Unit 2
Page Page 1 o f 92
Syllabus
Topic
Pages
28 - 35
Mechanica Mechanicall & Electromagnetic Waves
3 - 21
36 - 38
Refraction Refraction
22 - 26
39 - 40
Pol Po larisation
27 - 29
41
Diffrac Diffracti tion on
30 - 32
42 - 43
Wave Nature of Electrons
33 - 34
44 – 44 – 49 49
Pulse Echo Techniques Techniques
35 - 40
50
Charge and Current
41 – 41 – 44 44
51- 56
Voltage, Voltage, Current, Resistance & Power
45 – 45 – 58 58
57
Resistivity
59 – 59 – 61 61
59
EMF EMF & Internal Internal Resistance
62 - 65
58, 60 - 62
Potenti Po tential al Divider Divider
66 – 66 – 72 72
63 -67
Wave/Particle Wave/Particle nature of light
73 – 73 – 81 81
68
Spe Sp ectra & Energy Energy Levels
82 – 82 – 88 88
69 – 69 – 70 70
Radiation Radiation Flux
89 - 92
Page Page 2 o f 92
Syllabus
Topic
Pages
28 - 35
Mechanica Mechanicall & Electromagnetic Waves
3 - 21
36 - 38
Refraction Refraction
22 - 26
39 - 40
Pol Po larisation
27 - 29
41
Diffrac Diffracti tion on
30 - 32
42 - 43
Wave Nature of Electrons
33 - 34
44 – 44 – 49 49
Pulse Echo Techniques Techniques
35 - 40
50
Charge and Current
41 – 41 – 44 44
51- 56
Voltage, Voltage, Current, Resistance & Power
45 – 45 – 58 58
57
Resistivity
59 – 59 – 61 61
59
EMF EMF & Internal Internal Resistance
62 - 65
58, 60 - 62
Potenti Po tential al Divider Divider
66 – 66 – 72 72
63 -67
Wave/Particle Wave/Particle nature of light
73 – 73 – 81 81
68
Spe Sp ectra & Energy Energy Levels
82 – 82 – 88 88
69 – 69 – 70 70
Radiation Radiation Flux
89 - 92
Page Page 2 o f 92
Waves Mechanical waves These would be set up within a solid, a liquid or a gas due to the vibration of the molecules olec ules there. there. They are tran tra nsmitted by both inter interm molec ular ular forces and by coll collision isionss between between the molecules. In solids solids they could be eith eit her longitudinal ongitudina l or transv trans verse vibrations while in a liquid or a gas only longitudinal vibrations are really possible. Examples of these waves would be the P and S waves in the Earth's crust due to an earthquake earthquake and so und und waves in air. Earthquake waves.SWF
Electromagnetic waves
They all travel at the same speed 3.00 x 10 8 m s-1 .
They do not require req uire a med medium ium through whi wh ic h to pass.
The The y are ge nerated by accelerat ing char ged particles.
Electromagnetic waves.SWF waves.SWF
Gamma X-rays 10-15
10-9
U-V
Visiblee Visibl
I-R
4x10-7 4x10-7 - 7x10-7 10-6
Microwa Micr owav ves
Radio Radi o waves
10-4 - 10 -1
0.1 – 10 0.1 – 103 m
Electromagnetic waves share many of the general wave properties of mechanical waves:they transfer energy from place to place can be re flected flected can be re fracted can be superposed can be diffracted
ripple.exe
Refraction.SWF
Page Page 3 o f 92
Wave motion A wave motion is the transmission of energy from one place to another through a material or a vacuum. Wave motio n may occur in many forms such as water waves, sound waves, radio waves and light waves, but the waves are basically of only two types:
Sound in Helium.avi
(a) transverse waves - the oscillation is at right angles to the direction of propagation of the wave (Figure 1(a)). Examples of this type are water waves and most e lectromagnetic waves.
Wavelength ()
y0
y0
Figure 1(a)
Transverse wave generation.SWF
Wavelength ()
Waves, peaks & troughs.SWF
http://www.acoustics.salford.ac.uk/feschools/waves/waves.htm (b) longitudinal waves - the oscillation is alo ng the direction of propagation of the wave (Figure 1(b)). An example of this type is sound waves in air. wavelength wavelength wavelength
Figure 1(b)
Longitudinal wave creation.SWF
http://www.acoustics.salford.ac.uk/feschools/waves/waves.htm
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Basic definitions: Wavelength:
the distance between any two successive corresponding points that are vibrating in phase
Displacement: the distance from the mean, central, undisturbed position at any point on the wave (y) Amplitude:
the maximum displacement (y0 )
Frequency:
the number of vibrations per seco nd made by eac h particle/wave
Period:
the time taken for one complete oscillation (T= 1/f)
Phase:
the „delay‟ between the oscillations of neighbours
Wave terms.SWF
Amplide, wavelength & phase.SWF
Wave Speed The number of oscillations per second of each part of the medium is the FREQUENCY f . It also equals the number of complete waves passing any place in one second. If f waves per second go past a place, and the wavelength is , then the distance travelled by the waves per second ( i.e. the WAVE SPEED, c) is given by the equation
c f Sound in Helium.avi
Graphical representation of longitudinal waves
Longitudinal wave particle vibrations.GIF
Longitudinal wave pause.SWF
1
Longitudinal waves.GIF
Longitudinal wave.SWF
At large sports meetings the cro wds sometimes produce a „Me xican wave‟. (a)
What would be a better name for this manoeuvre?
(b)
Describe what the crowd would need to do to produce such a wave and suggest values for its frequency and amplitude.
Page 5 o f 92
2
The diagram shows a trans verse wave on a rope. It is moving to the right. Severa l particles on the rope have bee n labelled.
(a) (b) (c)
On the diagram draw arrows to show the direction in which particles P, R and T are moving. What can you say about the motion of Q and S? Mark on your diagram two particles that are (i) in phase, i.e. moving together - call them P and P ‟ (ii) in antiphase, i.e. moving oppositely- call them A and A '.
3
Describe, with the aid of a sketch, an electromagnetic wave.
4
Who will hear the singer first - a person who is 45 m from the stage, or a person watching the concert on TV at home 2400 km away? Assume that the microphone is very close to the singer and that the viewer is sitting close to the TV. Speed of electromagnetic waves = 3.00 x 108 ms-1 Speed of sound in air = 340 ms-1
5
Diagram ( i) represents part of a stretched spring. Diagram ( ii) represents the same section of the spr ing at o ne instant of time when a sinusoida l longitudinal wave is travelling along it.
(a) Use the diagram ( ii) to determine the wavelength of the longitudinal wave. (b) The wave speed is 2.00 m s-1 . Calculate the frequency of this wave. (c) Describe qualitatively the motion of an individual coil of this spring as the longitudinal wave travels along the spring.
Page 6 o f 92
6
The diagram shows the shape of a wave o n a stretched rope at one instant of time. The wave is tra velling to the right.
(a) Determine the wavelength of the wave. (b) Mark on the dia gram a point on the rope whose motion is e xactly out of phase with the motion at point A. Label this point X. (c) Mark on the dia gram a po int on the rope which is at rest at the instant shown. Label this point Y. (d) Draw an arrow on the diagram at point C to show the direction in which the rope at C is moving at the instant shown. (e) The wave speed is 3.2 m s-1 . After how long will the rope next appear exactly the same as in the diagram above? 7
A microwave generator produces p lane po larised electromagnetic waves of wavelength 29 mm. (a)
(i) (ii)
Calculate the frequency of this radiation. Complete the diagram of the electromagnet ic spectrum below by adding the names of the parts of the electromagnetic spectrum.
(iii)
State a typical value for the wavele ngth of radiation at boundary X.
Page 7 o f 92
8
The table below summar ises so me features of the electromagnet ic spectrum. Complete the table by filling in the missing types o f radiation, wa velengths and sources.
Radiation
Typical wavelength
Visible light
Source
Very hot objects
Gamma
100 m
-6
10 m
Page 8 o f 92
High frequency oscillator
Standing waves A stationary or standing wave is one in which the amplitude varies from place to place along the wave. Figure 1 is a diagram o f a stationary wave. There are places where the amplitude is zero and, halfway between, places where the amplitude is a maximum; these are known as nodes and antinodes respectively. A
A
X
a1
N
A
A
A
A
a3
a2 N
N
N
N
Y N
N
Fig 1
A node is a place of zero amplitude An antinode is a lace of maximum am litude Any stationary wave can be formed by the addition of two travelling waves moving in opposite directions. http://www.ngsir.netfirms.com/englishhtm/TwaveStatA.htm (Transverse Standing wave)
string1.swf
string3.swf
string2.swf
string4.swf
Stationary waves 2.SWF
Standing waves 1.SWF
string5.swf
string6.swf
string8.swf
string9.swf
http://www.enm.bris.ac.uk/anm/tacoma/tacoma.html#mpeg (Tacoma Narrows Bridge)
Page 9 o f 92
string7.swf
Tacoma Narrows.SWF
9
The diagram shows a wire with a mass of 1.30 kg at one end and a vibrator at the other e nd. 0.90 m N
1.30 kg
Vibrator
(a) (b) (c) (d) (e) (f) (g)
10
How many wavelengths does the diagram show? Use your answer to calculate the wavele ngth of the stationary wave. The frequency of the vibrator is 250 Hz. Calculate the speed of the waves. Point N in the diagram is a node. On the diagram, mark with an A an antinode. State one difference between a node and an antinode. Explain how a node is formed from two progressive waves. In energy ter ms, what is the difference betwee n a standing wave and a progressive wave?
(a)
Sketch three stationary patterns that could be formed on t he cord in the diagram. Mark the nodes and antinodes in each sketch. (b)
11
If in one of your sketches the distance between nodes is 0.55 m when the signal generator frequency is 40 Hz, what is the speed of the waves on the string?
The natural frequencies f of vibration of the string in Q2 can be found by putting n = 1,2,3, etc. in the formula f
n
T
2
where is the length of the string, T the tension in the string and its mass per unit length. (a) Show that the units of the right hand side of the formula are s -1 or Hz. (b)
Using the arrangement in the diagram explain how you would show exper imentally that f was inversely proportional to for a fixed value of n, e.g. n = 2.
Page 10 of 92
12
(a)
(b)
State two differences between a stationary wave and a progressive wave. Difference 1 Difference 2 Spiders are almost completely depe ndent on vibrations transmitted through their webs for receiving information about the location of their prey. The t hreads of the web are under tension. Whe n the threads are disturbed b y trapped prey, progressive transverse waves are transmitted along the sections o f thread and stationary waves are formed. Early in the morning droplets of moisture are seen evenly spaced along the thread when prey has been trapped.
(i) (ii)
13
Explain why droplets for m only at these points. The speed of a progressive transverse wave se nt by trapped prey along a thread is 9.8 cm s -1 . Use the diagram to help you determine the frequenc y of the stationary wave.
A stationary wave is produced on a stretched string by a vibration generator attached to one end. The graph shows part of the wave. The two full lines represent the extreme positions o f the string.
(a) (b) (c)
(d) (e)
State the wavelength of this wave Mark a letter A on the graph to label an antinode. The stationary wave is formed by the superposition of two waves travelling along the string in opposite directions. The frequency of the vibrator is 36.0 Hz. Calculate the speed of the travelling waves. State the phase relationship between the two travelling waves at an antinode. Deter mine the amplitude of each of the travelling waves.
Page 11 of 92
14
(a) (b) (c)
15
Describe, with the aid of a d iagram, an experiment to demonstrate stationary waves using microwaves. Using the idea of wave superposition, explain what is observed in your experiment. Describe how you could use the experiment to measure the wave length of microwaves.
A piece of string is connected to a variable frequency vibration generator. The fundamenta l frequency of this system is 60 Hz.
(a) Complete the table to show what would be observed as the frequency is gradually increased from 40 Hz to 180 Hz.
16
The cello is a stringed musical instrument that may be played either by stroking the strings with a bow or by plucking the strings with the fingers.
Page 12 of 92
(a)
One of the attached strings on the cello has a vibrating length of 0.80 m. The string is made to oscillate as a stationary wave by means of a bow and the following pattern of oscillations is seen. The position of the string at two different times is shown.
(i) (ii) (iii)
(b)
Explain how the movement of the bow causes this wa ve pattern. Using the diagram calculate the wavelength of the wave. State two differences between the wave on the string and the sound wave it produces.
The cello string is then plucked and the waveform of the resulting sound is analysed by an oscilloscope. It is found to consist of two frequencies of different amplitudes. The frequency spectrum is shown below,
The waveform of the 200 Hz wave has been dra wn on the axes below. On the same axes sketch the waveform of the 1000Hz wave.
Page 13 of 92
Superposition When two groups of waves (called wave trains) meet a nd over lap t hey „interfere‟ with each other. The resulting amplitude will depend on the amplitudes of both the waves at that point. If the crest of one wave meets the crest of the other the waves are said to be in phase and the resulting intensity will be large. This is known as constructive interference. If the crest of one wave meets the trough of the other they are said to be out of phase by
1 2
then the resulting intensity will be less/zero (if the waves have equal
amplitudes). This is known as destructive interference. http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=19
(Superposition Principle)
Superposition principle.GIF
This phase difference may be produced by allowing the two sets of waves to travel different distances - this difference in distance of travel is called the path difference between the two waves.
doubleslit.exe
The diagrams in Figure 1 below show two waves of equal amplitudes with different phase a nd path differences betwee n them. The first pair has a p hase difference of
1 2
or 180o and a path difference of an odd number of half-wavelengths. The second pair have a phase difference of zero and a path difference of a whole number of wavelengths, including zero.
+
=
destructive interference
+
Figure 1
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=
constructive interf erence
To obtain a static interference patter n at a point (that is, one that is constant with time) we must have two sources of the same wavelength, and two sources which have a co nstant phase differ ence between them. Sources with synchronised phase changes between are called coherent sources and those with ra ndom phase c hanges are called incoherent sources. This condition is met by two speakers connected to a signal generator because the sound waves that they emit are continuous – there are no breaks in the waves. However two separate light sources cannot be used as sources for a static interference pattern because although they may be monochromatic the light from them is emitted in a random series of pulses of around 10 -8 s duration. The phase difference that may exist between one pair of pulses emitted from the source may well be quite different from that between the next pair of pulses (Figure 2).
Pulses from source A
Pulses from source B
Figure 2
Therefore although an interference pattern still occurs, it changes so rapidly that you get the impression of uniform illumination. Another problem is that the atoms emitting the light may collide with each other so producing phase changes within one individual photon. We must therefore use one light source and split the waves from it into two. Minimum – crest meets trough
Maximum – crest meets crest
Minimum – crest meets trough
S1 Maximum – crest meets crest
S2 Minimum – crest meets trough
Maximum – crest meets crest
Figure 3
Minimum – crest meets trough
Page 15 of 92
Diagrams in Figure 3-7 show two sources S 1 and S 2 emitting waves - they could be light, sound or microwaves. The plan view of the waves in Figure 3 shows waves coming from two slits and „interfering‟ with each other. The lines along which the path differences will give maxima or minima.
This type of arrangement is like that produced in a ripple tank or in the double slits exper iment with light (see later). Figure 4(a) shows light interfering as it passes through two slits. In Figure 4(a) the appearance of the interference pattern on a sc reen placed in the path of the beam is shown.
Figure 4 (a)
You can see the maxima and minima and the way in which the intensity changes from one to the other. Changing the wavelength of the light (its wavelength), the separation of the slits or the distance of the slits from the screen will all give changes in the separation of the maxima in the interfere nce pattern.
Two dippers Ripple tank.GIF
Two source Max & Min.SWF
Young's slits Max & Min.SWF
Figure 4 (b)
minimum maximum
maximum
Figure 5 maximum
minimum
Figure 5 shows the interference effects of two speakers. The sound waves spread out all round the speakers and a static interference pattern is formed. (Not all the maxima and minima are labelled). You can hear this by setting up two speakers in the lab connected to one signal generator and then simply walking round the room. You will hear the sound go from loud to soft as you pass from maximum to minimum.
Page 16 of 92
In Figures 6 and 7 you ca n see that at the different points o n the screen the waves from S1 have travelled a different distance from those from S2 . In Figure 6 the path difference is zero, in Figure 7 it is half a wavelength
Path diff erence = /2
Path difference = 0
S1
S1 S2
screen
S2 screen Figure 6
Figure 7
Interference
The effect of amplitude If two sources have very different amplitudes, then you will not be able to observe superposition patterns. The wa ve with the larger amplitude will dominate, beca use it makes little difference to the total whether the wave with the s maller a mplitude is in phase or out of phase with it.
Wavefronts All points on a wavefront vibrate in phase. If you dip your finger in the water of a ripple tank you will notice that circular ripples spread out.
This is because waves travel at the same speed in all directions .
If we tilt the tank so that the depth of the water now varies: The wa vefront has an oval s hape beca use waves travel faster in deep water than in shallow water.
shallow
Page 17 of 92
deep
Phase Differences
x
x 2 Places marked and oscillate with a phase difference http://www.acoustics.salford.ac.uk/feschools/waves/super.htm
Phase difference.SWF
speed of light as = 3.00 x 10 8 ms-1 speed of sound in air = 330 ms -1 . 17
(a) (b)
18
The diagram shows an experimental arrangement for investigating the superposition of so und waves from two sources S 1 and S 2 . The sources are in phase a nd produce sound of wavelength 80 mm.
(a)
When S1 M is 800 mm the trace on the oscilloscope is a maximum. Suggest three possible values for S2 M.
(b)
19
What is meant by the principle of superposition of waves? Are there any types of wave which do not obey the pr inciple or any circumstances in which it does not apply?
Without moving any of the apparatus the leads to one of the speakers are reversed, i.e. the sources are now in antiphase. Explain the meanings of in phase and in antiphase and describe how the trace on the oscilloscope changes when the leads are reversed.
Why can‟t you get a static interference pattern with two light bulbs while it is possible with two loudspeakers.
Page 18 of 92
20
A motorist drives along a motorway at a steady speed of 30 ms -1 . There are radio transmitters at each end of the motorway. She is listening to the car radio and as she travels along she notices that the radio signal varies in strength, 5s elapsing between success ive maxima. Explain this effect and calculate the wavelength of the radio signal that she is tuned to.
21
The diagram shows an arrangement with sound waves.
A loudspeaker connected to a signal generator is mounted, pointing downwards, above a horizonta l bench. The sound is detected by a microphone connected to an oscilloscope. The height of the trace on the oscilloscope is proportional to the amplitude of t he sound waves at the microphone. When the vertical distance x between the microphone and the bench is varied, the amplitude of the sound waves is found to vary as shown on the grap h.,
(a) Explain why the amplitude of the sound has a number of maxima and minima. (b) The frequency of the sound waves is 3.20 kHz. Use this, together with information from the graph, to determine a value for the speed of sound in air. (c) The contrast between the maxima and minima becomes less pronounced as the microphone is raised further from the surface of the bench. Suggest an explanation for this.
Page 19 of 92
22
The diagram is a plan view of an experiment to meas ure the wavelength of microwaves. The diagram is to scale but one third of full size .
(a) As a microwave detector is moved around the arc from A to B, alternate maxima and minima of intensity are observed. Explain why.
Page 20 of 92
(b) A maximum is observed at point 0, and the next maximum at point X. By means of suitable measurements o n the diagram, determine the wavelength of the microwaves. (c) A teacher demonstrating this experiment finds that, even at the maxima, the wave intensity is small. A st udent suggests making the slits wider to let more energy through. Explain why this might not be a good idea. (c) For an interference pattern to be observed between waves from two sources, the sources must be coherent. Explain what is meant by coherent, and what makes the two so urces in this experiment coherent.
Page 21 of 92
Refraction Change of Speed For all refracted waves the path is deviated away from the normal when the speed increases and towards the normal when the wave is slowe d down .
Medium 2
2
Speed c 2
2 1
1 Speed c1 Medium 1
Light moves slower in medium 1 than in medium 2.
1
2
sin 1 sin 2
c1 c2
Since the path of the light is reversible sin 2 c2 2 1 sin 1 c1
Refraction occurs for all waves. Sound can be deviated as it passes from warm air to cooler air and microwaves ca n be refracted b y wax.
Page 22 of 92
Refraction – wave fronts
Refraction.SWF
Wavefronts animation The following diagram shows the refraction of a plane wave at a plane interface. The position of the refracted wave is formed using the idea of secondary wavelets
vat A
air
a g
vgt
D
glass
C
One side of the wave moves from A to C (a distance vgt) in glass in the same time that the other side of the wave front moves the form B to D (a distance v at) in the same time in air. The wave front recombines at CD. va t AD va a g v g sin g v g t AD
sin a
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Total internal reflection and critical angle
medium one medium tw o
c
>c
Total internal ref lection
Figure 1
When light passes from a material such as water into one of lower refractive index such as air it is found that there is a maximum angle of incidence in the water that will give a refracted beam in the air, that is, the angle of refraction is 90 o . The angle of incidence in the denser medium corresponding to an angle of refraction of 90 o in the less dense medium is known as the critical angle (c) (Figure 1). The reason for this is clear if we consider the formulae. For an angle of refraction of 90 o we have: sin 2 sin c 1 sin c 2 1 sin 1 sin 90 1 2 Example problem The refractive indices from air to glass and from air to water are 1.50 a nd 1.33 respectively. Calculate the critical angle for a water-glass surface. a
Air Glass
g g w
Water
w a
a g
g w
sin a sin g sin g sin w
, a w
Air
sin a sin w
sin g sin a 1 1 . .a w g a . a w 1.13 sin a sin w 1 . 50 a g
Page 24 of 92
Therefore the critical angle for light passing from glass to water is sin c
g w
1.13
1
1
sin 90 1.50 c 62.5 For an air-glass boundary sin c a
g
1.50
c 42
And for an air-water boundary
1
sin c a
w
1 1.13
c 48.5 For angles of incidence greater than the critical angle all the light is reflected back into the optically more dense material, that is, the one with the greater refrac tive index. This is known as total internal reflection and the normal laws of reflection are obeyed. Total internal reflection explains the shiny appearance of the water surface of a swimming pool when viewed at an angle from below. The phenomenon is used in prismatic binoculars (Mirages are ca used by continuous interna l reflection.)
glass
Figure 2
1 Blue light is deviated more than red light when it enters a glass block because: A it has a longer wavele ngth B it has a lower frequency C it travels at a greater speed in glass than red light D it travels at a lower speed in glass than red light 2
The diagram shows how a narrow beam of light strikes a layer of oil on the surface of a tank of water at an angle of 58.0 . 58
Refractive index of oil Refractive index of water
= 1.28 = 1.34
Calculate (a) The angle of refraction in the oil (b) The angle of refraction in the water (c) The angle of refraction in the water if the layer of oil was removed.
Page 25 of 92
3
Draw a diagram showing wavefro nts when light passes from air into glass. at incident angle of 45. a g 1.50
4
(a)
A ray of light enters one side of a rectangular glass block at incident angle of 40. Calculate the angle of refraction a g 1.55.
(b)
The opposite side of the block is immersed in a clear liquid. The angle of refraction is 28 when the ray passes into the liquid. Calculate the refractive index of the liquid
5
6
a
.
Calculate the critical angle for the interface between the core of an optical fibre with refractive index 1.60 and the cladding with refract ive index 1.52.
Rainbows are caused when sunlight is dispersed by raindrops. The differe nt colours follow separate paths. The diagram shows some of the rays of light passing through a raindrop. Sunlight A B
Violet Raindrop Red
(a) Name the process which occurs at A. (b) The ray at B is act ually only partially reflected at t he surface o f the water. Continue the ray to show the path of the red light which is not reflected. (c) Explain the condition that would be required to prevent the red light from emerging at B. Light changes its direction at A because of a c hange of speed on entering the water. (d) Red light has a frequency of 4.2 × 10 14 Hz. Calculate its wavele ngth in a raindrop. Speed of red light in water = 2.2 × 10 8 m s – 1.
Page 26 of 92
Polarisation If you get hold of one end of a rubber rope, tie the other end to a post, stretch it and then send a series of pulses down the rope the vibration travels down the rope. Although each success ive pulse may be sent in a different plane each pulse only vibrates in one direction. A wave in which the plane of vibration is constantly changing is called an unpolarised wave.
Magne tic field (B)
Figure 1
Electric field (E) Unpolarised radiation
Plane polarise d radiation
However if the vibrations of a transverse wave are in one plane only then the wave is said to be plane polarised. Electric field
Magnetic field e.m. radiation
Figure 2
Light is plane-polarised when the vibrations are made to occur in one plane only. Light is a transverse electromagnetic wave with the vibrations of an electric and a magnetic field occurring at right angles to each other and in any plane at right angles to the direction of travel of the light. Polarisation is easily observed with the rubber rope experiment described above but it can also be s hown with electromagnetic waves such as microwaves, TV, radio a nd light.
It is important to realise that transverse waves can be polarised while longitudinal waves cannot.
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Effects of polarisation with light.
Sunglasses Polar isation by reflection - glare/shine off roads Polar isation of scattered sunlight
Car windscreens Optical activity Stresses in materials
The effect of a polar iser and an analyser is shown in the following diagrams.
Polarisation of transverse waves.SWF
Polarisation of microwaves.SWF
Sensation of light polarization
Our own eyes are poor at sensing light polarization but very good at sensing its colour and brightness; however, sensation of light polarization is not at all unusual in the animal kingdom. It is found in some insects, crustaceans, fish, birds, and particularly in cephalopods, a class of molluscs that includes squids, octopuses a nd cuttlefish. A key requirement for polarization sensitivity is the excitation of two or more classes of visual pigments that have different alignments in the eye (technically, different axes of maximal excitation). Compariso n of the ne ural inputs from these two rows allows the animal to se nse polarized light.
Stomatopod1.JPG
Stomatopod2.GIF
The images in the animation were taken using a polarizing filter - we wouldn‟t see the flashing red and white signals with only the naked eye. Light reflected off shiny surfaces may be polarised. http://www.colorado.edu/physics/2000/applets/polarized.html Photoelastic stress analysis
When polarised light passes through some transparent materials, the plane of polarisation is rotated. If the material is put under stress, the amount of stress affects the degree of polarisation. If the incident polarised light is white, each of its component colours is rotated by a different amount, creat ing a pattern of coloured fringes when viewed through the analyser polaroid. Engineers make models of structural components out of materials such as Perspex. The model is then stressed, and the pattern of coloured fringes is examined in order to ide ntify regions of high stress.
Page 28 of 92
1
What is meant by plane polarized light?
2
Explain whether the human eye can distinguish polarized light from unpolarised light.
3
How could you tell if a television signal was plane polarized and also find out the direction of polarization?
4
5
Why do yachtsmen wear Polaroid sunglasses?
(a)
Explain with the aid of a diagram why transverse waves can be plane polarised but longitudinal waves cannot be plane polarised.
(b)
(i)
A filament lamp is observed directly and then thro ugh a sheet of Polaroid. Describe and explain the effect of the sheet of Polaroid on the intensity of the light seen.
(ii)
The sheet of Polaroid is now rotated in a plane perpendicular to the direction of tra vel of the light. What effect, if any, will this have on the intensity of the light seen?
6
Describe, with the aid of a diagram, how you would demonstrate that these microwaves were plane polarised.
7
(a)
(b)
Describe how you would de monstrate e xperimentally that light waves can be polarised, using either light or microwaves. Include a diagram of the apparatus you would use. What does the experiment tell you about the nat ure of electromagnetic waves?
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Diffraction When a wave hits an obstacle it does not simply go straight past, it bends round the obstacle. The same type of effect occurs at a hole - the waves spread out the other side of the hole. This phenomenon is known as diffraction and examples of the diffraction of plane waves are shown in the diagram. The effects of diffraction are much more noticeable if the size of the obstacle is small (a few wavelengths across), while a given size of obstacle will diffract a wave of long wavelength more than a shorter o ne. Diffraction can be easily demonstrated with sound waves or microwaves. It is quite easy to hear a sound even if there is an obstacle in the direct line between the source and your ears. By using the 2.8 cm microwave apparatus very good diffraction effects may be observed with obstacles a few centimetres across. One of the most powerful pieces of evidence for light being some for m of wave motion is that it also shows diffraction. The problem with light and that which led Newton to reject the wave theory is that the wavelength is very small and therefore diffraction effects are hard to observe. You can observe the dif fraction of light, however, if you know just where to look. The coloured rings round a street light in frosty weather, the coloured bands viewed by reflection from a record and the spreading of light round your eyelashes are all diffraction effects. Looking through the material of a stretched pair of tights at a small torch bulb will also show very good diffraction. A laser will also show good diffraction effects over large distances because of the coherence of laser light.
small g ap
large gap
small wavelength
diffraction at an edge
Diffraction is essentially the effect of removing some of the information from a wave front; the new wave front will be altered by the obstacle or aperture. Huygens' theory explained this satisfactorily. . diffraction round an obstacle
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http://www.launc.tased.edu.au/online/sciences/physics/diffrac.html http://www.ngsir.netfirms.com/englishhtm/Diffraction.htm http://lectureonline.cl.msu .edu/~mmp/kap 27/ Gary -Diffraction/ap p.htm
Huygens' principle1.SWF
Diffraction Intro.SWF
Diffraction&Huygens.SWF
Diffraction through a slit.SWF
Diffraction variable slit.SWF
Diffraction around an object.SWF
Diffraction of radio waves.SWF
Diffraction & resolving power.SWF
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1
Red monochromatic light falls on a narrow slit. Describe what happens to the diffraction pattern as the slit is slowly opened.
2
How would the diffraction pattern in Q1 be affected if blue light were used instead of red?
3
Which would be easier to receive in hilly areas and why, television or radio?
4
Why is the diffraction of light much more difficult to observe than the diffraction of microwaves?
5
Each of the diagrams below shows a series o f wavefronts, one wavelength apart, approac hing a gap betwee n two barriers in a ripple tank.
(a)
What is a wavefront?
(b)
Add further wavefronts to each diagram to show what happens as the waves pass through each gap.
(c)
The station BBC Radio 4 broadcasts both on the Long Wave band at 198 kHz and on VHF at approximately 94 MHz. In mountainous parts of the country, reception is better on Long Wave than on VHF. Suggest why.
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Wave Particle Duality for Electrons
Electrons scattered by a particle.SWF
electron gun
graphite screen
The electron beam is accelerated through a p.d. of a few kV and it hits a thin film of graphite in an evacuated tube. Electrons scattered by a crystal.SWF
Many of the electrons arrive near the centre of the screen but others arrive at other distances from the centre of the screen. So in addition to a bright spot in the c entre, two bright concentric rings are formed as well as other faint rings. http://cst-www.nrl.navy.mil/lattice/struk.jmol/a9.html
The electrons appear to be behaving like waves (diffracting ) If the acce lerating volta ge of the gun is increased the rings become smaller. Fast moving electrons (when regarded as particles) appear to have a short wavelength (when regarded as waves). After the discovery of the photoelectric effect it was realised that waves can possess particle- like properties, and a search was made to see if the reverse was true, could particles behave like waves. http://www.colorado.edu/UCB/AcademicAffairs/ArtsSciences/physics/PhysicsInitiative/Physi cs2000/applets/twoslitsb.html http://www.launc.tased.edu.au/online/sciences/physics/debroglie.html (But don‟t bother about the formulae!)
What are electrons?
When we are dealing with the forces on particles like electrons, and their energy changes, we can treat them as particles.
When we want to know where they are, we have to use wave ideas.
The phrase wave-particle duality is used to describe the t wo-sided nature of electromagnetic radiat ion, and the two-s ided nature of particles.
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http://chaos.nus.edu.sg/simulations/Modern%20Physics/Interference/interference.html
Two slit interference electrons & light.SWF
..\..\..\Multimedia\Mov ies\Electron Waves .mov
The location of an electron inside an atom may be described in terms of a wave whose amplitude at any place determines the probability of finding the electron at that position.
Particle or wave.SWF
Electrons in atoms
The amplitude of the „electron wave‟ at a p lace determines the probability of finding the electron at that place. Only particular standing waves are possible and these depend on the energy state of the electron. Animations\Schrodinger.xls
Quantised orbits
The simple Rutherford model of the atom had one serious disadvantage concerning the stability of the orbits. Bohr showed that in such a model the electrons would spiral into the nucleus in about 10 -10 s, due to electrostatic attraction. He therefore proposed that the angular momentum of the electron s hould be quantised, in line with Planck's quantum theory of radiation.
1
Which experiment shows particles behaving like waves?
2
Write down one device that uses the idea that particles have wave properties.
Scanning tunneling electron microscope image of graphite.GIF
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Pulse-echo techniques One method of finding the speed of sound, v, in air is to bang a drum while standing a measured distance, d , (at least 100 m) from the wall of a large building and measuring the time, t , between striking the drum and hearing the echo.
v
2d t
d Sonar and radar are methods during the Second World War that are still widely used to gauge the position of ships and aircraft. They ac hieve this by sending out pulses of sound and radio waves and noting the time and direction of the reflected pulses. Bats and dolphins are examples of animals that emit and receive high- frequency sounds.
What is Ultrasound? Ultrasound is the name given to high frequency sound - defined as sound with a frequency over 20 000Hz. Sounds with this freq uency are too high in pitch to be heard by the human ear. These waves can be transmitted in beams (like light) and are used to produce live 2-D images of the internal organs. Recent ly it has become possible to generate 3-D images by means of ultrasound. The ultrasound pulse travels through the body and ec hoes off the interna l orga ns. These ultrasound echoes are then recorded and displayed as a live image. It is used across a wide range of medical specialties including obstetr ics, gynaecology, cardiology, surgery, and gastroenterolo gy. Ultrasound is favoured in these areas as it is a safe and relatively inexpensive imaging method. For medical diagnostic purposes, frequencies used in Ultrasound scanning are in the range of 2.5-10MHz (2.5 to 10 million Hz). Very short bursts of sound lasting around one millionth of a seco nd are transmitted into the patient approximately 500 - 1000 times a second. As the sound tra vels in the body, it is reflected at the junction of differe nt the tissues, to produce ec hoes which are picked up by the transducer . The sound is reflected because different t issues carry sound at different rates. The fraction of sound that is reflected depends on the differe nce in a property known as the acoustic impedance of the tissue on each side of the interface. The acoustic impedance depends on the density of the medium so a much bigger reflection occurs at a tissue-bone boundary than a tissue-muscle interface. These echoes need to be electronically amplified in the scanner. The echoes that come from deep within the body are more attenuated (energy is absorbed or scattered
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within the body) than those from more superficial (or shallower) parts and therefore require more amplificatio n. When the echoes return to the transducer, it is possible to reconstruct a 2-D map of all the tissues that have been exposed to the ultrasound pulse. The information is stored in a computer and t hen displayed as an image on a television monitor. Stronger echoes appear as brighter dots on the screen. Ultrasound waves are produced by a piezoelectric transducer which is capable of changing electrical signals into mechanical waves ( ultrasound). The same transducer can also receive the reflected ultrasound and c hange it back into electric signals. This effect - known as the piezoelectric effect - is displayed by a number of naturally occurring materials Ultrasound is an extre mely useful means of diagnostic imaging. It is widely used in obstetrics, for foetal imaging and to guide diagnostic procedures, as well as gynaecolo gy. As fluid is a good conductor of sound, ultrasound is particularly good for d ifferentiating betwee n cysts and solid structures and viewing fluid- filled structures such as the bladder, or the foetus in the sac of amniotic fluid. Doppler Ultrasound
When ultrasound is transmitted towards a stationary reflector, the reflected waves will be of the same frequency as those originally transmitted. However, if the re flector is moving towards t he transmitter, the reflected frequency will be higher than the transmitted frequency. On the other hand, if the reflector is moving away from the transmitter, the reflec ted frequency will be lower than the transmitted frequency. This phenomenon is called the Doppler Effect, after its d iscoverer, C hristian Doppler. The difference between the frequencies is called the Doppler shift . This may sound very complicated, but it's a surprisingly everyday effect, most commonly illustrated by a siren on an ambulance or police car. As the car approaches, the sound appears higher and higher pitched - until the moment where it passes and the tone is heard to drop sharply. In medical practice, this is used in particular to measure the flow of blood through vessels and within the heart. As ultrasound is non-invasive and involves no ionising radiation. It is an extremely safe method of medical imaging. At the energies and doses currently used in diagnostic ultrasound, no harmful effects on any tissues have ever bee n demonstrated, over a very long period of use. This is another reason why it is ideal for foetal imaging, and indeed, is usually considered the only scan safe for pregnant women. Apart from its safety, ultrasound is also the method of choice for seeing many internal organs. Ultrasound provides high qua lity images of internal organs that some other scans such as MRI's cannot ac hieve. Finally, compared to many other techniques, it is relatively inexpensive - meaning it is very widely available.
Microsoft Office PowerPoint 97-2003
http://www.virtualcancercentre.com/investigations.asp?sid=8 The principles of medical Ultrasound http://www.mrcophth.com/commonultrasoundcases/principlesofultrasound.html#acoustic
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Thickness measurement (includes a good animation) http://www.ndt-ed.org/EducationResources/HighSchool/Sound/ultrasound.htm The amount of detail in a scan is affected by diffraction effects.
Microsoft Office Excel 97-2003 Works
The smaller the wavelength of sound used in an ultrasound scan, the smaller the finest detail that can be distinguished. However there is a contrary trend that the shorter the wavelength the sooner a wave pulse will be absorbed. Medical ultrasound scans generally compromise these competing factors using wavelengths in the range 0.075 mm to 1.5 mm. As a general guide the smallest detail that can be resolved will be about the size of a wavelength. To make out a foetal thumb that is 0.5 mm wide would require the use of ultrasound of wavelength 0.5 mm or less. There is an additional constraint that the resolution will be half the length of a pulse . A pulse may be a few wavelengths and its size may be calculated thus:
length speed time pulse is on Example An ultrasound system for examining the eye sends out a pulse of ultrasound waves with a frequency of 6 MHz. The pulse duration is 0.6 s. The speed of sound in the eye a vera ges 1510 m s-1 . What is the smallest detail that can distinguished?
1 2
Wavelength method Pulse length method
v f
1510 m s 1 6 10 s 6
1
2.5110 4 m 0.251mm
pulse length 1510 m s 1 resolution
0.6 10
half pulse length
6
s
0.906 10
3
m
0.453 mm
The worse resolution is 0.453 mm and finer details than this could not be seen in the image produced.
1
Ultrasound is preferred to X-rays for some diagnostic images because : A it gives a more detailed image B it is a longitudina l wave C it is less harmful to the patient D it penetrates the body more eas ily
2
A firework accelerates upwards and emits a constant high-p itched sound. An observer will hear: A a constant higher pitched sound B a constant lower pitched sound C a continually decreasing pitch D a continually increasing pitch
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3
A recorder at the finish line of a 100 m race sees the flash of the starting pistol and starts her stopwatch. A second recorde r fails to see the flash and starts his watch on hearing the bang. The winner‟s time differs by 0.3 s on the two watches. Explain the likely reason for this and use the differe nce to estimate the speed of sound in air.
4
A trawlerman uses sonar to detect shoals of fish. A strongly reflected pulse is received 1.60 s after it was transmitted. If the speed of sound in water is 1500 m s-1 , how far from the boat are the fish?
5
Give on advantage and on disadvantage for using high frequency ultraso und for diagnostic images in medicine.
6
A space probe sends radio signals back to Earth. W hy may the r adio rece iver at ground control have to be retuned after the launch of the probe?
7
A radar measurement of the distance to the Moon gives a round-trip time of 2.57 s. Calculate how far away the Moon is. (Speed of radio waves = 3.00 x 108 m s-1 ).
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Doppler Effect
Waves bunched together – wavelength shortened
Waves spread out – wavelength increased
Figure 1
The Doppler Effect is the apparent change of frequency and wavelength when a source of waves and an observer move relative to each other. These effects were first explained by Doppler in 1842 as a bunching up and a spreading out of waves. Looking at a d uck swimming in a pond would s how you that the waves it generates in the direction it is swimming are bunched while those behind it are spread o ut. (Figure 1) To demonstrate his theory he persuaded a group of trumpeters to stand and play in an open railway carriage while the carriage travelled across the Dutch countryside. Observers on the ground heard a change of pitch as the truck passed them. One of the most important applications of the Doppler Effect is in the study of the expansion of the Universe. Galaxies have their light shifted towards the red due to their speed of recession and when we receive the light at the Earth we describe it as Red Shifted. http://lectureonline.cl.msu.edu/~mmp/applist/doppler/d.htm Red shift.SWF
When a source of light is moving toward someone, the light will appear „bluer‟. If a source of light is moving away, the light will appear „redder‟. These two frequency shifts are called blue shift and red shift . Doppler effect.SWF
Blue shift a nd red shift are used to measure the velocity and rotation of stars and galaxies. Spiral Galaxy Red Blue Shift.MOV
Red shifts and blue shifts have been used to measure the orbital velocity of the Earth, to detect stars and quasars, and to detect the rotation of other galaxies. They have also been used to determine the speed of dust clouds in the Milky Way, and have helped to prove that our galaxy is rotating. Astronomers have discovered that all the distant galaxies are moving away from us, and by measuring their red shifts their speeds may be estimated.
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The furthermost galaxies have been estimated to have speeds approaching the speed of light. Hubble photographed the spectra of galaxies a nd detected that certain characteristic absorption lines were shifted towards the red end of the spectrum. http://zebu.uoregon.edu/nsf/hub.html The effect can be also be observed in the following uses and applications of the Doppler Effect (a) (b) (c) (d) (e) (f) (g)
Change in the pitch of a buzzer when it is whirled around your head Change in pitch of a train hooter or whistle as it passes through a station Shift of the freq uency of the light from the two sides of the solar disc due to the Sun's rotation Variation in the frequency of the light from spectroscopic binaries Police radar speed traps Doppler broadening of spectral lines in high temperature plasmas Measurement of the speed of the blood in a vein or artery
The Doppler Effect is commonly used in medicine. An ultraso und tra nsducer coupled to the body near an artery emits pulses that are reflected by moving blood cells within the arter y. A blockage due to a blood clot (thrombosis) or a constriction caused by a thickening of the arterial wall is detected by a s udden change in the shift frequency.
8
What is meant by the Doppler Effect?
9
Write down four uses or effects of the Doppler Effect.
10
If the light from a star is observed to be blue shifted as seen from the Earth what does this tell you about the motion of the star?
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Electron flow in a conductor Electric current
Current & electron flow 1.SWF
When you turn on an electric light an electric current flows in the wire. Do not think of it like water coming from a tap – the electricity current does not flow out from the switch – electric charge is already in the wire – connecting the lamp to the power supply via the switch simply gives the charged particles the energy to flow. This energy can come from a variety of sources – kinetic as in a dynamo, a chemical reaction in a cell, light falling on a photoelectric cell, heating the junction of two metals in a thermocouple, sound in a microphone or mechanical stress in a piezoelectric crystal. When an electric current flows electrical energy is converted to other forms of energy such as „heat‟, light, che mical, magnetic and so on. Consider a piece of metal wire - a very much enlarged view of which is shown in Figure 1 . atom/ion electron Figure 1
A piece of wire is made of a huge number of atoms and each one of these has its own cloud of electrons. However in a metal there are a large number of electrons that are not held around particular nuclei but are free to move at high speed and in a random way through the metal. These are known are „free‟ electrons and in a metal there are always large numbers of these. It is when these free electrons are all made to move in a certain direction by the application of a voltage across the metal that we have an electric current. The differe nce between a metal (a large and co nstant number of free electrons), a semiconductor (a few free electrons, the number of which varies with temperature) and an insulator (no/very few free electrons) is shown in Figure 2.
semiconduct
insulator
conductor
Figure 2
Each electron has a very small amount of electric charge, and it is more convenient to use a larger unit when measur ing charge. This unit is the coulomb.
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The charge o n one electron is -1.6 x 10-19 C. This is usually written as e. You would need about 6x1018 electrons to have a charge of one coulomb! The electrical charge passing any one point in a circuit in one second is called the electric current, a nd it is measured in Amperes (A). The Amp can be defined in the following way:
A current of 1 A flows in a wire if a charge of 1C passes any point in the wire each second. Current Rate of flow of charge
Rate of flow of charge.SWF
Velocity of free electrons in a wire
The free electrons in a metal have three distinct velocities associated with them:
a random velocity ( about 10 5 ms-1 )
a velocity with which electrical energy is transferred along the wire (about 108 ms-1 )
a drift velocity of the electrons as a whole when a current flows through the wire (this depends on the applied voltage but is us ually a few mms-1 for currents of a few amps in normal connecting leads).
Electron drift velocity
Microsoft Office PowerPoint 97-2003
I nAQv
Current & electron flow 2.SWF
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The table below shows some free electron concentrations Metal Lithium Sodium Silver Copper
Free electron concentration (m- ) (at 300 K) 4.7x10 2.7x10 5.9x10 8.5x10
Distinction between metals, semiconductors and insulators. Metals are good conductors because the char ge is carried by free electrons. For a good conductor like copper the number of free electrons per unit volume is n 1029 m-3 .
n is VERY BIG. Semi-conductors , like silicon, have fewer charge carriers per unit volume compared with metals. n is medium size. Insulators contain hardly any charge carriers at a ll (n is very small)
1
What is meant by the free electron concentratio n in a metal?
2
Why might a good electrical conductor also be a good thermal conductor?
3
Calculate the current flowing in a copper wire of cross- sectional area 2x10-7 m2 if the drift velocity is 3.5x10 -4 ms-1 .
4
Calculate the drift velocity in a silver wire of diameter 0.26 mm if a current of 20 mA flows through it.
5
A table tennis ball oscillates between two charged vertical metal plates. A sensitive ammeter connected between the plates records a c urrent of 3 A when the period of oscillation of the ball is 2 s. Calculate: (a) the charge carried between the plates by the ba ll in each osc illation (b) the number of electrons carried betwee n the plates by the ball in each oscillation
6
In a television tube 0.25 m long the electron velocity is 5x10 7 ms-1 . If the current in the tube is 1.5 mA calculate: (a) the number of electrons reaching the screen every second (b) the number of electrons in 1 cm of the bea m
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7
Diagra m 2 shows the results of an experiment to find the velocity of c harged ions. 0V It shows their movement during 50 s.
mm
Calculate: 10
(a)
the ion drift velocity
(b)
the charge reaching the positive electrode per second if the current is 1.25 mA
(c)
the number of ions reaching the positive electrode per second if there are each doubly charged
filter paper crystal
30 microscope slide
Diagram 2
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20
+75V crocodile clip
Potential and Potential Difference As a charge moves round a circuit from the positive to the negative it loses energy. There is a problem here. As you know an electric current is a flow of negatively charged electrons and these flow away from the negative terminal of a supply, round the circuit and back to the positive terminal. However the „traditional‟ view of current flow is from positive to negative and we will take that view when looking at the energy of electrical c harge. Figure 1
Electron flow direction
Traditional current direction
- to +
+ to -
We define the amount of electrical potential energy t hat a unit charge has as: The electrical potential energy of a unit charge at a point in a circuit is called the potential at that point. The next set of diagrams (Figure 2) show how the potential varies round some basic circuits. To simplify the treatment we are going to assume that the energy lost in the connecting wires is neglibgible and we are going to ignore it. This means that the energy of the charge at one end of a connecting wire is the same as that at the other end. The bigger the energy change the bigger the difference in potential. We call the difference in electrical potential between two points in the circuit the potential difference between those two places.
The potential differe nce between two points is defined as: Potential difference between two points in a circuit is the work done in moving unit charge (i.e. one coulomb) from one point to the other
http://regentsprep.org/Regents/physics/phys03/apotdif/default.htm http://www.rkm.com.au/ANIMATIONS/animation-electrical-circuit.html The units for potential difference are therefore Joules per coulomb, or volts. (1 volt = 1 Joule/coulomb).
So if a char ge Q moves between two points in a circuit that have a potential difference of V volts between them the energy gained (or lost) by the charge is given by the formula: Electrical energy = Charge x Potential difference(Voltage)
Joules = Coulombs x Volts = Amps x Time x Volts Electrical energy = ItV
Energy = VIt.SWF
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8
Calculate how much electrical energy is supplied by a 1.5V battery when: (a) a charge char ge o f 1000C passes passe s through it (b) a curren curre nt o f 2.5A flows from it fo r 2s
9
How much energy is drawn from a 12V car battery if it is used to supply 200A for 1.5s to the starter motor.
10
10 ide ide ntical tica l torch torch bulbs are connected in series across a 12 V d.c s upply. (a) What is the p.d across acro ss each eac h b ulb (b) What is is the potential pote ntial at the jo jo in o f the second seco nd a nd third b ulbs ulbs fro fro m the negative terminal?
11
(a) (b) (c)
Write Write t he word equati equat ion that defines pote ntial tia l differe nce. The unit o f pote po ten ntia l d iffere nce is the volt. Express the volt in ter ms o f base units units only o nly.. A 6.0 V battery batte ry of negligible interna interna l resista nce is connected to a filamen filame nt lamp. The c urrent urre nt in the lamp is 2.0 A.
Calculate how much energy is transferred in the filament when the battery is connected for 2.0 minutes. 12
(a) (b)
State the word word equation that is used to define char ge. A 9.0 V battery batte ry of negligible intern internaa l resista nce is is connected to a light light bulb. bulb.
Calculate the energy transferred in the light bulb when 20 C of charge flows through it. it. 13
A thick wire is co nnected nnected in series with a thin wire of the same material and a battery battery (a) (b)
In which wire do the e lectrons ec trons have the gre gre ater d rift ve loc ity? Expla in your answer. A battery is connected across a lar ge resistor and a small sma ll resistor is is connected connected in para llel. llel. The The currents curre nts through the resistors are differe nt. Which Whic h resistor res istor has the the highe highe r d is s ipation pa tion o f power? po wer? Explain Expla in your answer.
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Var Va riati at io n of curre curre nt with applie applie d voltage voltage There There are several severa l wa wa ys in which the the current through a de vice vice can be altered. Elast Elastic ic strain, temperat temperat ure and light light are exampl examp les of these. F igure igure 1 shows examples examples of of curre curren nt-v t- voltag olta ge curves for a number of d ifferent ifferent situ sit uation atio ns. Voltage-curr Voltage-current ent characteristics
The p.d. applied to the device is varied over a suitable range and the the current curre nt is measured. measured. The device is turned round to see if it behaves the same in both bot h direction directio ns.
A
Device
Graphs Graphs of c urrent against voltage are p lotted. Filament lamp
V
I
I
R R increases with temperature
V
V
V
Resistor I I
R
V
R is constant
V
V
Semiconductor Semiconductor Diod Dio de The diode The diode allows allows current to flow free reely ly in one dire dire ction only. The current increases rapidly for ‘forward’ ‘forward’ voltages vo ltages greater than 0.5 V.
I
V IV Graphs.SWF
http://micro.magnet.fsu.edu/electromag/java/filamentresistance/ (Resistance (Resistance at mol molecular ec ular level)
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Resistance and temperature When a mater ia l is is heated its resistance resista nce ma ma y c ha nge. This is is due d ue to the ther ma l moti ot ion of th t he atom a tomss within w ithin the spec spec imen. For a metal the temperature coefficient of resistance is positive - in other words and increase increase in the temperat temperatu ure gives gives an increase in resi res istan sta nce. This This can be explained by the motio motio n o f the atoms ato ms a nd free free e lectrons within the solid. so lid. At low low tempe temperatures ratures the thermal vibration is small and electrons can move easily within the lattice but at high temperatures the motion increases giving a much greater chance of collisions between the cond cond ucti uct io n e lectrons and the latt lattiice and so impeding imped ing their moti mot io n. In a light b ulb ulb o the filament is at about 2700 C when it is working and its resistance when hot is about ten times that when when col co ld. (For a typica typica l domestic light bulb the resistance measured at r oom oom temperature was 32 Ω and this rose to 324 Ω at its working temperature). However in non-metals such as semiconductors an increase in temperature leads to a drop in resi res istan sta nce. This can be explained by el e lectron ectro ns gaining energy and moving into the conduction band - in fact changing from being bound to a particular atom to being able to move freely - an increase increase in the number number of free electrons. The temperature coefficient of resistance and also that of the temperature coefficient of resistivity is therefore negative. 14
Why does th t he res istan sta nce of a metal eta l increase increase as its temperat temperatu ure is ra ised?
15
Why does the resistance of a semiconductor decrease as its temperature is raised?
16
Explain in terms of its „secret‟ composition, how the resistance of a fixed resistor may remain consta constan nt as its te te mpera mpera ture is raised?
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Resistance The free electrons in a metal are in constant random motion. As they move about they collide with each other and with the atoms of the metal. If a potential difference is now applied across the metal the electrons tend to move towards the positive connection. As they do so Figure 1 electron metal ato m their progress is interrupted by collisions. These collisions impede their moveme nt and this property of the materia l is ca lled its resistance. If the temperature of the metal is raised the atoms vibrate more strongly and the electrons make more violent collisions with them and so the resista nce o f the metal increase. The electron dr ift velocity v (in the equation I = nAQv) decreases. The resistance of any conducting material depends on the following factors: the material itself (actually how many free electrons there are per m3 ) its length its cross-sectional area and its temperature
Resistivity of warm wire.GIF
http://regentsprep.org/Regents/physics/phys03/bresist/default.htm The resistance of a given piece of material is related to the current flowing through it
and the potential difference between its ends by the equation: R
V I
The units of resistance are ohms (Ω). If the ratio of p.d to current remains constant for a series of different p.d.s the material is said to obey Ohm's Law. This is true for a metallic conductor at a constant temperature. This means that although we can always work out the resistance of a specimen knowing the curre nt through it and the p.d across it. However if these quantities are altered we can only PREDICT how it will behave under these new conditions if it obeys Ohm's law. It is also vital to realise that the resistance is simply the ratio of the voltage and current at a particular point and NOT generally the gradient of the V I c urve. http://micro.magnet.fsu.edu/electromag/java/filamentresistance/ (Resistance at molecular level)
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It is important to realise that Ohm‟s Law only holds for a metallic conductor if the temperature is constant. This means that if the temperature of the metal Voltage is held steady at say 15 o C the variation of (V) 75oC current and voltage will be linear. However if the temperature of the metal changes (as in the filament of a light bulb) then the resistance will 15oC also change. The collisions betwee n the electrons and the atoms will occur more often and be more violent. So if the wire is raised to 75 o C a second set of Current (I) readings can be taken – they will still be linear but the resistance of the wire (the ratio of V to I ) Figure 3 will be greater (see Figure 3). It is worth having a look at two graphs that show how the resistance of two types of material change when their temperature is changed. The first is a metal wire (Figure 4(a)), and the second is a (negative temperature coefficient) thermistor (Figure 4 (b). e g a t l o V
e g a t l o V
Current Figure 4 (a)
Current Figure 4 ( b)
In the case of the metal wire the resistance increases as the temperature increases, you can see this because the ratio of pairs of points on the V-I graph increases at high currents (hot wires). In the case of the thermistor the resistance decreases as the temperature increases, you can see because the ratio of pairs of points on the V-I graph decreases at high currents (high temperatures.). Although the gradients of the graphs suggest a change in resistance do not be tempted to use the gradient to work out the resistance. You must still deal with the voltage/current ratio only . The reason that the thermistor decreases is because t he thermistor is a semiconductor and more free electrons are produced as the temperature is raised. The number n (in the equation I = nAQv) increases. (In fact more electrons are raised to the conduction band of the material.)
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17
Calculate the current through the following resistors: (a) (b) (c) (d)
18
What is the resistance of the following? (a) (b)
19
120 Ω connected to 240V 4700 Ω connected to 12V 10kΩ connected to 6V 2.5MΩ connected to 25V
a lamp that draws 2A from a 12V supply a kettle that draws 4A from a 240V supp ly
The two graphs below show a specimen of metal wire at two different temperatures. I (A)
(a)
Calculate the resistance of the wire at each temperature
(b)
Which graph temperature?
(c)
Does the material disobey Ohm's Law? Explain your answer.
shows
the
higher
10
5
V (V) 20
20
Why is it more likely for the filament in a light bulb to break when you switch it on rat her than when it has been on for some time? Explain your answer.
21
The circuit shows a battery of negligible internal resista nce connected to three resistors.
(a) (b) (c)
Calculate the potential differe nce across the 15 resistor. Calculate the current I 1 in the 4.0 resistor. Calculate the current 12 and the resistance R.
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22
The circuit diagram shows a 12 V d.c. s upply o f negligible interna l resistance connected to an arrangement of resistors. The current at three places in the circuit and the resistance o f two of the resistors are given on the diagram.
(a) (b) (c)
Calculate the potential differe nce across the 4.0 resistor. Calculate the resistance of resistor R 2 . Calculate the resistance of resistor R 1 .
23
Two filament lamps are designed to work from a 9.0 V supply but they have different characteristics. The graph shows the current-potential difference relationship for each lamp.
(a)
The lamps are connected in parallel with a 9.0 V supply as shown.
(i) (ii) (iii)
Which lamp is brighter? Give a reason for your answer. Deter mine the current in the supply. Calculate the tota l resistance of the two lamps when they are connected in parallel.
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(b)
The lamps are now connected in series to a variable supply which is adjusted until the current is 0.8 A.
Compare and comment on the brightness of the lamps in this circuit. 24
A student connects the circuit as s hown in the diagram.
(a) (b)
The reading on the voltmeter is 1.8 V. Calculate the current in the resistor. Calculate the resistance of the thermistor.
The graph shows how the resistance of the thermistor depends on its temperature
(c) (d)
Deter mine the temperature of the thermistor. If the e.m.f, of the supply were doubled, would the reading on the voltmeter double? Explain your answer.
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25
Use the axes to draw the current- voltage characteristics of a diode and a filament lamp I
I
V
V
Diode
Filament lamp
Electrical Power Power is the rate at which work is done or energy changed fro m one form to another and so:
Electrical Power
Energy
V Q
t time Energy Power time V I t
V
Q t
V I
Example problems 1.
Calculate the power of a 12V light bulb using 2.5 A. Power = VI = 12V x 2.5A = 30 W
2.
Calculate the current used by a 12V immersion heater that is designed to deliver 30000J in 5 minutes Energy = Power x Time = 30000 30000 = Power x 300 Power = 100W. Current I = P /V = 100W/12V = 8.3A
3.
Calculate the energy transformed by a 12V car batter y that delivers a current of 200A for 3 s. Energy = Power x time = VIt = 12 x 200 x 3 = 72 000 J
Two alternative formulae for electrical power We can combine the formula R
V I
with the basic formula for electrical power to
give two alternative formulae: Since power = VI, we can write 2
Power V I I R
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V 2 R
Example problems 1 Calculate t he resistance of a 100W light bulb if it takes a current of 0.8 A P 100W P I 2 R R 156 2 2 I 0.8 A 2 Calculate the power of a 12V immersion heater with a res istance of 10
P
V 2 R
12 2 10
14.4W
http://www.ukpower.co.uk/running-costs-elec.asp (Electricity Running Cos ts Calculator)
26
Calculate the power loss in an electrical transmission cable, 15 km long, carrying a c urrent of 100A at a potential of 200 kV. The res istance per km of the cable is 0.2 Ω.
27
What power is supplied to the heater of an electric bar fire with a resistance of 50 Ω connected to the mains 240V supply?
28
What is the power loss down a copper connecting lead 50cm long with a resistance of 0.005 Ω per metre when it carries a current of 1.5A?
29
A 240 V mains lamp draws a current of 2 A from the supply when operating normally. Calculate: (a) the resistance of the lamp when hot (b) the power of the lamp when operating normally (c) the number of electrons passing through the lamp filament eac h second (d) the energy transferred to each coulomb by the supply (e) the energy transferred to each electron by the supply
30
What is the power of an electrical heater operating from 110 V if the resistance of the heater is 15 ?
31
What is the power loss down a copper connecting lead 75cm long with a resistance of 0.13 per metre when a current of 4.5 A flows through it?
Page 55 of 92
Two resistors in series (Figure 1)
Series circuit current.SWF
Series circuit voltage.SWF
The current (I) flowing through R 1 and R 2 is the same and so the potential differences across them are V1 = IR 1 and V2 = IR 2 V V 1 V 2
R is the effective series resistance of the two resistors. So:
I
R1
R2
V1
V2 Figure 1
V IR IR1 IR2 R R1 R2
Two resistors in parallel (Figure 2)
Parallel circuit current.SWF
Parallel circuit voltage.SWF
The potential difference (V) across each of the two resistors is the same and the current (I) flowing into junction A is equal to the sum of the currents in the two branches, there fore : I I 1 I 2 V IR1 IR2 I
V R
V R1
1 V R R2
1 R1
R1
I
I1
A
I2
V
R2
1
Figure 2
R2
where R is t he effective resistance of the two res istors in parallel. Notice that
two res istors in series always have a larger effect ive resistance than either of the two resistors on their own
two in parallel always have a lower resista nce than either of the two resistors on their own
This means that connecting two or more resistors in parallel, such as the use of a mains adaptor, will increase the current drawn from a supply.
Resistances in parallel.mdl
http://schools.matter.org.uk/Content/Resistors/Default.htm
Page 56 of 92
Examples 1
2
3
32
Calculate the resistance of the following combinations: (a) (b)
100 Ω and 50 Ω in series 100 Ω and 50 Ω in parallel
(a)
R = R 1 + R 2 = 100 + 50 = 150 1 1 1
(b)
R 100 50 R 33
Calculate the current flowing through the following when a p.d of 12V is applied across the ends : (a) (b)
200 Ω and 1000 Ω in series 200 Ω and 1000 Ω in parallel
(a) (b)
R = 1200 R = 167
I = V/R = 12/1200 = 0.01 A = 10 mA I = V/R = 12/167 = 0.072 A = 72 mA
You are given one 100 Ω resistor and two 50 Ω resistors. How would you connect a ny combination of them to give a combined resistance o f: (a) 200 Ω, (b) 125 Ω (a)
100 in series with both the 50
(b)
the two 50 in parallel and this in series with the 100
Four 10 resistors are co nnected as shown in the diagram
(a) Calculate t he total resistance of the co mbination. (b) Comment on your answer and suggest why such a combination of resistors might be used. 33
Complete each of the following statements in words: (a) The resistance of an ammeter is ass umed to be (b) The resistance of a voltmeter is assumed to be (c) Calculate the total resistance of four 5.0 resistors connected in parallel.
Page 57 of 92
34
An electric room heater consists of three heating elements connected in parallel across a power supply.
Each element is made from a metal wire of resistivity ( ) 5.5 x 10 -5 m at room temperature. The wire has a cross-sectional area (A) 8.0 x 10 -7 m2 and length 0.65 m. The heater is controlled by two switches, X and Y. (a) Show that the res istance of one heating element at room temperat ure is approximately 45 . R A
(b)
Calculate the tota l resistance of the heater for the following combinations of switches at the moment the switches are closed.
Switch X
Switch Y
Open
Closed
Closed
Open
Closed
Closed
Resistance of heater/
(c)
Calculate the maximum power output from the heater immediately it is connected to a 230 V supply.
(d)
After being connected to the supply for a few minutes the power output falls to a lower steady value. Explain why this happens.
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Resistivity There are three factors that affect the resistance of a specimen of material:
the temperature the dimensions of the specimen - the smaller the cross sectional area and the longer the specimen the larger the resistance the material from which the specimen is made
A The property of the material that affects its resistance is called the resistivity of the material and is given the symbol . Resistivity is related to resistance of a specimen of length l and cross sectional area A by the formula: R
A
RA
The units for resistivity are Ωm
Resistivity & Area.GIF
http://regentsprep.org/Regents/physics/phys03/bresist/default.htm
The following table gives the resistivities of a number of common materials. -
Material Copper Nichrome Aluminium Eureka Lead Manganin
Resistivity (x 10 1.69 130 3.21 49 20.8 44
Ωm)
Material Non metals Insulators Germanium Silicon Carbon Silver
Resistivity (Ωm) 10 10 - 10 0.65 2.3x1033 - 185x10 1.6x10-
The resistivities of solutions cannot be quoted generally because they depend on the concentrations and are therefore var iable quantities. As an example however the resistivity of pure water is about 2.5 x105 m and that o f a saturated solution of sodium chloride about 0.04 m at 20o C. The reciprocal o f resistivity is known as the conductivity of the material () 1.
Calculate the resistance of a 1.5 m long piece of eureka wire of diameter 0.5 mm
R
2.
A
49 10 8 m 1.5m
0.25 10
3
m
2
3.7
A piece of wire needed for a heater is to be made of Manganin. It is to have a cross -7 2 sectional area of 1.5x10 m and a resistance of 5 Ω. How long must it be?
RA
5 1.5 10 7 m 2 44 10 8 m
1.7 m
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Data: Resistivity of copper Resistivity of constantan Resistivity of germanium
1.7 x 10-8 m 47 x 10-8 m 0.65 m
35
Calculate the length of a copper wire of cross sectional area 0.65 x10 -7 m2 that has a resistance of 2Ω
36
Calculate the resistivity of a material if a 2.6 m length of wire of that material with a diameter of 0.56 mm has a resistance of 3Ω.
37
Calculate the resistance between the large faces of a slab of germanium of thickness 1 mm and area 1.5 mm 2 .
38
Calculate the length of a section of constantan wire of diameter 0.32 mm that has a resistance o f 10 Ω
39
(a)
(b) (c) (d)
40
A student is asked to carry out an experiment to find the resistivity of the material of a length of resistance wire. Draw an appropriate circuit diagram. List all the measurements the student should take to find the res istivity. How should these meas urements be used to find the resistivity? Suggest two precautions the student should take to ensure an accurate result.
Lord Kelvin discovered that the electrical resistance of iron wire changed when the wire was stretched or compressed. This is the principle on which a resista nce strain gauge is based. Such a gauge consists of a length of very fine iron wire cemented between two very thin sheets of paper.
(a) The cross-sectional area of the wire is 1.1 x 10 -7 m2 and the gauge length as shown in the diagram is 2.4 x 10 -2 m. The resistivity of iron is 9.9 x 10 -8 m. Calculate the resistance of the strain gauge. (b) When this gauge is stretched its length is increased by 0.1% but its crosssectional area remains the same. What is the change in the resistance of the gauge? (c) Explain the effect that stretching the wire will have on the drift velocity of electrons in the wire. Assume that the other physical dimensions of the wire remain unchanged and that there is a constant potential difference across the wire.
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41
The diagram shows a t ype of resistor commonly used in electronic circuits.
It consists of a thin film of carbon wrapped around a cylindrical insulator. The diagram be low (not to scale) shows a typical film of carbon, resistivity p, before it is wrapped round the insulator.
(a) Show that the resistance R of the carbon film is given by R
(b)
(c)
w t This film has length l = 8.0 mm, width w = 3.0 mm and thickness t = 0.0010 mm (i.e. t = 1.0 x 10 -6 m). If the resistivity of carbon is 6.0 x 10-5 m, calculate the resistance of t he carbo n film.
Show that the resistance of a square piece of carbon film of uniform thickness is independent of the length of the sides of the square.
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Electromotive force and internal resistance When current flows round a circuit energy is transformed in both the external resistor but also in the cell itself. All cells have a resistance of their own and we call this the internal resistance of the cell. The voltage produced by the cell is called the electromotive force or e.m.f for short and this produces a p.d across the cell and across the external resistor. E
The e.m.f of the cell can be defined as the maximum p.d that the ce ll can produce across its t erminals or the open circuit p.d since when no current flows from the cell then no electrical energy can be lost within it.
r R
E V Ir IR Ir
R
r
Consider the circuit shown in Figure 1. If the e.m.f of the cell is E and the internal resistance is r and the cell is connected to an external resista nce R then:
Ir E
The quantity of useful electrical energy available outside the cell is IR and Ir is the energy transformed to other forms within the cell itself.
IR
Figure 1
We usually require the internal resistance of a cell to be small to reduce the energy transformed within the cell; however it is sometimes helpful to have a rather larger internal resistance to prevent large currents from flowing if the cell terminals are shorted. The more correct definition of emf is
energy transformed in the complete circuit charge I 2 Rt I 2 rt
It IR Ir
V Ir
Hence V E Ir , and E = V when I = 0, i.e. when the load R is infinite (VERY BIG). We say that the battery (cell) is on open circuit .
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Example problem
A cell of e.m.f 12V and inter nal resistance 0.1Ω is used in two circuits. Calculate the p.d between its terminals when it is connected to (a) 10 Ω and (b) 0.2 Ω . (a) Total resistance = 10 + 0.1 = 10.1 Ω 12 V Therefore c urrent = 1.19 A 10.1 Loss of e nergy per coulomb in the ce ll = 1.19 x 0.1 = 0.119V P.d between terminals = 12 - 0.119 = 11.88V = 11.9 V (b) Total resistance = 0.2 + 0.1 = 0.3 Ω 12V Therefore c urrent = 40 A 0.3 Loss of e nergy per coulomb in the cell = 40 x 0.1 = 4V P.d between terminals = 12 - 4 = 8V
Short circuit current When a power supply is short-circuited by connecting the terminals together with a low resistance (R), the total resistance of the circuit is almost entirely due to the internal res istance (r) of t he power supply. Car batteries need to supply currents up to 200 A, so they are designed to have very low internal resistance. High-voltage power supplies are designed to have large internal resistance so as to prevent them from supplying dangerously high currents.
Experiment to measure internal resistance R may be varied and a range of values of V and I are measured. Keep a record of values of R for later.
E V
R
r
E IR Ir V IR V E Ir rI E
V E
A graph of V against I is plotted. The graph is a straight line which has a gradient - r and a „y intercept‟ E
Gradient = -r
I
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The power transformed in the load R is P VI How can we maximise the power? Make V and I as large as possible. What is the largest possible value of V ? E How big is the current when V E ? Zero! What happens to V if we try to increase the current? V decreases. Maximum power is delivered to the load when the load resistance R equals the internal res istance r of the supply.
Microsoft Office Excel 97-2003 Works
42
What is meant by: (a) the E.M.F. of a cell (b) the internal resistance of a cell?
43
How does the internal resistance of a cell affect the current drawn from it?
44
A cell of e.m.f 3.0 V is connected to a resistor of 2700 and when a voltmeter of very high resistance is connected across its terminals the voltmeter reads 2.8 V (a) (b) (c)
45
Explain the differe nce betwee n these two voltages Calculate the internal res istance of the cell Calculate the new reading of the voltmeter if the voltmeter has a resistance of 1500 .
A student wants to measure the internal resistance of a cell and he connects it in series with an ammeter and a variable resistor. The student then connects a voltmeter across the variable resistor and, for different settings of the variable resistor, obtains the following readings: V/V
1.43
1.41
1.39
1.33
1.20
/mA
143
176
231
333
600
Plot a graph of V against and deduce the internal resistance and e.m.f. of the cell. (V = E - r or V = -r + E )
Page 64 of 92
46
A resistance substitution box R is connected to a power supply. For various values of R the current is measured. R/
8.0
6.0
4.0
2.0
1.0
0.80
0.60
/mA
167
214
300
500
750
833
938
P/mW (a) (b) (c) (d) (e)
47
Calculate the power of the resistor for each value of R. Predict the power of the resistor whe n R is zero and when R is infinitely large. Plot a graph of the power, P (d issipated in R) against resistance R. For what value of R is the power a maximum? The internal resis tance of the power supply is known to be 1.0 . Comment.
Why are car batter ies des igned to have negligible internal res istance whereas high voltage (EHT) laboratory supplies are manufactured with a very large internal resistance?
Page 65 of 92
Potential divider The basic circuit is shown in the first circuit diagram. The output vo ltage across AB is given by:
V R2 R2 V V 2 I R2 R R R R 2 2 1 1
R1
Note t hat the input voltage (V) in this case supplied by the battery is constant. The current flowing through both resistors is the same (series circuit) and so the output voltage across one of them depends simply on the two resistance values and the input voltage.
V
A
R2
V2 B
Measuring the output
If we now attempt to act ually measure the output voltage things may change. (a)
Firstly consider using a digital voltmeter with very high (if not virtually infinite) resistance (R V). The resistor R 2 and the voltmeter are connected in parallel and so their combined resistance (R) is given by the equat ion;
1 1 1 R R2 RV huge – almost 1 infinite and so 0 and RV can be ignored. This 1 1 means that and R R2
R1
RV is
V R2
V2 VO
Digital voltmeter
so R R2 .
The output voltage (V 0 ) measured by the meter rea lly is that across R 2 , in other words V 2 .
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(b)
A moving coil meter. These meters have a much lower resistance than a digital meter, usually some t ens of k .
R1
This means that the combined resistance of R2 and RV is affected by the resistance of the voltmeter and is actually lower than R2 .
V R2
V2
VO Moving coil voltmeter
The proportion of the input voltage ( V ) dropped across R2 therefore falls and so the output voltage (V 0 ) is less than that measured with a digital meter. Replacing R 1 with a Light dependent resistor (LDR)
The LDR is a component that has a resistance that changes when light falls on it. As the intensity of the light is increased so the resistance of the LDR falls. If the LDR is connected as part of a potential d ivider as shown in the diagram then as the light level is increased its resistance falls and the proportion of the input voltage dropped across it will also fall.
R1
V LDR
V2
So in the light V2 is low and in the dark V2 is high.
Replacing R 1 with a thermistor Something very similar happens if R 2 is replaced by a thermistor. As the temperature of the thermistor rises its resistance falls and so the voltage dropped across it falls. When the thermistor is hot V 2 is low and when the thermistor is cold V2 is high.
R1
V V2 thermistor
http://www.crocodile-clips.com/absorb/AP5/sample/020201.html (Practice questions)
Page 67 of 92
48
Complete the following table showing the readings of a digital voltmeter of infinite resistance for the output voltage (V) for a series of different resistances and input voltages.
Input voltage (V o ) 12 6 24 6
R1 /
R2 /
100 k 25 k 5k 250
200k 10 k 20 k 100
Output voltage (V )
R1
Vo
R2
49
Repeat question one but this time assume that the meter connected to measure the output voltage is an analogue meter with a resistance of 200 k . Input voltage (V 0 ) 12 6 24 6
50
V
R1 / 100 k 25 k 5k 250
R2 / 200k 10 k 20 k 100
Now assume that res istor R1 is replaced by a thermistor T 1 (one where the resistance decreases as the temperature rises). If the values of the resistance R2 and the thermistor are equal at the start what will happen to the output potential difference (V ) as the thermistor is cooled?
Output voltage (V )
T1
Vo
R2
V
51
The thermistor is now replaced by an LDR. What happens to the output potential difference (V ) as the light intensity falling on the LDR is increased?
52
Assuming that the voltmeter used to measure V in Q51 has an almost infinite resistance what happens to the current through R2 as the light intensity falling on the LDR is decreased?
Page 68 of 92
53
A student connects a 9.0 V battery in series with a res istor R, a thermistor and a milliammeter. He connects a voltmeter in parallel with the res istor. The reading on the voltmeter is 2.8 V and the reading on the milliammeter is 0.74 mA.
(a)
(b)
(i)
Show that the resistance of R is approximately 4000 .
(ii)
Calculate the resistance of the thermistor
The thermistor is mounted on a plastic base that has steel sprung clips for secure connection in a circuit board.
Another student is using an identical circuit except that the bare metal pins of his thermistor are twisted together.
(c)
Suggest an explanation for how the reading on this student's milliammeter will compare with that of the first student.
Page 69 of 92
54
(a)
A student sets up a circuit and accidentally uses two voltmeters V1 and V2 instead of an ammeter and a voltmeter. The circuit is shown below.
(i)
Circ le the voltmeter which should be an ammeter.
(ii)
Both voltmeters have a resistance of 10 M. The student sees that the reading on V2 is 0 V. Explain why the potential differe nce across the 100 resistor is effect ively zero.
(b)
The student replaces the 100 resistor with another resistor of resistance R. The reading on V2 then beco mes 3.0 (i)
Complete the circuit diagram below to show the equivalent resistor network following this change. Label the resistor R.
(ii)
Calculate the value of R.
Page 70 of 92
V.
55
A light-emitting light-e mitting d iode (LED) is a d iode that t hat e mits light when it conducts. co nducts. Its circuit symbol is A student connects the circuit shown below.
She notices that the reading on the high resistance voltmeter remains at 0 Vas she she s lides lides th t he contact contac t between ter ter minals minals A an a nd B. (a) Explain Explain this t his obser obser vation as fully as you can. (b) The student then disconnects the LED and reconnects the circuit as shown below. below. S he intends to vary the intensity of the light emitted by t he LED b y sliding the the contact betwee n termin ter minaa ls A an a nd B.
The st udent cannot ca nnot detect dete ct a ny light emitted by the LED. Briefly Brie fly e xpla xpla in why the LED is so dim. (c) Draw the circuit that the t he student should have connected using thi t hiss apparatus in order to vary the brightness of the LED and measure the potential differe nce across it.
Page 71 of 92
56
The The following follow ing c ircuit can be used as a light meter.
(a) The maximum value of resistance of the light-dependent resistor (LDR) is 950 k . What What is th t he reading on the the voltmeter for this t his resista resista nce? (b) The minimum value of resistance of the LDR is 1.0 k . What is is t he reading on the voltmeter for this resista nce? (c) For this light meter the voltmeter is connected across the 10 k resistor, rather than the the LDR. Explain Expla in how the readings read ings o n the voltmeter voltmete r e nable this circuit circ uit to to be used as a light meter. 57
Two Two resistors of o f res res istan sta nce 2.0 M and 4.0 are connected in series across a supply voltage of 6.0 V. Toget Toget her they form form a simple potential divider divider circuit. State the potential difference across eac h resistor. P.d. across the 2.0 M resistor
=
P.d. across the 4.0 resistor
=
A second second pote ntial tia l divider circuit uses a re sistor and a diode connected in series series with w ith the sam sa me su s upply. Cal Ca lculate culat e the poten pote ntial tia l differe nce across each component when the resistance of the resistor and diode are 45 and 5.0 respectively.
P.d. across the 45 resistor = P.d. across the diode
=
Page 72 of 92
Nature of light The Phot P hotoe oellectri ec tricc Eff Effec ectt When is a wave not a wave - when it's a particle! In 1887 Heinrich Hertz noticed that sparks would jump between two spheres when their surfaces surfaces were illuminated by light from from anothe anothe r spark. This e ffect was stu st udied mo re carefully care fully in the following years b y Hallwac hs a nd Len Le nard. They called ca lled the effect photoelectr investigate photoelectric ic e mission and a very simple experiment can be used to investigate it.
no effect
no eff ect ect
Leaf falls immediately
In the diagram shown above a clean zinc plate is fitted to the top of a gold leaf electroscope and then given a positive charge (you can do this either with a charged glass rod or a n EHT s upply. The next ne xt thing is to s hine hine some radi rad ia tion on it, it, using a n ordin ord inar ary y la la mp, a he he lium- neon eo n lase laserr (giving (giving out intense inte nse red light) or an ultra violet light ha ha s absolutely abso lutely no e ffec ffect. t. The e le ctroscope ctrosc ope stay sta ys c harged ar ged and the leaf sta ys up. up. However if the plate is given a negative charge to start with (using say a charged polythe polythe ne rod ) th t here is a d iffe iffe ren re nce. Using the la la mp and e ven the lase laserr ha ha s no no effect, but but when ultra ultra violet violet light light is shone on the plate the leaf falls immediately: immediately : the e lectroscope ect roscope has bee be e n d ischar sc harged. ged. (Doing (Do ing the e xperiment in a vacuum vac uum p roves ro ves that tha t it is not ions in the air that are cau ca using the discharge.) No effect can b e produ prod uced with radiation radiat ion of o f long lo nger er wavelength ( lower frequency) frequency) no matter how long the radiation is shone on the plate. The plate was emitting electrons when the ultra violet radiation fell on it and this explained why the leaf only fell when it had an initial negative charge - when it was positive the the electrons were were attracted at tracted back to the plate. The The research researc he rs found found five importan importa nt facts abo ut th t he experiment:
no electro ns were emitted from fro m the pl p la te if it was pos itive the number of electrons emitted per second depended on the intensity of the incident incident radiati rad iation on the energy of the electrons depended on the frequency of frequency of the incident radiation there was a minimum frequency ( f 0 ) below which no electrons were emitted no matter how long radiation fell on the surface If elec elec trons are emitted this occurs immediately
Page 73 of 92
This minimum frequency is called the threshold frequency for that material. Photons with a lower frequency will never cause electron emission. This can be explained as follows The free electrons are held in the metal in a "hole" in the electric field, this is called a potential well. Energy has to be supplied to them to enable t hem to escape from the surface. Think of a person down a hole with very smooth sides. They can only escape if they can jump out of the hole in one go. They cannot get half way up and then have a rest - it's all or nothing! This is just like the electrons. The deeper the "hole" the more tightly bound the electrons are and the greater energy and therefore the greater is the frequency of radiation that is needed to release them. The quantum theory of Max Planck is needed to explain the photoelectric effect. In trying to explain the variation of energy with wavelength for the radiation emitted by hot objects he came to the conclusion that all radiation is emitted in quanta and the ene rgy of one quantum or photon is given by the e quation:
Photon Energy hf The amount of energy needed to just release a photoelectron is known as the work function for the metal. This can also be expressed in terms of the minimum frequency that will cause photoelectric emiss ion.
Work Function hf 0 The table below gives the work function for a number of surfaces - both in joules and in electron volts. The threshold frequency for each surface is also included. Work function Sodiu m Caesiu m Lithium Calcium Magnesium Silver Platinu m
-19
/ 10
3.8 3.0 3.7 4.3 5.9 7.6 10.0
J /eV 2.40 1.88 2.31 2.69 3.69 4.75 6.75
Frequency /1014 Hz 5.8 4.5 5.6 6.5 8.9 11.4 15.1
Wav elength /nm 520 666 560 462 337 263 199
Another way of looking at it is to think of a fa irground coconut shy. A brother and sister are trying to knock the coconuts off their stands. The boy has a large box of table tennis balls which he is throwing at the coconuts, with little effect. No matt er how many of the table tennis balls he throws at a coconut it will still stay in place – the table tennis balls represent the “red” quanta. However his sister has a pistol! This represents the violet quanta. A single shot from the pistol will knock off a coconut and it will do it immediately. As we saw in the previous experiment we could illuminate the zinc plate all day with a high powered laser and the leaf of the electroscope would not fall. However as soon as we shone the ultra violet light on the plate the leaf dropped. This is because the ultra violet light has a high enough frequency and therefore each quantum of ultra violet has sufficient energy. One quantum has enough energy to kick out an electron in one go. The photoelectric effect is therefore very good evidence for the particulate nature of light.
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Measuring the Planck Constant (h = 6.6 x 10-34 J s)
lens
* lamp
filter collector electron V
sodium emitter A
Stopping
For each coloured filter the positive potential („stopping‟) potential on the cathode is increased until the ammeter reading is zero - this means that no photoelectrons are reaching the detector.
Metal 1
A graph can then be plotted of stopping potential (electron energy) against frequency.
Metal 2
f o
Theory
If hf (or f
Frequenc
), f is the threshold frequency for that metal, and electrons are h just able to be removed. However they will have no K.E. after they have been emitted. If hf , the surplus photon energy hf is given to the electron as kinetic energy. A stopping p.d. is applied so that the collector is negative with respect to t he emitter and t his makes it diffic ult for the electrons to reach the collector. As this reverse voltage V is gradually increased the ammeter reading is reduced. Eventually when eV kemax the ammeter reading will be ZERO and electrons will no longer be able to reach the collector.
eV hf V
h e
f
e
A graph of V v f is plotted, and gradient =
h e
y intercept =
e
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x intercept =
h
If we change the sensitive surface to another metal (metal 2) with a higher work h function then since the gradient of the line is constant and we simply get a e second line parallel to the first but shifted to the right. This simple experiment is very good evidence that light sometimes behaves like a stream of particles. http://www.launc.tased.edu.au/online/sciences/physics/photo-elec.html http://lectureonline.cl.msu.edu/~mmp/kap28/PhotoEffect/photo.htm Charge on the electron = (-)1.6 x 10 -19 C Planck‟s constant = 6.63 x 10 -34 Js Speed of light in free space = 3.00 x 10 8 ms-1 Mass of an electron = 9.11 x 10 -31 kg 1
What is meant by a q uantum or photon?
2
Write down Planck‟s equation for the energy of a qua ntum of radiation.
3
Which has the greater energy, a quantum of yellow light or a quantum of violet light?
4
Which experiment shows waves behaving like partic les?
5
What is meant by the “stopping potential” in the photoelectric effect?
6
What is meant by the “work function” in the photoelectric effect?
7
What is the energy of a quantum of radiation that has a wavelength of 500 nm?
8
What is meant by an “electron volt”?
9
If the frequency of the incident radiation that falls on a metal surface is increased what happens to the photoelectrons that are emitted?
10
If the intensity of the incident radiatio n that falls on a metal surface is increased what happens to the photoelectrons that are emitted?
11
(a)
Light of wavelengt h 480 nm just gives photoelectric emission from the certain metal surface. What is the work function of that surface in Joules?
(b)
If the wavele ngth is reduced to 400 nm what is the energy of the electrons emitted?
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12
The diagram shows monochromatic light falling on a photocell.
The photocell is connected so that there is a reverse potential difference across the cathode and the anode. (a)
(b)
13
(a)
Explain the following observations. (i) Initially there is a current which is measured by the microammeter. As the reverse potential difference is increased the current reading on the microammeter decreases. (ii) When the potential difference reaches a certain value V, the stopping potential, the current is zero. What would be the effect on the value of the stopp ing potential V S of (i) increasing the intensity of the incident radiation whilst keeping its frequency constant. (ii) increasing the frequency of the incident radiation whilst keeping its intensity constant?
Photoelectrons are e mitted from the s urface of a metal when radiation above a certain frequency, f o , is incident upon it. The maximum kinetic energy of the emitted electrons is E K . On the axes below sketch a graph to show how EK varies with frequency f . E
f
(b)
State how the work function, , of the metal can be obtained from the graph.
(c)
Explain why this graph always has the same gradient irrespective of the metal used.
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14
The diagram shows a coulombmeter (an instrument for measuring charge) set up to demonstrate the photoelectric effect.
The clean zinc plate is negatively charged. Ultraviolet light is shone onto the zinc plate and the plate discharges. The coulombmeter reading gradually falls to zero. When the experiment is repeated with red light the plate does not discharge. (a) (b)
(c)
15
Explain these effects in terms of the particle theory of light. What would happen to the charged plate if (i) the inte nsity of the red light were increased (ii) the intensity of the ultraviolet light were increased? Zinc has a work function of 3.6 eV. Ca lc ulate the maximum kinetic energy of the photoelectrons when the zinc is illuminated with ultraviolet light of wavelength 250 nm.
The photoelectric effect supports a partic le theory of light b ut not a wa ve theory of light. Below are two features of the photoelectric e ffect. For each feature explain why it supports the particle theory and not the wave theory.
(a)
Feature 1:
The emission of photoelectrons from a metal surface can take place instantaneously.
Explanation (b)
Feature 2:
Incident light with a frequency below a certain threshold frequency cannot release electrons from a metal surface.
Explanation
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16
(a)
Define the intensity of a n electromagnetic wa ve.
(b)
Two beams of monochromatic electro magnet ic radiation, A and B, have equal intensities. Their wavelengths are: Beam A 300 nm Beam B 450 nm
In the table below, E denotes the energy of a photon and N denotes the number of photons passing per second through unit area normal to the beam. The subscripts A and B refer to the two beams. In the second column of the table, state the value of each ratio, and in the third column explain your answer. Ratio E A
N A
(c)
Value
Explanation
E B
N B
The table below gives the work functions of four metals. Metal
Work function/eV
Potassium
2.26
Magnesium
3.68
Tungsten
4.49
Iron
4.63
Define the term work function. (d)
A metal plate made from one of these metals is exposed to beams A and B in turn. Beam A causes electro ns to be emitted from the plate, but beam B does not. Calculate the photon energies in each beam and hence deduce from which metal the plate is made.
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17
A monochromatic light source is placed 120 mm above the cathode of a photocell.
(a)
The light source consumes 6 W of power and is 15% efficient. Calculate the light intens ity at the cathode. State an assumption that you made.
A potential difference is applied between the cathode and the anode of the photocell and the sensitive ammeter detects the current.
The table below shows the currents that are obtained w ith this apparatus for two differe nt intensities and two differe nt wavelengths of light, using two differe nt cathode materials. Work funct ion energies are given. Wavelength of incident radiation/nm
Cathode material
Work function/eV
320
Aluminium
640
Photocurrent/A when intensity of incident radiation is 1 W m-2
5 W m-2
4.1
0
0
Aluminium
4.1
0
0
320
Lithium
2.3
0.2 x 10-12
1.0 x 10-12
640
Lithium
2.3
0
0
(b)
Show that the incident photons of = 320 nm and) = 640 nm have energies of approximately 4 eV and 2 eV respectively.
(c)
Account for the photocurrent readings shown in the table.
(d)
Calculate the stopping potential for the photoelectrons released by lithium when irradiated by light of wavelength 320 nm.
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18
The table be low shows the results o f an exper iment like Millikan‟s using sodium as the metal plate.
Stopping voltage V s /V
Frequency of light f / 1014 Hz
0.43
5.49
1.00
6.91
1.18
7.41
1.56
8.23
2.19
9.61
3.00
11.83
(a) Plot a graph of V s against f . The following equation applies to t he photoelectric effect:
hf
eV S
where is the work function of the metal and e is the charge on an electron. (b) What infor mation about the elec trons emitted does the value of the term eV S give?
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Spectra and energy levels The atoms in the tube are excited and as a result, they emit radiation.
Narrow slit
When the radiation from excited atoms is viewed through a diffraction grating lines are seen at different angles. These correspond to images formed by light of different wavelengths (and frequencies).
Diffraction Grating Gas discharge tube containing e.g. sodium vapour
The emission line spectrum pattern is characteristic of the gas in the tube. http://jersey.uoregon.edu/elements/Elements.html
Line spectra & electron transitions.SWF
Electron energy levels Electrons are held in atoms at only certain energy levels. Normally an electron will lie in its lowest energy state - the ground state. (13.6) (12.75) (12.1) (10.2)
(0)
Ionised state -0e V -0.85 n 2 excited state -1.5 e V s 1 excited state -3.4 eV
Ground state -13.6 eV
The electron may remain above the ground state temporarily, but it will usually drop back to the ground state, either directly or via another energy level, giving out energy as it does so. Animations\BohrModel.swf
The energy levels for atoms are fixed. Only certain changes in energy are possible. To move between any two of these levels an electron needs to give out or receive a definite amount of energy. Small jumps mean small energy changes, which correspond to low frequencies of radiation and large .
E3 E = E3 – E2 = h f
E2
E = E2 – E1 = h f E1
When an electron is given sufficient energy to rise to one of the higher energy states the atom is said to be „excited‟.
Big jumps mean large energy changes, which correspond to high frequencies of radiation and small .
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The atoms of eac h element have a characterist ic set of energy levels, and so e mit a characteristic set of frequencies when they are excited. You can identify an element from the frequencies of radiation it emits when excited.
Emission spectra of different atoms.SWF
R
G
V
White V
G
R
http://www.dartmouth.edu/~chemlab/info/resources/mashel/MASHEL.html
Example problem Calculate the frequency and wavelength of a quantum of radiation emitted when an electron in level 4 falls to level 2. -19
Energy of level 4 = - 1.36 x 10 J -19 Energy of level 2 = - 5.42 x 10 J -19 Energy difference (E) = + 4.06 x 10 J Therefore
f
E h
4.06 10 19 6.63 10 34
3.00 108 14
6.12 10
6.12 1014
4.9 10 7 m 490 nm / blue green
A typical absorption spectrum
Emission & absorption spectra.SWF
http://www.dartmouth.edu/~chemlab/info/resources/mashel/MASHEL.html http://phys.educ.ksu.edu/vqm/html/emission.html http://www.launc.tased.edu.au/online/sciences/physics/linespec.html http://www.launc.tased.edu.au/online/sciences/physics/photo-elec.html http://lectureonline.cl.msu.edu/~mmp/kap28/PhotoEffect/photo.htm
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-19
(1 eV = 1.60 x 10 19
J)
The diagram shows some of the energy levels for an atom of hydrogen. Photons are emitted when an electron moves down from one level to another. (a) (i) (ii) (iii) (iv) (v)
(b) (c)
When an electron moves from level 2 to level 1, what is its loss of energy in eV its loss of energy in J the frequency of the emitted photon the wavelength of the emitted photon the part of the electromagnetic spectrum in which this radiation occurs?
Repeat part (a) for an electron moving from leve1 3 to level 2. Repeat part (a) for an electron moving from leve1 4 to level 3.
20
The two lowest excited states of a hydrogen atom are 10.2 eV and 12.1 eV above the ground state. (a) Calculate three wavelengths of radiation that could be produced by transitions between these states and the gro und state. (b) In which parts of the spectrum would you expect to find these wavelengths?
21
The figure shows an energy level diagram. Sketch a possible line spectrum for the light emitted when electrons make the tra nsitions shown. Label the lines, using the letters shown in the diagram, and indicate on your spectrum diagram which end corresponds to the higher frequency.
22
The four lowest energy levels for an atom consist of the ground state and three levels above that. How many transitions are possible between these four levels?
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23
The figure shows three energy levels for a particular atom. When an electron moves from level1 to the ground state the light emitted is blue. In what part of the spectrum would you expect to find the radiation emitted when an electron moves from level 2 to the ground state?
24 .
Refer to the d iagra m in Q19
25
(a)
If the atom is in the ground state, how much energy must be given to it to ionise it?
(b)
Suppose an electron of energy 2.2 eV collides with the atom. Explain the possible results if (i) the atom is in the ground state (ii) its electron is at the -3.41 eV level
(c)
What is the wavelength of the photon that could raise an electron from the -0.849 eV level to the -0.545 eV level?
(d)
If an electron retur ns from the -0.849 eV level to the ground state, what is the wavelength of the photon emitted?
Some of the energy levels for the sodium atom are -1.51 eV, -1.94 eV, -3.03 eV (two levels very close together) and -5.14 eV; which is the ground state. Draw a labelled diagram for these levels, and describe and explain what might happen if cool sodium vapour (i.e. sodium whose atoms are in the ground state) is bombarded with (a)
electrons whose k.e. is 2.00 eV
(b)
electrons whose k.e. is 2.50 eV
(c)
light of wavele ngth 590 nm.
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26
Four of the ener gy levels of a lithium atom are s hown below.
(a) Draw on the diagram all the possible transitions which the atom could make when going from the -3.84 eV level to the -5.02 eV level. (b) Photons of energy 3.17 eV are shone onto atoms in lithium vapour. Mark on the diagra m, and label with a T, the transition which could occur. (c) One way to study the energy levels of an atom is to scatter electrons from it and measure their kinetic energies before a nd after the collision. If an electron of kinetic energy 0.92 eV is scattered from a lithium atom which is initially in the -5.02 eV level, the scattered electron can have only two possible kinetic energies. State these two kinetic energy values, and explain what has happened to the lithium atom in each case. (You should assume that the lithium atom was at rest both before and after the collision.) Kinetic energy 1 Explanation Kinetic energy 2 Explanation
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27
The diagram shows the lowest four energy levels of atomic hydrogen.
(a) Calculate the ionisation energy in joules for atomic hydrogen. (b) On the diagram above draw (i) (ii)
a transition marked with an R which shows a photo n released with the longest wavelength, a transition marked with an A which shows a photon absorbed with the shortest wavelength.
(c) Describe how you would produce and observe the emission spectrum of hydrogen in the laboratory. (d) What would such a spectrum look like?
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28
The diagram shows some of the energy levels of a mercury atom.
a. Calculate the ionisation energy in joules for an electron in the -10.4 eV level. b. A proton of kinetic ener gy 9.2 eV collides with a mercury atom. As a result, an electron in the atom moves from the -10.4 eV level to the 1.6 eV level. What is the kinetic energy in eV of the proton after the collision? c. A transition between which two energy levels in the mercury atom will give rise to an emission line of wavelength 320 nm?
Page 88 of 92
Conservation of energy for waves Inverse square law A poi nt source of waves emits energy equally in all directions
If energy is conserved then as the waves spread out t he same e nergy is spread over a larger area. Inverse square law.SWF
http://mort.isvr.soton.ac.uk/SPCG/Tutorial/Tutorial/Tutorial_files/Web-basics pointsources.htm
The energy flux or intensity is defined as
power area
P
4 r 2
This relationship is called the inverse square law. When the distance is doubled the energy flux is red uced by a factor of four. 29
(a) (b)
Explain what is meant by the inverse square law of electroma gnetic waves such as visible light. Explain how this inverse square law is consistent with the law of conservation of energy.
30
A light with an output in the visible range of 50 W is switched on at night at the top of a high tower. Calculate the intensity of the light from the tower at distances of (a) l00 m (b) 200 m (c) 300 m.
31
A 60 W light bulb converts electrica l energy to visible light with an efficie ncy of 8%. Calculate the visible light intensity 2 m away from the light bulb.
32
The minimum intensity that can be detected by a given radio receiver is 2.2 10-5 W m-2 . Calculate the maximum distance that the receiver can be from a 10 kW transmitter so that it is just able to detect the signal.
33
In listening to a person talking to you who is standing 4.0 m away the intensity of the sound at your ear is 1.2 W m-2 . What is the power output power of the speaker's voice?
34
A radio-operated garage door opener responds to signals with intensity greater than 20 W m-2 . For a 250 mW transmitter unit that broadcasts equally in all directions, what is the maximum distance from the garage at which the transmitter will open the door?
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35
36
A communication satellite is in orbit above the Earth‟s surface. (a)
The satellite‟s-e lectrical system is powered by 20 000 photovoltaic cells, each of area 10 cm2 . The intensity of the sunlight falling on the cells is 1.4 kWm-2 . The ce lls produce 5.0 kW of electrical power. Calculate the efficiency of the cells in transferring solar energy to electrical energy.
(b)
(i)
The satellite generates a signal of power 5.0 kW and orbits at a height of 3.6 x 104 km above the Eart h‟s surface. Calculate the intensity which is detected at the Earth‟s surface if the satellite transmits uniformly in all directions. Assume there is no absorption of the signal along its path.
(ii)
In practice, reflec tors on the satellite focus all the 5.0 kW of transmitted power onto a small area of the Earth‟s surface. If this area is a circ le of dia meter 1000 km, calc ulate the intensity that would be detected there. Assume there is no absorption of the signal a long its pat h.
A leaf of a plant tilts towards the Sun to receive solar radiation of intensity 1.1 kW m-2 , which is incident at 50 to the surface of the leaf.
(a) The leaf is almost circular with an average radius of 29 mm. Show that the power of the radiation perpendicular to the leaf is approximately 2 W. (b) Calculate an approximate value for the amount of solar energy received by the leaf during 2.5 hours of sunlight.
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37
The graph shows how the intensity of light from a light-emitting diode (LED) varies with distance from the LED.
(a) Use data from the graph to show that the intensity obeys an inverse square law. (b) What does this suggest about the amount of light absorbed by the air?
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