P3 Student Book Answers

7/25/2019 P3 Student Book Answers

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© Pearson Education 2011. Edexcel GCSE Extension Units Teacher and Technician Planning GuideThis document may have been altered from the original.

Answers – P3: Applications of physics

P3.1 Radiation in medicine

Student Book

1 Several of them could be described as being the oddone out. The important thing is the justification.Light – the only one that is visible.

X-rays – the only one in which the radiation travels fromoutside the body on one side to outside the body on theother side.Gamma rays – the only one in which the rays detectedare emitted from inside the body.Ultrasound – the only one that doesn’t useelectromagnetic radiation.

2 Both X-rays and gamma rays are ionising.3 The rays could damage the fetus and result in

abnormalities.4

Type of radiation Uses of radiation

light High intensity (and coherent) light is usedto correct defects in vision by modifyingthe shape of the cornea

ultrasound Ultrasound can be used to treat kidneystones by breaking them up into smallerpieces

H5 =P/ A = 100/0.0005 (there are 10 000 cm2 in 1 m2)

= 200 000 W/m2

H6 = P/ A = 0.001/0.000 001 = 1000 W/m2

H7 = P/ A P = I A = 6 4 = 24 W

8 Ultrasound waves are transmitted into the leg andpartially reflected by the different tissues in the leg. Thereflected waves are used to form an image. X-rays aretransmitted through the leg from one side and the onesthat penetrate are detected on the other.

Skills spotlight

Benefit is the ability to diagnose medical conditions so they

can be treated.Risks include the possible risk of cancer from the scan.

Activity Pack

P3.1a Radiation matching

A, B, H, N, OF, G, R, D, TK, Q, M, I, EP, L, C, S, J

P3.1b Radiation in medicine

1 Any three sensible suggestions, such as when a womanis pregnant, if a person is complaining of pains in part oftheir body, if the doctor suspects a tumour or othercondition, if the patient has swallowed something

harmful like a pin, etc.2 Light can only be used to look into 'open' parts of the

body that are connected to the outside by tubes. Thisincludes the mouth, throat, stomach, bowel, vagina, etc.If the suspected problem is inside tissues light cannot beused because it will not penetrate these tissues.

3 X-rays are a form of ionising radiation, so they can harmtissues. For most uses of X-rays, the possible harm thatmay be caused by the ionising radiation is outweighedby the benefit to the patient of having their conditiondiagnosed.

4 a Ear phones produce sound when a part of themvibrates. The vibrations spread out through the air (andthe head) as waves.b Chemical reactions as the gas burns convert chemical

energy to heat energy. This spreads out as infraredradiation.c The TV produces light and sound energy. The lightenergy spreads out by transverse waves. The soundenergy spreads out by longitudinal waves in the air.

5 X-rays are a form of ionising radiation. The rays inquestion 4 are not.

6 Intensity is the power transferred by radiation divided bythe area over which the radiation is spread.

7 Glass is denser than air, so the intensity of the radiationdecreases more if it passes through glass.

8 The beam of light spreads out as it gets further from the

torch. This means that the radiation is being spread overa greater area when the light is further from the source,so the intensity decreases.

9 The intensity will be so low that the light cannot bedetected.

H P3.1c Medical imaging

1 Light is used to look at the inside of the throat, becauseit can pass through the air inside the mouth and isreflected by tissues.X-rays are used to produce images of tissues that arenot open to the air, as light cannot pass through tissues.X-rays can be used to make images of broken bones orto look for tumours or other problems.Gamma-rays can be used to produce images by giving a

patient a radioactive substance. The camera detectswhere the radioactive substance has collected, and canshow where there are unusual levels of activity in thebody.Ultrasound is used to make images of the inside of thebody, particularly unborn babies. Ultrasound is used forfetal scanning because it is safer than using X-rays.

2 Gamma rays and X-rays are forms of ionising radiation,and their use can harm the patient.

3 Any three from: sound from radios, sound and/or lightfrom TVs, light from light bulbs, heat from toasters,cookers or heaters.

4 X-rays are ionising.5 Glass is denser than air, so the intensity of the radiation

decreases more if it passes through glass.

6 a intensity = power/area = 1000 W/25 m

2

= 40 W/m

2

b i If the distance doubles the area increases by a factorof 4. New area = 100 m2.ii Intensity = 1000 W/100 m2 = 10 W/m2 c 1000 W/20 m2 = 50 W/m2

7 It would spread out so much that its intensity would betoo low to be detected.

8 Equipment needed should include a source of soundand a means of detecting sound intensity (reference to amicrophone and datalogger is sufficient). The planshould include making the same noise repeatedly, andmeasuring the intensity at different distances. If a graphof 1/intensity2 is plotted against distance it should give astraight line. Ambient noise is likely to be the biggestproblem in producing reliable results.

P3.2 How eyes workStudent Book

1

2 Hold an object at arm’s length and focus on it. Graduallybring it closer; when it gets to the point where you can’tfocus on it measure the distance between it and youreyes.



© Pearson Education 2011. Edexcel GCSE Extension Units Teacher and Technician Planning Guide

This document may have been altered from the original.

3 If you are focusing on objects different distances awayyour muscles have to work. The lens needs to bestronger to focus on a nearer object so the ciliarymuscles have to contract and then relax again whenlooking at a distant object.

4 This involves a lot of small precise movements of theeye and thus for the muscles controlling the position ofthe eye.

5 The ciliary muscles contract to let the lens become fatterwhen a near object is being focused on and relax tomake the lens thinner to focus on a distant object.

6 The iris in a human’s eye forms a circular pupil, however

large or small, whereas that in a cat’s eye narrows thepupil to a slit. However the function is the same; in bothcases the iris controls the amount of light entering theeye.

7 Light enters the eye through the cornea, where it isrefracted, to start to make the rays converge. It thenpasses through the pupil; the size of this is controlled bythe iris to make sure the right amount of light enters theeye. The light then enters the lens and is refracted sothat the rays are focused onto the retina to form animage. The lens can change shape to allow the eye tofocus on objects different distances away.

Skills spotlight

Ideas about refraction help to explain how the cornea andlens focus light on the retina.Ideas about light as a wave that transfers energy, and ideasabout types of energy and energy transfers help to explainhow the retina converts light into electrical signals.

Activity Pack

P3.2a Parts of the eye

1, 3 Labels in correct positions. 2 Ciliary muscles – control shape of lens.

Cornea – protects eye and focuses light.Iris – changes size to change the size of the pupil, socontrols the amount of light that can get into the eye.Lens – focuses light onto the retina.Pupil – a hole that allows light to enter the eye.

P3.2c How the eye works

Part of theeye

What does it do? What features help it doits job?

Iris Changes size to controlthe size of the pupil.

It can open and close.

Lens Focuses light onto theretina.

It can change shape,and springs back to itsthin shape when theciliary muscles relax.

It is transparent.

Ciliarymuscles

Change the shape ofthe lens so the eye canfocus on objects atdifferent distances.

Can contract and relax.

Arranged in a ringaround the lens.

Retina Converts light intoelectrical signals.

Connections toneurones which carryimpulses to the brain.

P3.2d Functions of parts of the eye

1 See Figure B on page 182 of the Student Book.2 Light is refracted as it passes through the cornea, which

makes the rays converge. Light is refracted again as itpasses through the lens. The ciliary muscles contract orrelax to change the shape of the lens to allow the eye tofocus on objects at different distances.

3 a If the average adult tries to look at an object closer tothem than about 25 cm, they will not be able to focus onit properly.b The average adult can see distant objects clearly evenif they are a long way away.

4 Answers may vary, but a possible answer is:The model should have a model iris that can changesize, so the optician can demonstrate how the eye reactsto light.The model could have a flexible lens and a way ofmaking it change shape, or a set of different lenses tomodel the lens in the eye when it is different shapes.The model could have some LDRs or other componentssensitive to light in the retina, so when a light is shoneinto the eye the output on a screen or meter can be seento change.

P3.3 Sight problemsStudent Book

1 So that they form a clearly focused image.2 So that it can change shape to focus rays coming from

objects different distances away.3 Because they could clearly focus on near objects but not

on distant ones.

H4 Labelled diagrams similar to those used in Figure E on

page 185 of the Student Book.

H5 a Power = energy transferred/time taken

= 1 × 103/10 × 10 –9

= 0.001/0.000 00001= 100 000 W (= 100 kW)

b = P/ A

= 100 000/(100

10

–6

100

10

-6

)= 105/(10 –4 10 –4)

= 1 1013 W/m2 6 They may feel that it looks better or they may find

spectacle frames inconvenient or impractical.

Skills spotlight

The key point is that light rays are shown as straight lines,with arrows to show the direction. Students may also mentionsome of the following: the light rays change direction wherethey would be refracted by the cornea or lens, the diagramsshow where light is focussed by the way two separate rayscome together, light from distant objects is represented bytwo parallel rays.

Activity Pack

P3.3b Corrective treatment for eyes

1 Long sight2 Suitable labelled diagram showing lens surrounded by

ciliary muscles.3 a Reading a book, reading a computer screen, etc.

b So they can change shape to focus on objects atdifferent distances.

4

5

6 a Short sightb Jo can see distant things clearly but not close ones.Her supervisor can see close things clearly but notdistant things.

7 a They are thin, flexible lenses that sit directly on thecornea.b They both refract light and change the angle of thelight as it enters the eye.





P3.3d Sight problems

1 Diagram similar to the top left part of Figure B on page184 of the Student Book. Light should be shownchanging direction as it passes through the cornea, andas it enters and leaves the lens. The light rays shouldconverge on the retina.

2 Diagram similar to the top right part of Figure B onpage 184.

3 Converging lens – sketch showing lens thicker in themiddle than at the edges, with light rays being bent sothey converge as they pass through.Diverging lens – sketch showing lens thicker at the

edges than in the middle, with light rays being bent sothey diverge as they pass through.

4 Diagram similar to bottom right part of Figure C onpage 184.

5 People with short sight can focus on close objects butnot distant ones.

6 Diverging7 Diagram similar to bottom left part of Figure C on

page 184.8 Contact lenses are thin, flexible lenses that sit directly on

the cornea.9 They both refract light and change the angle of the light

as it enters the eye.

H P3.3e Problems with vision

1 Diagrams similar to the top part of Figure B on page 184of the Student Book. Light should be shown refracting asit passes through the cornea, and as it enters and leavesthe lens.

2 Converging lens – sketch showing lens thicker in themiddle than at the edges, with light rays being bent sothey converge as they pass through.Diverging lens – sketch showing lens thicker at theedges than in the middle, with light rays being bent sothey diverge as they pass through.

3 Diagram similar to the bottom right part of Figure C onpage 184.

4 Diagram similar to the bottom left part of Figure C onpage 184.

5 Both work by refracting the light before it reaches the

cornea. Spectacles are glass or plastic lenses held on aframe in front of the face. Contact lenses are thin,flexible pieces of plastic that are placed directly on thecornea.

6 Explanations should include the fact that the corneadoes some of the focusing in the eye, and the focusingpower of a lens depends on its shape.

7 They show only two rays of light from an object, whereasin reality there would be many rays of light spreading outfrom each object, and rays of light from many differentobjects all entering the eye at once.

8 If all the possible rays of light were shown, the diagramwould be far too complicated to understand.

9 Light rays are shown as straight lines with an arrow toshow the direction of travel.

P3.5 Different lensesStudent Book

1 Metres, centimetres or millimetres.2 Rays that travel in the same direction and never meet or

move away from each other.3 The more gently curving lens. It doesn’t refract the rays

as much so they will meet further away from the lens.4 P = 1/f

P = 1/0.1 = 10 DP = 1/0.2 = 5 DP = 1/0.05 = 20 D

H5 f = 1/P = 1/60 = 0.0167 m (i.e. 1.67 cm)

H6 1/u + 1/v = 1/f

1/20 + 1/v = 1/10

1/v = 1/10 – 1/20 = 0.1 – 0.05 = 0.05Therefore v = 1/0.05 = 20 cm

H7 1/u + 1/v = 1/f

1/100 + 1/v = 1/51/v = 1/5 – 1/100 = 0.2 – 0.01 = 0.19Therefore v = 1/0.19 = 5.3 cm

H8 a 1/u + 1/v = 1/f

1/3 + 1/v = 1/51/v = 1/5 – 1/3 = -0.133Therefore v = 1/–0.133 = –7.52 cmb It has a negative value.

9 The lens makes the parallel rays from the Sun converge.If the paper is at the focal point of the lens the energywill be concentrated here.

H Skills spotlight

The lens equation helps us to work out the position of animage using the object distance and the focal length. If thedistance of the image has a negative value, this means it ison the same side of the lens as the object and so must be avirtual image. If it has a positive value, it is a real image.

Activity Pack

H P3.5b Lens calculations

1 u = 10 cm, v = 10 cm, 1/f = 0.1 + 0.1 = 0.2, f = 5 cm.2 u = 20 cm, v = 10 cm, 1/f = 0.05 + 0.1 = 0.15, f = 6.67 cm.3 u = 20 cm, f = 5 cm, 1/v = 1/f – 1/u = 0.2 – 0.05 = 0.15,

v = 6.67 cm4 u = 100 cm, f = 5 cm, 1/v = 1/f – 1/u = 0.2 – 0.01 = 0.19,

v = 5.26 cm.5 f = 5 cm, v = 25 cm, 1/u = 1/f – 1/v = 0.2 – 0.04 = 0.16,

u = 6.25 cm6 f = 5 cm, v = 50 cm, 1/u = 1/f – 1/v = 0.2 – 0.02 = 0.18,

u = 5.55 cm.7 f = -5 cm, u = 20 cm, 1/v = 1/f – 1/u = -0.2 – 0.05 = -0.25,

v = -4 cmThe image will be 4 cm from the lens, on the same sideas the object.

8 f = -10 cm, u = 10 cm, 1/v = 1/f – 1/u = -0.1 – 0.1 = -0.2,v = -5 cmThe image will be 5 cm from the lens, on the same sideas the object.

9 f = -25 cm, v = -5 cm, 1/u = 1/f – 1/v = -0.04 – (-0.2) = 0.16,u = 6.25 cm.

The object was 6.25 cm from the lens.10 f = u = 10 cm, 1/v = 1/f – 1/u = 0, so v is at infinity.

P3.5d Image formation and focal length

1 a Realb Virtualc Reald Virtuale Real

2 a Converging lenses bring rays of light together,diverging lenses make them spread out.b They both change the path of light rays.

3 a Convergingb Realc 2.5 cmd 4 cm

e 7 cm4 It is smaller (diminished) and upside down (inverted).5 a 3.33 D

b 10 Dc 4 D

6 A is more powerful than B.7 – 10 D8 f = 10 cm, u = 25 cm, 1/v = 0.1 – 0.04 = 0.06, v = 16.7 cm9 u = 10 cm, v = 5 cm, 1/f = 0.1 + 0.2 = 0.3, f = 3.33 cm

P3.5e Calculating the power of a lens

1 C, A, B

2 B, it has the highest power/it is the most curved.3 C, as it has the longest focal length. Diagram similar to

Figure D on page 189 of the Student Book.

4 a 5 Db 4 Dc 20 D

5 It is less





6 Diagrams completed to show rays converging anddiverging, respectively.Converging, refracts, converge/come together, the raysmeet where the image forms.Diverging, refracts, diverge/spread out, the image is at apoint that the rays appear to come from.

H P3.5f The power of a lens and the use of the lens

formula

1 a 5 Db 3.57 Dc 14.29 D

d -10 D2 it is less3 a 0.1 m (or 10 cm)

b 0.05 m (or 5 cm)c – 0.05 m (or – 5 cm)

4 a Diagram should have focal point twice as far awayfrom the lens.b Diagram should have focal point half the distance fromthe lens.

5 14 cm6 8 cm7 1/f = 1/14 + 1/8 = 0.196, f = 5.09 cm, this is similar to the

focal length stated in the question, but not exactly thesame.

8 1/v = 1/f – 1/v = 0.1875, v = 5.33 cm9 1/u = 1/f – 1/v = 0.1667, u = 6 cm10 1/f = 1/16 + 1/16, f = 8 cm11 a 1/v = 0.2 – 0.5 = -0.3, v = -3.33

The virtual image is 3.33 cm from the lens, on the sameside as the object.b It is negative.c It shows the image is virtual, and on the same side ofthe lens as the object.

12 1/v = 1/-5 – 1/8 = -0.325, v = -3 cm

P3.6 Reflection and refraction

Student Book

1 The angle of reflection will increase as well.2 It will be reflected back along the same path in the

opposite direction.

3 Where the ray is travelling from air into glass the angleof refraction is less than the angle of incidence. Wherethe ray is travelling from glass into air the angle ofrefraction is more than the angle of incidence.

4 Sketch based on Figure C but showing, in both cases,less of a change of direction. The ray is bent, and in thesame direction, but not by as much. The label “glass”should be changed to “water” in each case.

5 a It continues in the same direction, though its speed willchange.b Yes. It has entered a different medium and this hascaused its speed to alter.

H6 sin i / sinr = n2/n1

sin 20o/sin r = 1/1.331.33 sin 20o = sin r

1.33 0.342 = 0.455 = sin r r = 27.1o

7 As the light travels from the water into the air it speedsup. As it leaves the water it is refracted away from thenormal and into the person’s eyes.

Skills spotlight

Diagrams should look like this:

waves rays

The rays of light are at right angles to the waves. When youare a long way from the source, the circle made by the wavestravelling out from the source will be very big, and so a smallpart of the circle will almost look like a straight line. If thewave front is almost straight then two rays of lightrepresenting the same light, at right angles to the wave front,will be parallel.

Activity Pack

P3.6c Waves and refraction

1 They become more nearly parallel to the boundarybetween the two materials.

2 At right angles to the boundary.3 They have been deflected towards the normal.4 Diagram similar to that on the worksheet, with the rays

and waves emerging from the block in the samedirection as the original incident rays/waves.

5 Diagram similar to that on the worksheet, with the wavesinside the block almost horizontal.

6 Both represent refraction quite well.7 In the ripple tank it is fairly easy to see the direction

change as the waves cross the boundary, but thechange in speed is not easy to see. With the line ofpeople it is probably easier to see the change in speed,but it is likely to be harder to see the change in directionof the waves (it would rely on all the students changingtheir speed by the same amount).

8 Models can help to simplify a situation and/or helpscientists to think about things that are too small to beseen directly.

P3.6d The tricks our eyes play

1

2

3 The light rays from a point on the bottom of the poolappear to be coming from a place closer to the surfacethan the bottom really is.

4 a One ray would only show that the direction in which wesee a point on the bottom of the pool has changed, itwould not show that the bottom appeared to be higherthan it is.b Several rays would make the diagram confusing, andonly two are needed to explain the effect.

5 Rays of light must be shown to refract towards the fishas they hit the surface of the water.

F





P3.6e Reflection and refraction

1 a

b, c

2 a, b

3

H P3.6f Explaining reflection and refraction

1 a, b, c

d The light rays are not really coming from the image,they only appear to be.

2 a, b

c Light speeds up as it goes from the water into the air.3 a 1.33

b 18.5ºc 13.4º

4

5 With a smooth surface, rays of light from a sourcespread out, and are reflected at consistent angles so thatthey appear to be coming from a point behind a mirror.With a rough surface, rays from a point source can bescattered in different directions.Students may reproduce a diagram similar to that for

question 4, with an additional one showing light beingscattered by a rough surface.

P3.9 Critical angle

Student Book

1 If the angle of incidence is equal to or greater than thecritical angle the light won’t be refracted.

2 a Increasedb Water has a smaller refractive e index than glass, sothe light ray will be refracted less.

H3 sin c = 1/n1 = 1/1.48 = 0.676

c = 42.5o

H4 a sin c = n2/n1 = 1.33/1.5 = 0.887

c = 62.5o

Note that a critical angle only applies when a ray istravelling from a slower medium to a faster one, i.e. inthis case travelling from glass towards water. Thereforeglass is ‘medium 1’ and air is ‘medium 2’.b It is greater because there is less of a differencebetween the refractive indices – less of a differencebetween the speeds of the ray in the two media.

5 Light is scattered at the air/water boundary. Answershould also include a ray diagram similar to the thirddiagram in Figure B on page 196 of the Student Book.

Skills spotlight





Activity Pack

P3.9a Prisms

6 It has 5 sides.7

8 a The light can pass in and out of the pentaprism if it hitsthe interface at an angle less than the critical angle.b The light is trapped inside when it hits an interface withan angle of incidence greater than the critical angle, as itis then internally reflected.

H P3.9b Snell’s law

1 a 0.087b 0.063c 1.38

2 Yes3 14.3º4 21.2º5 38.8º6 0.7167 36.2º8 40.1º9 45.7º10 Sketch showing light arriving at the glass to air interface

and being refracted along the interface.11 It will be reflected inside the block.12 Total internal reflection

P3.9c Changing paths of light raysFirst diagram:

Incident ray hits the interface at an angle greater thanthe critical angle.

Angle of reflection = angle of incidenceSecond diagram:

Ray slows down as it enters the block.

Ray refracted towards the normal as it enters the block.

Ray speeds up as it leaves the block.

Ray refracted away from the normal as it leaves theblock.

Third diagram:

Light slows down as it enters the glass.

Light refracted towards the normal as it enters the glass.

Light hits the edge of the glass at an angle greater than

the critical angle. Light is totally internally reflected 3 times within the

glass.

Light speeds up as it leaves the glass.

Light is refracted away from the normal as it leaves theglass.

P3.9d Refraction and total internal reflection

1 a It meets the interface at 90ºb It meets the interface at an angle greater than thecritical angle, so it is totally internally reflected.c It meets the interface at 90ºd It slows down as it enters the prism and speeds up asit leaves it.

2 a

b

3 a, b

4 Place the block on a flat surface and shine a ray of lightat one face. Vary the angle of incidence, and it will beseen that the ray is reflected inside the block at someangles.

5 a

b A converging lens has a continuous curve on eachside, and would focus all the light from a parallel beaminto a point. This arrangement only appears to focus lightbecause only three rays are used. If more parallel rayswere drawn in between the ones in the diagram, theones hitting the rectangular block would go straightthrough and not converge on the same point as the threedrawn. More parallel rays hitting the prism would bedispersed and spread out, so they would not meet at thepoint either.





H P3.9e Calculating critical angle

1 a Where the ray meets the long face of the prism.b As the light enters and leaves the prism (the light ischanging speed, but the angle does not change as thelight hits the interface at 90°.

2 a

b

3

4 a sin i/sin r is always the same for the same materials.From the table, sin i/sin r = 0.68For i = 30°, r = 47.3°b 14.75°c c = 42.8°

5 The light going through the prisms will converge at apoint beyond the model lens, and this is a good modelfor a lens if you only consider the three rays of light

shown. However a converging lens has a continuouscurve on each side, and would focus all the light from aparallel beam into a point. This arrangement onlyappears to focus light because only three rays are used.If more parallel rays were drawn in between the ones inthe diagram, the ones hitting the rectangular block wouldgo straight through and not converge on the same pointas the three drawn. More parallel rays hitting the prismwould be dispersed and spread out, so they would notmeet at the point either.

P3.10 Using reflection and refraction

Student Book

1 When a ray approaches a boundary at an angle greater

than the critical angle, it is completely reflected at theboundary, instead of being partially refracted andreflected.

2 One set is to guide the light into the body and another toguide it out again.

3 For an endoscope to work it has to be flexible so that itcan be inserted and follow what could be a curved ortwisting route. Total internal reflection means that the lightwill be guided in or out without escaping no matter whatthe route is.

4 The waves are partially reflected when they meet adifferent type of tissue.

5 Ultrasound is less invasive, so more comfortable for thepatient. Recovery time is faster than recovery from surgery.

6 High intensity ultrasound sets up vibrations causing thestones to break up.

7 They are similar in that they both produce images that can

be used by medical staff to examine a patient anddiagnose a condition. They both produce images ‘in realtime’ – as the instrument as moved so the image alters.However, endoscopy uses light and involves inserting aprobe inside the body whereas ultrasound uses highfrequency sound and doesn’t involve anything going intothe body.

Skills spotlight

Intense ultrasound can be used to break up kidney stones,and less intense ultrasound can treat injured muscles. Thismeans that the ultrasound does have some effect on thetissues it passes through, and so could harm the foetus insome way and so cause decreased birth weight. However thebenefits from the scans, in terms of detecting any healthproblems in the foetus before birth, probably far outweigh thepotential risks.

Activity Pack

P3.10a Endoscopes and ultrasound

1 A possible answer is:Light travels in straight lines unless it is refracted orreflected. If light inside glass hits the edge of the glass atan angle bigger than the critical angle it is reflected andstays inside the glass.

2 a Diagram correctly labelled. b The objective lens and the fibres are put down thepatient's throat and down to their stomach. One bundle ofoptical fibres transmit light into the stomach, and some ofthis light is reflected back into the objective lens. Theother bundle of fibres takes this light to the eyepiece lens,

where an image is formed.3 a Sounds too high for humans to hear.

b Ultrasounds are sent into a patient and some arereflected when they pass into different kinds of tissue. Theechoes are detected and used to make an image of theinside of the body.c Any sensible example. The examples given in the bookare fetal scanning and locating kidney stones.

P3.10b Using reflection and refraction

1 Diagram should show the outline of a car, with a labelledoptical fibre going from the roof to the headlamps.Some light from the surroundings will enter the opticalfibres on the top of the car and will be transmitted alongthe fibre by total internal reflection to the headlamps. This

has the advantage that there are no batteries to run out orbulbs that may break. However the optical fibres may notgather much light unless the car is directly under a lightsource, and so the headlamps will not be very bright.

2 Diagram of an endoscope, similar to Figure C on p198 ofthe Student Book.The objective lens and the fibres are put down thepatient's throat and down to their stomach. One bundle ofoptical fibres transmit light into the stomach, and some ofthis light is reflected back into the objective lens. Theother bundle of fibres takes this light to the eyepiece lens,where an image is formed.

3 Ultrasounds are sent into a patient and some are reflectedwhen they pass into different kinds of tissue. The echoesare detected and used to make an image of the inside of

the body.Images formed by ultrasound are not as detailed or clearas those obtained using X-rays, but ultrasound is notionising radiation and so carries almost no risk of harm tothe patient.





4 The kidney stones absorb high-energy ultrasound wavesthat are directed at them, and break up. This method ofdealing with kidney stones does not involve surgery.

5 The new version would be more comfortable for thepatient, because the tether is much narrower than thebundle of fibres in a normal endoscope. It will becheaper to use, as with a normal endoscope, patientsare given a sedative and anaesthetic, which increase thecost of the procedure. If the procedure is cheap enough,it can be used for routine screening to detect cancers inearly stages while they are still treatable.

P3.11 X raysStudent Book

1 To attract electrons.2 A heated cathode (or negative electrode) that emits

electrons.3 It absorbs a lot of heat energy from the electrons

colliding with it.4 By increasing the size of the potential difference

H5 a 3.2 × 10 –5 A

b 1.5 mA

H6 a 1.6 × 10 –14 J

b 1.87 × 108 m/s7 A good answer will include the following points:

there is an evacuated tube containing an anode and a

cathode the cathode is an electron gun

electrons are accelerated to the anode

electrons collide with the anode

most of the kinetic energy of the electrons is transferredto thermal energy

some of the kinetic energy is transformed to X-rays.

Skills spotlight

Students should use both forms of the equation for the currenton the student book pages to show that as the number ofelectrons increases, the current increases.

Activity Pack

H P3.11a X-ray calculations

1 a 4.8 × 10

–3

Ab 1.28 × 10 –3 Ac 9.28 × 10 –4 Ad 1.328 × 10 –1 A or 0.1328 A

2 a 4.8 mAb 1.28 mAc 0.928 mAd 132.8 mA

3 a 2.9 × 1018 b 8.1 × 1018

4 a 1.2 × 10 –16 Jb 5.6 × 10 –16 Jc 8 × 10 –15 Jd 9.6 × 10 –14 J

5 1.62 × 107 m/s

P3.11b X-ray machines1 Parts labelled correctly and X-rays added correctly.2 Paragraphs should be numbered 4, 2, 5, 3, 1.3 Most of the electrons hitting the anode transfer their

kinetic energy to thermal energy.4 Its high melting point.5 It gives more electrons enough energy to escape from

the cathode.6 It gives the electrons more kinetic energy as they are

accelerated towards it.

3.11c Making X-rays

1 a A: cathode (with correct description), B anode (withcorrect description).b C electron, negatively charged; D X-rays, high energy.

c High melting point, conduct heat well, because 99% ofthe electrons hitting the anode transfer their kineticenergy to thermal energy.

2 Otherwise the electrons would collide with particles inthe air.

3 Increase the temperature of the cathode.4 Increase the potential difference between the cathode

and anode.5 Their frequency increases.

H6 11 mA (0.011 A)

H7 2 × 10 –14 J

8 9.07 × 10 –31 kg

P3.12 Using X rays

Student Book

1 a They absorb more X-rays.b They do not absorb X-rays.

2 An image produced using X-rays that shows slicesthrough the body.

3 A device that produces moving pictures of the inside ofthe body using X-rays.

4 A doctor looks for very dark or very bright areas whichcan indicate that there is a tumour.

5 Organs in babies and children are more easily damagedby X-rays than in adults.

6 Lung; tumour shows up as brighter area.7 A good answer will contain the following points:

X-rays are fired at a body

X-rays are absorbed by some parts of the body more

than others X-rays are detected on the opposite side of the body

from the source

Where more X-rays are detected, the image is darker

Where fewer X-rays are detected the image is lighter

An area which has no blood supply does not absorbmany X-rays

It will appear as a very dark area on the image.

Skills spotlight

Student’s own conclusion, but it should be backed up with agood argument.

Activity Pack

P3.12c Radiation doses

1 Your torso is bigger than your head.2 A CAT scan is more than just one image – it is a wholeseries of images.

3 a Cardiac angiogramb Up to 130 times.

4 The procedure takes much longer.5 As the distance doubles, the strength decreases to a

quarter.6 Thicker parts absorb more X-rays.

H P3.12d Using X-rays

1 a There is a much higher exposure to X-rays.b A CAT scan can show a whole series of ‘slices’through the patient, whereas an X-ray is just a singleimage.

2 a To minimise their exposure to X-rays, because they

are dealing with them all the time.b The thicker the screen, the more X-rays are absorbed.

3 They absorb less X-rays than other parts of the body.4 a It would spread out.

b When the distance doubles, the strength decreases toa quarter.

5 Beam of radiation should be parallel so that it does notspread out; patient needs to be kept still so that theycan’t move and accidentally expose another area to theradiation. The part of the patient that is to be exposed isput into a rigid container that is transparent to X-rays sothat the part cannot move.

6 Only some of the X-rays go through organs other thanthe tumour. The tumour gets the full dose where thebeams combine, but other parts of the body only get part

of the dose.





P3.13 ECGs and pulse oximetry

Student Book

1 A graph showing the potential differences produced bythe heart.

2 An electrical signal that tells the heart muscles to contract.3 The body contains salt dissolved in water, which

conducts electricity.4 80 beats per minute.5 By detecting changes in the amount of red light and

infrared radiation transmitted through/reflected by tissue.6 Different types of blood absorb different amounts of red

light and infrared radiation.7 The light needs to travel through the body and be

detected – the light cannot be detected if it has to travelthrough too much of the body.

8 A good answer will contain the following points:

regular pattern which is repeated every heart beat

P part of ECG relates to the action potential spreadingacross both atria

Q part relates to action potential at atrio-ventricular node

RS parts relate to action potential spreading up theventricle walls

T part relates to contraction spreading up from base ofventricles

heart that is not beating properly will not show a regularpattern

can tell from the ECG which part is not working properly.

Skills spotlight

Student’s own answers but could include greater accuracy,you don’t have to find an artery where you can feel the pulse,can also find out how much oxygen the blood is carrying, canbe used to monitor pulse rates continuously.

Activity Pack

P3.13a Reading ECGs

1 A: 0.8 s, 1.25 Hz, 75 beats per minuteB: 1 s, 1 Hz, 60 beats per minuteC: A: 0.8 s, 1.25 Hz, 75 beats per minute

2 PQRST labels added correctly as in Figure C on page204 of the Student Book.

3 A. B shows a lower and longer T part; C is irregular.P3.13b Pulse oximetry and ECGs

1 To measure the pulse rate and how much oxygen theblood is carrying.

2 Suitable labels based on Figure C on page 205 of theStudent Book.

3 Correct numbering is: 2, 4, 3, 1, 5.4 a Records the electrical signals produced by the heart.

b The pattern on the ECG is not regular.5 Detects action potentials, amplifies them and then

transmits them to other parts of the heart.6 a 1.25 Hz

b 75 beats per minute

H P3.13c Electricity in the heart

1 a Action potentials in the heart.b Action potential spreading across both atria.c Contraction spreading up from the base of theventricles.

2 a 0.88 sb 1.14 Hzc 68 beats per minute

3 a 2.08 Hzb 0.48 s

4 Transmission pulse oximeter.5 Correct description, based on Student Book description.6 a No, because the pattern shown is not the normal

PQRST pattern.b Fit a pacemaker. A pacemaker works by detecting theaction potentials from the sino-atrial node, and then

amplifies them and transmits them correctly to otherparts of the heart.

7 0.5 mV, between P and Q; 4.3 mV, between R and S.

P3.14 Beta and positron radiation

Student Book

1 a 13p, 14n, 13eb 16p, 16n, 16ec 8p, 8n, 8e

2 a Neutral (0)b Positive (+1)c Negative (−1)

3 (+1)

H4 A proton becomes a neutron, so the total positive charge

in the nucleus becomes one less, and a positron, which

has a positive charge, is emitted.H5 a The atomic number becomes one less as one of the

protons has become a neutron.b The total number of protons and neutrons isunchanged.

6 a The emission of an electron is caused by a neutronturning into a proton, so the number of protons hasincreased.b This change results in one less neutron but one moreproton, so the mass number is unchanged.

7 a It becomes less.b The pressure is increased.c It is strong enough to penetrate paper but weakenough to be reduced if the paper is thicker.

8 If the source is on one side of the bottles and the

detector on the other side, the detector will record thelevel of emissions travelling through the bottle. Morebeta particles will get through an empty bottle so ahigher level of radiation indicates emptiness of bottle.

Skills spotlight

The use of models helps us to explain and understandbehaviour of, for example, sub-atomic particles. Students mayalso mention that models need to be refined and improved asscientists gain more knowledge.

Activity Pack

P3.14a Bits of an atom

1 Much lower mass.2 The only one with a positive charge.

3 The only one with no charge.Change Result

Atom loses an electron Very little change to mass butbecomes a positive ion

Atom gains an electron Very little change to mass butbecomes a negative ion

Atom gains a proton Increases in mass and becomespositively charged.

Atom gains a neutron Increases in mass but no changeto charge.

P3.14b Making paper

1 a It would be too thin.b It would be too thick.

2 a Suitable for purpose. E.g. paper for a computer printerneeds to be thick enough to be strong but not too thickor it will get jammed.b Paper that’s too thin will be weak and causecomplaints. Paper that’s too thick will also be unsuitablebut will waste material, increasing costs.

3 Doesn’t need human intervention.4 a Alpha radiation would be stopped by paper.

b Gamma radiation would not be stopped even by thickpaper.

5 No6 It’s cheaper and more responsive.7 a It would decrease the pressure on the rollers as the

paper is too thin.b It would increase the pressure on the rollers as thepaper is too thick.

8 The ionising properties make the particles easy todetect.





P3.14c Beta emissions

1 Nitrogen 14Carbon 14Beta particle (electron)

2 a Thin paper or several cm of air.b Thin sheet of metal, such as aluminium.

3 They should as they can travel some distance and areionising.

H P3.14d β− and β+ emissions

1 a They are equal in number.b The atom wouldn’t be neutral – it would be an ion.

2 a Ascending mass: electrons and positrons (equal)protons and neutrons (equal).b Decreasing charge: protons and positrons (equal)neutrons, electrons.

3 a Similarities: Penetrating and ionising abilities.Differences: β− radiation emits an electron, which has anegative charge and β+ emits a positron, which has apositive charge.b β− and β+ radiation are safe as long as appropriateprecautions are taken. It is straightforward to shieldpeople from these sources. There are hazardsassociated with being in close contact.c β− radiation emits an electron and β+ emits a positron.Both can be stopped by thin aluminium.They are less ionising than α but more than γ.

4 Caesium 137 undergoes β- decay. When this occurs:a a nucleus splits into an electron and a protonb the atomic number increases by onec the mass number stays the same.

5 Sodium 22 undergoes β+ decay. When this occurs:a a proton changes into a neutron and a positronb the atomic number decreases by onec the mass number stays the same.

P3.15 Alpha and gamma radiation

Student Book

1 a 2+b i 4 times as large.ii 8000 times as large.

2 Both have two neutrons and two protons.3 They are not very penetrating as they are relativemassive, but this mass means that if they do collide withatoms they can easily dislodge electrons to turn theminto ions.

4 a Because there are now fewer nucleons in the nucleus.b Because the number of protons is unaffected.

H5 a Uranium-235. Mass number is down four and atomic

number down two.b Ruthenium 99. Mass number is same and atomicnumber up one.c Argon-37. Mass number is same and atomic numberdown one.

6 Alpha particles ionise the air and allow a tiny current toflow, which keeps the alarm off. If smoke gets into thedetector it stops the alpha particles from getting to the airand ionising it. Therefore the current stops flowing andthe alarm goes off.

Skills spotlight

Alpha and beta radiations are represented as particles thatcan move through space and have defined mass and charge.

Activity Pack

P3.15a Smoke detector 1 It means that particles are struck and become charged

ions.2 Positive ions are attracted to a negative electrode and

negative ions towards a positive one.3 The opposite sign of the charges.4 The α radiation has a very limited range.

5 It is capable of ionising the air but has a limited range.

H P3.15b Nuclear equations

1 a i Nitrogen-14ii Thorium-234

b 92

238 U 90

234 Th +2

4 He +0

0 γ

2 Neptunium-2383 a Alpha

b Beta minus

P3.15c Decay processes

1 +2 as opposed to +14 as opposed to 1

2 Thorium 234 ( 90234 Th)

3 a It is emitted as a high energy electron – a β particle.b It remains in the nucleus.

4 It is very penetrating. Although it isn’t as ionising as α orβ it does have ionising capabilities and is harder tocontain.

H P3.15d Decay processes

1 a It consists of particles, the particles are charged andthey are ionising.b Alpha particles are much more massive, they all havea positive charge and they are less penetrating.c They have a limited range but they are more ionisingwhen they do come in contact with atoms so it’s notnecessarily safer.

2 a It decreases by two.b It decreases by four.

3 Gamma radiation is a form of energy.4 a Its ionising capabilities are less.

b Its penetrating capabilities are more.c It has no mass.d It has no charge.

5 a It doesn’t affect it.b It doesn’t affect it.

6 a It doesn’t affect it.b It decreases by one.

7 a53

131 I 54

131 Xe +1

0

e

b87

211 Fr 85

207 At +2

4 He

c14

27 Si 13

27 Al +1

0

e

d26

59 Fe 27

59 Co +1

0

e

8 Americium-2449 Phosphorus-31

10 a 7

14 N +2

4 He 8

17 O +1

1 H

b

7

14 N +0

1 n 6

14 C +1

1 p

P3.16 The stability curve

Student Book

H1 Because it has a small number of protons and neutrons

compared with most elements.

H2 They have more neutrons and are above the N=Z line,

on which elements with equal numbers of protons andneutrons lie.

H3 By subtracting the atomic number (the number of

protons) from the mass number (the number of protonsand neutrons).

H4 a For lighter elements the number of neutrons is similar

to the number of protons.b For heavier elements the number of neutrons isgreater than the number of protons.

H5 a Uranium-234. Mass number is down four and atomic

number down two.b Neon-22. Mass number is same and atomic numberdown one.c Barium 137. Mass number is same and atomic numberup one.

H6 It is above the line. It has 74 protons and 106 neutrons.





H7 Hafnium-176 is on the stability line and is stable.

H8 It will undergo alpha decay, being a heavy element and

to the right of the N –Z curve.

Skills spotlight

Scientists can use the N –Z curve by finding the position ofisotopes on the curve, and relating this to stability.

Activity Pack

H P3.16a β− and β+ decay

1 a It increases it by one.b It decreases it by one.

c It increases it by one.d It remains the same.

2 a It decreases it by one.b It increases it by one.c It decreases it by one.d It remains the same.

H P3.16b Large-scale stability curve

1 x-axis represents atomic number.2 a Isotopes are different forms of an atom. They have the

same number of protons but different numbers ofneutrons.b They have the same atomic number.c They have the same mass number.

3 a Isotopes on this curve are stable.

b All isotopes beyond the curve are unstable.4 a β –b β+

5 a Isotopes above the line have too many neutrons to bestable. These isotopes decay by β – emission, so thereare no β+ emissions for these isotopes.b Isotopes below the line have too many protons to bestable. These isotopes decay by β+ emission, so therewill be no β – decays.c Isotopes below the line are already 'short of' neutrons(they have too many protons to be stable), so theycannot attain stability by getting rid of neutrons.

H P3.16c Studying the curve

1 Isotopes are different forms of an atom. They have thesame number of protons but different numbers of

neutrons.2 A stable isotope is one that won’t decay into a different

isotope.3 They will decay, sooner or later, into other isotopes. In

some cases these isotopes will be of other atoms.4 Any two from: H, He, C, N, O, Ne, Mg, Si, S, Ca.5 Heavier elements don’t have the same number of

protons and neutrons, even if they are stable. (Theyhave more neutrons than protons.)

6 β – decay results in an isotope which has one moreproton and one fewer neutron (on the graph it movesone step to the right and one step down).

7 β+ decay results in an isotope which has one fewerproton and one more neutron (on the graph it moves onestep to the left and one step up).

H P3.16d Working with the curve 1 a A stable isotope is one that won’t decay into a different

isotope.b For lighter elements the N –Z curve is on or close tothe N=Z line showing that stable isotopes of lightelements have a similar number of protons andneutrons.For heavier elements the N –Z curve is not close to theN=Z line, showing that stable isotopes of heavierelements have more neutrons than protons.It stops short of the top of the graph, showing that theheaviest elements don’t have stable isotopes.c They are unstable and will decay.

2 They are unstable and will decay.3 a They are to the left of the curve.

b β- decay.c It will increase the atomic number by one but leave themass number unchanged.

4 a They are to the right of the curve.b Either α or β+ decay.c α decay reduces the atomic number by two and themass number by four. β+ decay decreases the atomicnumber by one but leave the mass number unchanged.

5 a On the graph it moves it one step to the right and onestep down.b α decay moves the nucleus two steps to the left andtwo steps down. β+ decay moves the nucleus one stepto the left and one step up.

P3.17 Quarks

Student BookH1 a 1

b 1

H2 a 1

b Zero

H3 a No difference.

b Difference of one.

H4 An up quark needs to become a down quark.

H5 a It will decrease it by one.

b No difference.

H6 a It will increase it by one.

b No difference.

H7 a

55

137 Cs 56

137 Ba +1

0

e

This is β- decay. A down quark has changed into an upquark.b

11

22 Na 10

22 Ne +1

0

e

This is β+ decay. An up quark has changed into a downquark.

H Skills spotlight

Beta decay can be modelled by an up quark changing to adown (beta plus) or a down to an up (beta minus).

Activity Pack

H P3.17a Quark kit

1 1, +1, proton

2 1, zero, neutron3 a No change.

b From zero to one – it rises by one.c An electron.

4 a No change.b From one to zero – it drops by one.c A positron.

5 a To emphasise that they are combined in threes.b They are both positively charged.c They are both negatively charged.

6 a It shows relative charge, mass and how they combineto form particles.b It doesn’t indicate the actual mass, charge, size orappearance.

P3.18 Dangers of ionising radiationStudent Book

1 No level of radiation is entirely safe; the higher the levelthe greater the risk.

2 It may cause burns, it may damage the structure of theDNA in cells or it may damage or kill the cell itself.

3 The risk varies according to the type of source and theprecautions that are needed to keep it safe.

4 Alpha particles have a limited range and so there is littlerisk from sources some distance away.

5 Concrete is dense and good at absorbing radioactiveemissions.

6 It needs to be controlled in order to minimise the risk topeople and to allow preventative action to be taken ifexposure levels are too high.





7 The methods include increasing the distance from thesource, the containment of the source in suitablematerials such as lead and concrete, shielding fromsources in use and the time spent in proximity tosources. This needs to be varied according to thesource as the sources vary both in their penetratingproperties (depending on the type) and the strength(depending on the amount and age).

Skills spotlight

Risks are reduced by monitoring and controlling dosage, andby using shielding.

Activity PackP3.18b Dosimeters

1 If it was in a pocket some of the radiation wouldn’t reachthe badge and wouldn’t be recorded.

2 They may not follow the safe working practices, or thepractices may not be safe.

3 a Alpha will only register on the open window, whereasbeta will register on both. b It won’t, but doesn’t need to.

4 This will record high energy gamma rays.5 A: Most beta and all gamma

B: Some beta and all gammaC: Alpha, beta and gammaD: Some gamma

6 Other types are also dangerous – they may be lesspenetrating but they’re more ionising.

7 The badge is intended to record the radiation dosagereceived by each individual to see if it’s within safe limits.If badges are swapped it won’t do this.

8 They could:

use weaker sources

use better shielding

stay further from the sources

use the sources less often.9 It’s true that radiation is both natural and ever-present,

but not necessarily at the levels that someone would beexposed to in certain occupations.

P3.18c Radiation homework

1 It reduces them.

2 a Depending on the level of radiation it may causereddening of the skin, burns and increased likelihood ofcancer.b Radiation can affect the DNA of cells. If this happens itcan affect the functioning of the body or the formation oftissues and organs.

3 a Shielding makes sure that radiation only reaches partsof the body intended and the time of exposure ismonitored.b Personnel are shielded from sources, the amount oftime they use them is controlled, they keep a distancefrom the sources and their exposure is monitored.

4 a They have enabled images to be produced to allowbetter diagnoses of conditions, such as the functioning oforgans and the positioning and size of tumours. They

have also enabled procedures such as the treatment ofcancer.b The sources present a hazard to staff and they canaffect patients, including the risk of causing the growth ofcancer.

P3.18d Dangers of ionising radiation

1 Radiation is weakened by travelling through air andbecomes less dangerous, either because it has lessenergy or fails to reach far. This is especially true ofalpha radiation.

2 It makes most difference with alpha and least withgamma.

3 Alpha needs little shielding to completely stop it. Betatakes rather more and gamma takes a lot to prevent itfrom passing through. It is not only the thickness of thematerial but also the type of material that affects howmuch penetrates.

4 For patients it is important that the time is carefullymonitored so that the exposure is sufficient for thediagnosis or treatment but not excessive. Shielding willalso be important so that radiation only reaches the partsof the body that it is intended to.

5 By explaining that the isotope is short lived so that it isunlikely to do any lasting harm, but also that theinformation from the image will enable more effectivetreatment to be carried out and that this will outweigh therisks from the isotope.

6 a They are likely to release some of the energy. b They are likely to release gamma rays to reduce the

energy level.

P3.19 Radiation in hospitals

Student Book

1 So that it kills or damages cells in the tumour anddoesn’t turn healthy cells cancerous.

2 They are sufficiently penetrating to reach nearbytumours.

3 Different tracers are absorbed by different organs in thebody, so the tracer used will be selected according to theorgan being studied.

4 The isotopes have a short half-life, so they have to bemanufactured close to the place where they are to beused.

5 A cure disposes of a condition whereas palliative caremakes the condition more tolerable and less painful.

6 The scan will indicate how well or how quickly the organis, for example, absorbing and processing certainchemicals. It will indicate, for example, brain diseasessuch as dementia or the spread of cancer.

Skills spotlight

A doctor might not recommend radiation for diagnosis ortreatment if a patient were pregnant or if a cancer has spreadthrough a large part of the body.

Activity Pack

P3.19a How a PET scanner works

1 So that gamma rays travelling in all directions can bedetected.

2 There is no need for an isotope that will stay active and itwould be more dangerous for the patient.

3 Placing the source internally involves small scalesurgery but means that the source can be positionednext to the tumour so it can be most effective. Externalsources involve no surgery and can be carefullytargeted.

4 Radiotherapy can be used to shrink tumours, eventhough a complete cure may not be possible.

5 The β+ decay causes the emission of positrons. When apositron hits an electron it destroys both of them andcauses gamma rays to fly off in opposite directions.These are detected by the scanner; the machine canlocate the source from which the gamma rays wereemitted.

6 It means that the tumours are good at absorbing glucoseand therefore absorb the tracer with the fluorine-18.

7 Those parts of the brain would be less evident on thescan.

P3.20 Collaboration and circular motion

Student Book

1 It is important to publish the findings of scientificresearch in academic journals so that any findings in theoriginal experimentation can be verified by otherscientists who check the validity of the method to ensurethe quality of the data is good and may also repeat theexperiments. The publication of material in journals alsorequires peer review (critical appraisal by other expertsin the field). New theories and models can only emerge

once the original findings from research has beenverified by the wider scientific community.





2 The centripetal force is the resultant inward acting forceacting on an object that has circular motion.

3 In this case the inward acting force which causes thebung to move in a vertical circle is provided by theresultant of the tension in the string attached to the bungand the weight of the bung. At the top of the circle theweight and the tension both act downwards so theresultant centripetal force is tension + weight. At thebottom of the circle the tension acts upwards and theweight acts downwards so the resultant centripetal forceis tension - weight.

4 The magnetic field has to be at right angles to the path

of the moving charge so that the force on the charge dueto the magnetic field is an inward acting force. Withoutan inward acting resultant force the charge would notchange direction from moving in a straight line to movingin a circular path.

5 By bombarding a stable isotope’s nucleus with highenergy protons. This causes the nucleus of the stableelement to become unstable and start to decay byemitting ionising radiation (alpha, beta or gammaradiation). The unstable isotope is known as aradioactive isotope.

6 A good answer will contain the following points:

the LHC is a very large and sophisticated set ofexperiments

consequently it is extremely expensive

in addition the experiment draws from a range ofscientific disciplines

therefore need to involve expertise from around theworld and from different engineering/technical/scientificbackgrounds to suggest solutions to problems

the manufacture of specialised equipment and computersoftware creates many business opportunities acrossthe world

in addition it would enable university departmentsthrough the world to be involved in the design andconstruction of specialised apparatus and equipment,generating research funding and income for the variousdepartments

peer review is a fundamental aspect of scientificexperimentation and development of models and

theories – involving the wider scientific community in theproject enables findings to be shared amongst manyacademic institutions and will speed up the process ofverification; partners in the collaboration may alsosuggest new experiments that could be designed.

Skills spotlight

The findings could be shared with the scientific community ina number of ways:

Publication of findings in academic, peer-reviewed journals.

Presentations at international conferences for experts in thefield or to learned societies, such as IOP or the RoyalSociety giving details of the experimental details and thefindings from experiments.

Use of the internet and mass media to highlight the findings. Additional Points that could be made:

Findings shared with the wider scientific community throughscientific papers and at conferences.

Peer review.

Experiments repeated to see if results replicated.

Activity Pack

P2.20a The building blocks of matter

1 The atom.2 A particle that cannot be broken down into smaller parts.3 In 1968 they were discovered to be made up from

quarks.4 Radioactive processes such as the emission of alpha

particles or beta particles showed that changes wereoccurring inside the atom with the emission of ‘particles’,

and hence that the atom was not ‘indivisible’. Theparticle(s) emitted from inside the atom could existseparately.

5 They bombard matter with high-speed particles in orderto break down atoms into smaller (sub-atomic) particles.

6 To date the electron has not been broken down intosmaller parts, suggesting that it is a fundamental particle.

7 a Large Hadron Collider – a particle acceleratordesigned to produce very high energy particles movingclose to the speed of light.b The higher the particle’s speed, the higher the energy;higher energies are needed to ‘smash’ into atoms ornuclei to break them into smaller, subatomic particles.

8 Lack of data: Other particle accelerators are not able toproduce the large energies required. It is anticipated that

the energies produced by collisions at the LHC will besufficient.

9 Which of the various theories of matter are likely to becorrect (e.g. ones that require/do not require the Higgsboson)/what the nature of dark matter is. There isinsufficient evidence to support these theories atpresent.

P3.20b Particle accelerators and circular motion

1 To ‘smash’ into atoms or nuclei to break them intosmaller, sub-atomic particles.

2 Very high energies are needed to break apart the atomsor nucleus.

3 A centripetal force.4 a It would continue in a straight line.

b Kinetic energy.5 An ion is any charged particle.6 The force produced by the magnetic field on the moving

ion, which acts inwards towards the centre of the circle.7 It will increase.8 It will continue in a straight line because the inward

acting force due to the magnetic field has been removed.9 a Radioactive isotope.

b To produce radioactive isotopes for medical purposes /for PET scanners.

P3.20c The cyclotron

1 The ion gains kinetic energy.2 a The centripetal force.

b It acts towards the centre of the circular path followedby the ion.

c The magnetic field exerts a force on a moving chargedparticle, and the force is perpendicular to the directionthe charge is moving in.

3 An isotope of an element that is radioactive andrandomly emits ionising radiation.

4 a Nucleus of oxygen-18 (8 protons) gains one proton,becoming the nucleus of the element fluorine (with 9protons).b It is unstable as too many protons.

5 The final energy of the emerging ion could be increasedby increasing the accelerating voltage or by increasingthe size of the ‘dees’.Increasing the voltage transfers more energy to the ion(does more work on it) giving it greater kinetic energy.Increasing the size of the ‘dees’ means the ion will be in

the magnetic field for longer and be accelerated acrossthe gap more times than previously.

P3.22 Collisions

Student Book

1 a We know this because the velocity of the red ball afterthe collision is the same as that of the grey ball beforecollision. Kinetic energy depends on mass and velocity.Since the first ball stops and the second ball has thesame speed and mass after the collision that the greyball had before the collision, all the kinetic energy hasbeen transferred from the grey to the red ball.b Yes – since kinetic energy has been conservedc Momentum depends only on mass and velocity. Thevelocity after is the same as before and in the same

direction. The mass was also the same in each case.Therefore since momentum is equal to m x v thenmomentum has been conserved in the collision.





2 a The total momentum after collision is zero since bothtrolleys have come to a halt.b The total momentum of the trolleys before collisionwas also zero. This follows from total momentum= +mv – mv = 0 (or qualitatively, the velocities were thesame but in opposite directions, so the momenta are thesame but in opposite directions). The total momentumbefore and after the collision has remained the same(zero) therefore momentum has been conserved.

3 During the collision the fronts of the cars have beenbadly deformed. The energy required to do this hascome from the kinetic energy of the cars prior to collision

(some of the kinetic energy has been transferred aswork done in deforming the car’s bodywork). During theimpact, some of the kinetic energy has also beentransformed as sound energy and heat energy.

H4 a Kinetic energy before = ½ 500 52 + ½ 400 22

= 7050 J

Kinetic energy after = ½ 500 32 + ½ 400 4.52 = 6300 J

Since the kinetic energy is less after the collision thanbefore, kinetic energy has not been conserved.b Since total energy is always conserved some of thekinetic energy has been transformed into other forms ofenergy.

H5 a Sum of momenta before the collision

= 0.4 2.0 + 0.6 0 = 0.8 kg ms

Sum of the momenta after the collision= 0.4 v + 0.6 1.6 kg ms

Apply the conservation of momentum:

0.4 v + 0.6 1.6 = 0.8

0.4 v = 0.8 – 0.96 = –0.16v = –0.4 ms –1 Velocity of 400 g glider after the collision is 0.4 ms –1 tothe left.b Kinetic energy before the collision

= 0.5 0.4 (2.0)2 = 0.8 JKinetic energy after the collision

= 0.5 0.4 (0.4)2 + 0.5 0.6 (1.6)2= 0.032 J + 0.768 J= 0.8 JNo kinetic energy has been lost during the collision so

collision is elastic.6 A good answer will contain the following points:

need to know the mass of both cars

know the maximum speed of the first car (less than thespeed limit)

so can find the maximum momentum of the first car

momentum after the collision is zero (both cars stop)

momentum conserved in all collisions

hence can find the maximum momentum of the secondcar before the collision

and from its mass, find its maximum velocity

and hence whether it was travelling within the speedlimit.

Skills spotlight

The mass of the club is much greater than the mass of thegolf ball. In the collision between the golf club and the golf ballmomentum is conserved. Assuming all the momentum of thegolf club is transferred to the golf ball (note however that thiswill not be entirely true as the golfer will ‘follow through’) theball must leave the club with much greater velocity ifmomentum is to be conserved. For example if the golf clubwas 500 times the mass of the ball, then the ball would leavethe club with a velocity of 500 times that of the club.

Activity Pack

H P3.22b Momentum calculations

1 Momentum of first ball = 0.01 kg x 12 m/s = 0.12 kg m/sVelocity of 2nd ball = 0.12 kg m/s / 0.02 kg = 6 m/s

2 Momentum before collision = 80 kg x 4 m/s = 320 kg m/s

Total mass after collision = 120 kgVelocity after collision = 320 kg m/s / 120 kg = 2.66 m/s

3 Momentum before collision = 150 kg x 2 m/s = 300 kg m/sMass after collision = 250 kgVelocity after collision = 300 kg m/s / 250 kg = 1.2 m/s

4 Momentum before collision = (200 x 3) + (130 x -2)= 340 kg m/sIf the cars stick together again:Velocity after collision = 340 kg m/s / 330 kg = 1.03 m/s

5 Momentum before impact = 0.005 kg x 444 m/s= 2.22 kg m/sVelocity after impact = 2.22 kg m/s / 0.505 kg = 4.4 m/s

6 Initial momentum = 100 kg x 10 m/s = 1000 kg m/sFinal velocity = 1000 kg m/s / 125 kg = 8 m/s

7 Momentum of rocks = 1 000 000 000 kg x 200 m/s= 2 x 1011 kg m/sMass of mountain = 1 x 1012 kgRecoil velocity = 1 x 1012 / 2 x 1011 = 5 m/s

P3.22c True or false?

1 False2 False3 False4 True5 False6 True7 False8 True9 False10 False11 False12 False13 True14 False

P3.22d Analysis of collisions

1 ElasticKinetic energyInelasticMomentum

2 a –10 kg m/s (direction reversed)b The collision is perfectly elastic so no kinetic energyhas been lost.

3 a Before: total momentum = 25 + 10 = 35 kg m/s After: total momentum = 35 kg m/s

Ball B must have = 35 – 5 = 30 kg m/sb It must have been greater otherwise it would not havecollided with ball B.c The masses and the velocities so that the total kineticenergy before and after could be calculated.d The balls would have remained in contact otherwise.

4 a Momentum depends on mass and velocity.b The trolley moving east since it has the most massand velocity.c They will move in the easterly direction. This isbecause there is a net amount of momentum in theeasterly direction.

5 a On collision it transfers some of its kinetic energy tothe surroundings.b It is partly elastic since a small fraction of the kinetic

energy has been transferred.H P3.22e Momentum and energy calculations

1 a Momentum before = 0.5 3 = 1.5 kg m/sb Momentum after = 1.5 kg m/s

c 1.5 = 0.6 v; v = 2.5 m/s2 a Momentum before = mv

b v m/s. Since perfectly elastic collision no kinetic energyis transferred.c Change in momentum = mv –(–mv) = 2mv

3 a Momentum before = Momentum after

75 5 = 130 v v = 2.88 m/s

b Kinetic energy before = ½ 130 2.882 = 539.14 JLoss in kinetic energy = 937.5 – 539.14 = 398.36 Jc Some of the kinetic energy of the first skater has beentransferred elsewhere, for example as thermal energy toheat the colliding skaters, or as sound.





4 a Momentum before = 0.01 1 = 0.01 kg m/sMomentum after = 0.01 kg m/s

0.01 = 0.01 0.4 + 0.005 v v = 1.2 m/s

b Kinetic energy before = ½ 0.01 12 = 0.005 J

= 5 10 –3 J

Kinetic energy after = ½ 0.01 0.42 + ½ 0.005 1.22

= 4.4 10 –3 J5 Momentum before = Momentum after

0.75 v = 0.045 54v = 3.24 m/s

The velocity of the golf ball is very much larger than thevelocity of the golf club. The reason for this is that themass of the ball is very much less than the club andsince momentum is conserved the velocity of the ballmust be very much more. The ratio of the mass of theclub to the ball is 16.7:1. The ratio of the ball’s velocity tothe club’s velocity is 16.7:1.

6 Consider that the Earth can move (and hence havemomentum). Take as a frame of reference, a frame inwhich the Earth has zero velocity initially. Then toconserve momentum, the combined momentum of thePlasticine + Earth after impact must be non-zero, and bein the direction the Plasticine was initially moving. So thePlasticine + Earth move downwards. We do not notice theEarth moving because of its enormous mass, making its

velocity minute.

P3.23 PET scanners

Student Book

1 The conservation of energy means that energy has notbeen created or destroyed. In this context the totalenergy before the interaction has been transferred to theenergy of the gamma rays after the interaction. There isno loss of energy from the system.

2 The electron and positron masses have been destroyedand transferred to energy in the form of gamma rays.

3 The total of the observed gamma ray energies is equalto the total mass-energy equivalent of the electron andproton before they annihilate, using E = mc

2. This

confirms that an energy equivalent of mass can be foundfrom the equation.4 Mass-energy is conserved in the electron–-positron

annihilation since the sum of the kinetic energies of theelectron and positron before collision is equal to the sumof the energies of the two gamma rays after collision.

5 By having pairs of gamma rays produced simultaneouslyat 180 degrees through annihilation it is possible tolocate (by finding the intersection of the lines connectingeach pair of gamma ray detections) the source of theannihilation and to build a picture of the internal tissuestructure. If an isotope that produced gamma rays wasused instead, locating the exact position would be moredifficult.

6 a Although in this scenario energy would be conservedsince the mass energy equivalent before and after arethe same, the momentum would be different. Producingonly one gamma ray would mean there was a netamount of momentum in one direction after the eventeven though the total momentum before was zero. Thiswould contradict the conservation of momentum and istherefore impossible.b In this case the conservation of energy would apply, aswould the conservation of charge. However if thegamma rays were moving in the same direction thenthey would have momentum in the same direction whichmeans a net amount of momentum after the event. Thiswould contradict the conservation of momentum andtherefore is not possible.

Skills spotlight

According to the mass–energy equivalence idea, if themasses of the matter and antimatter particles are greater theywould have greater energy equivalent than the electron–positron interaction. It follows from the conservation of energythat in the annihilation process the gamma rays must acquiregreater energy than with smaller masses of matter andantimatter.The speed of the gamma rays will not be affected since it is aconstant (the speed of light) so with higher energy the gammarays may have a higher frequency in this case.

Activity PackP3.23a Anti-matter

1 They annihilate each other, releasing energy.2 They may be able to detect positrons, or detect the

radiation produced when some of the anti-matter cameinto contact with normal matter.

3 Positron4 It is converted to energy in the form of gamma rays.5 The gamma rays could be made to push on part of the

rocket, or they could be made to transfer their energy toa propellant.

6 How to store it, and how to make it without using farmore energy than will be obtained from its laterannihilation.

P3.23b The PET scanner1 a They are destroyed, and their mass is converted to

energy.b Annihilation

2 So that the patient is only subject to radioactivity for ashort period of time.

3 Beta particles are normal matter, and so there would beno gamma rays produced.

4 a The point at which the gamma rays are producedmight not be central between the two detectors.b The one that has the shortest distance to travel.c The position of the source.d By combining information from lots of gamma raypairs.

P3.23c Positrons and PET scanners

accelerators, radioactive, anti-matter, oppositeannihiliate, gamma, mass, energy, chargepositron, electron, image

H P3.23d Electron–positron annihilation

1 a A positron has a positive charge and an electron has anegative charge, so the total charge is zero both beforeand after the annihiliation.b Mass and energy

2 a E = mc2 = 9.11 x 10-31 x (3 x 108)2 = 8.2 x 10-14 Jb 8.2 x 10—14 Jc Total energy before annihilation is 2 x energy ofelectron (or positron)Energy is conserved, so total energy after is the same,divided between two gamma rays, so each gamma ray

has an energy of 8.2 x 10—14 J.d So momentum can be conserved.

3 It would annihilate when it came into contact with normalmatter.

4 The two particles must have equal but oppositemomentum before the collision.

5 Some possible responses are:The calculations involved in working out where theannihilation took place must take into account anydifference in the speed of the gamma rays caused by thedifferent tissues through which they pass.The number of pairs of gamma rays produced willreduce with time, so the detector needs to detect enoughgamma rays to be able to build up a detailed image. Itmay need to detect many each second when the isotope

is first injected.





P3.24 Kinetic theory

Student Book

1 The particles in solids vibrate around fixed positions. Theparticles in liquids and gases move around.

2 The forces caused by particles of the gas hitting the wallof a container.

3 As the temperature increases the particles move faster,so they hit the walls of the container harder.

4 a 500 K – 273 K = 227 °Cb 100 K – 273 K = –173 °C

5 It doubles.

6 A good answer will contain the following points: the Kelvin scale relates temperatures to absolute zero

absolute zero is the temperature at which the pressureof a gas becomes zero

because the particles are no longer moving

the size of the degrees on the Kelvin scale is the sameas the degrees on the Celsius scale

so we can convert between the two scales easily

the Kelvin scale is useful to scientists because thekinetic energy of the particles in a gas is proportional tothe temperature on the Kelvin scale.

Skills spotlight

Advantages: it keeps the model relatively simple, easy todraw, easy to represent physically using real balls/marbles

etc.Disadvantages: most gases consist of molecules with morethan one atom joined together and so are not spherical; evensingle atom gases are not really rigid, solid balls.

Activity Pack

P3.24a Kinetic cards

Solid: C, D, I, K, F or LLiquid: A, E, F or L, J, NGas: B, G, H, M, O

P3.24c Kinetic questions

1 a C, Eb B, F

2 The particles in a solid are close together and in a fixedarrangement. The particles in a gas are far apart andmoving around quickly.

3 F, as the particles are moving faster.4 C5 a 273 K

b 100 °Cc 293 Kd 268 Ke –100 °C

6 The pressure of the gas inside will increase when it getshot, and may make the can explode.

P3.24d Particles and pressures

1 In a solid the particles are in fixed positions and can onlyvibrate. In liquids the particles can move around but arestill very close to one another. In a gas the particles are

far apart and move around fast.2 a The one that has been in the oven will feel

stiffer/harder to squeeze than the one that has been leftat room temperature.b The bottle in the oven will contain air that is hotter thanthe bottle that has been left at room temperature. Thegas in this bottle will therefore be at a higher pressureand so the bottle will feel stiffer/harder. Accept answersthat describe how hard or fast particles hit the walls atdifferent temperatures.

3 a The temperature at which particles stop moving/thepressure of a gas is zero, −273 °C.b Measure the pressure of a gas at differenttemperatures and plot a graph. Extend theline/extrapolate to find the temperature at which the

pressure would be zero.

4 a 0 °C, 273 Kb 100 °C, 373 Kc 293 Kd 268 Ke –100 °C

5 a B. The kinetic energy of the particles doubles when theKelvin temperature of the gas doubles.b They should specify that the temperature is measuredin kelvin.

6 The pressure of the gas inside will increase when it getshot, and may make the can explode.

7 The can cools down and the steam inside will condense

to form liquid water. There are therefore fewer moleculesof gas inside the can, so the pressure due to the gas willreduce. The pressure will also drop because the airinside the can is cooling down. The pressure inside thecan is no longer enough to balance the atmosphericpressure outside the can, which pushes the sides of thecan in.

P3.27 Calculating volumes and pressures

Student Book

1 Any two sensible uses, such as giving patients oxygen,pain relief, divers, welding, etc.

2 V2 = V1P1/P2

= 0.15 m3 230 000 000 Pa/100 000 Pa= 345 m3

3 P2 = V1P1/V2

= 0.000 025 m3 100 000 Pa/0.000 010 m3 = 250 000 Pa

H4 P1V1/T1 = P2V2/T2

V1 = V2, so P1/T1 = P2/T2 and P2 = P1T2/T1

20 °C = 293 K and 40 °C = 313 K

P2 = 230 000 000 Pa 313 K/293 K = 245 699 659 Pa5 A good answer will contain the following points:

the amount of gas needed would take up a large volumeat atmospheric pressure

but if the pressure is increased the volume is reduced

so gases don't take up so much storage space at highpressures

the pressure and volume of a gas are related by the

equationV

1P

1 =V

2P

2 (Higher tier students mayrearrange to V2 = V1P1/P2)

so if engineers know the maximum pressure that acylinder can safely withstand and its volume

at the same temperature

they can work out the volume of gas at atmosphericpressure that can be stored in it.

Skills spotlight

a The statement about the paint tin is qualitative. Thestatement about the gas cylinder is quantitative.b The statement about the gas cylinder needs to bequantitative, so that engineers know how much gas they canput into the cylinder without causing it to break/explode. Forthe warning about the paint tin, it doesn't really matter exactlyhow much the gas will expand, you just need to know that itcan expand enough to make the tin explode.

Activity Pack

P3.27a Calculation cards

A, d, h, 4, 12B b, i, 16, 11 C, e, j, 17, 15D, a, f, 18, 6E, c, g, 13, 5



© Pearson Education 2011 Edexcel GCSE Extension Units Teacher and Technician Planning Guide

P3.27b Question loops

In which state of matter are theparticles furthest apart?

gas

In which state of matter are theparticles fixed in position?

solid

What happens to the averagespeed of particles if a gas isheated?

gets faster

What causes gas pressure? moving particles hittingthe wall of a container

Pressure increases withtemperature because… particles move fasterwhen the temperature ishigher

If the volume of a gas is increasedat a fixed pressure…

its temperaturedecreases

If the volume of a gas is decreasedat a fixed pressure…

its temperatureincreases

If the pressure of a gas is doubledat a fixed temperature…

its volume doubles

If the pressure of a gas is halved ata fixed temperature…

its volume is halved

What are the units for thetemperature scale that starts atabsolute zero?

kelvin

What is the Kelvin temperature forthe melting point of ice?

273 K

What is the Kelvin temperature forthe boiling point of water?

373 K

What is the Celsius temperature forabsolute zero?

-273 °C

What is 50 K in Celsius? -223 °C

What is 50 °C in Kelvin? 323 K

What are the units for pressure? pascals

What units must be used fortemperature when usingV1=V2T1/T2 ?

kelvin

In which state or states of matter

are the particles close together?

solids and liquids

What is the equation for calculatinga new volume of gas if the pressurechanges?

V1P1 = V2P2

What is the equation for calculatinga new volume of gas if thetemperature changes?

V1=V2T1/T2

P3.27c Pressure, volume and temperature questions

1 a Pascalsb Kelvin

2 Kelvin3 The gas would take up too much space at atmospheric

pressure. It takes up much less space when it is at ahigh pressure. Accept similar explanations.

4 a Biggerb V2 = V1P1/P2

c 0.2 m3 d 15 000 000 Pae 100 000 Pa

f V2 = 0.2 m3 15 000 000 Pa/100 000 Pag 30 m3

h Depends on student's answers5 a Larger

b V2 = V1T2/T1 c V1 = 1000 m3, T1 = 283 K, T2 = 343 K

d V2 = 1000 m3 343 K/283 K = 1212 m3

e Depends on student's answers.6 This answer assumes that students have followed the

method outlined in earlier questions.The pressure in the tyre should be higher.P2 = P1V1/V2 P1 = 100 000 Pa, V1 = 200 cm3, V2 = 50 cm3

P2 = 100 000 Pa 200 cm3/50 cm3 = 400 000 Pa7 The final volume should be smaller.

V2 = V1P1/P2

V1 = 30 cm3, P1 = 100 000 Pa, P2 = 1 000 000 Pa

V2 = 30 cm3 100 000 Pa/1 000 000 Pa = 3 cm3

P3.27d Using the gas equations

1 The gas would take up too much space at atmosphericpressure. It takes up much less space when it is at a

high pressure. Accept similar explanations.2 V2 = V1P1/P2

V2 = 0.2 m3 15 000 000 Pa/100 000 Pa = 30 m3

3 V2=V1T2/T1

V2 = 1000 m3 343 K/283 K = 1212 m3 4 a V2 = V1P1/P2

V2 = 30 cm3 100 000 Pa/400 000 Pa = 7.5 cm3

H b 32 °C = 303 K,

T2 = P2V2T1/P1V1

= 375 000 Pa V2 303 K / (400 000 Pa V1 )= 284 K= 11 °CIt is assumed that the volume of the tyre remains thesame.

5 a P2 = P1V1/V2

P2 = 100 000 Pa 200 cm3

/ 80 cm3

= 250 000 PaH b 0 °C = 273 K, 20 °C = 293 K

P2 = P1V1T2/V2T1

= 250 000 Pa 80 cm3 293 K/(80 cm3 273 K)= 268 000 Pa

H6 If the pressure and volume are doubled, the term P V

will have gone up four times. If the final value of PV/T isto be the same, then the temperature must have goneup by a factor of 4.(Accept similar explanations, or answers involving trialcalculations.)

7 Absolute pressure is the 'real' pressure, gauge pressureis the pressure above atmospheric pressure. A tyrepressure gauge reads gauge pressure, so it would readzero when there was only air at atmospheric pressure

inside the tyre.

Documents

P3 Student Book Answers