1. Coursebook A Level Physics

Chapter 13: Waves

End-of-chapter questions [3]

1 Figure 13.17 shows the screen of an oscilloscope. The time-base of the oscilloscope is set at 500 μs div−1.
Calculate the time period of the signal and hence its frequency.

Figure 13.17 For End-of-chapter Question 1. [2]
[1]
2 a State two main properties of electromagnetic waves. [1] 189
b State one major difference between microwaves and radio waves. [1]
c i Estimate the wavelength in metres of X-rays.
ii Use your answer to i to determine the frequency of the X-rays. [1]
[4]
3 A student is sitting on the beach, observing a power boat moving at speed on the sea. The boat has a [1]
siren emitting a constant sound of frequency 420Hz.
The boat moves around in a circular path with a speed of 25ms−1. The student notices that the pitch [1]
of the siren changes with a regular pattern. [1]
a Explain why the pitch of the siren changes, as observed by the student. [2]
b Determine the maximum and minimum frequencies that the student will hear. [2]
c At which point in the boat’s motion will the student hear the most high-pitched note? [2]
(Speed of sound in air = 330ms−1)

4 Figure 13.18 shows some air particles as a sound wave passes.

P

Figure 13.18 For End-of-chapter Question 4.

a On a copy of the diagram, mark:
i a region of the wave which shows a compression – label it C
ii a region of the wave which shows a rarefaction – label it R.

b Describe how the particle labelled P moves as the wave passes.
c The sound wave has a frequency of 240Hz. Explain, in terms of the movement of an individual particle,

what this means.
d The wave speed of the sound is 320ms−1. Calculate the wavelength of the wave.

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5 The diagram represents a wave on the surface of water at one instant of time. The speed of the wave is 7.0cms–1.
Calculate the amplitude, wavelength and time period for one oscillation.

12 mm

20 mm

9.0 cm

Figure 13.19 For End-of-chapter Question 5.

6 a Light is referred to as a type of electromagnetic wave. Explain what is meant by the term
electromagnetic wave. [2]

b i Two stars emit radiation with the same power. Star A is twice as far from the Earth as star B. [2]
Explain how the intensities of the radiations compare to an observer on the Earth. [2]

ii State how the amplitudes of the two signals received by the observer compare. [2]

190 c The main signal from star A is in the ultraviolet region of the spectrum with a wavelength
of 7.5×108m. Calculate the wavelength of this radiation.

7 Figure 13.20 shows a loudspeaker producing a sound and a microphone connected to a
cathode-ray oscilloscope.

loudspeaker microphone

Figure 13.20 For End-of-chapter Question 7.

a Sound is described as a longitudinal wave. Explain how the trace on the oscilloscope relates [3]
to the movements of the air particles in the sound wave. [2]
[2]
b The time-base on the oscilloscope is set at 5msdiv−1. Calculate the frequency of the wave.
c The wavelength of the wave is found to be 1.98m. Calculate the speed of the wave.

Chapter 13: Waves

8 Figure 13.21 shows the traces of two sound waves displayed on an oscilloscope screen.

Figure 13.21 For End-of-chapter Question 8. [2]
[3]
a State and explain how the wavelengths of the two waves compare. [2]
b Explain the meaning of the term phase difference. Illustrate your answer by estimating the [2]

phase difference between the two waves shown on the screen. 191
c Calculate the ratio of the intensities of the two waves. [3]

9 a State what is meant by the Doppler effect. [2]
b A star is moving away from the Earth at a speed of 6.4 × 105ms−1. Light of frequency 6.500×1014Hz [1]
emitted by the star is received on Earth. Calculate the change in frequency caused by the
Doppler effect. Give your answer to 2 significant figures. [1]
c When looking at distant stars and galaxies, the further a star or galaxy is from Earth, the [3]
more the light emitted from the star is shifted towards the red end of the spectrum. [3]
i State and explain whether the stars and galaxies are moving towards or away from the Earth. [1]
ii Explain what the greater shift in frequency for stars that are further away implies.

10 The Doppler effect can be used to measure the speed of blood. Ultrasound, which is sound of high
frequency, is passed from a transmitter into the body, where it reflects off particles in the blood.
The shift in frequency is measured by a stationary detector, placed outside the body and close
to the transmitter.
In one patient, particles in the blood are moving at a speed of 30cms−1 in a direction directly
away from the transmitter. The speed of ultrasound in the body is 1500cms−1.

This situation is partly modelled by considering the particles to be emitting sound of frequency
4.000MHz as they move away from the detector. This sound passes to the detector outside the
body and the frequency measured by the detector is not 4.000MHz.
a i State whether the frequency received by the stationary detector is higher or lower than the

frequency emitted by the moving particles.
ii Explain your answer to i.
b Calculate the difference between the frequency emitted by the moving particles and the frequency
measured by the detector.
c Suggest why there is also a frequency difference between the sound received by the particles and
the sound emitted by the transmitter.

192

Chapter 14:
Superposition of
waves

Learning outcomes

You should be able to:

■■ explain and use the principle of superposition of waves
■■ explain experiments that show diffraction and

interference
■■ solve problems involving two-slit and multiple-slit

interference

Chapter 14: Superposition of waves

Combining waves

Light travels as waves and can produce beautiful,
natural effects such as the iridescent colours of a
butterfly’s wing (Figure 14.1). However, these colours
do not come from pigments in the wing. Instead, they
arise when light waves, scattered from different points
on the wing, meet in your eye and combine to produce
the colours that we see.

Figure 14.1  The iridescent colours on a butterfly’s wing
demonstrate the beauty of nature. In this chapter we will
study the effect known as interference, which leads to the
production of these glorious colours.

The principle of superposition very different from the behaviour of particles. Two bullets 193
of waves meeting in mid-air would ricochet off one another in a
very un-wave-like way. If we look carefully at how two sets
In Chapter 13, we studied the production of waves and the of waves interact when they meet, we find some surprising
difference between longitudinal and transverse waves. In results.
this chapter we are going to consider what happens when
two or more waves meet at a point in space and combine When two waves meet they combine, with the
together (Figure 14.2). displacements of the two waves adding together. Figure
14.3 shows the displacement–distance graphs for two
So what happens when two waves arrive together at sinusoidal waves (blue and green) of different wavelengths.
the same place? We can answer this from our everyday It also shows the resultant wave (red), which comes from
experience. What happens when the beams of light waves combining these two. How do we find this resultant
from two torches cross over? They pass straight through displacement shown in red?
one another. Similarly, sound waves pass through one
another, apparently without affecting each other. This is Consider position A. Here the displacement of both
waves is zero, and so the resultant must also be zero.

At position B, both waves have positive displacement. The
resultant displacement is found by adding these together.

Displacement

0 Distance
AB

C

Figure 14.2  Here we see ripples produced when drops Figure 14.3  Adding two waves by the principle of
of water fall into a swimming pool. The ripples overlap to superposition – the red line is the resultant wave.
produce a complex pattern of crests and troughs.

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At position C, the displacement of one wave is positive Diffraction of waves
while the other is negative. The resultant displacement
lies between the two displacements. In fact, the resultant You should be aware that all waves (such as sound and light)
displacement is the algebraic sum of the displacements of can be reflected and refracted. Another wave phenomenon
waves A and B; that is, their sum, taking account of their that applies to all waves is that they can be diffracted.
signs (positive or negative). Diffraction is the spreading of a wave as it passes through a
gap or around an edge. It is easy to observe and investigate
We can work our way along the distance axis in diffraction effects using water waves, as shown in Box 14.1.
this way, calculating the resultant of the two waves by

algebraically adding them up at intervals. Notice that, for Diffraction of sound and light
these two waves, the resultant wave is a rather complex
wave with dips and bumps along its length. Diffraction effects are greatest when waves pass through a
gap with a width roughly equal to their wavelength. This
The idea that we can find the resultant of two is useful in explaining why we can observe diffraction
waves which meet at a point simply by adding up the readily for some waves, but not for others. For example,
displacements at each point is called the principle of sound waves in the audible range have wavelengths from a
superposition of waves. This principle can be applied to few millimetres to a few metres. Thus we might expect to
more than two waves and also to all types of waves. A observe diffraction effects for sound in our environment.
statement of the principle of superposition is shown below: Sounds, for example, diffract as they pass through

When two or more waves meet at a point, the resultant doorways. The width of a doorway is comparable to the
displacement is the algebraic sum of the displacements
of the individual waves. wavelength of a sound and so a noise in one room spreads
out into the next room.

Visible light has much shorter wavelengths (about
5 × 10−7 m). It is not diffracted noticeably by doorways

194 because the width of the gap is a million times larger
than the wavelength of light. However, we can observe
QUESTION diffraction of light by passing it through a very narrow

1 On graph paper, draw two ‘triangular’ waves like slit or a small hole. When laser light is directed onto a
those shown in Figure 14.4. (These are easier to slit whose width is comparable to the wavelength of the
work with than sinusoidal waves.) One should incident light, it spreads out into the space beyond to form
have wavelength 8 cm and amplitude 2 cm; the a smear on the screen (Figure 14.5). An adjustable slit
other wavelength 16 cm and amplitude 3 cm. allows you to see the effect of gradually narrowing the gap.
Use the principle of superposition of waves to
determine the resultant displacement at suitable You can see the effects of diffraction for yourself by

points along the waves, and draw the complete making a narrow slit with your two thumbs and looking
resultant wave.
through the slit at a distant light source (Figure 14.8). By
gently pressing your thumbs together to narrow the gap

between them, you can see the effect of narrowing the slit.

Displacement 0 Distance

Figure 14.4  Two triangular waves – for Question 1.

Figure 14.5  Light is diffracted as it passes through a slit.

Chapter 14: Superposition of waves

BOX 14.1: Observing diffraction in a ripple tank they arrive at the gap, however, they pass through and
A ripple tank can be used to show diffraction. Plane spread out into the space beyond. It is this spreading
waves are generated using a vibrating bar, and move out of waves as they travel through a gap (or past the
towards a gap in a barrier (Figure 14.6). Where the edge of a barrier) that is called diffraction.
ripples strike the barrier, they are reflected back. Where
The extent to which ripples are diffracted depends
Figure 14.6 Ripples, initially straight, spread out into the on the width of the gap. This is illustrated in Figure 14.6.
space beyond the gap in the barrier. The lines in this diagram show the wavefronts. It is as
if we are looking down on the ripples from above, and
ab drawing lines to represent the tops of the ripples at
some instant in time. The separation between adjacent
wavefronts is equal to the wavelength λ of the ripples.

When the waves encounter a gap in a barrier, the
amount of diffraction depends on the width of the gap.
There is hardly any noticeable diffraction when the gap
is very much larger than the wavelength. As the gap
becomes narrower, the diffraction effect becomes more
pronounced. It is greatest when the width of the gap is
roughly equal to the wavelength of the ripples.

c λ 195
λ

λ
λλλ
Figure 14.7 The extent to which ripples spread out depends on the relationship between their wavelength and
the width of the gap. In a, the width of the gap is very much greater than the wavelength and there is hardly any
noticeable diffraction. In b, the width of the gap is greater than the wavelength and there is limited diffraction.
In c, the gap width is approximately equal to the wavelength and the diffraction effect is greatest.

Figure 14.8 You can see the effects of diffraction by looking
through a narrow slit. What happens when you make the slit
narrower? What happens to the amount of diffraction when
you put different coloured filters in front of the lamp? What
does this tell you about the wavelengths of the different
colours?

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Diffraction of radio and microwaves the effect of an infinite number of ripples, we can say that
in some directions the ripples add together while in other
Radio waves can have wavelengths of the order of a directions they cancel out.
kilometre. These waves are easily diffracted by gaps in
the hills and by the tall buildings around our towns and ripples from A
cities. Microwaves, used by the mobile phone network, ripples from B
have wavelengths of about 10 cm. These waves are not ripples from C
easily diffracted (because their wavelengths are much
smaller than the dimensions of the gaps) and mostly travel A
through space in straight lines. B
C
Cars need external radio aerials because radio waves
have wavelengths longer than the size of the windows, so Figure 14.10  Ripples from all points across the gap contribute
they cannot diffract into the car. If you try listening to a to the pattern in the space beyond.
radio in a train without an external aerial, you will find
that FM signals can be picked up weakly (their wavelength
is about 3 m), but AM signals, with longer wavelengths,
cannot get in at all.

QUESTION Interference
2 A microwave oven (Figure 14.9) uses microwaves
Adding waves of different wavelengths and amplitudes
with a wavelength of 12.5 cm. The front door of results in complex waves. We can find some interesting
the oven is made of glass with a metal grid inside; effects if we consider what happens when two waves of the
the gaps in the grid are a few millimetres across. same wavelength overlap at a point. Again, we will use the
Explain how this design allows us to see the food principle of superposition to explain what we observe.
196 inside the oven, while the microwaves are not
allowed to escape into the kitchen (where they BOX 14.2: Observing interference
might cook us).

Interference of sound waves

A simple experiment shows what happens when
two sets of sound waves meet. Two loudspeakers
are connected to a single signal generator (Figure
14.11). They each produce sound waves of the same
wavelength. Walk around in the space in front of
the loudspeakers; you will hear the resultant effect.

signal generator

Figure 14.9  A microwave oven has a metal grid in 500 Hz
the door to keep microwaves in and let light out.

Explaining diffraction Figure 14.11  The sound waves from two loudspeakers
combine to give an interference pattern. This
Diffraction is a wave effect that can be explained by the experiment is best done outside so that reflections of
principle of superposition. We have to think about what sounds (or echoes) do not affect the results.
happens when a plane ripple reaches a gap in a barrier
(Figure 14.10). Each point on the surface of the water in the
gap is moving up and down. Each of these moving points
can be thought of as a source of new ripples spreading out
into the space beyond the barrier. Now we have a lot of
new ripples, and we can use the principle of superposition
to find their resultant effect. Without trying to calculate

Chapter 14: Superposition of waves

BOX 14.2: Observing interference (continued) Explaining interference
A naive view might be that we would hear a sound
twice as loud as that from a single loudspeaker. Figure 14.14 shows how interference arises. The
However, this is not what we hear. At some points, loudspeakers in Figure 14.11 (Box 14.2) are emitting
the sound is louder than for a single speaker. At waves that are in phase because both are connected
other points, the sound is much quieter. The space to the same signal generator. At each point in front
around the two loudspeakers consists of a series of the loudspeakers, waves are arriving from the two
of loud and quiet regions. We are observing the loudspeakers. At some points, the two waves arrive in
phenomenon known as interference. phase (in step) with one another and with equal amplitude
Interference in a ripple tank (Figure 14.14a). The principle of superposition predicts
The two dippers in the ripple tank (Figure 14.12) that the resultant wave has twice the amplitude of a single
should be positioned so that they are just touching wave. We hear a louder sound.
the surface of the water. When the bar vibrates, each
dipper acts as a source of circular ripples spreading a4 Displacement / 10–4 m Displacement / 10–4 m Displacement / 10–4 m resultant 197
outwards. Where these sets of ripples overlap, we 3
observe an interference pattern. Another way to 2 Time
observe interference in a ripple tank is to use plane 1
waves passing through two gaps in a barrier. The 0 resultant
water waves are diffracted at the two gaps and then –1
interfere beyond the gaps. Figure 14.13 shows the –2 Time
interference pattern produced by two vibrating –3
sources in a ripple tank. –4

Figure 14.12  A ripple tank can be used to show how b
two sets of circular ripples combine. 2
1
Figure 14.13  Ripples from two point sources produce 0
an interference pattern. –1
–2

c3 resultant
2
1 Time
0
–1
–2
–3

Figure 14.14  Adding waves by the principle of superposition.
Blue and green waves of the same amplitude may give
a constructive or b destructive interference, according to
the phase difference between them. c Waves of different
amplitudes can also interfere constructively.

At other points, something different happens. The
two waves arrive completely out of phase or in antiphase
(phase difference is 180°) with one another (Figure 14.14b).
There is a cancelling out, and the resultant wave has zero
amplitude. At this point, we would expect silence. At other
points again, the waves are neither perfectly out of step nor
perfectly in step, and the resultant wave has amplitude less
than that at the loudest point.

Where two waves arrive at a point in phase with one
another so that they add up, we call this effect constructive
interference. Where they cancel out, the effect is known as
destructive interference. Where two waves have different

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amplitudes but are in phase (Figure 14.14c), constructive to a phase difference of 180°. The waves interfere
interference results in a wave whose amplitude is the sum destructively because they are in antiphase. In general, the
of the two individual amplitudes. conditions for constructive interference and destructive
interference are outlined below. These conditions apply
QUESTION to all waves (water waves, light, microwaves, radio waves,
sound, etc.) that show interference effects. In the equations
3 Explain why the two loudspeakers must produce below, n stands for any integer (any whole number,
sounds of precisely the same frequency if we are including zero).
to hear the effects of interference.
■■ For constructive interference the path difference is a whole
number of wavelengths:

How can we explain the interference pattern observed path difference = 0, λ, 2λ, 3λ, etc.
in a ripple tank (Box 14.2)? Look at Figure 14.15 and or path difference = n λ
compare it to Figure 14.13. Figure 14.15 shows two sets of ■■ For destructive interference the path difference is an odd
waves setting out from their sources. At a position such number of half wavelengths:
as A, ripples from the two sources arrive in phase with path difference =  21 λ, 1 21 λ, 2 12 λ, etc.
one another, and constructive interference occurs. At B, or path difference = (n +  12 ) λ
the two sets of ripples arrive out of phase, and there is
destructive interference. Although waves are arriving at B, BOX 14.3: Interference of radiation
the surface of the water remains approximately flat.
Interference of light

Here is one way to show the interference effects

produced by light. A simple arrangement involves

directing the light from a laser through two slits

198 (Figure 14.16). The slits are two clear lines on a black

slide, separated by a fraction of a millimetre. Where

the light falls on the screen, a series of equally

spaced dots of light are seen (see Figure 14.21 on

page 201). These bright dots are referred to as
A B interference ‘fringes’, and they are regions where

Figure 14.15  The result of interference depends on the path
difference between the two waves.

Whether the waves combine constructively or screen
destructively at a point depends on the path difference slide with
of the waves from the two sources. The path difference is double slit
defined as the extra distance travelled by one of the waves
compared with the other. Figure 14.16  Light beams from the two slits interfere
in the space beyond.
At point A in Figure 14.15, the waves from the red
source have travelled 3 whole wavelengths. The waves
from the yellow source have travelled 4 whole wavelengths.
The path difference between the two sets of waves is
1 wavelength. A path difference of 1 wavelength is
equivalent to a phase difference of zero. This means that
they are in phase, so they interfere constructively.

Now think about destructive interference. At
point B, the waves from the red source have travelled
3 wavelengths; the waves from the yellow source have
travelled 2.5 wavelengths. The path difference between the
two sets of waves is 0.5 wavelengths, which is equivalent

Chapter 14: Superposition of waves

BOX 14.3: Interference of radiation (continued) narrow, more intense beam. This famous experiment is
called the Young double-slit experiment (see page 200),
light waves from the two slits are arriving in phase with although Thomas Young had no laser available to him
each other, i.e. there is constructive interference. The when he first carried it out in 1801.
dark regions in between are the result of destructive
interference. Interference of microwaves

Safety note Using 2.8 cm wavelength microwave equipment (Figure
If you carry out experiments using a laser, you 14.17), you can observe an interference pattern. The
should follow correct safety procedures. In microwave transmitter is directed towards the double
particular, you should wear eye protection and gap in a metal barrier. The microwaves are diffracted
avoid allowing the beam to enter your eye at the two gaps so that they spread out into the region
directly. beyond, where they can be detected using the probe
receiver. By moving the probe around, it is possible
These bright and dark fringes are the equivalent of to detect regions of high intensity (constructive
the loud and quiet regions that you detected if you interference) and low intensity (destructive
investigated the interference pattern of sounds from interference). The probe may be connected to a meter,
the two loudspeakers described in Box 14.2. Bright or to an audio amplifier and loudspeaker to give an
fringes correspond to loud sound, dark fringes to quiet audible output.
sound or silence.
microwave
You can check that light is indeed reaching the probe
screen from both slits as follows. Mark a point on the
screen where there is a dark fringe. Now carefully 199
cover up one of the slits so that light from the laser is
only passing through one slit. You should find that the microwave meter
pattern of interference fringes disappears. Instead, a transmitter metal
broad band of light appears across the screen. This sheets
broad band of light is the diffraction pattern produced
by a single slit. The point that was dark is now light. Figure 14.17  Microwaves can also be used to show
Cover up the other slit instead, and you will see the interference effects.
same effect. You have now shown that light is arriving at
the screen from both slits, but at some points (the dark
fringes) the two beams of light cancel each other out.

You can achieve similar results with a bright light
bulb rather than a laser, but a laser is much more
convenient because the light is concentrated into a

QUESTION Coherence
4 Look at the experimental arrangement shown
We are surrounded by many types of wave – light, infrared
in Figure 14.17. Suppose that the microwave radiation, radio waves, sound, and so on. There are waves
probe is placed at a point of low intensity in the coming at us from all directions. So why do we not observe
interference pattern. What do you predict will interference patterns all the time? Why do we need
happen if one of the gaps in the barrier is now specialised equipment in a laboratory to observe these
blocked? effects?

In fact, we can see interference of light occurring in
everyday life. For example, you may have noticed haloes
of light around street lamps or the Moon on a foggy night.

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You may have noticed light and dark bands of light if you sudden change
look through fabric at a bright source of light. These are of phase
interference effects.
a
We usually need specially arranged conditions to
produce interference effects that we can measure. Think double slit
about the demonstration with two loudspeakers. If they b
were connected to different signal generators with slightly
different frequencies, the sound waves might start off in Figure 14.19  Waves must be coherent if they are to produce a
phase with one another, but they would soon go out of clear interference pattern.
phase (Figure 14.18). We would hear loud, then soft, then
loud again. The interference pattern would keep shifting
around the room.

Displacement in phase out of phase QUESTION

0 Time 5 Draw sketches of displacement against time to
Figure 14.18  Waves of slightly different wavelengths (and illustrate the following:
therefore frequencies) move in and out of phase with one a two waves having the same amplitude and in
another. phase with one another
b two waves having the same amplitude and
with a phase difference of 90°
c two waves initially in phase but with slightly
different wavelengths.

200 By connecting the two loudspeakers to the same signal Use your sketches to explain why two coherent
generator, we can be sure that the sound waves that they sources of waves are needed to observe
interference.
produce are constantly in phase with one another. We

say that they act as two coherent sources of sound waves

(coherent means sticking together). Coherent sources emit

waves that have a constant phase difference. Note that The Young double-slit
the two waves can only have a constant phase difference if experiment
their frequency is the same and remains constant.
Now we will take a close look at a famous experiment
Now think about the laser experiment. Could we have which Thomas Young performed in 1801. He used this
used two lasers producing exactly the same frequency and experiment to show the wave nature of light. A beam
hence wavelength of light? Figure 14.19a represents the of light is shone on a pair of parallel slits placed at right
light from a laser. We can think of it as being made up of angles to the beam. Light diffracts and spreads outwards
many separate bursts of light. We cannot guarantee that from each slit into the space beyond; the light from the two
these bursts from two lasers will always be in phase with slits overlaps on a screen. An interference pattern of light
one another. and dark bands called ‘fringes’ is formed on the screen.

This problem is overcome by using a single laser
and dividing its light using the two slits (Figure 14.19b).

The slits act as two coherent sources of light. They are Explaining the experiment
constantly in phase with one another (or there is a
constant phase difference between them). In order to observe interference, we need two sets of waves.
The sources of the waves must be coherent – the phase
If they were not coherent sources, the interference difference between the waves emitted at the sources must
pattern would be constantly changing, far too fast for our remain constant. This also means that the waves must have
eyes to detect. We would simply see a uniform band of the same wavelength. Today, this is readily achieved by
light, without any definite bright and dark regions. From passing a single beam of laser light through the two slits.
this you should be able to see that, in order to observe A laser produces intense coherent light. As the light passes
interference, we need two coherent sources of waves. through the slits, it is diffracted so that it spreads out into the

space beyond (Figure 14.20). Now we have two overlapping

Chapter 14: Superposition of waves

laser 1

double slit interference aD A bright
in this region 2 x B dark

Figure 14.20  Interference occurs where diffracted beams double slit C bright
from the two slits overlap.
D
Figure 14.21  Interference fringes obtained using a laser and a
double slit. E

sets of waves, and the pattern of fringes on the screen shows screen
us the result of their interference (Figure 14.21).
Figure 14.22  Rays from the two slits travel different distances
How does this pattern arise? We will consider three to reach the screen.
points on the screen (Figure 14.22), and work out what we
would expect to observe at each. time, it has travelled an extra distance equal to a whole 201
wavelength λ. The path difference between the rays of
Point A light is now a whole wavelength. The two rays are in phase
at the screen. They interfere constructively and we see a
This point is directly opposite the midpoint of the slits. bright fringe.
Two rays of light arrive at A, one from slit 1 and the other
from slit 2. Point A is equidistant from the two slits, and so The complete interference pattern (Figure 14.21) can be
the two rays of light have travelled the same distance. The explained in this way.
path difference between the two rays of light is zero. If we
assume that they were in phase (in step) with each other QUESTION
when they left the slits, then they will be in phase when
they arrive at A. Hence they will interfere constructively, 6 Consider points D and E on the screen, where
and we will observe a bright fringe at A. BC = CD = DE. State and explain what you would
expect to observe at D and E.
Point B
Determining wavelength λ
This point is slightly to the side of point A, and is the
midpoint of the first dark fringe. Again, two rays of light The double-slit experiment can be used to determine the
arrive at B, one from each slit. The light from slit 1 has wavelength λ of light. The following three quantities have
to travel slightly further than the light from slit 2, and to be measured:
so the two rays are no longer in step. Since point B is at
the midpoint of the dark fringe, the two rays must be in ■■ Slit separation a – This is the distance between the centres
antiphase (phase difference of 180°). The path difference of the slits, which is the distance between slits 1 and 2 in
between the two rays of light must be half a wavelength Figure 14.22.
and so the two rays interfere destructively.
■■ Fringe separation x – This is the distance between the
Point C centres of adjacent bright (or dark) fringes, which is the
distance AC in Figure 14.22.
This point is the midpoint of the next bright fringe, with
AB = BC. Again, ray 1 has travelled further than ray 2; this ■■ Slit-to-screen distance D – This is the distance from the
midpoint of the slits to the central fringe on the screen.

Once these three quantities have been measured, the
wavelength λ of the light can be found using the relationship:

λ = ax
D

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WORKED EXAMPLE

1 In a double-slit experiment using light from a helium– Step 2  Substitute the values of a, x and D in the
neon laser, a student obtained the following results: expression for wavelength λ:
width of 10 fringes 10x = 1.5 cm
separation of slits a = 1.0 mm λ =  ax
slit-to-screen distance D = 2.40 m D

Therefore:

Determine the wavelength of the light. λ =  1.0 × 10−3 × 1.5 × 10−3 = 6.3 × 10−7 m
2.40

Step 1  Work out the fringe separation: Hint: Don’t forget to convert all the distances into metres.

fringe separation x =  1.5 × 10−2 = 1.5 × 10−3 m So the wavelength is 6.3 × 10−7 m or 630 nm.
10

QUESTION
7 If the student in Worked example 1 moved the

screen to a distance of 4.8 m from the slits, what
would the fringe separation become?

BOX 14.4: Using Young’s slits to measure λ

202 single slit, and the fringes are observed on a screen a
The Young double-slit experiment can be used to

determine the wavelength of light λ. Here we look at metre or so away. The experiment has to be carried out

a number of practical features of the experiment and in a darkened room, as the intensity of the light is low

consider how the uncertainty in the value of λ can be and the fringes are hard to see.

reduced. There are three important factors involved in the

One way to carry out the double-slit experiment is way the equipment is set up:

shown in Figure 14.23. Here, a white light source is used, ■■ All slits are a fraction of a millimetre in width. Since
rather than a laser. A monochromatic filter allows only the wavelength of light is less than a micrometre
one wavelength of light to pass through. A single slit (10−6 m), this gives a small amount of diffraction in the
diffracts the light. This diffracted light arrives in phase space beyond. If the slits were narrower, the intensity
at the double slit, which ensures that the two parts of of the light would be too low for visible fringes to be
the double slit behave as coherent sources of light. The achieved.
double slit is placed a centimetre or two beyond the
■■ The double slits are about a millimetre apart. If they

monochromatic were much further apart, the fringes would be too
filter
close together to be distinguishable.
■■ The screen is about a metre from the slits. The fringes

produced are clearly separated without being too dim.

Measuring a, x and D

Measuring slit separation a: a travelling microscope

shield around single double screen is suitable for measuring a. It is difficult to judge the
bright light source slit slit position of the centre of a slit. If the slits are the same

Figure 14.23  To observe interference fringes with white width, the separation of their left-hand edges is the
light, you must use a single slit before the double slit. same as the separation of their centres.

Chapter 14: Superposition of waves

BOX 14.4: Using Young’s slits to measure λ (continued)

Measuring fringe width x: it is best to measure A laser has a second advantage. The light from a
across several fringes (say, ten) and then to calculate laser is monochromatic; that is, it consists of a single
the average separation later. A metre rule or travelling wavelength. This makes the fringes very clear, and they
microscope can be used. are present in large numbers across the screen. With
white light, a range of wavelengths is present. Different
Measuring the slit-to-screen distance D: this can be wavelengths form fringes at different points across the
measured using a metre rule or a tape measure. screen, smearing them out so that they are not as clear.

Reducing percentage errors Using white light with no filter results in a central
fringe which is white (because all wavelengths are
Why use a laser rather than white light? With a laser, in phase here), but the other fringes show coloured
the light beam is more concentrated, and the initial effects, as the different wavelengths interfere
single slit is not necessary. The greater intensity of the constructively at different points. In addition, only a few
beam means that the screen can be further from the fringes are visible in the interference pattern.
slits, so that the fringes are further apart; this reduces
the percentage error in measurements of x and D, and
hence λ can be determined more accurately.

QUESTIONS

8 Use λ =  ax to explain the following observations: 9 Yellow sodium light of wavelength 589 nm is used in
D the Young double-slit experiment. The slit separation
a With the slits closer together, the fringes are is 0.20 mm, and the screen is placed 1.20 m from
further apart. the slits. Calculate the separation of neighbouring 203
fringes formed on the screen.
b Interference fringes for blue light are closer
together than for red light. 10 In a double-slit experiment, filters were placed in
front of a white light source to investigate the effect
c In an experiment to measure the wavelength of of changing the wavelength of the light. At first, a red
light, it is desirable to have the screen as far from filter was used instead (λ = 600 nm) and the fringe
the slits as possible. separation was found to be 2.40 mm. A blue filter was
then used instead (λ = 450 nm). Determine the fringe
separation with the blue filter.

Diffraction gratings Figure 14.24  A CD acts
as a reflection diffraction
A transmission diffraction grating is similar to the slide grating. White light is
used in the double-slit experiment, but with many more slits reflected and diffracted
than just two. It consists of a large number of equally spaced at its surface, producing
lines ruled on a glass or plastic slide. Each line is capable a display of spectral
of diffracting the incident light. There may be as many as colours.
10 000 lines per centimetre. When light is shone through
this grating, a pattern of interference fringes is seen.

A reflection diffraction grating consists of lines made
on a reflecting surface so that light is both reflected and
diffracted by the grating. The shiny surface of a compact
disc (CD) or DVD is an everyday example of a reflection
diffraction grating. Hold a CD in your hand so that you
are looking at the reflection of light from a lamp. You
will observe coloured bands (Figure 14.24). A CD has

Cambridge International AS Level Physics

thousands of equally spaced lines of microscopic pits on a b
its surface; these carry the digital information. It is the
diffraction from these lines that produces the coloured
bands of light from the surface of the CD.

Observing diffraction with a
transmission grating
In Figure 14.25, monochromatic light from a laser is ray 6
incident normally on a transmission diffraction grating. grating grating ray 5
In the space beyond, interference fringes are formed. ray 4
These can be observed on a screen, as with the double slit. ray 3
However, it is usual to measure the angle θ at which they ray 2
are formed, rather than measuring their separation. With ray 1
double slits, the fringes are equally spaced and the angles
are very small. With a diffraction grating, the angles are Figure 14.26 a Waves from each slit are in phase in the
much greater and the fringes are not equally spaced. straight-through direction. b In the direction of the first-order
maximum, the waves are in phase, but each one has travelled
one wavelength further than the one below it.

The fringes are also referred to as maxima. The central source, which is far away. All the rays with θ = 0 come
fringe is called the zeroth-order maximum, the next together at the back of your eye, where an image is formed.
fringe is the first-order maximum, and so on. The pattern It is here that interference occurs.
is symmetrical, so there are two first-order maxima, two
second-order maxima, and so on. The first-order maximum forms as follows. Rays of
light emerge from all of the slits; to form a bright fringe,

all the rays must be in phase. In the direction of the first-

204 screen n = –2 order maximum, ray 1 has travelled the smallest distance
(Figure 14.26b). Ray 2 has travelled an extra distance
n = –1 equal to one whole wavelength and is therefore in phase

n = 0 with ray 1. The path difference between ray 1 and ray 2 is

θ n = +1 equal to one wavelength λ. Ray 3 has travelled two extra

wavelengths and is in phase with rays 1 and 2. In fact, the

rays from all of the slits are in step in this direction, and a

di raction n = +2 bright fringe results.

grating QUESTION

Figure 14.25 The diffracted beams form a symmetrical
pattern on either side of the undiffracted central beam.
11 Explain how the second-order maximum arises.
Use the term path difference in your explanation.

Explaining the experiment Determining wavelength λ with a grating

The principle is the same as for the double-slit experiment, By measuring the angles at which the maxima occur, we
but here we have light passing through many slits. As it can determine the wavelength of the incident light. The
passes through each slit, it diffracts into the space beyond. wavelength λ of the monochromatic light is related to the
So now we have many overlapping beams of light, and angle θ by:
these interfere with one another. It is difficult to achieve
constructive interference with many beams, because they d sin θ = nλ
all have to be in phase with one another.
where d is the distance between adjacent lines of the
There is a bright fringe, the zeroth-order maximum, grating and n is known as the order of the maximum; n
in the straight-through direction (θ = 0) because all of can only have integer values 0, 1, 2, 3, etc. The distance d
the rays here are travelling parallel to one another and in is known as the grating element or grating spacing. This is
phase, so the interference is constructive (Figure 14.26a). illustrated in Worked example 2.

Imagine if you could look through the grating at the
source of light. Your eye would be focused on the light

Chapter 14: Superposition of waves

WORKED EXAMPLE QUESTION

2 Monochromatic light is incident normally on a 14 A student is trying to make an accurate
diffraction grating having 3000 lines per centimetre. measurement of the wavelength of green light
The angular separation of the zeroth- and first-order from a mercury lamp (λ = 546 nm). Using a double
maxima is found to be 10°. Calculate the wavelength slit of separation 0.50 mm, he finds he can see ten
of the incident light. clear fringes on a screen at a distance of 0.80 m
from the slits. The student can measure their
Step 1  Calculate the slit separation (grating overall width to within ±1 mm. He then tries an
spacing) d. Since there are 3000 slits per centimetre, alternative experiment using a diffraction grating
their separation must be: that has 3000 lines per centimetre. The angle
between the two second-order maxima can be
d  = 1 cm = 3.33 × 10−4 cm = 3.33 × 10−6 m measured to within ±0.1°.
3000 a What will be the width of the ten fringes that
he can measure in the first experiment?
Step 2  Rearrange the equation d sin θ = nλ and b What will be the angle of the second-order
substitute values: maximum in the second experiment?
θ  = 10.0°, n = 1 c Suggest which experiment you think will give
the more accurate measurement of λ.
λ =  d sin θ  =  3.36 × 10−6 × sin 10°
n 1

λ = 5.8 × 10−7 m = 580 nm

QUESTIONS Diffracting white light
A diffraction grating can be used to split white light up
12 a For the case described in Worked example 2, into its constituent colours (wavelengths). This splitting 205
at what angle would you expect to find the of light is known as dispersion, shown in Figure 14.27. A
second-order maximum (n = 2)? beam of white light is shone onto the grating. A zeroth-
order, white maximum is observed at θ = 0°, because all
b Repeat the calculation of θ for n = 3, 4, etc. waves of each wavelength are in phase in this direction.
What is the limit to this calculation? How
many maxima will there be altogether in this On either side, a series of spectra appear, with violet
interference pattern? closest to the centre and red furthest away. We can see
why different wavelengths have their maxima at different
13 Consider the equation d sin θ  = nλ. How will the angles if we rearrange the equation d sin θ = nλ to give:
diffraction pattern change if:
a the wavelength of the light is increased? sin θ = nλ
b the diffraction grating is changed for one with d
more lines per centimetre (slits that are more
closely spaced)?

BOX 14.5: Diffraction gratings versus double slits

It is worth comparing the use of a diffraction grating to ■■ With a diffraction grating, there are many slits per
determine wavelength with the Young two-slit experiment. centimetre, so d can be measured precisely. Because
the maxima are widely separated, the angle θ can
■■ With a diffraction grating, the maxima are very sharp. also be measured to a high degree of precision. So an
experiment with a diffraction grating can be expected
■■ With a diffraction grating, the maxima are also very to give measurements of wavelength to a much
bright. This is because rather than contributions from higher degree of precision than a simple double-slit
only two slits, there are contributions from a thousand arrangement.
or more slits.

■■ With two slits, there may be a large inaccuracy in the
measurement of the slit separation a. The fringes
are close together, so their separation may also be
measured imprecisely.

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white light n=0 From this it follows that the greater the wavelength λ, the
greater the value of sin θ and hence the greater the angle θ.
di raction Red light is at the long wavelength end of the visible
grating spectrum, and so it appears at the greatest angle.

screen QUESTION
Figure 14.27  A diffraction grating is a simple way of
separating white light into its constituent wavelengths. 15 White light is incident normally on a diffraction
grating with a slit separation d of 2.00 × 10−6 m.
a Calculate the angle between the red and
violet ends of the first-order spectrum. The
visible spectrum has wavelengths between
400 nm and 700 nm.
b Explain why the second- and third-order
spectra overlap.

Summary

■■ The principle of superposition states that ■■ For destructive interference the path difference is an

206 when two or more waves meet at a point, the odd number of half wavelengths:
resultant displacement is the algebraic sum of the
displacements of the individual waves. path difference =  21  λ, 1 12  λ, 2 12  λ, etc.  or 

■■ When waves pass through a slit, they may be path difference = (n  +  21 ) λ
diffracted so that they spread out into the space
beyond. The diffraction effect is greatest when the ■■ When light passes through a double slit, it is diffracted
wavelength of the waves is similar to the width of and an interference pattern of equally spaced light
the gap. and dark fringes is observed. This can be used to
determine the wavelength of light using the equation:

■■ Interference is the superposition of waves from two λ = ax
coherent sources. D

■■ Two sources are coherent when they emit waves that This equation can be used for all waves, including
have a constant phase difference. (This can only sound and microwaves.

happen if the waves have the same frequency or ■■ A diffraction grating diffracts light at its many slits

wavelength.) or lines. The diffracted light interferes in the space

■■ For constructive interference the path difference is a beyond the grating. The equation for a diffraction
whole number of wavelengths: grating is:

path difference = 0, λ, 2λ, 3λ, etc.  or  d sin θ  = nλ

path difference = nλ

Chapter 14: Superposition of waves

End-of-chapter questionsDisplacement [2]
[1]
1 a Copy the waves shown in Figure 14.28 onto a sheet of graph paper and use the principle of
superposition to show the resultant wave. 207

b How does the wavelength of the resulting wave compare with that of the component waves? [1]
[1]
0 [2]
Distance [2]
[2]
Figure 14.28 For End-of-chapter Question 1. [3]
2 Figure 14.29 shows a ripple tank being used to demonstrate diffraction of water waves.

Figure 14.29 For End-of-chapter Question 2.
Suggest how the diffracted wave shape will change if:
a the wavelength of the incident wave is increased
b the wavelength of the incident wave is decreased.
3 Explain why, in remote mountainous regions, such as the Hindu Kush, radio signals from
terrestrial transmitters can be received, but television reception can only be received from
satellite transmissions.
4 Damita and Jamal are organising a disco. Damita suggests that feeding the sound from the music
centre to a second loudspeaker will increase the loudness of the music. Jamal says it won’t work as
there will be places where the sound will be very loud, due to constructive interference, and places
where it will be much quieter, due to destructive interference. State who is correct and explain your
reasoning.
5 The constant frequency signal from a signal generator is fed to two loudspeakers placed 1.5m apart.
A girl, who is 8metres away from the speakers, walks across in a line parallel to the line between the
speakers. She finds that there is a distance of 1.2m between successive spots where the sound is very
quiet. Calculate the wavelength of the sound.
6 Two signal generators feed signals with slightly different frequencies to two separate loudspeakers.
Suggest why a sound of continuously rising and falling loudness is heard.

Cambridge International AS Level Physics

7 A hydrogen discharge lamp produces a spectral line of wavelength 656nm. Calculate the angles [5]
at which a diffraction grating, with 5000 lines cm−1, would produce the first and second maxima for [2]
this light.
[3]
8 a Explain what is meant by the term superposition. [2]
b In a Young double-slit experiment using yellow light of wavelength 590nm from a sodium [2]
discharge tube, a student sets up a screen 1.8m from the double slit. He measures the distance
between 12 fringes as 16.8mm. Calculate the separation of the slits. [2]
c State the effect of:
i using slits of narrower width (but the same separation)
ii using slits with a smaller separation but of the same width

9 a A laser light is described as producing light which is both highly coherent and highly monochromatic.
Explain what is meant by the terms coherent and monochromatic.

b Figure 14.30 shows the setup used to analyse the spectrum of a sodium discharge lamp using a
diffraction grating with 5000 lines cm−1. Figure 14.31 shows the spectral lines observed in the
developed photographic film.

collimator di raction grating

208 θ A

sodium
lamp

photographic
film
B
Figure 14.30 Experimental setup for End-of-chapter Question 9.

violet blue green yellow red violet blue green yellow red

A B
Figure 14.31 Spectral lines produced.

i Explain why two spectra are observed. [2]
ii Study the two spectra and describe two differences between them. [2]
iii The green maximum near end A is at an angle θ of 19.5°. Calculate the wavelength of the green light. [3]
iv Calculate the angle produced by the second green line. [2]

Chapter 14: Superposition of waves

10 a Explain what is meant by the term destructive interference. [2]
b A student sets up an experiment to investigate the interference pattern formed by microwaves of
wavelength 1.5cm. The apparatus is set up as in Figure 14.17 on page 199. The distance between the [3]
centres of the two slits is 12.5cm. The detector is centrally placed 1.2m from the metal plates where [2]
it detects a maximum. The student moves the detector 450cm across the bench parallel to the plates. [2]
Calculate how many maxima the detector will be moved through.
c Calculate the frequency of these microwaves.

11 a Explain what is meant by the diffraction of a wave.
b Figure 14.32 shows waves, in a ripple tank, spreading out from two slits.

209

Figure 14.32 For End-of-chapter Question 11. [1]
[1]
On a copy of Figure 14.32, draw: [1]
i a line showing points along the central maximum – label this line 0
ii a line showing the points along first maximum – label this line 1 [3]
iii a line showing points along one of the first minima – label this line min.
c The centres of the slits are 12cm apart. At a distance of 60cm from the barrier, the first maxima are
18cm either side of the central maximum. Calculate the wavelength of the waves. You may assume
that the formula developed for light waves is applicable in this example.

210

Chapter 15:
Stationary waves

Learning outcomes

You should be able to:

■■ demonstrate and explain the formation of stationary
waves

■■ identify nodes and antinodes on a stationary wave
■■ determine the wavelength of sound waves using

stationary waves

Chapter 15: Stationary waves

The bridge that broke this happened in October 1940 to the Tacoma
Narrows Bridge in Washington State, USA. High
Figure 15.1a shows the Normandy Bridge under winds caused the bridge to vibrate with increasing
construction in France. When designing bridges, amplitude until it fell apart (Figure 15.1b).
engineers must take into account the possibility of the
wind causing a build-up of stationary waves, which b
may lead the bridge to oscillate violently. Famously,

a

Figure 15.1  a A suspension bridge under construction. b One that failed – the Tacoma Narrows Bridge. 211

From moving to stationary stable pattern like one of those shown in Figure 15.3. Alter
the frequency in order to achieve one of the other patterns.
The waves we have considered so far in Chapters 13 and
14 have been progressive waves; they start from a source You should notice that you have to move the end of
and travel outwards, transferring energy from one place to the spring with just the right frequency to get one of these
another. A second important class of waves is stationary interesting patterns. The pattern disappears when the
waves (standing waves). These can be observed as follows. frequency of the shaking of the free end of the spring is
Use a long spring or a slinky spring. A long rope or piece slightly increased or decreased.
of rubber tubing will also do. Lay it on the floor and fix
one end firmly. Move the other end from side to side so Amplitude Amplitude Amplitude fixed free
that transverse waves travel along the length of the spring end end
and reflect off the fixed end (Figure 15.2). If you adjust the 0
frequency of the shaking, you should be able to achieve a

antinode Distance
node Distance

0

Figure 15.2  A slinky spring is used to generate a stationary 0
wave pattern.
Distance
Figure 15.3  Different stationary wave patterns are possible,
depending on the frequency of vibration.

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Nodes and antinodes T = period of wave

What you have observed is a stationary wave on the long Displacement λ resultant
spring. There are points along the spring that remain t=0 0
(almost) motionless while points on either side are Distance
oscillating with the greatest amplitude. The points that s wave moving to right
do not move are called the nodes and the points where
the spring oscillates with maximum amplitude are called t = T 0 x ‘Snapshots’
the antinodes. At the same time, it is clear that the wave 4 wave moving to le of the
profile is not travelling along the length of the spring. waves
Hence we call it a stationary wave or a standing wave. s x over a
time
We normally represent a stationary wave by drawing t = T 0 x of one
the shape of the spring in its two extreme positions 2 period, T.
(Figure 15.4). The spring appears as a series of loops,
separated by nodes. In this diagram, point A is moving s
downwards. At the same time, point B in the next loop
is moving upwards. The phase difference between points t = 3T 0
A and B is 180°. Hence the sections of spring in adjacent 4
loops are always moving in antiphase; they are half a cycle
out of phase with one another. s x
t=T 0
Aλ
λ
212 2 profile at t = 0 and T
B NN NNN N
profile at t = T and 3T
Distance 4 4

Amplitude Figure 15.4  The fixed ends of a long spring must be nodes in A A A A A profile at t = T
the stationary wave pattern. 2
λ
2

Key
wave moving to right
wave moving to le
resultant wave

Formation of stationary waves Figure 15.5  The blue-coloured wave is moving to the left
and the red-coloured wave to the right. The principle of
Imagine a string stretched between two fixed points, for superposition of waves is used to determine the resultant
example a guitar string. Pulling the middle of the string displacement. The profile of the long spring is shown in green.
and then releasing it produces a stationary wave. There
is a node at each of the fixed ends and an antinode in the ■■ After a time equal to one-quarter of a period (t  =  T4), each
middle. Releasing the string produces two progressive wave has travelled a distance of one quarter of a
waves travelling in opposite directions. These are reflected wavelength to the left or right. Consequently, the two
at the fixed ends. The reflected waves combine to produce waves are in antiphase (phase difference = 180°). The waves
the stationary wave.
■■ cwAofatmevrebasinateirmedebeasetcqrkuuiacnltptivoheoalyns,eeg-aihvgaianlifgno.zfTeahropeyedrioisonpdclae(ctae =gm aT2ein )n,ctt.ohme tbwinoe
Figure 15.3 shows how a stationary wave can be set up ■■ cAwofatnevsretasruatircmetieivneealqynu.taipl htoastheraegea-qinu.aTrhteeryscoofmabpienreioddes(ttr =u  c34tTi v),etlhye,
using a long spring. A stationary wave is formed whenever
two progressive waves of the same amplitude and with the resultant wave showing zero displacement.
wavelength, travelling in opposite directions, superpose. ■■ After a time equal to one whole period (t  = T), the waves
Figure 15.5 uses a displacement–distance graph (s–x) to
illustrate the formation of a stationary wave along a long combine constructively. The profile of the spring is as it was
spring (or a stretched length of string): at t  =  0.

■■ At time t  =  0, the progressive waves travelling to the left
and right are in phase. The waves combine constructively,
giving an amplitude twice that of each wave.

Chapter 15: Stationary waves

This cycle repeats itself, with the long spring showing It is worth noting that a stationary wave does not travel
nodes and antinodes along its length. The separation and therefore has no speed. It does not transfer energy
between adjacent nodes or antinodes tells us about the between two points like a progressive wave. Table 15.1
progressive waves that produce the stationary wave. shows some of the key features of a progressive wave and
its stationary wave.
A closer inspection of the graphs in Figure 15.5 shows
that the separation between adjacent nodes or antinodes wavelength Progressive wave Stationary wave
is related to the wavelength λ of the progressive wave. The frequency λ λ
important conclusions are: speed f f

separation between two adjacent nodes λ v zero
2
(or between two adjacent antinodes) = Table 15.1  A summary of progressive and stationary waves.

separation between adjacent node and antinode = λ QUESTION
4
1 A stationary (standing) wave is set up on a
The wavelength λ of any progressive wave can be vibrating spring. Adjacent nodes are separated
by 25 cm. Determine:
determined from the separation between neighbouring a the wavelength of the stationary wave
b the distance from a node to an adjacent
nodes or antinodes of the resulting standing wave pattern. antinode.
sλ2p.)eTedhivs
(This separation is = can then be used to or its
determine either the of the progressive wave

frequency f  by using the wave equation:

v = f λ

BOX 15.1: Observing stationary waves 213

Here we look at experimental arrangements for The pulley end of the string cannot vibrate; this is
observing stationary waves, for mechanical waves on a node. Similarly, the end attached to the vibrator can
strings, microwaves, and sound waves in air columns. only move a small amount, and this is also a node. As
the frequency is increased, it is possible to observe
Stretched strings – Melde’s experiment one loop (one antinode), two loops, three loops and
more. Figure 15.7 shows a vibrating string where the
A string is attached at one end to a vibration generator, frequency of the vibrator has been set to produce two
driven by a signal generator (Figure 15.6). The other end loops.
hangs over a pulley and weights maintain the tension in the
string. When the signal generator is switched on, the string A flashing stroboscope is useful to reveal the motion
vibrates with small amplitude. Larger amplitude stationary of the string at these frequencies, which look blurred to
waves can be produced by adjusting the frequency. the eye. The frequency of vibration is set so that there
are two loops along the string; the frequency of the
pulley vibration stroboscope is set so that it almost matches that of the
generator vibrations. Now we can see the string moving ‘in slow
motion’, and it is easy to see the opposite movements of
weights the two adjacent loops.
signal

generator

Figure 15.6  Melde’s experiment for investigating Figure 15.7  When a stationary wave is established, one
stationary waves on a string. half of the string moves upwards as the other half moves
downwards. In this photograph, the string is moving too
fast to observe the effect.

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BOX 15.1: Observing stationary waves (continued)

This experiment is known as Melde’s experiment, tuning
and it can be extended to investigate the effect of fork
changing the length of the string, the tension in the
string and the thickness of the string.

Microwaves air

Start by directing the microwave transmitter at a metal 4

plate, which reflects the microwaves back towards the

source (Figure 15.8). Move the probe receiver around

in the space between the transmitter and the reflector

and you will observe positions of high and low intensity.

This is because a stationary wave is set up between the

transmitter and the sheet; the positions of high and low

intensity are the antinodes and nodes respectively. water
If the probe is moved along the direct line from

the transmitter to the plate, the wavelength of the

microwaves can be determined from the distance

between the nodes. Knowing that microwaves travel

at the speed of light c (3.0 × 108 m s−1), we can then Figure 15.9  A stationary wave is created in the air in the
determine their frequency f using the wave equation: tube when the length of the air column is adjusted to the
correct length.
c  =  f λ

214 of a phenomenon called resonance. The experiment
reflecting described here is known as the resonance tube.
sheet
For resonance to occur, the length of the air column

must be just right. The air at the bottom of the tube is

probe unable to vibrate, so this point must be a node. The air

at the open end of the tube can vibrate most freely, so

this is an antinode. Hence the length of the air column

must be one-quarter of a wavelength (Figure 15.10a).

(Alternatively, the length of the air column could be

set to equal three-quarters of a wavelength – see

microwave Figure 15.10b.) a b
transmitter meter antinode

Figure 15.8  A stationary wave is created when λ
microwaves are reflected from the metal sheet. 4

An air column closed at one end node λ
antinode 4
A glass tube (open at both ends) is clamped so that
one end dips into a cylinder of water. By adjusting its Figure 15.10  Stationary node
height in the clamp, you can change the length of the wave patterns for air
column of air in the tube (Figure 15.9). When you hold in a tube with one end
a vibrating tuning fork above the open end, the air closed.
column may be forced to vibrate, and the note of the
tuning fork sounds much louder. This is an example

Chapter 15: Stationary waves

BOX 15.1: Observing stationary waves (continued) should hear a note whose pitch depends on the length 215
of the tube. Now cover the bottom of the tube with the
Figure 15.11  a The standard palm of your hand and repeat the process. The pitch of
representation of a standing the note now produced will be about an octave higher
sound wave may suggest that it than the previous note, which means that the frequency
is a transverse wave. b A sound is approximately twice the original frequency.
wave is really a longitudinal wave,
so that the particles vibrate as It is rather surprising that a standing wave can be set
shown. up in an open column of air in this way. What is going
ab on? Figure 15.12 compares the situation for open and
Take care! The representation of standing sound closed tubes. An open-ended tube has two open ends,
waves can be misleading. Remember that a sound wave so there must be an antinode at each end. There is a
is a longitudinal wave, but the diagram we draw is more node at the midpoint.
like a transverse wave. Figure 15.11a shows how we
normally represent a standing sound wave, while Figure For a tube of length l you can see that in the closed
15.11b shows the direction of vibration of the particles tube the standing wave formed is one-quarter of a
along the wave. wavelength, so the wavelength is 4l, whereas in the
open tube it is half a wavelength, giving a wavelength
Open-ended air columns of 2l. Closing one end of the tube thus halves the
wavelength of the note and so the frequency doubles.
The air in a tube which is open at both ends will vibrate
in a similar way to that in a closed column. Take an l
open-ended tube and blow gently across the top. You
ab
Figure 15.12  Standing wave patterns for sound waves in
a a closed tube, and b an open tube.

QUESTIONS 3 a Draw a stationary wave pattern for the microwave
experiment in Box 15.1. Clearly show whether
2 Look at the stationary (standing) wave on the string there is a node or an antinode at the reflecting
in Figure 15.7 on page 213. The length of the vibrating sheet.
section of the string is 60 cm.
a Determine the wavelength of the stationary b The separation of two adjacent points of high
wave and the separation of the two neighbouring intensity is found to be 14 mm. Calculate the
antinodes. wavelength and frequency of the microwaves.

The frequency of vibration is increased until a 4 Explain how two sets of identical but oppositely
stationary wave with three antinodes appears on the travelling waves are established in the microwave and
string. air column experiments described in Box 15.1.
b Sketch a stationary wave pattern to illustrate the
appearance of the string.
c What is the wavelength of this stationary wave?

Cambridge International AS Level Physics

Stationary waves and musical l = length of string wavelength frequency
instruments
fundamental N A N λ = 2l
The production of different notes by musical instruments
often depends on the creation of stationary waves (Figure f0
15.13). For a stringed instrument such as a guitar, the two
ends of a string are fixed, so nodes must be established at second NA N AN λ=l 2f0
these points. When the string is plucked half-way along harmonic NA N A N AN 3f0
its length, it vibrates with an antinode at its midpoint. third λ = 2 l
This is known as the fundamental mode of vibration of harmonic 3
the string. The fundamental frequency is the minimum
frequency of a standing wave for a given system or Figure 15.14  Some of the possible stationary waves for a
arrangement. fixed string of length l. The frequency of the harmonics is a
multiple of the fundamental frequency f0.

A AA
N N
l = length N A
of air
A N
column A
NN

fundamental second third
harmonic harmonic

wavelength λ = 4l λ = 4l λ = 4l
3 5

Figure 15.13  When a frequency f0 3f0 5f0
guitar string is plucked,
216 the vibrations of the Figure 15.15  Some of the possible stationary waves for an air
strings continue for some column, closed at one end. The frequency of each harmonic is
an odd multiple of the fundamental frequency f0.
time afterwards. Here you
can clearly see a node
close to the end of each
string. Determining the wavelength

Similarly, the air column inside a wind instrument is and speed of sound

caused to vibrate by blowing, and the note that is heard Since we know that adjacent nodes (or antinodes) of a
depends on a stationary wave being established. By
changing the length of the air column, as in a trombone, stationary wave are separated by half a wavelength, we can
the note can be changed. Alternatively, holes can be use this fact to determine the wavelength λ of a progressive
uncovered so that the air can vibrate more freely, giving a wave. If we also know the frequency f of the waves, we can
different pattern of nodes and antinodes. find their speed v using the wave equation v = f λ.

In practice, the sounds that are produced are made closed end
up of several different stationary waves having different

patterns of nodes and antinodes. For example, a guitar glass tube A
string may vibrate with two antinodes along its length.
This gives a note having twice the frequency of the N
fundamental, and is described as a harmonic of the
fundamental. The musician’s skill is in stimulating the signal loudspeaker A
string or air column to produce a desired mixture of generator N
frequencies.
A
The frequency of a harmonic is always a multiple of the
N dust piles
A up at nodes

fundamental frequency. The diagrams show some of the Figure 15.16  Kundt’s dust tube can be used to determine the
modes of vibration of a fixed length of string (Figure 15.14) speed of sound.
and an air column in a tube of a given length that is closed
at one end (Figure 15.15).

Chapter 15: Stationary waves

One approach uses Kundt’s dust tube (Figure 15.16). A QUESTIONS 217
loudspeaker sends sound waves along the inside of a tube.
The sound is reflected at the closed end. When a stationary 5 a For the arrangement shown in Figure 15.17,
wave is established, the dust (fine powder) at the antinodes suggest why it is easier to determine accurately
vibrates violently. It tends to accumulate at the nodes, the position of a node rather than an antinode.
where the movement of the air is zero. Hence the positions
of the nodes and antinodes can be clearly seen. b Explain why it is better to measure the distance
across several nodes.
BOX 15.2: Using stationary sound waves to
determine λ and v 6 For sound waves of frequency 2500 Hz, it is found
that two nodes are separated by 20 cm, with three
This method is shown in Figure 15.17; it is the same antinodes between them.
arrangement as used for microwaves (Box 15.1 a Determine the wavelength of these sound
above). The loudspeaker produces sound waves, waves.
and these are reflected from the vertical board. b Use the wave equation v  =  f λ to determine the
The microphone detects the stationary sound wave speed of sound in air.
in the space between the speaker and the board,
and its output is displayed on the oscilloscope. BOX 15.3: Reducing and eliminating errors
It is simplest to turn off the time-base of the
oscilloscope, so that the spot no longer moves The resonance tube experiment (Figure 15.9 on
across the screen. The spot moves up and down the page 214) can be used to determine the wavelength
screen, and the height of the vertical trace gives a and speed of sound with a high degree of accuracy.
measure of the intensity of the sound. However, to do this, it is necessary to take account
of a systematic error in the experiment.
By moving the microphone along the line
between the speaker and the board, it is easy Look at the representation of the stationary
to detect nodes and antinodes. For maximum waves in the tubes shown in Figure 15.10 on
accuracy, we do not measure the separation of page 214. In each case, the antinode at the top of the
adjacent nodes; it is better to measure the distance tube is shown extending slightly beyond the open
across several nodes. end of the tube. This is because experiment shows
that the air slightly beyond the end of the tube
oscilloscope vibrates as part of the stationary wave. This is shown
more clearly in Figure 15.18.
loudspeaker
c

l
4

to signal microphone 3l
generator 4
(2 kHz) Figure 15.18  The antinode
at the open end of a
reflecting resonance tube is formed
board at a distance c beyond the
Figure 15.17  A stationary sound wave is established open end of the tube.
between the loudspeaker and the board.

Cambridge International AS Level Physics

BOX 15.3: Reducing and eliminating errors (continued)

The antinode is at a distance c beyond the end So, although we do not know the value of c, we can
of the tube, where c is called the end-correction. make two measurements (l1 and l2) and obtain an
Unfortunately, we do not know the value of c. It cannot accurate value of λ. (You may be able to see from Figure
15.18 that the difference in lengths of the two tubes is
be measured directly. However, we can write: indeed equal to half a wavelength.)

for the shorter tube, λ  =  l1 + c The end-correction c is an example of a systematic
4 error. When we measure the length l of the tube, we
are measuring a length which is consistently less than
for the longer tube, 3 λ  =  l2 + c the quantity we really need to know (l + c). However,
4 by understanding how the systematic error affects
the results, we have been able to remove it from our
Subtracting the first equation from the second equation measurements.
gives:
You will find more detailed discussion of systematic
3 λ  −  4λ = (l2 + c) − (l1 + c) errors in Chapter P1: Practical skills at AS level.
4

Simplifying gives:

λ  =  l2 −  l1
2

Hence:
λ  = 2(l2 −  l1)

218

QUESTION

7 In a resonance tube experiment, resonance is
obtained for sound waves of frequency 630 Hz
when the length of the air column is 12.6 cm and
again when it is 38.8 cm. Determine:
a the wavelength of the sound waves causing
resonance
b the end-correction for this tube
c the speed of sound in air.

Summary ■■ An antinode is a point of maximum amplitude.

■■ Stationary waves are formed when two identical waves ■■ Adjacent nodes (or adjacent antinodes) are separated
travelling in opposite directions meet and superpose. by a distance equal to half a wavelength.
This usually happens when one wave is a reflection of
the other. ■■ We can use the wave equation v  = f λ to determine the
speed v or the frequency f of a progressive wave.
■■ A stationary wave has a characteristic pattern of nodes The wavelength λ is found using the nodes or
and antinodes. antinodes of the stationary wave pattern.

■■ A node is a point where the amplitude is always zero.

Chapter 15: Stationary waves

End-of-chapter questions

1 Figure 15.19 shows a stationary wave on a string.

vibrator
Figure 15.19 For End-of-chapter Question 1.

a On a copy of Figure 15.19, label one node (N) and one antinode (A). [1]
b Mark on your diagram the wavelength of the standing wave and label it λ. [1]
c The frequency of the vibrator is doubled. Describe the changes in the standing wave pattern. [1]
[1]
2 A tuning fork which produces a note of 256Hz is placed above a tube which is nearly filled with water. [2]
The water level is lowered until resonance is first heard. [4]
a Explain what is meant by the term resonance.
b The length of the column of air above the water when resonance is first heard is 31.2cm. 219
Calculate the speed of the sound wave.
[2]
3 a State two similarities and two differences between progressive waves and stationary waves. [3]
b Figure 15.20 shows an experiment to measure the speed of a sound in a string. The frequency of the [2]
vibrator is adjusted until the standing wave shown in Figure 15.20 is formed.

vibrator 75 cm pulley

slotted masses
Figure 15.20 For End-of-chapter Question 3.

i On a copy of the diagram, mark a node (label it N) and an antinode (label it A).
ii The frequency of the vibrator is 120Hz. Calculate the speed at which a progressive wave would

travel along the string.
c The experiment is now repeated with the load on the string halved. In order to get a similar standing

wave the frequency has to be decreased to 30Hz. Explain, in terms of the speed of the wave in the
string, why the frequency must be adjusted.

Cambridge International AS Level Physics

4 Figure 15.21 shows a standing wave, of frequency 400Hz, produced by a loudspeaker in a closed tube.

59.0 cm loudspeaker

piston to signal generator

AB

Figure 15.21 For End-of-chapter Question 4.

a Describe the movement of the air particles at: [2]
iA [1]
ii B.
[4]
b The piston is slowly moved outwards. The next resonant peak is heard when the length of [1]
the column of air is 99.4cm. [2]
Calculate: [2]
i the speed of the sound
ii the end correction which must be applied when using this tube. [2]
[5]
5 a Explain what is meant by: [2]
i a coherent source of waves.

220 ii phase difference.
b A student, experimenting with microwaves, sets up the arrangement shown in Figure 15.22.

microwave microwave metal
transmitter receiver plate

42.0 cm

A B
Figure 15.22 For End-of-chapter Question 5.

With the metal plate at position A there is a very small signal. He slowly moves the plate back,
leaving the receiver in the same position. As he does so he finds that the intensity initially
rises until it becomes a maximum, then falls back to a minimum. This cycle repeats a total
of five times until the plate reaches position B, where once again there is a minimum.

i Explain why a series of maxima and minima are heard.

ii Calculate the frequency of the microwaves.

c Explain why there was a minimum when the plate was at position A, next to the detector.

Chapter 15: Stationary waves

6 Figure 15.23 shows an experiment to measure the speed of sound in air.
P Q RS T

loudspeaker pile of dust glass tube
P
Q

90 cm

signal generator [3] 221
[3]
Figure 15.23 For End-of-chapter Question 6.
[1]
A small amount of dust is scattered along the tube. The loudspeaker is switched on. When the frequency
is set at 512Hz the dust collects in small piles as shown in Figure 15.23.
a Determine the wavelength of the sound wave and calculate the speed of sound in the air in the tube.
b On a copy of the diagram, show the movement of the air particles at positions P, Q, R, S and T.
c Mark two points on your diagram where the movements of the air particles are 180° out of phase with

each other – label them A and B.

222 Chapter 16:
Radioactivity

Learning outcomes

You should be able to:

■■ describe the nuclear model of the atom and the evidence
for it

■■ represent nuclides using their nucleon and proton
numbers

■■ use a simple quark model to describe protons and
neutrons

■■ appreciate that electrons and neutrinos are leptons
■■ show an understanding of the nature and properties of

-, - and -radiations

Chapter 16: Radioactivity

Radioactivity at work

Radioactive substances have many uses, for example
in engineering and medicine. They must be handled
with great care to ensure that no-one becomes
contaminated and so exposed to the radiation which
comes from these substances (Figure 16.1). In this
chapter we will look at the nature of radioactive
substances and the different types of radiation they
produce.

Figure 16.1  A worker at a nuclear power station is checked for
any radioactive material on his body.

Looking inside the atom These α-particles were known to be smaller than 223
atoms, and had relatively high kinetic energies. Hence
The idea that matter is composed of very small particles they were useful in experiments designed to discover the
called atoms was first suggested by the Greeks some 2000 composition of atoms.
years ago. However, it was not until the middle of the 19th
century that any ideas about the inside of the atom were Figure 16.2  Ernest Rutherford (on the right) in the
proposed. Cavendish Laboratory, Cambridge, England. He had a loud
voice that could disturb sensitive apparatus and so the
It was the English scientist J.J. Thomson who suggested notice was a joke aimed at him.
that the atom is a neutral particle made of a positive charge
with lumps of negative charge (electrons) in it. He could
not determine the charge and the mass of the negative
particles separately, but it was clear that a new particle,
probably much smaller than the hydrogen atom, had been
discovered. Since atoms are neutral and physicists had
discovered a negatively charged part of an atom, it meant
that there were both positive and negative charges in an
atom. We now call this the plum pudding model of the
atom (positive pudding with negative plums!).

Other experiments show that the electron has a mass
of approximately 9.11 × 10−31 kg (me) and a charge of
−1.60 × 10−19 C (−e). Today we use the idea of the electron
to explain all sorts of phenomena, including electrostatics,
current electricity and electronics.

Alpha-particle scattering and
the nucleus

Early in the 20th century, many physicists were
investigating the recently discovered phenomenon of
radioactivity, the process whereby unstable nuclei emit
radiation. One kind of radiation they found consisted of
what they called α-particles (alpha-particles).

Cambridge International AS Level Physics

In 1906, while experimenting with the passage of ■■ The α-particles were detected when they struck a solid
α-particles through a thin mica sheet, Ernest Rutherford ‘scintillating’ material. Each α-particle gave a tiny flash of
(Figure 16.2) noticed that most of the α-particles passed light and these were counted by the experimenters (Geiger
straight through. This suggested to him that there might be and Marsden).
a large amount of empty space in the atom, and by 1909 he
had developed what we now call the nuclear model of the ■■ The detector could be moved round to detect α-particles
scattered through different angles.

atom. Geiger and Marsden had the difficult task of observing and

In 1911 Rutherford carried out a further series of counting the tiny flashes of light produced by individual

experiments with Hans Geiger and Ernest Marsden at the α-particles striking the scintillation screen. They had to

University of Manchester using gold foil in place of the spend several minutes in the darkened laboratory to allow

mica. They directed parallel beams of α-particles at a piece the pupils of their eyes to become dilated so that they
of gold foil only 10−6 m thick. Most of the α-particles went could see the faint flashes. Each experimenter could only

straight through. Some were deflected slightly, but about stare into the detector for about a minute before the strain

1 in 20 000 were deflected through an angle of more than was too much and they had to change places.

90°, so that they appeared to bounce back off the foil. This Explaining α-scattering
helped to confirm Rutherford in his thinking about the How can we explain the back-scattering of α-particles by

atom – that it was mostly empty space, with most of the the gold atoms?

mass and all of the positive charge concentrated in a tiny If the atom was as Thomson pictured it, with negatively

region at the centre. This central nucleus only affected the charged electrons scattered through a ‘pudding’ of positive

α-particles when they came close to it. charge, an individual α-particle would pass through it like

Later, Rutherford wrote: ‘It was quite the most a bullet, hardly being deflected at all. This is because the

incredible event that has happened to me in my life. It was
almost as incredible as if you fired a 15-inch shell at a piece α-particles are more massive than electrons – they might
push an electron out of the atom, but their own path would
224 of tissue paper and it came back and hit you.’ In fact, he was be scarcely affected.
not quite as surprised as this suggests, because the results
On the other hand, if the mass and positive charge
confirmed ideas he had used in designing the experiment.
of the atom were concentrated at one point in the atom,
scintillation
gold foil screen as Rutherford suggested, an α-particle striking this part
would be striking something more massive than itself and
source of microscope
α-particles with a greater charge. A head-on collision would send the
α-particle backwards.

The paths of an α-particle near a nucleus are shown in

Figure 16.5. Rutherford reasoned that the large deflection

to vacuum pump of the α-particle must be due to a very small charged

nucleus. From his experiments he calculated that the
diameter of the gold nucleus was about 10−14 m. It has since

been shown that the very large deflection of the α-particle

side view is due to the electrostatic repulsion between the positive

Figure 16.3  The apparatus used for the α-scattering charge of the α-particle and the positive charge of the
experiment. The microscope can be moved round to detect nucleus of the atom. The closer the path of the α-particle

scattered radiation at different angles. gets to the nucleus, the greater will be this repulsion. An

Figure 16.3 shows the apparatus used in the α-particle making a ‘head-on’ collision with a nucleus
α-scattering experiment. Notice the following points: is back-scattered through 180°. The α-particle and
nucleus both experience an equal but opposite repulsive

■■ The α-particle source was encased in metal with a small electrostatic force F. This force has a much greater effect on

aperture, allowing a fine beam of α-particles to emerge. the motion of the α-particle than on the massive nucleus

■■ Air in the apparatus was pumped out to leave a vacuum; of gold.
α-radiation is absorbed by a few centimetres of air.

■■ One reason for choosing gold was that it can be made into
a very thin sheet or foil. Rutherford’s foil was only a few
hundreds of atoms thick.

Chapter 16: Radioactivity

BOX 16.1: An analogy for Rutherford scattering From the α-particle scattering experiment, Rutherford
deduced the following.
A very simple analogy (or model) of the experiment
is shown in Figure 16.4. When you roll a ball- ■■ An α-particle is deviated due to the repulsive force
bearing down a slope towards the ‘cymbal’, it may between the α-particle and the positive charge in the
be deflected, but even if it is rolled directly at the atom.
cymbal’s centre, it does not come back – it rolls over
the centre and carries on to the other side. However, ■■ Most α-particles have little or no deviation – so most
using the ‘tin hat’ shape, with a much narrower but of an atom is empty space.
higher central bulge, any ball-bearings rolled close
to the centre will be markedly deflected, and those ■■ A very few α-particles are deviated more than 90° –
rolled directly towards it will come straight back. so most of the mass of an atom is concentrated in
a small space (the nucleus) and most of the atom is
‘cymbal’ empty space.

‘tin hat’ QUESTIONS
also known
as 1/r hill 1 Rutherford’s scattering experiments were done
in an evacuated container. Explain why this is
necessary. 225

Figure 16.4  An analogy for Rutherford’s experiment. 2 In Rutherford’s experiment, α-particles were
directed at a thin gold foil. A small fraction of the
The shape of the cymbal represents the shape α-particles were back-scattered through 180°.
of the electric field of an atom in the ‘plum pudding’
model: low central intensity and spread out. The Describe and explain how the fraction back-
‘tin hat’ represents the shape of the electric field scattered changes if each of the following changes
for the nuclear model: high central intensity and are (separately) made.
concentrated. a A thicker foil is used.
b Faster α-particles are used.
c A silver foil is used – a silver nucleus has less
positive charge than a gold nucleus.

α-particles force A simple model of the atom

nucleus After Rutherford had presented his findings, the nuclear
force model of the atom gained rapid acceptance. This
was partly because it helped chemists to explain the
Figure 16.5  Possible paths of an α-particle near a nucleus. phenomenon of chemical bonding (the way in which
The nucleus and the α-particle both experience electrostatic atoms bond together to form molecules). Subsequently,
repulsion. the proton was discovered. It had a positive charge, equal
and opposite to that of the electron. However, its mass was
too small to account for the entire mass of the atom and it
was not until the early 1930s that this puzzle was solved by
the discovery of the neutron, an uncharged particle with
a similar mass to that of the proton. This suggests a model
for the atom like the one shown in Figure 16.6:

■■ Protons and neutrons make up the nucleus of the atom.

■■ The electrons move around the nucleus in a cloud, some
closer to and some further from the centre of the nucleus.

Cambridge International AS Level Physics

nucleus So the proton has a density of roughly 1018 kg m−3. This is
also the density of a neutron, and of an atomic nucleus,
Key because nuclei are made of protons and neutrons held
= n, neutron closely together.
= p, proton
= e, electron Compare the density of nuclear material with that of
water whose density is 1000 kg m−3 – the nucleus is 1015
Figure 16.6  A simple model of the atom. If the nucleus were times as dense. Nuclear matter the size of a tiny grain of
drawn to scale, it would be invisible (and the electrons are sand would have a mass of about a million tonnes! This is
even smaller!). a consequence of the fact that the nucleus occupies only
a tiny fraction of the volume of an atom. The remainder
is occupied by the cloud of orbiting electrons whose mass
makes up less than one-thousandth of the atomic mass.

QUESTION

From this model it looks as though all matter, including 3 Gold has a density of 19 700 kg m−3.A mass of
ourselves, is mostly empty space. For example, if we scaled 193 g of gold contains 6.02 × 1023 atoms. Use this
up the hydrogen atom so that the nucleus was the size of information to estimate the volume of a gold
a 1 cm diameter marble, the orbiting electron would be a atom, and hence its radius. State any assumptions
grain of sand some 800 m away! you make.

The scale of things Nucleons and electrons

It is useful to have an idea of the approximate sizes of We will start this section with a summary of the particles
typical particles: mentioned so far (Table 16.1). All nuclei, except the lightest
226 ■■ radius of proton ~ radius of neutron ~ 10−15 m form of hydrogen, contain protons and neutrons, and
each nucleus is described by the number of protons and
■■ radius of nucleus ~ 10−15 m to 10−14 m neutrons that it contains.

■■ radius of atom ~ 10−10 m

■■ size of molecule ~ 10−10 m to 10−6 m.

(Some molecules, such as large protein molecules, are very ■■ Protons and neutrons in a nucleus are collectively called
large indeed – compared to an atom!) nucleons. For example, in a nucleus of gold, there are 79
protons and 118 neutrons, giving a total of 197 nucleons
The radii of nuclear particles are often quoted in altogether.
femtometres (fm), where 1 fm = 10−15 m.
■■ The total number of nucleons in a nucleus is called the
Nuclear density nucleon number (or mass number) A.

We can picture a proton as a small, positively charged ■■ The nucleon number is equal to the sum of the number of
sphere. Knowing its mass and radius, we can calculate its neutrons in the nucleus, the neutron number N, and the
density: number of protons, the proton number (or atomic number)
Z, i.e.
mass of proton mp = 1.67 × 10−27 kg A = N + Z

radius of proton r = 0.80 fm = 0.80 × 10−15 m Particle Relative mass (proton = 1)(a) Charge(b)

(In fact, the radius of the proton is not very accurately proton (p) 1 +e
known; it is probably between 0.80 × 10−15 m and
0.86 × 10−15 m.) neutron (n) 1 0

volume of proton = 43 πr 3 = 43 π × (0.80 × 10−15)3 electron (e) 0.0005 −e
= 2.14 × 10−45 m3 ≈ 2.1 × 10−45 m3
alpha-particle (α) 4 +2e

density = mass (a)The numbers given for the masses are approximate.
volume (b)e = 1.60 × 10−19 C.

density = 1.67 × 10−27 ≈ 7.8 × 1017 kg m−3 Table 16.1  Summary of the particles that we have met so far
2.14 × 10−45 in this chapter. The α-particle is in fact a helium nucleus (with
two protons and two neutrons).

Chapter 16: Radioactivity

The nucleus of any atom can be represented by the Isotopes

symbol for the element along with the nucleon number Although atoms of the same element may be identical
chemically, their nuclei may be slightly different. The
and proton number, as shown below: number of protons in the nucleus of an atom determines
what element it is: helium always has 2 protons, carbon
nucleon nnuummbbeerr element symbol   ZAX 6 protons, oxygen 8 protons, neon 10 protons, radium 88
proton protons, uranium 92 protons, and so on.

oxygen 186O   gold 19797 Au   uranium 29382 U However, the number of neutrons in the nuclei for a
A specific combination of protons and neutrons in a given element can vary. Take neon as an example. Three
different naturally occurring forms of neon are:
nucleus is called a nuclide.
2100 Ne     1210 Ne     2120 Ne
The proton and nucleon numbers of some common The first has 10 neutrons in the nucleus, the second 11
neutrons and the third 12 neutrons. These three types of
nuclides are shown in Table 16.2. neon nuclei are called isotopes of neon. Each isotope has
the same number of protons (for neon this is 10) but a
QUESTIONS different number of neutrons. The word ‘isotope’ comes
from the Greek isotopos (same place), because all isotopes
4 Table 16.2 shows the proton and nucleon numbers of the same element have the same place in the Periodic
of several nuclei. Determine the number of Table of elements.
neutrons in the nuclei of the following elements
shown in the table: Isotopes are nuclei of the same element with different
numbers of neutrons but the same number of protons.
a nitrogen d gold

b bromine e mercury.

c silver

5 State the charge of each of the following in terms
of the elementary charge e:

a proton d molecule

b neutron e α-particle. 227

c nucleus

Element Nucleon Proton Element Nucleon Proton
number A number Z number A number Z
hydrogen 1 1 bromine 79 35
helium 4 2 silver 107 47
lithium 7 3 tin 120 50
beryllium 9 4 iodine 130 53
boron 11 5 caesium 133 55
carbon 12 6 barium 138 56
nitrogen 14 7 tungsten 184 74
oxygen 16 8 platinum 195 78
neon 20 10 gold 197 79
sodium 23 11 mercury 202 80
magnesium 24 12 lead 206 82
aluminium 27 13 bismuth 209 83
chlorine 35 17 radium 226 88
calcium 40 20 uranium 238 92
iron 56 26 plutonium 239 94
nickel 58 28 americium 241 95

Table 16.2  Proton and nucleon numbers of some nuclides.

Cambridge International AS Level Physics

Any atom is electrically neutral (it has no net positive or

negative charge), so the number of electrons surrounding

the nucleus must equal the number of protons in the

nucleus of the atom. If an atom gains or loses an electron,

it is no longer electrically neutral and is called an ion.

For an atom, the number of protons (and hence the

number of electrons) determines the chemical properties

of the atom. The number of protons and the number of

neutrons determine the nuclear properties. It is important

to realise that, since the number of protons, and therefore

the number of electrons, in isotopes of the same element

are identical, they will all have the same chemical

properties but very different nuclear properties. Figure 16.8  The Horsehead Nebula in Orion. The large
(deuHteyrdiruomge)nanhdas13Hthr(tereitiimumpo)r(tFaingtuirseot1o6p.7e)s.,1111HH,a21nHd coloured regions are expanses of dust and gas, mostly
deuterium occur naturally, but tritium has to be made. hydrogen, that are ionised by nearby stars so that they emit
light. The dark ‘horse head’ is where the areas of gas and dust
remain in atomic form and block out the light from behind.
Deuterium and tritium form the fuel of many fusion

research reactors. Hydrogen is the most abundant element

in the Universe (Figure 16.8), because it consists of just one Table 16.3 gives details of some other commonly
proton and one electron, which is the simplest structure occurring isotopes.
possible for an atom.
Element Nucleon Proton Neutron
The different numbers of neutrons in the isotopes of number A number Z number N

an element means that the isotopes will have different

228 relative atomic masses. There are differences too in some hydrogen 1 1 0
of their physical properties, such as density and boiling 2 1 1

point. For example, heavy water, which is water containing carbon 12 6 6
deuterium, has a boiling point of 104 °C under normal 14 6 8
atmospheric pressure.
oxygen 16 8 8

18 8 10

neon 20 10 10

21 10 11

potassium 39 19 20

40 19 21

strontium 88 38 50

11H (sometimes called ‘protium’) 12H, deuterium 90 38 52

caesium 135 55 80

137 55 82

lead 206 82 124

Key radium 208 82 126
= n, neutron uranium 226 88 138
= p, proton 228 88 140
= e, electron 235 92 143

13H, tritium 238 92 146

Figure 16.7  The isotopes of hydrogen. Table 16.3  Some commonly occurring isotopes.

Chapter 16: Radioactivity

QUESTIONS tightly, and this can make the nucleus unstable. The more
protons there are in a nucleus, the greater the electric
6 Uranium has atomic number 92. Two of its forces between them, and we need a few extra neutrons to
common isotopes have nucleon numbers 235 help ‘keep the protons apart’. This is why heavy nuclei have
and 238. Determine the number of neutrons for more neutrons than protons.
these isotopes.
The proton and neutron numbers for some common
7 There are seven naturally occurring isotopes of nuclides are shown in Table 16.3. You can see that for light
mercury, with nucleon numbers (and relative elements these two numbers are the same, but they become
abundances) of 196 (0.2%), 198 (10%), 199 (16.8%), very different for heavy elements. Adding more neutrons
200 (23.1%), 201 (13.2%), 202 (29.8%) and helps to keep the nucleus stable, but when the number of
204 (6.9%). protons is greater than 83, adding more neutrons is not
a Determine the proton and neutron numbers enough. Elements with a proton number greater than 83
for each isotope. are all unstable – they undergo radioactive decay.
b Determine the average relative atomic mass
(equivalent to the ‘average nucleon number’) Most atoms that make up our world have stable nuclei;
of naturally occurring mercury. that is, they do not change as time goes by, which is quite
fortunate really! However, some are less stable and give
8 Group the following imaginary elements A–H into out radiation. Whether or not an atom is unstable depends
isotopes and name them using the Periodic Table on the numbers of protons and neutrons in its nucleus.
in the back of the book (Appendix 3). Hydrogen-1 (1p), helium-4 (2p, 2n), carbon-12 (6p, 6n)
and oxygen-16 (8p, 8n) are all stable – but add or subtract
ABCDE FGH neutrons and the situation changes.

Proton 20 23 21 22 20 22 22 23 For example, add a neutron to helium-4 and you
number get helium-5, a very unstable nucleus – it undergoes
radioactive emission. (There is much more about
Nucleon 44 50 46 46 46 48 50 51 radioactive decay later in this chapter.) 229
number

Forces in the nucleus QUESTION

As you know from earlier in this chapter, there are two 9 State which of the following forces act between
kinds of particle in the nucleus of an atom: protons, which protons and neutrons in a nucleus.
each carry positive charge +e; and neutrons, which are a gravitational
uncharged. It is therefore quite surprising that the nucleus b electrostatic
holds together at all. You would expect the electrostatic c strong nuclear.
repulsions from all those positively charged protons to
blow it apart. The fact that this does not happen is very Fundamental particles?
good evidence for the existence of an attractive force
between the nucleons. This is called the strong nuclear Chemistry is very complicated because there are literally
force. It only acts over very short distances (10−14 m), and it billions of different molecules that can exist. The discovery
is what holds the nucleus together. of the Periodic Table simplified things because it suggested
that there were roughly 92 different elements whose atoms
Diluting the protons could be arranged to make these various molecules. The
idea that atoms are made up of just three types of particle
In small nuclei the strong nuclear force from all the (protons, neutrons and electrons) seemed to simplify
nucleons reaches most of the others in the nucleus, but things still more, and scientists were very happy with it
as we go on adding protons and neutrons the balance because it seemed to provide a very simple explanation of
becomes much finer. The longer-range electrostatic force a complex world. Protons, neutrons and electrons were
affects the whole nucleus, but the short-range strong thought of as fundamental particles, which could not be
nuclear force of any particular nucleon only affects those subdivided further.
nucleons around it – the rest of the nucleus is unaffected.
In a large nucleus the nucleons are not held together so

Cambridge International AS Level Physics

However, in the middle decades of the 20th century,
physicists discovered many other particles that did not fit
this pattern. They gave them names such as pions, kaons,
muons, etc., using up most of the letters of the Greek
alphabet.

These new particles were found in two ways:
■■ by looking at cosmic rays, which are particles that arrive at

the Earth from outer space
■■ by looking at the particles produced by high-energy

collisions in particle accelerators (Figure 16.9).

Figure 16.10  Particle accelerators have become bigger and
bigger as scientists have sought to look further and further into
the fundamental nature of matter. This is one of the particle
detectors of the Large Hadron Collider (LHC), as it was about to
be installed. The entire collider is 27 km in circumference.

fundamental questions about this family of particles. In

Figure 16.9  Particle tracks in a bubble chamber detector. 2013, they announced the discovery of the Higgs boson, a
A particle has entered from the left and then struck another particle which was predicted 50 years earlier and which is
230 particle just to the right of the centre. Four new particles fly required to explain why matter has mass.

out from the point of impact. Inside hadrons

To sort out the complicated picture of the hadron family

The discovery of new particles with masses different of particles, Murray Gell-Mann in 1964 proposed a new
from those of protons, neutrons and electrons suggested model. He suggested that they were made up of just a few
that these were not fundamental particles. Various different particles, which he called quarks.
attempts were made to tidy up this very confusing picture.
Figure 16.11 shows icons used to represent three
In principle, we can never know for certain whether quarks, together with the corresponding antiquarks. These
a particle such as the electron is truly fundamental; the are called the up (u), down (d) and strange (s) quarks.
possibility will always remain that a physicist will discover Gell-Mann’s idea was that there are two types of hadron:
some deeper underlying structure. baryons, made up of three quarks, and mesons, made up
of two quarks. In either case, the quarks are held together

Families of particles by the strong nuclear force. For example:
■■ A proton is made up of two up quarks and a down quark;
Today, sub-atomic particles are divided into two families:
proton = (uud).

■■ Hadrons such as protons and neutrons. These are all ■■ A neutron is made up of one up quark and two down quarks;
particles that are affected by the strong nuclear force. neutron = (udd).

■■ Leptons such as electrons. These are particles that are ■■ A pi+ meson is mmeasdoenu=p(oufd–a).n up quark and a down
unaffected by the strong nuclear force. antiquark; pi+

The word ‘hadron’ comes from a Greek word meaning ■■ Aanpthisitmraensgoenqisuamrka;dpehuipmoefsaonst=ra(ns–sg)e. quark and an
‘bulky’, while ‘lepton’ means ‘light’ (in mass). It is certainly
true that protons and neutrons are bulky compared to Antiquarks are shown with a ‘bar’ on top of the letter
electrons. for the quark. Antiquarks are needed to account for the
existence of antimatter. This is matter that is made of
At the Large Hadron Collider (Figure 16.10) at the antiparticles; when a particle meets its antiparticle, they
CERN laboratory in Geneva, physicists are experimenting annihilate each other, leaving only photons of energy.
with hadrons in the hope of finding answers to some

Chapter 16: Radioactivity

Up Down Strange In fact, there are two types of β-radiation. The more
familiar is beta-minus (β−) radiation, which is simply
Quarks an electron, with negative charge of −e. However, there
are also many unstable nuclei that emit beta-plus (β+)
Antiquarks radiation. This radiation is in the form of positrons,
similar to electrons in terms of mass but with positive
Figure 16.11  Icons representing three ‘flavours’ of quark, charge of +e. Positrons are a form of antimatter. When a
up, down and strange, and their antiquarks. positron collides with an electron, they annihilate each
other. Their mass is converted into electromagnetic energy
Discovering radioactivity in the form of two gamma photons (Figure 16.13).

The French physicist Henri Becquerel (Figure 16.12) is e– γ
credited with the discovery of radioactivity in 1896. He
had been looking at the properties of uranium compounds γ
when he noticed that they affected photographic film – he
realised that they were giving out radiation all the time e+
and he performed several ingenious experiments to shed
light on the phenomenon. e– + e+ γ+γ
electron positron gamma-ray photons

Figure 16.13  Energy is released in the annihilation of matter
and antimatter.

Table 16.4 shows the basic characteristics of the 231
different types of radiation. The masses are given relative
to the mass of a proton; charge is measured in units of e,
the elementary charge.

Radiation Symbol Mass Charge Typical speed
-particle (relative +2e ‘slow’ (106 m s−1)
−-particle , 42He to proton) −e ‘fast’ (108 m s−1)
+-particle , −, e,  –01e
4 +e ‘fast’ (108 m s−1)
-ray , +, e+,  1
+01e 1840 0 speed of light
1 (3 × 108 m s−1)
1840

0

Figure 16.12  Henri Becquerel, the discoverer of radioactivity, Table 16.4  The basic characteristics of ionising radiations.
in his laboratory. His father and grandfather had been
professors of physics in Paris before him. Note the following points:

Radiation from radioactive ■■ α- and β-radiation are particles of matter. A γ-ray
substances is a photon of electromagnetic radiation, similar to
an X-ray. (X-rays are produced when electrons are
There are three types of radiation which are emitted by decelerated; γ-rays are produced in nuclear reactions.)
radioactive substances: alpha (α), beta (β) and gamma (γ)
radiations come from the unstable nuclei of atoms. Nuclei ■■ An α-particle consists of two protons and two
consist of protons and neutrons, and if the balance between neutrons; it is a nucleus of helium-4. A β−-particle is
these two types of particles is too far to one side, the simply an electron and a β+-particle is a positron.
nucleus may emit α- or β-radiation as a way of achieving
greater stability. Gamma-radiation is usually emitted after ■■ The mass of an α-particle is nearly 10 000 times that of
α or β decay, to release excess energy from the nuclei. an electron and it travels at roughly one-hundredth of
the speed of an electron.

Cambridge International AS Level Physics

Discovering neutrinos In α decay, an alpha particle (two protons and two
neutrons) is emitted by a nucleus. Although these nucleons
There is a further type of particle which we need to are now outside the nucleus, the equation shows that there
consider. These are the neutrinos. When β decay was first is the same number of nucleons after the decay (218 + 4) as
studied, it was realised that β-particles were electrons before the decay (222). So nucleon number A is conserved.
coming from the nucleus of an atom. There are no Similarly, proton number Z is conserved (84 + 2 = 86).
electrons in the nucleus (they ‘orbit’ outside the nucleus),
so the process was pictured as the decay of a neutron to The conservation of nucleon number and proton
give a proton and an electron. number are important laws in nuclear physics. They apply
to all nuclear changes, not just to α and β decay.
It was noticed that β-particles were emitted with a
range of speeds – some travelled more slowly than others. There is a third quantity that is conserved. You
It was deduced that some other particle must be carrying might expect mass to be conserved, but this is not so.
off some of the energy and momentum released in the For example, in the α decay equation given above, the
decay. This particle is now known as the antineutrino (or, combined mass of the polonium nucleus and the alpha
more correctly, the electron antineutrino), with symbol –ν. particle is slightly less than that of the original radon
The decay equation for β− decay is written as: nucleus. The ‘lost’ mass has become energy – this is where
the fast-moving alpha particle gets its kinetic energy. The

relationship between mass m and energy E is given by
Einstein’s equation E = mc2, where c is the speed of light
beta-minus (β−) decay:  10n → 11p +  −01e + –ν in free space. So, instead of saying that mass is conserved

in nuclear processes, we have to say that mass–energy is

Neutrinos are bizarre particles. They have very little mass conserved. There is much more about this in Chapter 31.

(much less than an electron) and no electric charge, which
makes them very difficult to detect. The Austrian physicist Fundamental families
232 Wolfgang Pauli predicted their existence in 1930, long
before they were first detected in 1956. Electrons and neutrinos both belong to the family of
In β+ decay, a proton decays to become a neutron and fundamental particles called leptons. These are particles
an electron neutrino (symbol ν) is released: that do not feel the strong nuclear force. Recall from
page 230 that particles that experience the strong force

beta-plus (β+) decay:  11p → 10n +  +01e + ν are hadrons, and that these are made up of fundamental

particles called quarks.

So we have two families of fundamental particles,

The two equations highlighted above show two important quarks and leptons. How can we understand β decay in
terms of these particles?
features of radioactive decay. Firstly, nucleon number
A is conserved; that is, there are as many nucleons after Consider first β− decay, in which a neutron decays. A
the decay as there were before. In β– decay, a neutron has neutron consists of three quarks (up, down, down or u d
become a proton so that the total number of nucleons is d). It decays to become a proton (u u d). Comparing these
unchanged. In β+ decay, a proton becomes a neutron, so shows that one of the down quarks has become an up quark.
again A is conserved. In the process, it emits a β-particle and an antineutrino:

Secondly, proton number Z is also conserved. In β– d → u + −10e + –ν
decay, we start with a neutron (Z = 0). After the decay, we
have a proton (Z = +1) and a β– particle (Z = –1). Together In β+ decay, a proton decays to become a neutron. In this
these have Z = 1 – 1 = 0. Since Z tells us about the charge of case, an up quark becomes a down quark:

u → d + +01e + ν

each particle, we would be surprised if we had a different

amount of charge after the decay than before the decay. A Fundamental forces
similar analysis shows that Z is conserved in β+ decay.
Do these conservation laws apply to α decay? Here is an The nucleus is held together by the strong nuclear force,
acting against the repulsive electrostatic or Coulomb
equation that represents a typical α decay:
force between protons. This force explains α decay, when
28226Rn → 28184Po + 24He a positively charged α-particle flies out of the nucleus,

leaving it with less positive charge.

Chapter 16: Radioactivity

However, the strong force cannot explain β decay. Alpha-radiation is the most strongly ionising, because 233
Instead, we have to take account of a further force within the mass and charge of an α-particle are greater than
the nucleus, the weak interaction, also known as the weak those of a β-particle, and it usually travels more slowly.
nuclear force. This is a force that acts on both quarks and This means that an α-particle interacts more strongly with
leptons. The weak interaction is responsible for β decay. any atom that it passes, and so it is more likely to cause
ionisation. Beta-particles are much lighter and faster, and
QUESTIONS so their effect is smaller. Gamma-radiation also causes
ionisation, but not as strongly as α- and β-particles, as
10 The equation 11p → 10n + +01e + ν represents β+ decay. γ-rays are not charged.
Use the equation to explain why the neutrino ν can
have no charge and very little mass. QUESTION

11 What are the differences between a proton, a 14 a Explain why you would expect β−-particles to
positron and a photon? You can describe how travel further through air than α-particles.
their masses differ, how their charges differ, or
whether they are particles or antiparticles. b Explain why you would expect β−-particles to
travel further through air than through metal.
12 State the names of:
a all the hadrons that are mentioned in this Behaviour of radiations in electric and
chapter magnetic fields
b all the leptons that are mentioned in this
chapter. Because α-, β−- and γ-radiations have different charges,
or no charge, they behave differently in electric and
13 State two differences between hadrons and magnetic fields. This can be used to distinguish one kind
leptons. of radiation from another.

Properties of ionising radiation Figure 16.15 shows the effect of an electric field. A
mixture of α-, β−- and γ-radiations is passing through the
Radiation affects the matter it passes through by causing gap between two parallel plates; the electric field in this
ionisation. Both α- and β-particles are fast-moving space is uniform (Chapter 8). Since α- and β−-particles
charged particles, and if they collide with or pass close are charged, they are attracted to the plate that has the
to atoms, they may knock or drag electrons away from opposite charge to their own. β−-particles are deflected
the atoms (Figure 16.14). The resulting atoms are said to more than α-particles, since their mass is so much less.
be ionised, and the process is called ionisation. In the Gamma-rays are undeflected since they are uncharged.
process, the radiation loses some of its kinetic energy.
After many ionisations, the radiation loses all of its energy +
and no longer has any ionising effect.
   β−  
γ
α

– – –
– α-particle Figure 16.15  An electric field can be used to separate α-, β−-
and γ- radiations. (The deflection of the α-radiation has been
greatly exaggerated here.)

Figure 16.14  As an α-particle passes through a material, it
causes ionisation of atoms.

Cambridge International AS Level Physics

Figure 16.16 shows the effect of a magnetic field. In this
case, the deflecting force on the particles is at right angles
to their motion.

α

γ

magnetic field β−
into paper in
this region

Figure 16.16  A magnetic field may also be used to separate
α-, β−- and γ- radiations. The deflection of the α-radiation has
been greatly exaggerated here.

QUESTION

15 a Some radioactive substances emit α-particles Figure 16.17  Alpha-particle tracks show up in this photograph
having two different speeds. Draw a diagram of a cloud chamber. Notice that all the particles travel roughly
similar to Figure 16.16 to show how these the same distance through the air, indicating that they all have
particles would move in a uniform electric roughly the same initial kinetic energy.
field. Label your diagram to show the tracks
Alpha-particles can also be detected by a solid-state
234 of the faster and slower α-particles. detector, or by a Geiger–Müller (GM) tube with a thin
b A β−-emitting radioactive substance emits end-window (Figure 16.18) connected to an electronic
β−-particles with a range of speeds. Add to counter. By moving the source back and forth in front of
the diagram you drew in a to show how these the detector, it is simple to show that the particles only
particles would behave in a uniform electric penetrate 5 or 6 cm of air. Similarly, with the source close
field. to the detector, it can be shown that a single sheet of paper
is adequate to absorb all of the α-radiation.
Radiation penetration

Safety note paper GM tube
When working with radioactive sources, it is essential to
follow the relevant safety regulations, which your teacher α-source
will explain to you.
Figure 16.18  Alpha-radiation can be absorbed by a single
Alpha-radiation sheet of paper.

Because α-radiation is highly ionising, it cannot penetrate Beta-radiation
very far into matter. A cloud chamber can be used to show
the tracks of α-particles in air (Figure 16.17). The tracks A Geiger–Müller tube can detect β−-radiation. The source
are very dense, because of the dense concentration of ions is placed close to the tube, and different materials are
produced, and they extend for only a few centimetres positioned between source and tube. Paper has little effect;
into the air. By the time the α-particles have travelled a denser material such as aluminium or lead is a more
this far, they have lost virtually all of their kinetic energy. effective absorber. A few millimetres of aluminium will
The α-particle, which is a nucleus of helium-4, grabs two almost completely absorb β−-radiation.
drifting electrons in the air and becomes a neutral atom of
helium gas.

Chapter 16: Radioactivity

Gamma-radiation When an electron (with a charge of magnitude 1.60 ×
10–19 C) travels through a potential difference, energy is
Since γ-radiation is the least strongly ionising, it is the transferred. The energy change W is given by:
most penetrating. Lead can be used to absorb γ-rays. The
intensity of the radiation decreases gradually as it passes W = QV = 1.60 × 10–19 × 1 = 1.60 × 10–19 J
through the lead. In principle, an infinite thickness of So we define the electronvolt as follows:
lead would be needed to absorb the radiation completely;
in practice, a couple of centimetres of lead will reduce the One electronvolt (1 eV) is the energy transferred when an
intensity by half and 10 cm will reduce the intensity to a electron travels through a potential difference of one volt.
safe level in most situations.
Therefore:
The different penetrating properties of α-, β−- and 1 eV = 1.60 × 10–19 J
γ-radiations can be summarised as follows:
There is more about the electronvolt and its use in energy
■■ α-radiation is absorbed by a thin sheet of paper or a calculations in Chapter 30.
few centimetres of air.
QUESTIONS
■■ β−-radiation is absorbed by a few millimetres of 16 Explain why the most strongly ionising radiation
metal.
(α-particles) is the least penetrating, while the
■■ γ-radiation is never completely absorbed but a few least ionising (γ-rays) is the most penetrating.
centimetres of lead, or several metres of concrete, 17 A smoke detector (Figure 16.20) uses a source of
greatly reduces the intensity. α-radiation to detect the presence of smoke in
the air. Find out how the smoke detector works
This is illustrated in Figure 16.19. and suggest why an α source is more suitable for
this than a β− or γ source.
~1 mm ~2 mm ~2 cm 235

α

β−

γ

paper aluminium lead

Figure 16.19  A summary of the penetrating powers of α-, β−-
and γ-radiations. The approximate thickness of the absorbing
material is also shown.

The electronvolt (eV) Figure 16.20  A smoke detector that uses the absorption
of α-radiation as the principle of its operation.
Alpha and beta particles move quickly; gamma photons
travel at the speed of light. These types of radiation all have
energy, but the energy of a single particle or photon is very
small and far less than a joule. So we use another, much
smaller unit of energy, the electronvolt, when considering
the energy of individual particles or photons.

Cambridge International AS Level Physics

Summary ■■ Isotopes are nuclei of the same element with a
different number of neutrons but the same number
■■ The α-particle scattering experiment provides of protons.
evidence for the existence of a small, massive and
positively charged nucleus at the centre of the atom. ■■ Different isotopes (or nuclides, if referring to the
nucleus only) can be represented by the notation ZAX,
■■ Most of the mass of an atom is concentrated in its where X is the chemical symbol for the element.
nucleus.
■■ There are three types of ionising radiation produced
■■ The nucleus consists of protons and neutrons, and is by radioactive substances: α-particles, β-particles and
surrounded by a cloud of electrons. γ-rays.

■■ The number of protons and neutrons in the nucleus of ■■ In radioactive decay, the following quantities are
an atom is called its nucleon number A. conserved: proton number, nucleon number and
mass–energy.
■■ The number of protons in the nucleus of an atom is
called its proton number (or atomic number) Z. ■■ The most strongly ionising, and hence the least
penetrating, is α-radiation. The least strongly ionising
■■ Hadrons (e.g. the neutron) are particles that consist of is γ-radiation.
quarks and hence are affected by the strong nuclear
force. Leptons (e.g. the electron) are particles that are ■■ Because of their different charges, masses and speeds,
unaffected by the strong nuclear force. the different types of radiation can be identified by the
effect of an electric or magnetic field.
■■ The weak interaction acts between quarks and is
responsible for β-decay.

236

End-of-chapter questions

1 Before Rutherford’s model, scientists believed that the atom was made up of negatively charged [3]
electrons embedded in a ‘plum pudding’ of positive charge that was spread throughout the atom.
Explain how the α-particle scattering experiment proved that this old model of the atom was [1]
incorrect. [5]

2 A nucleus of strontium has a nucleon number of 90 and a proton number of 38. Describe the [3]
structure of this strontium nucleus.
[1]
3 State the changes that take place in a nucleus when it emits an α-particle and then two β−-particles. [2]

4 The nuclide of iodine with a nucleon number of 131 and a proton number 53 emits a β−-particle. [2]
Write a nuclear equation for this decay. [1]

5 An isotope of carbon 14 C emits a β−-particle and changes into an isotope of nitrogen (N).
6

a What are β−-particles?

b Write a nuclear decay equation for the decay.

c Draw a graph with the y-axis representing nucleon numbers between 10 and 16 and the x-axis
representing proton numbers between 4 and 10. On your graph mark:
i the isotope 164C
ii the daughter nucleus produced in the decay.

Chapter 16: Radioactivity

6 The uranium isotopes U-236 and U-237 both emit radioactive particles. A nucleus of uranium-237 [4]
may be written as 29327U and emits a β−-particle. A nucleus of uranium-236 emits an α-particle. The [2]
number of protons in a nucleus of uranium is 92. [2]
a Describe the differences between an α-particle and a β−-particle.
b Explain how uranium can exist in a number of different isotopes. [2]
c Write down the nuclear equation for the decay of U-236. [3]

7 Approximate values for the radius of a gold atom and the radius of a gold nucleus are 10−10m and [2]
10−15m, respectively. [2]

a Estimate the ratio of the volume of a gold atom to the volume of a gold nucleus. 237
b The density of gold is 19000kgm−3. Estimate the density of a gold nucleus, stating any [3]

assumptions that you make in your answer.

8 The nuclide of lead 28102Pb decays in three separate stages by α- and β− emission to another
lead nuclide, 20862Pb.
210
a Describe the structure of a nucleus of 82 Pb.

b α- and β−-particles are known as ionising radiations. State and explain why such radiations
can be described as ionising.

c The two lead nuclides are shown in the graph in Figure 16.22 of nucleon number A against
proton number Z.

Copy the graph and on your copy draw three arrows to represent one possible route for the
three decays between the two isotopes of lead. Label each arrow to show whether an α-particle
or a β−-particle is emitted.

A

210 28102 Pb

208

206 20862 Pb

79 80 81 82 83 84 85 Z
Figure 16.22 For End-of-chapter Question 8.

Cambridge International AS Level Physics

9 Geiger and Marsden carried out an experiment to investigate the structure of the atom. In this [2]
experiment α-particles were scattered by a thin film of gold. [3]
a When Rutherford analysed their results, what conclusions did he draw about the distribution
of mass and charge in the atom? [3]
b Describe and explain the experimental observations that led to these conclusions. [1]
[1]
10 Beta decay occurs as either β+-decay or β−-decay. An isotope of calcium 4206Ca decays by [1]
β-emission into the isotope 4216Sc, and an isotope of magnesium 2123Mg decays by β-emission into
the isotope 2113Na. [2]
a Copy and complete the following decay equations for the calcium and magnesium isotopes. [2]
i decay of calcium: 4260Ca → ……e + ……Sc+ + … [1]
ii decay of magnesium: 2123Mg → ……e + ……Na+ + … [2]
b State what happens in each type of β-decay in terms of the quark model of nucleons.
i β−-decay [2]
ii β+-decay.

c Name the force responsible for β-decay.

11 a A quark is a fundamental particle but a neutron is not. Explain what this statement means.

b A proton and a neutron each contain three quarks, either up or down quarks.

i Copy and complete the table to show the charge on a proton and a neutron and the quarks
that they contain.
238
Charge Quarks

proton

neutron

ii Using information from your table, suggest why some quarks must have a positive charge
and some quarks a negative charge.

c State what interaction is responsible for holding the nucleus together.
d When a neutron decays it produces an electron and two other particles. Copy and complete the

decay equation for a neutron.
01n →
e The electron and the neutron belong to different groups of particles.

Copy and complete the table to show the group of particles to which the electron and neutron
belong and state the name of another member of each group.

Group to which it belongs Another particle in the same group

electron
neutron