PHYSICS TUTORIAL NOTES

 

From Ideas To Implementation

 

SYLLABUS TOPIC 9.4

 

 

These notes are meant as a guide only and are designed to focus your thoughts on the dot points mentioned in the syllabus. They give a very brief overview of the topic and should be used in conjunction with your class notes, your textbooks and your research from other areas such as the library and the Internet.

 

 

 

Notes compiled by:

CARESA EDUCATION SERVICES

 

FROM IDEAS TO IMPLEMENTATION

 

“1. Increased understandings of cathode rays led to the development of television” (syllabus)

 

Explain why the apparent inconsistent behaviour of cathode rays caused debate as to whether they were charged particles or electromagnetic waves” (syllabus)

 

When cathode rays were first discovered scientists such as Hertz thought that they were electromagnetic waves. Hertz adopted this view based on the fact that he did not detect any deflection of cathode rays by electric fields (the ones he used were too weak), cathode rays cause fluorescence in some reactions, and cathode rays were able to penetrate thin layers of metal without damaging the metal.

 

Other scientists such as Crookes thought that they were particles. This was because cathode rays were deflected by magnetic fields, the speed of cathode rays was measured as being less than that of light, cathode rays can cause a small paddle wheel to rotate, showing that they have momentum and a target that had been bombarded with cathode rays attained a negative charge.

 

Explain that cathode ray tubes allowed the manipulation of a stream of charged particles” (syllabus)

 

A cathode ray tube consists of an evacuated tube containing a cathode and an anode with a high potential difference between them. The high potential difference causes electrons to be emitted from the cathode and accelerate towards the anode. By passing the electrons (cathode rays) between suitable electric and magnetic fields it is possible to manipulate their path so that they strike a screen at any point.

 

Identify that moving charged particles in a magnetic field experience a force” (syllabus)

 

A charged particle experiences a force when it cuts lines of magnetic flux. This occurs when it moves through a magnetic field and it also occurs when a changing magnetic field moves past the charged particle. A stationary charged particle in a magnetic field does not experience a force.

 

Identify that charged plates produce an electric field” (syllabus)

 

From the preliminary course you will remember that there is an electric field between two charged plates. The electric field is equal to the force experienced by a unit positive charge and can be represented diagramatically by showing the lines of force.

The lines of force flow from positive to negative.

 

 

Describe quantitatively the force acting on a charge moving through a magnetic field F = qvB sinq (syllabus)

 

The force acting on a charge moving through a magnetic field is proportional to its rate of cutting lines of magnetic flux. When the charged particle is moving at a right angle to the magnetic field this is given by F = qvB. When it is moving at an angle q to the magnetic field the perpendicular component of the velocity, v sinq, is used and the equation becomes f = qvB sinq.

 

Discuss qualitatively the electric field strength due to a point charge, positive and negative charges and oppositely charged parallel plates” (syllabus)

 

A charged particle has a region of influence around it in which another charged particle will experience a force. This is known as an electric field. There are two types of electric charge: positive and negative. Like charges (two positive charges or two negative charges) will repel each other while unlike charges (a positive and a negative charge) will attract each other.

 

Electric fields can be shown diagramatically as lines of force. These show the direction of the force on a positive charge and therefore the arrows point from positive to negative.

Lines of force between two charged parallel plates.

 

Lines of force around a small positively charged particle.

 

           

Lines of force around a small particle with a high positive charge.

Note that a strong electric field has more lines of force than a weak one.

 

 

                       

Lines of force around a small negatively charged particle.

 

                       

Lines of force between a positive and a negative point charge.

The strength of the electric field is represented by the number of lines of force; the more lines of force, the stronger the electric field.

 

Describe quantitatively the electric field due to oppositely charged parallel plates” (syllabus)

 

The electric field between two oppositely charged parallel plates is given by: E = V/d

e.g. What is the electric field strength between two parallel plates separated by 1.0 cm with a potential difference of 500V between them?

E = V/d = 500/0.01 = 50 000 Vm-1 = 50 000 NC-1

Note that the units Vm-1 and NC-1 are equivalent.

 

Outline Thomson’s experiment to measure the charge/mass ratio of an electron” (syllabus)

 

THOMSON

 

On the basis of the work of Sir William Crookes it appeared as if cathode rays consisted of a stream of particles. His paddle wheel experiment showed that they had momentum and hence mass and he also showed that they were deflected by magnetic fields and hence had a charge. In 1897 Joseph John Thomson developed an instrument to determine the charge to mass ratio of cathode rays. He set up an electric field and a magnetic field at right angles to each other and fired the cathode rays through them. He adjusted the electric and magnetic fields so that they exerted forces of equal magnitude but opposite direction on the cathode rays. The rays then passed through the fields undeflected. By equating the electric force FE to the magnetic force FB is was possible to determine the speed of the cathode rays.

            FE = FB

            QE = QvB

            v = E / B

Thomson then turned off the electric field and measured the deflection of the cathode rays by the magnetic field only. By equating the magnetic force to the centripetal force Thomson was able to determine the charge to mass ratio.

            QvB = mv2 /r

            Q/m = v / rB

Thomson proposed that cathode rays consisted of small negatively charger particles (electrons) that were constituent of all matter. On the basis of this he proposed that atoms consisted of spheres of positive fluid in which were embedded a number of negatively charged electrons. This became known as the plum pudding model of the atom.

 

In 1909 Robert Millikan determined the charge on the electron, and by use of Thomson’s charge to mass ratio, the mass of the electron. He sprayed a fine mist of oil droplets between two charged plates. As he did so, some of the oil drops attained a charge due to friction with the atomiser. Millikan concentrated on one oil drop and adjusted the electric field strength between the plates so that the gravitational force on the oil drop was just balanced by the electrostatic force and the oil drop remained stationary between the plates. By timing the oil drop to fall a measured distance through air and by use of Stokes’ equation for the mass of an object falling through a viscous medium, Millikan was able to determine the mass of the oil drop.

            QE = mg

            Q = mg/E

Millikan now had a value for the charge on the oil drop. He repeated the experiment hundreds of times and found the highest common factor of the charge. This value                   (1.6 x 1019 C) was determined to be the fundamental unit of charge.

 

Questions.

(1) Suggest a way that Thomson could set up (a) an electric field of known strength

 

Two parallel plates connected to a source of potential difference have a uniform electric field between them. The strength of the electric field can be found from the formula: E = V/d

 

(b) a magnetic field of known strength.

The magnetic field inside a long, closely wound solenoid can be determined. The formula B = 2pknI/l (k = Coulomb’s constant = 2.0 x 10-7 NA-2) is beyond the scope of the HSC course. By using magnetic coils resembling a solenoid with just a small gap for the cathode rays to travel between the two halves of the solenoid, Thomson was able to accurately control the strength of the magnetic field.

 

(2) How did the work of Thomson refute Dalton’s atomic theory?

 

Dalton’s atomic theory proposed that atoms were the smallest particles of matter. Thomson’s discovery of the electron showed that there was at least one type of particle smaller than the atom.

 

 

Outline the role of:

-          electrodes in the electron gun

-          deflection plates or coils

-          the fluorescent screen

in the cathode ray tube of conventional TV displays and oscilloscopes” (syllabus)

 

The electron gun has a high potential difference that accelerates the electrons (cathode rays) as they travel from the cathode to the anode. As electrons travel from the negative to the positive electrode they gain kinetic energy. This is outlined in the equations;

W = qV = ½ mv2

where W = work done, q = charge of electron, v = speed of electron, V = potential difference between electrodes, m = mass of electron,

 

The deflection plates or coils provide the electric and magnetic fields that guide the cathode rays to sweep the screen as they form an image on the screen.

 

The fluorescent screen is coated with phosphors. The phosphors absorb energy from the cathode rays and emit it as light. We observe the light as patterns on the screen.

 

 

perform an investigation and gather first-hand information to observe the occurrence of different striation patterns for different pressures in discharge tubes” (syllabus)

DISCHARGE TUBES

Early attempts to get a current to flow through a discharge tube were unsuccessful because the vacuum pumps of the day were inefficient and the air or gas in the tube would prevent the current (electrons as we know now,) from travelling from one electrode to the other.

In 1855, Heinrich Geissler developed a more efficient vacuum pump that enabled a very low pressure to be obtained. His friend, Julius Plucker used this to produce low- pressure discharge tubes and found that by using a high potential difference, he was able to get a current to flow through the tube. He also found that the glass at the positive electrode (anode) glowed green. Plucker also observed that the rays that were produced were able to be deflected by a magnetic field. From these observations he deduced that the rays emanated from the cathode. Eugene Goldstein repeated Plucker’s work and named the rays “cathode rays”.

 

Plucker and others at the time noted the effects of gradually reducing the pressure in discharge tubes while maintaining a high potential difference. When a potential difference of about 1000 volts is applied to the discharge tube the following observations can be made:

1.      At atmospheric pressure, no current will pass through the tube.

2.      At about 1/50 atmosphere violet streamers appear. These are accompanied by a crackling noise that disappears as the pressure decreases.

3.      At about 1/100 atmosphere the whole of the gas glows with a colour that depends on the type of the gas in the tube. This glow is known as the “positive column” and is the source of light in sodium street lights (they are the yellow fog lights), as well as in various advertising signs.

4.      Further reduction in pressure causes the positive column to become detached from the cathode, and a dark region called the “Faraday Dark Space” to appear between the positive column and the cathode. The cathode itself is surrounded by a blue glow called the “Negative Glow”. It is blue regardless of the gas in the tube.

5.      At about 1/5000 atmosphere the positive column moves towards the anode and breaks up into bands of colour and darkness called “striations”. The Faraday dark space and the negative glow meanwhile move towards the anode to form another dark region called “Crookes’ Dark Space”.

6.      At about 1/10000 atmosphere the positive column and negative glow disappear and Crookes’ dark space fills the whole tube. The walls of the tube fluoresce with a colour depending on the type of glass that makes the tube.

Before trying to explain the patterns formed in the discharge tubes we will look at what happens when an atom absorbs energy. The electrons in the atom originally orbit in the ground state, i.e. the state of minimum energy. They can absorb energy and lose an electron. Chemists will recognise this as ionisation energy. They can also absorb a lesser, but fixed, amount of energy that will make the electron orbit at a higher energy level. The electron at this level is unstable and will drop back to its ground state, emitting a photon of light as it does so. The photon has a definite frequency as governed by Planck’s equation; E = hf and we see it as light of a specific

colour.

 

 

In the following explanations I will refer to cathode ray electrons and atom electrons. They are exactly the same sort of electrons, identical in every respect, but the distinction will be made to describe their origin.

 

A cathode ray electron can hit a gaseous atom. The atom’s electron will gain energy and proceed to a higher energy level where it will be unstable. It will then revert to its original energy level and give off a quantum of energy of a specific wavelength as it does so. The colour of the light emitted is characteristic if the gas in the tube.

 

Why is the negative glow blue?

When atoms are ionised some of the positive ions recombine with electrons. Two attracting bodies have potential energy and this is converted into other forms of energy as the bodies come together. As the ion and the electron come together a lot of energy is released so it is released as light of a short (high energy) wavelength i.e. blue. So the light of the negative glow is due to the energy released when positive ions and electrons combine and emit blue light in the process.

 

The potential drop across the discharge tube is uneven with most of it occurring close to the cathode. Electrons are accelerated rapidly and attain a high velocity as they move through Crookes’ dark space. The average distance that the cathode rays travel before colliding with atoms depends on the amount of matter and so the pressure in the tube. Crookes’ dark space becomes longer as the concentration of matter decreases when the pressure is reduced.

 

As the cathode rays move through Crookes’ dark space they gain sufficient energy to ionise the gas in the tube and produce the blue negative glow. They are slowed down by collisions in this region and don’t have much energy as they enter Faraday’s dark space.

 

They accelerate through the Faraday dark space until they have sufficient energy to excite atoms rather than ionise them. The atom’s electrons attain a higher energy level and then return to the unexcited state, emitting a photon of light as they do so. This is seen as a specific colour that is characteristic of the gas in the tube. The cathode rays lose energy in their collision with an atom and are accelerated until they again have sufficient energy to excite an atom. Thus the positive column can consist of a number of bright regions where the cathode rays have excited the atoms, separated by a number of dark regions. In this way the positive column can appear striped or striated.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

perform an investigation to demonstrate and identify properties of cathode rays using discharge tubes:

-          containing a maltese cross

-          containing electric plates

-          with a fluorescent display screen

-          containing a glass wheel and analyse the information gathered to determine the sign of the charge on cathode rays. ” (syllabus)

 

 

 

 

 

 

 

Solve problems and analyse information using:

F = qvb sinq

F = qE                        and

E = V/d ” (syllabus)

 

Q.1.     An electron is fired between two parallel plates separated by 5 cm with a potential difference of 400V between them. What is the force on the electron?

 

E = V/d           = 400/0.05 = 8000 NC-1

F = qE = 1.6x10-19 x 8000      

= 1.28 x 10-15 N towards positive plate

 

 

Q.2.     An electron travelling at 106 ms-1, enters a magnetic field of flux density 10-2T at an angle of 60o to the normal.

 

 

 

 

 v = 106ms-1

                               60o

 

 

 


                                                                        B = 10-2T

 

 


Calculate the force acting on the electron.

 

F = qvB sinq = 1.6 x 10-19 x 106 x 10-2 sin 30o

= 8 x 10-16 N into page

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

“2. The reconceptualisation of the model of light led to an understanding of the photoelectric effect and black body radiation” (syllabus)

 

Describe Hertz’s observation of the effect of a radio wave on a receiver and the photoelectric effect he produced but failed to investigate” (syllabus)

 

In 1887 Heinrich Hertz produced and detected radio waves. He used an induction coil to produce a high potential difference between two spheres separated by a small gap. The high potential difference caused a spark between the spheres which in turn caused the electrons in the circuit to oscillate. This caused electric fields to be emitted which cut the receiver and caused the electrons in the receiver to oscillate at the same frequency. This causes a spark to be produced across the gap between the spheres in the receiver loop.

The frequency of these waves was around 100 MHz which is in the region now called radio waves. Hertz showed that the radiation emitted from this apparatus had the same properties as electromagnetic waves. He used the interference properties of the radiation to determine its wavelength and to show that its velocity was the same as that of light.

Hertz used an induction coil to obtain the high voltage to produce a spark across a gap. He used a parabolic mirror to reflect any waves towards the receiving antenna a few metres away. The receiving antenna consisted of a wire loop with a spark gap at the focus of a second parabolic mirror. The spark across the transmitting gap produced waves that caused sparks to be produced across the receiving gap.

 

 

Outline qualitatively Hertz’s experiments in measuring the speed of radio waves and how they relate to light waves” (syllabus)

 

Hertz was able to determine the frequency of the waves by the plate size of the capacitors and the current inductance. He then used a reflecting sheet to compare the path difference between a wave that travelled directly from the transmitter to the receiver and one that was reflected from a metal sheet in its path from transmitter to receiver. By examining interference patterns formed by these two waves Hertz was able to determine the wavelength of the waves. By substituting the frequency and wavelength into the equation v = fl Hertz determined the velocity of the waves.

 

 

 

 

 

Identify Planck’s hypothesis that radiation emitted and absorbed by the walls of a black body cavity is quantised” (syllabus)

 

A black body is one that will absorb all radiation that hits it. At equilibrium the amount of incident radiation equals the amount of emitted radiation so the temperature remains constant.

A good way to examine black body radiation is to use a cavity radiator.

Cavity Radiator

A cavity radiator is a block of hollowed out metal with a small opening to the outside. Just as a long tunnel or deep hole appears black because it does not reflect any light so too will the cavity of the cavity radiator appear black because it does not reflect any radiation. Radiation entering the hole will bounce around inside the cavity until it is absorbed. Planck proposed that the atoms inside the cavity oscillated back and forth, emitting radiation, similar to the way a radio antenna works. He proposed that the atoms could only absorb and emit specific amounts of energy and that emitted radiation comes in whole number multiples of this specific quantity; no fraction values are allowed.

Classical physics assumes that light and other electromagnetic waves are a continuous quantity and are infinitely divisible into smaller and smaller parts.

Quantum physics assumes that light and other electromagnetic waves are multiples of discrete small amounts of energy i.e. that the waves are quantised.

While the classical model was the preferred model before 1900, the work of Planck with hot bodies led to the adoption of the quantum model after 1900.

 
 

 

 

 

 

 

 

 

 

 

 

 


Identify Einstein’s contribution to quantum theory and its relation to black body radiation” (syllabus)

 

Einstein extended Planck’s quantum idea to explain the photoelectric effect. He assumed light consisted of specific “bundles” of energy that he called “photons”.

He verified this with his work on the photoelectric effect for which he won the Nobel Prize in 1921.

 

 

 

 

Explain the particle model of light in terms of photons with particular energy and frequency” (syllabus)

 

This is verified by the photoelectric effect.

The photoelectric effect refers to the emission of electrons when light falls onto a particular surface. Metals are the most common materials to exhibit the photoelectric effect although there are some other substances.

 

When light falling onto the metal surface produces a current it is found that increasing the intensity of the light increases the current flowing through the galvanometer. However when monochromatic light of different frequencies is used it is found that some frequencies produce the photoelectric effect while others do not. Apparently there is a minimum frequency, the threshold frequency, below which photoemission of electrons will not take place. Classical theory predicts that there should be photoemission at all frequencies.

To measure the energy of the electrons emitted from the cathode a potential difference was supplied to oppose the photocurrent. The anode was given a negative charge so that as the photoelectrons approached the anode they would be repelled by the like charge. The voltage was gradually increased until the applied voltage stopped the flow of photocurrent. The voltage at which the photocurrent stopped was called the stopping voltage.

 

                                   

It was found that at the threshold frequency, the stopping voltage was zero but as the frequency of the light increased, so too did the stopping voltage. This indicates that more energetic photoelectrons are emitted at higher frequencies.  Increasing the intensity of the light did not change the stopping voltage. Classical theory predicts that increasing the intensity of the light should also increase the energy of the photoelectrons and so increase the stopping voltage.

 

 

Einstein proposed that light was quantised and could only come in small, fixed amounts given by Planck’s equation E = hf. He called these units photons. This explained why high frequency light would produce a photocurrent but low frequency light would not. The high frequency light had more energetic photons and these were able to knock electrons from the metal surface while low energy photons could not.

The minimum energy required to release a photoelectron is given by E = hf0 where f0 is the threshold frequency. The energy of these photons is called the “work function, F” and F = hf0.

Consider an incoming photon with energy hf. The energy required to knock an electron from the metal surface is hf0.

If hf < hf0 then no photoelectrons will be emitted.

If hf = hf0 then photoelectrons will be emitted but no energy is left over.

If hf > hf0 then the energy hf0 will be used to knock out the electron and an amount of hf – hf0 will be left over. This is the kinetic energy of the photoelectrons and is given by the product of the charge and stopping voltage.

Ek = hf - F = qVstop

Einstein plotted the kinetic energy of photoelectrons against frequency of light for different metals and obtained straight lines with the same gradient. The gradient Ek/f was equal to Planck’s constant, 6.6 x 10-34 Js.

                     Ek

             

 

 

                                                           f

 

 

 

 

                                                  

 

 

Identify the relationships between photon energy, frequency, speed of light and wavelength:

E = hf             and

c = fl ” (syllabus)

 

The energy of a photon is given by Planck’s equation: E = hf                       where

E = energy,      f = frequency,             h = Planck’s constant = 6.6 x 10-34 Js

 

The speed of a wave is given by v = fl and this relationship extends to electromagnetic waves where c = fl

c = speed of light,       f = frequency of wave,           l = wavelength

 

 

 

perform an investigation to demonstrate the production and reception of radio waves” (syllabus)

 

Tune a radio midway between stations on the A.M. band. Hold a wire against the negative terminal (the case) of a dry cell battery. Touch the wire to the positive terminal of the battery.

As the wire makes and breaks contact with the battery it will produce radio waves and these will be recorded as static on the radio.

 

identify data sources, gather, process and analyse information and use available evidence to assess Einstein’s contribution to quantum theory and its relation to black body radiation” (syllabus)

 

identify data sources, gather, process and present information to summarise the use of the photoelectric effect in:

-          solar cells

Solar cells consist of a layer of n-type semiconductor in contact with a layer of p-type semiconductor. When the n-type semiconductor is exposed to light it releases electrons due to the photoelectric effect and this produces a potential difference at the junction between the semiconductor layers.

Q. Why are solar cells put in the syllabus before p and n type semiconductors? Sorry. No answer to that question.

 

 

-          Photocells” (syllabus)

 

A photocell uses the photoelectric effect to convert light into electrical energy. They can be used in such things as photographic light meters and automatic switches to turn street lights on and off. Photocells have limited application since the threshold frequency of most materials is such that only high frequency radiation such as blue light and ultra violet rays are suitable. Also their large size and fragile nature has caused them to be largely replaced by solar cells using semiconductors.

 

 

solve problems and analyse information using:

E = hf             and

c = fl ” (syllabus)

 

1.                  What is the energy of a photon of red light of frequency 5.0 x 1014 Hz?

 

E = hf = 6.6x10-34 x 5 x1014 = 3.3 x 10-19 J

 

2.                  What is the wavelength of the light in Q.1?

 

 c = fl;   l = c/f   = 3 x 108 / 5 x 1014 =6 x 10-7 m.

 

3.                  What is the mass of a photon of red light of frequency 5.0 x 1014 Hz?

 

E = hf = mc2 ;  m = hf / c2  = 6.6 x 10-34 x 5 x 1014 / (3 x 108)2 = 3.7 x 10-36 kg.

 

 

process information to discuss Einstein and Planck’s differing views about whether science research is removed from social and political forces” (syllabus)

 

Planck and Einstein did not let the social forces of their tragic family lives detract from their science research. They maintained a strong friendship and shared a love of music, quite independently of their scientific research.

They held different political views during WWI when they worked together in Berlin with Planck supporting the war and Einstein distributing pacifist material. This did not detract from their friendship and research. However in WWII both became involved in political actions in separate countries. Planck in Germany opposed the anti-Semitic policies of the Nazis and Einstein in the U.S.A.  supported  research into the atom bomb. In these cases it appears that their research overlapped with the political forces of the time.

 

 

Max Planck was born in Kiel, a port on the Baltic Sea in 1858. Kiel belonged to Denmark in 1858 but it became part of Prussia, (now part of Germany) in 1866. He won the Nobel Prize in 1918 for his work on quantum physics. Albert Einstein was born in Ulm, Germany in 1879, and was awarded the Nobel Prize in 1921 for his work on the photoelectric effect. Both lived in Germany in the late 19th century. Einstein worked in Switzerland in the early 1900s and published papers on Brownian motion, the photoelectric effect and his theories of special and general relativity.

Both Planck and Einstein had tragic family lives. Planck lost two daughters at childbirth and his first wife died young. His younger son was killed in the First World War and his elder son was executed in the Second World War. Einstein had an illegitimate child who was given up for adoption before his first marriage. He had two sons from his first marriage, one who became a professor of engineering, the other was schizophrenic.  Einstein went through a difficult divorce from his first wife and his second wife died comparatively young, leaving him as a widower for his last 20 years.

In 1914 Einstein moved to Berlin to work with Planck. They became good friends despite their different views about the First World War, which raged at the time. Both played the violin and they sometimes played together. Einstein was a pacifist and distributed pacifist literature in Berlin while Planck strongly supported the German cause. Between the two world wars Planck held influential positions at the Prussian Academy of Sciences while with the rise of the anti-Semitism movement in Germany, Einstein, who was Jewish, moved to the U.S.A.

Despite being a pacifist, Einstein signed a letter to President Roosevelt in 1939 that recommended developing the atomic bomb although he did not become part of the Manhattan project that developed it. He spent the post-war period of his life campaigning against the use of nuclear weapons and a week before his death in 1955 he signed a pacifist, anti-nuclear manifesto that was being circulated by the philosopher Bertrand Russell.

Planck was opposed to the anti-Semitic policies of the Nazis and his son was executed for being involved in a plot to kill Hitler. Planck died in 1947.

 

 

 

 

“3. Limitations of past technologies and increased research into the structure of the atom resulted in the invention of transistors.” (syllabus)

 

Identify that some electrons in solids are shared between atoms and move freely” (syllabus)

Metals have a lattice of positive ions with the valency electrons loosely held in such a way that they are shared by all of the atoms. The electrons are in constant random motion and in the presence of an electric field will drift against the field i.e. towards the positive region.

 

Describe the difference between conductors, insulators and semiconductors in terms of band structures and relative electrical resistance” (syllabus)

 

From your junior science course you will remember that the electrons in the outer shell of an atom are called valency electrons. Atoms conduct electricity when electrons can move along from one atom to the next. Because noble gases have complete outer shells they do not conduct electricity. However, most non metals undergo covalent bonding to fill their valency shells. Consequently non metals are poor conductors. Metals, on the other hand, have a region of delocalised electrons around them and so are good conductors.

The atoms of solids are close together and because of this the outer shell electrons of one atom influence the outer shell electrons of the next atom. The regions where these influences take place are called energy bands with the highest energy levels called the valency band. In a conductor the valency band is only partially full so it can easily gain an electron from an adjacent atom. Because of this it is also called the conduction band. In an insulator the valency bands are full so energy has to be put in to raise one of the electrons to a higher level. At the higher energy level the band is only partially full so an electron can move to the next atom. By gaining energy it has moved from the valency band to the conduction band. The energy required to do so is called the energy gap

In conductors the valency band and the conduction band overlap so there are always free electrons in the conduction band that can move. In a semiconductor there is a small energy gap between the valency band, which is almost filled, and the conduction band. An insulator has a large energy gap between the valency band, which is completely filled, and the conduction band.

 

Identify absences of electrons in a nearly full band as holes, and recognise that both electrons and holes help to carry current” (syllabus)

 

As a semiconductor heats up, electrons move from the valency band to the conduction band. This will leave holes in the valency band and these act as a flow of positive current moving in the opposite direction to the electron flow.

 

The hole current flows towards the negative terminal as electrons fill holes, leaving further holes where they came from which will be filled by electrons leaving further holes and so on.

 

Electrons can flow through the overlapping conduction bands of adjacent atoms whereas hole current must pass from one atom to the next. Consequently electron current is faster than hole current.

 

Compare qualitatively the relative number of free electrons that can drift from atom to atom in conductors, semiconductors and insulators” (syllabus)

 

Conductors have a large number of free electrons that can drift from atom to atom. Semiconductors only have a few free electrons while insulators have very, very few.

An insulator such as sulfur has a trace number of free electrons that can drift from atom to atom. A semiconductor such as silicon would have around a trillion (1012) times as many as sulphur while a conductor such as copper would have around a trillion times as many as silicon.

 

Identify that the use of germanium in early transistors is related to lack of ability to produce other materials of suitable purity” (syllabus)

 

Germanium was used in the development of early transistors but it suffered from the problem of not handling heat well. At high temperatures it would lose its semiconducting properties and the electronic device would break down. This led to the development of ways to purify silicon, which is more abundant and less prone to temperature damage. Ways were found to form large, single crystals of silicon, which could then be cut into wafers for use in solid state devices.

 

Describe how ‘doping’ a semiconductor can change its electrical properties” (syllabus)

 

Doping a semiconductor involves adding a small amount of an impurity to improve the conducting properties of the semiconductor. Doping can produce semiconductors of two types; p-type and n-type.

 

A p-type semiconductor can be made by introducing a small amount of a group III element such as boron or aluminium to the silicon or germanium. This creates an unfilled covalent bond or “hole” and conduction takes place as an adjacent electron moves to fill the “hole” and in the process, leaves a hole where it came from.

 

An n-type semiconductor can be made by introducing a small amount of a group V element such as phosphorus or arsenic to the silicon or germanium. Four of the five valency electrons form covalent bonds with the semiconductor while the fifth valency electron is available to conduct current.

 

Identify differences in p and n-type semiconductors in terms of the relative number of negative charge carriers and positive holes” (syllabus)

 

The three valency electrons in the doping atom for p-type semiconductors will be used up in forming bonds with the silicon or germanium. Since silicon and germanium have 4 valency electrons there will be one bond unsatisfied. This “hole” attracts electrons near it and as it captures an electron it leaves a hole somewhere else. The movement of holes across the crystal can be considered as a movement of positive charges. A group V element on the other hand will bond with all 4 silicon or germanium valency electrons and have an electron left over which is promoted to the conduction band. Because of the negative charges it is called an n-type semiconductor.

 

Describe differences between solid state and thermionic devices and discuss why solid state devices replaced thermionic devices” (syllabus)

Solid state devices have a more compact size than thermionic devices, are less fragile or prone to breakdown than thermionic devices, are cheaper to mass produce than thermionic devices, have no warm-up time and have less power consumption than thermionic devices. Consequently they have replaced thermionic devices in most electronic equipment used in the home or industry. However, thermionic devices are less prone to variations with temperature and are able to handle more power than solid state devices so they are still used in some hi-fi sound systems and in situations involving high temperatures.

 

Perform an investigation to model the behaviour of semiconductors, including the creation of a hole or positive charge on the atom that has lost the electron and the movement of electrons and holes in opposite directions when an electric field is applied across the semiconductor” (syllabus)

 

Place 10 eggs in a carton, leaving the two spaces at one end vacant. Move one egg forward into the adjacent hole. This will fill the hole but leave another hole where the transported egg used to be. Move the adjacent egg forward into the hole. In this way the eggs move one way along the carton and the holes move the other.

 

Gather, process and present secondary information to discuss how shortcomings in available communication technology lead to an increased knowledge of the properties of materials with particular reference to the invention of the transistor” (syllabus)

 

Thermionic devices such as the diode and triode valves were brittle and bulky so scientists sought alternatives that would do the job better. Research into semiconductors led to the development of the germanium diode and germanium based transistors. Although these were an improvement on thermionic devices they tended to break down in hot conditions. Germanium was favoured over silicon at the time because technology had not been developed that would produce silicon in sufficiently pure form for solid-state use. However as methods of purifying silicon were developed, this became the preferred semiconductor for solid-state devices because it retained its semiconducting properties at higher temperatures than germanium. As research into crystal growth and etching techniques were developed, scientists were able to miniaturize components and develop the integrated circuit where hundreds of transistors and other components are included in a chip that is smaller than a fingernail.

 

Identify data sources, gather, process, analyse information and use available evidence to assess the invention of transistors on society with particular reference to their use in microchips and microprocessors” (syllabus)

 

Transistors have led to the development of the microchip (a small piece of silicon containing an integrated circuit) and the microprocessor (an integrated circuit containing memory, processing and control circuits).

In the 1960s computers would occupy a large room and cost millions. Today the average family has their own personal computer with more power than the 1960s model, at a fraction of the cost and small enough to fit on a desktop. This has been made possible by the development of microchips and microprocessors. Navigation systems can now be preset making navigation easier and more accurate. This has led to safer aircraft operation. Microprocessors in automatic teller machines and at EFTPOS outlets enable money to be transferred in an instant while still keeping accurate records of accounts. In biotechnology it has led to the mapping of the human genome- impossible without the use of microprocessors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

“4. Investigations into the electrical properties of particular metals at different temperatures led to the identification of superconductivity and the exploration of possible applications.” (syllabus)

 

Outline the method used by the Braggs to determine crystal structure” (syllabus)

 

In 1915 Sir William Henry Bragg and Sir William Lawrence Bragg won the Nobel Prize for their study of crystal structure with X-rays. They shone x-rays onto the surface of crystals and found that at certain angles they could produce diffraction patterns. By applying the equation; 2d sinq = nl (Bragg’s Law), they were able to determine the spacing between atoms in a crystal.

The x-rays were detected by a photographic plate that had been wrapped in lightproof paper.

 

Bragg’s law of x-ray diffraction is: 2d sin q = nl. where

d = distance between successive layers of atoms in crystals.

q = diffraction angle i.e. angle between the x-ray beam and the crystal surface when diffraction occurs.

l = wavelength of x-rays

n = integer

 

The numerical example below illustrates the method used by the Braggs.

A beam of x-rays of wavelength 1.5 x 10-10 m shines onto a crystal at an angle of 15o and forms a diffraction pattern.

(i)                 What is the distance between the atoms in the crystal?

(ii)               At what angle will the second order diffraction pattern appear?

 

(i) 2d sin q = nl          d = nl / 2sinq = 1 x 1.5 x 10-10 / 2sin15 = 2.9 x 10-10 m.

(ii) 2 x 2.9 x 10-10 sinq = 2 x 1.5 x 10-10          sinq = 1.5/2.9  q = 31o           

 

Identify that metals possess a crystal lattice structure” (syllabus)

 

Metals have a lattice of positive ions with the valency electrons loosely held in such a way that they are shared by all of the atoms. (Chemists will recognise this as metallic bonding)

 

Describe conduction in metals as a free movement of electrons unimpeded by the lattice” (syllabus)

 

In a metallic conductor, e.g. a length of wire, the delocalised electrons are in constant, random motion, with equal numbers moving in opposite directions along the conductor at any given time. When a potential difference is applied it creates an electric field and the electrons drift in the opposite direction to the electric field. While the speed of the electrons is very fast (in the order of 106 ms-1) the drift velocity of the cloud of electrons is very slow (in the order of 1 mms-1).

 

 

 

 

Identify that resistance in metals is increased by the presence of impurities and scattering of electrons by lattice vibrations” (syllabus)

 

As electrons move along the metal their progress is impeded by collisions with the ions in the crystal lattice. As a metal heats up, the ions in the lattice vibrate more vigorously and so more effectively impede the progress of the electrons along the conductor i.e. they increase the resistance. Impurities in the crystal lattice also impede the free movement of electrons and so increase the resistance.

As the electric field pushes the electrons through the lattice it does work. The work to move a unit of charge is the potential difference. If it takes one joule of work to move one coulomb of charge from one point to the other the potential difference is one volt.

 

Describe the occurrence in superconductors below their critical temperature of a population of electron pairs unaffected by electrical resistance” (syllabus)

 

Superconductivity refers to the disappearance of electrical resistance, which occurs in many substances at very low temperatures. It occurs because vibrations of the crystal lattice cause the charge on electrons to be distributed unevenly so that electrons are able to “pair up” (Cooper pairs) and travel through the conductor unaffected by electrical resistance.

 

Discuss the BCS theory” (syllabus)

 

The BCS theory of superconductors refers to the passage of electrons through superconductors, as described by John Bardeen, Leon Cooper and John Schrieffer who developed the model to explain how superconductivity works.

 

A phonon is a quantum of thermal energy due to the vibrations of ions within a crystal. It has a value of hf where h = Planck’s constant and f = the frequency of vibration.The distortion of the crystal lattice emits a phonon that forms a region of positive charge behind the electron. This attracts the electron behind it to form a pair known as a Cooper pair. There is a transfer of a phonon from the first to the second electron and it is this transfer of phonon energy that keeps the Cooper pairs together. They are not permanent bonds and Cooper pairs keep breaking up and reforming.

 

Discuss the advantages of using superconductors and identify limitations to their use” (syllabus)

 

Superconductors have no resistance so there is no energy loss in their transmission from one place to another. They can also carry large currents without heat loss so they are able to generate strong magnetic fields.

Superconductors can carry a maximum current called the “critical current density”. The lower the temperature the higher the critical current density. Above this current the substance loses its superconducting properties. Superconductors still require very low temperatures, approaching absolute zero for metals. The superconductors that operate at slightly higher temperatures than others (but still lower than 100oC below zero) are ceramics and can’t be drawn into wires. The technology is still very expensive and the very low temperatures involved represent an extra safety hazard in the case of accidents.

 

Process information to identify some of the metals, metal alloys and compounds that have been identified as exhibiting the property of superconductivity and their critical temperatures” (syllabus)

 

In 1911 Heike Kamerlingh Onnes carried out a series of experiments in which he measured the resistance of mercury at very low temperatures. He found that the resistance dropped as expected as the temperature decreased, but at around 4.2K the resistance suddenly dropped to virtually zero. A graph of resistance against temperature for mercury is shown below. Note the sudden drop in electrical resistance at 4.2K.

                       

He repeated his experiment by cooling other metals and found that a number of them had virtually no resistance at temperatures approaching absolute zero.

The temperature at which a metal’s resistance becomes zero is known as its critical temperature and this varies from metal to metal. The critical temperatures of some metals are: mercury 4.2K., tin 3.7K., lead 7.2K., aluminium 1.2K.

Metal alloys were found to have slightly higher critical temperatures than metals.

The big breakthrough came with the discovery of the superconductivity of some ceramics with critical temperatures above 77K (the boiling point of liquid nitrogen). One of the most widely used superconductors is the ceramic YBa2Cu3O7, abbreviated YBCO. This has a critical temperature of 90K.

 

Perform an investigation to demonstrate magnetic levitation” (syllabus)

 

A simple demonstration of magnetic levitation can be performed by placing two magnets, like poles together, in a glass tube.

                       

Magnetic levitation involving superconductors requires low temperatures. Since there are safety issues involving the use of liquid nitrogen in schools (a thermos flask of liquid nitrogen spilt over someone could kill them) magnetic levitation involving superconductors is best done as a demonstration at a university.

Analyse information to explain why a magnet is able to hover above a superconducting material that has reached the temperature at which it is superconducting” (syllabus)

 In 1933 Hans Meissner found that a magnetic field was unable to penetrate a superconductor. As a magnetic field is brought near a superconductor it induces a current on the surface of the superconductor and this current produces a magnetic field to just balance and oppose the one that would have penetrated. This effect, known as the Meissner effect allows a magnet to hover above a superconductor.

 

Magnetic levitation using a superconductor can be demonstrated by the following method:

A hollow container made of a superconductor such as YCBO (Critical temp. 90K) is filled with liquid nitrogen (B.P. 77K). A magnet placed on top of the superconductor will float in mid-air.

The magnetic field is prevented from entering the superconductor. As the lines of flux cannot enter the superconductor they cause the magnet to lift so that they can complete their path from the north to the south poles.

 

Above the critical temperature the magnetic lines of flux from the magnet are able to penetrate the ceramic such as YCBO and the magnet will rest on it as normal.

                       

A magnet resting on a superconductor above its critical temperature.

 

Below the critical temperature the ceramic becomes a superconductor and excludes the lines of flux. This will cause the magnet to hover above the superconductor.

                       

A magnet hovering above a superconductor below its critical temperature.

 

 

 

 

 

 

 

 

 

 

 

 

 

Gather and process information to describe how superconductors and the effects of magnetic fields have been applied to develop a maglev train” (syllabus)

 

The E.D.S. (electrodynamic suspension) system in Japan uses the opposing force between superconducting magnets on the train and conventional electromagnets in the guideways (rails) to levitate it. It begins to levitate once it reaches 40 km/h.

The E.M.S. (electromagnetic suspension) system in Germany uses conventional electromagnets under the train and also in the guideways. The like polarities cause the train to lift.

Both trains are propelled by the continually changing polarity of alternate magnets along the track.          

 

For more information on maglev trains have a look at these websites;

http://www.rtri.or.jp/rd/maglev/html/english/maglev_frame_E.html

http://www.o-keating.com/hsr/maglev.htm

http://travel.howstuffworks.com/maglev-train.htm

http://www.calmaglev.org/

 

 

Process information to discuss possible applications of superconductivity and the effects of those applications on computers, generators and motors and transmission of electricity through power grids” (syllabus)

 

Superconductors can speed up the rate at which computers are able to process information. Switching devices using superconductors do not generate heat and are many times faster than the equivalent semiconductor switch. Josephson junctions consist of two superconductors separated by a thin layer of insulator. Their use in transistors would speed up processing of signals resulting in faster computers.  

 

Motors and generators using superconductors would be more efficient than their present-day counterparts. Superconductors would not have any resistance in the coils and so would not lose energy due to heat loss. They would not require a soft iron core and consequently would be a lot smaller and lighter than present motors and generators with the same output.       

 

The use of superconductors in power transmission would reduce energy loss as no energy would be lost as heat in the transmission lines.

 

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