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9.2 – Space 9.2.1 – The earth has a gravitational field that exerts a force on objects both on it and around it. 9.2.1.1 – Define weight as the force on an object due to a gravitational field. On Earth, objects have a weight force equivalent to their mass time the acceleration due to gravity. W = mg Where: W = weight (N) m = mass (kg) g = acceleration due to gravity (ms-2) On Earth, the value for g is taken to be 9.8 ms-2. (There are subtle variations possible which will be discussed later). On other planets etc. g becomes a different value and can be calculated using the equation: mp “g = G ” r 2p Where: g = acceleration due to gravity (ms-2) G = gravitational constant (6.67 x 10-11 Nm2kg-2) r2p = radius of the planet (m) mp = mass of planet (kg) Calculations 1) If Dean’s has a mass of 95.0 kg. Calculate his weight on Earth and on the moon where the g value is approximately 1/6 that of Earth. W = mg = 95 x 9.8 = 931 N on Earth W = mg = 95 x (9.8 x 1/6) = 1.6 N on the Moon 2) Lee has a weight of 15500 N on an unknown planet while his weight on Earth is 647 N. Calculate his mass and the g on this unknown planet. m = W/g = 647/9.8 = 66.0 kg is Lee’s mass g = W/m = 15500/66 = 234 ms-2 is the gravity on the unknown planet. a = 34.0 ms-2 Fr Fg Scales Fnet = Fr - Fg Ma = Fr – mg 70 x 34 = Fr – 70 x 9.8 Fr = 70 x (34 + 9.8) = 3066 N is the apparent weight = 312 kgf is the apparent mass In orbit FR = 0 because of a freefall situation where there is zero g. Landing on the moon, decelerating at 5.0 ms-2: Stationary on the moon: Fnet = Fr - Fg Ma = Fr – mg Fr = m x (a + g) = 60 x (5 + 1.6) = 396 N is the apparent weight = 40.4 kgf is the apparent mass Fnet = Fg - Fr Ma = mg - Fr Fr = mg – Ma = 60 x 1.6 – 60 x 0 = 96 N is the apparent weight = 9.8 kgf is the apparent mass 9.2.1.2 – Explain that a change in gravitational potential energy is related to work done. On Earth: GPE = mgh Where: GPE = Gravitational Potential Energy (J) m = mass (kg) g = acceleration due to gravity h = height (m) 1 70.0kg 20.0m 2 35.0m 3 70.0m 57.0m 4 0.0m 5 Situation 1: GPE = mgh = 70 x 9.8 x 70 = 48020 J = 48000 J (to 3 sig fig) = 4.8 x 104 J Total Energy = GPE + KE Situation 2: GPE = mgh = 70 x 9.8 x 50 = 34300 J ET = GPE + KE 48020 = 34300 + KE KE = 13720 J KE = ½ x mv2 13720 = ½ x 70 x v2 v2 = 13720/35 v = √392 = 19.8 ms-1 Situation 3: GPE = mph = 70 x 9.8 x 35 = 24010 J ET = GPE + KE 48020 = 24010 + KE KE = 24010 KE = ½ x mv2 24010 = ½ x 70 x v2 v = √24010/35 = 26.2 ms-1 On earth if we wish to lift an object up higher, we must do work on it. W = Fs Where: W = work (J) F = force (N) s = distance (m) Here F must be equal to or greater than Fg (weight) = mg & s = height lifted = h Therefore: W = mgh or GPE = mgh In this case we chose the ground as our zero level where Ep (potential energy) = 0 J. When objects are raised to say point x the value of Ep at this point is greater than zero. In Space: On the larger scale we need to look at the situation differently. Using the of universal gravitation (9.2.3), the force of attraction between a planet and an object will drop to zero only at infinity (∞). For this reason the ∞ point is chosen to represent zero potential energy (at ∞, Ep = 0). Due to this choice, since work must be done on an object to move it from point x to ∞ so that it gains Ep. This means that all other points have negative Ep values. ∞ x Figure 1.5 9.2.1.3 – Define gravitational potential energy as the work done to move an object from a very large distance away to a point in a gravitational field. Ep G m1 m2 r Where: Ep = Gravitational Potential Energy (J) G = Gravitational Constant (6.67 x 10-11) m1 = Mass of significant object (kg) m2 = Mass of other object (kg) r = distance between objects (m) Problem: mearth = 5.97 x 1024 kg mmoon = 7.35 x 1022 kg msun = 1.99 x 1030 kg rearth – moon = 3.84 x 108 m rearth – sun = 1.50 x 1011 m (a) Calculate the gravitational potential energy of: i. The moon within the earth’s gravitational field ii. The earth within the suns gravitational field iii. Kate the astronaut’s gravitational potential energy from the earth if she is halfway to the moon if she weighs 100 kg. i. Ep G me mm r Ep 6.67 10^ 11 (5.97 10^ 24) (7.35 10^ 22) 3.84 10^8 = -7.62 x 1028 J ii. Ep G me ms r Ep 6.67 10^ 11 (5.97 10^ 24) (1.99 10^30) 1.50 10^11 = -5.28 x 1033 J iii. Ep G me mk r Ep 6.67 10^ 11 = -2.07 x 108 J (5.97 10^ 24) 100 (3.84 10^8) 0.5 9.2.2 – Many factors have to be taken into account to achieve a successful rocket launch, maintain a stable orbit and return to Earth. 9.2.2.1 – Describe the trajectory of an object undergoing projectile motion within the Earth’s gravitation field in terms of horizontal and vertical components. An object thrown into the air near to the Earth’s surface will follow a parabolic path. Its horizontal component’s velocity will remain constant (neglecting air resistance), while the vertical component will undergo a constant negative acceleration of 9.8ms − 2. The overall velocity is the sum of these components. These two components are vectors and can be summed by adding them head to tail, or as they are vertical and horizontal, using Pythagoras theorem. 9.2.2.2 – Describe Galileo’s analysis of projectile motion. Galileo was the first to propose that all projectiles took a parabolic path. He also proposed that two objects (a stone and a feather) would reach the ground at the same time if they were dropped from the same height at the same time provided there was no air resistance. He also said that the horizontal and vertical components of projectile motion occurred both independently and simultaneously. An alternate interpretation: Galileo’s analysis of projectile motion is that the motion of a body can be split into two components, horizontal and vertical. Where the horizontal motion will have no acceleration acting on it, and the vertical motion will have acceleration due to gravity acting on it. 9.2.2.3 – Explain the concept of escape velocity in terms of the: Gravitational constant Mass and radius of the planet By considering the kinetic and gravitational potential energy of a projectile, a mathematical formula can be derived. vescape = 2Gm( planet ) r ( planet ) Where: vescape = Escape velocity (ms-1) G = Gravitational constant (6.67 x 10-11) mplanet = Mass of planet (kg) rplanet = Radius of planet (m) Example: Using previously supplied data find the escape velocity for: a) Earth 2Gm( planet ) vescape = r ( planet ) (2 6.67 10^ 11) 5.97 10^ 24 6.38 10^6 = 11172.6 ms-1 = 11200 ms-1 (3 sig fig) b) Mercury 2Gm( planet ) vescape = r ( planet ) = (2 6.67 10^ 11) 3.3 10^ 23 2440000 = 4247.6 ms-1 = 4250 ms-1 (3 sig fig) c) Venus 2Gm( planet ) vescape = r ( planet ) = (2 6.67 10^ 11) 4.9 10^ 24 6052000 -1 = 10392.7 ms = 10400 ms-1 (3 sig fig) = d) Io vescape = 2Gm( planet ) r ( planet ) (2 6.67 10^ 11) 8.9 10^22 1815000 = 2557.6 ms-1 = 2560 ms-1 (3 sig fig) e) Jupiter 2Gm( planet ) vescape = r ( planet ) = (2 6.67 10^ 11) (5.97 10^ 24 318) (6.38 10^6 11) -1 = 60072.0 ms = 60100 ms-1 (3 sig fig) = Since ‘G’ is a constant, although it determines the escape velocity it cannot change it. The planets mass and radius can, however, alter the required velocity needed to escape the planet’s gravitational field. As m increases, the escape velocity increases and as r increases the escape velocity will decrease. Also as density increases, escape velocity also increases. 9.2.2.4 – Outline Newton’s concept of escape velocity. Isaac Newton wrote that it should be possible to launch a projectile fast enough so that it achieves an orbit around the Earth. Through this principle it was found that escape velocity is the initial velocity required by a projectile to rise vertically and just escape the gravitational field of a planet. If this specified escape velocity is exceeded slightly then the object will follow an elliptical orbit around the Earth. If the specified velocity is exceeded further still, then the object will follow a parabolic or hyperbolic path away from the Earth. This is the manner in which space probes depart the Earth and head off into space. 9.2.2.5 – Identify why the term ‘g forces’ is used to explain the forces acting on an astronaut during launch. Your body is a mass lying somewhere in a gravitational field, and therefore experiences a true weight. The sensation of weight that you feel derives from your apparent weight, which is equal to the sum of the contact forces resisting you true weight. This includes the normal reaction force of the floor and the thrust of a rocket engine. The term ‘g force’ is used to express a person’s apparent weight as a multiple of his/her normal true weight. Hence: g force = apparent _ weight normal _ true _ weight (Figure 2.18) During the launch of an astronaut into space the astronaut’s body is exerting a downward weight force on the floor, and the floor meets this with an upward reaction force equal to mg. In addition, the floor is exerting an upward accelerating force equal to ma. Therefore: mg ma 9 .8 m ga = 9.8 g force = Where: g = acceleration due to gravity at altitude (ms-2) (+ve away from earth, -ve towards earth) m = mass of astronaut (kg) 9.2.2.6 – Discuss the effect of Earth’s orbital motion and its rotational motion on the launch of a rocket. The fuel required to launch a rocket into orbit can be minimised if it is launched to the east and as close to the equator as possible. This is because the Earth rotates from west to east, so a rocket launched to the east will get a boost in speed from this rotation, and since the surface speed is fastest at the equator this is where the largest boost will be obtained. Transfer orbits between the Earth and other planets in the solar system can be achieved by using the Earth’s orbital velocity around the Sun. If the rocket is launched against the motion of the Earth’s orbit it will have an overall velocity of the Earth’s orbital velocity minus the escape velocity, and it will consequently transfer to an orbital path nearer to the sun than that of the Earth. If the rocket is launched with the Earth’s orbit it will be travelling at the Earth’s orbital velocity plus the escape velocity and will hence transfer to an orbit further away from the sun than that of the Earth Because the Earth rotates, when a rocket is launched from Earth the horizontal velocity of the rocket will be the same as the rotational speed of the Earth plus the orbital motion of earth (depends where on earth you are). This will affect the path of the rocket in outer space. It will also contribute to the rockets speed. The same applies for the Earth’s orbital motion. So where on earth the rocket is launched from and what time of day and year will affect where the rocket will initial be heading. So to fully utilise this and minimise fuel usage the location on Earth of the rocket and what time of day and year it is is chosen to get the rocket to head in a particular way. Remember, the axis of Earth’s rotation on its own axis is from pole to pole. Hence launching from the equator will result in the most velocity boost from Earth’s rotation. Earth rotates on its own axis one revolution per 24 hours, and around the sun once every year. 9.2.2.S1 - Solve problems and analyse information to calculate the actual velocity of a projectile from its horizontal and vertical components using : v2x = u2x v = u + at v2y = u2y + 2ay∆y ∆x = uxt ∆y = uyt + ½ ayt2 s = ut + ½ at2 Where: x = horizontal y = vertical u = initial velocity (ms-1) v = final velocity (ms-1) a = acceleration (ms-2) t = time (s) ∆x = horizontal displacement ∆y = vertical displacement s = displacement There is no acceleration in the horizontal component, i.e. ux or vx is constant. Therefore ∆x = uxt. Vertically there is constant acceleration due to gravity, hence the equations of motion. Example: Kate drives a car at 30.0 ms-1 (horizontally), off the edge of an 895m cliff. Find the following: a) Time taken to reach the ground b) Final vertical velocity c) The range or distance from base of the cliff, and; d) The final velocity a) ∆y = 895m ux = 30.0 ms-1 ay = 9.8 ms-2 uy = 0.0 ms-1 ∆y = uyt + ½ ayt2 895 = 0.0 + 4.9t2 895 t= 4 .9 t = 13.5 s b) vy = uy + ayt = 0 + 9.8 = 132 ms-1 c) ∆x = uxt = 30 x 13.5 = 405 m tanθ = 132/30 Θ = 77.1° d) 30m 132m Examples: Lee standing on top of a bell tower throws an orange at 8.50 ms-1. It takes the orange 3.05s to strike the ground. Find: a) Height of the tower b) Range c) Its actual final velocity Answers: a) ∆y = uy + ½ at2 = 0.0 + 4.9 x 3.052 = 45.6 m b) ∆x = uxt = 8.5 x 3.05 = 25.9 m c) vy = uy + ayt =0 + 9.8 x 3.05 = 29.9 ms-1 v = 8.5^ 2 29.9^ 2 = 31.08 ms-1 A ball is thrown upwards from a 63.0 m cliff so that it reaches a maximum height of 139 m (from base). Calculate: a) Initial velocity b) Total time a) v2y = u2y + 2ay∆y 0.0 = u2y – 19.6 x 76 u = 1489.60 = 38.6 ms-1 b) ∆y = uyt + ½ ayt2 76 = 38.6t + 4.9t2 b b ^ 2 4ac t 2a 38.6 38.6^ 2 4 4.9 76 t 2 4.9 t =3.94s 9.2.2.7 – Analyse the changing acceleration of a rocket during launch in terms of the: Law of Conservation of Momentum Forces experienced by astronauts A rocket engine is a little different to most other engines in that it must carry/contain both the fuel and the oxygen supply (assuming a typical combustion reaction). Modern rockets can use either solid or liquid propellants. The forward motion of the rocket can be understood in terms of either Newton’s third law of motion (for every action there is an equal and opposite reaction [Fab = Fba]); and the Law of Conservation of Momentum ( Mbefore = Mafter). mrur + meue = mrvr + meve 0 = mrvr + meve mrvr = -meve Mrocket = ∆Mengine Since ∆M or ∆p = Ft Frocket = Fengine ie. Newton’s 3rd Law We assume that the rocket provides a relatively constant thrust and therefore the reaction force propelling the rocket upwards should also be relatively constant. However the mass of the rocket is continually decreasing due to the propulsion of exhaust gases; this means that as mass decreases the acceleration increases using Newton’s Second Law (F = ma). To more economically place a satellite into space and to minimize g-forces experienced by astronauts, multi stage rockets are used. This creates a series of changes as shown in figure 2.2 pg 31 – In stage 1 the rocket accelerates at an increasing rate as the mass of the propellant decreases in the first booster (peaks at around 4g; when stage 1 breaks away, the g-forces drop dramatically to 0; at this point the second stage rocket ignites and a similar pattern of gradual increase in acceleration to about 1.8g occurs before again it drop to 0 as the second booster is jettisoned; similar for stage three. Forces experienced by Astronauts As we discussed earlier in g-forces, an astronaut feels an apparent weight equivalent to ma + mg where a is the acceleration away from the Earth’s surface. Since g-force = ga apparent _ weight then g-force = . 9.8 real _ weight Referring to figure 2.2 on page 31, the maximum g-force experienced by Apollo Astronauts, using the Saturn V rockets, was around 4g (at 4g’s most people begin to loose some of their colour and peripheral vision). Soon after this most people will black out (individual differences). In the first manned space flight by the US, astronaut Alan Shepard had to tolerate a peak g-force of 6.3g. Rocket design has improved significantly and the current space shuttle launch does not exceed 3g’s. To minimize some of the effects of these g-forces on humans careful design has been implemented. These include: That a transverse (lying down) position coped better with g-forces since the blood was not as easily forced away from the brain. An ‘eyeballs-in’ application in g loads is easier to tolerate than an ‘eyeballs-out’, ie. On lift off the astronauts should be facing up and also reenter backwards facing up so that the g-forces are always directed upwards. Supporting the body in as many places as possible, most using a contoured couch built of molded fiberglass specific to that astronaut. Calculations actually show that even larger forces can be involved on reentry. Alan Shepard experienced a maximum reentry force of 11.6g’s. 9.2.2.8 – Analyse the forces involved in uniform circular motion for a range of objects, including satellites orbiting the Earth. v (orbital speed/velocity) Fc r Centripetal Force (Fc) = mv^ 2 r Where: Fc = Centripetal force (N) and is always directed to the center of the circle. m = Mass (kg) v = velocity (ms-1) For an object on the end of a string swung around our head in a horizontal circle, the force causing Fc is in fact the tension in the string (see prac on circular motion). If the string suddenly breaks, the object on the end ‘fly’s off’ at a tangent. For celestial objects (planets and stars) or for natural and artificial satellites the Fc is Gm1m 2 mv^ 2 Gm1m 2 supplied by Fg (Fg = ). Therefore, Fc = Fg, where m1 is the d ^2 r d ^2 larger mass and m2 is the smaller mass undergoing circular motion. Also r = d. m2v ^ 2 Gm1m2 Gm1 2r and finally v ^ 2 . Furthermore, v . d d d T 9.2.2.9 – Compare qualitatively low Earth and geo-stationary orbits. Low Earth orbits are above the Earth’s atmosphere but below the Van Allen radiation belts. Most of these are polar (north to south) elliptical orbits. They are a lower orbit than geostationary orbits and thus have a faster orbital velocity. Some examples of uses for low Earth orbits are to survey the surface of the Earth for minerals, weather and imaging satellites and as an orbit for the space shuttle. Geostationary orbits have an orbital period equal to the rotational period of the Earth. They orbit above the equator. Because of this the satellites footprint remains over the same place on the surface of the Earth for its whole period. These satellites are used as weather and communications satellites. This means that for satellite TV, the receiver (satellite on your house) can just point in the same place above the sky and the satellite will always be there. 9.2.2.10 - Define the term orbital velocity and the quantitative and qualitative relationship between orbital velocity, the gravitational constant, mass of the central body, mass of the satellite and the radius of the orbit using Kepler’s Law of Periods. r ^3 GM T ^ 2 4 ^ 2 Where: r = Radius (m) T = Period (s) G = Universal Gravitational Constant M = Mass of central object (kg) Calculate the period an orbital velocity for the following satellites: a) Altitude 250 km b) Altitude 400 km c) Altitude 40000 km Earth’s average radius = 6.38 x 106 m Earth’s average mass = 5.97 x 1024 kg G = 6.67 x 10-11 Answers: (6.38 10^6 250000)^3 ((6.67 10^ 11) (5.97 10^ 24)) a) T ^2 4 ^ 2 ((6.38 10^6 250000)^3)4 ^ 2 ((6.67 10^ 11) (5.97 10^ 24))T ^ 2 ((6.38 10^6 250000)^3)4 ^ 2 T ^2 ((6.67 10^ 11) (5.97 10^ 24)) T 28893500.25 T 5375.27s 2r T 2 (6.38 10^6 250000) v 5375.27 v 7750 ms-1 = 27900 km/h v b) (6.38 10^6 400000)^3 ((6.67 10^ 11) (5.97 10^ 24)) T ^2 4 ^ 2 ((6.38 x10^6 400000)^3)4 ^ 2 ((6.67 10^ 11) (5.97 10^ 24))T ^ 2 ((6.38 10^6 400000)^3)4 ^ 2 T ^2 ((6.67 10^ 11) (5.97 10^ 24)) T 30899300.88 T 5558.71s 2r T 2 (6.38 10^6 400000) v 5558.71 v 7660 ms-1 = 27600 km/h v c) (6.38 10^6 40000000)^3 ((6.67 10^ 11) (5.97 10^ 24)) T ^2 4 ^ 2 ((6.38 x10^6 40000000)^3)4 ^ 2 ((6.67 10^ 11) (5.97 10^ 24))T ^ 2 ((6.38 10^6 40000000)^3)4 ^ 2 T ^2 ((6.67 10^ 11) (5.97 10^ 24)) T 9891264254 T 99454.84s 2r T 2 (6.38 10^6 40000000) v 99454.84 -1 ms = 10 550km/h v 2930 v 9.2.2.11 – Account for Orbital decay of Satellites in Low Earth Orbit. When a satellite is in a low Earth orbit it will collide with small particles in the atmosphere and will hence experience drag. This causes the satellites orbit to decay (or lose altitude). As the altitude is reduced the atmosphere becomes denser and the process of orbital decay is accelerated as drag is increased. 9.2.2.12 - Discuss issues associated with safe reentry into the Earth’s atmosphere and landing on the Earth’s surface. As a spacecraft enters the Earth’s atmosphere an immense amount of heat is produced. This is because the craft is traveling at an extremely high velocity, and is colliding with particles in the atmosphere. If this heat is not dissipated the craft may burn up. This can be combated by using heat shields spread over large surfaces (so that the heat is spread evenly). It can also be minimised by increasing the time taken for re-entry. As the spacecraft enters the atmosphere and heat builds up around the craft atoms in the air become ionised and surround the craft, resulting in a phenomenon known as ionisation blackout. These ions deflect any radio signals sent to or from the craft. This can be dangerous if there is contact needed between the craft and the Earth at this phase of the flight. The deceleration during re-entry can cause high g forces to be experienced by the astronaut, which can cause blood to rush out of the brain, causing loss of consciousness or even death. This loss of blood can be minimised by reclining the astronaut and applying these forces perpendicular to the astronaut’s long axis. The g forces can be minimised by increasing the re-entry time, slowing the rate of descent. The human body can stand g forces of up to about 4g without comfort, with 10g being tolerable for short periods of time along the long axis of the body. The spacecraft must touch down softly on the surface of the Earth. This can be achieved by using parachutes to slow the craft down and splashing into the ocean, or by landing on an airstrip. 9.2.2.13 - Identify that there is an optimum angle for safe re-entry for a manned spacecraft into the Earth’s atmosphere and the consequences of failing to achieve this angle. The angle of re-entry must be between 5°-7°. If the angle is too shallow the spacecraft will bounce off the atmosphere and back into orbit. If it is too steep the craft will burn up due to excessive drag. 9.2.3 – The Solar System is held together by Gravity. 9.2.3.1 – Describe a gravitational field in the region surrounding a massive object in terms of its effects on other masses in it. Any field is a region of influence cause by such things as charge, magnetism or mass. These fields are vector quantities (the have strength/magnitude and direction). A gravitational field is a little different to others in that it is always directed towards the large mass which causes the gravitational field. 9.2.3.2 – Define Newton’s Law of Universal Gravitation. F G m1m2 d^2 9.2.3.3 – Discuss the importance of Newton’s Law of Universal Gravitation in understanding and calculating the motion of satellites. 9.2.3.4 – Identify that a slingshot effect can be provided by planets for space probes. 9.2.4 – Current and emerging understanding about time and space has been dependent upon earlier models of the transmission of light. 9.2.4.1 – Outline the features of the aether model for the transmission of light 9.2.4.2 – Describe and evaluate the Michelson-Morley attempt to measure the relative velocity of the Earth through the aether 9.2.4.3 – Discuss the role of the Michelson-Morley experiments in making determinations about competing theories 9.2.4.S1 – Gather and process information to interpret the results of the MichelsonMorley experiment The analogy below can be used to model the concept of aether that Michelson and Morley were trying to prove. In this analogy the aether is represented by the water current; boat A is traveling in the same direction/plane as the aether whilst boat B is at 90º. B Current Boat A and B both travelA2km one way and 2km back. They also both travel at 5km/h in still water. Boat A Upstream trip Speed = 5 – 3 = 2 km/h Time = distance/speed = 2/2 = 1 hr Downstream trip Speed = 5 + 3 = 8 km/h Time = distance/speed = 2/8 = 0.25 hrs Total trip = 1 hr + 0.25 hrs = 1.25 hrs Boat B For boat B to travel straight across, it has to aim upstream. 3 km/h 4 km/h 5 km/h By Pythagora’s theorem, the measured speed straight across the river is 4 km/h, therefore the total round trip: Conventional Mirror Time = distance/speed V (aether wind) 2a/b Original Light Ray M 1a/b Conventional = 4/4 = 1 hr In the above simplified diagram, light from the light source traveled horizontally to the special mirror (M) (half silver mirror) which splits the light into two separate ways. Half passes straight through (1a) where it strikes a conventional mirror, reflects back to M (1b) and in turn is reflected back to the observer (1c). The other half of the original ray is reflect by M upwards (2a); reflects from this conventional mirror (2b) and passes straight through M to reach the observer (2c). If, in fact, there is an aether wind as shown (V) then 2c should reach the observer slightly more quickly than 1c (see previous boat analogy). This means that 1c and 2c will be out of phase with each other (to some extent). In 1887 when the experiment was conducted, this piece of apparatus was capable of picking up minute differences in the time or path length. Small differences in path length may have been attributable to the pieces of equipment eg. Phase changes on reflection, thicknesses of mirrors etc. However, when the apparatus is rotated so that one of the rays was no longer exactly in the same direction as the suspected aether wind, there should have been a subsequent change in the interference pattern (since the path difference would also have changed). In fact there were no changes even when the apparatus was rotated through a variety of angles up to 360º. This was therefore a nil result – it failed to detect the aether. The experiment was repeated numerous times at different times of the day and year and some groups repeated a similar experiment with even more sensitive apparatus. No evidence could ever be found. The scientific community was not quick to abandon the aether model and may tried to adapt or modify the theory. One such suggestion was that a large object like a planet could drag the aether along with it; another was that objects contract in the direction of the aether wind. None of these modifications, however, survived careful scrutiny. It was not until Einstein in 1905 provided an alternative which did not require the aether construct. This alternative explanation for light and the inability to detect aether eventually led to this aether wind theory being rejected and Einstein’s special relativity became the new accepted theory. 9.2.4.4 – Outline the nature of internal frames of reference. 300 hundred years before Einstein, Galileo proposed the ‘principle’ of relativity – ‘all steady motion is relative and cannot be detected without reference to an outside point’. The principle of relativity applies only for non-accelerated motion (at rest or moving with constant velocity). This is referred to as an inertial frame of reference. Situations involving acceleration are called non-inertial frames of reference. This principle states that within an inertial frame of reference you cannot perform any mechanical experiment or observation that would reveal to you whether you were moving with uniform velocity or standing still. 9.2.4.5 – Discuss the principle of relativity. (See earlier dot point [9.2.4.4] on Galileo and the principle of relativity) The belief in aether posed a real problem for the principle of relativity, since the aether was supposed to be stationary and light had a fixed velocity relative to the aether. This would mean that if a scientist tried to measure the speed of light from the back of a train carriage to the front and the train was traveling into the aether it should be a slower value than moving away. This means a ‘simple’ optical experiment would violate the principle of relativity. 9.2.4.6 – Describe the significance of Einstein’s assumption of the constancy of the speed of light. At around the turn of the 20th century, Einstein puzzled over this apparent violation of the principle of relativity posed by the aether model. He presented many of his explanations as thought experiments (‘Gaedaken’). He used the analogy of a train traveling at the speed of light, holding up a mirror in front and posing the question would he see his own reflection. If the aether model was right, there would never be a reflection since the light would not be able to catch up to the mirror (this would violated the principle of relativity since the absence of a reflection would be a way of detecting motion). On the other hand, if the principle of relativity were not to be violated the reflection must be seen normally BUT this would mean an observer on an embankment would see light traveling at twice its normal speed (using conventional physics). Einstein believed firmly in the principle of relativity which meant that the aether could not exist. To solve the two dilemmas, Einstein decided that both the train rider and the person on the embankment must both observe the light traveling at its normal speed (3 x 108 ms-1) but if this is to occur since speed = distance/time, than the distance and time witnessed b both observers must be different or relative. 9.2.4.7 – Identify that if c is constant then space and time become relative. (See previous dot point [9.2.4.6]) Einstein’s ideas were published in a 1905 paper ‘On the Electrodynamics of Moving Bodies’. The main ideas were: The laws of physics are the same in all frames of reference (ie. The principle of relativity always holds) The speed of light in empty space always has the same value c independent of the motion of the observer (ie. Everyone always observes the same speed of light regardless of their motion) He stated that the luminiferous aether is superfluous This allows us to generate another series of equations for the length an time based on the relative velocity based on the frames of reference. 9.2.4.8 – Discuss the concept that length standards are defined in terms of time in contrast to the original meter standard. Originally defined in 1973 9.2.4.9 – Explain qualitatively and quantitatively the consequence of special relativity in relation to: The relativity of simultaneity The equivalence between mass and energy Length contraction Time dilation Mass dilation Relativity of Simultaneity Einstein contended that if an observer sees two events to be simultaneous then any other observer in relative motion to the first would not judge them to be simultaneous. Einstein offered the following modified thought experiment/explanation. The operator of a lamp sits in the middle of a train carriage. The doors at either end are light operated and another observer on the embankment (stationary) is exactly in line with the lamp operator when they switch the lamp on to open the doors. The operator of the lamp in the moving train will see both doors open simultaneously because light travels at the same speed forward and backward (c). The observer on the embankment, however, sees something different. Light from the lamp travels at a constant speed forward and backward but the light moving forward travels a longer distance to reach the front door because the door has moved forward whilst the train is moving. Similarly, the light will reach the back door earlier since the back door ahs moved forwards ie. The stationary observer sees the back door open before the front door. Both observers are correct from their different frames of reference – this is a direct consequence of the constancy of the speed of light. The Equivalence Between Mass and Energy As a consequence of mass dilation, then there is also equivalence between mass and energy. As force is applied to create acceleration, this acceleration leads to higher velocities that eventually create in creases in mass. As mass increases towards infinity, then an infinite force would be needed to cause the acceleration – this is why no object can be accelerated beyond the speed of light. If a force is applied on an object then work is done or energy has been given to an object (in this case kinetic energy [Ek]). However, near the speed of light the object does not acquire velocity but rather mass (or inertia). The amount of energy or work the has been applied has been used to increase the objects mass. This leads to the equation: E = mc2 Where: E = energy (J) m = mass (kg) c = speed of light (3 x 108 ms-1) Example Calculate the energy equivalent of the typical uranium atom with a mass of 238 amu (1 amu = 1.661 x 10-27 kg). E = mc2 = (238 x [1.661 x 10-27]) x (3 x 108)2 = 3.558 x 10-8 J Length Contraction Again as a consequence of the constancy of c, length/distance is also relative and will be perceived differently by two observers in relative motion to each other. A derivation of the equation shown below can be found in the textbook (pg 77-80) but it is complex. lv = l0 1 v^2 c^2 Where: l0 = Normal length as seen by someone within the moving frame (or if the object was stationary) (m) lv = Length perceived by the other observer where the object is moving relative to them (m) v2 = Velocity of the moving object/frame c2 = Speed of light (3 x 108 ms-1) Example When stationary, the carriages on a very fast train are 20.00m long. How long would these appear to a person standing on a station platform if the train is traveling at: a) 330 ms-1 b) 50% the speed of light c) 95% the speed of light v^2 a) lv = l0 1 c^2 lv = 20.00 x 1 330^2 (3 10^8)^2 = 20.00 m v^2 lv = l0 1 c^2 b) lv = 20.00 x 1 0.5 [(3 10^8)^2] (3 10^8)^2 = 17.32 m v^2 lv = l0 1 c^2 c) lv = 20.00 x 1 0.95 [(3 10^8)^2] (3 10^8)^2 = 6.24 m by another train and the observer in this train estimates Lee’s width to be Lee is passed 15 cm. How fast was this train going? lv = l0 1 v^2 c^2 0.15 = 0.4 x 1 v^2 (3 10^8)^2 v^2 1 - (0.375)2 = (3 10^8)^2 v2 = -0.140625 x (9 x 1016) v = 1.0265625 10^17 = 2.828 x 108 ms-1 Time Dilation Similarly, tie is also relative, and the equation which relates to this is: tv = to v^2 1 c^2 Where: tv = Time as measured by a frame moving relative to the object (s) t0 = Time measured within the frame of reference (as if the frame were stationary) Example A train driver sneezes just as the train passes a station platform. A person on the train measures the time interval for the sneeze to be 1.000 seconds. What time will an observer standing on the platform measure this sneeze to be if the train was traveling at: a) 50% of c b) 2.95 x 108 ms-1 a) tv = tv = to v^2 1 c^2 1.000 1 (0.5)^2 = 1.155 s b) tv = to v^2 1 c^2 tv = 1.000 1 (2.95 10^8)^2 (3 10^8)^2 = 1.155 s Mass Dilation Example Calculate the mass of an electron at the following speeds if its rest mass is 9.109 x 10-31 kg: a) 5.65 x 104 ms-1 b) 80% the speed of light c) 99.9% the speed of light a) mv = mv = mo v^2 1 c^2 9.109 10^31 (5.65 10^4)^2 1 (3 10^8)^2 = 9.109 x 10-31 kg b) mv = mv = c) 9.109 10^31 0.8 ((3 10^8)^2) 1 (3 10^8)^2 = 1.518 x 10-30 kg (This is approximately 1.7 times the rest value) mv = mo v^2 1 c^2 mo v^2 1 c^2 mv = 9.109 10^31 0.999 ((3 10^8)^2) 1 (3 10^8)^2 = 2.037 x 10-29 kg (This is approximately 22 times the rest value) 9.3 – Motors and Generators 9.3.1 – Motors use the Effect of Forces on Current-carrying Conductors in Magnetic Fields A current-carrying wire in a magnetic field experiences a force. This is the motor effect. In electric motors, electrical energy is transformed into mechanical energy. 9.3.1.1 – Discuss the Effect on the Magnitude of the Force on a Current-carrying Conductor variations in: The strength of the magnetic field in which it is located The magnitude of the current in the conductor The length of the conductor in the external magnetic field The angle between the direction of the external magnetic field and the direction of the length of the conductor. 9.3.1.S3 – Solve problems and analyse information about the force on current carrying conductors in magnetic field using: F = BIlsinØ I B Figure 1.1 – Right Hand Palm Rule The direction of the force acting on a current carrying wire is perpendicular to both the direction of the current and the direction of the magnetic field. If the fingers of the right hand point in the direction of the magnetic field and the thumb points in the direction of current, then in the force. Where: the conventional the palm points direction of the F = force on the wire in Newtons (N) B = magnetic field in Tesla (T) I = current in Ampere (A) l = length of the wire in B in meters (m) Ø = angle between the wire and B (directions determined by right hand rule) It becomes obvious then that the force acting on the wire will increase if: Magnetic field strength is increased (using permanent or electromagnets) The current is increased by increasing voltage across the wire Increase in the length of the wire within B by using multiple coils. Increasing the angle between l and B up to 90° (If the angle is 0°, then there is no F) We can use the above equation to solve problems. 9.3.1.2 – Describe qualitatively and quantitatively the force between long parallel current carrying conductors: F I1I2 K l d 9.3.1.S1 – Solve problems using the above. Where: F I1I2 K l d F = force of current carrying wires (N) between two parallel lines l = Common length of the two wires (m) K = magneticforce constant I1 = current in wire 1 or 2x10-7 NA-2 I2 = current in wire 2 (A) d = distance between two wires parallel only (m) As can be seen from the above equation, the following factors will increase the magnitude of F between the two wires: Increase current in I1 or I2, decrease distance between wires and increase the length. As can be seen below, ‘Like’ current attract and ‘unlike’ current repel. 9.3.1.3 – Define torque as the turn moment of force using: τ= Fd F = force applied ‘at the edge’(N) d = distance from turning point (m) Where: τ= torque in Newton meters (Nm) This principle was used in the junior school when dealing with the turning effect or moment of a lever. In the case of a first order lever, eg. A see-saw, where the two moments or turning effects are balanced. 9.3.1.4 – Identify that the motor effect is due to the force acting on a current carrying conductor in a magnetic field. 9.3.1.5 – Describe the forces experienced by a current carrying loop in a magnetic field and describe the net result of the forces. Initially B C b l A D +ve -ve Axle/Spindle Forces on AB only: Due to the motor effect F = BIlsin Ø. In this case F = BIl and F is into the page Forces on CD only: Similarly F = BIl where F is out of the page Forces on BC: F = 0 since Ø = 0, sin Ø = 0 Forces on AD: Similarly F = 0 B After Half Turn (180°) C B B b l D A -ve +ve Axle/Spindle Forces on AB: F = BIl and is into page Forces on CD: F = BIl and is out of the page Forces on CB and AD: F = 0 This coil as simply flop rotating. DC device to direction coil. The is a split means that the pictured will oscillated or flip rather than To convert this apparatus into a motor we need a alter the relative of current in the simplest of these ring commutator. As the coil and split ring commutator rotate, the direction of the current relative to the coil reverses each half turn (180°). This will the result in the coil rotating rather than oscillating. There is still a problem of changing magnetic force as the coil undergoes each rotation. The force on each of section AB and CD is a maximum when each section of the wire is perpendicular to the magnetic wire or when the plane of the coil itself is parallel to the magnetic field (B). If the coil part way (90°) from the diagram so that it is perpendicular to the magnetic filed then there is no force acting on the coil. Part way in between there is some force but not the maximum. This therefore creates variable torque. To overcome this issue of variable torque, many motors and generators use a radial magnetic field. This means that a uniform magnitude magnetic field (but not direction) is produced so that the force on the coil at almost any angle is also of constant magnitude (ie. Torque is constant). 9.3.1.S4 – Solve problems and analyse information about simple motors using: τ= nBIAcosθ Where: τ= torque (Nm) B = magnetic field (T) n = number of coils I = current (A) A = area of coil (m2) Θ= angle between the plane of the coil and B B C B b l A D +ve -ve Axle/Spindle Net Forces: FAB = BIlsin Θ = BIl (into) FCD = BIl (out of) Net Torque = Fd τ = τAB + τCD b b = BIl BIl 2 2 = BIlb and b x l = Area of the coil τ= BIA for 1 coil only In a typical motor there is not just one coil but many, each making its own contribution. The number of coils is shown by the symbol n. Similarly, as discussed earlier the force and therefore torque will vary as the angle between the coil and the magnetic field changes (it is a maximum as shown parallel to the field and a minimum when perpendicular). All of these factors combine for the equation: τ = nBIAcos Θ To overcome this problem of variable torque a radial magnetic field is used as mentioned earlier. 9.3.1.6 – Describe the main features of a DC motor and the role of each feature. 9.3.1.7 – Identify that the required magnetic fields in a DC motor can be produced either by current carrying coils or permanent magnets. Feature Magnets Role Coil/Coils Current (power source) Split Ring Commutator Creates B Permanent or Electromagnet Radial Magnet Carries I Coil is insulated (usually copper) Usually a large number of coil Different planes Eg. Transformer, battery etc. V I R Relative to the coil, it reverses the direction of the current each half turn Brushes Axle/Spindle Armature/Stator Completes the circuit I from power to coil Made from wire, often graphite Provides rotation for coils (low friction) Stationary – motor body and magnets Rotating frame made of ferromagnetic material. 9.3.2 – The relative motion between a conductor and a magnetic field is used to generate an electrical voltage. 9.3.2.1 – Outline Michael Faraday’s discovery of the generation of an electrical current by a moving magnet. After Oersted proved that a magnetic field could be produced from an electrical current people were sure that the reverse could also be done. This was referred to as electromagnetic induction which involves the conversion of mechanical energy to electrical energy. This discovery was landed to Michael Faraday who was successful in noticing that a changing magnetic field was necessary to produce electricity. Faraday explained electromagnetic induction as whenever there is relative motion between a conductor and a magnetic field such that field lines are ‘cut’ and emf (voltage) is induced and current will flow if provided. 9.3.2.2 – Define magnetic field strength B as magnetic flux density. 9.3.2.3 – Describe the concept of magnetic flux in terms of magnetic flux density and surface area. The number of magnetic lines of force emerging through an imaginary surface in a magnetic field is called the magnetic flux of the field. Magnetic Flux density is the magnetic flux per unit of area and is the unit of measure for magnetic field strength/intensity. The larger the flux density the more intense the field in that region. Therefore the magnetic flux density (φ) is equal to the area (A) times the magnetic field strength (B). Mathematically: φ = Bacosθ 9.3.2.4 – Describe generated potential difference as the rate of change of magnetic flux through a current. Experiments show that the size of the induced emf (potential difference) depends upon: The value of B The speed at which the conductor ‘cuts’ the flux lines; and The number of conductors. This is stated in Faraday’s Law which says that the induced emf is proportional to the rate of change of flux through a circuit. 9.3.2.5 – Account for Lenz’s Law in terms of conservation of energy and relate it to the production of back emf in motors. 9.3.2.6 – Explain that, in electric motors, back emf opposes supply emf. Lenz’s Law states that the direction of the induced emf is such that the current it produces creates a magnetic field opposing the change that produced this emf. With reference to the conservation of energy whereby energy cannot be created nor destroyed the induced current must the change that gives rise to it ie. Back emf. A back emf will be produced in the coil to oppose the motion of the coil as given by Lenz’s Law. This back emf helps limit the current in the coil and hence its speed. As the applied current increases, the motor speeds up. This produces increasing emf that is in opposition to the applied voltage. 9.3.2.7 – Explain the production of eddy currents in terms of Lenz’s Law. Eddy currents are caused by a moving (or changing) magnetic field intersects a conductor, or vice-versa. The relative motion causes a circulating flow of electrons, or current, within the conductor. These circulating eddies of current create electromagnets with magnetic fields that oppose the effect of the applied magnetic field as per Lenz’s Law. Eddy currents can be used for various applications including induction cooking and electromagnetic braking (refer to 9.3.2.S3 & 9.3.2.S4) the latter of which can be seen below. 9.3.3 – Generators are used to provide large scale power production. A generator is a device for producing electrical energy from mechanical energy (the reverse of an electric motor). 9.3.3.1 – Describe the main components of a generator. 9.3.3.2 – Compare the structure and function of a generator to an electric motor. The prime mover is the mechanical device that is used to drive the generator. It could be a steam turbine driven by steam produced by burning coal, oil or natural gas or one driven by moving water ie. A water turnbine. The other main components of a generator include the armature, a field structure which can be from an electro or permanent magnet, slip rings and brushes to take the induced current away, a spindle, a stator, and, if it is a DC generator only, a commutator. A generator can function as a motor and, similarly, a motor can function as a generator. 9.3.3.3 – Describe the differences between AC and DC generators. The essential difference between an AC and a DC generator is the nature of the connection between the rotor coils and the external circuit. In an AC generator, the brushes run on slip rings which maintain a constant connection between the rotating coil and the external circuit as can be seen in the above diagram. This means that as the induced emf changes polarity with every half-turn of the coil, the voltage in the external circuit varies like a sine wave and the current alternates direction. In a DC generator, the brushes run on a split-ring commutator which reverses the connection between the coil and the external circuit for every half-turn of the coil refer to syllabus point 9.3.1.5. This means that as the induced emf changes polarity with every half-turn of the coil, the voltage in the external circuit fluctuates between zero and a maximum while the current flows in one constant direction. 9.3.3.4 - Discuss the energy losses that occur as energy is fed through transmission lines from the generator to the consumer. Power stations are usually located large distances from major population areas (in NSW most are located near Newcastle and in the Hunter Valley; a ready coal supply and still close to Sydney and water supplies). This presents problems associated with loss of energy as the current flows through the transmission lines due to Ohmic heating. R = ρL / A ρ = RA / L = Ωm Where: p is resistivity R is resistance L is length A is the cross sectional area And V = IR and P (power) = VI combining P = I²R This is the power loss due to heating of the wire/conductor. Example Q1. A power station generates electricity at 120 KW. It sends this power to a town 10KM away through transmission lines with a total resistance of 0.40Ω. If the power is transmitted at 240V, calculate: a) I b) The voltage drop across the transmission lines c) The voltage available to the town d) Power loss a) I = P / V I = 120000 / 240 = 500 A b) V = IR = 500 x 0.4 = 200 V c) Vtown = 240 – 200 = 40 V d) P = I2R = 5002 x 0.4 = 100000 W = 100 KW Obviously, this forms problems as most of the power is lost before reaching the town. To overcome this transformers can be used to step up the voltage (V) before transmission so that a lower current (I) flows through the transmission lines and thus reducing dramatically the heating losses. Eg. If V is made 100 times larger, I becomes 100 times smaller and P becomes 10000 times smaller. In NSW the Bayswater Power Station in the upper hunter has 4 x 660 mW generators each with an output voltage of 23 kV. It is then stepped up with transformers at a transmission sub-station to 330 kV. Near the end of transmission lines it stepped down to 66 kV then to 22 and 11 kV and finally the pole transformers step down from higher values our 240 V domestic standard. 9.3.3.5 – Assess the effects of the development of AC generators on society and the environment. AC Generators What is it? Why AC? o Most widely used, o transformed/change in V to reduce I to minimise heat losses (P = I2R) Societal impacts? o Large scale electrical production, o lifestyle ease with appliances, o city development/urbanization/jobs/employment, o changes in lifestyle eg shift work, o overdependence on AC, o unskilled labour Environmental impacts? o Coal burning/CO2/greenhouse, o global warming, rising sea levels, changing climate eg droughts etc., o light/noise pollution, o issues associated with coal mining, o air/water pollution, o laying of cables, o hydroelectricity 9.3.4 – Transformers 9.3.4.1 – Describe the purpose of transformers in electrical circuits. Transformers are devices that change (increase or decrease) an AC voltage. They may be designed to be step up transformers (eg in CRT sense) and step down transformers (eg computers, cordless phones etc.) They consist of two coils of insulated copper wire – the primary and secondary wrapped around a soft iron core. See figure 8.18 pg 144. They are designed so that when an AC flows in the primary coil (due to an external AC voltage source) constantly changing magnetic flux is created which induces an AC voltage at the terminals of the secondary coil. The factor determining any changes in voltage is the ratio between the number of turns/coils in the primary relative to the secondary. 9.3.4.2 – Compare step-up and step-down transformers. Step–up transformer – Here the voltage in the secondary coil is greater than that in the primary. VS > VP (nS > nP where n is the number of coils) Primary – 12V Secondary – 24V VP Vs This is determined by the number of turns in each of the coils. Step-down transformer – As above however vice versa. 9.3.4.3 – Identify the relationship between the relationship between the number of turns in the primary and secondary coils and the ratio of primary to secondary voltage. 9.3.4.S2 – Solve problem and analyse information about transformers using: Vp np Vs ns Where: VP = Voltage in primary coils (V) “Input” VS = Voltage in secondary coils (V) “Output” nP = Number of turns in primary coil nS = Number of turns in secondary coil This equation applies to an ideal transformer assuming that all the energy/power in the primary coil is transferred to the secondary coil. Therefore in a step-up transformer not only is Vp < VS but nP < nS. Example Q1. The transformer in an electric piano converts 240V AC into 12V AC. If there are 30 turns in the piano coil, how many in the primary? Vp np Vs ns 240 np 12 30 12nP = 7200 nP = 600 Q2. A neon sign needs 12kV to operate. If the primary coil has 100 turns and mains electricity is supplied, how many turns in the secondary coil? Vp np Vs ns 240 100 12000 ns 240nS = 1200000 nS = 5000 9.3.4.4 - Explain why voltage transformations are related to the conservation of energy. Using the law of conservation of energy we know that in theory the total energy input must equal the total energy output. The equations and derivation below show how this conservation of energy can be applied to modify the equation from the previous syllabus point. P = VI W P= t W = Pt Where: W = Work P = Power W = VIt or E = VIt Energy in primary (p) and secondary (s) EP = ES VPIP = VSIS Vp Is np Vs Ip ns Example Q1. A power station generates electric power at 120 kW. It sends this power to town 10 km away through transmission lines that have a total resistance of 0.40Ω. If the power is transmitted at 5.00 x 105V, calculate: a) the current in the transmission lines P = IR 120000 I= 500000 = 0.24 A b) the voltage loss V = IR = 0.24 x 0.4 = 0.096 V c) the voltage available to the town = 500000 – 0.096 = 5 x 105 V d) the power lost due to heating P = I2R = 0.242 x 0.4 = 0.023 W 9.3.4.S3 – Gather, analyse and use available evidence to discuss how difficulties of heating caused by eddy currents in transformers may be overcome. Eddy current o 3D o Solids o Conductors o Cause – induction (Lenz’s Law) o Changing B Core o Why have iron core? Reducing Eddy currents o Lamination (reduces currents to one plane) o Ferrite Material (intensifies B but not very conductive) Cooling 9.3.4.5 – Explain the role of transformers in electricity sub-stations. 9.3.4.S4 – Gather and analyse information to discuss the need for transformers in the transfer of electrical energy from a power station to its point of use. There is usually a substation in each regional center and in each municipality of a city. Here the voltage is stepped down to values of 11 kV and 22 kV using transformers. Power substations can perform three tasks including: Step down the voltage using transformers Split the distribution voltage to go in different directions Enable, using circuit breakers and switches, the disconnection of the substation from the transmission grid to be switched on and off. Transformers are used for the transfer of energy from the power station to the point of use as it assists in overcoming the problems faced with power losses due to heating and transmitting electrical energy at low voltages over long distances. As mentioned above, by increasing the voltage transmitted the current decreases hence reducing the power loss as Ploss = I2R. Using transformers enables electricity to be supplied over large distances without wasting too much energy. If transformers were not used power station would have to be built in towns and cities or the users of electricity would have to live near to the source leading to increased societal pollution and increased health issues. Power Station Approx. 20000 V Step-up Substation High V Transmission Wires to Grid Approx. 250000 V Step-down Home/Household 240 V Step-down Town/Suburb Approx. 11000 V Step-down District Area Distribution Approx. 132000 V 9.3.4.6 - Discuss why some electrical appliances in the home that are connected to the mains domestic power supply use a transformer. Many appliances inside the home still require a transformer since they work more efficiently and/or a designed to operate at a voltage different to the 240 V AC domestically supplied (this 240 V is the RMS value, it actually fluctuates between ±339 V). CRT TV’s have a picture tube that requires a voltage of about 1500 V to operate. It therefore contains a step-up transformer. Most electronic circuits (involving microprocessors, transistors etc.) use voltages between 3 and 12 V DC. Such devices have a step-down transformer and also incorporate a rectifying unit to convert AC into DC. For example, laptops, cordless phones etc. 9.3.4.7 - Discuss the impact of the development of transformers on society. This syllabus point fits closely to 9.3.3.5 on AC supply. What is a transformer? o Step-up/step-down o AC o Primary/secondary coil o Faraday/Induction o Eddy currents o Annotated diagram o Pp = Ps Vp np o Vs ns Pros o Reduce heating loss in transmission o P = I2R o Power to remote location o Don’t need as many power generators near to town centers (less pollution) o Cost/Economics on society o Appliances need high or low V and even DC can be used Cons o Extra danger with high V o Extra danger with AC – affect heart rhythm o Still heating losses in wires o Heating losses in metal towers due to induction o Aesthetic spoiling of environment o Loss of unskilled labor/reliance on TV and multimedia/reliance on electricity 9.3.5 – AC Motors 9.3.5.1 - Describe the main features of an AC electric motor. There are two types of electric motors – the universal motor (which is essentially similar in construction to a DC motor and can run on either AC or DC) and the induction motor. Both are described in more detail from page 160-165 in our text. The Induction Motor – consists of a stator which is a series of wire coils wound on soft iron cores that surround the rotor. They provide usually three pairs of opposite radial fields. They are connected to a central external power supply in such a way that the magnetic field polarity B rotates at constant speed in one direction (uses either the three phases of power supply or manipulates the one phases using capacitors). The rotor consists of coils wound on a laminated iron armature mounted on an axle. These rotor coils are not connected to an external power supply and there is no need for commutators or brushes. Eddy currents are induced in the rotor coils by the rotating magnetic field or the stator. The eddy currents in turn produce magnetic fields which interact with the rotating stator fields to produce a torque on the rotor in the same direction as the rotation of the stator field. The rotor coils are often simplified to single copper bars (which can carry a large current) embedded in the surface of the armature connected at the ends by a ring or disc of copper which allows current to flow in a loop between opposite bars. This physically resembles a ‘squirrel cage’ (like the exercise unit for mice). As mentioned earlier, because there is no need for an external current to be supplied to the rotor, there is no need for slip-rings, brushes or commutators. This simplifies the motor, reducing maintenance, wear and tear, etc. The motor works only on AC and rotates at a constant speed equivalent to the speed/frequency of the AC supply (50 Hz or 3000 rpm). If faster or slower speeds are needed, gear or pulleys are used. The simplicity and reliability of the induction motor greatly outweighs the limitation of speed and it is estimated that AC induction motors make up 95% of all motors. (see figures 9.3, 9.5, 9.6, 9.7, 9.8 and 9.9 in text). 9.3.5.S2 - Gather, process and analyse information to identify some of the energy transfers and transformations involving the conversion of electrical energy into more useful forms in the home and industry. Electricity is such a widely used form of energy since it can be so easily converted into other energy forms. Electrical energy is transferred from the primary to the secondary coil in a transformer. Similarly it is transferred when it passes into any electrical conductor such as a lead etc. It is then transformed (changed) into another useful form of energy. The electrical devices which do this do so with variable ease and efficiency. In Home Electrical energy is transformed into radiant energy. Light in light globe Heat in toaster and kettle Microwaves in microwave ovens Radio/microwaves in In Industry Electrical energy is transformed into radiant energy. Light in laser circuit printing Heat in induction ovens Microwaves in wood curing Radio/microwaves in communications Electrical energy is transformed into mechanical energy. Rotation in a food blender (motor) Vibration in a stereo system Electric motor in lawn mower Electrical energy is transformed into chemical energy. Recharging batteries communication X-rays in medical Electrical energy is transformed into mechanical energy. Rotation in industrial motors Vibration in TV speaker Kinetic energy and GPE in fun park rides Electrical energy is transformed into chemical energy. Electro plating/electrolysis 9.4 – From Ideas to Implementation 9.4.1 – Cathode Rays to TV 9.4.1.1 – Explain why the apparent inconsistent behaviour of cathode rays caused debate as to whether they were charged particles or electromagnetic waves. In 1855 a German physicist Geyssler refined a vacuum pump capable of lowering the pressure inside a glass tube to 0.01% (a near vacuum). His friend Plucker placed electrodes at either end and applied a high voltage and produced a green glow in the tube. This green glow appeared to be coming from the glass at the anode (opposite the cathode). Debate occurred as to what caused these fluorescents and in fact to the properties of the cathode rays as they became known. Further experiments by a variety of scientists but in particular William Crookes (1875) are summarised below: [Figure 10.2 pg 171] As mention there was debate as to whether cathode rays were electromagnetic waves or streams of charged particles. The properties of cathode rays were: [Dot points on pg 180] 9.4.1.2 – Explain that cathode ray tubes allowed the manipulation of a stream of charged particles. Most of the experiments mentioned above and in our practicals (9.4.1.S1 and 9.4.1.S2) involved the manipulation of the charged particles (cathode rays). They could be affected by: Pressure of the gas inside the discharge tube They could be blocked by pieces of metal etc. (eg. Maltese Cross) or restricted to a narrow beam by a slit in the metal Can be view more readily using fluorescent screens Deflected by a magnetic field (B) (Cathode rays/electrons travel in the opposite direction to conventional current) Cathode Ray 1 X X X X X X X x x x x x x x x x x x x x x x x x x x x x x x x x x x x Cathode Ray 2 x x x x x x x x x x x x x x x x x x x x x Deflected by an electric field - ve Cathode Ray 1 E + ve It is also possible to initiate chemical reactions with cathode rays eg. Darkening of silver salts. 9.4.1.3 – Identify that moving charged particles in a magnetic field experience a force. As we can see from the previous syllabus point, electrons and other charged particles eg. Protons, have a force exerted on them by a magnetic field (B). This is a maximum when the velocity (v) and B are perpendicular to each other and is zero when they are parallel. We can express this mathematically as: F = qvBsinθ Where: F = Force (N) direction from right hand palm rule q = Charge (Coulombs [C]) v = Velocity of the charged particle (ms-1) B = Magnetic field strength (Tesla [T]) θ = Angle between B and v Example: An electron of charge -1.6 x 10-19 C is projected into a magnetic field of strength 2.0 x 10-2 T into the page while its velocity is 2.5 x 104 ms-1 due east. Draw a diagram to show the path of the electron and calculate the force acting on the electron. If this force is exerted for 1.00 seconds, calculate the final velocity of the electron as it leaves the field at mass 9.109 x 10-21 kg. Electron @ -1.6 x 10-19 C v = 2.5 x 104 ms-1 X X X X X X X X x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x B = 2.0 x 10-2 T F = qvB = (-1.6 x 10-19) x (2.5 x 104) x (2.0 x 10-2) = -8.0 x 10-17 N South F = ma 8.0 x 10-17 = (9.109 x 10-21) x a a = (8.0 x 10-17)/(9.109 x 10-21) = 8.78 x 1013 ms-2 v = u + at = 0 + 8.78 x 1013 = 8.78 x 1013 ms-1 Similarly, the magnetic field would have to be huge in dimensions if the particle were trapped for one second horizontally as at first assumed. 9.4.1.4 – Identify that charged plates produce an electric field. d E E = V/d Where: E = Electric field strength (NC-1) V = Potential Difference/Voltage (V) d = Distance between plates (m) Also from earlier work, Where: q = Charge (C) F = qE V And, W = Fd and therefore, W = qEd Where: W = Work done on or by the charge (J) F = Force (N) d = Distance F is applied in the same direction as the force (m) When work is done on a charge, it will acquire energy – in this case KE (EK = ½ mv2). Combining these equations then, for a charge entering an electric field due to a parallel plate capacitor, it experiences a force and gains energy given by W = qV = ½ mv2. Example: Find E between two parallel plates separated by 5.0 mm if the V applied is 48 Volts. E V d E 48 0.005 E = 9600 NC-1 How much work is done moving a charge of 3.6 micro Coulombs through a potential difference of 15 Volts. W = qV W = 3.6 x 10-6 x 15 W = 0.000054 J Two parallel plates are 5.0 mm apart with a V of 200 Volts across them. A small object of mass 1.8 x 10-12 kg and charge of 12 micro Coulombs is released from rest near the positive plate (ignore gravitational effects). Calculate the velocity gained as it strikes the negative plate. Example: Two parallel plates 10 cm apart with a potential difference of 20.0 V between the plates. Calculate the electric field, if a charge of +2.0 x 10-3 C is placed in the field calculate the force acting on it, if this charge is released from rest near the positive plate, calculate it’s acquired maximum velocity and if this same charge is travelling horizontally at 750.0 ms-1 near the positive plate and the plates are 25.0 cm in length, calculate the path (final velocity) of the charge and determine whether it passes through the plates. E V d qV = ½ mv2 E 20 0.1 0.002 x 20 = ½ x 1.8 x 10-12v2 4.44444444444 10^10 E = 200 NC-1 v= F = qE v = 2.11 x 105 ms-1 towards negative plate F = 2.0 x 10-3 x 200 v2 = u2 +2as (2.11 x 105)2 = 0 + 2 x 0.1 x a F = 0.4 N a = 2.23 x 1011 ms-2 v = u + at 2.11 x 105 = 0 + 2.23 x 1011t t = 9.47 x 10-7 s Range = uxt = 750 x 9.47 x 10-7 = 7.1 x 10-4 m Therefore the charge would descend within 7 mm of entering the electric field. 9.4.1.5 – Describe quantitatively the force acting on a charge moving through a magnetic field. F = qvBsinØ [See 9.4.1.3] Example: A charge of 5.25 mC moving with a velocity of 300 ms-1 northeast enters a uniform magnetic field of strength 0.310 T directed vertically downwards into the page. Calculate the magnetic force on the charge. F = qvBsinØ F = 5.25 x 10-3 x 300 x 0.310 x 1 F = 4.88 x 10-1 N Northwest Example: An electron travelling at an unknown speed enters a magnetic field of strength 0.666 T east and experiences a force of 3.80 x 10-7 N initially south. Calculate the magnitude and direction of the electron velocity. F = qvBsinØ 3.80 x 10-7 = 1.602 x 10-19 x v x 0.666 v = 3.57 x 1012 ms-1 Out of the page* *Physical impossibility as it is faster than c 9.4.1.6 - Discuss qualitatively the electric field strength due to a point charge, positive and negative charges and oppositely charged parallel plates. [Pg 173 in text] 9.4.1.7 – Describe quantitatively the electric field due to oppositely charged parallel plates. See earlier notes. E = V/d or V = Ed 9.4.1.8 – Outline Thomson’s experiment to measure the charge:mass ratio of an electron. In the 1890’s, J. J. Thomson carried out a series of brilliant experiments to investigate the properties of cathode rays and in particular to design an intricate experiment to measure the charge to mass ratio of these cathode rays (or electrons as they become known as). As discussed earlier, Thomson was able to cause cathode rays to be deflected by electric fields produced by parallel plate capacitors. He then went on to examine the effect of B and E in combination and alone – in doing so was able to derive an expression for q:m or q . m He did this is two stages: 1. By combining and adjusting magnetic field and electric field he was able to balance B and E so that the cathode rays passed through without being deflected. This meant that FE = FB. Therefore FE = qV . . . . . . qE or and Cathode Ray e d . . . . . . E FB = qvBsinØ B . . . . . . Pass through undeflected + 2. Thomson then removed the E while maintaining the same B This means if B was strong enough the path of the cathode ray/electron would be circular and that the force due to B (FB = qvBsinØ) is supplying the centripetal force (Fc = mv2/r). By combining equations from both experiments he was able to determine a mathematical value for q/m (charge to mass ratio). From experiment 1: FE = qE FB = qvBsinØ (Ø = 90º) But FE = FB Therefore, qE = qvB V = E/B From experiment 2: FB = qvB FC = mv2/r But FB = FC Therefore, qvB = mv2/r q/m = v/Br But from experiment 1: V = E/B Therefore, q/m = E/B2r E can be calculated/measured (E = V/B). Similarly B can be calculated/measured. r can be measured from the curvature from experiment 2. Thus, Thomson could calculate the charge to mass ratio (q/m or q:m). Thomson found exactly the same ratio regardless of the material the cathode was made from and also that this value was incredibly large (1.76 x 1011 C/kg-1). This value was around 1800 times greater than the ratio obtained by using the lightest know element and producing irons by electrolysis (H+). This means that it is likely to be smaller (less mass) than the lightest known element and since it was common to all cathode materials it must be a subatomic particle. Thomson is therefore accredited for discovering the electron. *Outside course* It was not until 1909 that an American physicist Robert Milicon, performed an experiment to actually calculate the charge on an electron (qe). Here he balanced the electrical force (FE = qE) on an oil drop with gravity (Fg = mg). Equating these two equations mg = qE and q = mg/E. E and g could or were known and m could be calculated using diameter, density and fluid physics equations. This allowed Milicon to calculate the charge on one oil drop. He repeated this for thousands of oil drops and found each charge was a multiple of 1.6 x 1019 C – this was accepted as the charge on an electron. 9.4.1.9 – Outline the role of – the electrodes in the electron gun, the deflection plates or coils, the fluorescent screen – in the cathode ray tube of conventional TV displays and oscilloscopes. [Excel textbook] 9.4.2 – Black Body Radiation/Photoelectric 9.4.2.1 – Describe Hertz’s observation of the effect of a radio wave on a receiver and the photoelectric effect he produced but filed to investigate. 9.4.2.2 – Outline qualitatively Hertz’s experiments in measuring the speed of radio waves and how they relate to light waves. Early history – Maxwell’s theory of EM waves Based on observations that a changing magnetic field produced/induces and electric field around a magnet and that a current carrying wire induces a magnetic field, Maxwell concluded an interrelationship electric and magnetic fields. The following sequence explains how an EM wave propagates: A time varying electric field in one region produces a time and space varying magnetic field at all points around it. This varying B similarly produces a varying E in it’s neighbourhood Thus, an EM disturbance starting at one location (eg. Vibrating charges in a hot gas or in an antenna) creates a disturbance which travels out through mutual generation of E and B The E and B propagate through space in the form of an EM wave (ie. Self propagating) Hertz’s Experiments with Radio Waves Two of the more important predictions of Maxwell and his equations were: EM waves could exist with many different frequencies All EM waves would propagate through space at the speed of light In 1886, Heinrich Hertz conducted a series of experiments verifying these predictions. Hertz reasoned he should be able to produce EM waves of different f values by creating rapidly oscillating electric field with an induction coil. He used the induction coil to produce sparks between the spherical electrodes of the transmitter (see figure 11.6/7/8). He observed that when a small length of wire was bent in a loop with a small gap between the ends (and held in the right plane), a spark would jump across this detector loop even though this loop was not connected to any source of current. Hertz concluded that electromagnetic waves had travelled from the transmitter to the detector. He then showed that these EM waves could be reflected from a metal mirror and refracted as they passed through a prism made from pitch ie. They had wave properties not light. He was also able to show that these ‘new’ EM waves could be polarised. If the detector loop was perpendicular to the transmitter gap, the radio waves produced no spark; however if it was placed parallel to the spherical electrodes of the transmitter the spark was at a maximum. See figures 11.7 (perpendicular, no spark) and figure 11.8 (parallel, maximum spark). Hertz reasoned that if the EM waves travelled from the transmitter to the detector that there should be a small delay between the transmitted spark and the detector spark. He measured this speed in 1888 using a predetermined frequency from an oscillating circuit and determined it’s wavelength by measuring it’s interference effects (refer to laser refracting experiment done later). These measurements allowed Hertz to calculate the velocity of this wave (v = ƒλ) and he achieved an answer identical to the speed of light. Hertz’s experiments and procedures were the first involving EM waves of a know frequency. His waves are now classified as radio waves. Hertz never transmitted his radio waves over more than a few hundred meters. It was not until the later work of Marconi (late 1890s to early 1900s), that radio waves became used in communication. He found that longer wave lengths penetrated further than short wave lengths and that tall aerials were more effective for producing the highly penetrating radio waves (long wave length) than short aerials. (1895 sent a radio message about 3 km and by 1901 across the Atlantic from England to Canada). Different radio frequencies can be generated easily and precisely by oscillating electric currents (AC) and the length/s of the aerial. Hertz also noticed the photoelectric phenomena but did not carry out any further research on it. He noted that if the detector loop was darkened/enclosed then the maximum spark length became decidedly smaller (ie. Illuminating the spark gap with UV light gives stronger sparks). 9.4.2.3 – Identify Planck’s hypothesis that radiation emitted and absorbed by the walls of a black body cavity is quantised. When an object such as a light filament is heated it glows with different colours (black, red, yellow, blue/white), as it gets hotter. To explain the emission of radiation from such objects explaining how the landa/frequency, vary with temperature, we use objects called black bodies. [] The modification was seen by Planck as a small correction to a ‘classical’ thermodynamics but it finished up being one of the most significant steps in the development of a new branch of physics – quantum theory. Essentially he had developed a mathematical ‘trick’ to explain the results of black body radiation and he failed to accept the quantisation of radiation until much later in his career. Classical physics in broad terms is described as the physics up until the end of the 19th century. It relied on Newton’s mechanics and included Maxwell’s theories of electromagnetism (This is still largely of use today for typical situations encountered on earth). Classical physics predicted that the emission of EM radiation is continuous were as quantum physics are stated that the energy occurred in discreet packets or quanta. 9.4.2.4 – Identify Einstein’s contribution to quantum theory and its relation to black body radiation. As mentioned earlier, Hertz was the first to notice this photoelectric effect when studying the spark in the detector loop. In 1899, J. J. Thompson set up some apparatus similar to that shown in figure 11.14. He found that UV light caused electrons to be emitted from the metal surface of the cathode (in this case zinc). The new feature of this experiment was that the electrons were ejected from the metal by the radiation rather than the strong electric field used in cathode ray tubes. It was shown that these ejected particle (electrons) were identical to cathode ray particles. In 1902, Von Leonard studied how the energy of emitted photoelectrons varied with the intensity of light used. He observed that: Doubling the light intensity double the number of electrons emitted (greater current) There was no change in the maximum KE when the light intensity was increased (measured by the voltage required to stop the electrons reaching the other electrodes [Vstop]) The maximum KE of the electrons depended on the frequency of the light illuminating the metal. Classical physics was unable to explain these results. It predicted that energy from the light could be absorbed slowly over time until the electrons gain sufficient energy to leave the metal surface. The existence of a threshold frequency (f0) could not be explained. The table below summarises the classical predictions and the actual experimental results (which Einstein’s model was able to explain). Einstein successfully explained the photoelectric effect using Planck’s theory in which the particles of light (photons) carried energy that could be transferred to matter. He proposed that: 1. Light exists as photons, each with an energy (E = hf) 2. Light intensity depends on the number of photons 3. Photons with the highest energy correspond to light with the highest frequency 4. *To produce the photoelectric effect (freeing an electron from the surface of the metal) the energy in the photon must be equal to or greater than the energy holding the electron to the surface. The energy require to release an electron from the surface is called the work function (value is different for different metals) 5. *If the energy of the photon is greater than the work function, then this difference provides kinetic energy to the photoelectron (ie. A high frequency photon is more likely to pass on a higher difference in energy and therefore the photoelectrons released will have a higher maximum kinetic energy. This KEmax can be measured by changing the voltage across the plates until the flow of photoelectrons just ceases [this is known as the stopping voltage or Vstop]) Example: Q1. A beam of monochromatic light (lambda = 5.0 x 10-7 m) hits a perfect black body and imparts 0.10 mW of power to it. Calculate: a. f b. Energy per photon c. Number of photons striking per second a. c = fλ 3.8 x 108 = f x 5.0 x 10-7 f = 6.0 x 1014 Hz b. E = hf E = 6.63 x 10-34 x 6.00 x 1014 = 3.98 x 10-19 J c. P = 0.10 mW = 1 x 10-4 W P = W/t W = Pt = 1.0 x 10-4 x 1 = 1.0 x 10-4 J 1.0 10^4 3.98 10^19 = 2.52 x 1014 Therefore, No. Photons = Q2. a. f = 7.0 x 1015 Hz KEmax = 9.0 x 10-19 J W = KEmax W = qVstop 9.0 x 10-19 = -1.6 x 10-19 x V V = -5.62 Volts (to repel photoelectrons) b. E = hf = WF + KEmax hf = WF + KEmax 6.626 x 10-34 x 7.0 x 1015 = WF + 9.0 x 10-19 WF = 3.74 x 10-18 J c. hf0 = WF (threshold frequency = f0) 3.74 10^18 f0 = 6.626 10^34 = 5.64 x 1015 Hz 9.4.3 – Semiconductors and Transistors 9.4.3.1 – Identify that some electrons in solids are shared between atoms and move freely. Different materials vary greatly in their ability to conduct electricity (as a solid). Their conductivity depends on the ease with witch the electrons are able to move within the crystal lattice or arrangement of atoms/ions. Covalently bonded substances (usually between non-metallic elements) involve the sharing of electrons. These electrons are therefore held tightly and are therefore usually unavailable for electrical conductivity/electron flow. Such substances (which do not conduct electricity easily) are referred to as electrical insulators. Ionic compounds for an intricate lattice held together by the attraction of positive and negative ions (the metals loose electrons to form positive ions while the non-metals gain electrons to form negative ions). Ionic compounds will conduct electricity when molten or dissolved because the ionic lattice is broken and the ions are free to move. However, in the solid state the ions are fixed and so no conductivity occurs. Metals, on the other hand, consist of an orderly array (a lattice) of positive metal ions surrounded by the de-localised valance or other electrons. These de-localised electrons or sea or electrons are able to move randomly throughout the lattice and will display a net movement when an electric field (E) or voltage is applied. This means that metals are good conductors of electricity. [Refer to figure 12.2, pg 209] 9.4.3.2 – Describe the difference between conductors, insulators and semiconductors in terms of band structures and relative electrical resistance. When atoms of any type of substance (conductors, semiconductors and insulators) are very close together as in a solid, there highest electron energy levels (valance levels) form a band called the valance band. In a conductor, this valance band is only partly filled and the valance band and conduction band above overlap. In an insulator, electrons have been shared to completely fill the valance band and so there is a very large energy gap between this valance band and the conduction band (energy level needed to ‘escape’ the atom and therefore conduct). In the case of semiconductors, there is an energy gap between the valance band and conduction band but it is much smaller than the case with the insulator. [figure 12.5,Band table Conduction 12.1]. Conduction Band Valance Band Valance Band Energy Gap Valance Band With metals, increasing temperature increases resistance (decreases conductivity). Increasing temperature increases lattice vibration which ‘upsets’/restricts the flow of electrons. In the case of semiconductors, however, an increase in temperature results in a decrease of electrical resistance (increase in conductivity). The increase in temperature allows some electrons to move across the energy gap from valance to conductive and once there are free to move under the influence of E. At low temperatures eg. Absolute zero (273ºC) a semiconductor would act like an insulator. 9.4.3.3 – Identify absences of electron in a nearly full band as holes, and recognise that both electrons and holes help to carry current. The semiconductor material widely used today is silicon (historically germanium was the first used). Silicon is in group 4 of the periodic table and has 4 outer (or valance) electrons. To achieve chemical stability the silicon covalent bonds with neighbouring silicon atoms to produce a network covalent solid (an intricate lattice). .. .. .. : Si : Si : Si : .. .. .. .. .. .. : Si : Si : Si : .. .. .. 9.4.3.5 Early diodes and transistors were made from germanium because suitable industrial techniques were developed to purify germanium to the ultra pure levels required to act as a semiconductor during WWII. The major problem with Germanium, however, is that it becomes too good of a conductor when it gets too hot – allows too much current to pass through which can damage the electronic equipment. (The resistance to electric current flow that makes a semiconductor useful also generates heat). Silicon was the other element with semi conducting properties predicted to be ideal as a semiconductor. Silicon in fact is quite common (most sand is made of Silicon Dioxide [SiO2]). It is however much more difficult to purify and extract. However, once Silicon is purified to the ultra pure level needed, it is affected much less by higher temperatures while maintaining it’s performance level. The first transistors were made in 1957 by Gordon Teal (Texas Instruments). From the 1960’s onwards Silicon has become the material of choice. 9.4.3.6 – Describe how ‘doping’ a semiconductor can change its electrical properties. 9.4.3.7 – Identify differences in p and n-type semiconductors in terms of the relative number of negative charge carriers and positive holes. As mentioned earlier, pure semiconductors contain the precise number of electrons to fill their valance band. They can only conduct if electrons are introduced into the conduction band or are removed from the valance band to produce holes (intrinsic semiconductors). To enhance the conductivity of a semiconductor, a tiny amount of an impurity atom can be introduced into the crystal lattice to alloy with the material. This is called ‘doping’. There are two types of extrinsic semiconductors – n type (negative) and p type (positive). n-type These are formed when a group 5 impurity atoms such as Phosphorus and Arsenic is substituted into the crystal lattice. Group 5 elements have 5 valance electrons (compared to Silicon which has 4); 4 of the 5 Arsenic electrons will pair up with corresponding Silicon atoms in the lattice but this leaves 1 ‘extra’ electron left over. This extra electron is easily promoted to the conduction band. This increases the number of negative charged carriers in the conduction band and such semiconductors that produce excess of negative charge carriers are called n-type semiconductors. (figure 12.8 pg 215) p-type Here an atom from group 3 such as Boron or Gallium is substituted into the crystal lattice. This atom has 3 valance electrons which means that there is one electron short in the lattice thus creating a hole. 9.4.3.8 – Describe the differences between solid state and thermionic devices and discuss why solid state devices replaced thermionic devices. In electrical devices and circuits there is often a need to control the direction of current flow by either converting AC to DC (rectifying); switch current flow on or off or amplify a current. Prior to the invention of semiconductors and the associated solid state devices, thermionic devices accomplished the task. Thermionic devices use heated filaments and terminals/electrodes set in vacuum tubes (eg. The valves in old radio sets). The filament in the vacuum valve is heated by an electric current, causing it to liberate electrons and act as a cathode (negative electrode). These electrons are then accelerated by a high potential difference towards an anode. The simplest of such a valve is the diode shown below. (Figure 12.11 pg 216) The cathode may be heated directly (with current flowing through the cathode) or indirectly (with a separate filament). With the negative terminal of the high voltage source attached to the cathode, electrons can flow through the diode creating an electric current. If this battery or high voltage source were connected in reverse, no current would flow. Such ‘unidirectional conduction’ makes a diode suitable as an ‘electronic switch’ and for converting AC into DC current (rectification). Later scientists (around 1900) added a positively charged plate or grid within the vacuum tube which enabled the apparatus to amplify the input signal (known as a triode). [Put Ron’s notes here] Electrons can only flow in one direction using such combinations when an electric field is applied – they can only move from n to p (with an n semiconductor, there is an excess of electrons which can flow across the junction into the p-type to fill the positive holes available when the field is applied; electrons cannot flow from the p to the n). This ensures current flows in one direction only so that a simple p/n combination could act as a diode to ensure unidirectional current. Transistors usually are used to amplify current/readings and use two junctions, usually either npn or pnp (see pg 220 for further explanation). 9.4.4 – Superconductivity 9.4.4.1 – Outline the methods used by the Braggs to determine crystal structure. 9.4.4.2 – Identify that metals possess a crystal lattice structure. Interference/Diffraction Introduction We know from earlier prelim studies that one of the wave properties is that they undergo diffraction and show interference patterns. For example, when water waves interfere from 2 source points (coherent), the waves spread out, interfering with each other to produce a regular interference pattern consisting of regions of maximum amplitude (constructive interference) and ‘zero’ amplitude (destructive interference). For constructive interference, the waves from each source must arrive in phase; for destructive interference they must be exactly out of phase (half lambda, one and a half lambda etc.). It can be shown that that light waves undergo similar interference, producing similar patterns. Experiment/Exercise Lambda (nm) 612 555 492 428 f (Hz) x 1014 4.9 5.4 6.1 7.0 Colour Red Orange Yellow Blue Frequency vs. Back Voltage Back Voltage/Vstop (V) Back Voltage/Vstop (V) 0.349 0.552 0.777 1.141 [Insert Figure 13.3 Here] It can be mathematically shown that: n = yd L Where: n = integer (which maxima 0, 1, 2, 3, 4….) = wavelength (m) to n maxima (m) y = distance from central maxima d = distance between the two sources (m) L = distances from sources to screen (m) Example: d = 3.333333333 x 10-6 y = 0.3675 L = 1.727 n yd L n 3.333333 106 0.3675 1.727 = 7.09 x 10-7 = 709 nm Light and/or other waves either bend around an obstacle or spread into a shadow zone provided the gap or barrier is comparable to the wavelength of the wave. X-ray Diffraction X-rays were discovered by Röntgen in the 1890’s and were shown to be electromagnetic waves with wavelengths much shorter then that of visible light (in the order of 10-10 m). Within a short period of time they could be reliably produced with a specific frequency. In 1912, the German physicist Max Von Laue proposed that the regular spacing of a crystal such as sodium chloride, might form a natural 3D diffraction gradient for x-rays. His colleagues, Friedrich and Knipping obtained a diffraction pattern on 3D film when they bombarded a crystal of zinc sulphide. 9.4.4.3 – Describe conduction in metals as free movement of electrons unimpeded by the lattice. 9.4.4.4 – Identify that resistance in metals is increased by the presence of impurities and scattering of electrons by lattice vibrations. 9.4.4.5 – Describe the occurrence in superconductors below their critical temperature of a population of electron pairs unaffected by electrical resistance. The explanation for how superconductivity is occurs will be covered I the next syllabus point on BCS theory (here pairs of electrons). 9.4.4.6 – Discuss the BCS theory 9.4.4.7 – Discuss the advantages of using superconductors and identify limitations to their use. 9.4.4.S5 – 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. The major problem (limitation) with wide scale use of superconductivity is obtaining and maintaining the very low temperatures (tc) needed. Those superconductors which operate at ‘higher temperatures’ eg. Metal ceramics; do not have suitable properties for durability/longevity and is often brittle and will react with oxygen. Further research however may be able to produce superconducting materials with more of the desired metal-like characteristics, at temperatures which can be more economically achieved. There are numerous current applications and potentially limitless applications if these limitations problems are overcome. Current can potentially be infinitely stored in circular circuits (no joule heating effects [Ploss = I2R]); transport of electricity (DC) large distances without loss of energy as well. Also hug super magnets can also be produced with their own resulting applications. Some current and potential applications are listed below: Power transmission – traditional transmission lines (Copper and/or aluminium) loss an appreciable amount of energy due to the resistance of the wires and the heating effect (at this stage set as high V AC to reduce this effect). Superconducting wires could carry 5 or more time the current (DC). At this stage however the ‘high temperature’ superconductors are far too brittle to be of use. One current experimental technique is to use a ‘High Temperature’ Superconductor (HTS) (wound around a hollow helium liquid core). Power generation – could use superconducting magnets requiring a much smaller iron core (fraction of the size and mass of present generators). Potentially more efficient – uses less fossil fuels to generate similar amounts of electricity. Power storage – power stations cannot store electricity easily – most is used immediately after production. This means that power stations/generators cannot work at their optimum efficiency levels but are rather ‘turned up or down’ to meet demand. Similarly other alternative energy sources for large-scale electricity production are often inconsistent in their output eg. Solar, wind, tidal etc. Superconducting Magnetic Energy Storage (SMES) is one possible solution. The DC current produced by whichever method can be passed into the SMES’s circular interior path and flow indefinitely without energy loss until it is needed. Magnetically Levitated Trains (Maglev) – see assessment task. 2 types currently used: EMS in Germany using conventional electromagnets to lift the train above the track and EDS in Japan which uses superconducting magnets. Electronics – computers are limited to some extent by generation of heat due to resistance and the speed with which signals can be sent. Potentially superconductors may solve some of these problems. Already in use is the Josephson Junction a very fast switching device with current uses in sensitive detectors used in geology. Medical diagnosis – currently used to produce the intense magnetic fields used in Magnetic Resonance Imaging (MRI) instruments. A superconducting alloy of tin – niobium is used capable of producing magnetic fields greater to or equal to 4 T. MRI machines effectively measures the concentration of hydrogen atoms and from this a measure of the soft tissue of the patient can be computer generated. 9.8 – From Quanta to Quarks 9.8.1 – Early Models of the Atom 9.8.1.1 – Discuss the structure of the Rutherford model of the atom, the existence of the nucleus and electron orbits. Our current model of the atom has evolved over time. One of the earlier scientists to propose/discuss atoms was John Dalton. Later contributors around the turn of the 20th century included Thompson, Lenard and Rutherford. After his ‘discovery’ of the electron, J. J. Thompson went on to propose a model for the atom (commonly know as his ‘plum pudding’ model). He visualised a spherical positive mass with electrons embedded like plums in a pudding (positive and negative charges cancel out). He even made an attempt to place these embedded electrons in shell type structures to represent periodicity/electron configuration. However, experimental evidence failed to support Thompson’s and later Lenard’s models. In 1907, Ernest Rutherford (Kiwi) moved to Manchester in England and returned to investigations of the scattering of alpha particles, this time by very thin metal foils. He set work for 2 scientists/undergraduate student (Geiger and Marsden). The approximate apparatus and set up is shown below. Lead Box Scintillation scope Alpha source Think metal sheet with hole Zinc of sulfide screen UV thin sheet of metal (gold) foil Geiger and Marsden found that the majority of alpha particles passed straight through the thin foil as expected (red arrow) based on the then current theories. However, a significant number were found to be deflected at much larger angles. It was Rutherford that took these results and developed a new atomic theory to account for these deflections. He proposed that the atom must be made of almost empl 9.8.1.2 – Analyse the significant of the hydrogen spectrum in the development of Bohr’s model if the atom. The spectrum of hydrogen was well known and each spectral lines frequency and wavelength had been accurately measured and recorded. In fact, several mathematicians had derived several equations to match these measured wavelengths. Balmer derived the n2 equation: = b( 2 2 ) n 2 Where: b is a constant N is an integer > 2 This was modified by Rydberg to produce the more familiar equation: 1 1 1 RH ( 2 2 ) nf ni Where: is the wavelength of the spectral line RH is the Rydberg constant level/shell the electron finishes at nf is the electron energy ni is the electron energy shell/level that the e originally has. Bohr knew that the electrons in atoms must somehow produce the characteristic radiation of the spectrum for each element. He was also familiar with Planck’s earlier description of ‘atomic oscillators’ (see photoelectric effect). Bohr attempted to introduce the quantum ideas of Planck to the atom and once he was mad aware or Balmer’s equation, was able to explain the spectrum for hydrogen. Exercise: Calculate and compare to the accepted values the first four lines in the Balmer series (nf = 2). n.d. RH = 1.097 x 107 1 1 RH ( 1 1 ) 2 nf ni2 1.097 107 ( 1 1 ) 22 32 6.56 107 1 1.097 107 ( 1 1 ) 22 4 2 4.86 107 1 1.097 107 ( 1 1 ) 22 52 4.34 107 1 1.097 107 ( 1 1 ) 22 62 4.10 107 Calculate frequency for each wavelength and E for each result. n=6 n=5 Paschen Series (infra-red) n=4 n=3 n=2 n=1 Brakett Series (infra-red) Lyman series (ultraviolet) Pfund series (infra-red) Balmer series (visible light) Principle Quantum Number (n) Energy E (eV) 7 6 5 -0.28 -0.38 -0.54 4 -0.85 3 -1.51 2 -3.4 Pfund series Paschen series Balmer series Brakett series 1 Lyman series -13.6 9.8.1.S1 9.8.1.S2 Aim: To observe and match the spectra of hydrogen (emission) and to explain the Balmer series diagrammatically. Method: Enclosed in Box Induction Coil H2 Lamp Spectroscope 6V DC The apparatus was set up as per above. Care taken with induction coil (x-rays, high voltage, damage if not connected to 6 volt DC). Similar x-ray dangers with the hydrogen spectral tube. If necessary the H2 tube was enclosed by a cardboard box with viewing slits to eliminate background light. The spectroscopes used were made from diffraction gradients which spread the spectra when viewing and provide a scale to estimate the wavelength of spectral lines observed. The lines observed were recorded in a table and a diagram; and compared to the accepted values for hydrogen from a spectral chart or the text on pg 423 figure 22.9. Diagram/s representing the Balmer series were also drawn. Results: Observed Wavelength Red Blue Not observed Not observed 4.10 Theoretical Wavelength Red Blue Violet Violet 6.60 x 10-7 m 4.90 x 10-7 m 4.35 4.85 6.55 x 10-7 m 4.85 x 10-7 m 4.35 x 10-7 m 4.10 x 10-7 m 6.55 n=6 Principle Quantum Number (n) n=5 Energy E (eV) 7 6 5 n=4 n=3 4 n=2 3 n=1 2 Balmer series 1 Balmer series (visible light) Eg. The 1st line is the Balmer series. 9.8.3 – Nucleus to Nuclear Energy 9.8.3.1 – Define the components of the nucleus (protons and neutrons) as nucleons and contrast their properties. All particles found within the nucleus (protons and neutrons) are known as nucleons. The table below summarises and contrasts their properties: Particle Symbol/Representation Charge Mass Mass (approx) (accurate) -19 1 Proton +1 or +1.6 x 10 1.0 amu 1.673 x 10-27 kg 1p [Reflected by E and B] Neutron 1 0 n 0 [No deflection in E or B] 1.0 amu 1.675 x 10-27 kg Location Penetration Nucleus (Nucleon) Reasonably high Nucleus (Nucleon) Very high *Introductory Discovery of Radioactivity* In the 1890s, Becquerel was studying the phosphorescence of minerals (exposed to sunlight) then reemitted radiation in the dark later. On a particularly cloudy day he threw a mineral sample (without being exposed to sunlight) into a draw containing a photographic plate. The mineral produced radiation that penetrated the black paper surrounding the photographic plate (which would normally block any light and darkened the plate. The properties of these new rays seemed similar to recently discovered x-rays of Roenteng. Rutherford was able to show in 1898 that this particular mineral emitted two types of radiation (alpha and beta); and the third gamma was discovered in 1900 by Villard. As we investigated in the preliminary course, alpha, beta and gamma have their own unique properties. Name Symbol Charge Mass Penetration Ionising Ability 4 Alpha ( ) 2 He +2 4 amu Poor Excellent (due to (10-20cm air, thin size and charge) metal, paper) +1 Reasonable but Reasonable ) 01 e 0 ( Beta annihilates when [Positron] contact with normal electron 0 -1 Reasonable Reasonable 0 Beta ( ) 1 e [Electrons ] Gamma 0 0 (Wave) Very High Very Poor (lack or hf photon of mass) () 9.8.3.2 – Discuss the importance of conservation laws to Chadwick’s discovery of the neutron. After the discovery of the nucleus it seemed logical to assume that the nucleus contained protons and electrons (which would explain the emission of alpha and beta) however there were major problems with this concept and in 1920 Rutherford proposed a neutral party (neutron) with a mass comparable to a proton. In 1919, Rutherford bombarded nitrogen gas with alpha particles (from Bi-214). A positively charged particle more penetrative than an alpha was produced and later identified as a proton. This was the first artificially induced transmutation. In 1930, Bothe and Becker fired alpha particles at Beryllium and found that a highly penetrating radiation was produced. The radiation seemed to be similar to gamma but even more penetrative and had a much higher energy value (10 MeV) than any previously observed gamma ray. Frederick Joliot and his wife Irene Curie allowed this radiation to fall on a block of paraffin wax (a hydrocarbon rich in hydrogen atoms – protons. Some leading physicists were adamant that they would but others including Bohr thought otherwise (it wasn’t until 1936 that Bohr dropped his ideas of non-conservation of energy). The diagram and equations below show/summarise the overall development. Beryllium Paraffin Wax 1 0 1 0 5 Conventional Mirror n? n? source 9 4 Be 24 He 126 C 01n ? 9.8.3.3 – Define the term ‘transmutation’. 9.8.3.4 – Describe nuclear transmutations due to natural radioactivity. Transmutation is the process where a new element (daughter) is formed from a different element (parent). In the preliminary course wee covered much of this and wrote equations for alpha and beta decay. We also need to incorporate neutrinos and antineutrinos associated with beta+ and beta- decay (see later syllabus points). 4 U 234 90Th 2 He (Alpha decay) 238 42 _ 0 Bi 212 84 Po1e(electron) U (antineutrino) (Beta- decay) 212 83 O 157 N 01e( positron) (neutrino) (Beat+ decay) 15 8 Uranium-238 undergoes both alpha and beta (assume beta-) decay until a stable isotope of Pb-206 is reached. Calculate the number of alphas and betas. He-4 each decay reduces mass by 4. Total mass lost = 238 – 204 = 32 Therefore, No. of Alpha = 32/4 =8 Each alpha decay reduces charge by 2 (8 x -2 16) But each beta- increases each charge by 1 92 – (8 x alpha) + (? X beta) = 82 92 – 16 + b = 82 b=6 Elements are naturally radioactive for usually one of three reasons: Too many neutrons compared to the number of protons (neutron turns into proton and ejects an electron/beta-) Too many protons for the number of neutrons (proton turns into a neutron and ejects positron/beta+) The element is too big/heavy – elements above atomic No. 83 (Bi) are likely to undergo alpha decay Gamma radiation often accompanies either alpha or beta as the nucleus de-excites. Gamma is not a transmutation since a new element is not formed. As well as natural radioactivity, artificial transmutation can also be induced (see Rutherford’s first artificial transmutation in the previous section on the discovery of neutrons). 27 13 10 5 9 4 28 Al 12H 14 Si 01n B 24He 136 C11p Be11p 36 Li 24He As can be seen from the artificial transmutations above, alpha and beta particles can and were used to supply an understanding of physics atomic principles by probing other atoms/nucleon. The resulting protons and neutrons sometimes produced could also be used as a probing/diagnostic tool, particularly neutrons which have excellent penetration (due to its neutral charge) but still with a significant KE/momentum. 9.8.3.5 - Describe Fermi’s initial experimental observation of nuclear fission. After the discovery of the neutron, Fermi and his co-workers in Rome carried out many experiments with the neutron bombardment of many elements. They noticed that for some of the heavy elements there was often a delayed emission of a beta particle resulting in an element with a higher atomic number. Uranium (atomic no. 92) is the heaviest naturally occurring element and when Fermi’s team fired neutrons at it they were able to make the first transuranic element. In a similar fashion, Np-239 can be converted to Pu-240. 1 0 0 n238 239 239 92U 92U 93 Np1e 1 0 0 n 239 240 240 93Np 93 Np 94 Pu1e This can continue until eventually Am-241. Transuranic elements heavier than Americium cannot be made by neutron capture (nowadays this neutron capture is achieved by placing a sample of suitable material in or near a fission nuclear reactor where there is an abundance of neutrons); for elements with atomic number greater than 95, particle accelerators must be used (see 9.8.4). We now know that when Fermi and his associates bombarded Uranium with neutrons, they must have produced nuclear fission. The production of transuranic elements would also have occurred but the observations of radioactivity that Fermi assumed was solely due to transuranic element formation was in fact partly due to the production of fission products. The first observations confirming nuclear fission were made by Otto Frisch using an ionisation chamber to detect emitted particles. During the course of their neutron bombardment experiments, Fermi’s co-workers (Amaldi and Pontecorvo) discovered that the same experiment yielded different results when performed in different parts of the room (the radioactivity was much greater after the substances had been irradiated with neutrons on a wooden rather than marble table). Fermi decided to investigate this anomaly by placing a paraffin block between the neutron source and the target - he discovered that slow neutrons were much more likely to cause a reaction than fast ones (the De Broglie wavelength of slow neutrons was longer and meant there was a greater possibility of capture by the nucleus). The paraffin was effectively acting as a moderator (see nuclear reactors lately). Fermi was awarded the noble prize in 1938 for the discovery of transuranics (and associated use of slow neutrons) - it allowed he and his Jewish wife to accept the prize in Sweden, but they ‘escaped’ to the USA. Near the end of 1938, Lise Meitner, Otto Hahn and Fritz Strassmann studies the new heavier elements produced by neutron bombardment and discovered Barium as one of the products rather than the previously identified Radium. The group began to consider Bohr’s ‘liquid drop’ model of the nucleus. They considered that if the nucleus was like a liquid drop, it could become unstable and split in two, the parts forced apart but electrostatic repulsion. Using this model, the velocities of each main part should have had an energy of about 200MeV. Meitner calculated this energy using mass defects and E = mc2 and showed that an energy of about 200MeV would be released. He had proved mathematically that fission had occurred. In fact we can write an equation for this (several possibilities). Slow Neutron Begin to Nucleus 1 236 141 92 U n U Ba 36 Kr 301n 92 56separate 0 unstable Split (2 fragments) with high 235 1 236 147 87 1 velocity 92U 57 La 35Br 20 n 92U 0 n due to electrostatic Frisch returned to Copenhagen and informed Bohr of their theory just as he was to travel repulsion to the US. Frisch performed the first experiment that actually confirmed the fission of fission reached the US even before Frisch and Meitner had published reaction. News their papers and many others performed similar experiments. 235 92 9.8.3.6 - Discuss Pauli’s suggestion of the existence of neutrino and relate it to the need to account for the energy distribution of electrons emitted in beta-decay. Even though scientists confirmed that beta- decay was in fact an electron (and that therefore a neutron had decayed into a proton, in the process ejecting and electron) they knew that an electron could not possibly be confined to the nucleus (De Broglie wavelength of the electron was much larger than the radius of the nucleus). Prior to WWI Chadwick was able to show that beta particles were emitted with a wide spread of energies (figure 24.10, pg 460 in text shows these). There was a dilemma as how the same beta decay from the same nucleus (to produce the same new nucleus) with different energy values. To resolve the paradox, Wolfgang Pauli predicted there must be another sub-atomic particle - if the laws of conservation were to remain valid, the expulsion of beta particles must be accompanied by a very penetrating radiation of neutral particles which had not yet been observed. He referred to the yet unknown particle as a neutron - it was later renamed the neutrino to avoid confusion with Chadwicks neutron. By late 1933 Fermi had completed his famous paper on beta decay. He included Pauli’s suggestion of the neutrino (along with the electron) and the suggestion by Heisenberg that the nucleus contained only proton and neutrons. He proposed that the number of electrons and neutrinos was not constant and that they could be created or disappear just like photons. In a complex series of quantum aspects outside of this course he could produce the shape of the beta decay spectrum (it assumed the mass of the neutrino was zero or very close to it. By the 1950s, beta decay was believed to be associated with the weak nuclear force. This joined gravitational force, electromagnetic force and the strong nuclear force as a fundamental force of nature.