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Page 1 of 92
28 - 35
Mechanica Mechanicall & Electromagnetic Waves
3 - 21
36 - 38
22 - 26
39 - 40
Pol Po larisation
27 - 29
Diffrac Diffracti tion on
30 - 32
42 - 43
Wave Nature of Electrons
33 - 34
44 – 49
Pulse Echo Techniques
35 - 40
Charge and Current
41 – 44
Voltage, Voltage, Current, Resistance & Power
45 – 58
59 – 61
EMF EMF & Internal Internal Resi Res istance
62 - 65
58, 60 - 62
Potenti Po tential al Divider Divider
66 – 72
Wave/Particle nature of light
73 – 81
Spe Sp ectra & Energy Energy Levels
82 – 88
69 – 70
89 - 92
Page 2 of 92
28 - 35
Mechanica Mechanicall & Electromagnetic Waves
3 - 21
36 - 38
22 - 26
39 - 40
Pol Po larisation
27 - 29
Diffrac Diffracti tion on
30 - 32
42 - 43
Wave Nature of Electrons
33 - 34
44 – 49
Pulse Echo Techniques
35 - 40
Charge and Current
41 – 44
Voltage, Voltage, Current, Resistance & Power
45 – 58
59 – 61
EMF EMF & Internal Internal Resi Res istance
62 - 65
58, 60 - 62
Potenti Po tential al Divider Divider
66 – 72
Wave/Particle nature of light
73 – 81
Spe Sp ectra & Energy Energy Levels
82 – 88
69 – 70
89 - 92
Page 2 of 92
Waves Mechanical waves These would be set up within a solid, a liquid or a gas due to the vibration of the molecules th t here. They are tran tra nsmitted by both inter inter molecul olec ular ar forces and by collisions collisions between the molecules. In so so lid lid s they they could could be either eit her lo lo ngitudina l or trans 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 and sound waves in air. Earthquake waves.SWF
They all travel at the same speed 3.00 x 10 8 m s-1 .
They do not require a medium through which to pass.
They are generated by accelerating charged particles.
Electromagnetic waves.SWF waves.SWF
Gamma X-rays 10-15
4x10-7 4x10-7 - 7x10-7 10-6
Microwave Microwa vess
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 p lace to p lace can be reflected can be refracted can be su s uperposed can be diffracted
Page 3 of 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 electromagnetic waves.
Transverse wave generation.SWF
Waves, peaks & troughs.SWF
http://www.acoustics.salford.ac.uk/feschools/waves/waves.htm (b) longitudinal waves - the oscillation is along the direction of propagation of the wave (Figure 1(b)). An example of this type is sound waves in air. wavelength wavelength 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 )
the number of vibrations per seco nd made by eac h particle/wave
the time take n for one complete oscil lation (T= 1/f)
the „delay‟ between the oscillations of neighbours
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
At large sports meetings the crowds sometimes produce a „Mexican wave‟. (a)
What would be a better name for this manoeuvre?
Describe what the crowd would need to do to produce such a wave and suggest values for its frequency and amplitude.
Page 5 of 92
The diagram shows a transverse wave on a rope. It is moving to the right. Several particles on the rope have been 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 '.
Describe, with the aid of a sketch, an electromagnetic wave.
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 sitt ing close to the TV. Speed of electromagnetic waves = 3.00 x 10 8 ms-1 Speed of sound in air = 340 ms-1
Diagram ( i) represents part of a stretched spring. Diagram ( ii) represents the same section of the spring at one instant of time when a sinusoidal 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 of 92
The diagram shows the shape of a wave on a stretched rope at one instant of time. The wave is travelling to the right.
(a) Determine the wavelength of the wave. (b) Mark on the diagram a point on the rope whose motion is exactly out of phase with the motion at point A. Label this point X. (c) Mark on the diagram a point 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 plane polarised electromagnetic waves of wavelength 29 mm. (a)
Calculate the frequency of this radiation. Complete the diagram of the electromagnetic spectrum below by adding the names of the parts of the electromagnetic spectrum.
State a typical value for the wavelength of radiation at boundary X.
Page 7 of 92
The table below summarises some features of the electromagnetic spectrum. Complete the table by filling in the missing types of radiation, wavelengths and sources.
Very hot objects
Page 8 of 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 of 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 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)
The diagram shows a wire with a mass of 1.30 kg at one end and a vibrator at the other end. 0.90 m N 1.30 kg
(a) (b) (c) (d) (e) (f) (g)
How many wavelengths does the diagram show? Use your answer to calculate the wavelength 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 ener gy terms, what is the difference between a standing wave and a progressive wave?
Sketch three stationary patterns that could be formed on the cord in the diagram. Mark the nodes and antinodes in each sketch. (b)
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
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 diagra m explain how you would show experimentally that f was inversely proportional to for a fixed value of n, e.g. n = 2.
Page 10 of 92
State two differences between a stationary wave and a progressive wave. Difference 1 Difference 2 Spiders are almost completely dependent on vibrations transmitted through their webs for receiving information about the location of their prey. The threads of the web are under tension. When the threads are disturbed by trapped prey, progressive transverse waves are transmitted along the sections of 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.
Explain why droplets form 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 frequency 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 of the string.
(a) (b) (c)
State the wavelength of this wave Mark a letter A on the graph to label an antinode. The stationary wave is formed by the superpos ition 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. Determine the amplitude of each of the travelling waves.
Page 11 of 92
(a) (b) (c)
Describe, with the aid of a diagram, 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 wavelength of microwaves.
A piece of string is connected to a variable frequency vibration generator. The fundamental 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.
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
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)
Explain how the movement of the bow causes this wave 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 drawn 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 and overlap they „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
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?to pic=19
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.
The diagrams in Figure 1 below show two waves of equal amplitudes with different phase and path differences between them. The first pair has a phase difference of o
or 180 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.
Page 14 of 92
constructive interf erence
To obtain a static interference pattern 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 constant phase difference between them. Sources with synchronised phase changes between are called coherent sources and those with random phase changes 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
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
Minimum – crest meets trough
Page 15 of 92
Diagrams in Figure 3-7 show two sources S 1 and S2 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 experiment 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 screen 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 interference pattern.
Two dippers Ripple tank.GIF
Two source Max & Min.SWF
Figure 4 (b)
Figure 5 maximum
Page 16 of 92
Young's slits Max & Min.SWF
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.
In Figures 6 and 7 you can see that at the different points on the screen the waves from S1 have travelled a different distance from those from S 2 . In Figure 6 the path difference is zero, in Figure 7 it is half a wavelength
Path difference = /2
Path difference = 0
S2 screen Figure 6
The effect of amplitude If two sources have very different amplitudes, then you will not be able to observe superposition patterns. The wave with the larger amplitude will dominate, because it makes little difference to the total whether the wave with the smaller amplitude 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 wavefront has an oval shape because waves travel faster in deep water than in shallow water.
Page 17 of 92
x Places marked and oscillate with a phase difference 2 http://www.acoustics.salford.ac.uk/feschools/waves/super.htm
speed of light as = 3.00 x 10 8 ms-1 speed of sound in air = 330 ms -1 . 17
The diagram shows an experimental arrangement for investigating the superposition of sound waves from two sources S 1 and S2 . The sources are in phase and produce sound of wavelength 80 mm.
When S1 M is 800 mm the trace on the oscilloscope is a maximum. Suggest three possible values for S 2 M.
What is meant by the principle of superposition of waves? Are there any types of wave which do not obey the principle 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
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 successive maxima. Explain this effect and calculate the wavelength of the radio signal that she is tuned to.
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 the 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 graph.,
(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
The diagram is a plan view of an experiment to measure 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 on 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 student 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 sources 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 .
1 Speed c1 Medium 1
Light moves slower in medium 1 than in medium 2. 1 2
sin 1 sin 2
Since the path of the light is reversible sin 2 c2 2 1 sin 1 c1
Refractio n occurs for all waves. Sound can be deviated as it passes from war m air to cooler air and microwaves can be refracted by wax.
Page 22 of 92
Refraction – wave fronts
Wavefronts a nimation 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
One side of the wave moves from A to C (a distance v gt) 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 t vg sin g AD
Page 23 of 92
Total internal reflection and critical angle
medium one medium tw o
Total internal ref lection
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 and 1.33 respectively. Calculate the critical angle for a water-glass surface. Air
g g w w
sin a sin g sin g sin w
, a w
sin a sin w
sin g sin a 1 1 g a . a w 1.13 . .a w 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
sin 90 1.50 c 62.5 For an air-glass boundary sin c a
And for an air-water boundary
sin c a
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 caused by continuous internal reflection.)
1 Blue light is deviated more than red light when it enters a glass block because: A it has a longer wavelength B it has a lower freq uency 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
Draw a diagram showing wavefronts when light passes from air into glass. a g 1.50 at incident angle of 45.
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.
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 a .
Calculate the critical angle for the interface between the core of an optical fibre with refractive index 1.60 and the cladding with refractive index 1.52.
Rainbows are caused when sunlight is dispersed by raindrops. The different 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 actually only partially reflected at the surface of 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 change of speed on entering the water. (d) Red light has a frequency of 4.2 × 10 14 Hz. Calculate its wavelength 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 successive 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.
Magnetic field (B)
Electric field (E) Unpolarised radiation
Plane polaris ed 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
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 shown with electromagnetic waves such as microwaves, TV, radio and light.
It is important to realise that transverse waves can be polarised while longitudinal waves cannot.
Page 27 of 92
Effects of polarisation with light.
Sunglasses Polarisation by reflection - glare/shine off roads Polarisation of scattered sunlight
Car windscreens Optical activity Stresses in materials
The effect of a polariser 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 and 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). Compar iso n of the neural inputs from these two rows allows the animal to sense polarized light.
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 s urfaces 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, creating 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 identify regions of high stress.
Page 28 of 92
What is meant by plane polarized light?
Explain whether the human eye can distinguish polarized light from unpolarised light.
How could you tell if a television signal was plane polarized and also find out the direction of polarization?
Why do yachtsmen wear Polaroid sunglasses?
Explain with the aid of a diagram why transverse waves can be plane polarised but longitudinal waves cannot be plane polarised.
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.
The sheet of Polaroid is now rotated in a plane perpendicular to the direction of travel of the light. What effect, if any, will this have on the intensity of the light seen?
Describe, with the aid of a diagram, how you would demonstrate that these microwaves were plane polarised.
Describe how you would demonstrate experimentally 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 nature of electromagnetic waves?
Page 29 of 92
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 one. 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 apparat us 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 form 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 obser ve the diffraction 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.
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 apert ure. Huygens' theory explained this satisfactorily. . diffraction round an obstacle
Page 30 of 92
http://www.launc.tased .edu.au/online/s ciences/p hysics/d iffrac.ht ml http://www.ngs ir.netfirms.co m/englishhtm/Diffraction.ht m http://lectureonline.cl.msu.edu/~mmp/kap 27/ Gary -Diffraction/ap p.htm
Diffraction through a slit.SWF
Diffraction variable slit.SWF
Diffraction around an object.SWF
Diffraction of radio waves.SWF
Diffraction & resolving power.SWF
Page 31 of 92
Red monochromatic light falls on a narrow slit. Describe what happens to the diffraction pattern as the slit is slowly opened.
How would the diffraction pattern in Q1 be affected if blue light were used instead of red?
Which would be easier to receive in hilly areas and why, television or radio?
Why is the diffraction of light much more difficult to observe than the diffraction of microwaves?
Each of the diagrams below shows a series of wavefronts, one wavelength apart, approaching a gap between two barriers in a ripple tank.
What is a wavefront?
Add further wavefronts to each diagram to show what happens as the waves pass through each gap.
The station BBC Radio 4 broadcasts both on the Long Wave band at 198 kHz and on VHF at approximately 94 MHz. In mountaino us parts of the country, reception is better on Long Wave than on VHF. Suggest why.
Page 32 of 92
Wave Particle Duality for Electrons
Electrons scattered by a particle.SWF
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 ce ntre, two bright concentric rings are formed as well as other faint rings. http://cst-www.nrl.navy.mil/lattice/s truk.jmo l/a9.html
The electrons appear to be behaving like waves ( diffracting) If the accelerating voltage 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 proper ties, 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 two-sided nature of electromagnetic radiation, and the two-sided nature of particles.
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
Elect rons 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
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 should be quantised, in line with Planck's q uantum theory of radiation.
Which experiment shows particles behaving like waves?
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.
d Sonar and radar are methods during the Second World War that are still widely used to gauge the position of ships and aircra ft. 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. Recently it has become possible to generate 3-D images by means of ultrasound. The ultrasound pulse travels through the body and echoes off the internal organs. These ultraso und echoes are then recorded and displayed as a live image. It is used across a wide range of medical specialties including obstetrics, gynaecology, cardiology, surgery, and gastroenterology. Ultraso und 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 second are transmitted into the patient approximately 500 - 1000 times a second. As the sound travels in the body, it is reflected at the junction of different the tissues, to produce echoes which are picked up by the transducer. The sound is reflected because different tissues carry sound at different rates . The fraction of sound that is reflected depends on the difference 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 bo undary 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 ec hoes ret urn 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 then displayed as an image on a television monitor. Stro nger 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 change it back into electric signals. This effect - known as the piezoelectric effect - is displayed by a number of naturally occurring materials Ultrasound is an extremely useful means of diagnostic imaging. It is widely used in obstetrics, for foetal imaging and to guide diagnostic procedures, as well as gynaecology. As fluid is a good conductor of sound, ultrasound is particular ly good for differentiating between cysts and solid str uctures 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 reflector is moving towards the 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 reflected frequency will be lower than the transmitted frequency. This phenomenon is called the Doppler Effect, after its discoverer, Christian 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 been 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 quality 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 Of fice PowerPoint 97-2003
http://www.virtualcancercentre.com/investigations.asp?sid=8 The princ iples 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 wavele ngth. 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 averages 1510 m s-1 . What is the smallest detail that can distinguished?
Wavelength method Pulse length method
1510 m s 1 6 10 s 6
2.5110 4 m 0.251mm
pulse length 1510 m s resolution
half pulse length
The worse resolution is 0.453 mm and finer details than this could not be seen in the image produced.
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 easily
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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A recorder at the finish line of a 100 m race sees the flash of the starting pistol and starts her stopwatch. A second recorder 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 difference to estimate the speed of sound in air.
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 so und in water is 1500 m s-1 , how far from the boat are the fish?
Give on advantage and on disadvantage for using high frequency ultrasound for diagnostic images in medicine.
A space probe sends radio signals back to Earth. Why may the r adio receiver at ground control have to be retuned after the launch of the probe?
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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Waves bunched together – wavelength shortened
Waves spread out – wavelength increased
The Doppler Effect is the apparent change of frequency and wavelength when a source of waves and a n 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 duck swimming in a pond would show you that the waves it generates in the direction it is swimming are bunched while those behind it are spread out. (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 and 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 and 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 aro und your head Change in pitch of a train hooter or whistle as it passes thro ugh 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 ultrasound transducer coupled to the body near an arter y emits pulses that are re flected 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 sudden change in the shift frequency.
What is meant by the Doppler Effect?
Write down four uses or effects of the Doppler Effect.
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 c urrent flows e lectrical energy is converted to other forms of energy such as „heat‟, light, chemical, 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 eac h 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 difference between a metal (a large and constant 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.
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 on 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, and it is measured in Amperes (A). The Amp can be defined in the following way:
A current of 1A 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 e lectrons 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 usually a few mm s-1 for currents of a few amps in normal connecting leads).
Electron drift velocity
Microsoft Office PowerPoint 97-2003
Current & electron flow 2.SWF
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The table below shows some free electron concentrations Metal Lithium Sodium Silver Copper
Distinction between metals, semiconductors and insulators. Metals are good conductors because the charge 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 all ( n is very small)
What is meant by the free electron concentration in a metal?
Why might a good electrical conductor also be a good thermal conductor?
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 .
Calculate the drift velocity in a silver wire of diameter 0.26 mm if a current of 20 mA flows through it.
A table tennis ball oscillates between two charged vertical metal plates. A sensitive ammeter connected between the plates records a current 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 oscillation (b) the number of electrons carried between the plates by the ball in each oscillation
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 beam
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Diagram 2 shows the results of an experiment to find the velocity of charged ions. 0V It shows their movement during 50 s.
the ion drift velocity
the charge reaching the positive electrode per second if the current is 1.25 mA
the number of ions reaching the positive electrode per second if there are each doubly charged
filter paper cr stal
30 microscope slide
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+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 charge. Figure 1
Electron flow direction
Traditional cu rrent direction
+ to -
- to +
We define the amount of electrical potential energy that 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 difference 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 charge 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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Calculate how much electrical energy is supplied by a 1.5V battery when: (a) a charge o f 1000C passes through it (b) a current o f 2.5A flows from it fo r 2s
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 identical torch bulbs are connected in series across a 12 V d.c supply. (a) What is the p.d across each b ulb (b) What is is the potent ial at the jo jo in o f the second and third b ulbs ulbs fro fro m the negative terminal?
(a) (b) (c)
Write the word equation that defines potential difference. The unit o f potentia po tentiall d ifferenc iffere ncee is the volt. Express the volt in term ter ms o f base units only. A 6.0 V batter ba ttery y o f negligible internal resistance resis tance is conne con nected cted to a filamen filame nt lamp. The current cu rrent 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
State the word equ eq uation atio n that is used to define de fine char ge. A 9.0 V batter ba ttery y o f negligible internal resistance resis tance is conne con nected cted to a light bulb.
Calculate the energy transferred in the light bulb when 20 C of charge flows through it. 13
A thick wire is connected in series with a thin wire of the same material and a battery (a) (b)
In which wire do the electron electro ns have the greater greater d rift ve locity? loc ity? Expla in your answer. A battery batter y is conne con nected cted across a lar ge resistor resisto r and a small sma ll resistor resis tor is connected in parallel. The currents through the resistors are different. Which resistor resisto r has has the higher higher dissi diss ipation pa tion o f power? po wer? Explain Expla in your answer.
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Variat Variatiio n of curr c urree nt with applied voltage There are severa l wa wa ys in which the current cu rrent through a d e vice vice can b e altered. Elastic strain, temperat ure and light are examp les o f these. F igure igure 1 shows shows examples o f current-voltage curves for a number of different situations. Voltage-c Voltage-curr urree nt ch c haracte ristics
The p.d. applied to the device is varied over a suitable range and the current is measured. The device is turned round to see if it behaves the same in both directions.
Graphs of current against voltage are plotted. Filament lamp
R R increases with temperature
Resistor I I
R is constant V
Semiconductor Diode The diode allows current to flow freely in one direction only. Thee curre Th curre nt increases rapidly rapidly for ‘forward’ ‘forward’ voltages vo ltages gre gre ate ate r than 0.5 V.
Resistance and temperature When a materi mater ial is heated its resistance may c ha nge. nge. This is is due d ue to the ther ther mal motion of the atoms within the specimen. For a metal the temperature coefficient of resistance is positive - in other words and increase in the temperatu temperat ure gives gives an increase in resi res istance sta nce.. This can be explained exp lained by the motio motio n o f the atoms and free electrons electro ns within the solid. At low temperatures temperat ures 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 uction electrons and the latt lattiice and so so impeding imped ing their moti mot io n. In a light light bulb o the filament is at about 2700 C when it is working and its resistance when hot is about te te n times that when cold. (For (Fo r a typ typ ical domestic do mestic light b ulb the resistance measured at room room temperature was 32 Ω and this rose to 324 Ω at its working temperature). However in no no n-m n- metal eta ls such s uch as semiconductors a n inc inc rease in temperatur temperat uree leads to a drop in resi res istance sta nce.. This can be explained exp lained b y e lectrons ectro ns ga ining energy and moving into the conduction band - in fact changing from being bound to a particular atom to being be ing abl ab le to mo mo ve freely - an increase in the number of free free e lectrons. lectro ns. The temperature coefficient of resistance and also that of the temperature coefficient of resistivity is therefore negative. 14
Why does the resistance of a metal increase as its temperature is raised?
Why does the resistance of a semiconductor decrease as its temperature is raised?
Explain in terms of its „secret‟ composition, how the resistance of a fixed resistor may remain constant as its temperature 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 pote ntial difference is now applied across the metal the electrons tend to move towards the positive connection. As they do so Figure 1 electron metal atom their progress is interrupted by collisions. These collisions impede their movement 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 viole nt collisions with them and so the resistance of the metal increase. The electron drift 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
The units of resistance are ohms (Ω). If the ratio of p.d to c urrent remains constant for a series of d ifferent 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 curve. http://micro.magn et.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 curre nt 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 between 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 the 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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Calculate the current through the following resistors: (a) (b) (c) (d)
What is the resistance of the following? (a) (b)
120 Ω connected to 240V 4700 Ω connected to 12V 10kΩ connected to 6 V 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)
Calculate the resistance of the wire at each temperature
Which graph temperature?
Does the material disobey Ohm's Law? Explain your answer.
V (V) 20
Why is it more likely for the filament in a light bulb to break when you switch it on rather than when it has been on for some time? Explain your answer.
The circuit shows a battery of negligible internal resistance connected to three resistors.
(a) (b) (c)
Calculate the pote ntial difference 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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The circuit diagram shows a 12 V d.c. supply of negligible internal resistance connected to an arrangement of resistors. The curre nt at three places in the circuit and the resistance of two of the resistors are given on the diagram.
(a) (b) (c)
Calculate the pote ntial difference across the 4.0 resistor. Calculate the resistance of resistor R2 . Calculate the resistance of resistor R1 .
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.
The lamps are connected in para llel with a 9.0 V supply as shown.
(i) (ii) (iii)
Which lamp is brighter? Give a reason for your answer. Determine the current in the supply. Calculate the total resistance of the two lamps when they are connected in parallel.
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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 shown in the diagram.
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
Determine the temperature of the thermistor. If the e.m.f, of the supply were doubled, would the read ing on the voltmeter double? Explain your answer.
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Use the axes to draw the current-voltage characteristics of a diode and a filament lamp I
Electrical Power Power is the rate at which work is done or energy changed from one form to another and so:
t time Energy Power time V I t
Example problems 1.
Calculate the power of a 12V light bulb using 2.5 A. Power = VI = 12V x 2.5A = 30 W
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
Calculate the ener gy transformed by a 12V car battery 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
with the basic formula for electrical power to
give two alternative for mulae: Since power = VI, we can write 2
Power V I I R
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V 2 R
Example problems 1 Calculate the resistance of a 100W light bulb if it takes a current of 0.8 A P 100W P I 2 R R 156 I 2 0.8 A2 2 Calculate the power of a 12V immersion heater with a resistance of 10 P
V 2 R
12 2 10
http://www.ukpower.co.uk/running-costs-elec.asp (Electricity Runn ing Costs Calculator)
Calculate the power loss in an electrical transmission cable, 15 km long, carrying a current of 100A at a potential of 200 kV. The resistance per km of the cable is 0.2 Ω.
What power is supplied to the heater of an electric bar fire with a resistance of 50 Ω connected to the mains 240V supply?
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?
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
What is the power of an electrical heater operating from 110 V if the resistance of the heater is 15 ?
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?
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Two resistors in series (Figure 1)
Series circuit current.SWF
Series circuit voltage.SWF
The current (I) flowing through R1 and R2 is the same and so the potential differences across them are V1 = IR1 and V2 = IR2 V V 1 V 2
R is the effective series resistance of the two resistors. So:
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, therefore: I I 1 I 2 V IR1 IR2 I
1 V R R2
where R is the effective resistance of the two resistors in parallel. Notice that
two resistors in series always have a larger effective resistance than either of the two resistors on their own
two in parallel always have a lower resistance 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.
Calculate the resistance of the following combinations: (a) (b)
100 Ω and 50 Ω in series 100 Ω and 50 Ω in parallel
R = R1 + R2 = 100 + 50 = 150 1 1 1
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 para llel
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 any combination of them to give a combined resistance of: (a) 200 Ω, (b) 125 Ω (a)
100 in series with both the 50
the two 50 in parallel and this in series with the 100
Four 10 resistors are connected as shown in the diagram
(a) Calculate the total resistance of the combination. (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 assumed 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.
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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 resistance of one heating element at room temperature is approximately 45 . R A
Calculate the total resistance of the heater for the following combinations of switches at the moment the switches are closed.
Resistance of heater/
Calculate the maximum power output from the heater immediately it is connected to a 230 V supply.
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 re lated to resistance of a specimen of length l and cross sectional area A by the formula: R
The resistivities of solutions cannot be quoted generally because they depend on the concentrations and are therefore variable quantities. As an example however the resistivity of pure water is about 2.5 x105 m and that of a saturated solution of sodium chloride about 0.04 m at 20o C. The reciprocal of 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
49 10 8 m 1.5m
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?
5 1.5 10 7 m 2 44 10
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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
Calculate the length of a copper wire of cross sectional area 0.65 x10 -7 m2 that has a resistance of 2Ω
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.
Calculate the resistance between the large faces of a slab of germanium of thickness 1 mm and area 1.5 mm 2 .
Calculate the length of a section of constantan wire of diameter 0.32 mm that has a resistance of 10 Ω
(b) (c) (d)
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 measurements 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 resistance 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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The diagram shows a type of resistor commonly used in electronic circuits.
It consists of a thin film of carbon wrapped around a cylindrical insulator. The diagram below (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
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 the carbon film.
Show that the res istance 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 cell can produce across its terminals or the open circuit p.d since when no current flows from the cell then no electrical energy can be lost within it.
E V Ir IR Ir
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 resistance R then:
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.
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 2
I Rt I rt
It IR 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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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 12V Therefore current = 1.19 A 10.1 Loss of energy per coulomb in the cell = 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 current = 40 A 0.3 Loss of energy 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 resistance (r) of the 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 IR Ir V IR V E Ir rI E
A graph of V against I is plotted. The graph is a straight line which has a gradient - r
Gradient = - r
and a „y intercept‟ E
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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 resistance r of the supply.
Microsoft Office Excel 97-2003 Works
What is meant by: (a) the E.M.F. of a cell (b) the internal resistance of a cell?
How does the internal resistance of a cell affect the current drawn from it?
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)
Explain the difference between 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
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 )
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A resistance substitution box R is connected to a power supply. For various values of R the current is measured. R/
P/mW (a) (b) (c) (d) (e)
Calculate the power of the resistor for each value of R. Predict the power of the resistor when 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 resistance of the power supply is known to be 1.0 . Comment.
Why are car batteries designed to have negligible internal resistance whereas high voltage (EHT) laboratory supplies are manufactured with a very large internal resistance?
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Potential divider The basic circuit is shown in the first circuit diagram. The output voltage across AB is given by:
V R2 R2 V V 2 I R2 R R R R 2 2 1 1
Note that 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.
Measuring the output
If we now attempt to actually 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 R2 and the voltmeter are connected in parallel and so their combined resistance (R) is given by the equation; 1 R
huge – almost 1 infinite and so 0 and RV can be ignored. This 1 1 means that and R R2
so R R2 .
The output voltage (V 0 ) measured by the meter really is that across R 2 , in other words V 2 .
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A moving coil meter. These meters have a much lower resistance than a digital meter, usually some tens of k .
This means that the combined resistance of R2 and RV is affected by the resistance of the voltmeter and is actually lower than R2 .
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 divider 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.
So in the light V2 is low and in the dark V2 is high.
Replacing R1 with a thermistor Something very similar happens if R2 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 t hermisto r is hot V2 is low and when the thermistor is cold V2 is high.
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
100 k 25 k 5k 250
200k 10 k 20 k 100
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
Output voltage (V )
R1 / 100 k 25 k 5k 250
R2 / 200k 10 k 20 k 100
Now assume that resistor 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 )
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?
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?
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A student connects a 9.0 V battery in series with a resistor R, a thermistor and a milliammeter. He connects a voltmeter in para llel with the res istor. The reading on the voltmeter is 2.8 V and the reading on the milliammeter is 0.74 mA.
Show that the res istance of R is approximately 4000 .
Calculate the resistance of the thermistor
The ther mistor 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.
Suggest an explanation for how the reading on this student's milliammeter will compare with that of the first student.
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A student sets up a circuit and accide ntally uses two voltmeters V 1 and V2 instead of an ammeter and a voltmeter. The circuit is shown below.
Circle the voltmeter which should be an ammeter.
Both voltmeters have a resistance of 10 M. The student sees that the reading on V2 is 0 V. Explain why the potential difference across the 100 resistor is effectively zero.
The student rep laces the 100 resistor with another resistor o f resistance R. The reading on V2 then becomes 3.0 V. (i)
Complete the circuit diagra m below to show the equivalent resistor network following this change. Label the resistor R.
Calculate the value of R.
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A light-emitting light-e mitting diode diode (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 slides the contact between terminals A and B. (a) Explain this observation as fully as you can. (b) The student then disconnects the LED and reconnects the circuit as shown bel be low. S he intends intends to vary the intensity o f the the light emitted emit ted b y the LED LED b y sliding the contact between terminals A and B.
The student st udent cannot ca nnot detect any light emitted by the LED. Briefly Brie fly expl exp lain why the LED is so dim. (c) Draw the circuit that the student should have connected using this apparatus in order to vary the brightness of the LED and measure the potential difference across it.
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The following circuit can be used as a light meter.
(a) The maximum value of resistance of the light-dependent resistor (LDR) is 950 k . What is the reading on the voltmeter for this resistance? (b) The minimum value of resistance of the LDR is 1.0 k . What is is t he reading on the voltmeter for this resistance? (c) For this light meter the voltmeter is connected across the 10 k resistor, rather than the LDR. Explain Expla in ho ho w the readings on o n the voltmeter voltmete r enable enable this circuit to be used as a light meter. 57
Two resistors of resistance 2.0 M and 4.0 are connected in series across a supply voltage of 6.0 V. Togeth Toget her they for form m a simple simp le potent ial divider circuit. State the potential difference across each resistor. P.d. across the 2.0 M resistor
P.d. across the 4.0 resistor
A second potential divider circuit uses a resistor and a diode connected in series with the same supply. Cal Ca lculate cu late the potentia po tentiall d ifferen iffere 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 d iode
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Nature of light The Photoelectric Effect 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 illumina ted b y light from anothe anothe r spark. This e ffect was studi st udied ed mo re careful ly in the following follow ing years years b y Hallwac Hallwac hs a nd Len Le nard. They called the photoelectr ctric ic em e miss ission ion and a very simple experiment can be used to investigate effect photoele it.
no eff 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 an EHT supply. supp ly. The next ne xt thing is to shine some radi rad ia tion on it, it, using an ordin ord inary ary lamp, a helium- neon eo n laser laser (giving out o ut intense red red light) or an ultra ultra violet light ha ha s absolutely abso lutely no e ffect. ffect. The electroscope electro scope sta sta ys charged charged an a nd the leaf stay sta ys up. However if the plate is given a negative charge to start with (using say a charged polythene rod) there is a difference. Using the la la mp and and even the lase lase r ha ha s no no effect, but when ultra violet light is shone on the plate the leaf falls immediately : the electroscope has been be en dischar dischar ged. (Doing (Do ing the the experiment in a vacuum vac uum p rove ro vess tha tha t it is is not ions in the air that are causing the discharge.) No effect can be produced with radiation of longer wavelength (lower 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 electrons were attracted back to the plate. The researchers found five important facts about the experiment:
no electrons were emitted from the plate if it was positive the number of electrons emitted per second depended on the intensity of the incident incident radiation radiation the energy of the electrons depended on the 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 electrons are emitted this occurs immediately
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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 them 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 tight ly 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 energy of one quantum or photon is given by the equation:
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 emission.
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
/ 10-19 J /eV
Sodiu m Caesiu m Lithium Calcium Magnesium
3.8 3.0 3.7 4.3 5.9
2.40 1.88 2.31 2.69 3.69
Frequen cy 14 /10 Hz 5.8 4.5 5.6 6.5 8.9
Wav elength /nm 520 666 560 462 337
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)
filter collector electron V
sodium emitter A
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.
A graph can then be plotted of stopping potential (electron energy) against frequency.
If hf (or f
), 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 the emitter and this makes it difficult 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
A graph of V v f is plotted, and gradient =
y intercept =
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x intercept =
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 quantum or photon?
Write down Planck‟s equation for the e nergy of a qua ntum of radiation.
Which has the greater energy, a quantum of yellow light or a quantum of violet light?
Which experiment shows waves behaving like particles?
What is meant by the “stopping potential” in the photoelectric effect?
What is meant by the “work function” in the photoelectric effect?
What is the energy of a quantum of radiation that has a wavelength of 500 nm?
What is meant by an “electron volt”?
If the frequency of the incident radiation that falls on a metal surface is increased what happens to the photoelectrons that are emitted?
If the intensity of the incident radiation that falls on a metal surface is increased what happens to the photoelectrons that are emitted?
Light of wavelength 480 nm just gives photoelectric emission from the certain metal sur face. What is the work function of that surface in Joules?
If the wavelength is reduced to 400 nm what is the energy of the electrons emitted?
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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)
Explain the following observations. (i) Initially there is a current which is measured by the microammeter. As the reverse potential differe nce is increased the current reading on the microammeter decreases. (ii) When the potent ial 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 emitted from the surface 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
State how the work function, , of the metal can be obtained from the graph.
Explain why this graph always has the same gradient irrespective of the metal used.
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The diagram shows a coulombmeter (an instrument for measuring charge) set up to demonstrate the photoelectric effect.
The clean zinc plate is negat ively charged. Ultravio let 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)
Explain these effects in terms of the particle theory of light. What would happen to the charged plate if (i) the intensity of the red light were increased (ii) the intensity of the ultraviolet light were increased? Zinc has a work function of 3.6 eV. Calculate the maximum kinetic energy of the photoelectrons when the zinc is illuminated with ultraviolet light of wavelength 250 nm.
The photoelectric effect supports a particle theory of light but not a wave theory of light. Below are two features of the photoelectric effect. For each feature expla in why it supports the particle theory and not the wave theory.
The emission of photoelectrons from a metal sur face ca n take place instantaneously.
Incide nt light with a frequency below a certain threshold frequency cannot release electrons from a metal surface.
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Define the intensity of an electromagnetic wave.
Two beams of monochromatic electromagnetic 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 subscr ipts 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
The table below gives the work functio ns of four metals. Metal
Work functio n/eV
Define the term work function. (d)
A metal plate made from one of these metals is exposed to beams A and B in tur n. Beam A causes electrons 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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A monochromatic light source is placed 120 mm above the cathode of a photocell.
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 a nd the sensitive ammeter detects the current.
The table below shows the currents that are obtained with this apparatus for two different intensities and two different wavelengths of light, using two different cathode materials. Work funct ion energies are given. Wavelength of incident radiation/nm
Photocurrent/A when intensity of incident radiation is 1 W m-2
5 W m-2
0.2 x 10-12
1.0 x 10-12
Show that the incident photons of = 320 nm and) = 640 nm have energies of approximately 4 eV and 2 eV respectively.
Account for the photocurre nt readings shown in the table.
Calculate the stopp ing potential for the photoelectrons released by lithium when irradiated by light of wavelength 320 nm.
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The table be low shows the results of an exper iment like Millikan‟s using sodium as the metal plate.
Stopping voltage V s /V
Frequency of light f / 1014 Hz
(a) Plot a graph of V s against f . The following equation applies to the photoelectric effect:
where is the work function of the metal and e is the charge on an electron. (b) What information about the electrons 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.
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)
Ionised state -0 e V -0.85 n 2 excited state -1.5e V st 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 .
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 each element have a characteristic set of energy levels, and so emit a characteristic set of frequencies when they are excited. You can identify an element from the frequencies of radiation it emits when excited.
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)
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.
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 ground state. (b) In which parts of the spectrum would you expect to find these wavelengths?
The figure shows an energy level diagram. Sketch a possible line spectrum for the light emitted when electrons make the transitions shown. Label the lines, using the letters shown in the diagram, and indicate on your spectrum diagram which end corresponds to the higher frequency.
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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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?
Refer to the diagram in Q19
If the atom is in the ground state, how much energy must be given to it to ionise it?
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
What is the wavelength of the photon that could raise an electron from the -0.849 eV level to the -0.545 eV level?
If an electron returns 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
electrons whose k.e. is 2.50 eV
light of wavelength 590 nm.
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Four of the energy levels of a lithium atom are shown 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 diagram, 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 and 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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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 photon 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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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 energy 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?
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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 the same energy is spread over a larger area. Inverse square law.SWF
This relationship is called the inverse square law. When the distance is doubled the energy flux is reduced by a factor of four. 29
Explain what is meant by the inverse square law of electromagnetic waves such as visible light. Explain how this inverse square law is consistent with the law of conservation of energy.
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.
A 60 W light bulb converts electrical energy to visible light with an efficiency of 8%. Calculate the visible light intensity 2 m away from the light bulb.
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.
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?
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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A communication satellite is in orbit above the Earth‟s surface. (a)
The satellite‟s-electrical system is powered by 20 000 photovoltaic cells, each of area 10 c m2 . 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.
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.
In practice, reflectors 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 circle of diameter 1000 km, calculate the intensity that would be detected there. Assume there is no absorption of the signal along its path.
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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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?