm

. (a) Find its frequency and the energy of its photons in joules and in electron-volts. (b) A student says that blue light of wavelength

450 nm

has less energy because

450

is a smaller number than

550

. Find the energy of blue light and decide whether she is right.

Given

$λ_{g} = 550 nm = 5.5 \times 1 0^{- 7} m$
$λ_{b} = 450 nm = 4.5 \times 1 0^{- 7} m$
$h c = 1.989 \times 1 0^{- 25} J m$
$1 eV = 1.6 \times 1 0^{- 19} J$

To find

$f_{g}$ , $E_{g}$ (J and eV), $E_{b}$

(a) Frequency of green light, from (20.2).
$f_{g} = \frac{c}{λ _{g}} = \frac{3 \times 1 0 ^{8}}{5.5 \times 1 0 ^{- 7}} = 5.45 \times 1 0^{14} Hz$
Energy from (20.3).
$E_{g} = \frac{h c}{λ _{g}} = \frac{1.989 \times 1 0 ^{- 25}}{5.5 \times 1 0 ^{- 7}} = 3.62 \times 1 0^{- 19} J$
Check with $E = h f$ .
$6.63 \times 1 0^{- 34} \times 5.45 \times 1 0^{14} = 3.61 \times 1 0^{- 19} J ✓$
The small difference comes from rounding $f_{g}$ .
Change joules to electron-volts by dividing by the charge of one electron, $1.6 \times 1 0^{- 19}$ .
$E_{g} = \frac{3.62 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} = 2.26 eV$
(b) Energy of blue light.
$E_{b} = \frac{1.989 \times 1 0 ^{- 25}}{4.5 \times 1 0 ^{- 7}} = 4.42 \times 1 0^{- 19} J = \frac{4.42 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} = 2.76 eV$
The student is wrong. Energy is inversely proportional to wavelength, so the shorter wavelength (blue) has the greater energy: $2.76 eV > 2.26 eV$ . The ratio is $E_{b} / E_{g} = 550/450 = 1.22$ , which agrees with $4.42/3.62 = 1.22$ .

Photon energy against wavelength

350

400

450

500

550

600

650

700

750

800

0.5

1.5

2.5

3.5

λ (nm)

E (eV)

400

750

blue

green

E = h c / λ

The curve

E = h c / λ

with the energy in electron-volts. Between the dashed lines is the visible range. The curve falls as the wavelength increases, so blue light (

450 nm

) lies higher than green light (

550 nm

f_{g} = 5.45 \times 1 0^{14} Hz, E_{g} = 3.62 \times 1 0^{- 19} J = 2.26 eV; E_{b} = 2.76 eV (greater)

20.2Applications of electromagnetic waves

The electromagnetic spectrum has a wide range of waves, from very long radio waves to very short gamma rays. These waves play an essential role in scientific, medical and technological applications in our everyday lives. The sections that follow take each region in turn.

Summary of applications by region
Region	Applications described in the book
Radio waves	Radio and television broadcasting; radio astronomy
Microwaves	Satellite television; mobile phones; Bluetooth; microwave oven
Infrared	Remote controllers; intruder alarms; optical fibres; thermal imaging; night vision
Visible light	Photography; vision
Ultraviolet	Security marking; detecting counterfeit bank notes; sterilising water
X-rays	Medical imaging; security scanners; killing cancerous cells; detecting cracks in metals
Gamma rays	Medical treatment; sterilising food and medical equipment; engineering applications

20.2.1Applications of radio waves

Radio waves have the longest wavelengths in the electromagnetic spectrum. Without them it would not be possible to send and receive information as easily as it is done today. Radio waves are used in the working of television, mobile phones, wireless networks, GPS receivers, wireless clocks and even small items such as door openers.

A. Radio and television broadcasting

Radio and television broadcasting are essential parts of modern communication. They use radio waves to carry audio and video signals from broadcasting stations directly into our homes. The radio waves carry coded information (audio and video signals) from the broadcasting antennas to our devices, which decode it and show the content.

How a broadcast reaches you

The content (news, music, entertainment, educational programmes) is generated at the radio or TV station.
The audio and video signals are encoded: they are converted into a series of coded information suitable for transmission.
The encoded signals are transmitted from broadcasting antennas that are placed to cover wide geographical areas. The antennas emit radio waves.
Radios, televisions and smartphones receive the waves. Each device has a tuner that detects the specific frequency of the broadcasting station.
The device's internal circuitry decodes the received signal and converts it back into audio and video.

Figure 20.2 (radio transmission): Process of radio signal transmission. The microphone gives an audio-frequency signal. The mixer combines it with the radio-frequency output of the oscillator, and the transmitting antenna sends the result out as radio waves to the reception antenna and radio.

B. Astronomy

Astronomical objects with changing magnetic fields can emit radio waves. The study of these radio waves from space is known as radio astronomy. By analysing them, scientists learn about the composition, structure and motions of celestial bodies in the universe.

Moon sighting for Ramadan or Eid is done with optical telescopes, but sometimes the Moon cannot be seen because of clouds or bright sunlight. Radio telescopes (Figure 20.3) have an advantage here: their observations are not affected by sunlight, clouds or rain.

One limitation is that radio waves are much longer than visible light waves. To get image details (resolution) comparable to those of optical telescopes, radio telescopes need to be significantly larger.

20.2.2Applications of microwaves

Microwaves have a higher frequency and a shorter wavelength than radio waves, but they are still low-energy radiations, which makes them generally safe for humans. Their high frequencies allow faster transmission of information, so they are ideal for communication, especially in satellite technology.

A. Satellite television communication

Microwaves have a high frequency and short wavelength, so they can travel easily through the Earth's atmosphere. In satellite television communication, television signals are sent by microwaves to a satellite in space. The satellite amplifies the signals and retransmits them to Earth, so areas that terrestrial transmission cannot reach are covered. Dishes and other receiving antennas pick up the signals (Figure 20.4 (a)). More than one satellite is often needed to give uninterrupted coverage across different regions.

Figure 20.4 (a): Satellite television communication. The ground station sends the TV signal to the satellite by microwaves; the satellite amplifies it and retransmits it to the dish on the house.

B. Mobile phones

A mobile phone communicates with a mobile phone tower using radio waves. The towers communicate with satellites using microwaves, because microwaves can pass through the atmosphere. The signal can be sent to a satellite and used to communicate around the world: mobile signals are sent to the satellite by microwaves and the signals are interchanged between satellites and Earth stations (Figure 20.4 (b)).

Waves used in mobile communication
Link	Type of wave
Mobile phone and phone tower	radio waves
Tower and satellite (and satellite to Earth station)	microwaves

C. Bluetooth

Bluetooth uses radio waves and microwaves to send information between two devices directly. It uses weaker waves, which need less power to generate, so it is particularly useful for battery-powered devices. These weaker waves also mean that Bluetooth works only over short distances, less than 30 feet (about 9\,\mathrm{m}). A Bluetooth connection stays active as long as the devices remain within range, without a router or any other device in between (Figure 20.4 (c)).

D. Microwave oven

Microwave ovens cook and heat food evenly and efficiently. Microwaves make the molecules in the food vibrate, which creates the heat that cooks the food (Figure 20.4 (d)). A typical consumer microwave oven uses a frequency of 2.45\,\mathrm{GHz} and a wavelength of 12.2\,\mathrm{cm}.

20.2.3Applications of infrared

The infrared frequencies extend up to the lower limit of visible light, just below red. That is why they are called infra-red, which means "below red". Infrared radiation is commonly absorbed or emitted during molecular vibrations and rotations. Therefore all objects emit infrared radiation because of their temperature.

Many common household items use infrared radiation: kitchen toasters, mobile phones, electric heaters, microwave ovens and Wi-Fi devices. Some applications of infrared are listed below.

A. Remote controllers

Hand-held remotes change channels and settings on TVs, adjust air conditioners and control sound systems. They usually send signals through an infrared beam (Figure 20.5 (a)). Remote controllers use different infrared frequencies for different functions. When you press a button, you send a specific frequency to the TV, which receives the signal and reacts accordingly.

B. Intruder alarms

An intruder alarm emits a loud siren when it detects a significant increase in infrared radiation, typically due to the presence of a human. These alarms use sensors to identify infrared radiation from heat-producing sources. Door and window sensors detect movement when opened, and the alarms can also work as fire alarms by detecting the large amount of infrared radiation emitted by fire. Some alarms use an LED and photodiode circuit: the continuous infrared beam is interrupted by movement, which breaks the circuit and triggers the alarm (Figure 20.5 (b)).

Figure 20.5 (b): An intruder alarm. The infrared LED sends a continuous beam to the photodiode. When movement breaks the beam, the circuit triggers the siren.

C. Optical fibres

Optical fibres are thin strands of glass or plastic with two different refractive indices (a core and a cladding). Light or infrared radiation entering the core is reflected at the core-cladding boundary by total internal reflection, so it travels through the core at the speed of light. Ordinary light, laser light and infrared radiation are all used in optical fibres. Infrared is particularly advantageous because its wavelength has low absorption in glass fibres, which allows long-distance data transmission with minimal power loss (Figure 20.5 (c)).

Figure 20.5 (c): An optical fibre. The ray travels along the core by total internal reflection at the core-cladding boundary.

D. Thermal imaging

Thermal imaging converts the infrared radiation (heat) emitted by objects into visible images using a thermal camera. It is very effective at penetrating smoke, dust, aerosols and clouds, so it is useful in many conditions, both day and night (Figure 20.5 (d)).

E. Night vision

Infrared radiation makes night vision possible. Night vision scopes detect the infrared emitted by warm objects, including humans, and convert it to visible light (Figure 20.6).

The Sun radiates with a surface temperature of 6000\,\mathrm{K}. About half of the solar radiation reaching the Earth is in the infrared region, most of the rest is in the visible part of the spectrum, and a relatively small amount is in the ultraviolet region.

20.2.4Applications of visible light

Visible light is the narrow segment of the electromagnetic spectrum to which the normal human eye responds. The colours of light are associated with pure wavelengths, as shown in Figure 20.7. Red light has the lowest frequency and the longest wavelength, and violet has the highest frequency and the smallest wavelength in the visible spectrum.

Figure 20.7: The visible light spectrum. Red light has the longest wavelength (about 700\,\mathrm{nm}) and violet the shortest (about 400\,\mathrm{nm}).

Visible light is the most prominent part of electromagnetic radiation in our lives, and the beauty of nature is enjoyed through it. The book describes two uses.

A. Photography

In photography, visible light plays a pivotal role in capturing images. Cameras are designed to copy the human eye's ability to detect and focus visible light. Photographic sensors and film capture the light reflected or emitted by the subject, which is then processed to create an image. Different wavelengths of visible light are responsible for the colours seen in photographs. Advances in photographic technology have given high-resolution, colour-accurate images that are used in art, media, scientific research and documentation (Figure 20.8).

B. Vision

The ability to see something is called vision. Without light there would be no sight. The visual ability of humans and other animals is the result of the complex interaction of light, eyes and brain. Light enters the eye, is focused onto the retina and is converted into neural signals, which the brain reads to detect the appearance, location and movement of the objects we see.

20.2.5Applications of ultraviolet waves

Ultraviolet means "above violet". The frequency of ultraviolet radiation extends upwards from violet, which has the highest frequency in the visible spectrum. Solar ultraviolet is subdivided into three regions, ranked from long to short wavelengths:

Regions of solar ultraviolet
Region	Wavelength range
Ultraviolet-A	320\,\mathrm{nm} to 400\,\mathrm{nm}
Ultraviolet-B	290\,\mathrm{nm} to 320\,\mathrm{nm}
Ultraviolet-C	220\,\mathrm{nm} to 290\,\mathrm{nm}

Most ultraviolet-B and all ultraviolet-C are absorbed by the ozone (\mathrm{O_3}) molecules in the upper atmosphere. As a result, 99\% of the solar ultraviolet radiation reaching the Earth's surface is ultraviolet-A. Some practical applications of UV light are given below.

Fluorescence: A process in which a material absorbs radiation of high energy and frequency, such as ultraviolet, and re-emits radiation of lower energy, such as visible light.

A. Security marking

An ultraviolet watermark is included in many sensitive documents such as credit cards, driver's licences and passports to help prevent fraud. The watermark can be seen only under an ultraviolet-emitting light source. The ink used does the fluorescence: it absorbs ultraviolet radiation and then emits visible light that the human eye can see (Figure 20.9 (a)).

B. Detecting counterfeit bank notes

The currencies of various countries, including Pakistan's bank notes, have an image and many multicolour fibres that are visible only under ultraviolet light, to stop counterfeiters. The process is the same as in security marking by fluorescence: under UV light the fibres and images on a note can be seen, which shows that it is genuine (Figure 20.9 (b)).

C. Sterilising water

To sterilise water is to purify it of harmful micro-organisms such as bacteria and viruses. An ultraviolet purifier treats micro-biologically unsafe water with germ-killing ultraviolet light (Figure 20.9 (c)). Ultraviolet sterilisation is considered one of the most effective water treatment methods for disinfection: it sterilises water quickly, needs no additional chemicals, is cost effective, and there is no danger of overdosing.

20.2.6Applications of X-rays

The low-energy range of X-rays overlaps with the ultraviolet, and high-energy X-rays overlap with gamma rays (in their low-energy region). X-rays have many medical, technological and social applications.

A. Medical imaging

The widest use of X-rays is for imaging objects that are opaque to visible light, such as the human body. The ability of X-rays to penetrate matter depends on density, so an X-ray image can show very detailed density information. Fractures of bones can be found with X-rays (Figure 20.10 (a)). The use of X-ray technology in medical science is called radiology. X-rays can affect a photographic film: when a person stands in front of a photographic film and X-rays are directed at the body, an image is formed.

B. Security scanners

X-ray scanners are used to detect threats such as weapons or explosives that a person could be carrying under clothing. The scanners used for this purpose use very low energy X-rays that are reflected back to the scanner, which shows any metal object. X-ray scanning is also used in many industrial and commercial applications, most notably in security and health care (Figure 20.10 (b)).

C. Killing cancerous cells

Radiation therapy is a cancer treatment that uses high doses of radiation to kill cancer cells and shrink tumours (Figure 20.10 (c)). At low doses, X-rays are used to see inside the body; at high doses, radiation therapy kills cancer cells or slows their growth by damaging their DNA. Cancer cells whose DNA is damaged beyond repair stop dividing or die.

D. Detecting cracks in metals

Radiographic testing (RT) is a non-destructive testing (NDT) method that uses X-rays or gamma rays to examine the internal structure of manufactured components and identify flaws or defects. The test part is placed between the radiation source and the film (or detector). A crack makes the metal thinner at that place, so more X-rays pass through the crack and the region where the crack exists appears dark on the film, while the region without a crack appears light. The dark and light regions show the exact size and location of the crack (Figure 20.11).

Figure 20.11: Detecting a crack by radiographic testing (redrawn). More X-rays pass through the thinner metal at the crack, so the film is darker there.

20.2.7Applications of gamma rays

Gamma rays are the most energetic form of electromagnetic radiation and have the shortest wavelength. They are a type of nuclear radiation and are emitted from the nucleus of a heavy atom. Because of their high energy, gamma rays are highly penetrating and can pass through many materials that block other types of radiation.

A. Medical treatment

Gamma rays are useful in medicine for both diagnosing and treating diseases. In nuclear medicine, radioactive materials that emit gamma rays are introduced into the body by injection or swallowing in small amounts. The radioactive substances then collect in specific organs or tissues, depending on their chemical properties. Specialised imaging equipment, such as gamma cameras or positron emission tomography (PET) scanners, detects the gamma rays emitted by these materials (Figure 20.12 (a)). This technique lets doctors see the internal structures of the body in real time and gives vital information about the functioning of various organs.

Gamma rays are also used to treat cancer by radiation therapy. High-energy gamma rays are aimed precisely at cancerous cells to destroy them or stop their growth. The treatment can be given externally, with machines that produce gamma rays (Figure 20.12 (b)), or internally by brachytherapy, in which radioactive sources are placed directly in or near the tumour. Because gamma rays penetrate tissues, they are particularly effective at targeting tumours while minimising damage to the surrounding healthy cells.

B. Sterilising food and medical equipment

Gamma rays are used to sterilise both medical equipment and food. In gamma sterilisation, the products are exposed to gamma rays. The radiation source normally used is the cobalt-60 isotope, which emits gamma radiation in a process that destroys any micro-organism present. For food, cobalt-60 is again used, to kill bacteria, mould, parasites and insects in the food. The most important point is that the food itself does not become radioactive (Figure 20.13).

C. Engineering application

Gamma rays are used in engineering for non-destructive testing (NDT) and radiographic testing (Figure 20.14). They inspect industrial components, welds and structures for defects, cracks and other flaws without damaging the materials. The process is similar to X-ray testing; the only difference is that gamma rays have more penetrating power than X-rays and so give a better test for hard materials. In radiography, gamma rays penetrate the material to show hidden flaws.

20.3Risks associated with electromagnetic waves

Electromagnetic waves, along with their many applications, have some hazards associated with them. Every type of electromagnetic wave has its own particular hazards. The majority of electromagnetic waves in small amounts are safe for humans and the atmosphere, but at higher intensities they can damage living organisms on Earth and its atmosphere.

Table 20.2: Electromagnetic waves, a hazard analysis
Wave	Potential hazards	Potential hazards (continued)
Radio waves	High-energy radio waves can cause tissue heating, leading to burns or heat-related injuries.	Continuous exposure can interfere with biological systems and with medical devices such as pacemakers and hearing aids.
Microwaves	Microwaves can cause tissue damage, especially in sensitive organs such as the eyes, potentially leading to cataracts.	Exposure to high-powered microwaves may result in burns or heat stress and can interfere with implanted medical devices.
Infrared	Prolonged exposure to infrared waves can cause thermal damage, resulting in burns or overheating.	High-intensity infrared radiation over time can cause eye damage, skin ageing and diseases.
Visible light	Direct viewing of intense visible light over long periods can cause eye strain and retinal damage.	Long exposure to bright artificial lighting can disrupt biological rhythms and, in extreme cases, cause burns or permanent eye injuries.
Ultraviolet	Long-term exposure to UV waves damages skin cells, causing sunburn and increasing the risk of skin cancers.	UV radiation can harm the eyes, leading to cataracts and other vision impairments, and weaken the immune system.
X-rays	X-rays cause ionisation that can lead to tissue damage, cell mutations and a high risk of cancer.	Frequent or high-dose exposure can cause burns, cataracts and acute radiation sickness affecting overall health.
Gamma rays	Gamma rays cause cell mutations, leading to cancer or death because of their penetrative and ionising nature.	Prolonged exposure leads to acute radiation sickness and poses environmental and health threats through nuclear accidents or improper disposal.

Ionising radiation

Ionising radiation: Electromagnetic waves that have enough energy to remove tightly bound electrons from atoms and so create ions. This can lead to chemical and biological changes in living tissues.

Among the electromagnetic waves, ultraviolet rays, X-rays and gamma rays are ionising radiation. They differ mainly in their energy levels and wavelengths.

Ultraviolet rays have wavelengths shorter than visible light but longer than X-rays. Prolonged exposure can cause skin cancer, cataracts and other health problems.
X-rays have shorter wavelengths and higher energy than UV. Their ionising power can cause cell damage and increases the risk of cancer with excessive exposure.
Gamma rays are at the high-energy end of the spectrum, with the shortest wavelengths and highest frequencies. They have great ionising power and can penetrate almost any material. This makes them both a powerful tool in medical treatments such as radiotherapy for cancer and a considerable hazard. Their intense ionising effect can cause severe cell and DNA damage, so strict safety rules are needed where gamma rays are present.

Lower-energy radiation such as visible light, infrared, microwaves and radio waves does not have enough energy to ionise atoms. It is generally considered non-ionising, and so it is less harmful in causing immediate biological damage.

Protection

It is essential to protect oneself from excessive exposure to ionising radiation, especially X-rays and gamma rays. The important preventive measures are:

shielding from the source,
limiting the time of exposure,
keeping a safe distance from radiation sources.

Worked example: which of these waves are ionising?

Four electromagnetic waves have frequencies 1\times10^{12}\,\mathrm{Hz}, 6\times10^{14}\,\mathrm{Hz}, 2\times10^{16}\,\mathrm{Hz} and 3\times10^{18}\,\mathrm{Hz}. Use Table 20.1 to name the region of each, find its wavelength and photon energy, and say which are ionising. Which is the more dangerous to living tissue, and why?

Given

f = 1\times10^{12},\ 6\times10^{14},\ 2\times10^{16},\ 3\times10^{18}\,\mathrm{Hz}
c = 3\times10^{8}\,\mathrm{m\,s^{-1}}
h = 6.63\times10^{-34}\,\mathrm{J\,s}

To find

region, \lambda, E and ionising or not, for each wave

Use \lambda=c/f for each wavelength and E=hf for each photon energy. For example, for the first wave:
\begin{aligned}\lambda&=\dfrac{3\times10^{8}}{1\times10^{12}}=3\times10^{-4}\,\mathrm{m}\\ E&=6.63\times10^{-34}\times1\times10^{12}=6.63\times10^{-22}\,\mathrm{J}\end{aligned}
In the same way, 6\times10^{14}\,\mathrm{Hz} gives \lambda=\dfrac{3\times10^{8}}{6\times10^{14}}=5\times10^{-7}\,\mathrm{m}=500\,\mathrm{nm} and E=3.98\times10^{-19}\,\mathrm{J}.
For 2\times10^{16}\,\mathrm{Hz}: \lambda=\dfrac{3\times10^{8}}{2\times10^{16}}=1.5\times10^{-8}\,\mathrm{m}=15\,\mathrm{nm} and E=6.63\times10^{-34}\times2\times10^{16}=1.33\times10^{-17}\,\mathrm{J}.
For 3\times10^{18}\,\mathrm{Hz}: \lambda=\dfrac{3\times10^{8}}{3\times10^{18}}=1\times10^{-10}\,\mathrm{m}=0.1\,\mathrm{nm} and E=6.63\times10^{-34}\times3\times10^{18}=1.99\times10^{-15}\,\mathrm{J}.
Compare each frequency with the ranges of Table 20.1 to name the regions, then use the rule that UV, X-rays and gamma rays are ionising.

f (Hz)	Region (Table 20.1)	\lambda	E (J)	Ionising?
1\times10^{12}	microwave	0.3\,\mathrm{mm}	6.63\times10^{-22}	no
6\times10^{14}	visible	500\,\mathrm{nm}	3.98\times10^{-19}	no
2\times10^{16}	ultraviolet	15\,\mathrm{nm}	1.33\times10^{-17}	yes
3\times10^{18}	X-rays	0.1\,\mathrm{nm}	1.99\times10^{-15}	yes

The four waves placed on a frequency scale. The boundary at about 10^{15}\,\mathrm{Hz} (the lower limit of ultraviolet in Table 20.1) separates the non-ionising waves from the ionising ones.

The X-rays are the more dangerous of the two ionising waves. Their photon energy is \dfrac{1.99\times10^{-15}}{1.33\times10^{-17}}\approx150 times that of the UV photon, so they can remove tightly bound electrons from atoms and damage cells and DNA much more strongly.
The microwave and visible waves do not have enough energy to ionise atoms, so they are less harmful in causing immediate biological damage, although intense exposure can still heat or injure tissue (Table 20.2).

\text{microwave and visible: non-ionising; UV and X-rays: ionising; X-rays have the larger energy }(1.99\times10^{-15}\,\mathrm{J})

20.4Scattering of lightOften examined

Scattering of light: The bouncing of light waves off the particles of a medium in different directions.

The amount of scattering depends on the size of the particles and on the wavelength of the light. Shorter wavelengths scatter more, while longer wavelengths scatter less and travel straighter. For effective interaction, the wavelength of the light must be similar in size to the particles.

Sunlight contains almost all the wavelengths of the visible spectrum: shorter wavelengths such as blue and violet, and longer wavelengths such as orange and red.

Scattering and wavelength
Light	Wavelength	Scattered by air molecules
Blue, indigo, violet	shorter	strongly, in all directions
Red, orange, yellow	longer	weakly, they travel almost straight on

20.4.1Blue colour of the skyOften examined

The light coming from the Sun has many wavelengths, among which the blue colour is present in the major percentage of the lower wavelengths. As the wavelength of the light decreases, the scattering efficiency of the small molecules in the atmosphere increases.

When the Sun's light enters the atmosphere of the Earth it is scattered by air molecules, dust particles and water vapour. The major components of the atmosphere are gases such as oxygen and nitrogen, and these cause most of the scattering. The molecules of these gases are small and comparable in size to the smaller wavelengths of light, mainly blue with indigo and violet, so these wavelengths interact most with the gas molecules and are scattered around.

The larger wavelengths of visible light (red, orange, yellow) are larger than the molecules, so they do not suffer such scattering and pass straight through the atmosphere. In the upper atmosphere the scattered smaller wavelengths are seen, and they give the appearance of blue colour: the blue sky. The scattered blue light fills the sky with a blue hue (Figure 20.15).

Figure 20.15: Blue colour of the sky. The shorter wavelengths (blue, indigo, violet) are scattered in all directions by the air molecules, while the longer wavelengths (red, orange, yellow) pass through the atmosphere without appreciable scattering.

Case study: a clear summer day and a beautiful sunset

Imagine a summer day at the beach. The sky is bright blue, clear and vibrant. But as the Sun sets, the sky turns shades of orange, pink and red. Why? Sunlight contains all colours, so it is a mixture of different wavelengths. During the day, the sunlight reaching the Earth's atmosphere interacts with the tiny gas molecules in the air, which scatter it in all directions. Blue light, with its shorter wavelength, scatters more than other colours such as red and green, and this scattered blue light reaches our eyes from all directions, making the sky appear blue. At sunset, the Sun is low on the horizon, so sunlight has to pass through more of the atmosphere to reach us. By the time it reaches our eyes, most of the blue and violet light has been scattered out in different directions, leaving more of the red and orange wavelengths, which are longer and scatter less. This creates the warm colours of sunset. This effect is known as Rayleigh scattering, and it explains the different colours seen in the sky at different times of day.

20.4.2Red sky at duskOften examined

In the evening and in the morning, the Sun's light has to travel a greater distance in the atmosphere because the Sun is at a low angle with the surface of the Earth. In this situation the smaller wavelengths cannot reach us, because they suffer more scattering and are scattered away. The longer wavelengths, such as red and orange, can travel the greater distance and reach us. This gives the reddish hues seen at sunset, when the Sun's light passes through more air molecules before reaching our eyes (Figure 20.16). The same happens in the early morning, at sunrise, when the sky appears reddish.

Figure 20.16: Red sky in the evening. At a small angle above the horizon the sunlight crosses a long path through the air. The blue light is scattered out of the beam on the way, and the red light reaches the observer.

Worked example: drawing the path of sunlight at noon and at sunset

Draw a diagram of the path of sunlight through the atmosphere for an observer at noon and for an observer at sunset. Use it to explain why the Sun does not look red at noon but looks red at sunset.

Draw the Earth's surface as a line, with the atmosphere as a layer above it, and mark the observer on the surface.
(a) Noon. The Sun is overhead, so the ray comes straight down. It crosses the layer by the shortest route, a short path through the air.
(b) Sunset. The Sun is near the horizon, so the ray arrives at a small angle to the ground. It has to cross the same layer along a slanting route, a long path through the air. Mark this longer part of the ray.
Explanation. At noon only a little blue light is scattered out of the direct beam, so the Sun does not look red. At sunset the beam passes through many more air molecules, so most of the blue and violet light is scattered out of it. What remains has more of the longer wavelengths, red and orange, so the Sun and the sky near it look red.

The path of sunlight through the atmosphere (a) at noon and (b) at sunset. The part of the ray inside the atmosphere (coloured) is much longer at sunset.

\text{noon: short path, little scattering; sunset: long path, blue scattered out, red remains}

20.5Nature of lightOften examined

Imagine light as a stream of tiny, super-fast bullets. These "bullets" are called photons. Unlike regular bullets, photons have no mass: they are incredibly light, almost weightless. Even though photons are massless, they still carry energy and momentum. Light may seem like a continuous wave, but it is actually made up of these tiny massless particles. Photons are the building blocks of light (Figure 20.17).

Figure 20.17: Particle nature of light. A source of light sends out a stream of photons that travel at the speed of light.

Wave nature and particle nature

Up to the start of the 20th century light was thought of as a wave only, because there was a lot of evidence for it: interference, reflection, refraction, diffraction and polarisation. At the beginning of the 20th century, Max Planck introduced the concept of light having a particle nature. He suggested that the behaviour of radiation emitted from objects could be explained only if light were considered as tiny packets of energy, which he called "quanta of light".

Photon: A tiny massless particle of light. It carries some energy and momentum but no charge. A photon is a fundamental particle and a quantum of the electromagnetic field, and it always travels at the speed of light in a vacuum.

The energy of a photon is given by equation (20.1):

E=hf

and its momentum is given by the same equation:

p=\dfrac{h}{\lambda}=\dfrac{hf}{c}

(20.4)

As the basic building blocks of light, photons are classified as elementary particles. This raises the question whether light is a wave or a particle. The puzzle is addressed by Louis de Broglie's theory of the dual nature of light, which states that light shows both properties.

Two aspects of light
Wave nature	Particle nature
Interference, reflection, refraction, diffraction and polarisation	Light is made of photons that carry energy (E=hf) and momentum (p=h/\lambda)

Example 20.2: a radiation monitor

You are using a radiation monitor which gives you a signal of some radiations received. The wavelength of those radiations was 200\,\mathrm{nm}. Find the frequency, momentum and energy of the radiation signals.

Given

\lambda = 200\,\mathrm{nm}=200\times10^{-9}\,\mathrm{m}
c = 3\times10^{8}\,\mathrm{m\,s^{-1}}
h = 6.63\times10^{-34}\,\mathrm{J\,s}

To find

frequency f, momentum p, energy E

Frequency from (20.2).
f=\dfrac{c}{\lambda}=\dfrac{3\times10^{8}}{200\times10^{-9}}=1.5\times10^{15}\,\mathrm{Hz}
Momentum from (20.4), p=hf/c.
p=\dfrac{6.63\times10^{-34}\times1.5\times10^{15}}{3\times10^{8}}=3.315\times10^{-27}\,\mathrm{kg\,m\,s^{-1}}
Check using p=h/\lambda.
\dfrac{6.63\times10^{-34}}{2\times10^{-7}}=3.315\times10^{-27}\,\mathrm{kg\,m\,s^{-1}}\ \checkmark
Energy from (20.3).
E=\dfrac{hc}{\lambda}=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{200\times10^{-9}}=9.945\times10^{-19}\,\mathrm{J}
The frequency 1.5\times10^{15}\,\mathrm{Hz} lies in the ultraviolet range (10^{15} to 10^{17}\,\mathrm{Hz} in Table 20.1; the wavelength 200\,\mathrm{nm} is also in the 400\,\mathrm{nm} to 1\,\mathrm{nm} range), so the radiation detected by the monitor is ultraviolet.

f=1.5\times10^{15}\,\mathrm{Hz},\ p=3.315\times10^{-27}\,\mathrm{kg\,m\,s^{-1}},\ E=9.945\times10^{-19}\,\mathrm{J}\ (\text{ultraviolet})

Worked example: momentum of a red photon and of an X-ray photon

A red photon has a wavelength of 650\,\mathrm{nm} and an X-ray photon has a wavelength of 0.1\,\mathrm{nm}. Find the momentum of each and say how many times greater the momentum of the X-ray photon is. Photons have no mass, so how can they have momentum?

Given

\lambda_r = 650\,\mathrm{nm}=6.5\times10^{-7}\,\mathrm{m}
\lambda_x = 0.1\,\mathrm{nm}=1\times10^{-10}\,\mathrm{m}
h = 6.63\times10^{-34}\,\mathrm{J\,s}

To find

p_r, p_x and p_x/p_r

The momentum of a photon in terms of its wavelength is given by (20.4).
p=\dfrac{h}{\lambda}
Red photon.
p_r=\dfrac{6.63\times10^{-34}}{6.5\times10^{-7}}=1.02\times10^{-27}\,\mathrm{kg\,m\,s^{-1}}
X-ray photon.
p_x=\dfrac{6.63\times10^{-34}}{1\times10^{-10}}=6.63\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}
Ratio of the momenta.
\dfrac{p_x}{p_r}=\dfrac{6.63\times10^{-24}}{1.02\times10^{-27}}\approx6500
Check using the wavelengths: since p\propto1/\lambda, the ratio is \lambda_r/\lambda_x=\dfrac{6.5\times10^{-7}}{1\times10^{-10}}=6500, which agrees.
A photon has no mass, but it carries energy and it moves at the speed of light, and (20.4) gives it a momentum h/\lambda that depends only on its wavelength. The shorter the wavelength, the greater the momentum.

p_r=1.02\times10^{-27}\,\mathrm{kg\,m\,s^{-1}},\ p_x=6.63\times10^{-24}\,\mathrm{kg\,m\,s^{-1}};\ p_x\ \text{is about }6500\text{ times greater}

20.5.1Radiation pressure

Imagine tossing a tennis ball: it has weight and you can feel it. Now think of a tiny, fast bullet made of energy. This bullet has no weight, but it can still push things around. Photons are like these energy bullets. They have no mass, but they carry energy and momentum. That is why sunlight can warm your skin and why solar sails can move spacecraft.

Despite having no mass, photons interact strongly with matter. Because of the transfer of momentum, light can exert pressure on objects. This phenomenon is known as radiation pressure (Figure 20.18). The pressure depends on the type of surface and on the intensity of the light.

Radiation pressure: The pressure that light exerts on an object because of the transfer of momentum from its photons.

Figure 20.18: Radiation pressure. Photons (each of momentum p) strike a surface of area \Delta A and give their momentum to it, which pushes the surface.

Sunlight warms your hands by transferring tiny momentum, which is imperceptible because of the very high speed of light. Radiation pressure is too small to notice with ordinary light, but it is crucial in astronomical contexts, such as balancing gravitational forces in stars or pushing dust in comet tails. It also has practical applications, such as laser cooling and trapping, which manipulate atoms and small objects.

Project: studying radiation pressure

Apparatus: two very light concave mirrors, an inextensible light fibre (string), a very light pipe and a source of intense light (a torch).

Take two concave mirrors and connect them by a light rod or pipe.
Hang the mirrors-and-pipe system by an inextensible light string.
Bring the apparatus to rest and make sure there is no wind in the room.
Shine an intense torch on one mirror. The whole assembly rotates a little because of the pressure exerted by the light.
Change the intensity of the light and note how far the assembly rotates, then guess the relation between the intensity of light and the radiation pressure.

Conclusion: the light moved the mirror assembly, which is possible only if the incident light exerts a pressure (a force on unit area of the mirror). When the intensity of the light was increased the assembly moved further, which shows that the radiation pressure is directly proportional to the intensity of the light.

20.5.2Solar sails

One application of radiation pressure is the solar sail. A solar sail is a method of spacecraft propulsion that uses the momentum of photons. These particles of light have no mass, yet when they strike something they impart momentum and give it a tiny push.

Solar sail: A large but very light spacecraft sail made of reflective material that captures and reflects solar radiation to give a continuous thrust.

Imagine a sailing boat in space, where sunlight acts as the wind pushing against the sail and moving the spacecraft forward. The incident photons hit the sail, transfer their momentum to it, and are reflected at some angle. This gives a thrust to the sail (Figure 20.19).

Figure 20.19: A solar sail. Photons from the Sun strike the reflective sail and are reflected; the momentum they transfer gives the sail a thrust away from the Sun, along the sail normal.

Formula sheet

Speed of an electromagnetic wave

c=f\lambda

c: speed of light, 3\times10^{8}\,\mathrm{m\,s^{-1}} in a vacuum (and nearly the same in air); f: frequency; \lambda: wavelength (20.2)

Unit: \mathrm{m\,s^{-1}}

Planck's equation (photon energy from frequency)

E=hf

E: energy of the radiation (photon); h=6.63\times10^{-34}\,\mathrm{J\,s}: Planck's constant; f: frequency (20.1)

Unit: \mathrm{J}

Photon energy from wavelength

E=\dfrac{hc}{\lambda}

E: energy; h: Planck's constant; c: speed of light; \lambda: wavelength in metres (20.3)

Unit: \mathrm{J}

Photon momentum

p=\dfrac{h}{\lambda}=\dfrac{hf}{c}

p: momentum of the photon; h: Planck's constant; \lambda: wavelength; f: frequency; c: speed of light (20.4)

Unit: \mathrm{kg\,m\,s^{-1}}

Joules and electron-volts

1\,\mathrm{J}=\dfrac{1}{1.6\times10^{-19}}\,\mathrm{eV}=6.25\times10^{18}\,\mathrm{eV}

divide an energy in joules by the magnitude of the electron charge, 1.6\times10^{-19}\,\mathrm{C}, to get electron-volts

Unit: \mathrm{eV}

Key terms

Electromagnetic waves: Transverse waves produced by oscillating electric and magnetic fields that are perpendicular to each other; they do not need a material medium for their production and propagation.
Electromagnetic spectrum: The range of different frequencies and wavelengths of electromagnetic waves: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays.
Speed of light: The speed of electromagnetic waves in a vacuum, a universal constant denoted by c with the value 3\times10^{8}\,\mathrm{m\,s^{-1}}. In air it is approximately the same.
Planck's equation: The relation E=hf between the energy and the frequency of an electromagnetic wave, where h=6.63\times10^{-34}\,\mathrm{J\,s} is Planck's constant.
Electron-volt: A small unit of energy used for electromagnetic radiation at atomic levels; 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J}.
Spectroscopy: The branch of physics used to study the ways in which different electromagnetic waves interact with matter.
Radio waves: The electromagnetic waves of the longest wavelengths. They travel at the speed of light and are used in radio and television transmission and in astronomy.
Radio astronomy: The study of the radio waves that reach us from space; it tells us about the composition, structure and motions of celestial bodies.
Microwaves: Waves of higher frequency and shorter wavelength than radio waves, but still low in energy. They are used in satellite television, mobile phones, Bluetooth and microwave ovens.
Infrared radiation: Radiation produced by thermal motion, the vibration and rotation of atoms and molecules. It is used in household appliances, remote controllers, intruder alarms, thermal imaging, night vision and optical fibres.
Visible light: The narrow segment of the electromagnetic spectrum to which the normal human eye responds.
Ultraviolet radiation: Radiation whose frequency extends upwards from violet, the highest frequency of visible light. It is used in security marking, detecting counterfeit bank notes and sterilising water.
Fluorescence: A process in which a material absorbs radiation of high energy and frequency, such as ultraviolet, and re-emits radiation of lower energy, such as visible light.
X-rays: Electromagnetic waves whose low-energy range overlaps ultraviolet and whose high-energy range overlaps gamma rays. They are used in medical imaging, security scanners, killing cancerous cells and detecting cracks in metals.
Radiographic testing: A non-destructive testing method that uses X-rays or gamma rays to examine the internal structure of components and find flaws such as cracks.
Gamma rays: The most penetrating nuclear radiation, with the shortest wavelength and highest energy. They are used in medical treatment, sterilising food and equipment, and engineering.
Ionising radiation: Electromagnetic waves (ultraviolet, X-rays and gamma rays) with enough energy to remove tightly bound electrons from atoms and create ions.
Non-ionising radiation: Lower-energy radiation (visible light, infrared, microwaves and radio waves) that lacks enough energy to ionise atoms.
Scattering of light: The reflection of light waves in different directions from the particles of the medium.
Blue sky: The appearance of blue colour in the upper atmosphere, caused by the scattering of the smaller wavelengths of sunlight by air molecules.
Rayleigh scattering: The scattering of sunlight by the tiny gas molecules of the atmosphere, which explains the blue sky by day and the red colours at sunrise and sunset.
Photon: A tiny massless particle of light that carries energy and momentum but no charge, and travels at the speed of light in a vacuum.
Radiation pressure: The pressure exerted by light on an object because of the transfer of momentum from its photons.
Solar sail: A spacecraft propulsion method that uses the momentum of photons from sunlight, striking and reflecting from a large, light, reflective sail.

Multiple choice questionsp. 305–306

Identify the electromagnetic wave with the greatest ability to penetrate matter.

ARadio waves
BUltraviolet
CX-rays
DGamma raysAnswer

Why

Penetrating power grows with photon energy. Gamma rays have the shortest wavelength (less than 10^{-12}\,\mathrm{m}) and the highest frequency (Table 20.1), so each photon carries the most energy and they pass through the most material.

Radio waves (A) have the lowest energy, and ultraviolet (B) is easily absorbed by a thin layer of material. X-rays (C) penetrate well, but gamma rays penetrate further, which is why they are used in radiographic testing of thick metal parts.

Which type of radiation primarily carries heat energy?

ARadio waves
BUltraviolet
CInfraredAnswer
DVisible light

Why

Infrared radiation is produced by the thermal motion, vibration and rotation of atoms and molecules, and it is absorbed by objects as heat. This is why infrared lamps and quartz heaters warm us (book p. 291).

Radio waves (A) carry very little energy per photon and are not felt as heat. Ultraviolet (B) causes chemical changes such as sunburn rather than warmth. Visible light (D) does carry energy, but the book shows that about half of the Sun's radiation reaching Earth is infrared, which is the main heating part.

The wavelength of 10 cm is corresponding to spectrum of:

AMicrowavesAnswer
BUltraviolet
CInfrared
DX-rays

Why

The wavelength is \lambda=10\,\mathrm{cm}=0.10\,\mathrm{m}, so the frequency is f=\dfrac{c}{\lambda}=\dfrac{3\times10^{8}}{0.10}=3\times10^{9}\,\mathrm{Hz}. This is a microwave frequency, about 3 GHz.

Ultraviolet (B), infrared (C) and X-rays (D) all have wavelengths far below 1\,\mathrm{mm} (Table 20.1), so 10 cm cannot belong to them.

Note. Table 20.1 gives the microwave row as 1 mm to 25 \mum and the radio row as above 1 mm. Strictly, 10 cm falls in the table's radio row, but the question's options do not include radio. In the usual classification, microwaves cover wavelengths of about 1 mm to 1 m, so A is the intended answer.

The atomic structure of solids or cracks in solid structure can be investigated by:

AMicrowaves
BUltraviolet
CGamma rays
DX-raysAnswer

Why

X-rays are used to look inside solid materials. Their wavelength (1 nm to 1 pm) is small enough to reveal the arrangement of atoms in a solid, and in radiographic testing they show cracks in metal as dark lines (Figure 20.11).

Microwaves (A) are too long to resolve atoms, and ultraviolet (B) does not pass through solids. Gamma rays (C) can also reveal flaws in thick metal, but they are not the standard tool for studying atomic arrangement, so X-rays are the intended answer.

Which waves are used for treating muscle aches in physiotherapy?

AMicrowaves
BUltraviolet
CInfraredAnswer
DX-rays

Why

Infrared radiation is absorbed by the skin and the tissue just below it, where it is turned into heat. This warming relieves muscle aches, which is why infrared lamps and heaters are used in physiotherapy.

Ultraviolet (B) and X-rays (D) are harmful to tissue (book p. 297) and are not used to warm muscles. Microwaves (A) are used to heat food in ovens, not to treat aches at the skin's surface.

X-rays have a higher energy than visible light, because of its:

AShorter wavelengthAnswer
BLonger wavelength
CLower frequency
DHigher speed

Why

The energy of a photon is E=\dfrac{hc}{\lambda}, so energy is inversely proportional to wavelength. X-rays have much shorter wavelengths than visible light, so each X-ray photon has more energy.

Longer wavelength (B) and lower frequency (C) both mean less energy, because E=hf. Higher speed (D) is wrong: all electromagnetic waves travel at the same speed in a vacuum.

Electromagnetic waves are produced due to the motion of which of the following?

ANeutral particles
BCharged particles
CAccelerating neutral particles
DAccelerating charged particlesAnswer

Why

An electromagnetic wave is produced when a charge is accelerated. The changing electric field of the accelerating charge creates a changing magnetic field, and the two regenerate each other and travel outwards as a wave.

Neutral particles (A, C) have no charge, so they produce no electromagnetic wave even when accelerated. A charge moving at constant velocity (B) does not radiate a wave, because it is not accelerating.

If the frequency of an electromagnetic wave is 2\times10^{18}\,\mathrm{Hz}, to which spectrum it belongs to?

AMicrowaves
BUltraviolet
CInfrared
DX-raysAnswer

Why

Table 20.1 gives the X-ray range as 10^{17} to 10^{20}\,\mathrm{Hz}, and 2\times10^{18}\,\mathrm{Hz} lies inside it. The wavelength is \lambda=\dfrac{c}{f}=\dfrac{3\times10^{8}}{2\times10^{18}}=1.5\times10^{-10}\,\mathrm{m}, which is also in the X-ray range (1 nm to 1 pm).

Microwaves (A) are at about 10^{11} to 10^{13}\,\mathrm{Hz}, infrared (C) at about 10^{13} to 4\times10^{14}\,\mathrm{Hz}, and ultraviolet (B) at about 10^{15} to 10^{17}\,\mathrm{Hz}, so each of them is too low in frequency.

Light with longest wavelength in visible spectrum is:

AGreen
BViolet
COrange
DRedAnswer

Why

In the visible spectrum the wavelength increases from violet to red (Figure 20.7). Red light has the longest visible wavelength (about 700 nm) and the lowest frequency.

Violet (B) has the shortest visible wavelength. Green (A) and orange (C) lie between violet and red, and orange is shorter than red.

Electromagnetic waves exhibit particle-like behavior in quantum mechanics. These particles are called:

AAtom
BX-ray
CElectron
DPhotonAnswer

Why

The particles associated with electromagnetic waves are photons. A photon is a massless particle that carries energy E=hf and momentum p=\dfrac{h}{\lambda} (section 20.5).

Atoms (A) and electrons (C) are particles of matter, not of the wave itself. An X-ray (B) is an electromagnetic wave, not a particle.

The energy of an electromagnetic wave depends directly on its:

AWavelength
BAmplitude
CFrequencyAnswer
DSpeed

Why

The energy of a photon is E=hf, which is directly proportional to frequency. A higher frequency gives a higher energy.

Wavelength (A) affects energy inversely, E=\dfrac{hc}{\lambda}. Amplitude (B) changes the intensity, which is the number of photons, not the energy of each photon. Speed (D) is the same for all electromagnetic waves in a vacuum.

When an electromagnetic wave enters from one medium into the other, which quantity does not change?

AWavelength
BSpeed
CFrequencyAnswer
DEnergy

Why

The frequency is set by the source of the wave, and it does not change when the wave crosses into another medium. The speed changes, and since v=f\lambda, the wavelength changes with it.

Wavelength (A) and speed (B) both change. Energy (D) is also unchanged for a single photon, because E=hf depends only on frequency, but the question asks for the quantity that the medium cannot alter, and frequency is the standard answer.

Note. Strictly, the energy of a single photon also stays the same, so D is true as well. The book prints no key. C is the intended answer: the frequency is fixed by the source and is the quantity the question asks about.

On halving the area of a totally reflecting surface, the radiation pressure will:

ARemain the sameAnswer
BDouble
CHalve
DQuadruple

Why

Radiation pressure is force per unit area. For a totally reflecting surface it is P=\dfrac{2I}{c}, where I is the intensity of the light (power per unit area). Intensity and c do not depend on the size of the surface, so the pressure does not depend on the area.

Halving the area halves the total force, but the force is shared over half the area, so the pressure is unchanged. Options B, C and D treat the total force as if it were the pressure.

Solar sails are preferred for deep-space missions because they require:

AElectric energy
BFuel
CChemical energy
DNo fuelAnswer

Why

A solar sail is pushed by the momentum of sunlight photons that are reflected from its large, light surface. It needs no fuel to produce this thrust, so it can keep working for a long time on a long mission.

Fuel (B) and chemical energy (C) are what conventional rockets need, and a solar sail does not carry them. Electric energy (A) is not the reason solar sails are chosen.

Which of these electromagnetic waves has the least momentum?

AGamma rays
BUltraviolet
CInfrared
DRadio wavesAnswer

Why

The momentum of a photon is p=\dfrac{h}{\lambda}=\dfrac{hf}{c}. It is least for the longest wavelength and the lowest frequency, and radio waves have the longest wavelength of the four.

Gamma rays (A) have the shortest wavelength and the largest momentum. Ultraviolet (B) and infrared (C) lie between radio waves and gamma rays, so they have more momentum than radio waves.

Constructed response questionsp. 306

Take some colloidal solution of sulphur from your Chemistry lab and fill the solution in some transparent beaker. Assemble the beaker between a light source 'S' and slit 'C' as shown in figure and illuminate the beaker from one side with white light. Follow the directions to respond to the following questions.

Apparatus for question QII: light source S, lens, beaker T with the colloidal sulphur solution, slit C, lens, and screen N-M.

Solution

The three parts are the three directions of QII in the book. The book prints no answers for this activity, so the answers follow the book's explanation of scattering in section 20.4. The colour of a real sulphur colloid also depends on the size and concentration of its particles.

(a) What color of the solution appears if seen from above? Why?

The sulphur particles in the colloid are much larger than molecules, so they scatter light. Scattering is strongest for the shortest wavelengths, so the blue and violet part of the white light is scattered most strongly in all directions.

Light scattered upwards from the beam reaches the eye from above, and it is mainly blue. The solution therefore looks bluish when seen from above, while the beam itself is visible only where the particles scatter it.

\text{Bluish (blue) colour, because blue light is scattered most strongly by the particles.}

(b) What color spot appears on the screen 'N-M' after the light passes through the whole beaker?

The light that reaches the screen has passed through the solution. Its blue and violet components have been scattered out of the beam, so mainly the longer wavelengths remain (orange and red).

The slit and lenses then form a spot of this light on the screen, so the spot appears reddish (red to orange), the colour left after the blue has been removed.

\text{A reddish (red to orange) spot.}

The phenomenon is the scattering of light by the particles of the colloid. The particles reflect light in different directions, which makes the path of the beam visible, and the scattering depends on the wavelength of the light.

This is the same scattering that gives the blue colour of the sky (section 20.4): shorter wavelengths are scattered more, so the colour of the light that gets through changes.

\text{Scattering of light by the colloidal particles.}

Note. Book labels the parts Q.1, Q.2 and Q.3. The activity is one set of directions with one setup, so the three questions are given as parts (a), (b) and (c).

Short response questionsp. 306–307

13 marks

A leaking microwave oven in a home can sometimes cause interference with the home's Wi-Fi system. Why?

Solution

A microwave oven works at a frequency of about 2.45 GHz, which has a wavelength of 12.2 cm (book p. 289). Wi-Fi also uses microwaves in the same range of frequencies, close to 2.4 GHz.

When the oven leaks microwaves, they have almost the same frequency as the Wi-Fi signal, so they overlap with it and disturb it. The router and the devices then receive a distorted signal, and the connection can become slow or drop.

23 marks

Give an example of energy carried by an electromagnetic wave.

Solution

Sunlight is an example. The electromagnetic waves from the Sun carry energy to Earth, and when they are absorbed by your skin or by objects they are turned into heat. This is why sunlight warms your hands (book p. 303).

Microwaves in a microwave oven are another example: they carry energy into the food, which then heats up.

33 marks

Your friend says that more patterns and colors can be seen on the wings of birds if viewed in ultraviolet light. Would you agree with your friend? Explain.

Solution

Yes, the friend is right. Some patterns on feathers are invisible in ordinary light because they reflect ultraviolet, which the eye cannot see.

Under ultraviolet light, the pigments in the feathers absorb the ultraviolet and re-emit visible light. This is the same fluorescence used for ultraviolet security marks on bank notes (book p. 293), so the hidden patterns and colours become visible.

43 marks

Can human body detect electromagnetic radiations that are outside the visible region of the spectrum?

Solution

The eye responds only to the visible region, roughly 400 nm to 750 nm (Table 20.1). Radio waves, microwaves, ultraviolet and X-rays cannot be seen by the eye.

The skin can detect infrared, but only as heat. Infrared is absorbed by the skin, which warms up, so you feel it as warmth (book p. 291). Ultraviolet does damage the skin, such as sunburn, but it gives no sensation of its own.

53 marks

What property of light corresponds to loudness in sound?

Solution

The property of light that corresponds to loudness is its amplitude. A larger amplitude means a more intense wave, which is seen as brighter light, just as a larger amplitude sound wave is heard as louder.

63 marks

Certain orientations of a television antenna give better reception than others for a particular station. Explain.

Solution

A television signal is a transverse electromagnetic wave. Its electric field oscillates in one direction, set by the transmitting antenna of the station, and the receiving antenna picks up the wave through this electric field.

The rods of the receiving antenna respond best when they are parallel to the direction of the electric field, because the field then pushes the electrons in the rods along their length. When the rods are at right angles to this direction, the signal picked up is weak, so reception is poorer.

73 marks

Why microwaves are used for satellite TV and radio waves are used for terrestrial TV?

Solution

Microwaves have a high frequency and a short wavelength. They travel easily through the atmosphere, so the signal can be sent up to a satellite and back down to a dish on the ground (book p. 288).

Radio waves are used for terrestrial TV, which is broadcast from towers on the ground. Their longer wavelengths spread over a wide area around the broadcasting antenna, which covers the land around the station (book p. 287).

83 marks

Can we use x-rays and gamma rays for broadcasting radio and TV signals?

Solution

No. Radio and TV broadcasting uses radio waves, which have wavelengths of more than 1 mm (Table 20.1) and can be produced and received by ordinary antennas and tuners. X-rays and gamma rays have wavelengths below 1 nm, very high energies, and they are ionizing.

Because they ionize matter, they damage living tissue, and broadcasting would expose people to them along the way. They are also not suitable for carrying audio and video signals through the antennas and tuners that broadcasting uses.

93 marks

When you stand outdoors in the sunlight, why can you feel the energy that sunlight carries but not the momentum it carries?

Solution

Each photon carries momentum p=\dfrac{h}{\lambda}=\dfrac{hf}{c}, but this is very small because Planck's constant h is very small and the speed of light c is very large. A visible photon of wavelength 400 nm has a momentum of only about 1.7\times10^{-27}\,\mathrm{kg\,m\,s^{-1}} (question QV.6).

The energy of the photons is turned into heat in your skin, which you feel. The momentum of the photons is transferred to your body too, but the total push is far too small to notice, although radiation pressure is real (book p. 302).

103 marks

A newscaster in the studio speaks to a reporter from some remote place the voice of reporter is sometimes delayed. What causes this delay?

Solution

The reporter's voice is changed into a coded radio or microwave signal, which is sent over a long path to the studio, often through a satellite. Electromagnetic waves travel at a finite speed, c=3\times10^{8}\,\mathrm{m\,s^{-1}}, so the signal takes a measurable time to cover the path.

The encoding and decoding of the signal also take some time. Together, the travel time over the long path and the processing cause the delay that the viewers notice.

113 marks

Is the visible region a major portion of the electromagnetic spectrum? Explain.

Solution

No. The visible region is a very narrow band. It covers wavelengths from about 400 nm to 750 nm, and frequencies from about 4\times10^{14} to 7.5\times10^{14}\,\mathrm{Hz} (Table 20.1).

The whole spectrum extends from radio waves with wavelengths of more than 1 mm down to gamma rays with wavelengths below 10^{-12}\,\mathrm{m}, a range of many orders of magnitude (Figure 20.2). The visible part is only a tiny segment of it.

123 marks

As the light passes through the atmosphere it scatters. Does the same happen with infrared light and radio waves? Explain.

Solution

No, not in the same way. Scattering is strongest for short wavelengths and becomes weaker as the wavelength gets longer. Blue light is scattered much more than red light, which is why the sky is blue.

Infrared light has longer wavelengths than visible light, so it is scattered much less. Radio waves have wavelengths from millimetres to metres, which are far larger than the molecules of the air, so they are scattered very little and pass through the atmosphere almost unchanged (book p. 299).

133 marks

Why the radiation pressure of an electromagnetic wave is greater when it is perfectly reflected as compared to when it is absorbed by the surface?

Solution

Radiation pressure is due to the transfer of momentum from photons to the surface. The size of the change in momentum of each photon decides the force, and the reflected case gives a larger change.

Derivation Radiation pressure on a reflecting and an absorbing surface

Aim. Compare the force on the surface when the photons are absorbed and when they are perfectly reflected.

Momentum change of a photon on an absorbing surface (left) and on a perfectly reflecting surface (right).

Momentum of one photon arriving on the surface, moving along the normal:
p=\dfrac{h}{\lambda}
If the photon is absorbed, its momentum drops from p to zero. The change in momentum of the photon is:
\Delta p_{\text{absorbed}}=p-0=p
If the photon is perfectly reflected, it bounces back along the same line, so its momentum changes from p to -p. The change is:
\Delta p_{\text{reflected}}=p-(-p)=2p
By Newton's second law, the force on the surface equals the rate of change of momentum, which is the same as the rate of change of the photon momentum with the opposite sign. For N photons per second, the force is N times the change in momentum of one photon.
F_{\text{absorbed}}=Np, \qquad F_{\text{reflected}}=2Np
The pressure is the force divided by the area, so the reflected surface gets twice the pressure of the absorbing surface.
P_{\text{reflected}}=2P_{\text{absorbed}}

Result

The radiation pressure on a perfectly reflecting surface is twice the radiation pressure on a perfectly absorbing surface for the same light.

In the reflected case the photon is turned round, so its momentum is reversed. Turning it round needs twice the change of momentum that stopping it needs, so the surface is pushed twice as hard.

143 marks

How are the photons produced?

Solution

Photons are produced when a charged particle loses energy. In an atom, an electron that jumps from a higher energy state to a lower one emits a photon whose energy is equal to the difference of the two states, E=hf (book p. 292).

In hot objects, the thermal motion, vibration and rotation of the atoms and molecules emit infrared photons (book p. 290). Accelerating charged particles also produce electromagnetic waves, which are made of photons.

153 marks

Explain the reason that why clouds appear white in color?

Solution

The water droplets in a cloud are much larger than the wavelengths of visible light. Large particles scatter all the visible colours almost equally, instead of favouring the short wavelengths as the small air molecules do (section 20.4).

Because every colour is scattered in about the same amount and in all directions, the light that reaches the eye is a mixture of all the colours, which appears white.

Long response questionsp. 307–308

Break down the electromagnetic spectrum in terms of wavelength and frequency. Why is the speed of electromagnetic waves in air nearly the same as in vacuum?

Solution

The spectrum in terms of wavelength and frequency

The electromagnetic spectrum is the range of all electromagnetic waves, arranged by frequency or by wavelength. The two orders are reversed: as the frequency increases, the wavelength decreases. Radio waves have the lowest frequencies and the longest wavelengths, and gamma rays have the highest frequencies and the shortest wavelengths.

Table 20.1: Spectrum of electromagnetic waves
Type of radiation	Wavelength range	Frequency range (Hz)
Radio	more than 1\,\mathrm{mm}	less than 3\times10^{11}
Microwave	1\,\mathrm{mm} to 25\,\mu\mathrm{m}	3\times10^{11} to 3\times10^{13}
Infrared	25\,\mu\mathrm{m} to 2.5\,\mu\mathrm{m}	10^{13} to 4\times10^{14}
Visible	750\,\mathrm{nm} to 400\,\mathrm{nm}	4\times10^{14} to 7.5\times10^{14}
Ultraviolet	400\,\mathrm{nm} to 1\,\mathrm{nm}	10^{15} to 10^{17}
X-rays	1\,\mathrm{nm} to 1\,\mathrm{pm}	10^{17} to 10^{20}
Gamma rays	less than 10^{-12}\,\mathrm{m}	10^{20} to 10^{24}

The waves are linked by c=f\lambda, so \lambda=\dfrac{c}{f}. For example, a frequency of 10^{20}\,\mathrm{Hz} gives a wavelength of \dfrac{3\times10^{8}}{10^{20}}=3\times10^{-12}\,\mathrm{m}, which is in the gamma range.

Why the speed in air is almost the speed in vacuum

In a vacuum all electromagnetic waves travel at c=3\times10^{8}\,\mathrm{m\,s^{-1}}. In a material the wave is slowed down by its interaction with the particles of the medium, but in air the molecules are very far apart compared with the wavelength of the light, so the interaction is very weak. The speed in air is therefore nearly the same as in vacuum (book p. 283).

\text{Frequency and wavelength are inversely proportional: } \lambda=\dfrac{c}{f}. \text{ The speed in air is nearly } 3\times10^{8}\,\mathrm{m\,s^{-1}}.

Evaluate the importance of radio waves in communication systems and astronomy compared to other electromagnetic waves. Why are they uniquely suited for this purpose?

Solution

Radio waves in communication

Radio waves have the longest wavelengths and the lowest frequencies in the spectrum (more than 1\,\mathrm{mm} and less than 3\times10^{11}\,\mathrm{Hz}). They carry coded audio and video signals from broadcasting antennas to radios, televisions, mobile phones and wireless networks. Each device has a tuner that picks out the frequency of its station, and its circuits decode the signal (book p. 287).

Radio waves in astronomy

Astronomical objects with changing magnetic fields emit radio waves. Radio astronomy studies these waves from space, which gives information about the composition, structure and motion of celestial bodies (book p. 287). A radio telescope can observe even when clouds, rain or bright sunlight block an optical telescope (book p. 288).

Why radio waves are uniquely suited

Their long wavelength lets them spread over wide areas from a single broadcasting antenna, and they reach receivers over long distances.
They are not absorbed or scattered appreciably by the atmosphere, so clouds and rain do not stop them.
They have very low energy per photon, so they are not ionizing and are safe for everyday use, although very high intensity can heat tissue.
They can be produced by ordinary electric circuits and antennas, and they can be received by simple antennas and tuners.

Feature	Radio waves	Visible light	X-rays
Wavelength	more than 1\,\mathrm{mm}	400 to 750\,\mathrm{nm}	1\,\mathrm{nm} to 1\,\mathrm{pm}
Scattered by the atmosphere	hardly at all	strongly for blue light	hardly at all
Blocked by clouds and rain	no	yes	no
Ionizing	no	no	yes

Evaluation

Radio waves are the best choice for broadcasting and for radio astronomy because they combine long range, freedom from blocking by weather and safety. Their main limitation is that their long wavelength gives poor image detail, so a radio telescope must be much larger than an optical telescope to achieve the same resolution (book p. 288).

\text{Radio waves are uniquely suited because their long wavelength carries signals far, passes through the atmosphere and weather, and is safe.}

Compare the use of microwaves in household applications versus industrial or scientific uses. How do their characteristics make them versatile?

Solution

Household applications

Microwave ovens use microwaves of about 2.45 GHz, with a wavelength of 12.2 cm. The microwaves make the water molecules in food vibrate, and this vibration heats the food evenly and quickly (book p. 289).
Bluetooth uses radio waves and microwaves to link devices such as headphones and speakers. The signal is weak, so it works only over short distances, less than about 9 m (book p. 289).
Wi-Fi and the satellite dishes of home television also use microwaves.

Industrial and scientific uses

Satellite communication sends television and mobile signals from Earth stations up to satellites, which amplify them and send them back to other places on Earth (book p. 288).
Weather monitoring uses the fact that water absorbs microwaves. The weaker the signal reaching the detector, the more rain the microwaves have passed through, so the amount of rain can be estimated (book p. 289).
Radar uses microwaves in some systems, as the book notes in its summary of section 20.2.

Feature	Household use	Industrial or scientific use
Example	microwave oven, Bluetooth, Wi-Fi	satellite links, rainfall monitoring, radar
Distance	short, within a room or a house	long, from ground to satellite and back
Power	low (Bluetooth) or high (oven)	controlled to reach the satellite or detector

How the characteristics make microwaves versatile

High frequency: many cycles per second, so a large amount of information can be sent quickly.
Short wavelength: small antennas and dishes can focus the beam, and the waves pass through the atmosphere.
Low energy per photon: microwaves are safe for people in normal use, although a leaking oven can interfere with other devices.
Absorption by water: the same property heats food in an oven and gives a measure of rainfall in weather stations.

Because the same waves can heat, carry information and measure rain, microwaves are used in very different places, from the kitchen to the satellite.

\text{Microwaves are versatile because they have high frequency, pass through the atmosphere, are low in energy, and are absorbed by water.}

What are infrared radiations and how they can be produced? Write their characteristics?

Solution

What infrared radiation is

Infrared radiation is the part of the spectrum just below the red end of the visible light, which is why it is called infra-red (meaning below red). Its frequencies are about 10^{13} to 4\times10^{14}\,\mathrm{Hz}, and its wavelengths are about 25\,\mu\mathrm{m} to 2.5\,\mu\mathrm{m} (Table 20.1).

How it is produced

Infrared radiation is produced by the thermal motion of atoms and molecules, and by their vibration and rotation (book p. 290). Because of this, every object at a temperature above absolute zero emits infrared, and a hotter object emits more of it. Electric heaters, toasters and infrared lamps are everyday sources, and remote controllers and intruder alarms use infrared beams.

Characteristics

It is invisible to the human eye, but the skin feels it as heat, because it is absorbed by the body.
It is absorbed by molecules, which is why it produces a heating effect and why prolonged exposure can cause burns or overheating.
It travels at the speed of light in vacuum, like all electromagnetic waves.
It is transmitted with little loss through glass optical fibres, so it can carry data over long distances.
It is emitted by warm objects, so it can be detected by thermal imaging and night-vision scopes.
It makes up about half of the solar radiation that reaches Earth (book p. 291).

Application	How infrared is used
Remote controllers	A button press sends a coded infrared beam to the TV.
Intruder alarms	A continuous infrared beam is broken by movement, which triggers the alarm.
Optical fibres	Infrared carries data with little loss in glass.
Thermal imaging	Infrared from warm objects is converted into a visible image.

\text{Infrared is produced by the thermal motion of atoms and molecules. It has wavelengths of } 25\,\mu\mathrm{m} \text{ to } 2.5\,\mu\mathrm{m}, \text{ is felt as heat and travels at } c.

What is visible light? Why is visible light crucial for life on Earth?

Solution

What visible light is

Visible light is the narrow part of the electromagnetic spectrum to which the normal human eye responds. Its wavelengths are about 400 nm to 750 nm, and its frequencies are about 4\times10^{14} to 7.5\times10^{14}\,\mathrm{Hz} (Table 20.1).

Each colour of the visible spectrum has its own wavelength. Red has the longest wavelength and the lowest frequency, and violet has the shortest wavelength and the highest frequency (Figure 20.7). White light is a mixture of all these colours, and a prism separates it into its components.

Why visible light is crucial for life

Vision: light enters the eye, is focused on the retina and is converted into nerve signals that the brain interprets. This is how the appearance, location and movement of objects around us are detected (book p. 292).
Energy from the Sun: about half of the solar radiation reaching Earth is infrared, and most of the rest is visible light (book p. 291). The visible part is the main light by which living things see and move.
Photosynthesis: plants use visible light to make food, which supports almost all other living things and releases oxygen.

Visible light is therefore the part of the spectrum that lets living things sense their surroundings, and it also carries much of the energy that keeps life on Earth going.

\text{Visible light is the narrow part of the spectrum, from } 400 \text{ to } 750\,\mathrm{nm}, \text{ that the eye responds to. It is crucial for vision and for the energy that supports life.}

Ultraviolet radiation has both beneficial and harmful effects. Discuss its characteristics, applications, and risks. How can we balance its advantages (e.g., sterilization) with its dangers (e.g., skin cancer)?

Solution

Characteristics

Ultraviolet means 'above violet'. Its wavelengths are about 400 nm to 1 nm, and its frequencies are about 10^{15} to 10^{17}\,\mathrm{Hz} (Table 20.1).
It is invisible to the human eye, and it has more energy per photon than visible light.
Sunlight's ultraviolet is divided into three regions: UV-A (320 nm to 400 nm), UV-B (290 nm to 320 nm) and UV-C (220 nm to 290 nm). The ozone layer absorbs most of UV-B and all of UV-C, so about 99% of the solar ultraviolet reaching the surface is UV-A (book p. 293).

Applications

Security marking: an ultraviolet watermark is seen only under ultraviolet light, because its ink fluoresces. It is used on credit cards, driving licences and passports.
Detecting counterfeit bank notes: multicolour fibres and images are visible only under ultraviolet light (book p. 293).
Sterilizing water: ultraviolet kills bacteria and viruses in water, without chemicals and without the danger of overdosing.

Risks

Long-term exposure damages skin cells, causing sunburn and increasing the risk of skin cancer.
It can harm the eyes, leading to cataracts and other vision problems.
It can weaken the immune system (Table 20.2, book p. 297).

Balancing the advantages and dangers

The benefit and the harm depend on the dose and on the situation. Sterilization uses ultraviolet inside closed equipment, where people are not exposed, so the water is cleaned and no one is harmed. On the skin, the risk is controlled by limiting the time spent in strong sunlight, by covering the skin and using shade, and by protecting the eyes. The ozone layer already absorbs most of the most harmful ultraviolet, which is why it is important to protect it.

\text{Ultraviolet is useful for security marking, counterfeit detection and sterilization, and harmful to skin and eyes. Its use is balanced by controlling the dose and shielding people.}

How x-rays are different from the rest of electromagnetic radiations? Discuss in details.

Solution

Where X-rays lie in the spectrum

X-rays have wavelengths of about 1 nm to 1 pm and frequencies of about 10^{17} to 10^{20}\,\mathrm{Hz} (Table 20.1). They overlap with ultraviolet at their long-wavelength end and with gamma rays at their high-energy end (book p. 294).

Energy

The energy of an X-ray photon is E=\dfrac{hc}{\lambda}. At 1\,\mathrm{nm} the energy is \dfrac{6.63\times10^{-34}\times3\times10^{8}}{10^{-9}}\approx2.0\times10^{-16}\,\mathrm{J}, and at 1\,\mathrm{pm} it is about 2.0\times10^{-13}\,\mathrm{J}. This is far more energy per photon than visible light, which has about 5\times10^{-19}\,\mathrm{J} at 400\,\mathrm{nm}.

Penetration and imaging

X-rays penetrate matter, and how far they go depends on the density of the material. Dense material, such as bone, absorbs more X-rays than soft tissue, so an X-ray photograph shows the bones clearly (Figure 20.10 a). The same property shows cracks in metal, because a crack is thinner and lets more X-rays through to the film (Figure 20.11).

Ionizing effect

X-rays are ionizing: their photons have enough energy to remove electrons from atoms. This can damage cells and DNA, which is why exposure must be limited for imaging, and why high doses can kill cancer cells in radiation therapy.

Applications

Medical imaging (radiology): low doses show the inside of the body.
Security scanners: very low energy X-rays reveal metal objects under clothing.
Cancer treatment: high doses damage the DNA of cancer cells, which stop dividing or die (book p. 294).
Radiographic testing: X-rays detect cracks and flaws in manufactured metal parts.

Radiation	Wavelength	Relative energy per photon	Ionizing
Ultraviolet	400 to 1\,\mathrm{nm}	lower than X-rays	yes, at the high-energy end
X-rays	1\,\mathrm{nm} to 1\,\mathrm{pm}	high	yes
Gamma rays	less than 10^{-12}\,\mathrm{m}	highest	yes

So X-rays are different because they carry a very large energy per photon, which makes them more penetrating than ultraviolet, more ionizing than visible light, and useful for looking inside solid materials.

\text{X-rays have wavelengths of } 1\,\mathrm{nm} \text{ to } 1\,\mathrm{pm}, \text{ very high photon energy, strong penetration and an ionizing effect. They are used for imaging, scanning and cancer treatment.}

Gamma rays are the most energetic. How do their properties make them suitable for applications in nuclear science and astrophysics?

Solution

Properties of gamma rays

They have the shortest wavelengths (less than 10^{-12}\,\mathrm{m}) and the highest frequencies (10^{20} to 10^{24}\,\mathrm{Hz}), so each photon has the highest energy (Table 20.1).
They are a type of nuclear radiation, emitted from the nucleus of a heavy atom (book p. 295).
They are highly penetrating and can pass through many materials that stop other radiations.
They are ionizing, so they can cause cell mutations and DNA damage.

Nuclear science

Because gamma rays come from nuclei, radioactive materials that emit them are used in nuclear medicine. Gamma cameras and PET scanners detect the gamma rays from these materials to show the functions of organs (Figure 20.12). Their penetration is also used to kill cancer cells in radiation therapy, and Cobalt-60, a gamma source, sterilizes medical equipment and food without making the food radioactive (Figure 20.13).

In engineering, gamma rays test welds and structures for hidden flaws (radiographic testing, Figure 20.14). Their greater penetrating power than X-rays makes them better for testing hard and thick materials.

Astrophysics

Gamma rays are produced in very energetic events, far beyond the Earth. Because their photons carry so much energy, they bring information about these events to the detectors on Earth, which is why gamma rays are an important part of studying the universe alongside radio waves and visible light (book p. 282).

Their high energy also means they must be handled with strict safety, because they can damage living cells. This is why they are shielded in both laboratories and hospitals.

\text{Gamma rays are highly penetrating and very energetic, so they are used to image and treat in nuclear medicine, to sterilize, to test materials, and to study energetic events in space.}

What potential hazards are associated with various parts of the electromagnetic spectrum? Compare the risks posed by ionizing and non-ionizing radiation, and evaluate why ionizing radiation is more dangerous in terms of its impact on living tissues and biological systems.

Solution

Hazards of each part of the spectrum

Table 20.2: Hazards of the electromagnetic spectrum
EM wave	Potential hazards (Table 20.2)
Radio waves	Tissue heating, burns; interference with pacemakers and hearing aids.
Microwaves	Damage to sensitive organs such as the eyes (cataracts); burns; interference with implanted devices.
Infrared	Thermal damage, burns and overheating; eye damage and skin ageing with high intensity.
Visible light	Eye strain and retinal damage with intense light; disrupted biological rhythms from bright artificial light.
Ultraviolet	Sunburn, skin cancer, cataracts, weakened immune system.
X-rays	Ionization, tissue damage, cell mutations, cancer, burns, acute radiation sickness.
Gamma rays	Cell mutations, cancer or death; acute radiation sickness from prolonged exposure.

Ionizing and non-ionizing radiation

Ionizing radiation has enough energy per photon to remove tightly bound electrons from atoms, which creates ions. Ultraviolet at its high-energy end, X-rays and gamma rays are ionizing. Radio waves, microwaves, infrared and visible light are non-ionizing, because their photons lack the energy to remove electrons. Their main effect on tissue is heating (book p. 299).

Feature	Non-ionizing radiation	Ionizing radiation
Examples	radio, microwave, infrared, visible	X-rays, gamma rays, high-energy ultraviolet
Effect on atoms	makes molecules vibrate or rotate	removes electrons, creating ions
Main harm	heating and thermal burns	cell and DNA damage, mutations, cancer
Typical risk	mostly at high intensity	even at low doses, with long-term effects

Why ionizing radiation is more dangerous

Its photons carry enough energy to break chemical bonds in the molecules of living cells, so it can change the molecules directly rather than only heating them.
It can damage DNA. A change in the DNA can cause a cell to mutate, and the mutated cell may grow out of control and become cancerous.
The damage is cumulative and can be hidden, because the cell may not die immediately, so the harm can appear years later.
Non-ionizing radiation mainly heats tissue, which the body can remove or repair when the exposure is small, while the body cannot repair DNA damage in the same way.

Evaluation: non-ionizing radiation becomes dangerous at high intensity, and the risks depend on dose and exposure time. Even so, ionizing radiation is more dangerous because it changes the molecules of living tissue, which can cause mutations, cancer and radiation sickness, and this is why exposure to X-rays and gamma rays must be shielded and limited (book p. 299).

\text{Ionizing radiation (X-rays, gamma rays, high-energy UV) removes electrons and damages DNA, so it is more dangerous than non-ionizing radiation, which mainly heats tissue.}

Explain the scattering of light from the atmosphere and why the color of the sky appears different at different times in a day?

Solution

Scattering of light from the atmosphere

Scattering is the reflection of light in different directions by particles in its path. Sunlight contains all the visible colours. The molecules of the air, mainly oxygen and nitrogen, are much smaller than the wavelengths of visible light. Their scattering is strongest for the short wavelengths, so blue, indigo and violet light are scattered much more than red, orange and yellow (section 20.4).

The blue sky during the day

When the Sun is high, the sunlight passes through a relatively short layer of air. The blue light is scattered in all directions by the molecules, and some of it reaches your eyes from every direction of the sky. The sky therefore looks blue (book p. 299).

The reddish sky at sunrise and sunset

When the Sun is low on the horizon, its light travels a much longer path through the atmosphere. On this long path most of the blue and violet light is scattered out of the beam, away from your eyes. The longer wavelengths, red and orange, pass through and reach you, so the sun and the sky look reddish (book p. 300).

Sunlight at midday and at sunset (atmosphere thickness exaggerated, not to scale). The path through the atmosphere is much longer at sunset, so more blue light is scattered out of the beam before it reaches the observer.

Why the colour changes during the day

Midday: short path, so little scattering of blue light out of the beam; the sky is blue.
Sunrise and sunset: long path, so the blue is removed and the red and orange remain; the sky looks red or orange.
Between these times: the path length changes gradually, so the colour changes from blue through lighter shades to reddish.

\text{Air molecules scatter short wavelengths most. The sky is blue when the path through air is short, and reddish at sunrise and sunset when the path is long.}

Compare radiation pressure propulsion with chemical propulsion in space exploration. Which is more viable for long-term missions, and why?

Solution

Radiation pressure propulsion

A solar sail is pushed by the radiation pressure of sunlight. Photons carry momentum p=\dfrac{hf}{c}, and when they are reflected from a large, light surface they transfer twice their momentum to it, giving a small but continuous thrust (book p. 303). The sail needs no fuel: the thrust comes from sunlight, and it lasts as long as the sail receives light.

Chemical propulsion

A chemical rocket burns fuel and throws out the exhaust gases at high speed. The push of the exhaust gives a large thrust for a short time, so the spacecraft can accelerate quickly. However, the rocket must carry all the fuel it will use, and the fuel mass limits how long and how far it can go.

Comparison

Feature	Radiation pressure (solar sail)	Chemical propulsion (rocket)
Source of thrust	momentum of sunlight photons	exhaust gases from burning fuel
Fuel needed	none	large amount, carried on board
Size of thrust	very small	very large
Time of thrust	continuous while sunlight reaches the sail	short bursts, until the fuel runs out
Acceleration	gradual	rapid
Limit on the mission	weaker further from the Sun, but no fuel supply to run out	fuel mass, which is used up

Which is more viable for long-term missions

For a long-term mission, radiation pressure propulsion is the more viable choice, because it does not run out of fuel. Its thrust is small, but it acts continuously for a long time, so the spacecraft can gain a large speed over months or years. Chemical propulsion is better for launch and for quick changes of course, but its fuel limits how long a mission can last.

\text{Solar sails are more viable for long-term missions because they need no fuel and give continuous thrust, although the thrust is small.}

Numerical response questionsp. 308

Calculate the wavelength of an electromagnetic wave with a frequency of 5.0\times10^{14}\,\mathrm{Hz}.

Solution

Given

f = 5.0\times10^{14}\,\mathrm{Hz}
c = 3\times10^{8}\,\mathrm{m\,s^{-1}} speed of light

To find

\lambda (wavelength)

The wave speed, frequency and wavelength are related by c=f\lambda. Make \lambda the subject.
\lambda=\dfrac{c}{f}
Substitute the values.
\lambda=\dfrac{3\times10^{8}}{5.0\times10^{14}}
Evaluate.
\lambda=6.0\times10^{-7}\,\mathrm{m}=600\,\mathrm{nm}

\lambda=6.0\times10^{-7}\,\mathrm{m}=600\,\mathrm{nm}

Book answer. 600 nm

A gamma ray photon has a frequency of 2.0\times10^{20}\,\mathrm{Hz}. Calculate its energy.

Solution

Given

f = 2.0\times10^{20}\,\mathrm{Hz}
h = 6.63\times10^{-34}\,\mathrm{J\,s} Planck's constant

To find

E (energy of the photon)

The energy of a photon is given by E=hf.
E=hf
Substitute the values.
E=6.63\times10^{-34}\times2.0\times10^{20}
Evaluate.
E=1.33\times10^{-13}\,\mathrm{J}

E=1.33\times10^{-13}\,\mathrm{J}

Book answer. 1.33 × 10^-13 J

A laser emits light with a frequency of 4.0\times10^{14}\,\mathrm{Hz}. What is the momentum of each photon?

Solution

Given

f = 4.0\times10^{14}\,\mathrm{Hz}
h = 6.63\times10^{-34}\,\mathrm{J\,s} Planck's constant
c = 3\times10^{8}\,\mathrm{m\,s^{-1}} speed of light

To find

p (momentum of one photon)

The momentum of a photon is p=\dfrac{h}{\lambda}=\dfrac{hf}{c}, which avoids needing the wavelength.
p=\dfrac{hf}{c}
Substitute the values.
p=\dfrac{6.63\times10^{-34}\times4.0\times10^{14}}{3\times10^{8}}
Evaluate.
p=8.84\times10^{-28}\,\mathrm{kg\,m\,s^{-1}}

p=8.84\times10^{-28}\,\mathrm{kg\,m\,s^{-1}}

Book answer. 8.84 × 10^-28 kg m/s

What will be the momentum and energy of the photon from the radio waves with wavelength 100 mm?

Solution

Given

\lambda = 100\,\mathrm{mm}=0.100\,\mathrm{m}
h = 6.63\times10^{-34}\,\mathrm{J\,s} Planck's constant
c = 3\times10^{8}\,\mathrm{m\,s^{-1}} speed of light

To find

p (momentum)
E (energy, in joules and in electron-volts)

Momentum of the photon:
p=\dfrac{h}{\lambda}=\dfrac{6.63\times10^{-34}}{0.100}
Momentum evaluated:
p=6.63\times10^{-33}\,\mathrm{kg\,m\,s^{-1}}
Energy of the photon, from E=\dfrac{hc}{\lambda}:
E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{0.100}=1.99\times10^{-24}\,\mathrm{J}
Convert to electron-volts by dividing by 1.6\times10^{-19}\,\mathrm{J} per eV:
E=\dfrac{1.99\times10^{-24}}{1.6\times10^{-19}}=1.24\times10^{-5}\,\mathrm{eV}

p=6.63\times10^{-33}\,\mathrm{kg\,m\,s^{-1}},\quad E=1.24\times10^{-5}\,\mathrm{eV}

Book answer. 6.63 × 10^-33 kg m/s, 1.24 × 10^-5 eV

The surface of Earth is hit by many radiations of different frequency and wavelength at all the times mainly during the day times. The maximum energy of radiations received in your room during a sunny day is 4.7\,\mathrm{eV}. Find the frequency and wavelength of this radiation also tell the type of radiation.

Solution

Given

E = 4.7\,\mathrm{eV}=4.7\times1.6\times10^{-19}\,\mathrm{J}=7.52\times10^{-19}\,\mathrm{J}
h = 6.63\times10^{-34}\,\mathrm{J\,s} Planck's constant
c = 3\times10^{8}\,\mathrm{m\,s^{-1}} speed of light

To find

f (frequency)
\lambda (wavelength)
type of radiation

Frequency from E=hf:
f=\dfrac{E}{h}=\dfrac{7.52\times10^{-19}}{6.63\times10^{-34}}
Evaluate the frequency.
f=1.13\times10^{15}\,\mathrm{Hz}
Wavelength from c=f\lambda:
\lambda=\dfrac{c}{f}=\dfrac{3\times10^{8}}{1.13\times10^{15}}
Evaluate the wavelength.
\lambda=2.65\times10^{-7}\,\mathrm{m}=265\,\mathrm{nm}
Compare with Table 20.1. The ultraviolet range is 400\,\mathrm{nm} to 1\,\mathrm{nm} (frequency 10^{15} to 10^{17}\,\mathrm{Hz}), so this radiation is ultraviolet. It lies in the UV-C band (220 nm to 290 nm).

f=1.13\times10^{15}\,\mathrm{Hz},\quad \lambda=265\,\mathrm{nm},\quad \text{ultraviolet}

Book answer. 1.13 × 10^15 Hz, 265 nm, UV

Find the momentum and energy of the photon from the visible light of the smallest wavelength.

Solution

Given

\lambda_{\min} = 400\,\mathrm{nm}=4.0\times10^{-7}\,\mathrm{m} smallest visible wavelength in Table 20.1
h = 6.63\times10^{-34}\,\mathrm{J\,s} Planck's constant
c = 3\times10^{8}\,\mathrm{m\,s^{-1}} speed of light

To find

p (momentum)
E (energy)

Energy of the photon, from E=\dfrac{hc}{\lambda}:
E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{4.0\times10^{-7}}=4.97\times10^{-19}\,\mathrm{J}
Convert to electron-volts:
E=\dfrac{4.97\times10^{-19}}{1.6\times10^{-19}}=3.11\,\mathrm{eV}
Momentum of the photon, from p=\dfrac{h}{\lambda}:
p=\dfrac{6.63\times10^{-34}}{4.0\times10^{-7}}=1.66\times10^{-27}\,\mathrm{kg\,m\,s^{-1}}

E=3.11\,\mathrm{eV}\,(4.97\times10^{-19}\,\mathrm{J}),\quad p=1.66\times10^{-27}\,\mathrm{kg\,m\,s^{-1}}

Book answer. 3.27 eV, 1.74 × 10^-27 kg m/s

Note. The book's answer is not reproduced by Table 20.1. Its values (3.27 eV and 1.74 × 10^-27 kg m/s) correspond to a wavelength of about 380 nm, which is below the 400 nm limit that the book gives for visible light in Table 20.1 and Example 20.1. Using 400 nm, the energy is 3.11 eV and the momentum is 1.66 × 10^-27 kg m/s.

FBISE 2026 questions

Questions from the FBISE SSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

FBISE SSC-II 2026 Annual I, Set B, Q.2 (viii)3 marks

Name any six regions of the electromagnetic spectrum in order of increasing wavelength.

Solution

The regions of the electromagnetic spectrum in order of increasing wavelength are:

Gamma rays
X-rays
Ultraviolet
Visible light
Infrared
Microwaves
Radio waves

Table 20.1: wavelength of each region
Region	Wavelength range
Gamma rays	<10^{-12}\,\mathrm{m}
X-rays	1\,\mathrm{nm} to 1\,\mathrm{pm}
Ultraviolet	400\,\mathrm{nm} to 1\,\mathrm{nm}
Visible light	750\,\mathrm{nm} to 400\,\mathrm{nm}
Infrared	25\,\mu\mathrm{m} to 2.5\,\mu\mathrm{m}
Microwaves	1\,\mathrm{mm} to 25\,\mu\mathrm{m}
Radio waves	>1\,\mathrm{mm}

Gamma rays have the shortest wavelength (below 10^{-12}\,\mathrm{m}) and radio waves the longest (above 1\,\mathrm{mm}). Any six of the seven regions, in this order, earn the marks.

FBISE SSC-II 2026 Annual I, Set B, Q.2 (xi) OR1+2 = 3 marks

What is the color of sky on a clear day? How does scattering of light help in determining the color of the sky? Explain briefly.

Solution

Colour of the sky (1 mark)

On a clear day the sky is blue.

How scattering gives this colour (2 marks)

Scattering is the sending of light in different directions by the particles in its path. Sunlight contains all the colours, from red (longest wavelength) to violet (shortest). The air is mostly oxygen and nitrogen molecules, which are very small and comparable in size to the short wavelengths of light. Shorter wavelengths are scattered more than longer ones, so the molecules scatter blue (with indigo and violet) strongly in all directions.

The longer wavelengths (red, orange, yellow) are not scattered much and pass straight through. Wherever you look in the sky, some of the scattered blue light comes to your eyes, so the whole sky looks blue (Figure 20.15).

Blue light is scattered in all directions by air molecules, while red light passes straight through.

FBISE SSC-II 2026 Annual I, Set D, Q.2 (viii)3 marks

Enlist any six regions of electromagnetic spectrum in correct order (Higher to lower) of their wave lengths.

Solution

Written from the higher to the lower wavelength, the order is:

Radio waves
Microwaves
Infrared
Visible light
Ultraviolet
X-rays
Gamma rays

Table 20.1: wavelength of each region
Region	Wavelength range
Radio waves	>1\,\mathrm{mm}
Microwaves	1\,\mathrm{mm} to 25\,\mu\mathrm{m}
Infrared	25\,\mu\mathrm{m} to 2.5\,\mu\mathrm{m}
Visible light	750\,\mathrm{nm} to 400\,\mathrm{nm}
Ultraviolet	400\,\mathrm{nm} to 1\,\mathrm{nm}
X-rays	1\,\mathrm{nm} to 1\,\mathrm{pm}
Gamma rays	<10^{-12}\,\mathrm{m}

Radio waves have the longest wavelength (above 1\,\mathrm{mm}) and gamma rays the shortest (below 10^{-12}\,\mathrm{m}). Any six of the seven regions, in this order, earn the marks.

FBISE SSC-II 2026 Annual I, Set D, Q.2 (xi) OR2+1 = 3 marks

Briefly state the particle nature of light. Also mention its one application.

Solution

Particle nature of light (2 marks)

Light also behaves as a stream of tiny particles called photons. A photon is a massless packet (quantum) of light energy. It carries energy and momentum but no charge, and in a vacuum it always travels at the speed of light. Max Planck introduced this idea in the early 20th century, calling the packets quanta of light.

E=hf=\dfrac{hc}{\lambda}\qquad p=\dfrac{h}{\lambda}=\dfrac{hf}{c}

Here h=6.63\times10^{-34}\,\mathrm{J\,s} is Planck's constant. Light shows both wave properties (interference, diffraction, polarisation) and particle properties, which is its dual nature.

One application (1 mark)

Radiation pressure and solar sails. Because each photon has momentum p=h/\lambda, light that strikes a surface exerts a very small force on it. A large, light, mirror-like solar sail in space is pushed by the photons of sunlight and can move a spacecraft without fuel, which suits long deep-space missions.

FBISE SSC-II 2026 Annual I, Set M, Q.2 (viii)3 marks

Write the correct order of electromagnetic waves from radio waves to gamma rays.

Solution

Going from radio waves to gamma rays, the frequency and the energy increase while the wavelength decreases. The correct order is:

Radio waves
Microwaves
Infrared
Visible light
Ultraviolet
X-rays
Gamma rays

Table 20.1: wavelength of each region
Region	Wavelength range
Radio waves	>1\,\mathrm{mm}
Microwaves	1\,\mathrm{mm} to 25\,\mu\mathrm{m}
Infrared	25\,\mu\mathrm{m} to 2.5\,\mu\mathrm{m}
Visible light	750\,\mathrm{nm} to 400\,\mathrm{nm}
Ultraviolet	400\,\mathrm{nm} to 1\,\mathrm{nm}
X-rays	1\,\mathrm{nm} to 1\,\mathrm{pm}
Gamma rays	<10^{-12}\,\mathrm{m}

Radio waves have the lowest frequency and gamma rays the highest. Any six of the seven regions, in this order, earn the marks.

FBISE SSC-II 2026 Annual I, Set M, Q.2 (xi) OR1+2 = 3 marks

Why does the color of the sky change during sunset? Name the phenomenon and explain briefly.

Solution

Phenomenon (1 mark)

The change of colour is due to the scattering of light by the molecules of the atmosphere (shorter wavelengths are scattered more than longer ones).

Explanation (2 marks)

At sunset the Sun is low in the sky, so its light has to travel a much longer path through the atmosphere than at noon (Figure 20.16). On this long path it meets many more air molecules. The short wavelengths (blue and violet) are scattered more, so they are scattered out of the beam and do not reach the observer. The long wavelengths, red and orange, are scattered least, so they travel the long distance and reach the eyes. The Sun and the sky near it therefore look red or orange. The same happens at sunrise.

Figure 20.16: sunlight at midday and at sunset (atmosphere thickness exaggerated, not to scale). The path through the air is much longer at sunset.

Exam focus

State the regions of the electromagnetic spectrum in order of wavelength or frequency, with their ranges. (short)
Q.2 (viii) in all three 2026 sets (B: increasing wavelength, D: higher to lower wavelength, M: radio waves to gamma rays); also comprehensive question 1 and MCQs 3, 8 and 9 of the book.
Go to answer
Explain why the sky is blue by day and red at sunrise and sunset, naming the scattering of light. (short)
Q.2 (xi) OR in 2026 Set B (blue sky on a clear day) and Set M (colour at sunset); also comprehensive question 10 and crq 1.
Go to answer
Why does the sky look red at sunset? The long path through the atmosphere scatters the blue away. (short)
2026 Set M Q.2 (xi) OR; a standard one-diagram answer.
Go to answer
State the particle nature of light (photon, E=hf, p=h/\lambda) and give one application. (short)
2026 Set D Q.2 (xi) OR; the same ideas drive the book numericals 2 to 6.
Go to answer
Numericals on photon energy, momentum and wavelength using c=f\lambda, E=hf and p=h/\lambda. (numerical)
No numerical from this unit appeared in the 2026 papers, but the book gives six numericals and the formula list on the paper does not print E=hf, so a formula-based question is a real possibility.
Go to answer
Uses and hazards of each region of the spectrum (radio, microwave, infrared, visible, UV, X-rays, gamma rays). (long)
Book comprehensive questions 2 to 9; no long question from this unit was set in 2026, but 3-mark questions on applications are likely.
Go to answer
Why microwaves are used for satellite TV and radio waves for terrestrial TV; why clouds look white. (short)
Book short questions 7 and 15; short-answer style that matches the 3-mark Q.2 parts.
Go to answer
Radiation pressure and solar sails: why total reflection gives a greater pressure than absorption. (short)
Book short question 13 and MCQs 13 to 15 on the last sections of the unit; not asked in 2026.
Go to answer