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%ODFN%RG\5DGLDWLRQ
Factors that affect how an object absorbs, emits (radiates), and reflects EM radiation incident on them:
1) Nature of the surface: material, shape, texture, etc.
2) Color:
a) Light-colored or silvery objects: absorb little energy, reflect most energy
b) Dark objects: absorb most energy, reflect little energy
When the object is in thermal equilibrium with its surroundings,
energy absorbed = energy radiated
Pin = Pout
Iin = Iout
An object that acts as a “black-body” will . . . absorb all incoming radiation, not reflect any, then radiate all of it. %ODFNERG\UDGLDWLRQ: radiation emitted by a “perfect” emitter When heated, a low-pressure gas will . . .emit a discrete spectrum
When heated, a solid will . . . emit a continuous spectrum (PLVVLRQ6SHFWUDIRU%ODFN%RGLHV 1. Not all wavelengths of light will be emitted with equal intensity. 2. Emitted wavelength with highest intensity (Ȝmax ) is related to . . . temperature.
3. Area under curve is proportional to . . . total power radiated by body 4. As body heats up, Ȝmax . . . decreases and total power . . . increases 1
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Ȝmax Į 1/T (Wien’s law)
Use the axes at right to sketch the emission spectra for a black-body radiating at low and high temperature. Be sure to label the axes and indicate which curve represents which temperature.
7KH6WHIDQ%ROW]PDQQ/DZ RI5DGLDWLRQ relates intensity of radiation to the temperature of the body
1. How does the energy radiated by an object change if its
temperature doubles?
, V7 4 3 V7 4 $ 3 V$7 4
where V=Stefan-boltzmann constant : V=5.67 x 108 2 4 P .
3D 7 4 27 o 163
2. The supergiant star Betelgeuse has a surface temperature of about 2900 K and a radius of 3 x 1011 m. a) Determine how much energy Betelgeuse radiates each second. 4 x 1030 W
b) What is the intensity of Betelgeuse’s radiation at its surface?
c) What is the intensity of Betelgeuse’s radiation at a location that is 3 x 1011 m from its surface?
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d) What major assumption was made in calculating the power radiated by Betelgeuse? That it acts as a black-body
(PLVVLYLW\ H ±ratio of power emitted by an object to the power emitted by a black-body at the same temperature.
Formula:
H
3 3%%
3 H3%% 3 HV$7 4 e) Compute the power radiated by Betelgeuse if its emissivity is measured to be only 0.90.
3. Calculate the power emitted by a square kilometer of ocean surface at 100C if its emissivity is 0.65.
4. Calculate the power radiated by the Earth if it is taken to be a) a black-body at 300 K.
b) at 300 K with an effective emissivity of 0.62.
3
6RODU5DGLDWLRQ
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1. Calculate the power radiated by the Sun if it is taken to be a black body at 5778 K and a mean radius of 6.96 x 108 meters.
2. What is the intensity of the solar radiation at the Sun’s surface?
See pg. 3 of data booklet for 1.5 x 1011 m 3. What is the intensity of the solar radiation that reaches the upper atmosphere of Earth?
Solar constant: 1360-1370 W/m2 Rounded 1400
4. How much solar energy is incident on the Earth every second?
Take solar constant and multiply by area of disc as cross-section 1.75 x 1017 W
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5. What is the average intensity of the solar energy absorbed by the Earth? Average 1.75 x 1017 W over whole surface area of Earth = 4 pi r2 340 W/m2
$OEHGR Į ±ratio of total solar power scattered to total solar power incident
D Formula:
total scattered power total incident power 3 D UHIOHFWHG 3LQ
Meaning: fraction of the total incoming solar radiation that is reflected back out into space 6. What is the albedo of a black-body? 0
What is the emissivity of a black-body? 1
7. Use the diagram at right to determine the Earth’s average albedo. Atmosphere, clouds, and ground
Global annual mean albedo on Earth: 0.30 = 30%
The Earth’s albedo varies daily and is dependent on:
1. season 2. cloud formations 3. latitude 8. How much energy is actually absorbed by the Earth each second? 0.70 x 1.75 x 1017 W = 1.23 x 1017 W
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9. Use the results of your prior calculations to estimate the equilibrium temperature of the Earth and comment on your answer. Assume black-body emiss=1 Pin=Pout Use SB T= 255 k = -18oC too cold
10. At present, the average temperature of the Earth is measured to be 288 K. a) Calculate the average emissivity of the Earth.
b) Comment on why this might be.
Actually warmer since atmosphere absorbs some of the radiation emitted by Earth surface Start to separate “Earth” into its parts – atmosphere vs. ground
6
7KH*UHHQKRXVH(IIHFW
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*UHHQKRXVH*DVHV each has natural and man-made origins
1) Water Vapor (H2O): evaporation 2) Carbon Dioxide (CO2): product of photosynthesis in plants, product of fossil fuel combustion 3) Methane (CH4): product of decay and fermentation and from livestock, component of natural gas 4) Nitrous Oxide (N2O): product of livestock, produced in some manufacturing processes *UHHQKRXVH(IIHFW±
a) Short wavelength radiation (visible and short-wave infrared) received from the Sun causes the Earth’s surface to warm up. b) Earth will then emit longer wavelength radiation (long-wave infrared) which is absorbed by some gases in the atmosphere. c) This energy is re-radiated in all directions (scattering). Some is sent out into space and some is sent back down to the ground and atmosphere. d) The “extra” energy re-radiated causes additional warming of the Earth’s atmosphere and is known as the Greenhouse Effect.
1. What is the molecular mechanism by which greenhouse gases absorb infrared radiation?
5HVRQDQFH± a transfer of energy in which a system is subject to an oscillating force that matches the natural frequency of the system resulting in a large amplitude of vibration $SSOLFDWLRQWRWKHJUHHQKRXVHHIIHFW
The natural frequency of oscillation of the molecules of the greenhouse gases is in the infrared region (1 – 300 ȝm)
2. What do the following transmittance and absorption graphs show about the atmosphere? Sun radiates in visible
Earth radiates in infrared
Water vapor absorbs incoming solar radiation and
outgoing IR radiation
CO2 absorbs outgoing IR radiation
7
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2XWJRLQJHQHUJ\
The average albedo (reflectivity) of the Earth is about 0.3, which means that 30% of the incident
solar energy is reflected back into space, while 70% is absorbed by the Earth and reradiated as
infrared. The planet's albedo varies from month to month, but 0.3 is the average figure. It also
varies very strongly spatially: ice sheets have a high albedo, oceans low. The contributions from
geothermal and tidal power sources are so small that they are omitted from the following
calculations.
So 30% of the incident energy is reflected, consisting of: x x x
6% reflected from the atmosphere
20% reflected from clouds
4% reflected from the ground (including land, water and ice)
The remaining 70% of the incident energy is DEVRUEHG: x
51% absorbed by land and water, then emerging in the following ways: R 23% transferred back into the atmosphere as latent heat by the evaporation of water, called latent heat flux R 7% transferred back into the atmosphere by heated rising air, called Sensible heat flux R 6% radiated directly into space R 15% transferred into the atmosphere by radiation, then reradiated into space x 19% absorbed by the atmosphere and clouds, including:
R 16% reradiated back into space
R 3% transferred to clouds, from where it is radiated back into space
When the Earth is at thermal equilibrium, the same 70% that is absorbed is UHUDGLDWHG: x x
64% by the clouds and atmosphere
6% by the ground
Some energy balance climate models for the Earth
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Predict increase in planet’s temp using SB law like test question
9
IB 12 6XUIDFH+HDW&DSDFLW\ &6 ±energy required to raise the temperature of a unit area of a planet’s surface by 1 K.
Formula:
Units:
4 $'7 4 $&6'7
P2 .
&6 6XUIDFHKHDWFDSDFLW\RI(DUWK CS = 4.0 x 108 J m-2 K-1
1. How much solar energy is needed to increase the surface temperature of one square kilometer
of Earth’s surface by 2 K?
7HPSHUDWXUH FKDQJHIRUPXOD:
4
$'7 3 'W $'7
&6 &6
&6 '7
(, LQ ,RXW)'W
'7
(,LQ ,RXW)'W &6
2. If the Earth is in thermal equilibrium, it will emit as much radiation as is incident on it from the Sun (344 W/m2). Suppose a change causes the intensity of the radiation emitted by Earth to decrease 10%. a) Suggest a mechanism by which this might happen. Increased amounts of greenhouse gases cause more solar radiation to be trapped in atmosphere b) Calculate the new intensity of radiation emitted by Earth. 0.90(340) = 306 W/m2
c) Calculate the amount by which Earth’s temperature would rise over the course of a year as a result.
'7
(, LQ ,RXW)'W &6
(340 306)(365)(24)(3600) 4.0 x 108 '7 2.7. '7
10
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*OREDO:DUPLQJ records show that the mean temperature of Earth has been increasing in recent years. Global mean surface temperature anomaly relative to 1961–1990
In specific terms, an increase of 1 or more Celsius degrees in a period of one hundred to two hundred years would be considered global warming. Over the course of a single century, an increase of even 0.4 degrees Celsius would be significant. The Intergovernmental Panel on Climate Change (IPCC), a group of over 2,500 scientists from countries across the world, convened in Paris in February, 2007 to compare and advance climate research. The scientists determined that the Earth has warmed .6 degrees Celsius between 1901 and 2000. When the timeframe is advanced by five years, from 1906 to 2006, the scientists found that the temperature increase was 0.74 degrees Celsius.
The global average surface temperature range for each year from 1861 to 2000 is shown by solid red bars, with the confidence range in the data for each year shown by thin whisker bars. The average change over time is shown by the solid curve.
3RVVLEOHPRGHOVVXJJHVWHGWRH[SODLQJOREDOZDUPLQJ
1. changes in the composition of greenhouse gases may increases amount
of solar radiation trapped in Earth’s atmosphere
2. increased solar flare activity may increase solar radiation 3. cyclical changes in the Earth’s orbit may increase solar radiation 4. volcanic activity may increases amount of solar radiation trapped in
Earth’s atmosphere
A column of gas and ash rising from Mount Pinatubo in the Philippines on June 12, 1991, just days before the volcano’s climactic explosion on June 15.
In 2007, the IPCC report stated that: “0RVWRIWKHREVHUYHGLQFUHDVHLQJOREDOO\DYHUDJHGWHPSHUDWXUHVLQFHWKHPLGWK FHQWXU\LV YHU\OLNHO\GXHWRWKHLQFUHDVHLQDQWKURSRJHQLF>KXPDQFDXVHG@JUHHQKRXVHJDVFRQFHQWUDWLRQV.” (the HQKDQFHGJUHHQKRXVHHIIHFW) (QKDQFHG $QWKURSRJHQLF *UHHQKRXVH(IIHFW± Human activities have released extra carbon dioxide into the atmosphere, thereby enhancing or amplifying the greenhouse effect.
Major cause: the burning/combustion of fossil fuels Possible effect: rise in mean sea-level Outcome: climate change and global warming
11
(YLGHQFHOLQNLQJJOREDOZDUPLQJWRLQFUHDVHGOHYHOVRIJUHHQKRXVHJDVHV
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7KH.HHOLQJ&XUYH: Named after American climate scientist Charles David Keeling, this tracks changes in the concentration of carbon dioxide (CO2) in Earth’s atmosphere at a research station on Mauna Loa in Hawaii. Although these concentrations experience small seasonal fluctuations, the overall trend shows that CO2 is increasing in the atmosphere.
,QWHUQDWLRQDO,FH&RUH5HVHDUFK Between 1987 and 1998, several ice cores were drilled at the Russian Antarctic base at Vostok, the deepest being more than 3600 meters below the surface. Ice core data are unique: every year the ice thaws and then freezes again, forming a new layer. Each layer traps a small quantity of the ambient air, and radioactive isotopic analysis of this trapped air can determine mean temperature variations from the current mean value and carbon dioxide concentrations. The depths of the cores obtained at Vostok means that a data record going back more than 420,000 years has been built up through painstaking analysis.
Inspect the graphical representation of the ice core data and draw a conclusion. There is a correlation between Antarctic temperature and atmospheric concentrations of CO2 0HFKDQLVPVWKDWPD\LQFUHDVHWKHUDWHRIJOREDOZDUPLQJ
1. Global warming reduces ice and snow cover, which in turn reduces the albedo. This will result in an increase in the overall rate of heat absorption. 2. Temperature increase reduces the solubility of CO2 in the sea and increases atmospheric concentrations. 3. Continued global warming will increase both evaporation and the atmosphere’s ability to hold water vapor. Water vapor is a greenhouse gas. 4. The vast stretch of permanently frozen subsoil (permafrost) that stretches across the extreme northern latitudes of North America, Europe, and Asia, also known as tundra, are thawing. This releases a significant amount of trapped CO2. 5. Deforestation results in the release of more CO2 into the atmosphere due to “slash-and-burn” clearing techniques, as well as reduces the number of trees available to provide “carbon fixation.”
Smoldering remains of a plot of deforested land in the Amazon rainforest of Brazil. Annually, it is estimated that net global deforestation accounts for about 2 gigatons of carbon emissions to the atmosphere.
12
5LVHLQ6HDOHYHOV
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Generally, as the temperature of a liquid rises, it expands. If this is applied to water, then as the average temperature of the oceans increases, they will expand and the mean sea-level will rise. This has already been happening over the last 100 years as the sea level has risen by 20 cm. This has had an effect on island nations and low-lying coastal areas that have become flooded. &RHIILFLHQWRI9ROXPH([SDQVLRQ ȕ ±fractional change in volume per degree change in temperature
Formula:
'9 E 9R '7 '9
Units:
1 .
E 9R '7
1. The coefficient of volume expansion for water near 20o C is 2 x 10-4 K-1. If a lake is 1 km
deep, how much deeper will it become if it heats up by 20o C? 0.4 m
Precise predictions regarding the rise in sea-levels are hard to make for such reasons as: D $QRPDORXVH[SDQVLRQRIZDWHU: Unlike many liquids, water does not expand uniformly. From 00C to 40C, it actually contracts and then from 40C upwards it expands. Trying to calculate what happens as different bodies of water expand and contract is very difficult, but most models predict some rise in sea level. E 0HOWLQJRILFH: Floating ice, such as
the Arctic ice at the North Pole,
displaces its own mass of water so
when it melts it makes no difference.
But melting of the ice caps and
glaciers that cover land, such as in
Greenland and mountainous regions
throughout the world, causes water to
run off into the sea and this makes the
sea level rise.
Glaciers on land melting: raise sea level Sea ice glaciers melting: don’t raise sea level 13
3RVVLEOHVROXWLRQVIRUUHGXFLQJWKHHQKDQFHGJUHHQKRXVHHIIHFW
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3. Greater efficiency of power production. To produce the same amount of power would require less fuel, resulting in reduced CO2 emissions.
4. Replacing the use of coal and oil with natural gas. Gas-fired power stations are more efficient (50%) that oil and coal (30%) and produce less CO2.
5. Use of combined heating and power systems (CHP). Using the excess heat from a power station to heat homes would result in more efficient use of fuel.
6. Increased use of renewable energy sources and nuclear power. Replacing fossil fuel burning power stations with alternative forms such as wave power, solar power, and wind power would reduce CO2 emissions.
1. Use of hybrid vehicles Cars that run on electricity or a combination of electricity and gasoline will reduce CO2 emissions.
2. Carbon dioxide capture and storage (carbon fixation) A different way of reducing greenhouse gases is to remove CO2 from waste gases of power stations and store it underground.
,QWHUQDWLRQDOHIIRUWVWRUHGXFHWKHHQKDQFHGJUHHQKRXVHHIIHFW
1. ,QWHUJRYHUQPHQWDO3DQHORQ&OLPDWH&KDQJH ,3&& Established in 1988 by the World Meteorological Organization and the United Nations Environment Programme, its mission is not to carry out scientific research. Hundreds of governmental scientific representatives from more than 100 countries regularly assess the up-to-date evidence from international research into global warming and human induced climate change.
2. .\RWR3URWRFRO This is an amendment to the United Nations Framework Convention on Climate Change. In 1997, the Kyoto Protocol was open for signature. Countries ratifying the treaty committed to reduce their greenhouse gases by given percentages. Although over 177 countries have ratified the protocol by 2007, some significant industrialized nations have not signed, including the United States and Australia. Some other countries such as India and China, which have ratified the protocol, are not currently required to reduce their carbon emissions.
3. $VLD3DFLILF3DUWQHUVKLSRI&OHDQ'HYHORSPHQWDQG&OLPDWH $33&'& This is a non-treaty agreement between 6 nations that account for 50% of the greenhouse emissions (Australia, China, India, Japan, Republic of Korea, and the United States.) The countries involved agreed to cooperate on the development and transfer of technology with the aim of reducing greenhouse emissions.
14
'LJLWDO7HFKQRORJ\ 'HFLPDO1XPEHUV
Example:
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0= 1= 2= 3= 4= 5= 6= 7= 8= 9= 10 = 11 = 12 = 13 = 14 = 15 = 16 = 17 = 18 = 19 = 20 =
452 = 4 x 100 + 5 x 10 + 2 x 1
Symbols: 0-9 Place Values: multiples of ten
%LQDU\1XPEHUV
Examples: 10011 = 1 x 16 + 0 x 8 + 0 x 4 + 1 x 2 + 1 x 1 = 19
Symbols: 0 and 1 Place Values: multiples of two
Least-significant bit (LSB) Most-significant bit (MSB)
'LJLWDOVLJQDO: potential difference is either High (1) or Low (0) $QDORJXHVLJQDO: potential difference varies continuously with time
0HWKRGRIVWRUDJH
7\SLFDO LQIRUPDWLRQ
2YHUYLHZRISURFHVV
$QDORJXHRU 'LJLWDO
Photocopying
Text or pictures
Optics and electrostatics used to fix powder to paper
Analogue
LPs (vinyl)
Music or speech
Sound variations stored as grooves in vinyl
Analogue
Cassette tapes
Music or speech
Sound variations stored in magnetic fields on tape
Analogue
Floppy disks Hard disks
All forms
Bits stored as variations in magnetic fields on disk
Digital
CD, DVD
All forms
Bits stored as series of optical bumps to be read by laser
Digital 1
0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1110 1111 10000 10001 10010 10011 10100
6WRULQJ,QIRUPDWLRQRQ&'V
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A CD is a fairly simple piece of plastic, about four one-hundredths (4/100) of an inch (1.2 mm) thick. Most of a CD consists of an injection-molded piece of clear polycarbonate plastic. During manufacturing, this plastic is impressed with microscopic bumps arranged as a single, continuous, extremely long spiral track of data. Once the clear piece of polycarbonate is formed, a thin, reflective aluminum layer is sputtered onto the disc, covering the bumps. Then a thin acrylic layer is sprayed over the aluminum to protect it. The label is then printed onto the acrylic. A cross section of a complete CD (not to scale) looks like this:
The elongated bumps that make up the track are each 0.5 microns wide, a minimum of 0.83 microns long and 125 nanometers high. (A nanometer is a billionth of a meter.) Looking through the polycarbonate layer at the bumps, they look something like this: You will often read about "pits" on a CD instead of bumps. They appear as pits on the aluminum side, but on the side the laser reads from, they are bumps. The incredibly small dimensions of the bumps make the spiral track on a CD extremely long. If you could lift the data track off a CD and stretch it out into a straight line, it would be 0.5 microns wide and almost 3.5 miles (5 km) long.
Reflection from bump or pit: Constructive interference Interpreted as 0
Reflection from edge of pit: Destructive interference Interpreted as 1 Condition for destructive interference: height of bump/depth of pit = Ȝ/4 So path difference between two light beams is Ȝ/2
Example: What wavelength of laser light should be used to read the data shown encoded at right? 600 nm
2
$GYDQWDJHVRIGLJLWDOVWRUDJHRILQIRUPDWLRQRYHUDQDORJXHVWRUDJH 'LJLWDO 4XDOLW\
Output can be virtually indistinguishable from input
5HSURGXFLELOLW\
Use of laser ensures that each retrieval is virtually identical since light does not damage surface Very high speed – different section can be accessed randomly
5HWULHYDOVSHHG 3RUWDELOLW\RI VWRUHGGDWD 0DQLSXODWLRQRI GDWD HGLWLQJ
Miniaturization techniques ensure that large quantities of data can be stored in a small device (eg. flash drives) Easily achieved with little corruption of data (eg. Photoshop)
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$QDORJXH Output can be virtually indistinguishable from input but is more liable to damage or corruption (eg. scratches on LPs) Process of retrieval often affects quality of future retrievals (eg. needle may scratch LP) Slow retrieval speed – data needs to be retrieved in sequential order Storage devices usually take up much more space All manipulation increases possibility of data corruption
,PSOLFDWLRQVIRUVRFLHW\RIHYHULQFUHDVLQJGDWDVWRUDJHFDSDELOLWLHV
0RUDO(WKLFDO
6RFLDO
x Information that is potentially problematic can easily be shared (eg. terrorism, crime) x Issues concerning ownership of electronic data (eg. piracy) x Privacy concerns x Use in documenting abuses of human rights x Unequal access to the Internet x Control of information and opinions
(FRQRPLF
x Quicker access to information needed to make economic decisions (eg. price comparisons) x Rise of new businesses and decline of older ones
(QYLURQPHQWDO
x Reduction in use of paper and other materials used traditionally to stored information x Recycling of electronic junk is problematic due to more dangerous materials used in their manufacture
3
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&KDUJH&RXSOHG'HYLFH &&' a silicon chip divided into small area called pixels. Each pixel can be considered to behave as a capacitor (a device that stores charge). The CCD is used to electronically record an image focused onto its surface. ,PDJHFDSWXUH Incident light causes charge to build up within each pixel due to the photoelectric effect. An electrode then measures the potential difference developed across each pixel and converts it into a digital signal. The position of the pixel is also stored. ,PDJHUHWULHYDO Since both the location and potential difference of each pixel are recorded, all the information needed to store and reconstruct the image is saved. Each p.d. can be converted to a digital signal and the digital signals can be converted to an image. The intensity of the light at each pixel, and thus the image, can be reconstructed.
&DSDFLWDQFH: the ratio of charge to potential difference
Formula:
C = Q/V
Units: C/V or farad (F)
Type: Scalar
1. The potential difference measured across a 100 pF pixel is 25 mV. Determine the charge and number of electrons stored in the pixel.
4XDQWXP(IILFLHQF\ ratio of the number of photoelectrons emitted to the number of
photons incident on the pixel
2. If the quantum efficiency of the pixel described above is 90%, how many photons were incident on it?
4
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0DJQLILFDWLRQ±ratio of the length of the image on the CCD to the length of the object
3. A digital camera is used to photograph an object that is 3.0 x 10-1 m2 in area. The image that is focused onto the CCD is 4.5 x 10-3 m2. What is the magnification of the camera?
5HVROXWLRQ: the ability to distinguish between two sources of light Applied to CCD: two points on an object may be just resolved on a CCD if the images
of the points are at least two pixels apart
4. The CCD of a digital camera has a square image collection area that measures 25 mm on each side and a “resolution” of 5.0 megapixels. An object that is photographed by the camera has an area of 4.6 x 10-3 m2. The image formed on the CCD has an area of 1.0 x 10-4 m2. a) Calculate the magnification.
b) Estimate the length of a pixel on the CCD.
c) Two small dots on the object are separated by a distance of 0.20 mm. Deduce whether the images of the dots will be resolved.
5
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(IIHFWRQTXDOLW\RISURFHVVHGLPDJH 4XDQWXP HIILFLHQF\
The greater the quantum efficiency, the greater the sensitivity of the CCD.
0DJQLILFDWLRQ
A greater magnification means that more pixels are used for a given section of the image. The image will be more detailed.
5HVROXWLRQ
The greater the resolution, the greater the clarity of the image and the amount of detail recorded.
Some advantages of using CCDs compared with the use of film: a) Each photo does not require film and is thus cheaper and uses less resources/produces less waste. b) Traditional film has a quantum efficiency of less than 10% while CCDs can have quantum efficiencies of over 90%. This means that very faint images can be photographed with CCDs. c) The image is digital and can be stored and edited more easily. d) Images can be viewed immediately with no processing time delay. e) Storage, archiving, and retrieving a large number of photos is easy and efficient. CCDs are used for image capturing in a large range of the electromagnetic spectrum (not just visible light).
'LJLWDO FDPHUDV
Very convenient to take and share photographs, but image quality can be less than that of traditional film unless the camera is of high quality (more expensive).
Digitized images are usually better quality than analogue images stored on magnetic videotape and are easier to store and transport. It is possible to 9LGHR continuously record video without interruption during playback. Searches are FDPHUDV faster and easier to perform. Digital storage is fast and utilizes re-usable media, an advantage for security cameras. Sensitivity of CCDs is better than traditional film and allows for detailed analysis over a range of frequencies. CCDs also allow for remote operation of 7HOHVFRSHV telescopes, both ground-based and in orbit, like the Hubble space telescope. 0HGLFDO ;UD\ LPDJLQJ
Digital X-rays have better contrast and can be processed, allowing for enhancements and detailed study. Information can be quickly shared between hospitals and more easily stored and retrieved.
6
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In the electric circuit shown below, energy is transferred from the battery to the light bulb by charges that move through a conducting wire because of a potential difference set up in the wire by the battery. The circuit shown contains a typical 9-volt battery. a) What is the emf of the circuit?
b) How much energy does one coulomb of charge carry around the circuit?
Schematic c) How much energy do two coulombs of charge carry around the circuit?
d) How much energy does each coulomb of charge have at point B?
e) How much energy does each coulomb of charge have at point C?
f) What is VB?
What is VC?
mark potentials at each spot
g) What is ǻVBC?
What is ǻVCD?
What is ǻVDA?
(OHFWULF&XUUHQW Formula: I = ǻq/ǻt
Units: A (ampere) = C/s
Type: Scalar
Unofficial definition: rate of flow of electric charge 2IILFLDO'HILQLWLRQRI2QH$PSHUH $ RIFXUUHQW±DIXQGDPHQWDOXQLW One ampere is the amount of current flowing in each of two infinitely-long parallel wires of negligible cross-sectional area separated by a distance of one meter in a vacuum that results in a force of exactly 2 x 10-7 N per meter of length of each wire. 6KRUWIRUP– Current is defined in terms of the force per unit length between parallel current-carrying conductors.
Closed circuit: complete pathway for current
Open circuit: incomplete pathway for current – break in circuit – infinite resistance
Short circuit: circuit with little to no resistance – extremely high current – overheating
sketch sketch
1
5HVLVWDQFH
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5HVLVWDQFH: ratio of potential difference applied across a piece of material to the current through the material
Formula:
For a wire conductor:
Formula:
A short fat cold wire is the best conductor
R = ȡL/A
Units:
R=V/I
Type:
ohm (ȍ) = V/A
scalar
A long hot skinny wire has the most resistance
Unit: W = J/s
Power: energy per unit time Mechanical Power:
Electrical Power: P = E/t = (ǻq V)/ǻt
Alternate Formulas: Substitute V = IR
P=IV
P = I (IR) = I2 R
P = W/t = F s/t = F v
Type: scalar
P = (V/R)V = V2/ R
Meters in a circuit
Ammeter: measures current
Placement: Must be placed in series to allow current to flow through it Circuit must be broken to insert ammeter
Ideal ammeter: has zero resistance so it will not affect current flowing through it
Schematic diagram
Voltmeter: measures potential difference
Placement: Must be placed in parallel to measure potential difference between two points circuit does not to be broken
Ideal voltmeter: has infinite resistance so it will not allow any current to flow through it and disrupt circuit
2
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6HULHV&LUFXLW
3DUDOOHO&LUFXLW
one
More than one
Current
Same everywhere – same for all devices
Current splits – shared among devices
Potential Difference (Voltage)
Voltage shared among devices – voltage splits
Same for all devices
Overall resistance
high
low
Power
low
high
Number of pathways for current
Formulas:
VT = V1 + V2 + …
VT = V1 = V2 = . . .
IT = I1 = I 2 = …
IT = I1 + I 2 + . . .
RT = R1 + R2 + …
1/RT = 1/R1 + 1/R2 + . . .
PT = P1 + P2 + . . .
PT = P1 + P2 + . . .
6HULHV &LUFXLWV
3DUDOOHO &LUFXLWV
Voltage Ratio
Current Ratio
Power Ratio
Power Ratio
3
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Determine the current through and the voltage drop across each resistor in each circuit below.
1.
2.
3.
4.
3RWHQWLDO'LYLGHU Resistors in series act as a “potential (voltage) divider.” They split the potential of the source between them. 5. A 20ȍdevice requires 40 V to operate properly but no 40 V source is available. In each case below, determine the value of added resistor R1 that will reduce the voltage of the source to the necessary 40V for device R2.
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'
6. A mini light bulb is rated for 0.60 W at 200 mA and is placed in series with a variable resistor. Only a 9.0 volt battery is available to power it. To what value should the variable resistor be set to power the bulb correctly?
Bulb needs only 3 V Bulb has resistance of 15 ȍ at rated power Added resistance should be 30 ohms
4
7KH8VHRI6HQVRUVLQ&LUFXLWV
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1. /LJKW'HSHQGHQW5HVLVWRU /'5 or /LJKW6HQVRU: A photo-conductive cell made of semiconducting material whose resistance decreases as the intensity of the incident light increases. Describe how the LDR activates the light switch.
$XWRPDWLFOLJKWVZLWFK
As ambient light decreases, resistance of LDR increases Potential difference across LDR increases Switch needs minimum PD to turn on When light intensity drops to desired level, PD is high enough to turn on switch
2. 1HJDWLYH7HPSHUDWXUH&RHIILFLHQW 17& 7KHUPLVWRU or 7HPSHUDWXUH6HQVRU: A sensor made of semiconducting material whose resistance decreases as its temperature increases. Describe how the NTC thermistor activates the fire alarm.
)LUHDODUP
As external temperature increases, resistance of NTC decreases Potential difference across R2 increases Switch needs minimum PD to turn on When temperature increases to desired level, PD is high enough to turn on switch
3. 6WUDLQ*DXJH or)RUFH6HQVRU: A long thin metal wire whose resistance increases as it is stretched since it becomes longer and thinner. Describe how the strain gauge can measure the strain put on a
section of an airplane body.
As strain increases, resistance of strain gauge increases
Potential difference across R2 decreases
Voltmeter can read change in voltage which can be used
to determine amount of strain on part
5
&RPELQDWLRQ6HULHV3DUDOOHO&LUFXLWV
IB 12
1. Determine the current through and the voltage drop across each resistor.
2. The battery has an emf of 12 V and negligible internal resistance and the voltmeter has an internal resistance of 20 kȍ. Determine the reading on the voltmeter.
3. A cell with negligible internal resistance is connected to three resistors as shown. Compare the currents in each part of the circuit.
6
IB 12
4. Determine the current through and the voltage drop across each resistor.
5. A battery with emf ( and negligible internal resistance is connected in a circuit with three identical light bulbs. a) Determine the reading on the voltmeter when the switch is open and when it is closed. b) State what effect closing the switch has on the current through each bulb and the brightness of each bulb.
7
2KP¶V/DZ
IB 12
5HVLVWDQFH: ratio of potential difference applied across a piece of material to the current through the material
Formula: R=V/I
2KP¶V/DZ: for a conductor at constant temperature, the current flowing through it is proportional to the potential difference across it
Relationship:
VĮI
2KPLF'HYLFH a device that obeys Ohm’s law for a wide range of potential differences
Meaning: a device with constant resistance
Example: resistor
1. On the axes at right, sketch the ,9FKDUDFWHULVWLFV for a resistor. Resistance:
a) R = V/I at any point
b) related to slope of graph
(Reciprocal = resistance)
2. A resistor is connected to two 1.5 volt cells and has 0.40 ampere of current flowing through it. a) Calculate the resistance of the resistor.
R = V/I
R = 7.5 ȍ
b) If the voltage is doubled, what is the new current? V = IR for resistor
Resistance is constant so double current
8
1RQ2KPLF'HYLFH a device that does not obey Ohm’s law
Meaning: resistance is not constant
IB 12
Example: filament lamp
1. On the axes at right, sketch the ,9FKDUDFWHULVWLFV for a filament lamp.
Resistance:
a) R = V/I at any point
b) as current increases, wire filament heats up and resistance increases
c) Resistance is NOT related to the slope
d) except in initial linear region
2. A flashlight bulb is connected to two 1.5 volt cells and has 0.40 ampere of current flowing through it. a) Calculate the resistance of the bulb.
R = V/I
R = 7.5 ȍ
b) If the voltage is doubled, what is the new current? V = IR for bulb but resistance is not 7.5 ohms any
more – R increases with T so less than double current
3. Discuss how the resistance varies with increasing potential difference for devices X, Y, and Z.
X: resistance increases - ratio V/I increases Y: resistance is constant – ratio V/I is constant Z: resistance decreases – ratio V/I decreases
9
8VLQJD3RWHQWLRPHWHUWR0HDVXUH,9&KDUDFWHULVWLFV
IB 12
3RWHQWLRPHWHU a type of variable resistor with three contact points Common use: as a potential divider to measure the I-V characteristics of a device The schematic shows how a potentiometer can be used as a potential divider to measure the I-V characteristics of a filament lamp. It is placed in parallel with the lamp and the slider (center contact point) effectively splits the potentiometer into two separate resistors AB and BC. By moving the slider, the ratio of the voltage drops across the resistors AB and BC is varied. Redraw the schematic with an ammeter and a voltmeter correctly placed to measure the I-V characteristics of the filament lamp.
Comment on the circuit characteristics as the slider is moved from A to B to C. Slider at A:
Slider at B:
Slider at C:
10
IB 12
,QWHUQDO5HVLVWDQFHRI%DWWHULHV
A 6 volt battery is connected to a variable resistor and the current in the circuit and potential difference across the terminals of the battery are measured over a wide range of values of the resistor. The results are shown in the table. 5HVLVWDQFH
3UHGLFWHG &XUUHQW $
$FWXDO &XUUHQW $
9ROWDJHDFURVV EDWWHU\ 9
2000 200 20 2 0.2 0.02 0.002 0.0002
0.003 0.03 0.3 3 30 300 3000 30000
0.003 0.03 0.29 2.4 8.8 11.5 12.0 12.0
6.00 5.99 5.85 4.80 1.71 0.23 0.02 0.00
Why does the current seem to be limited to a maximum of 12.0 amperes and why does the voltage across the battery not remain constant at 6.0 volts? The battery has some internal resistance. As the
external resistance decreases, more and more of the
energy supplied by the battery is used up inside the
battery.
(OHFWURPRWLYHIRUFH HPI total energy per unit charge supplied by the battery
Symbol: İRU( Units: V = J/C
7HUPLQDO9ROWDJH 9WHUP potential difference across the terminals of the battery
Ideal Behavior: Vterm always equals emf since no internal resistance Real Behavior: 1) Think of battery as internal E and tiny internal resistor r
2) Vterm only equals the emf when no current is flowing 3) E is split between R and r 4) When R>>r, Vterm § emf
5) As R decreases, Vr increases and VR decreases 11
IB 12 5HODWLRQVKLSEHWZHHQHPIDQGWHUPLQDOYROWDJH Treat internal resistance as a series resistor
İ = I RT
İ = I (R + r)
İ = IR + Ir
Note that in the absence of internal resistance, İ = Vterm
1. A resistor is connected to a 12 V source and a switch. With the switch open, a voltmeter reads the potential difference across the battery as 12 V yet with the switch closed, the voltmeter reads only 9.6 V and an ammeter reads 0.40 A for the current through the resistor. Sketch an appropriate circuit diagram and calculate the internal resistance of the source.
2. Discuss the expected I-V characteristics for this battery and how they can be experimentally determined. R can be adjusted from 0 to its max value
A graph of Vterm vs. I can be drawn
Specific equation of graph can be compared to math model to derive
internal resistance Emf = Vterm + Ir Vterm = -Ir + emf
so slope = -r and y-intercept = emf
12
(OHFWURPDJQHWLVP
IB 12
Direction of magnetic field lines: the direction that the North pole of a small test compass would point if placed in the field (N to S)
Magnetic Field around a Bar Magnet
What is the cause of magnetic fields? Moving electric charges Therefore: current in a wire will produce a magnetic field
The Right Hand Rule for the Magnetic Field around a Wire
7KXPE direction of conventional current )LQJHUWLSV direction of magnetic field – tangent to circle
a) head-on view
b) side view
Draw concentric circles with increasing spacing and arrows in circular fashion.
Alternate Right Hand Rule for Loops )LQJHUWLSV direction of current
Component fields
Draw concentric circles around wire.
Resultant field
c) side view
Draw dots and crosses
Your turn
7KXPE points North Note that a wire loop acts like a:
bar
magnet
Solenoid: coil of wire – many loops
Note that a solenoid bar magnet acts like a:
Draw the magnetic field around this solenoid. Use alternate RHR to find North.
1
0DJQHWLF)RUFHRQD:LUH If a wire with current flowing through it is placed in an external magnetic field, it will experience a force. Why? The Right Hand Rule for the Magnetic Force on a Current-Carrying Conductor in a Magnetic Field
IB 12
Two magnetic fields – around wire and from external magnet – will either attract or repel
)ODW+DQG thumb and fingers at right angles )LQJHUV external B field – north to south 7KXPE current 3DOP force – “palm pushes” Maximum force occurs when: current is perpendicular to B field No force occurs when: current is parallel to B field
Use the right hand rule for forces to confirm the direction of the force in each case.
0DJQLWXGHRI WKHPDJQHWLF IRUFHRQDZLUH 'HILQLWLRQRIPDJQLWXGH RIPDJQHWLFILHOG B = F / (IL sin ș)
Find the magnitude and direction of the force on the wire segment confined to the gap between the two magnets as shown when the switch is closed. The strength of the magnetic field in the gap is 1.9 T.
Magnetic field strength Magnetic field intensity Magnetic flux density
) %,/VLQș Where ș is angle between current and B field
Units: Tesla (T)
The ratio of the magnetic force on a wire to the product of the current in the wire, the length of the wire and the sine of the angle between the current and the magnetic field
0.62 N up
2
0DJQHWLF)RUFHRQD0RYLQJ&KDUJHG3DUWLFOH Why is there a magnetic force on a charged particle as it moves through a magnetic field?
The Hand Rule for the Magnetic Force on a Charge moving in a Magnetic Field
IB 12
Moving charged particle creates its own magnetic field – two magnetic fields interact
)ODW+DQG thumb and fingers at right angles
)LQJHUV external B field – north to south
5LJKW+DQG positive charge
7KXPE velocity
/HIW+DQG negative charge
3DOP force – “palm pushes”
Maximum force occurs when: velocity is perpendicular to B field No force occurs when: velocity is parallel to B field Find the direction of the magnetic force on each particle below as each enters the magnetic field shown.
0DJQLWXGHRIWKHPDJQHWLFIRUFH RQDPRYLQJFKDUJHGSDUWLFOH 'HILQLWLRQRIPDJQLWXGH RIPDJQHWLFILHOG
) TY%VLQș Where ș is angle between v and B
The ratio of the force on a charged particle moving in a magnetic field to the product of the particle’s charge, velocity and sine of the angle between the direction of the magnetic field and velocity.
B = F / (qv sin ș)
A proton in a particle accelerator has a speed of 5.0 × 106 m/s. The proton encounters a magnetic field whose magnitude is 0.40T and whose direction makes an angle of T = 30.0° with respect to the proton's velocity. Find the magnitude of the magnetic force on the proton and the proton’s acceleration. How would these change if the particle was an electron?
a) 1.6 x 10-13 N b) 9.6 x 1013 m/s2 c) 1.8 x 1017 m/s2, same force but opposite direction 3
0RWLRQRID&KDUJHG3DUWLFOHLQD0DJQHWLF)LHOG
IB 12
1. A charged particle will follow a circular path in a magnetic field since the magnetic force is always perpendicular to the velocity.
2. The magnetic force does no work on the particle since the magnetic force is always perpendicular to the motion.
3. The particle accelerates centripetally but maintains a constant speed since the magnetic force does no work on it.
5DGLXVRI&LUFXODU3DWK
a) Sketch the paths of a slow and a fast
moving proton at constant speed.
b) Sketch the path of a proton that is slowing
down and one that is speeding up.
Ȉ F = m ac FB = m v2 /r Q v B = m v2 /r r = mv/ qB
c) How would the radius of the path change if the particle were an alpha particle?
&RPSDULQJ(OHFWULFDQG0DJQHWLF)LHOGVDQG)RUFHV
(OHFWULF)LHOG
0DJQHWLF)LHOG
draw paths for stationary, parallel and perp charges
4
(OHFWULF)LHOGVDQG0DJQHWLF)LHOGV 1. A proton is released from rest near the positive plate and leaves through a small hole in the negative plate where it enters a region of constant magnetic field of magnitude 0.10T. The electric potential difference between the plates is 2100 V.
IB 12
a) Describe the motion of the proton while in the electric field constant acceleration in a straight line
b) Describe the motion of the proton while in the magnetic field constant acceleration and constant speed – circular path
c) Find the speed of the proton as it enters the magnetic field.
Use conserv. Of energy
a) 6.3 x 105 m/s
d) Find the radius of the circular path of the proton in the magnetic field.
b) 6.6 x 10-2 m
2. A 9HORFLW\6HOHFWRU is a device for measuring the velocity of a charged particle. The device operates by applying electric and magnetic forces to the particle in such a way that these forces balance. a) Determine the magnitude and direction of an electric field that will apply
and electric force to balance the magnetic force on the proton.
perp to v and B – from bottom to top of page ȈF = 0
FB – Fe = 0
FB = Fe
qvB = Eq
E = vB
b) What is the resulting speed and trajectory of the proton? v = E/B in a straight line
5
7KH0DVV6SHFWURPHWHU
IB 12
A PDVVVSHFWURPHWHU is a device used to measure the masses of isotopes. Isotopes of the same element have the same charge and chemical properties so they cannot be separated by using chemical reactions but have different masses and so can be separated by a magnetic field. A common type of mass spectrometer is known as the %DLQEULGJHPDVVVSHFWURPHWHU and its main parts are shown below.
Ion Source: source of charged isotopes – same charge – different mass
Velocity selector: so all ions have the same speed
Magnetic deflection chamber: radius is proportional to mass
1. A singly charged ion with mass 2.18 x 10-26 kg moves without deflection through a region of crossed magnetic and electric fields then is injected into a region containing only a magnetic field, as shown in the diagram, where it is deflected until it hits a photographic plate. The electric field between the plates of the velocity selector is 950 V/m and the magnetic field in both regions is 0.930 T. Determine the sign of the charge and calculate where the ion lands on the photographic plate.
Sign: could be either in velocity selector Only positive in deflection chamber
velocity selector v = E/B
v = 1.0 x 103 m/s
magnetic chamber
r = mv/qB
r = 1.5 x 10-4 m d = 3.0 x 10-4 m
2. A hydrogen atom and a deuterium atom (an isotope of hydrogen) move out of the velocity selector and into the region of a constant 0.10 T magnetic field at point S, as shown below. Each has a speed of 1.0 x 106 m/s.
Calculate where they each hit the photographic plate at P.
Hydrogen = 0.20 m Deuterium = 0.41 m
6
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IB 12
In 1819, Hans Christian Oersted discovered that a magnetic compass experiences a force in the vicinity of an electric current – that is, that electric currents produce magnetic fields. Because nature is often symmetric, this led many scientists to believe that magnetic fields could also produce electric currents, a concept known as HOHFWURPDJQHWLFLQGXFWLRQ.
Why does moving a wire through a magnetic field induce a current in the wire?
Free electrons in the wire are charged particles moving through a magnetic field so there is a qvB force on them causing them to move resulting in a current.
'HULYDWLRQRIIRUPXODIRU(0)LQGXFHGLQDPRYLQJZLUH A straight conductor is moved at constant velocity perpendicular to a uniform magnetic field. 1. Electrons in the moving conductor experience a downward magnetic force and migrate to the lower end of the conductor, leaving a net positive charge at the upper end.
FB = qvB
2. As a result of this charge separation, an electric field is built up in the conductor.
E
3. Charge builds up until the downward magnetic force is balanced by the upward electric force due to the electric field. At this point, the charges stop flowing and are in equilibrium.
FB = Fe qvB = Eq E = vB
4. Because of this charge separation, a potential difference is set up across the conductor.
LHR for electrons to show direction of force
ǻV = E d = E Ɛ ǻV= vB Ɛ İ %ƐY
If the conductor is connected to a complete circuit, the induced emf will produce an induced current. … is equivalent to …
$PRXQWRI&XUUHQW
'LUHFWLRQRI&XUUHQW
The amount of induced current in the circuit is given by
The direction of the induced emf and induced current can be found from the right hand rule for forces to find the force on a positive charge in the conductor.
İ %ƐY İ=IR I = BƐv/R
7
, IB 12
7ZR2SSRVLQJ)RUFHV ,
Y
3DOPSXVKHVFXUUHQWXS )% TY% An applied force (Fapp) in the direction of the velocity induces an emf which causes current to be pushed upwards.
The magnetic force acts to oppose the applied force, like drag or friction.
)%
3DOPSXVKHVEDUEDFN )% %,O
At a constant speed, Fapp = FB = BIƐ
The induced current now generates a magnetic field around the moving bar that causes a magnetic force (FB) on itself.
Suppose a rod is moving at a constant speed of 5.0 m/s in a direction perpendicular to a 0.80-T magnetic field as shown. The rod has a length of 1.6m and negligible electrical resistance. The rails also have negligible resistance. The light bulb, however, has a resistance of 96 :. Find: a) the emf produced by the motion of the rod 6.4 V
(b) the magnitude and direction of the induced current in the circuit
e) How much external force is applied to keep the rod moving at this constant speed? 0.086 N
0.067 A CCW
c) the electrical power delivered to the bulb 0.43 W
f) How much work is done by the applied force in 60.0 seconds? 26 J
d) the energy used by the bulb in 60.0 s. 26 J
g) What happens to this work? Converted to electrical energy
8
0DJQHWLF)OX[
IB 12
0DJQHWLF)OX[
6\PERO: ĭ
Number of field lines
8QLWV Weber (Wb)
)RUPXOD: B = ĭ/ A
0DJQHWLF)OX['HQVLW\ ILHOGVWUHQJWKLQWHQVLW\
or
ĭ=BA
number of field lines per unit area
= T m2
6\PERO: B 8QLWV Wb/m2 = T (tesla)
$QJOH'HSHQGHQFHRI)OX[ What is the amount of magnetic flux if the field lines are not perpendicular to the cross-sectional area? Only the perpendicular component of the magnetic field contributes to the magnetic flux. 1RUPDOOLQH: line perpendicular to plane of cross-sectional area
)RUPXOD: ĭ = (B cos ș) A = B A cos ș $QJOH: ș = angle between normal line and field lines
0DJQHWLF)OX[ ĭ product of the magnetic field strength and a cross-sectional area and the cosine of the angle between the magnetic field and the QRUPDO to the area
)RUPXOD ĭ %$FRVș
8QLWV 7P
0DJQHWLFIOX[OLQNDJH PDJQHWLFIOX[OLQNLQJDFRLO product of magnetic flux through a coil of wire and the number of turns of the wire )RUPXOD 1ĭ 1%$FRVș
8QLWV 7P
1. A single loop of wire whose cross-sectional area is 0.50 m2 is located in a 0.20 T magnetic field as shown. Calculate the flux through the loop in each case.
a) 0.10 T m2
b) 0.050 T m2
c) 0 2. If the coil of wire in the above example consisted of 50 turns of the wire, calculate the amount of flux linking the coil in each case.
a) 5.0 Tm2
b) 2.5 T m2 c) 0 9
(0),QGXFHGE\D7LPH&KDQJLQJ)OX[
Moving a magnet towards a coil will increase the magnetic flux linking the coil and will induce an emf and a current in a certain direction.
Holding the magnet stationary will not change the amount of magnetic flux linking the coil and so will not induce an emf or current.
IB 12
Moving the magnet away from the coil will decrease the magnetic flux linking the coil and will induce an emf and a current in the opposite direction.
Methods of inducing an EMF by a time-changing flux
2. Move magnet or coil
2. Rotate coil
1. Vary magnetic field
)DUDGD\¶V/DZ an induced emf is proportional to the rate of change of the flux linkage
)RUPXOD İ = - N (ǻĭ/ǻt) 1. A coil of area 0.030 m2 with 300 turns of wire rotates as shown in 0.10 second in a magnetic field of constant 0.25 T strength. a) What is the magnitude of the induced emf? 11.3 V
b) What is the magnitude of the induced emf if the coil were stationary at 00 but the field strength changed from 0.25 T to 0.60 T in 0.10 second? 22.5 V
10
IB 12
2. A 50 turn coil of wire of area 0.20 m2 is perpendicular to a magnetic field that varies with time as shown by the graph. a) Determine the emf induced in the coil during each time interval. 3 V, 0 V, -1.5 V
b) Sketch a graph of the induced emf vs. time. Emf = - derivative of flux
/HQ]¶V/DZ±)LQGLQJWKH'LUHFWLRQRIWKH,QGXFHGHPI /HQ]¶V/DZ The direction of an induced emf is such that it produces a magnetic field whose flux opposes the flux change that induced it.
(An emf will be induced so as to keep the net flux constant.)
a) Original flux change – an increasing flux induces an emf and current.
b) Induced flux – opposes increasing flux by pointing in opposite direction thus current is in direction shown.
1. If the magnetic field linking this coil is decreasing with time, in which direction is the induced current?
c) Result - two magnetic fields acting to keep net flux constant.
2. The diagrams show a conducting ring that is placed in a uniform magnetic field. Deduce the direction of the induced current in each case if there is
(a) an increasing B field
(b) a decreasing B field
11
IB 12 3. If the current in the wire is increasing, in which direction will there be an induced current in the rectangular wire loop?
4. If the wire loop moves away from a steady current in the straight wire, in which direction will there be an induced current in the loop?
5. A conducting loop moves at a constant speed into and through a uniform magnetic field as shown in the diagram. Indicate the direction of the induced current. Graph the flux through the loop and the induced emf as a function of time.
6. If a clockwise current through the primary coil is increasing with time, what effect will this have on the secondary coil?
7. Determine the direction of the current in the solenoid in each case.
8. Determine the direction(s) of the induced current as the magnet falls through the loop.
12
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IB 12
Basic Operation: 1. coil of wire is turned by mechanical means in an external magnetic field
2. emf and current are induced in coil as coil cut flux lines
3. current varies in magnitude and direction as flux linkage changes – current and emf variations are sinusoidal 4. brushes and rings maintain contact with external circuit without getting tangled
5RWDWLRQRID&RLOLQD8QLIRUP0DJQHWLF)LHOGLQGXFHVDQ(0)
As the coil rotates, the flux linking it changes
3RVLWLRQ
0D[LPXP(0)DQG&XUUHQW 1. sides of coil cut field lines perpendicularly
2. plane of coil is parallel to field lines 3. normal to coil is perpendicular to field lines (900)
0LQLPXP(0)DQG&XUUHQW
3RVLWLRQ
Mark when the coil is in positions 1 and 2.
Sketch the graph of the induced current.
1.
sides of coil do not cut field lines perpendicularly (move parallel to them)
2.
plane of coil is perpendicular to field lines
3.
normal to coil is parallel to field lines (00) Sketch a graph of the induced emf for a coil with:
twice the frequency of rotation.
half the frequency of rotation.
13
$OWHUQDWLQJ&XUUHQW The output of an AC generator is an emf that varies sinusoidally with time.
IB 12
V0 = peak/ maximum voltage I0 = peak/ maximum current
V = Vo sin Ȧt
(where Ȧ = 2ʌf)
I = V/R so I = (Vo/R) sin Ȧt = Io sin Ȧt
The power output of an AC generator
P=IV P = (Io sin Ȧt)(Vo sin Ȧt) P = Io Vo sin2 Ȧt
Maximum Power Pmax = Io Vo
Average Power Pav = ½ Io Vo = (Io/ rad 2)(Vo / rad2)
Pmax = 2 Pav
Root-Mean-Squared values (RMS):
RMS Values Irms = Io / rad 2 Vrms = Vo / rad 2
= Irms Vrms
The rms value of an alternating current (or voltage) is that value of the direct current (or voltage) that dissipates power in a resistor at the same rate.
1. In the USA, most household voltage is stated as “120 V at 60 Hz.” This is the root-mean-square voltage and the frequency of the AC voltage. Calculate the maximum voltage and mark Vo, Vrms, on the graph. Vo = 170 V
2. In Europe, the “mains electricity” is rated at 230 V. What is the peak household voltage in Europe?
14
IB 12 Rating: rms values are given as the AC values to be used in calculations, as if they were DC values
Formula:
R = V0/I0 = Vrms/Irms
1. A stereo receiver applies an AC voltage of 34 V to a speaker. The speaker behaves approximately as if it has a resistance of 8.0 :, as the circuit figure indicates. Determine a) the maximum voltage,
b) the rms current,
c) the average power for this circuit.
a) 48 V
b) 4.25 A
c) 145 W
2. A 100 W light bulb is designed to operate from a 120 VAC mains. Determine:
3. A maximum alternating voltage of 170 V is applied across a 50 ȍ resistor. Determine:
a) the maximum power of the light bulb
a) the maximum current through the resistor
b) the maximum current drawn by the bulb
b) the average power dissipated by the resistor
a) 200 W
b) 1.2 A
a) 3.4 A
b) 289 W 15
7KH7UDQVIRUPHU
IB 12
According to Michael Faraday’s original experiment that first produced electromagnetic induction, an emf and current were only induced in the secondary coil when the switch in the primary coil was being opened or closed, that is, when the current in the primary coil was changing (increasing or decreasing). No emf or current was induced in the secondary coil while the switch was stationary in the open or closed position, that is, when the current was steady or off. Therefore, emf can only be induced in the secondary coil when the magnetic field from the current in the primary coil is building up or dying down, that is, while the magnetic flux is changing.
7UDQVIRUPHU a device that increases or decreases AC voltage.
6WUXFWXUHDQGRSHUDWLRQRIDWUDQVIRUPHU
1. An alternating potential difference (VP) applied across the primary coil creates an alternating current in the primary coil. 2. This creates an alternating magnetic field (time-changing flux) in the primary coil.
3. The soft iron core concentrates the magnetic flux from the primary coil and links it with the secondary coil.
4. The time-changing flux in the secondary coil induces a secondary alternating emf (VS).
7UDQVIRUPHUIRUPXOD
İP = VP = -NP (ǻĭ/ǻt)
6WHS8S7UDQVIRUPHU: If NS > NP, then VS > VP and voltage increases from primary to secondary
İS = VS = -NS (ǻĭ/ǻt) since flux changes are identical
93 96 1316
6WHS'RZQ7UDQVIRUPHU: If NS < NP, then VS < VP and voltage decreases from primary to secondary 16
Voltage and turns in same ratio
IB 12 How can the voltage increase or decrease without violating the conservation of energy principle? The power input at the primary equals the power output at the secondary. (This assumes 100% efficiency
and such a transformer is termed an LGHDOWUDQVIRUPHU.)
,GHDO7UDQVIRUPHU)RUPXOD PP = PS VP IP = VS IS VP/ VS = IS / IP Voltage and current in inverse ratio
1. A 120 VAC wall outlet is used to run a small electronic appliance with a resistance of 2.0 ȍ, as shown in the diagram. a) Is the transformer a step-up or step-down transformer? Cite evidence for your answer.
b) How much voltage does the device need?
c) If the current in the primary coil is 150 mA, how much current does the device use? Assume an ideal transformer.
a) down b) 6 V c) 3 A Real Transformers
Reasons for power losses in real transformers 1. resistance of wires in P and S coils causes heating of coils
Ps < PP eff = Ps / PP
2. not all flux from P coil is linked to S coil
3. core warms up as result of cycles of flux changes (hysteresis)
Modern transformers are up to 99% efficient
4. small currents are induced in core (eddy currents) – reduce by lamination 17
IB 12
2. The figure shows a step-down transformer used to light a filament lamp with a resistance of 4.0 ȍ under operating conditions. The secondary coil has an effective resistance of 0.2 ȍand the primary current is 150 mA. Calculate: a) the reading on the voltmeter with switch S open 12 V
d) the power taken from the mains supply 36 W
b) the current in the secondary coil with switch S closed 2.86 A
e) the efficiency of the transformer 95%
c) the power dissipated in the lamp and the secondary coil 3.27 W and 1.6 W
+HDOWKDQG6DIHW\&RQFHUQVDVVRFLDWHGZLWK+LJK9ROWDJH3RZHU/LQHV 1. Extra-low-frequency electromagnetic fields, such as those produced by electrical appliances and power lines, induce currents within a human body. Just as AC can induce emfs and currents in secondary coils, so to can they be induced in the human
body since it is a conducting medium
Changing magnetic field induces current in human body
2. Current research suggests that low-frequency fields do not harm genetic material.
f = 60 Hz individual photons of this frequency do not have enough energy to cause ionization in the body childhood leukemia clusters are suspected to have a link to living near overhead power cables 3. The risks attached to the inducing of current in the human body are not well-understood. Risks are likely to be dependent on current density, frequency, and length of exposure
18
3RZHU7UDQVPLVVLRQ
IB 12
Power loss in transmission lines When current flows through a wire, some energy is lost to the surroundings as the wire heats up due to the collisions between the free electrons in the current and the lattice ions of the wire. This is known as -RXOH KHDWLQJor UHVLVWLYHKHDWLQJ. Since the energy lost per second, or power loss, is proportional to the square of the current (P = I2 R), this energy loss is also know as “I2R loss.”
Methods of reducing I2R loss in power transmission lines 1. Reduce resistance: thicker cables – low resistivity material
Constraints: lengths are fixed, thicker cables are heavier and more expensive
2. Increase voltage: step voltage up to very high levels
Constraints: high voltages are dangerous – must be stepped back down for household use
For economic reasons, there is no ideal value of voltage for electrical transmission. Typical values are shown below. 1. AC power is generated at a power plant at 12,000 V and then stepped up to 240,000 V by step-up transformers. 2. The high-voltage, low-current power is sent via high-voltage transmission lines long distances. 3. In local neighborhoods, the voltage is stepped-down (and current is stepped-up) to 8000 V at substations. 4. This voltage is stepped-down even further at transformers on utility poles on residential streets.
An average of 120 kW of power is delivered to a suburb from a power plant that is 10 km away. The transmission lines have a total resistance of 0.40 ȍ. Calculate the power loss if the transmission voltage is a) 240 V
a) 240,000 V
I = 500 A P = 100 kW
I = 0.50 A P = 0.10 W
19
(OHFWURVWDWLFV
IB 12
1) electric charge: 2 types of electric charge: positive and negative
2) charging by friction: transfer of electrons from one object to another
3) positive object: lack of electrons
negative object: excess of electrons
4) Types of materials: a) &RQGXFWRUV: materials in which electric charges move freely (e.g. metals, graphite) b) ,QVXODWRUV: materials in which electric charges do not move freely (e.g. plastic, rubber, dry wood, glass, ceramic) c) 6HPLFRQGXFWRUV: materials with electrical properties between those of conductors and insulators (e.g. silicon) d) 6XSHUFRQGXFWRUV: materials in which electrical charges move without resistance (e.g. some ceramics at very low temperatures)
3URSHUWLHVRI$WRPLF 3DUWLFOHV H = elementary unit of charge (magnitude of charge on electron)
Particle
Mass
Electric Charge
Electron
PH [ NJ
T H T [ &
Proton
PS [ NJ
T H T [ &
Neutron
PQ [ NJ
T T &
H = 1.60 x 10-19 C
1. A balloon has gained 2500 electrons after being rubbed with wool. What is the charge on the balloon? What is the charge on the wool?
q = -4.0 x 10-16 C q = +4.0 x 10-16 C
2. A rubber rod acquires a charge of -4.5 ȝC. How many excess electrons does this represent?
2.8125 x 1013 e
&RQVHUYDWLRQRI(OHFWULF&KDUJH The total electric charge of an isolated system remains constant.
1
(OHFWULF)RUFH (OHFWURVWDWLF)RUFH&RXORPE)RUFH
IB 12
&RXORPE¶V/DZ The electric force between two point charges is directly proportional to the product of the two charges and inversely proportional to square of the distance between them, and directed along the line joining the two charges.
&RXORPE)RUFH
T1T2 )H N 2 U
k = Coulomb constant (electrostatic constant) k = 8.99 x 109 Nǜm2 C -2
NOTE: +-F denotes direction of force not sign of charge Point charge: a charged object that acts as if all its charge is concentrated at a single point Alternate formula for Coulomb force:
)H
1 T1T2 4SH 0 U2
)H
T1T2 4SH 0 U 2
k = 1/ 4ʌİ0 İ0 = permittivity of free space = 8.85 x 10-12 C2 N-1 m -2
Use the Coulomb force to estimate the speed of the electron in a hydrogen atom.
2
7KH3ULQFLSOHRI6XSHUSRVLWLRQ
IB 12
The net electric force acting on a charged particle is the vector sum of all the electric forces acting on it.
1. Determine the net electrostatic force on charge q1, as shown below.
2. Where can a third charge of +1.0 μC be placed so that the net force acting on it is zero?
3. Three point charges of
-2.0 μC are arranged as shown. Determine the magnitude and direction of the net force on charge T.
D = 2/3 m
3
(OHFWULF)LHOG
IB 12
Electric field: a region in space surrounding a charged object in which a second charged object experiences an electric force Test charge: a small positive charge used to test an electric field
1. Positively charged sphere
(OHFWULF)LHOG'LDJUDPV 2. Positive point charge
3. Negative point charge
Radial Field: field lines are extensions of radii
5. Two positive charges
8. Oppositely charged parallel plates
6. Two negative charges
7. Two unlike charges
3URSHUWLHVRI(OHFWULF)LHOG/LQHV
1. Never cross 2. Show the direction of force on a small positive test charge 3. Out of positive, into negative 4. Direction of electric field is tangent to the field lines
Uniform Field: field has same intensity at all spots Edge Effect: bowing of field lines at edges
5. Density of field lines is proportional to field strength (density = intensity) 6. Perpendicular to surface 7. Most intense near sharp points
4
(OHFWULF)LHOG6WUHQJWK
IB 12
(OHFWULF)LHOG6WUHQJWK ,QWHQVLW\ electric force exerted per unit charge on a small positive test charge
Electric Field:
)H ( T Units: N/C
Electric Force:
)H (T
Electric Field for a Point Charge:
N
(
4T U 2 N 4 1 4 T U2 4SH 0 U2
Units: N 6SKHULFDO &RQGXFWRU
3RLQW&KDUJH
1. a) Find the magnitude and direction of the electric field at a spot 0.028 meter away from a sphere whose charge is +3.54 microcoulombs and whose radius is 0.60 centimeters.
2. a) Find the magnitude and direction of the gravitational field at an altitude of 100 km above the surface of the Earth.
b) Find the magnitude and direction of the electric force acting on a -7.02 nC charge placed at this spot.
b)
Find the magnitude and direction of the gravitational force exerted on a 6.0 kg bowling ball placed at this spot.
c) Find the electric field strength at the surface of the sphere.
c) Find the gravitational field strength at the surface of the Earth.
5
3. a) Find the magnitude and direction of the net electric field halfway between the two charges shown below.
IB 12
b) Determine the electric force on a proton placed at this spot.
4. Two charged objects, $ and %, each contribute as follows to the net electric field at point 3: ($ = 3.00 N/C directed to the right, and (%= downward. What is the net electric field at 3?
E = 3.61 N/C Theta =33.70
5. a) Two positive point charges, T1 = +16 PC and T2 = +4.0 PC, are separated in a vacuum by a distance of 3.0 m. Find the spot on the line between the charges where the net electric field is zero.
6
6. A proton is released from rest near the positive plate. The distance between the plates is 3.0 mm and the strength of the electric field is 4.0 x 103 N/C.
IB 12
a) Describe the motion of the proton.
constant acceleration in a straight line b) Write an expression for the acceleration of the proton.
c) Find the time it takes the proton to reach the negative plate.
d) Find the speed of the proton when it reaches the negative plate.
7. A particle is shot with an initial speed through the two parallel plates as shown. a) Sketch and describe the path it will take if it is a proton, an electron, or a neutron. b) Which particle will experience a greater force?
c) Which particle will experience a greater acceleration?
d) Which particle will experience a greater displacement?
8. In the figure, an electron enters the lower left side of a parallel plate capacitor and exits at the upper right side. The initial speed of the electron is 5.50×106 m/s. The plates are 3.50 cm long and are separated by 0.450 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.
7
(OHFWULF3RWHQWLDO(QHUJ\
IB 12
*UDYLWDWLRQDO3RWHQWLDO(QHUJ\ (3
High amount of EP
Reason for EP: 1. Test object has mass (test mass = m) 2. Test mass is in a gravitational field (g) caused by larger object (M) 3. Larger object exerts a gravitational force on test mass (Fg = mg) Low amount of EP 4. Test mass has tendency to move to base level due to force 5. Work done moving object between two positions is path independent. Base level where EP = 0
Gravitational potential energy: EP = mgh W = ǻEP = mg ǻh
(OHFWULF3RWHQWLDO(QHUJ\ (3 Reason for EP: High amount of EP
1. Test object has charge (test charge = +q) 2. Test charge is in an electric field caused by larger object (Q) 3. Larger object exerts an electric force on test charge (FE = Eq)
Low amount of EP
4. Test charge has tendency to move to base level due to force 5. Work done moving object between two positions is path independent.
Electric potential energy: EP = Eq h W = ǻEP = Eq ǻh
Base level where EP = 0
(OHFWULF3RWHQWLDO(QHUJ\ (3 the work done in bringing a small positive test charge in from infinity to that point in the electric field Derivation for Point Charges
EP = 0
(Work done by field)
(3 : )V cosT (3 (T'V U N4 (3 ³ ( 2 T)GV f V
Electric Potential Energy due to a point charge Formula:
Units:
Type:
J
scalar
U
(3
N4T V f
(3
N4T
U
(3
N4T U
8
IB 12 (OHFWULF3RWHQWLDO 9 work done per unit charge moving a small positive test charge in from infinity to a point in an electric field. (OHFWULF3RWHQWLDOGXHWRDSRLQWFKDUJH
N4T (3 Formula: 9 U T T N4 4 9 U 4SH 0 U
Units:
(3 T9
% Higher potential
Lower potential
%
1. a) Calculate the potential at a point 2.50 cm away from a +4.8 ȝC charge.
J/C = volts(V)
$ Lower potential
Type: scalar
Zero potential
Zero potential
Higher potential
$
b) How much potential energy will an electron have if it is at this spot?
3. What is the potential where a proton is placed 0.96 m from a -1.2 nC charge?
9
IB 12
3RLQW&KDUJHV
4
4
4
4
(OHFWULF)LHOG
(OHFWULF3RWHQWLDO(QHUJ\
(OHFWULF 3RWHQWLDO
Two objects needed – interaction between the two
One object needed – property of that one object
Two objects needed – quantity possessed by the system
One object needed – property of the field
Magnitude: F = Eq
Magnitude: E = F/q
Magnitude: EP = qV
Magnitude: V = EP/q
(OHFWULF)RUFH
F = kQq/r2
E = kQ/r2
P
= kQq/r
V = kQ/r
Units: N
Units: N/C
Units: J
Units: J/C
Type: vector
Type: vector
Type: scalar (+/-)
Direction: likes repel, unlikes attract
Direction: away from positive, towards negative
Type: scalar (+/-) E Sign: use signs of Q and q
Sign: don’t use when calculating – check frame of reference
Sign: don’t use when calculating – check frame of reference
F = 0 where E = 0
Sign: use sign of Q
EP = 0 where V = 0 10
IB 12 1. a) Calculate the net electric field at each spot (A and B):
b) Calculate the net electric force on a proton placed at each spot.
2. a) Calculate the net electric potential at each spot (A and B):
b) Calculate the electric potential energy of a proton placed at each spot.
11
(OHFWULF3RWHQWLDODQG&RQGXFWRUV
IB 12
1. all the charge resides on the outside surface 2. the electric field is zero everywhere within
Value at surface = kQ/r2 Electric Potential
For a hollow or solid conductor,
Electric Field Strength
*UDSKVIRUDVSKHULFDOFRQGXFWRU
3. the external electric field acts as if all the charge is concentrated at the center 4. the electric potential is constant ( 0) everywhere within and equal to the potential at the surface
radius
radius Distance
Distance
A spherical conducting surface whose radius is 0.75 m has a net charge of +4.8 ȝC. a) What is the electric field at the center of the sphere?
b) What is the electric field at the surface of the sphere?
c) What is the electric field at a distance of 0.75 m from the surface of the sphere?
d) What is the electric potential at the surface of the sphere?
e) What is the electric potential at the center of the sphere?
f) What is the electric potential at a distance of 0.75 m from the surface of the sphere?
12
(TXLSRWHQWLDO6XUIDFHV
IB 12
(TXLSRWHQWLDOVXUIDFH: a surface on which the electric potential is the same everywhere
1. Locate points that are at the same electric potential around each of the point charges shown. 2. Sketch in the electric field lines for each point charge. 3. What is the relationship between the electric field lines and the equipotential surfaces?
Perpendicular
Field lines point in direction
of decreasing potential
(OHFWULF3RWHQWLDO*UDGLHQW
Formula:
The electric field strength is the negative of the electric potential gradient.
'9 ( '[
Units: N/c or V/m For each electric field shown, sketch in equipotential surfaces.
Sketch in equipotential surfaces for the two configurations of point charges below.
13 http://wps.aw.com/aw_young_physics_11/0,8076,898593-,00.html
http://www.surendranath.org/Applets.html
(OHFWULF3RWHQWLDO'LIIHUHQFH
IB 12
(OHFWULF3RWHQWLDO'LIIHUHQFH ǻ9 ±work done per unit charge moving a small positive test charge between two points in an electric field
Formula:
'9 '9
Units: J/C = V
: T '(3 T
'( 3 T'9
+LJKDQG/RZ3RWHQWLDO 1. a) Which plate is at a higher electric potential? positive b) Which plate is at a lower electric potential? negative c) What is the electric potential of each plate? Arbitrary – relative to base level d) What is the potential difference between the plates? Not arbitrary – depends on charge, distance between, strength of electric field, geometry of plates, etc.
Mark plates with example potentials, as well as spots within field Mark “ground” – mark equipotentials
e) Where will: a proton have the most electric potential energy?
an electron?
a neutron?
an alpha particle? Not arbitrary
2. An electron is released from rest near the negative plate and allowed to accelerate until it hits the positive plate. The distance between the plates is 2.00 cm and the potential difference between them is 100. volts. a) Calculate how fast the electron strikes the positive plate.
Eo = Ef
Ee = EK
qV = ½ mv2
v = sqrt (2qV/m)
v = sqrt(2(1.6 x 10-19)(100 V)
/(9.11 x 10-31))
v = 5.9 x 106 m/s
Formula: qV = ½ mv2 Ve = ½ mv2
b) Calculate the strength of the electric field.
Formula: E = -ǻV/ǻx E = V/d
14
7KH(OHFWURQYROW
IB 12
(OHFWURQYROW: energy gained by an electron moving through a potential difference of one volt
Derivation:
ǻEe = qǻV ǻEe = (1e)(1 V) = 1 eV
ǻEe = (1.6 x 10-19 C)( 1 V) ǻEe = 1.6 x 10-19 J Therefore: 1 eV = 1.60 x 10-19 J
1. How much energy is gained by a proton moving through a potential difference of 150. V?
150 eV
or 150(1.60 x 10-19) = 2.4 x 10-17 J
2. A charged particle has 5.4 x 10-16 J of energy. How many electronvolts of energy is this?
Factor-label (5.4 x 10-16 J) (1 eV/1.6 x 10-19) = 3375 eV 3. An electron gains 200 eV accelerating from rest in a uniform electric field of 150 N/C. Calculate the final speed of the electron.
4. In Rutherford’s famous scattering experiments (which led to the planetary model of the atom), alpha particles were fired toward a gold nucleus with charge +79H. An alpha particle, initially very far from the gold nucleus, is fired at 2.00 × 107 m/s directly toward the gold nucleus. Assume the gold nucleus remains stationary. How close does the alpha particle get to the gold nucleus before turning around? (the “distance of closest approach”)
2.74 x 10-14 m
15
(QHUJ\DQG3RZHU
IB 12
3RZHU*HQHUDWLRQLQDW\SLFDOHOHFWULFDOSRZHUSODQW a) Some fuel is used (coal, natural gas, oil, uranium) to release thermal energy which is used to boil water to make steam. b) *HQHUDWRU '\QDPR - Steam turns turbines attached to coils of wire which turn in a magnetic field inducing an alternating potential difference. c) Potential difference is stepped up by transformers in order to reduce I2R loss of power in transmission lines then stepped down for consumer use.
What are the energy transformations that take place?
Chemical (nuclear) energy in fuel
Thermal energy – hot gases go out chimney/stack
thermal energy in steam
rotational mechanical energy/kinetic energy in turbine
Thermal energy – radiation and convection from boiler
electrical
energy in
generator
Thermal energy – friction in the generator
Degraded energy: energy transferred to the surroundings that is no longer available to do useful work – can’t be converted into other forms
1
IB 12 Why does the generation of electrical power involve the degradation of energy? 1. Thermal energy can be completely converted to work in a single process.
2. A continuous conversion of thermal energy into work requires a cyclical process.
Example: isothermal expansion Q = ǻU + W ǻU = 0 so Q = W
6HFRQG/DZRI7KHUPRG\QDPLFV 1) The total entropy of the universe is increasing. 2) No cyclical process (engine) is ever 100% efficient. Some energy is transferred out of the system (lost to the surroundings) as unusable energy (degraded energy).
7\SLFDO (IILFLHQF\
)XHO
6DQNH\GLDJUDPV HQHUJ\IORZGLDJUDPV : used to keep track of
energy transfers and transformations
Coal
30-35%
1) Thickness of arrow is proportional to amount of energy. Natural Gas
50%
2) Degraded energy points away from main flow of energy. Oil
30-35%
3) Total energy in = total energy out.
Input 100% Chemical energy
Fuel
boiler
Output 30% Electrical energy
dynamo
10% Thermal energy Friction in dynamo
40% Thermal energy
exhaust gas out
chimney
20% Thermal energy
boiler
2
)XHOV
IB 12
Fuel: source of energy (in a useful form) How does a fuel work? A fuel releases energy by changing its chemical (or nuclear) structure. Chemical (or nuclear) bonds are broken reducing the fuel’s internal potential energy but increasing the kinetic energy of the substance’s particles which is seen macroscopically as an increase in the temperature of the substance. It is this thermal energy that is used to heat the water that will change to steam to turn the generator’s turbines.
Fossil fuels: coal, oil, natural gas, peat Origins of fossil fuels: organic matter decomposed under conditions of high temperature and pressure over millions of years
Non-renewable fuels: rate of production of fuel is much smaller than rate of usage so fuel will be run out - limited supply
Renewable fuels: resource that cannot be used up or is replaced at same rate as being used
7\SHRIIXHO
5HQHZDEOH"
&2 HPLVVLRQV"
Fossil fuels
No
Yes
Nuclear
No
No
Hydroelectric
Yes
No
Wind
Yes
No
Solar
Yes
No
Wave
Yes
No
This histogram shows the relative proportions of world use of the different types of energy sources, though it will vary from country to country.
NOTE: In most instances, the prime energy source for world energy is . . . the Sun. Exceptions: nuclear, tidal (Moon)
3
)RVVLO)XHO3RZHU3URGXFWLRQ
IB 12
Historical and geographical reasons for the widespread use of fossil fuels: 1. industrialization led to a higher rate of energy usage (Industrial Revolution) 2. industries developed near large deposits of fossil fuels (coal towns) Transportation and storage considerations: 1. Natural gas is usually transported and stored in pipelines. $GYDQWDJHV cost effective 'LVDGYDQWDJHV unsightly, susceptible to leaks, explosions, terrorist activities, political instability (withholding use of pipelines or terminals for political reasons) 2. Many oil refineries are located near the sea close to large cities. Oil is transported via ships, trucks, and pipelines. $GYDQWDJHV workforce and infrastructure in place, easy access to shipping 'LVDGYDQWDJHV oil spills and leakage, hurricanes, terrorist activities
3. Power stations using coal and steel mills are usually located near coal mines. $GYDQWDJHV minimizes shipping costs 'LVDGYDQWDJHV environmental impact (strip mining), mine cave-ins
8VHRIIRVVLOIXHOVIRUJHQHUDWLQJHOHFWULFLW\ $GYDQWDJHV
'LVDGYDQWDJHV
1. high energy density
1. combustion produces pollution, especially SO2 (acid rain)
2. relatively easy to transport
2. combustion produces greenhouse gases (CO2)
3. cheap compared to other sources
3. extraction (mining, drilling) damages environment
4. power stations can be built anywhere
4. nonrenewable
5. can be used in the home
5. coal-fired plants need large amounts of fuel
4
IB 12
(QHUJ\GHQVLW\RIDIXHO the ratio of the energy released from the fuel to the mass of the fuel consumed
Formula: De = E/m
)XHO Fusion fuel Uranium-235 Natural gas Gasoline (Petrol) Diesel Biodiesel Crude oil Coal Sugar Wood Cow dung Household waste
Units: J/kg
Use: to compare different types of fuels
How is choice of fuel influenced by energy density? Fuels with higher energy density cost less to transport and store
(QHUJ\'HQVLW\ 0-NJ 300,000,000 90,000,000 53.6 46.9 45.8 42.2 41.9 32.5 17.0 17.0 15.5 10
1. An oil-fired power station produces 1000 MW of power. a) How much energy will the power station produce in one day? ( W
( 3W 3
( (1000 x 106:)(24 [3600) ( 8.6 [1013 -
b) Estimate how much oil the power station needs each day. HII
useful out (RXW total in (LQ
'H
( P - 2.46 [1014 NJ P
8.6 [1013 - .35 (LQ
41.9 [106
(LQ 2.46 [1014 -
P 5.9 [106 NJ
5
IB 12 2. A 250 MW coal-fired power plant burns coal with an energy density of 35 MJ/kg. Water enters the cooling tower at a temperature of 350 K and leaves at a temperature of 293 K and the water flows through the cooling tower at a rate of 4200 kg/s. a) Calculate the thermal energy removed from the water in the cooling towers each second. 4 PF'7 4 (4200NJ )(4.19 [10 3
)(350 293) NJ.
4 1.0 [109 -
( W 1.0 [109 - 3
1V
3 1.0 [109: 10000: 3
b) Assuming the only significant loss of energy is this thermal energy of the water, calculate the energy produced by the combustion of coal each second. (LQ (RXW (LQ 1000 MJ + 250 MJ E LQ 12500-
c) Calculate the mass of coal burned each second. HII
useful out 3RXW total in 3LQ
2500: HII 12500: HII .20
( 'H P 1250 [106 - 6 - 35 [10
NJ P
P 36NJ
6
1XFOHDU(QHUJ\
IB 12
Most common source: fissioning of uranium-235 with conversion of some mass into energy Process: a) unstable uranium nucleus is bombarded with a
neutron and splits into two smaller nuclei and
some neutrons
Why use neutrons? Neutral, not repelled by nucleus b) rest mass of products is less than reactants so
some matter is converted into energy
Form of energy: KE of products (thermal energy) 235 92
92 1 8 10 Qo141 %D .U 3 56 36 0Q
c) released neutrons strike other uranium nuclei
causing further fissions
1) A particular nuclear reactor uses uranium-235 as its fuel source. When a nucleus of uranium-235 absorbs a neutron, the following reaction can take place: 235 92
90 1 8 10 Qo144 54 ;H 38 6U [0 Q
a) How many neutrons are produced in the reaction? 2 b) Use the information to show that the energy released in the reaction is approximately 180 MeV.
rest mas of
8 2.1895 x 105 MeV c 2
235 92
rest mas of 10 Q 939.56 MeV c 2 2 5 rest mas of 144 54 ;H 1.3408 x 10 MeV c 2 5 rest mas of 90 38 6U 8.3749 x 10 MeV c
7
IB 12
2. The energy released by one atom of carbon-12 during combustion is approximately 4 eV. The energy released by one atom of uranium-235 during fission is approximately 180 MeV. a) Based on this information, determine the ratio of the energy density of uranium-235 to that
of carbon-12. (Then, check your answer with the given table of energy densities.)
PDVV=
1 x molar mass 1 $
PDVV=
1 x .235 kg 6.02 [ 1023 mass=3.90x1025 NJ
1 x .012 kg 6.02 [1023 mass=1.99x1026 NJ
mass =
mass =
§ 180 [106 H9 · § 1.60 [1019 -· 11 ¨ ¸¨ ¸ 2.88 [10 - 1 1H9 © ¹© ¹
11
' H
2.88 [ 10 7.38 [1013 -/ NJ 25 3.90 [10 NJ
7.38 [107 0- / NJ
'H U-235 'H C-12
1 x molar mass
1 $
19 §4H9 · §1.60 [10 - · 19 ¸ 6.40 [10 ¨ ¸¨ © 1 ¹ © 1H9 ¹
6.40 [1019 - 'H 3.22 [107 -/ NJ 26 1.99 [10 NJ 32.20-/ NJ
7.38 [107 2.3[106 32.2
b) Based on your answer above, suggest one advantage of uranium-235 compared with fossil fuels. Higher energy density implies that uranium will produce more energy per kilogram – less fuel needed to produce the same amount of energy
8
1XFOHDU)XHODQG5HDFWRUV
IB 12
Naturally Occurring Isotopes of Uranium: 1) 8UDQLXP most abundant (99.3%) but not used for fuel since it has a very small probability of fissioning when it captures a neutron. 2) 8UDQLXP rare (0.3%) but used for fuel since it has a much greater probability of fissioning when captures a neutron but must be a low-energy neutron (thermal neutron). 7KHUPDO1HXWURQ low-energy neutron (§1eV) that favors fission reactions – energy comparable to gas particles at normal temperatures
)XHO(QULFKPHQW process of increasing proportion of uranium-235 in a sample of uranium
1) formation of gaseous uranium (uranium hexafluoride) from uranium ores 2) separated in gas centrifuges by spinning – heavier U-238 moves to outside 3) increases proportion of U-235 to about 3-5% of total (low enrichment) 4) This low enriched hex is compressed and turned into solid uranium-oxide fuel pellets which are packed into tubes called IXHOURGV which will be used in the core of a nuclear reactor. $GYDQWDJH More uranium is available for fission and a chain reaction can be sustained in a reactor to produce nuclear energy. 'LVDGYDQWDJH If the fuel is enriched to a high level (90% = weapons grade) it can be used in the core of a nuclear weapon. Possession of nuclear weapons is seen by many to be a threat to world peace.
9
IB 12
&KDLQ5HDFWLRQ± neutrons released from one fission reaction go on to initiate further reactions
Uncontrolled Chain Reaction
Uncontrolled nuclear fission: nuclear weapons Controlled nuclear fission: nuclear power production
1) Some material (control rod) absorbs excess neutrons before they strike another nucleus.
Controlled Chain Reaction
2) This leaves only one neutron from each reaction to produce another reaction. 3) If the total mass of uranium used is too small, too many neutrons will escape without causing further fissions so the reaction cannot be sustained. &ULWLFDO0DVV minimum mass of radioactive fuel (uranium) needed for a chain reaction to occur
7KH1XFOHDU5HDFWRU&RUH
)XHO5RGV enriched solid uranium
When neutrons are emitted from a fission reaction in the fuel rods, they have a very high kinetic energy and will pass right out of the fuel rod without colliding with another uranium nucleus to cause more fission. High energy neutrons cannot sustain a chain reaction. Therefore, a material is needed to slow them down. Typically, a material like water or graphite (called a “PRGHUDWRU”) is used to slow down these high-energy neutrons down to “thermal levels” (thermal neutrons § 1 eV) for use in further fission reactions to sustain the chain reaction. The high-energy neutrons slow down when they collide with the atoms in the moderator. To control the rate at which the thermal energy is produced, and therefore to control the temperature of the reactor core, FRQWURO URGV are used to speed up or slow down the chain reaction. These are rods made of a neutron-absorbing substance, like cadmium or boron. They are inserted in between the fuel rods and raised or lowered as needed. If the reaction is proceeding too fast (too hot) the rods are lowered and enough thermal neutrons are absorbed to slow down the reaction to the desired level. Conversely, if the reaction is too slow, the control rods are raised allowing more thermal neutrons to collide with uranium nuclei. 10
IB 12
How is the thermal energy released in the fission reactions used to generate electricity? The FRRODQW (which is often the same as the moderator) is fluid circulating around the fuel rods in the reactor core and is heated up by the thermal energy released in the fission chain reaction. This coolant in a closed loop (primary loop) flows through pipes in a tank of water known as the “KHDW H[FKDQJHU” Here the thermal energy of the hot coolant is transferred to cooler water in a secondary loop which turns it to steam. This steam expands against fan blades of turbines and turns a magnet is a coil of wire to generate electricity. 1. State the energy transformations in using nuclear fuels to generate electrical energy: Nuclear energy in fuel….thermal energy in coolant . . thermal energy in steam in heat exchanger…rotational mechanical energy/kinetic energy…electrical energy in turbines
2. Sketch a Sankey diagram for a typical nuclear power plant.
11
IB 12 3. Suppose the average power consumption for a household is 500 W per day. Estimate the amount of uranium-235 that would have to undergo fission to supply the household with electrical energy for a year. State some assumptions made in your calculation.
Assume plant is 100% efficient
Assume 200 MeV per fission
12
IB 12
4. A fission reaction taking place in a nuclear power station might be:
235 92
92 1
810 Qo141 56 %D 36 .U 3 0 Q
Estimate the initial amount of uranium-235 needed to operate a 600 MW reactor for one year assuming 40% efficiency and 200 MeV released for each fission reaction.
13
3OXWRQLXPDQG1XFOHDU5HDFWRUV
IB 12
Plutonium-239 is another nuclide used as nuclear fuel because of the energy it releases when it undergoes fission. However, it is not as naturally abundant as uranium and so it typically must be artificially produced as a by-product of uranium fission. In a uranium-fueled reactor, as the U-235 depletes over time, the amount of Pu-239 increases. This plutonium is then extracted (by reprocessing of the uranium fuel rods) for use in a plutonium reactor or in a nuclear warhead. How is plutonium-239 produced in a uranium reactor? It actually is produced from the nonfissionable isotope uranium-238 that occurs in large amounts in fuel rods. Uranium-238 doesn’t undergo nuclear fission but is considered “fertile” since it produces plutonium-239 by the following process. 238 92
239 239 810 Qo92 8 o93 1S 01 H Q 239 1So94 3X01 H Q
239 94
3X Qo %D 6U4 Q 140 56
96 38
239 94
3X 239.052157X
rest mass of 10 Q 1.008665X
239 93
1 0
rest mass of
rest mass of 140 56 %D 139.910581X 1 0
rest mass of 96 38 6U 95.921750X
1. Complete the nuclear reactions listed above. 2. Construct a nuclear energy level diagram for the series of nuclear reactions listed above.
3. Determine the amount of energy released in the fissioning of plutonium-239.
P PǻP 0.19383 u = 180 MeV
Some uranium reactors are even specially designed to produce (or “breed”) large amounts of plutonium and are known as EUHHGHUUHDFWRUV. They are designed so that the fuel rods are surrounded by a blanket of U-238 so that neutrons escaping from the U-235 fissions will induce the conversion of this U-238 to Pu-239. 14
6DIHW\,VVXHVDQG5LVNVLQWKH3URGXFWLRQRI1XFOHDU3RZHU
IB 12
8UDQLXP0LQLQJ x RSHQFDVWPLQLQJ environmental damage, radioactive waste rock (tailings) x XQGHUJURXQGPLQLQJ release of radon gas (mines need ventilation), radioactive rock is dangerous for workers, radioactive waste rock (tailings) x OHDFKLQJ Solvents are pumped underground to dissolve the uranium and then pumped back out. This leads to contamination of groundwater. 7KHUPDO0HOWGRZQ Overheating and melting of fuel rods may be caused by a malfunction in the cooling system or the pressure vessel. This overheating may cause the pressure vessel to burst sending radioactive material and steam into atmosphere (as in Chernobyl, Ukraine 1986). Hot material may melt through floor (as in Three Mile Island, Pennsylvania 1979), a scenario dubbed the “China syndrome.” The damage from these possible accidents is often limited by a containment vessel and a containment building. 1XFOHDU:DVWH x
/RZOHYHOZDVWH Radioactive material from mining, enrichment and operation of a plant must be disposed of. It’s often left encased in concrete.
x
+LJKOHYHOZDVWH a major problem is the disposal of spent fuel rods. Some isotopes have ½ lives of thousands of years. Plutonium’s is 240,000 years.
1) Some are stored under water at the reactor site for several years to cool off then sealed in steel cylinders and buried underground. 2) Some are reprocessed to remove any plutonium and useful uranium. The remaining isotopes have shorter ½ lives and the long-term storage need is reduced. 1XFOHDU:HDSRQV0DQXIDFWXUH x Enrichment technology could be used to make weapons grade uranium (85%) rather than fuel grade (3%) x Plutonium is most used isotope in nuclear weapons and can be gotten from reprocessing spent fuel rods &RPSDULQJ1XFOHDU)XHOWR)RVVLO)XHO $GYDQWDJHV 'LVDGYDQWDJHV
1. No global warming effect – no CO2 emissions
1. Storage of radioactive wastes
2. Waste quantity is small compared with fossils fuels
2. Increased cost over fossil fuel plants
3. Higher energy density
3. Greater risks in an accident (due to radioactive contamination)
4. Larger reserves of uranium than oil
15
1XFOHDU)XVLRQ
IB 12
1XFOHDU)XVLRQ Two light nuclei combine to form a more massive nucleus with the release of energy.
Naturally occurring fusion: main source of Sun’s energy – fusion of hydrogen to helium A probable mechanism for the Sun’s fusion is called the SURWRQSURWRQFKDLQ. 1 1
+11 + o12 +10 H Q
1 1
+12 + o3
2 +H J
3 2
+H32 +Ho42 +H11 +11 +
This chain is sometimes simplified to
4+ o+HHQHUJ\
1. If the total mass of four hydrogen nuclei is 6.693 x 10-27 kg and the mass of a helium nucleus
is 6.645 x 10-27 kg, determine the energy released in this simplified fusion reaction.
4.3 x 10-12 J
2. The Sun has a radius 5 of 7.0 x 108 m and emits energy at a rate of 3.9 x 1026 W. The nuclear reactions take place in the spherical core of the Sun of radius 0.255. Determine the number of nuclear reactions occurring per cubic meter
per second in the core of the Sun.
4.1 x 1012 m-3 s-1
16
IB 12
$UWLILFLDOO\LQGXFHGIXVLRQ Attempts have been underway since the 1950s to build fusion reactors. Experimental reactors have come very close to producing more energy than the amount of energy put in, but a commercial fusion reactor has yet to be built. 3ODVPD The fuel for a fusion reactor is known as a SODVPD. This is a high energy ionized gas in which the electrons and nuclei are separate. If the energy is high enough (that is, the plasma is hot enough), nuclei can collide fast enough to overcome Coulomb repulsion and fuse together. Heating the plasma to the required temperatures (10 million K) is challenging. The nuclei, since they are charged, are accelerated by means of magnetic fields and forces to high kinetic energies (high temperatures). 0DJQHWLFFRQILQHPHQW These charged particles are contained via magnetic fields and travel in a circle in a doughnut shaped ring called a “tokamak” which an acronym of the Russian phrase for “toroidal chamber with magnetic coils” (WRURLGDO QD\DNDPHUDVPDJQLWQ\PL NDWXVKNDPL). 3UREOHPVZLWKFXUUHQWIXVLRQWHFKQRORJ\ x
Maintaining and confining these very high-density and high-temperature plasmas for any length of time is very difficult to do.
x
Experimental reactors that currently can achieve fusion use more energy input than output which makes them not commercially efficient.
Comparison of Nuclear Fission and Nuclear Fusion
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17
6RODU3RZHU 6RODUKHDWLQJSDQHO DFWLYHVRODUKHDWHU converts light energy from Sun into thermal energy in water run through it
Use: heating and hot water
IB 12 3KRWRYROWDLFFHOO VRODUFHOO converts light energy from Sun into electrical energy
Use: electricity
Advantages of solar heating panel over solar cell: requires less (storage) area, less cost, more efficient
The amount (intensity) of sunlight varies with: a) time of day b) season (angle of incidence of sunlight – altitude of Sun in sky – Earth’s distance from Sun) c) length of day d) latitude (thickness of atmosphere) Which way should a solar panel or cell be facing in the Northern hemisphere? Why? South to receive Maximum radiation from the sun to provide maximum energy for whole day
$GYDQWDJHV
'LVDGYDQWDJHV 1. Large area needed to collect energy
1. Renewable source of energy 2. Only provides energy during daylight 2. Source of energy is free 3. No global warming effect – no CO2 emissions
3. Amount of energy varies with season, location and time of day
4. No harmful waste products
4. High initial costs to construct/install
18
IB 12
1. An active solar heater whose efficiency is 32% is used to heat 1400 kg of water from 200 C to 500 C. The average power received from the Sun in that location is 0.90 kW per m2. a) How much energy will the solar heater need to provide to heat the water?
b) How much energy will be needed from the Sun to heat the water?
c) Calculate the area of the solar heater necessary to heat the water in 2.0 hours.
2. A photovoltaic cell with an area of 0.40 m2 is placed in a position where the intensity of the Sun is 1.0 kW/m2. a) If the cell is 15% efficient, how much power does it produce?
b) If the potential difference across the cell is 5.0 mV, how much current does it produce?
c) Compare placing 10 of these solar cells in series and in parallel.
19
:LQG3RZHU
IB 12
Basic features of a horizontal axis wind turbine: a) Tower to support rotating blades. b) Blades that can be rotated to face into the wind. c) Generator. d) Storage system or connection to a distribution grid. Energy transformations: Solar energy heating Earth . . . Kinetic energy of air . . . kinetic energy of turbine . . electrical energy
1. Determine the power delivered by a wind generator:
2. Reasons why power formula is an estimate: a) Not all KE of wind is transformed into mechanical energy b) Wind speed varies over course of year c) Density of air varies with temperature d) Wind not always directed at 900 to blades
.( 1/ 2PY 2 P= W W 1§P· 2 3 ¨ ¸Y 2© W ¹ 1 § U9 · 2 3 ¨ ¸ Y 2© W ¹ 1 § U $G · 2 ¨ ¸Y 2 © W ¹ 1 3 UY Y2 2 1 $U Y3 3 2
3. Why is it impossible to extract this maximum amount of power from the air? a) Speed of air cannot drop to zero after impact with blades b) Frictional losses in generator and
turbulence around blades
3
4. Why are turbines not placed near one another?
a) Less KE available for next turbine b) Turbulence reduces efficiency of next turbine
20
IB 12
1. A wind turbine has a rotor diameter of 40 m and the speed of the wind is 25 m/s on a day when the air density is 1.3 kg/m3. Calculate the power that could be produced if the turbine is 30% efficient.
2. A wind generator is being used to power a solar heater pump. If the power of the solar heater pump is 0.50 kW, the average local wind speed is 8.0 m/s and the average density pf air is 1.1 kg/m3, deduce whether it would be possible to power the pump using the wind generator.
$GYDQWDJHV
1. Renewable source of energy
'LVDGYDQWDJHV
1. Large land area needed to collect energy since many turbines are needed
2. Source of energy is free 2. Unreliable since output depends on wind speed 3. No global warming effect – no CO2 emissions 3. Site is noisy and may be considered unsightly 4. No harmful waste products 4. Expensive to construct 21
:DYH3RZHU
IB 12
Energy can be extracted from water waves in many ways. One such scheme is shown here. 2VFLOODWLQJ:DWHU&ROXPQ 2:& ocean wave energy converter:
1. Wave capture chamber is set into rock face on land
where waves hit the shore.
2. Tidal power forces water into a partially filled
chamber that has air at the top.
3. This air is alternately compressed and decompressed
by the “oscillating water column.”
4. These rushes of air drive a turbine which generates
electrical energy.
Energy transformations: Kinetic energy of water . . . Kinetic energy of air . . . kinetic energy of turbine . . electrical energy
Determining the energy in each wavelength of the wave and the power per unit length of a wavefront
Energy in each wavelength of the wave PE = mgh PE=mgA
SRZHU
PE = (U V)gA
P
1 PE = (U ( O $/))gA 2 1 PE = $2 O gU / 2
1 2 $ O gU / 3 2 7 1 3 $2 YgU / 2 power per unit length
3( W
3 1 2 $ YgU / 2
How would this power estimate change if the waves were modeled as sine waves instead of square waves? 22
IB 12
1. Waves of amplitude 1.5 meter roll onto a beach with a speed of 10 m/s. Calculate: a) how much power they carry per meter of shoreline
b) the power along a 2 km stretch of beach.
2. Waves that are 6.0 meters high with a 100 meter wavelength roll onto a beach at a rate of one wave every 5.0 seconds. Estimate the power of each meter of the wavefront.
$GYDQWDJHV
'LVDGYDQWDJHV
1. Renewable source of energy
1. Can only be utilized in particular areas
2. Source of energy is free
2. High maintenance due to pounding of waves
3. No global warming effect – no CO2 emissions
3. High initial construction costs
4. No harmful waste products
23
+\GURHOHFWULF3RZHU
IB 12
There are many schemes for using water to generate electrical energy. But all hydroelectric power schemes have a few things in common. Hydroelectric energy is produced by the force of falling water. The gravitational potential energy of the water is transformed into mechanical energy when the water rushes down the sluice and strikes the rotary blades of turbine. The turbine's rotation spins electromagnets which generate current in stationary coils of wire. Finally, the current is put through a transformer where the voltage is increased for long distance transmission over power lines. By far, the most common scheme for harnessing the original gravitational potential energy is by means of VWRULQJZDWHULQODNHV, either natural or artificial, behind a dam, as illustrated in the top picture at right. A second scheme, called WLGDOZDWHUVWRUDJH, takes advantage of big differences between high and low tide levels in bodies of eater such as rivers. A barrage can be built across a river and gates, called sluices, are open to let the high-tide water in and then closed. The water is released at low tide and, as always, the gravitational potential energy is used to drive turbines to produce electrical energy. A third scheme is called SXPSHGVWRUDJHWater is pumped to a high reservoir during the night when the demand, and price, for electricity is low. During hours of peak demand, when the price of electricity is high, the stored water is released to produce electric power. A pumped storage hydroelectric power plant is a net consumer of energy but decreases the price of electricity. Energy transformations: Gravitational PE of water . . . Kinetic energy of water . . . kinetic energy of turbine . . electrical energy
$GYDQWDJHV
'LVDGYDQWDJHV
1. Renewable source of energy
1. Can only be utilized in particular areas
2. Source of energy is free
2. Construction of dam may involve land being buried under water
3. No global warming effect – no CO2 emissions 3. Expensive to construct 4. No harmful waste products 24
IB 12
1. A barrage is placed across the mouth of a river at a tidal power station. If the barrage height is 15 meters and water flows through 5 turbines at a rate of 100 kg/s in each turbine, calculate the power that could be produced if the power plant is 70% efficient. 2.6 x 104W Use average height for EP = mgh
2. A reservoir that is 1.0 km wide and 2.0 km long is held behind a dam. The top of this artificial lake is 100 meters above the river where the water is let out at the base of the dam. The top of the intake is 25 meters below the lake’s surface. Assume the density of water is 1000 kg/m3. a) Calculate the energy stored in the reservoir. 4.3 x 1013 J
b) Calculate the power generated by the water if it flows at a rate of 1.0 m3 per second through the turbine. 875 kW
25
8QLYHUVDO*UDYLWDWLRQ
IB 12
.HSOHU¶V7KUHH/DZVRI3ODQHWDU\0RWLRQ
/DZ All planets orbit the Sun in elliptical paths with the Sun at one focus.
/DZ An imaginary line joining any planet to the Sun sweeps out equal areas in equal time intervals.
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7 2DU 3 7 2 NU 3
/DZ The square of the orbital period of any planet is proportional to the cube of its average orbital radius.
7 DU
3 2
1HZWRQ¶V/DZRI8QLYHUVDO*UDYLWDWLRQ Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of the masses and that is inversely proportional to the square of the distance between them.
Two approximations used in deriving the law: 1. Masses are considered to be point masses. Point mass: infinitely small object (radius = 0) whose mass is P
2. The force between two spherical masses whose separation is large compared to their radii is the same as if the two spheres were point masses with their masses concentrated at the centers of the spheres. 6XQ
Mean Earth-Sun distance = 1.50 x 1011 m
(DUWK
Mean radius = 6.37 x 106 m Mean radius = 6.96 x 108 m
1
([WHQGHGVSKHULFDOERG\
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P1 P2 U2 * P1 P2 )J = U2 )J D
*UDYLWDWLRQDO&RQVWDQW G = 6.77 x 10-11 N m2/kg2
Re 2Re 3Re 4Re
1HZWRQ¶V'HULYDWLRQRI.HSOHU¶V7KLUG/DZ What provides the centripetal force for orbital motion? gravitation
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2.0 [1030 NJ
accepted value=1.99 [10 NJ 30
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What is the resultant gravitational force on the Earth from the Sun and Moon, as shown below? Average Earth-Sun distance = 1.50 x 1011 m
Average Earth-Moon distance = 3.84 x 108 m
6XQ
(DUWK
Mass = 1.99 x 1030 kg
Mass = 5.98 x 1024 kg
0RRQ Mass = 7.36 x 1022 kg
2
*UDYLWDWLRQDO)LHOG6WUHQJWK *UDYLWDWLRQDOILHOGVWUHQJWK at a point in a gravitational field:
IB 12
the gravitational force exerted per unit mass on a small/point mass
Symbol: g
Deriving formula for gravitational field strength at any point above the surface of a planet
3RLQWPDVV
Formula: g = Fg / m
g = Fg /m g = (GMm/r2)/m g = GM/r2
Units: N/kg (m/s2)
Type: vector
Deriving formula for gravitational field strength at the surface of a planet
1. What is the gravitational field strength of the Earth at its surface?
g = GM/r2 go = G Mp / Rp2
“g” at the surface of the Earth 2. What is the gravitational field strength at an altitude equal to the radius of Earth? 2
go = G ME/Re
([WHQGHGVSKHULFDOERG\
go
“g” ratio g/go = RE2/r2
Re 2Re 3Re 4Re
3
Average Earth-Moon distance = 3.84 x 108 m
(DUWK
IB 12
0RRQ
Mass = 5.98 x 1024 kg
Mass = 7.36 x 1022 kg
3. a) What is the resultant gravitational field strength at a point midway between the Earth and Moon?
b) What is the resultant gravitational force acting on a 1500. kg space probe at this location?
Average Earth-Moon distance = 3.84 x 108 m
(DUWK Mass = 5.98 x 1024 kg
0RRQ Mass = 7.36 x 1022 kg
4. a) Is there a point where the resultant gravitational field strength of the Earth and Moon is zero? If so, where?
b) What is the resultant gravitational force acting on a 1500. kg space probe at this location?
4
*UDYLWDWLRQDO3RWHQWLDO(QHUJ\
IB 12
Difference in gravitational potential energy between any two points in a gravitational field:
ǻEP = mgǻh
This difference is SDWKLQGHSHQGHQW. 1. Same ǻEP between any two points no matter what path is taken between them. 2. Work done in moving a mass between two points in a gravitational field is independent of the path taken. 3. ǻEP is zero between any two points at the same level no matter what path is taken. 4. ǻEP is zero for any closed path (a path that begins and ends at same point). “Old” formula for gravitational potential energy: Ep = mgh
discuss 2 problems with definition 1) “g” varies above surface 2) arbitrary base level
Base level: infinity
Gravitational PE at infinity: zero
EP = -400 J
Gravitational Potential Energy of a mass at a point in a gravitational field:
EP = -100 J
EP = 0
the work done in bringing a small point mass in from infinity to that point in the gravitational field Derivation of gravitational potential energy formula
5
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IB 12
Ep = - GMm/r EP = -Gm1m2/r Formula not valid inside planet
Ep at surface Ep = -GMpm/Rp
6\PERO 9 8QLWV - 7\SH VFDODU What is the gravitational potential energy of a 5000 kg satellite: a) on the surface of the Earth?
b) orbiting the Earth at an altitude of 200 km?
c) How much does the potential energy of the satellite increase when it is put into this orbit?
*UDYLWDWLRQDO3RWHQWLDO *UDYLWDWLRQDOSRWHQWLDO at a point in a gravitational field:
work done per unit mass to bring a small point mass in from infinity to that point in the gravitational field
)RUPXODV
*UDYLWDWLRQDO3RWHQWLDO YVGLVWDQFH
Difference in gravitational potential: ǻV = W/m
ǻV = ǻEp/m
Gravitational potential at a point: 6\PERO 9 V = Ep/m V = -GM/r
8QLWV -NJ 7\SH VFDODU
V at surface Vo = -GMp/Rp 6
IB 12
1. What is the gravitational potential due to the Earth’s gravitational field: a) at the surface of the Earth?
b) At a location three Earth radii from the center of the Earth?
c) What is the change in potential in moving from the surface to this new location?
d) What is the minimum amount of energy needed to lift a 5000 kg satellite to this location?
2. What is the net gravitational potential at a spot midway between the Earth and the Sun?
6XQ
(DUWK
3. Derive an expression for the gravitational potential at the surface of a planet in terms of the gravitational field strength.
7
(VFDSH6SHHG
IB 12
(VFDSH6SHHG: minimum initial speed an object must have at the surface of a planet in order to escape the gravitational attraction of the planet
Travel to infinity Just make it – means velocity is zero at infinity – means EK is zero at infinity as well as EP
Eo = Ep + Ek
Ef = Ep
3ODQHW
Assumptions: planet is isolated – ignore air resistance 'HULYDWLRQ Note: 1. Direction of travel is irrelevant – ǻEp is path independent
2. independent of mass of rocket
3. More speed (EK) is needed in real life since air friction is not negligible at lower altitudes
1. What is the escape speed for Earth?
2. If the Earth became a black hole, how large would it be?
8
6DWHOOLWH0RWLRQ
1DWXUDO6DWHOOLWHV Period of a satellite:
7 DU 2
72
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Acceleration of a satellite:
6)F PD F
v = 2ʌ r/T
v = 2ʌ r/T
ac = 4ʌ2 r/T2
3 2
7 DU “Weightlessness”
Orbital speed of a satellite:
ac = v2/r
3
NU 3
IB 12
ac = g = GM/r2
Free fall
Orbital motion
*0P PY2 U2 U *0 Y2 U *0 Y U Deep space
1. Compare the motion of satellites A and B. A – faster, less time (smaller period) B – slower, more time (longer period) T2/R3 = constant true for all satellites T = kR3/2
2. What happens to the required orbital speed if: a) the mass of the satellite increases? Nothing – speed is independent of mass
b) the satellite is boosted into a higher orbit? Satellite would orbit at a slower (tangential) speed
3. What would happen to a satellite if it encountered appreciable air resistance? Slow down, drop to lower orbit, and speed up, encounter even more air molecules (denser), cycle continues – spiral to Earth
9
(QHUJ\RI2UELWLQJ6DWHOOLWHV
IB 12
Compare the energies of the two orbiting satellites. Gravitational Potential Energy
Kinetic Energy
(QHUJ\'HULYDWLRQV
Total Energy
*UDSKVRIWKHHQHUJLHVRIDQRUELWLQJVDWHOOLWH
Gravitational Potential Energy
Kinetic Energy
Total Energy
RE
Comparisons:
A 1500 kg satellite is to be put into orbit around the Earth at an altitude of 200 km.
a) How much potential energy will the satellite have at this altitude?
b) How much kinetic energy will the satellite need to orbit at this altitude?
d) What is the orbital speed of the satellite?
e) What is the minimum amount of energy needed to lift the satellite from the surface of the Earth to this altitude?
c) What is the total amount of energy the satellite has at this altitude?
10
&RPSDULVRQV
IB 12
(TXLSRWHQWLDOVXUIDFH: a surface on which the potential is the same everywhere
1. The gravitational force does no work as a mass moves on along equipotential surface. 2. The work done in moving a mass between equipotential surfaces is path independent. 3. The work done in moving a mass along a closed path is zero. RQHSRLQWPDVV
WZRSRLQWPDVVHV 11
IB 12
On the diagram at right: a) Sketch the gravitational field around the point mass.
b) Sketch equipotential surfaces around the point mass. What is the relationship between the gravitational field and the equipotential surfaces? Perpendicular
Field lines point in direction
of decreasing potential
*UDYLWDWLRQDO3RWHQWLDO*UDGLHQW gradient: rate of change with respect to something “slope” or “derivative”
gravitational potential gradient: the gravitational field is the negative gradient of the gravitational potential with respect to distance
)RUPXOD
-70 J/kg
$ %
-80 J/kg
derive g = -ǻV/ǻr
What is the average gravitational field strength between equipotential surfaces A and B if they are 5.0 m apart?
12
3UDFWLFH4XHVWLRQV 1. a) Calculate the gravitational force the Sun exerts on the Earth
2. a) Calculate the strength of the gravitational field of the Sun at a location one million kilometers from the Sun.
3. a) Calculate the strength of the Sun’s gravitational field at the surface of the Earth.
IB 12
b) Compare this to the gravitational force that the Earth exerts on the Sun.
b) What is the Sun’s gravitational force at this point?
b) Explain why the net gravitational field strength at the surface of the Earth can be approximated as due solely to the Earth’s gravitational field.
13
IB 12 4. a) Calculate the resultant gravitational field at a spot midway between the Earth and Sun.
b) Compare the contributions from the Sun and the Earth to this resultant field.
c) What is the gravitational force acting on a 5000 kg space probe at this location?
5. A 5000kg satellite orbits Mars at a distance of 1000 km.
Mass of Mars: 6.42 x 1023 kg Mean planetary radius: 3.37 x 106 m
a) What is the gravitational potential at the surface of Mars?
0DUV
b) How much gravitational potential energy does the satellite have on the surface of Mars?
c) What is the gravitational potential at orbiting altitude?
d) How much gravitational potential energy does the satellite have at this altitude?
e) What is the minimum energy needed to lift the satellite to this altitude?
14
IB 12
6. A 5000. kg satellite is placed in a low altitude orbit. a) If the altitude is sufficiently low, what is the approximate radius of the satellite’s orbit?
b) Calculate the satellite’s orbital speed.
c) Calculate the orbital period of the satellite.
d) Calculate the gravitational potential energy of the satellite.
e) Calculate the kinetic energy of the satellite.
f) Calculate the total energy of the satellite.
g) What is the minimum amount of energy needed to lift the satellite into this orbit?
15
3K\VLFVDQG3K\VLFDO0HDVXUHPHQW
IB 12
(YHU\PHDVXUHGYDOXHKDVXQFHUWDLQW\
A child swings back and forth on a swing 10 times in 36.27s ± 0.01 s. How long did one swing take?
notice that multiple trials reduces uncertainty for a single repetition
(36.27 ± 0.01) / 10 = 3.627 s ± 0.001 s
Measurements of time are taken as: 14.23 s, 13.91 s, 14.76 s, 15.31 s. 13.84 s, 14.18 s. What value should be reported?
0HDQ: 14.37 s
*UHDWHVW5HVLGXDO 0.94
5HVLGXDOV: 14.37 – 13.84 = 0.53 15.31 – 14.37 = 0.94
5HSRUWHG9DOXH: 14.37 s ± 0.94 s = mean ± greatest residual
1. 0HDVXUHPHQW Record as many significant figures as the calibration of the measuring instrument allows SOXV one estimated digit.
Voltage ± uncertainty
V ± ǻV
2. 8QFHUWDLQW\ Record a reasonable uncertainty estimate that a) has one significant digit, and b) matches the measurement in place value (decimal place).
11.6 V ± 0.2 V
$EVROXWH8QFHUWDLQW\
)UDFWLRQDO8QFHUWDLQW\
3HUFHQWDJH8QFHUWDLQW\
ǻV
ǻV/V
ǻV/V · 100%
0.2 V
0.2 V / 11.6 V
0.2 V / 11.6 V · 100% = 1.7 %
&DOFXODWLRQVZLWK8QFHUWDLQWLHV $GGLWLRQ6XEWUDFWLRQ5XOH When two or more quantities are added or subtracted, the overall uncertainty is equal to the VXPRIWKHDEVROXWHXQFHUWDLQWLHV. Ex. 1: The sides of a rectangle are measured to be (4.4 ± 0.2) cm and (8.5 ± 0.3) cm. Find the perimeter of the rectangle.
4.4 + 8.5 + 4.4 + 8.5= 25.8 cm 0.2 + 0.3 + 0.2 + 0.3 = 1.0 cm 25.8 cm ± 1.0 cm
1
IB 12
0XOWLSOLFDWLRQ'LYLVLRQ5XOH
When two or more quantities are multiplied or divided, the overall uncertainty is equal to the VXPRIWKHSHUFHQWDJHXQFHUWDLQWLHV. Ex. 2: The sides of a rectangle are measured to be (4.4 ± 0.2) cm and (8.5 ± 0.3) cm. F ind the area of the rectangle.
4.4 x 8.5 = 37.4 cm2
0.2/4.4 = 4.55%
37.4 cm2 ± 8.08%
0.3/8.5 = 3.53%
37.4 cm2 ± 3.02192 cm2
Total = 8.08%
37 cm2 ± 3 cm2
3RZHU5XOH When the calculation involves raising to a power, PXOWLSO\WKHSHUFHQWDJHXQFHUWDLQW\E\WKHSRZHU. 1
(Don’t forget that [ [2 ) Ex. 3: The radius of a circle is measured to be 3.5 cm ± 0.2 cm. What is the area of the circle with its uncertainty?
DUHD SU 2 DUHD S(3.5)2 38.48FP 2
§0.2 · 2 ¨ u100% ¸ 2 5.71% 11.4% ©3.5 ¹
DUHD 38.48FP2 r11.4% DUHD 38.48FP2 r4.39FP2 DUHD 38FP 2 r4FP 2
2
([HUFLVHV
IB 12
1. Five people measure the mass of an object. The results are 0.56 g, 0.58 g, 0.58 g, 0.55 g, 0.59 g. How would you report the measured value for the object’s mass?
0.57 g ± 0.02 g
(mean ± greatest residual)
2. Juan Deroff measured 8 floor tiles to be 2.67 m ±0.03 m long. What is the length of one floor tile?
0.334 m ± 0.004 m
(0.33375 m ± 0.00375 m)
3. The first part of a trip took 25 ± 3 s, and the second part of the trip took 17 ± 2s. a. How long did the whole trip take?
42 s ± 5 s
b. How much longer was the first part of the trip than the second part? 8s±5s
4. A car traveled 600. m ± 12 m in 32 ± 3 s. What was the speed of the car?
18.75 m/s
12/600 = 2.000% and 3/32 = 9.375 % so total = 11.38% 18.75 m/s ± 11.38% 18.75 m/s ± 2.13375 m/s
Speed = 19 m/s ± 2 m/s 5. The time W it takes an object to fall freely from rest a distance G is given by the formula: where J is the acceleration due to gravity. A ball fell 12.5 m ± 0.3 m. How long did this take?
V W
1 § 0.3 · 1 u100% ¸ 2.4% 1.2% ¨ 2 ©12.5 ¹ 2
W
G J
W 1.60Vr1.2% W 1.60Vr0.0192V W 1.60Vr0.02V
3
$QDO\]LQJ'DWD*UDSKLFDOO\
IB 12
The masses of different volumes of alcohol were measured and then plotted (using *UDSKLFDO$QDO\VLV . Note there are three lines drawn on the graph – the best-fit line, the line of maximum slope, and the line of minimum slope. The slope and y-intercept of the best-fit line can be used to write the specific equation and the slopes and y-intercepts of the max/min lines can be used to find the uncertainties in the specific equation. The specific equation is then compared to a mathematical model in order to make conclusions.
*HQHUDO(TXDWLRQ: y= mx + b 6SHFLILF(TXDWLRQ M = (0.66 g/cm3)V + 0.65 g 8QFHUWDLQWLHV: slope: 0.66 g/cm3 ± 0.11 g/cm3 y-intercept: 0.65 g ± 3.05 g 0DWKHPDWLFDO0RGHO: D = M/V so M = DV
&RQFOXVLRQ3DUDJUDSK: 1. The purpose of the investigation was to determine the relationship between volume and mass for a sample of alcohol.
2. Our hypothesis was that the relationship is linear. The graph of our data supports our hypothesis since a best-fit line falls within the error bars of each data point. 3. The specific equation of the relationship is M = (0.66 g/cm3)V + 0.65 g.
4.
We believe that enough data points were taken over a wide enough range of values to establish this relationship. This relationship should hold true for very small volumes, although if it becomes too small for us to measure with our present equipment we won’t be able to tell, and for very large volumes, unless the mass becomes so large that the liquid will be compressed and change the density.
5. Zero falls within uncertainty range for y-intercept (0.65 g ± 3.05 g) so our results agree with math model and no systematic error is apparent
6. By comparison to the mathematical model we conclude that the slope of the graph represents the density. Therefore the density of the sample is 0.66 g/cm3 ± 0.11 g/cm3. 7. The literature value for the density of this type of alcohol is 0.72 g/cm3. Our results agree with the literature value since the literature value falls within the experimental uncertainty range of 0.66 g/cm3 ± 0.11 g/cm3 .
4
/LQHDUL]LQJ*UDSKV8VLQJ/RJDULWKPV
IB 12
This is a special linearizing (straightening) technique that works with general equations that are SRZHUIXQFWLRQV
3RZHU)XQFWLRQ y = c·xn 0HWKRGRIVWUDLJKWHQLQJ graph log y vs. log x
'HULYDWLRQ
“log-log plot”
y = cxn
Compare to y = mx + b
Take log of both sides
slope = n y-intercept = log c or
n
c = 10b
Log y = log (cx ) Log y = log c + log xn Log y = log c + n log x
([DPSOHV
m=2
y = cx2
y = cx-1
m = -1 or -2
y = kx-2
m = 0.5 y = kx
0.5
m=1 y = kx
1
5
/RJDULWKPLF6WUDLJKWHQLQJ
IB 12
Why use logarithms? use when you’re not sure what the type of relationship is –
use to check the exponent
5HVHDUFK4XHVWLRQ:
What is the relationship between kinetic energy and
speed for a uniformly accelerating object?
)LQGLQJ(UURU%DUVIRUWKH6WUDLJKWHQHG*UDSK 1. Error bars needed on only one axis – choose whichever axis has the most significant uncertainties. 2. Use the greatest residual for the data point with the highest percent uncertainty as the error bars on all data points. KE: log (5-2) = log 3 = .477 log 5 = .699 log (5+2) = log 7 = .845
residuals: .699 - .477 = .222 .845 - .699 = .146
greatest residual = .222 = .22
use for error bars on log KE axis
$QDO\VLV general equation: y = cxn slope = 1.980 = n
ORJ.(YVORJVSHHG
y-intercept= 0.7007 = log c so c = 100.7007 = 5.0 specific equation: KE = 5.0 v1.98
slope residuals: 1.98 – 1.40 = 0.58 2.60 – 1.98 = 0.62
slope: 1.98 ± 0.62
greatest residual: 0.62
3DUWLDO&RQFOXVLRQ
The purpose of the investigation was to determine the relationship between the kinetic energy and the speed of a uniformly accelerating object. Our hypothesis was that the relationship is quadratic and the graph of our original data supports our hypothesis since a best-fit parabola can be drawn within the error bars of all data points. The data was then linearized using logarithms. Using this graph, the specific equation for the relationship was found to be KE = 5.0 v1.98 . Since a value of 2 falls within the uncertainty range for the exponent of 1.98 ± 0.62, the data is consistent with a quadratic relationship between speed and kinetic energy. However, since the uncertainty range for the exponent is so large (30.%), the relationship might not be quadratic but some other power function.
6
([HUFLVHV±/LQHDUL]LQJ'DWDZLWK/RJDULWKPV
IB 12
In each example below, straighten each graph by logarithms. Then, write the specific equation for each relationship.
What is the most probable type of relationship in each case?
7LPH V V
'LVSODFHPHQW P
P
$FFHOHUDWLRQ PV PV
'LVWDQFH P P
Specific equation: d = (3.2)t1.9
0DVV NJ NJ
General Equation: y = c xn
)RUFH 1 1
Slope: 1.94 Y-intercept: 0.50 = log c c = 3.2
General Equation: y = c xn
Type of relationship: quadratic
Specific equation: a = (12)m-.99
Slope: -0.99 Y-intercept: 1.083 = log c c = 12.1
General Equation: y = c xn
Type of relationship: inverse
Specific equation: F = (15)d-1.9
Slope: -1.9 Type of relationship: inverse quadratic Y-intercept: 1.166 = log c c = 14.7
7
6FDODUVDQG9HFWRUV
IB 12
6FDODUV: quantities that have magnitude only
e.g. - Mass, time, volume, energy, distance, speed
9HFWRUV quantities that have magnitude and direction
e.g. - Velocity, displacement, acceleration, force, momentum, impulse, magnetic field strength, gravitational field strength, electric field strength
Notation: Bold italic ) or arrow hat
G )
$GGLQJ9HFWRUV Find the sum $ + %
6XEWUDFWLQJ9HFWRUV Find the difference $ - %
5HVROYLQJD9HFWRU LQWRLWV&RPSRQHQWV
Sin ș = y/r
Cos ș = x/r
y = r sin ș
x = r cos ș
Practice naming components
A sin ș
A sin 20 3. B cos 35 4.
A cos ș
A cos 20
2. B sin 35 62 sin 35 = 36 m
62 cos 35 = 51 m
8
IB 12
9
0HFKDQLFV
IB 12
.LQHPDWLFV±7KH6WXG\RI0RWLRQ
Symbols:
s = distance or displacement
v = final speed or velocity
u = initial speed or velocity
a = average acceleration
Equations:
Y
V XY also Y W 2
'Y Y X 'W W Y X DW D
VR V
YW
or
V
1 V XW W 2 2 2 2 Y X 2 DV
X Y W 2
Condition for applying equations for uniformly accelerated motion:
must be constant, smooth acceleration equations use average acceleration = instantaneous if acceleration is constant
average vs. instantaneous: over a period of time vs. at one instant
([DPSOH: Two friends bicycle 3.0 kilometers north and then turn to bike 4.0 kilometers east in 25 minutes.
a) What is their average speed?
distance/time 7.0 km / 25 min = 0.28 km/min x 60 min/hr = 16.8 = 17 km/hr
b) What is their average velocity?
displacement/time 5.0 km/ 25 min = 0.20 km/min x 60 min/hr = 12 km/hr Angle: 53o west of north
&RQVWDQW9HORFLW\
Time (s) Distance (m) Velocity (m/s) Acceleration (m/s2)
0 0 25 0
1 25 25 0
2 50 25 0
3 75 25 0
4 100 25 0
IB 12
&RQVWDQW$FFHOHUDWLRQ
Time (s) Distance (m) Velocity (m/s) Acceleration (m/s2)
a) What does the slope of a position-time graph represent? b) What does the slope of a velocity-time graph represent?
0
1
2
3
0 0 5
3 5 5
10 10 5
23 15 5
Instantaneous velocity - derivative Instantaneous acceleration - derivative
c) What does the area under a velocity time graph represent?
displacement - integral
'URSSLQJ 1. A stone is dropped from rest from the top of a tall building. After 3.00 s of free-fall, what is the displacement of the stone? What is its velocity?
s = ut + ½ at2
v = u + at
s = ½ at2 = -45 m
v = -30 m/s
discuss change of frame of reference – downward is positive
How would these graphs change in the presence of air resistance?
Terminal velocity: no acceleration – constant velocity – Fg = Fair
2
IB 12
7KURZLQJ8S A ball is thrown straight up in the air (shown here stretched out for clarity.) Sketch velocity and acceleration vectors at each instant.
A football game customarily begins with a coin toss to determine who kicks off. The referee tosses the coin up with an initial speed of 6.00 m/s. In the absence of air resistance, how high does the coin go above its point of release? How long is it in the air? v2 = u2 + 2 as 02 = (6)2 + 2(-10)s s = 1.8 m
v = u + at 0 = 6.00 + (-10)t t = 0.6 s x 2 = 1.2 s
discuss change of frame of reference – downward is positive
3URMHFWLOH0RWLRQ resultant of two independent components of motion
1. Vertical: constant acceleration (in absence of air resistance)
2. Horizontal: constant velocity – no horizontal acceleration
+RUL]RQWDO3URMHFWLOH
A ball is shot horizontally off a cliff that is 100. m high at a speed of 25 m/s. How long does it take to hit the ground? How far away from the base of the cliff does it land?
y-direction:
x-direction:
s = ut + ½ at2 -100 = 0 + ½(-10)t2 t = 4.5 s
s = ut + ½ at2 s = 25 (4.5) + 0 s = 113 m
3
$QJOHG3URMHFWLOH
IB 12
1. Break initial velocity into horiz and vert components
2. maximum height occurs after ½ time 3. maximum range occurs after fulltime and at 450
4. air resistance: not as high nor as far – show on diagram A football was kicked with a speed of 25 m/s at an angle of 30.0o to the horizontal. Determine how high it went and where it landed. Components
time:
Height:
x: cos 300 = vi/25 m/s
v = u + at
s = ut + ½ at2
vi = 21.7 m/s
0 = 12.5 m/s + (-10)(t)
s = (12.5 m/s)(1.25 s) + ½ (-10 m/s2)(1.25 s)2
t = 1.25 s
s = 7.8 m
y: sin 300 = vi/25 m/s total time = 2(1.25) = 2.5 vi = 12.5 m/s
Range: s = ut + ½ at2 s = (21.7 m/s)(2.5 s) + 0 s = 54 m
6WDWLFVDQG'\QDPLFV±7KH6WXG\RI)RUFHV 1HZWRQ¶V/DZVRI0RWLRQ 1. An object at rest remains at rest and an object in motion remains in motion at a constant speed in a straight line (constant velocity) unless acted on by unbalanced forces. (An object continues in uniform motion in a straight line or at rest unless a resultant (net) external force acts on it.) 2. When unbalanced forces act on an object, the object will accelerate in the direction of the resultant (net) force. The acceleration will be directly proportional to the net force and inversely proportional to the object’s mass. (The resultant force on an object is equal to the rate of change of momentum of the object.) 3. For every action on one object, there is an equal and opposite reaction on another object. (When two bodies A and B interact, the force that A exerts on B is equal and opposite to the force that B exerts on A.) 4
1HZWRQ¶V6HFRQG/DZ
F = ǻp / ǻt
For constant mass
F = ǻ(mv) / ǻt
F = m (ǻv / ǻt)
IB 12
F=ma
QG/DZRUUG /DZ" Net force on ball: not zero so it accelerates – not in equilibrium Fnet = Fg = mg
Action-Reaction pairs: Earth pulls ball down Ball pulls Earth up FEB = -FBE mA = -Ma
Net force on block: zero at rest – in equilibrium Fnet = FN - Fg = 0
Action-Reaction pairs: Earth pulls block down Block pulls Earth up block pushes down on table table pushes up on block
)J
Translational equilibrium: net force acting on object is zero – no acceleration
1. Find the resistive force F caused by the drag of the water on the boat moving at a constant velocity in the diagram shown.
5
IB 12
2. Find the tension in each cable supporting the 600 N cat burglar pictured.
600 997
796
3. A 20.0-kg floodlight in a park is supported at the end of a horizontal beam of negligible mass that is hinged to a pole, as shown. A cable at an angle of 30.0° with the beam helps to support the light. Find (a) the tension in the cable and (b) the horizontal and vertical forces exerted on the beam by the pole.
7
+ 9
4. How would your answers change if the mass of the beam shown above was not negligible? T increase show how reaction force is oriented
a)
400N
b) Rx = 346 Ry = 0 1
G
5. Indicate the direction of the reaction force from the floor and the reaction force from the wall for the situation shown below.
FW
FN Fg1 Fg2 6 Ff
:HLJKW0DVVDQGWKH1RUPDO)RUFH 0DVV 1) the amount of matter in an object 2) the property of an object that determines its resistance to a change in its motion (a measure of the amount of inertia of an object)
IB 12
:HLJKW the force of gravity acting on an object Property: Varies from place to place
Symbol : Fg or W Units : N
Property: Remains constant
Symbol : m
Units : kg
(OHYDWRUV In each case, the scale will read . . . the normal or reaction force, not the weight
Calculate the acceleration of the man in each case.
ȈF=0
Ȉ F = ma
ȈF=ma
ȈF=ma
FN – F g = 0
FN – F g = m a
FN – Fg = m a
– Fg = m a
FN – 700 = 0
1000 – 700 = 70 a
400 – 700 = 70 a
– 700 = 70 a
FN = 700 N
a = +4.3 m/s2
a = - 4.3 m/s2
a = -10 m/s2
,QFOLQHG3ODQH±Assume the box shown is in equilibrium and draw the . . . Free-body diagram
Head-to-tail vector diagram
Concurrent vector diagram Concurrent vector diagram with perpendicular components
FN Ff
Fg ș ș
7
IB 12 Calculate the force of friction acting on this box if it accelerates down the incline at a rate of 0.67 m/s2. 4.5 kg
Fnet = ma = 4.5 (0.67) = 3.0 N
Fg|| = mg sin ș = 15 N
Ff = 12 N
200
8QLIRUP&LUFXODU0RWLRQ
8QLIRUP&LUFXODU0RWLRQ constant speed and constant radius
3HULRG time take for one complete cycle symbol: T units: s
1. The direction of the object’s instantaneous velocity is always tangent to the circle in
the direction of motion.
2. Since the direction of the object’s motion is always changing, its velocity is always
changing therefore the object is always accelerating and is never in equilibrium.
3. Direction of net force – towards the center - centripetal
)RUPXODV
V bar = distance / time V bar = circumference / period V bar = 2ʌr / T
V = 2ʌr / T
V bar = 2ʌr / T ac = v2 / r
ac = (2ʌr / T)2 / r ac = (4ʌ2r2 / T2) / r
F = m a
Fc = m ac
Fc = mv2 / r
ac = 4 ʌ2 r / T2
The phrase “centripetal force” does not denote a new and separate force created by nature. The phrase merely labels the net force pointing toward the center of the circular path, and this net force is the vector sum of all the force components that point along the radial direction. 1. The model airplane shown has a mass of 0.90 kg and moves at a constant
speed on a circle that is parallel to the ground. Find the tension 7 in the
guideline (length = 17 m) for a speed of 19 m/s.
8
IB 12 2. At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of the riders “plastered” against the wall. For a particular ride with a radius of 8.0 m and a top speed of 21 m/s, calculate the reaction force and the friction force from the wall acting on a 60. kg rider. Which of these is the centripetal force?
3. A 2100-kg demolition ball swings at the end of a 15-m cable on the arc of a vertical circle. At the lowest point of the swing, the ball is moving at a speed of 7.6 m/s. Determine the tension in the cable.
:RUN3RZHUDQG(IILFLHQF\ :RUN product of force and displacement in the direction of the force )RUPXOD W = (F cos ș) s
8QLWV N m or Joules (J)
W = F s cos ș ș is angle between F and s
3RZHU 1) the rate at which work is done
)RUPXOD P = W/ t = E/t = Q/t
$OWHUQDWH)RUPXOD P=W/t = (F cos șd) / t = F v cos ș
2) the rate at which energy is transferred (IILFLHQF\: 1) ratio of useful work done by a system to the total work done by the system 2) ratio of useful energy output of a system to the total energy input to the system 3) ratio of useful power output of a system to total power input to the system
7\SH Scalar but can be positive or negative 8QLWV J/s = Watts (W) 7\SH Scalar )RUPXOD e = useful out/ total in 9
1. A 45.0-N force is applied to pull a luggage carrier an angle T = 50° for a distance of 75.0 m at a constant speed.IB 12 Find the work done by the applied force. W = F s cos ș WA = (45.0 N)(75 m) cos 50o = 2170 J
2. a) How much work is done dragging the 5.00 kg box to the top of the hill shown if the hill exerts an average friction force of 5.0 N?
b) Compare your answer to the amount of work done lifting the box straight up to the top of the hill.
c)
Calculate the power expended if the box is dragged to the top in 15 seconds.
d) Calculate the efficiency of dragging the box rather than lifting the box.
'HWHUPLQLQJ:RUN'RQH*UDSKLFDOO\
Work = area under curve W=½bh W=½Fs W = (avg force) (displacement)
Work = area under curve W=bh W=fs
1. Work done by a constant force
2. Work done by a constantly varying force ex. stretching a spring
10
(QHUJ\ 1.
2.
3.
IB 12
4.
7\SHVRI(QHUJ\ 1. Kinetic energy
5.
)RUPXODV 1. EK = ½ mv2
(energy of motion)
2. Gravitational Potential energy (energy of position)
2. EP = mgh
3. Elastic potential energy
3. Eelas = ½ kx2
4. Internal energy (thermal energy)
4. Q = mcǻt
5. Chemical Potential energy (stored in chemical bonds) Electrical energy Light energy
5. Ee = Pt = VIt = I2Rt = V2/R t
Q = mL
&RQVHUYDWLRQRI(QHUJ\3ULQFLSOH ,QDQLVRODWHGV\VWHPWKHWRWDODPRXQWRIHQHUJ\UHPDLQVFRQVWDQW 1. A motorcyclist is trying to leap across a canyon by driving horizontally off the cliff at a speed of 38.0 m/s. Ignoring air resistance, find the speed with which the cycle strikes the ground on the other side.
2. What is the speed of the box at the bottom of the incline if an average frictional force of 15 N acts on it as it slides?
160 N
20 meters 300
11
/LQHDU0RPHQWXPDQG,PSXOVH /LQHDU0RPHQWXP: the product of an object’s mass and velocity
A 1500. kg car is traveling east at a speed of 25.0 m/s.
inertia (inertial mass) m = 1500 kg scalar
Compare its inertia, momentum, and kinetic energy.
IB 12
)RUPXOD
8QLWV
p = mv
kg m/s
momentum
kinetic energy
p = (1500) (25) = 3.75 x 104 kg m/s, east vector
Ek = ½ (1500)(25)2 = 468750 J = 4.69 x 105 J scalar
Alternate formula for kinetic energy:
How does the momentum of an object change? A net external force acts for a finite amount of time ,PSXOVH: (the change in momentum of a system) the product of the average force and the time interval over which the force acts
8QLWV
7\SH
Ns or
vector
Usually . . . Force is not instantaneous and is not constant
kg m/s
'HULYDWLRQRI,PSXOVH)RUPXOD
'S 'W
'S )'W
'(PY) ) 'W P'Y )'W (if mass is constant) 'Y - 'S ) 'W P )
Graphically, the impulse is J = Favg ǻt = area
If force is linear: J = ½ Fmax ǻt
12
%RXQFLQJDQG,PSXOVH
IB 12
A 0.50 kg basketball hits the floor at a speed of 4.0 m/s and rebounds at 3.0 m/s. Calculate the impulse applied to it by the floor. Calculation:
In general:
K K 'S P'Y K K K 'S P Y I YL K 'S P Y I YL
'S
'S
P'Y NJ( 4.0 (3.0)) 0.50 'S 7.0NJP / V
Velocity vs. time graph for bounce
Force
Velocity
Force vs. time graph for bounce
Time
Time
7KH3ULQFLSOHRI&RQVHUYDWLRQ RI/LQHDU0RPHQWXP
7KHWRWDOPRPHQWXPRIDQLVRODWHGV\VWHPUHPDLQVFRQVWDQW SEHIRUH SDIWHU 7\SHVRI,QWHUDFWLRQV
%RXQF\
6WLFN\
([SORVLRQ
13
&ROOLVLRQV (ODVWLFFROOLVLRQ: a collision in which the total kinetic energy is conserved
IB 12
,QHODVWLFFROOLVLRQ: a collision in which the total kinetic energy is not conserved
Where does some of the mechanical energy go in an inelastic collision? energy of deformation, internal energy, sound energy
1. A freight train is being assembled in a switching yard, and the figure below shows two boxcars. Car 1 has a mass of P1 = 65×103 kg and moves at a velocity of Y01 = +0.80 m/s. Car 2, with a mass of P2 = 92×103 kg and a velocity of Y02 = +1.3 m/s, overtakes car 1 and couples to it. Neglecting friction, find the common velocity Yf of the cars after they become coupled.
2. Is this collision elastic or inelastic? Justify your answer.
3. A EDOOLVWLFSHQGXOXP is sometimes used in laboratories to measure the speed of a projectile, such as a bullet. A ballistic pendulum consists of a block of wood (mass P2 = 2.50 kg) suspended by a wire of negligible mass. A bullet (mass P1 = 0.0100 kg) is fired with a speed Y01. Just after the bullet collides with it, the block (with the bullet in it) has a speed Yf and then swings to a maximum height of 0.650 m above the initial position (see part Eof the drawing). Find the speed Y01 of the bullet, assuming that air resistance is negligible.
14
1XFOHDU3K\VLFV
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1XFOLGH: a particular type of nucleus 1XFOHRQ: a proton or a neutron $WRPLFQXPEHU = SURWRQQXPEHU number of protons in nucleus 0DVVQXPEHU $ QXFOHRQQXPEHU number of protons + neutrons 1HXWURQQXPEHU 1 number of neutrons in nucleus (N = A – Z) ,VRWRSHV: nuclei with same number of protons but different numbers of neutrons x 1 u = 1.661 x 10-27 kg x 1 u = 1 g/mol x 1 u = 931.5 MeV/c2
8QLILHGDWRPLFPDVVXQLW X 1/12th the mass of a carbon-12 nucleus $WRPLFPDVV§ A * u
$ =
;
56 26
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27 13
$O
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&
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&
$WRPLF1XPEHU
26
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56 u
27 u
12 u
14 u
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56 g
27 g
12 g
14 g
How big are atomic nuclei? 10-15m – 10-14 m How do we know this? Alpha-particle scattering experiments (Geiger-Marsden)
How do we know that neutrons exist? By the existence of isotopes How do we know that isotopes exist? By measurements in the mass spectrometer
Formulas: ET = ET EK = Ee ½ mv2 = qV ½ mv2 = q(kQ/r)
Formulas: v = E/B Note that most nuclei have approximately the same . . .density = 2 x 1017 kg/m3 = 1014 times denser than water
r = mv/qB 1
1XFOHDU6WDELOLW\
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What interactions exist in the nucleus? 1. *UDYLWDWLRQDO (long range) attractive but very weak/negligible 2. &RXORPERU(OHFWURPDJQHWLF (long range) repulsive and very strong between protons 3. 6WURQJQXFOHDUIRUFH (short range) attractive and strongest – between any two nucleons 4. :HDNQXFOHDUIRUFH (short range) involved in radioactive decay
Why are some nuclei stable while others are not?
The Coulomb force is a long-range force which means that every
proton in the nucleus repels every other proton. The strong
nuclear force is an attractive force between any two nucleons
(protons and/or neutrons). This force is very strong but is short
range (10-15 m) which means it only acts between a nucleon and
its nearest neighbors. At this range, it is stronger than the
Coulomb repulsion and is what holds the nucleus together.
Neutrons in the nucleus play a dual role in keeping it stable. They
provide for the strong force of attraction, through the exchange of
gluons with their nearest neighbors, and they act to separate
protons to reduce the Coulomb repulsion.
Each dot in the plot at right represents a stable nuclide and the
shape is known as the “band (or valley) of stability.” With few
exceptions, the naturally occurring stable nuclei have a number 1 of
neutrons that equals or exceeds the number = of protons. For small
nuclei (Z < 20), number of neutrons tends to equal number of
protons (N = Z).
As more protons are added, the Coulomb repulsion rises faster than the strong force of attraction since the Coulomb force acts throughout the entire nucleus but the strong force only acts among nearby nucleons. Therefore, more neutrons are needed for each extra proton to keep the nucleus together. Thus, for large nuclei (Z > 20), there are more neutrons than protons (N > Z). After Z = 83 (Bismuth), adding extra neutrons is no longer able to counteract the Coulomb repulsion and the nuclei become unstable and decay in various ways. Nuclei above (to the left of) the band of stability have too many neutrons and tend to decay by alpha or beta-minus (electron) emission, both of which reduce the number of neutrons in the nucleus. Nuclei below (to the right of) the band of stability have too few neutrons and tend to decay by betaplus (positron) emission which increases the number of neutrons in the nucleus.
2
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The total mass of a nucleus is always less than the sum of the masses its nucleons. Because mass is another
manifestation of energy, another way of saying this is the total energy of the nucleus is less than the combined
energy of the separated nucleons.
0DVVGHIHFW PDVVGHILFLW ǻP Difference between the mass of the nucleus and the sum of the masses of its individual nucleons 1XFOHDUELQGLQJHQHUJ\ ǻ(
)RUPXODV
1. energy released when a nuclide is assembled from its individual components 2. energy required when nucleus is separated into its individual components
PQXFOHXV ǻP PQXFOHRQV '( 'P F
Different nuclei have different total binding energies. As a general trend, as the atomic number increases . . . the total binding energy for the nucleus increases. 3DUWLFOH 3URWRQ 1HXWURQ (OHFWURQ
(OHFWULF&KDUJH (OHFWULF&KDUJH H & +1 0 -1
+1.60 x 10-19 0 -1.60 x 10-19
5HVW0DVV NJ
5HVW0DVV X
5HVW0DVV 0H9F
1.673 x 10-27 1.675 x 10-27 9.110 x 10-31
1.007276 1.008665 0.000549
938 940 0.511
1. The most abundant isotope of helium has a 24He nucleus whose mass is 6.6447 × 10-27 kg. For this nucleus, find the mass defect and the binding energy.
helium: 0.050404 x 10-27 kg
4.53636 x 10-12 J
2. Calculate the binding energy and mass defect for 816O whose measured mass is 15.994915 u.
Oxygen: 0.132613 u 123.5 MeV
3
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To see how the nuclear binding energy varies from nucleus to nucleus, it is useful to compare the binding
energy for each nucleus on a per-nucleon basis, as shown in the graph below.
a) This graph is used to compare the energy states of different nuclides and to determine what nuclear reactions are energetically feasible. As binding energy per nucleon increases so does the stability of the nucleus. +LJKHU binding energies represent ORZHU energy states since more energy was released when the nucleus was assembled. b) Binding energy per nucleon increases up to a peak at 2656Fe then decreases, so 2656Fe is the most stable nuclide. Most nuclides have a binding energy per nucleon of about 8 MeV. Lighter nuclei are held less tightly than heavier nuclei. c) Nuclear reactions, both natural (radioactive decay) and artificial/induced (fission, fusion, bombardments) occur if they increase the binding energy per nucleon ratio. Fusion occurs for light nuclei (below 2656Fe) and fission occurs for heavy nuclei (above 2656Fe). d) For both natural and induced nuclear reactions, the total rest mass of the products is less than the total rest mass of the reactants since energy is released in the reaction. Also, the products are in a lower energy state since energy was released in the reaction and so the products have a greater binding energy per nucleon than the reactants. 1. Use the graph above to estimate the total binding energy of an oxygen-16 nucleus. 8 MeV x 16 = 128 MeV
7\SHVRI1XFOHDU5HDFWLRQV $UWLILFLDO ,QGXFHG 7UDQVPXWDWLRQ A nucleus is bombarded with a nucleon, an alpha particle or another small nucleus, resulting in a nuclide with a different proton number (a different element). 1XFOHDU)XVLRQ Two light nuclei combine to form a more massive nucleus with the release of energy. 1XFOHDU)LVVLRQ A heavy nucleus splits into two smaller nuclei of roughly equal mass with the release of energy. 1DWXUDO5DGLRDFWLYLW\ When an unstable (radioactive) nucleus disintegrates spontaneously, the nucleus emits a particle of small mass and/or a photon.
5HOHDVHRIHQHUJ\LQQXFOHDUUHDFWLRQV:
P PǻP
Energy is usually released in the form of kinetic energy for the products. %LQGLQJHQHUJ\SHUQXFOHRQ greater for product nuclei than for original nuclei since energy is released. 4
5DGLRDFWLYH'HFD\5HDFWLRQV
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$OSKD'HFD\ Alpha particle: helium nucleus, Į, 24He Example reaction:
226 88 $ =
222 5Do86 5Q 24 +HHQHUJ\
$4 ; o=2 <24 +H HQHUJ\
Where does the kinetic energy come from? Rest mass of nucleons Result: nucleus is in a more stable state with higher binding energy and higher BE per nucleon since it released energy 1. A radium nucleus, initially at rest, decays by the emission of an alpha particle into radon in the reaction described above. The mass of 88226Ra is 226.025402 u and the mass of 86222Rn is 222.017571 u and the mass of the alpha particle is 4.002602 u. a) Calculate the energy released in this decay.
ǻm = 0.005229 u 4.87 MeV
b) Compare the momenta, speeds, and kinetic energies of the two particles produced by this reaction.
c) If the kinetic energy of the alpha particle is 4.77 MeV, calculate its speed.
d) Calculate the recoil speed of the radon nucleus. Use momentum – no need to change units
5
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Beta-minus particle: electron, ȕ ,
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0 -1 e
+
Beta-plus particle: positron, ȕ , +10e Consider the following two “mysterious” results of beta decay: a) Observe the before and after picture of beta decay.
What’s wrong?
b) Inspect the graph of kinetic energy carried away by
the beta particles. Why are so few beta particles
leaving with the majority of the kinetic energy?
Where did this missing kinetic energy go?
Conclusion: there is third particle involved with beta decay that carries away some KE and momentum – virtually undetectable
Neutrino and anti-neutrino: fundamental particles - no charge – very small mass %HWDPLQXVGHFD\
%HWDSOXVGHFD\
Example reaction:
Example reaction:
14 6
0 0 &o14 7 11 H 0 Q HQHUJ\
0 0 1o12 6 &1 H0 Q HQHUJ\
12 7
General equation:
$ =
General equation:
$ ; o=1 <01 H 00 Q HQHUJ\
$ ; o =1 <01 H00 Q HQHUJ\
$ =
How does this happen? Weak force
How does this happen? Weak force 1 0
Qo11 S01 H 00 Q HQHUJ\
1 1
S o10 Q01 H00 Q HQHUJ\
*DPPD'HFD\ Gamma particle: high energy photon, J Example reaction:
12 6
&* o12 6 & b HQHUJ\
General equation:
$ =
;* o = $ ; b HQHUJ\
Before Decay
After Decay
Where does the photon (energy) come from? Rest mass of nucleons 6
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The nucleus itself, like the atom as a whole, is a quantum system with allowed states and discrete energy levels. The nucleus can be in any one of a number of discrete allowed excited states or in its lowest energy relaxed state. When it transitions between a higher energy level and a lower one, it emits energy in the form of alpha, beta, or gamma radiation. When an alpha particle or a gamma photon is emitted from the nucleus, only discrete energies are observed. 7KHVHGLVFUHWHHQHUJ\VSHFWUDJLYHHYLGHQFHWKDWD QXFOHXVKDVHQHUJ\OHYHOV. (However, the spectrum of energies emitted as beta articles is continuous due to its sharing the energy with a neutrino or antineutrino in any proportion.)
Importance: discrete energy spectra give evidence for nuclear energy levels
$OSKDVSHFWUD
%HWDVSHFWUD
*DPPD VSHFWUD
Discrete
continuous
discrete
,RQL]LQJ5DGLDWLRQ ,RQL]LQJ5DGLDWLRQ – As this radiation passes through materials, it “knocks off” electrons from neutral atoms thereby creating an ion pair: free electrons and a positive ion. This LRQL]LQJSURSHUW\ allows the radiation to be detected but is also dangerous since it can lead to mutations in biologically important molecules in cells, such as DNA.
Į
ȕ
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3DUWLFOH
helium nucleus
Electron or positron
high-energy photon
3HQHWUDWLRQDELOLW\
low
medium
high
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less than 1 meter
effectively infinite
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a few cm
'HWHFWLRQRI5DGLDWLRQ WKH*HLJHU0XOOHUWXEH *HLJHUFRXQWHU The Geiger counter consists of a gas-filled metal cylinder. The Įȕor Ȗ rays enter the cylinder through a thin window at one end. Gamma rays can also penetrate directly through the metal. A wire electrode runs along the center of the tube and is kept at a high positive voltage (1000-3000 V) relative to the outer cylinder. When a high-energy particle or photon enters the cylinder, it collides with and ionizes a gas molecule. The electron produced from the gas molecule accelerates toward the positive wire, ionizing other molecules in its path. Additional electrons are formed, and an avalanche of electrons rushes toward the wire, leading to a pulse of current through the resistor 5. This pulse can be counted or made to produce a "click" in a loudspeaker. The number of counts or clicks is related to the number of disintegrations that produced the particles or photons.
%LRORJLFDO(IIHFWVRI,RQL]LQJ5DGLDWLRQ Alpha and beta particles have energies typically measured in MeV. To ionize an atom requires about 10 eV so each particle can potentially ionize 105 atoms before they run out of energy. When radiation ionizes atoms that are part of a living cell, it can affect the ability of the cell to carry out its function or even cause the cell wall to rupture. In minor cases, the effect is similar to a burn. If a large number of cells that are part of a vital organ are affected then this can lead to death. Alternatively, instead of causing the cell to die, the damage done by ionizing radiation might just prevent cells from dividing and reproducing. Or, it could be the cause of the transformation of the cell into a malignant form. If these malignant cells continue to grow then this is called cancer. The amount of harm that radiation can cause is dependent on the number and energy of the particles. When a gamma photon is absorbed, the whole photon is absorbed so one photon can ionize only one atom. However, the emitted electron has so much energy that it can ionize further atoms, leading to damage similar to that caused by alpha and beta particles. On a positive note, rapidly diving cancer cells are very susceptible to the effects of radiation and are more easily killed than normal cells. The controlled use of the radiations associated with radioactivity is of great benefit in the treatment of cancerous tumors.
7
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5DGLRDFWLYHGHFD\:
1) 5DQGRPSURFHVV: It cannot be predicted when a particular nucleus will decay, only the
probability that it will decay.
2) 6SRQWDQHRXVSURFHVV It is not affected by external conditions. For example, changing the pressure or temperature of a sample will not affect the decay process. 3) 5DWHRIGHFD\GHFUHDVHVH[SRQHQWLDOO\ZLWKWLPH Any amount of radioactive nuclei will reduce to half its initial amount in a constant time, independent of the initial amount. +DOIOLIH 7
Units: time = s or hr or d or yr
- the time taken for ½ of the radioactive nuclides in a sample to decay the time taken for the activity of a sample to decrease to ½ of its initial value
1 = number of radioactive nuclei originally present 1 = number of radioactive nuclei present at any one time
Radioactive X ĺ stable Y + particle Show how amount of daughter Y mirrors X
A nuclide X has a half-life of 10 s. On decay the stable nuclide Y is formed. Initially a sample contains only atoms of X. After what time will 87.5% of the atoms in the sample have decayed into nuclide Y? 30 s
8
$FWLYLW\
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$FWLYLW\ $ – the number of radioactive disintegrations per unit time (decay rate) 8QLWV )RUPXOD
'1 $ 'W
6WDQGDUG XQLWV:
decays/time = s-1 or hr-1or d-1 or yr-1
Becquerel (Bq) 1 Bq = 1 decay per second
1. A sample originally contains 8.0 x 1012 radioactive nuclei and has a half-life of 5.0 seconds. Calculate the activity of the sample and its half-life after:
a) 5.0 seconds 8.0 x 1011 Bq 5.0 s
b) 10. seconds 6.0 x 1011 Bq 5.0 s
c) 15 seconds 4.7 x 1011 Bq 5.0 s
2. Samples of two nuclides X and Y initially contain the same number of radioactive nuclei, but the half-life of nuclide X is greater than the half-life of nuclide Y. Compare the initial activities of the two samples.
Activity of sample Y is greater
$FWLYLW\
$D 1 $ O 1 ,QLWLDO$FWLYLW\
$0
O 10
7KH5DGLRDFWLYH'HFD\/DZ The rate at which radioactive nuclei in a sample decay (the activity) is proportional to the number of radioactive nuclei present in the sample at any one time.
[As the number of radioactive nuclei decreases, so does the average rate of decay (the activity).] The initial activity (A0) is directly related to the number of radioactive nuclei originally present (N0) in the sample.
3. The isotope Francium-224 has a half-life of 20 minutes. A sample of the isotope has an initial activity of 800 disintegrations per second. What is the approximate activity of the sample after 1 hour?
N = 1/8 N0
So A = 1/8 A0
A = 100 disintegrations s-1
9
IB 12 8QLWV inverse time
'HFD\FRQVWDQW Ȝ – constant of proportionality between the decay rate (activity) and the number of radioactive nuclei present.
= s-1 or hr-1or d-1 or yr-1
- probability of decay of a particular nuclei per unit time.
'HULYLQJWKH5DGLRDFWLYH'HFD\/DZ
'1 O 1 $ 'W G1 O 1 GW
5HODWLQJWKH'HFD\&RQVWDQWDQG+DOIOLIH 1 2
At W 71/ 2 then 1 1 0 The decay equation becomes 1 10H OW 1 10 2
A solution to this equation is an exponential function
10HO71/ 2
ª1 «¬ 2
H
O71/ 2
º ¼»
1
HO71/ 2
2
O7
ln ª 2 H 1/ 2 º ¬ ¼ ln 2 O71/ 2 ln 2 0.693 71/ 2 71/ 2
O
of the form 1 1 0 H OW
71/ 2
ln 2 0.693
O
O
also A = A o H OW O 1 0 H OW
1. The half-life of a certain radioactive isotope is 2.0 minutes. A particular nucleus of this isotope has not decayed within a time interval of 2.0 minutes. What is the probability of it decaying in: a) the next two minutes
b) the next one minute
c) the next second
2. A sample of a radioactive isotope X has the same initial activity as a sample of the isotope Y. The sample of X contains twice the number of atoms as the sample of Y. If the half-life of X is TX then the half-life of Y is 0.5 TX
10
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3. The half-life of a radioactive isotope is 10 days. Calculate the fraction of the sample that will be left after 15 days. 35%
4. The half-life of a radioactive substance is 10 days. Initially, there are 2.00 x 1026 radioactive nuclei present. a) What is the probability of any one particular nucleus decaying?
b) What is the initial activity?
c) How many radioactive nuclei are left after 25 days?
d) What is the activity of the sample after 25 days?
e) How long will it take for the activity to fall to 1.0 x 1024 dy-1?
11
*UDSKVRI5DGLRDFWLYH'HFD\
1 1 0 H OW
Radioactive nuclei vs. time
ln 1
Straightening by natural log
ln ª¬ 1 0 H OW º¼
ln 1
ln 1 0 O W
ln 1
10 O W ln \ P[ E
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ln(N0)
T1/2 = ln 2/(-slope)
Activity of sample vs. time
Slope = -Ȝ
Straightening by natural log
0HWKRGVRI'HWHUPLQLQJ+DOIOLIH ,IWKHKDOIOLIHLVVKRUW, then readings can be taken of activity versus time using a Geiger counter, for example. Then, either 1.
A graph of activity versus time would give the exponential shape and several values for the half-life could be read from the graph and averaged.
2.
A graph of ln (activity) versus time would be linear and the decay constant can be calculated from the slope.
OR
,IWKHKDOIOLIHLVORQJ, then the activity will be effectively constant over a period of time. If a way could be found to calculate the number of nuclei present chemically, perhaps using the mass of the sample and Avogadro’s number, then the activity relation or the decay equation could be used to calculate half-life.
12
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1. Cesium-138 decays into an isotope of barium. Measurements of the activity of a particular sample of cesium-138 were taken and graphed at right. a) Suggest how the data for this graph could have been obtained. Geiger counter
b) Use the graph to estimate the half-life of cesium-138. 35 minutes c) Use the graph to estimate the half-life of the barium isotope. Wait until 250/300 minutes when very little cesium left
90 minutes
2. A 2.0 mg sample of carbon-14 is measured to have an activity of 6.5 x 1010 Bq. a) Use this information to determine the half-life of carbon-14 in years. 5700 years
b) A student suggests that the half-life can be determined by taking repeated measurements of the activity and analyzing the data graphically. Use your answer to part (a) to comment on this method of determining the half-life.
3. The radioactive isotope potassium-40 undergoes beta decay to form the isotope calcium-40 with a half-life of 1.3 x 109 yr. A sample of rock contains 10 mg of potassium-40 and 42 mg of calcium-40. a) Determine the age of the rock sample.
3.1 x 109 yr
b) What are some assumptions made in this determination of age? No calcium-40 originally present = all potassium-40 No loss of either isotope in intervening years
13
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1XFOHDU)LVVLRQ A heavy nucleus splits into two smaller nuclei of roughly equal mass with the release of energy. 1XFOHDU)XVLRQ Two light nuclei combine to form a more massive nucleus with the release of energy.
5HOHDVHRIHQHUJ\LQQXFOHDUUHDFWLRQV:
P PǻP Energy is usually released in the form of . . . kinetic energy for the products.
%LQGLQJHQHUJ\SHUQXFOHRQ greater for product nuclei than for original nuclei since energy is released.
One Common Fission Reaction 235 92
236 810 Qo92 8* o;< neutrons
There are about 90 different daughter nuclei (X and Y) that can be formed. Here is a typical example: 235 92
92 1 8 10 Qo141 56 %D 36 .U 3 0 Q
1. Estimate the amount of energy released when a uranium nucleus fissions.
Use plot above Original: 7.5 MeV per nucleon Daughters: 8.5 MeV Energy difference: 1.0 MeV per nucleon x 235 nucleons = 235 MeV as KE of daughters and neutrons
2. A neutron collides with a nucleus of plutonium and the following fission reaction occurs. Determine the number of neutrons produced and calculate the amount of energy released.
239 94
96 3X10 Qo140 %D 56 38 6U
P PǻP
Masses: 239 Pu = 239.052157 u 94 96 38 Sr = 95.921750 u 140 Ba = 139.910581 u 56 1 0 n = 1.008665 u
0.19383 u = 180 MeV 14
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&KDLQ5HDFWLRQ± neutrons released from one fission reaction go on to initiate further reactions
Uncontrolled nuclear fission: nuclear weapons Controlled nuclear fission: nuclear power production 1) some material absorbs excess neutrons before striking nucleus Uncontrolled Chain Reaction
2) leaving only one neutron from each reaction to produce another reaction Critical Mass: if mass of uranium is too small, too many neutrons escape without causing further fission in uranium so the reaction cannot be sustained Controlled Chain Reaction
Thermal Neutron: low-energy neutron (§1eV) that favors fission reactions – energy comparable to gas particles at normal temperatures Naturally Occurring Isotopes of Uranium: 1) Uranium-238: most abundant, 99.3%, very small probability of fissioning when it captures a
neutron, not used for fuel, more likely to capture high energy neutron than low energy one
2) Uranium-235: 0.3%, 500 times greater probability of fissioning when captures a neutron but must be a low-energy (thermal) neutron, used for fuel
Fuel Enrichment: process of increasing proportion of uranium-235 in a sample of uranium 1) formation of gaseous uranium (uranium hexafluoride) from uranium ores 2) Separated in gas centrifuges by spinning – heavier U-238 moves to outside 3) increases proportion of U-235 to about 3% to be used as fuel in nuclear reactors Advantage: more uranium is available for fission and reaction can be sustained Disadvantage: enriched fuel can be used in the manufacture of nuclear weapons – threat to world peace – 85% = weapons grade 15
1XFOHDU5HDFWRUV
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Most nuclear reactors: thermal fission reactor using uranium-235 as fuel
Fuel Rods: enriched solid uranium Moderator: material (water, graphite) used to slow down high-energy neutrons emitted from fission reactions to thermal levels for use in further fission reactions to sustain the chain reaction - slow neutrons by collisions
Control Rods: inserted between fuel rods – made of neutron-absorbing cadmium or boron D used to control reactor temperature to prevent overheating – lowered if too many neutrons/reactions and excess thermal neutrons are absorbed
Heat Exchanger: hot fluid circulating around fuel rods (primary loop) is fed into tank of water – heat is transferred to water and makes steam – steam expands adiabatically against fan blades of turbines and turns a magnet is a coil of wire to generate electricity
16
1HXWURQ&DSWXUHDQG3OXWRQLXP Uranium-238 is a non-fissionable isotope but is considered “fertile” 238 92
239 239 810 Qo92 8 o93 1S01 H Q
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Neutron capture and Beta-minus decay
239 93
239 1So94 3X01 H Q
Beta-minus decay
239 94
96 1 3X10 Qo140 %D 6U3 56 38 0 Q
Fission reaction
Advantage: plutonium-239 used as fuel in “breeder reactors” Disadvantage: plutonium-239 used in nuclear weapons
6DIHW\,VVXHVDQG5LVNVLQWKH3URGXFWLRQRI1XFOHDU3RZHU
Uranium Mining: open-cast mining: environmental damage, radioactive waste rock (tailings) underground mining: release of radon gas (need ventilation), radioactive rock dangerous for workers, radioactive waste rock (tailings) leaching: solvents pumped underground to dissolve uranium and then pumped back out – contamination of groundwater Thermal Meltdown: overheating and melting of fuel rods – may be caused by malfunction in cooling system or pressure vessel – overheating may cause pressure vessel to burst sending radioactive material and steam into atmosphere (as in Chernobyl, Ukraine 1986) – hot material may melt through floor – “China syndrome” as in Three Mile Island – limited by containment vessel and containment building
Nuclear Waste: Low-level waste: radioactive material from mining, enrichment and operation of plant – must be disposed of – left untouched or encased in concrete High-level waste: disposal of spent fuel rods- some isotopes have ½ lives of thousands of years – plutonium 240,000 years 1) stored under water at reactor site for several years to cool of then sealed in steel cylinders, buried underground 2) reprocessed to remove any plutonium and useful uranium, remaining isotopes have shorter ½ lives and long-term storage need is reduced Nuclear Weapons Manufacture: Enrichment technology could be used to make weapons grade uranium (85%) rather than fuel grade (3%) Plutonium is most used isotope in nuclear weapons and can be gotten from reprocessing spent fuel rods 17
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1. Suppose the average power consumption for a household is 500 W per day. Estimate the amount of uranium-235 that would have to undergo fission to supply the household with electrical energy for a year. Assume that for each fission, 200 MeV is released.
2. A fission reaction taking place in a nuclear power station might be
235 92
92 1 810 Qo141 56 %D 36 .U 3 0 Q
Estimate the initial amount of uranium-235 needed to operate a 600 MW reactor for one year assuming 40% efficiency and 200 MeV released for each fission reaction.
18
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1XFOHDU)XVLRQ Two light nuclei combine to form a more massive nucleus with the release of energy.
1. Write the reaction equation for the fusion reaction shown at right.
2 1
+13 + o24 +H 10 Q
2. Calculate how much energy is released in this fusion reaction.
P PǻP
2 1
0.0189 u = 17.6 MeV
H (deuterium, 2.0141 u) 3 1 H (tritium, 3.0161 u) 4 2 He (4.0026 u) neutron (1.0087 u)
3. Calculate the energy released per nucleon and compare this with a fission reaction.
5 nucleons react = 17.6 MeV / 5 = 3.5 MeV
Compared with §1.0 MeV for fission
Important occurrence of fusion: main source of Sun’s energy – fusion of hydrogen to helium Suggested Mechanism: proton-proton cycle 1 1
0 +11 +o12 +1
H Q
1 1
+12 +o3
2 +H J
Then either: 1 1
4 0 +32 +H o 2 +H 1 H J
Or: 3 2
+H 32 +H o 42 +H 11 + 11 +
19
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Plasma: fuel for reactor – high energy ionized gas (electrons and nuclei are separate) – if energy is high enough (hot enough), nuclei can collide fast enough to overcome Coulomb repulsion and fuse together
Magnetic confinement: charged particles are contained via magnetic fields – travel in a circle in a doughnut shaped ring (tokamak)
Heating Plasma: accelerate nuclei by means of magnetic fields and forces = high temperatures (high kinetic energies)
Problems with current fusion technology: maintaining and confining very high-density and high-temperature plasmas – very difficult to do – uses more energy input than output – not commercially efficient
$UWLILFLDO ,QGXFHG 7UDQVPXWDWLRQ $UWLILFLDO ,QGXFHG 7UDQVPXWDWLRQ A nucleus is bombarded with a nucleon, an alpha particle or another small nucleus, resulting in a nuclide with a different proton number (a different element).
Requirement: the bombarding particle must have sufficient kinetic energy to overcome the Coulomb repulsion
1. In 1919, Ernest Rutherford discovered that when nitrogen gas is bombarded with alpha particles, oxygen and protons are produced. Complete the equation for this reaction.
14 7
17 1 1 42 +H o 8 21 +
2. Neutron bombardment of lithium can produce the radioactive isotope of hydrogen known as tritium. Complete the reaction.
6 3
/L10 Qo13 +42 +H
NOTE: isotopes produced
Importance: artificial isotopes produced are used in medical tests and therapies
20
4XDQWXP3K\VLFV
IB 12
What are photons? Quantum of energy – particle of light
Properties of photons: m = 0, v = c (in a vacuum), q = 0
(QHUJ\RID3KRWRQ E = hf
where h = Planck’s constant = 6.63 x 10-34 J s
1. A beam of monochromatic light has a frequency of 4.4 x 1014 Hz. Determine the energy of each photon of this light in both joules and electron-volts.
2. Light from a 2.5 mW laser has a wavelength of 670 nm. a) Find the energy of each photon in joules and electron-volts.
b) How many photons does it emit in 3.0 minutes?
c) The laser beam falls normally on a plane surface and appears as a small circle whose diameter is 1.5 mm. What is the intensity of the laser beam?
,QWHQVLW\±power per unit area
Formula: I = P/A Symbol: I Units: W/m2
1
IB 12
Energy of each photon
Total energy
Number of photons
If the frequency of the light is constant, as the intensity of the light increases….
remains the same
increases
increases
If the intensity of the light is constant, as the frequency of the light increases….
increases
remains the same
decreases
3. Which contains more photons – 1 joule of red light or 1 joule of blue light? Red 4. Which emits more photons per second – a 1 W laser of red light or a 1 W laser of green light? red
7KH3KRWRHOHFWULF(IIHFW the emission of electrons from a metal when electromagnetic radiation of high enough frequency falls on the surface
7KH([SHULPHQW 1. Light of varying frequencies and intensities are shone on a metal surface (photoemissive surface). 2. Light below a certain frequency will not emit electrons (photo-electrons) no matter how intense it is or how long it shines on the surface. Light at or above a certain frequency will immediately emit electrons no matter how intense it is. 7KUHVKROGIUHTXHQF\ IR : minimum frequency of light needed to eject electrons from the surface of the metal
How are these results in conflict with the classical theory about light? &ODVVLFDO7KHRU\says . . . light acts as a wave and the energy of a wave depends on its amplitude (intensity) not its frequency. &ODVVLFDOSUHGLFWLRQV
([SHULPHQWDOHYLGHQFH
Whether electrons are ejected or not depends on ...
Intensity of the light (If intense enough, electrons will be ejected no matter what the frequency)
Frequency of the light
The maximum kinetic energy of the ejected electrons depends on . . .
Intensity of the light
Frequency of the light
Takes time
Occurs instantaneously above threshold frequency 2 but never below certain frequency
At low intensities, ejecting electrons . . .
IB 12
Einstein’s explanation of the photo-electric effect: 4. Light acts like a particle (not a wave) in which its energy is proportional to its frequency. 5. Electrons at the surface of the metal need DPLQLPXPHQHUJ\LQRUGHUWREHHMHFWHGIURPWKH VXUIDFH, called the ZRUNIXQFWLRQan amount which varies from metal to metal. (Electrons under the surface of the metal need more energy to be emitted.)
(
:RUN)XQFWLRQ I
K I K I R
I
1. There is a one-to-one interaction in which one electron absorbs one photon. If the photon has enough energy (high enough frequency) to overcome the work function, the electron will leave surface immediately with no time delay. If not, the electron will still absorb the photon but will remain bound to the metal. 2. Any “extra” energy (above the work function) is retained by the electron in the form of kinetic energy. The maximum kinetic energy (Ekmax) is retained by electrons that were most loosely held on the very surface of the metal. 3. The number of photons arriving per second, and therefore the rate of emission of electrons, is determined by the intensity of the light, not its frequency. The intensity of the light plays no role in the energy each photon has.
(LQVWHLQ¶V
3KRWRHOHFWULF(IIHFW
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(7
(7
KI I (m ax KI
KI0 (max
1. Photons strike a metal surface whose work function is 2.1 electronvolts, ejecting electrons with a maximum kinetic energy of 7.5 electronvolts. a) Find the energy of the photons.
hf = 2.1 eV + 7.5 eV hf = 9.6 eV
b) Find the threshold frequency of the metal. 2.1 eV(1.6 x 10-19J/1 eV) = hfo 3.36 x 10-19 J = hfo fo = 5.07 x 1014 Hz
3
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Monochromatic light is incident on a metal surface in a photo-cell as shown. The frequency of the light is above the threshold frequency for this metal. The current in the photo-cell is measured using a microammeter. The potential difference of the voltage source is varied until the reading on the microammeter is a maximum (called the “saturation current.”)
1. Sketch a graph of how this maximum current varies with the intensity of light if the frequency of the light is kept constant.
Schematic of Experimental Apparatus
Explanation: Intensity of light is proportional to number of photons per second striking plate Each photon ejects an electron So current is proportional to intensity
2. Describe and explain what will happen to the current if the intensity is kept the same but the frequency of the light is increased. Sketch the resulting graph on the axes above. Less current since total energy is constant but the energy per photon has increased – so fewer photons are striking the metal per second and fewer electrons are ejected
A plot of the maximum kinetic energy of the ejected electrons versus frequency of the incident light is shown. Discuss the features of this graph. Sketch a graph of maximum kinetic energy versus wavelength. 0DWKHPDWLFDO0RGHO
KI I (m ax (max
KI I
(max
KI KI 0
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IB 12
3XUSRVH to test Einstein’s model of the photo-electric effect 0HWKRG 1) Make collecting plate (electrode) negative to repel electrons
emitted from the surface (reverse the normal polarity).
2) Increase the potential difference until the current drops to zero. 3) Electrons emitted from metal surface have a maximum energy.
If this maximum energy is less than the energy required for
electrons to move between plates (against the potential difference), electrons will not reach the collecting plate.
Two comparable schematics of the stopping potential experimental apparatus
6WRSSLQJ3RWHQWLDO 9V 1) minimum potential difference that stops all current 2) reading on voltmeter equal to KE max in electronvolts 0D[LPXPNLQHWLF HQHUJ\RIHMHFWHG HOHFWURQV (PD[
(H T9V
(Nmax (Nmax
H9V (Nmax 0DWKHPDWLFDO0RGHO
KI I (Nmax KI KI 0 H9V
H9V KI I 9V
Experimental Results
[LQWHUFHSW I 6ORSH KH \LQWHUFHSW ĭH
KI I
H H
Use the graph above to determine a value for the work function in electronvolts and for Planck’s constant.
5
IB 12 The apparatus shown is used to investigate the photo-electric effect. The potential difference 9 applied between the metal plates and electrode may be varied in magnitude and direction. In one particular experiment, the frequency and intensity of the light are held constant. The graph shows the variation with the potential difference of the current measured on the microammeter.
1. Discuss the features of the graph.
a) current reaches a maximum value with positive potential difference
b) current drops to zero with a negative potential difference = stopping potential
2. How would this graph change if the intensity of the light
increased at the same frequency? Sketch it on the axes.
Same Vs (same energy per photon = same Emax) – higher max current (more photons)
3. How would this graph change if the frequency of the light
increased at the same intensity? Sketch it on the axes.
More negative Vs (more energy) – lower max current (fewer photons so fewer electrons)
4. The potentiometer is adjusted to give the minimum voltage at which there is zero reading on the microammeter. State and explain what change, if any, will occur in the microammeter when a) the intensity of the incident light is increased but the frequency remains unchanged. No change – stopping potential depends on energy of each electron – no change in frequency so no change in photon energy so no change in electron energy
b) the frequency of the light is increased at a constant intensity. Reading increases from zero – photon energy increases so electron energy increases.
6
0DWWHU:DYHV
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Louis de Broglie (French physicist, 1892 – 1987) postulated in his doctoral dissertation that because light can have both wave and particle characteristics, perhaps all forms of matter have both characteristics. 'H%URJOLH+\SRWKHVLV (1924): All particles can behave like waves whose wavelength is given by Ȝ KS where h = Planck’s constant and p = the momentum of the particle
0DWWHUZDYH: All moving particles have a “matter wave” associated with them whose wavelength is the de Broglie wavelength.
:DYH3DUWLFOH'XDOLW\ Both matter and radiation have a dual nature. They exhibit both particle and wave properties.
'H%URJOLH ZDYHOHQJWK
K K O S PY
Sketch the relationship
between speed and the
de Broglie wavelength of a moving object
1. Determine the de Broglie wavelength for an electron moving at 6.0 × 106 m/s and a baseball (mass = 0.15 kg) moving at 13 m/s. Electron baseball Ȝ = 6.63 x 10-34 / (0.15 x 13) Ȝ = 6.63 x 10-34 / (9.11 x 10-31 x 6.0 x 106 ) Ȝ= 3.3 x 10-34 m Ȝ= 1.2 x 10-10 m
2. Why don’t we notice the wavelike nature of matter in everyday life? Wavelengths are too small
3. Compare the momentum of photons and particles. Which has more momentum – a red photon or a blue photon? blue Photon Momentum
K S
O
Particle Momentum
S
K
O
PY 7
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4. Give some experimental evidence to verify the de Broglie hypothesis. ([SHULPHQW Davisson-Germer experiment (electron diffraction) 0HWKRG directed beam of electrons onto a crystal of nickel and measured number of electrons scattered at various angles
Experimental Apparatus
Sample Results
5HVXOWV electrons in scattered beam are only detected at certain angles by the collector &RQFOXVLRQ electrons are scattered from two layers of atoms and interfere with each other as waves do Another Electron Diffraction Experiment A beam of electrons is sent at a target and the results are observed on a fluorescent screen. Notice that the resulting pattern looks very similar to that of light diffraction through a circular aperture. Experimental Apparatus
Sample Results
,PSRUWDQFH experimental confirmation of de Broglie hypothesis (matter waves)
Why are the wave-like properties of matter evident in these experiments but not in everyday life? De Broglie wavelength of electrons is comparable to the size of the spacing between atoms so noticeable diffraction and interference occurs
8
IB 12
( PF 2
5. Compare the energy of photons and particles.
( KI
Photon Energy
Particle Energy
1 (N PY 2 2
S2 2P
S 2
(N
2P S 2P(N
O (
N
K K S 2P(N K2 2PO2
E kD
1
O2
OD
1 (N
wavelength
6. Compare the kinetic energy of a particle and its de Broglie wavelength
Kinetic energy
7. An electron is accelerated through a potential difference of 1.00 kV. What is its resulting de Broglie wavelength?
E = V/d
Ee = qV
9
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1XFOHDU0RGHORIWKH$WRP 5XWKHUIRUG0RGHO±3ODQHWDU\0RGHO Simple model in which electrons are kept in orbit around the nucleus as a result of the electrostatic attraction between the electrons and the nucleus (YLGHQFHIRUWKH1XFOHDU0RGHO
([SHULPHQW Geiger-Marsden experiment (1909), alpha scattering experiment, Rutherford experiment
0HWKRG Alpha particles from radioactive source are directed at thin gold foil. Scattered alpha particles are detected by a glow on a fluorescent screen. 5HVXOWV 1. Most particles went straight through or were deflected at small angles. 2. A few were deflected at very large scattering angles. &RQFOXVLRQV 1. Most of the atom is empty space since most particles go straight through. 2. All positive charge and most of the mass are concentrated in a very small space called the nucleus.
/LPLWDWLRQRIWKHQXFOHDUPRGHORIWKHDWRP: According to classical physics, an orbiting electron is accelerating, and accelerating bodies radiate energy. This would mean that electrons would radiate energy as they orbit the nucleus. This contradicts observations for two reasons: 1. Electrons would lose energy and spiral into the nucleus. This would destroy all matter.
2. Electrons would radiate energy as light in a continuous spectrum of colors. This contradicts experimental observation since the emission spectra of atoms are observed to consist of only welldefined discrete wavelengths. &RQFOXVLRQ: Observations of atomic emission and absorption spectra indicate that: 1. electrons do not radiate energy when in stable orbits. Stable orbits only occur at certain radial distance from the nucleus. Thus, electrons in these orbits have a well-defined discrete amount of energy. 2. electrons only radiate or absorb energy only when they move (transition) between stable orbits. This energy is quantized and fixed by the energy differences between the allowed orbital levels. ,PSRUWDQFH: Atomic emission and absorption spectra provide evidence for the existence of atomic energy levels. 10
$WRPLF(PLVVLRQDQG$EVRUSWLRQ6SHFWUD Production of (PLVVLRQ6SHFWUD 1. Low pressure gas is energized by applying a potential difference across it causing it to heat up.
2. The hot gas emits light energy only at certain welldefined frequencies, as seen through a diffraction grating (spectroscope) or prism.
Experimental Apparatus
IB 12
Production of $EVRUSWLRQ6SHFWUD 1. Light is shone through a cool low pressure gas.
2. A diffraction grating or prism is used to determine what frequencies pass through the gas and which are absorbed.
Experimental Apparatus
Sample Results
Balmer Series
The VSHFWUDOOLQHV produced (emission or absorption) are characteristic of the particular element producing them. Sample Results
Note that emission and absorption spectral lines occur at the same locations for the same element.
How do atomic spectra provide evidence for the quantization of energy in atoms? 1. Electrons do not radiate energy when in stable orbits. Stable orbits only occur at certain radial distance from the nucleus. Thus, electrons in these orbits have a well-defined discrete amount of energy. 2. Electrons only radiate or absorb energy only when they move (transition) between stable orbits. This energy is quantized and fixed by the energy differences between the allowed orbital levels. 11
IB 12 Electron transitions to a higher energy level require the addition of energy – the basis of the absorption spectrum.
Electron transitions to a lower energy level involve the release of energy – the basis of the emission spectrum.
1. An electron is excited from the ground state to the n = 4 excited state. a) How many possible different photons may be emitted as the electrons relaxes back down to the ground state? Sketch them on the diagram. 6 transitions b) Which transition produces a photon with the most energy? 4 to 1 c) Which transition produces a photon with the highest frequency? 4 to 1 d) Which transition produces a photon with the longest wavelength? 4 to 3 e) Which has the highest wavelength? 4 to 3
7KH9LVLEOH(PLVVLRQ6SHFWUXPRI+\GURJHQ 2. Calculate the wavelength of the spectral line associated with an energy level transition from n = 3 to n = 2.
12
7KH³(OHFWURQLQD%R[´0RGHORIWKH$WRP
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How can the atomic energy levels be explained as quantized matter waves? )HDWXUHVRI0RGHO: 1. If the electron is thought to be confined to move in one dimension by a box, the de Broglie wavelength associated with it will be a standing wave that will only resonate at certain well-defined wavelengths. That is, the electron matter wave is a standing wave that fits certain boundary conditions, like a standing wave on a string fixed at both ends. Resonant Wavelengths
Standing waves on a
string fixed at both
ends must have a node
at each end. The
resonant modes are
then integral numbers
of ½ wavelengths.
L = nȜ/2 Ȝ = 2L/n where L = length of box and n = a positive integer
An electron matter wave has the same resonant modes as a standing wave on a string.
2. The kinetic energy of the electron in the “box” can be found from the de Broglie wavelength.
Calculation of first energy level (use radius = 0.53 x 10-10 m)
Derivation
2/ Q
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O
(.
(.
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(.
(2) 2 (6.63[1034 ) 2 8(9.11[1031 )(.53 [1010 ) 2
(. 2.18 [1018 (. 13.6H9
Q2 K2 8PH/2 13
7KH6FKUöGLQJHU 4XDQWXP0HFKDQLFDO 0RGHORIWKH$WRP
IB 12
(UZLQ6FKU|GLQJHU (Austrian physicist, 1887-1961) made use of de Broglie’s hypothesis to develop the first truly quantum theory of the atom using wave mechanics. )HDWXUHVRI0RGHO: 1. Electrons can be described as matter waves, rather than particles. The mathematical equation for this matter wave is called a “wave function.” :DYH)XQFWLRQ: a mathematical wave function (ȥ) is assigned to the electron.
<
§ QS [ · $sin ¨ ¸ © / ¹
These wave function equations are solutions to a second order differential equation known as Schrödinger’s equation:
G 2< G[2
8S 2 P 2 ( 9 < K
2. The position of the electron is undefined. But the square of the amplitude of the wave function is proportional to the probability of finding the electron at any particular location. “electron cloud” of probability for the first electron energy level
plot of the square of the wave function (probability) versus radial distance from the nucleus for the electron in its lowest energy state Most probably found at r = a = 53 x 10-10 m
+HLVHQEHUJ¶V8QFHUWDLQW\3ULQFLSOH :HUQHU+HLVHQEHUJ (German physicist, 1901-1976) won a Nobel prize in 1932 for the development of his uncertainty principle which identifies a fundamental limit to the possible precision of any physical measurement. 8QFHUWDLQW\3ULQFLSOH 1) Both the position and momentum of a particle cannot be precisely known at the same time. 2) Both the energy state of a particle and the amount of time it is in that energy state cannot be
precisely known at the same time.
FRQMXJDWHTXDQWLWLHV: position and momentum or energy and time ,PSOLFDWLRQV: a) The more you know about one of the conjugate quantities, the less you know about the other. b) If one of the conjugate quantities is known precisely, all knowledge of the other is lost. 14
IB 12 ǻx · ǻp h/4ʌ
Mathematical Representations of the Uncertainty Principle:
ǻE · ǻt h/4ʌ
** delta = +D not final - initial
Application to Electron Diffraction A beam of electrons passes through a double slit. If the electrons act like particles, they should only hit the screen in two locations as shown in figure (a). But if the size of the slits is comparable to the de Broglie wavelengths of the electrons, the electrons will exhibit wave properties. A series of bright and dark bands will show up on the screen instead, as seen in figure (b), indicating that the electrons have diffracted upon passing through the slits and interfered to produce the fringe pattern. How can the electrons “diffract and interfere?” One way to interpret this is to consider that while it is not possible to specify in advance where a particular electron will hit the screen after passing through one or the other slit, one can predict the probability of it hitting at a certain location. Bright fringes correspond to places where electrons have a high probability of landing, and thus over time many electrons do hit there as seen in figure (c), and dark fringes correspond to places where electrons have a low probability of landing. The de Broglie matter wave associated with each electron can thus be seen as a probability wave, as predicted by Schrodinger.
Derivation of Uncertainty Principle Electrons passing through a single slit can be diffracted up or down within the central maximum as far as the location of the first minimum (dark fringe) – neglecting the other bright fringes. This means that although the electron originally had no momentum in the vertical direction before entering the slit, now it may have a vertical momentum component as large as ǻpy. Thus, the uncertainty in its momentum is only in the vertical direction and is equal to ǻpy.. Its horizontal momentum component remains constant at px and so ǻpx = 0.
sin T
O
tan T
E 'S \ S[
sin T | tan T O 'S\ | E S[
O|
K S[
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15
IB 12
1. If the width of the slit is 1.5 x 10-11 m, find the minimum uncertainty in the: a) horizontal component of the momentum 0
b) vertical component of the momentum
ǻy · ǻpy = h/4ʌ ǻpy = h/(4ʌ ǻy) = h/(4 ʌ 1.5 x 10-11 m) = 3.5 x 10-24 kg m/s
Note that the uncertainty in the momentum is perpendicular to its original motion
2. How is the uncertainty principle related to the de Broglie hypothesis? If a particle has a uniquely defined de Broglie wavelength, then its momentum is known precisely. That means that all knowledge of the position of the particle is lost. $SSOLFDWLRQWRWKHK\GURJHQDWRP: If the wavelength of the electron’s matter wave is well-defined, then the position of the electron is unknown.
16
5HODWLYLW\
IB 12
)UDPHRI5HIHUHQFH: the point of view of an observer or a coordinate system against which measurements are made – x,y,z, axes and a clock
,QHUWLDO)UDPHRI5HIHUHQFH: 1. a frame of reference in which Newton’s law of inertia is valid, that is, a frame in which an object with no unbalanced forces will remain at rest or move at a constant velocity 2. a frame of reference that is at rest or moving with a constant velocity – not accelerating
*DOLOHDQ7UDQVIRUPDWLRQVDQG5HODWLYH9HORFLWLHV
1. What is the velocity of the ball as measured by the ground-based observer?
Stationary frame: ground based observer Moving frame: truck ux/ = velocity of object in x-direction as measured in moving frame ux = velocity of object in x-direction as measured in stationary frame
v = velocity of frame 2 in x-direction as measured in stationary frame *DOLOHDQWUDQVIRUPDWLRQ
G JG G X [ X c [ Y G G JG X c [ X [ Y
Example
G 8 X[ 15 G X [ 23P/ V
2. What is the velocity of the ball as measured by the ground-based observer?
G
8 X[ 15
G X [ 7 P / V
*DOLOHDQSULQFLSOHRIUHODWLYLW\: 1. laws of mechanics are the same in all inertial reference frames
2. there is no preferred frame of reference (absolute frame) for describing the laws of mechanics 1
What is the problem with the Galilean principle of relativity? x
different observers will measure different values for the speed
of light
x
but according to the laws of electromagnetism (Maxwell’s
equations), the speed of light in a vacuum is a fixed value
IB 12
How can this contradiction be resolved? Is the speed of light variable or is it fixed? Two possibilities exist: 1. The Galilean transformation laws are incomplete or incorrect. This means that the formulas for adding and subtracting relative velocities will need to be revised so that the speed of light is the same for all observers. 2. The laws of electromagnetism are not the same in all inertial reference frames. This means that there must exist a preferred reference frame in which the speed of light is a constant value but in other reference frames the speed of light can vary according to the Galilean transformations. Possible solution: Find an absolute frame of reference in which light travels at its predicted constant speed and then all other reference frames can be compared to this absolute frame using the Galilean transformations.
/XPLQLIHURXVHWKHU: a massless fluid that fills all space that is the medium through which light travels $EVROXWH)UDPH a frame of reference at rest with respect to the ether 0LFKHOVRQ0RUOH\H[SHULPHQW Aim: to detect the ether Apparatus: interferometer 1. a beam of light is split by a half-silvered mirror into two beams
2. the two beams reflect off mirrors and recombine
3. an observer looks at the interference patterns these two beams make Your turn
4. the apparatus is rotated 900 to see if the interference pattern changes Results: no change in interference pattern was ever noticed Conclusions: 1. the ether does not exist – there is no preferred frame of reference
2. result is consistent with the speed of light being constant
3. Galilean transformations must be revised
2
6SHFLDO5HODWLYLW\
IB 12
Special Theory of Relativity (1905): Einstein’s attempt to resolve the paradox about the speed of light and the laws of electromagnetism
Two Postulates of Special Relativity: 1. The laws of physics are the same in all inertial reference frames.
Consequence: There is no preferred frame of reference.
2. The speed of light in a vacuum is the same for all observers.
Consequence: The speed of light is independent of the speed of its source or the speed of any observer. Consequences of Special Relativity: A. rethinking simultaneity B. time dilation C. length contraction D. Twin paradox E. relativistic formulas for addition of velocities (revised Galilean transformations) F. relationship between mass and energy G. relativistic momentum and energy 3LRQGHFD\H[SHULPHQWV 3LRQ pi meson - neutral pion (uu or dd) Experiment: (CERN 1964) fast moving neutral pions converted (decayed) into two high energy gamma-ray photons pions moving at 99.9% speed of light emitted photons whose speed was still measured to be 3.00 x 108 m/s
Importance: Evidence supporting special relativity since it shows that the speed of light is independent of its source 3
$ 6LPXOWDQHLW\DQGWKH5HODWLYLW\RI7LPH
IB 12
Two events occurring at different points in space and which are simultaneous for one observer cannot be
simultaneous for another observer in a different frame of reference.
A train is traveling to the right with speed Y with respect to the ground when, at the moment observer O’ passes observer O, two bolts of lightning strike the ends of the car at A’ and B’. What does each observer notice and why? Inertial frames of reference Observer O: ground based observer at rest with respect to the ground halfway between A and B Observer O’: train based observer at rest with respect to the train in middle of car
Observer O: sees lightning strike each end of the car at the same time Reason: light has to travel equal distances to reach O at speed = c so reaches in equal times
Observer O’: sees lightning at B’ first and then at A’ Reason: while light from B’ is traveling toward him, he is moving toward light so light has less distance to travel at speed = c so takes less time – reverse for A’
Result: Observer O sees lightning strikes happen simultaneously but Observer O’ sees them happen at different times
Whose version of events is correct? both are – there is no preferred inertial reference frame
4
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IB 12
/LJKW&ORFN a beam of light reflected between two parallel mirrors used to measure the time interval between two events Beginning Event: the light pulse is emitted from the source
Ending Event: the light pulse is detected at the detector
Each observer uses a light clock to measure the time, as seen from their frame of reference, between the pulse being emitted and detected. When the space ship is at rest with respect to the observer on Earth, the two clocks measure the same amount of time. ǻt0 = time measured in the astronaut’s frame of reference ǻt = time measured in the earth observer’s frame of reference If the two frames of reference are at rest with respect to one another, then
F
' 'W
and
F
' 'W0
so 'W 'W0
If the spaceship moves to the right with a speed Y, the observer on Earth sees the light pulse travel a greater distance between the two events. Since each observer measures the same speed for the light pulse, if it traveled a greater distance then it must have taken a longer time. The observer on Earth thus measures a greater time interval between the two events than the astronaut does. If the astronaut’s frame of reference is moving with respect to the observer on Earth, then
'W ! 'W0
7LPHGLODWLRQ stretching of time – moving clock runs more slowly than stationary clocks
NOTE: situation is symmetric – astronaut sees Earth observer’s clock run more slowly since ship could be at rest and the earth observer moving in the opposite direction
3URSHUWLPHLQWHUYDO ǻW the time between events as measured in a frame where the events take place at the same point in space (in moving frame)
NOTE: The proper time is the shortest possible time that any observer could correctly record for the time between events. 5
IB 12
'HULYDWLRQRIWLPH
GLODWLRQIRUPXOD
/RUHQW]IDFWRU
J
1 Y 1 2 F 2
J t1
For an object at rest:
At low (non-relativistic) velocities:
At high (relativistic) velocities:
J 1
J |1
J o f
VR 'W 'W0
VR 'W | 'W0
VR 'W o f
6
relative velocity (v/c)
v = 1.50 x 108 m/s so v/c = 0.500 or v = 0.500 c
IB 12
v = 2.80 x 108 m/s so v/c =0 .833 or v = 0.833 c
Variation of Lorentz factor with velocity (v)
Variation of Lorentz factor with relative velocity (v/c)
([DPSOH: A certain particle created in an experiment has a lifetime of 2.2ȝs when measured in a reference frame in which the particle is at rest. a) Describe a reference frame in which the particle could be considered at rest. the laboratory if it is at rest with respect to the laboratory or its own frame of reference
b) What is the “proper lifetime?” ǻt0 = 2.2 x 10-6 s c) In another experiment, the particle is accelerated in a “particle accelerator” to a speed of 2.7 x 108 m/s. This is the speed of the particle as measured relative to a
stationary frame of reference. Give an example of such a frame of reference.
d) Calculate the Lorentz factor for this particle.
the laboratory
e) Calculate the lifetime of the particle as measured in the stationary reference frame.
f) What would be its lifetime if it traveled at 0.98c?
7
& /HQJWK&RQWUDFWLRQ
IB 12
Because of Special Relativity, observers moving at a constant velocity relative to each other measure different time intevals between two evetnts. Bt if VSHHG GLVWDQFHWLPH and the speed is the same for each observer, then the two observers must measure different distances or lengths as well. This effect is known as OHQJWKFRQWUDFWLRQ
OHQJWKFRQWUDFWLRQ according to a stationary observer, moving objects contract (shrink) in the direction of motion but not in perpendicular directions
For example, a ruler at rest appears to have a length of /. This is known as its SURSHU OHQJWK
For a stationary observer on Earth, a moving ruler would appear to be shorter but just as thick. It only shrinks in the horizontal direction.
SURSHUOHQJWK the length of an object recorded in a frame of reference where the object is at rest
NOTE: This is the greatest possible length for the object.
A rocket travels to Alpha Centauri at a speed of Y 0.95F, as measured by an Earth-based observer. Both observers agree on the relative speed since, to the astronaut, the Earth observer is moving the other way at Y = 0.95F. There is no preferred inertial frame of reference from which to measure absolute speed. However, to the Earth observer, the clock on board the space ship will appear to run more slowly and the ship will appear to shrink in the direction of motion. The situation is reversed for the astronaut. Relative to the astronaut, the clock on Earth will appear to run slowly and the width of the Earth, as well as the distance to the star, will appear to shrink. Both observers will agree, however, on the diameter of the ship and “height” of the Earth.
Earth-based observer’s frame of reference
astronaut’s frame of reference
'HULYDWLRQRIOHQJWKFRQWUDFWLRQIRUPXOD Earth observer:
astronaut:
Y
Y
/0 'W
/ 'W0
/0 / 'W 'W0 'W0 /0 'W 'W / / 0 0 J'W0
/
/
/0
J
8
A note on proper time and proper distance
IB 12
The proper time in this example is the time recorded by the astronaut because only in the astronaut’s frame of reference do the two events (leaving Earth and arriving at the star) occur at the same location (the door of the ship). To the astronaut, it’s as if the ship is at rest and the Earth and star are in motion in the other direction and pass by the door of the ship as they move. The correct frame of reference in which to measure the proper length, however, depends on what is being measured. If the distance from Earth to the star is being measured, then the correct frame of reference is the Earth-based observer’s since both the star and the Earth are at rest relative to this person. But if the length of the ship is to be measured, then the correct frame of reference is the astronaut’s since the ship is at rest relative to the astronaut.
EXAMPLE: An astronaut is set to go on a journey to Alpha Centauri, a nearby star in our galaxy that the astronaut measures from her observatory to be 4.07 x 1016 m away. The astronaut boards the ship at rest on Earth before take-off and uses a meter stick to measure the length of the ship as 82 m and the diameter as 21 m. After take-off, an observer on Earth notices the space ship traveling past him at a speed of Y= 0.950F in route to Alpha Centauri. a) How long does the trip to Alpha Centauri take as measured by: i) the Earth bound observer?
ii) the moving astronaut?
b) What is the distance between Earth and the star as measured by: i) the Earth bound observer?
ii) the moving astronaut?
c) While the ship is on its journey, what is the length of the ship as measured by: i) the Earth bound observer?
ii) the moving astronaut?
d) While the ship is on its journey, what is the diameter of the ship as measured by: i) the Earth bound observer?
ii) the moving astronaut?
9
&RVPLF5D\0XRQ([SHULPHQW
IB 12
0XRQ: unstable elementary particle Experiment: 1) can be produced by collisions of cosmic radiation with atoms in upper atmosphere
2) due to unstable nature should only survive for a short time before decaying – shouldn’t reach surface of earth
3) measurements of number of muons at top of mountain approximately same as at bottom of mountain
Question: Why do so many muons reach the ground before decaying? From Earth frame of reference: time runs slowly for muon so it has time to reach ground before decaying
From muon’s frame of reference: height of atmosphere contracts so has very little distance to travel
EXAMPLE: A muon having a lifetime of 2.2 ȝs as measured in its own frame of reference is created in the upper atmosphere and travels toward Earth at a speed of 0.99c. 1. How far can a muon travel before it decays, as measured in its own frame of reference?
2. What is the lifetime of the muon, as measured from the Earth?
3. How far will the muon travel through the atmosphere, as measured from the Earth?
10
' 7KH7ZLQ3DUDGR[
IB 12
According to special relativity, there is no preferred inertial reference frame so the time dilation effect is the same for all observers. Since each observer sees the other as moving past at a constant speed, each observer measures the other’s clock as running slowly – the effect is symmetric. But what about this? 7ZRWZLQV(LQDQG6WHLQJURZXS(LQEHFRPHVDQGDVWURQDXWDQG6WHLQEHFRPHV
DSK\VLFVWHDFKHU2QHGD\(LQVD\VJRRGE\HWRKLVEURWKHUDQGOHDYHVRQDVSDFH
YR\DJHWRDGLVWDQWVWDU6RPHWLPHODWHUZKHQKHUHWXUQVKRPHKHPHHWVKLV
EURWKHUDJDLQ +RZHYHUE\QRZKLVEURWKHULV\HDUVROGHUWKDQKHLV
PLJKWWKLQNWKDWWKLVLVEHFDXVHRIUHODWLYHPRWLRQ 7KHFORFNLQWKHVSDFHVKLSUXQV
PRUHVORZO\WKDQWKHFORFNRQWKH(DUWKVR(LQKDVDJHGOHVV%XWZKDWDERXWWKH
V\PPHWU\RIWKHWLPHGLODWLRQHIIHFW"$FFRUGLQJWRDVWURQDXW(LQKLVVKLSZDVDW
UHVWZKLOHEURWKHU6WHLQDQGWKH(DUWKPRYHGLQWKHRWKHUGLUHFWLRQ6LQFH6WHLQ¶V
FORFNLVQRZWKHPRYLQJRQHVKRXOGQ¶WKLVFORFNUXQPRUHVORZO\DQG(LQUHWXUQWR
(DUWKDVWKHROGHUEURWKHU":KRVHYLHZRIWKHVLWXDWLRQLVFRUUHFW" ,QIDFW
VKRXOGQ¶WWKHEURWKHUVVWLOOEHWKHVDPHDJHVLQFHWKHUHLVQRSUHIHUUHGLQHUWLDO
IUDPHRIUHIHUHQFH"
Explanation: situation is not symmetric since formulas for special relativity are only symmetrical when the two observers are in constant velocity relative motion -brother on space ship was not in an inertial frame of reference for the entire trip – he accelerated and decelerated and was acted on by external forces – brother on ground was not subject to forces or acceleration so his view of the situation is correct.
7KH+DIHOH.HDWLQJ([SHULPHQW
In 1971, experimenters J.C. Hafele and R.E. Keating from the U.S. Naval Observatory undertook an experiment to test time dilation. They made flights around the world in both directions, each circuit taking about three days. They carried with them four cesium beam atomic clocks, accurate to within ± 10-9 s. The researchers expected that the relative motion of the clocks would produce a measurable time dilation effect (“moving clocks run slow”). In a frame of reference at rest with respect to the center of the earth, the clock aboard the plane moving eastward, in the direction of the earth's rotation, is moving faster than a clock that remains on the ground, while the clock aboard the plane moving westward, against the earth's rotation, is moving slower. When they returned, they compared their clocks with a ground based clock at the Observatory in Washington, D.C. The time intervals measured by the clocks that had traveled on the aircraft differed from those time intervals measured by the ground based clocks and provided confirmation of the time dilation effects of relativity. In this experiment, both time dilation due to motion or kinematics (special relativity) and time dilation due to gravity (general relativity) are significant and had to be taken into account. Quote from their published paper: "During October, 1971, four cesium atomic beam clocks were flown on regularly
scheduled commercial jet flights around the world twice, once eastward and once
westward, to test Einstein's theory of relativity with macroscopic clocks. From
the actual flight paths of each trip, the theory predicted that the flying clocks,
compared with reference clocks at the U.S. Naval Observatory, should have lost
40+/-23 nanoseconds during the eastward trip and should have gained 275+/-21
nanoseconds during the westward trip ... Relative to the atomic time scale of the
U.S. Naval Observatory, the flying clocks lost 59+/-10 nanoseconds during the
eastward trip and gained 273+/-7 nanosecond during the westward trip, where the
errors are the corresponding standard deviations. These results provide an
unambiguous empirical resolution of the famous clock "paradox" with
macroscopic clocks."
Experimental Results
J.C. Hafele and R. E. Keating, Science 177, 166 (1972)
11
( 5HODWLYLVWLF)RUPXODVIRU$GGLWLRQRI9HORFLWLHV
IB 12
1. A motorcyclist drives past a stationary observer at a speed of 0.80F and throws a ball forward at 0.70F, as shown. How fast is the ball moving relative to the stationary observer?
ux/ = velocity of object in x-direction as measured in moving frame ux = velocity of object in x-direction as measured in stationary frame v = velocity of frame 2 in x-direction as measured in stationary frame Galilean transformation:
JG G G Xc[ X[ Y G 0.70F X [ 0.80F G X[ 1.50F 5HODWLYLVWLFWUDQVIRUPDWLRQ IRUPXOD
Xc[
X[ Y XY 1 [2 F
2. Suppose the motorcyclist in the above example shines a flashlight ahead of him. How fast does the stationary observer see the light beam travel?
X Y Xc[ [ XY 1 [2 F X 0.80F F [ X 0.80F 1 [ 2 F X[ F
Relativistic transformation:
X Y Xc[ [ XY 1 [2 F X 0.80F 0.70F [ X 0.80F 1 [ 2 F X[ 0.96F
3. Two bicyclists approach each other at a speed of 0.60F. What is their relative velocity of approach?
X Y Xc[ [ XY 1 [2 F 0.5F (0.5F) Xc[ 0.5F(0.5F) 1 F2 Xc[ 0.8F
12
) 5HODWLRQVKLS%HWZHHQ0DVVDQG(QHUJ\
IB 12
Newton mechanics: A constant force produces a constant acceleration. Implication: Therefore, any speed can be attained if the force is applied for enough time.
Relativistic mechanics: As the object’s speed approaches the speed of light, the acceleration decreases even if the force is constant. Implication: Its mass is increasing.
Newtonian mechanics
velocity
Relativistic mechanics
c
velocity c
time
time
5HVWPDVV P the mass of an object as measured in a frame of reference where the object is at rest NOTE: rest mass is an invariant quantity 0DVV P ³relativistic mass” – mass of moving object – resistance to acceleration – inertial mass Relationship:
P JP0 mass
mass
Mass versus actual speed line horizontal for v < 0.5c
Mass versus relative speed line horizontal for v/c < 0.5
m0 c
m0
v
1
v/c
Consequence: No object can ever attain the speed of light in a vacuum or go faster.
Explanation: The greater the speed of an object, the greater its mass. As mass increases, so does force needed to accelerate it. Mass becomes infinite as speed approaches c. Infinite amount of force or energy would be needed.
13
5HVWHQHUJ\ ( energy equivalent of the mass of an object at rest Relationship:
(0 P0 F 2
IB 12
1. What is the energy equivalent of a 0.20 kg golf ball at rest?
2. What is the rest energy of an electron?
$OWHUQDWHXQLWVIRUHQHUJ\ 1 eV = energy gained by one electron accelerated through a potential difference of 1 volt 1 eV = 1.60 x 10-19 J
1 MeV = 1 x 106 eV
3. What is the rest energy of an electron?
$OWHUQDWHXQLWVIRUPDVV 4. What is the rest mass of an object whose energy equivalent is 1 MeV?
5. What is the rest mass of an electron?
E0 = m0 c2
1MeV = m0 c2
m0 = 1 MeV/c2
= 1 MeV c-2
14
)RUPXODUHSUHVHQWLQJ HTXLYDOHQFHRIPDVVDQGHQHUJ\
(
PF
2
JP0 F
IB 12
2
For an object at rest: Ȗ = 1 so E = E0 For an object in motion: E > E0 since mass increases 6. What is the energy equivalent of an electron accelerated to a speed of 0.90F?
7RWDOHQHUJ\RIDPRYLQJREMHFW rest energy + kinetic energy
Derivation:
ET = E0 + EK
E = E0 + EK
5HODWLYLVWLFNLQHWLFHQHUJ\IRUPXOD
EK = (Ȗ-1)m0c 2
E = E0 + EK mc2 = m0c2 + EK EK = mc2 – m0c2 EK = Ȗm0c2 – m0c2 EK = (Ȗ-1)m0c2
don’t use EK = 1/2mv2 at high speeds
7. What is the kinetic energy of an electron accelerated to a speed of 0.90F? The rest mass of an electron is 0.51 MeV c-2.
15
IB 12 8. A proton is accelerated to a speed of 0.95F. Determine its energy, rest energy, and kinetic energy. look up rest mass on tables = 938 MeV c-2
3DUWLFOHDFFHOHUDWLRQ units for charge: derive E = m0c2 + qV
9. An electron is accelerated through a potential difference of 2.0 x 106 V. Calculate its energy, kinetic energy, and speed.
16
* 5HODWLYLVWLF0RPHQWXPDQG(QHUJ\ Newtonian momentum and kinetic energy
S PY
Relativistic momentum and kinetic energy units for Newtonian momentum
5HODWLYLVWLF WRWDOHQHUJ\
S JP0X
(.
1 PY 2 2
(.
S2 2P
(. (J 1)P0 2
units for relativistic momentum
kg m/s
IB 12
( JP0 F 2
MeV/c
= MeV c-1
2 4 ( 2 S 2 F 2 P0 F
1. The linear particle accelerator at Stanford University (SLAC) is 3.0 km long and can accelerate electrons to a speed of 0.999F. a) Find the magnitude of the relativistic momentum of such an electron.
b) Find the energy and kinetic energy of one of these electrons.
b) How long is the accelerator tunnel, as measured in the electron’s frame of reference?
c) How long will it take the electron to travel through the accelerator, as seen by:
i) an outside stationary observer?
ii) an observer moving with the electron?
17
IB 12
2. A proton is accelerated through a potential difference of 3.0 x 109 V. a) Calculate the energy of the proton after its acceleration.
b) Calculate the final momentum of the proton.
3. “Pair production” is a process by which antimatter pairs of particles are produced from energy. This can happen when a high energy gamma ray photon is in the vicinity of a heavy nucleus. For example, if a gamma photon is near a lead atom, the reaction pictured at right might occur, where the photon creates an electron-positron pair. If the energy of the photon is 3.20 MeV, calculate the following quantities. (Neglect the recoil of the lead atom and assume the energy is shared equally between the particles.) a) The energy and kinetic energy of each particle.
J oH H
c) The mass of each particle.
b) The speed of each particle.
d) The momentum of each particle.
18
*HQHUDO5HODWLYLW\
IB 12
*HQHUDO7KHRU\RI5HODWLYLW\
non-inertial reference frame:
,QHUWLDO0DVVYV*UDYLWDWLRQDO0DVV 1. ,QHUWLDO0DVV – WKHUDWLRRIWKHUHVXOWDQWIRUFHWRDFFHOHUDWLRQ. (The property of an object that determines how much it resists accelerating.)
mi
= F/A
2. *UDYLWDWLRQDO0DVV – WKHSURSHUW\RIDQREMHFWWKDW GHWHUPLQHVKRZPXFKJUDYLWDWLRQDOIRUFHLWIHHOVZKHQ QHDUDQRWKHUREMHFW
mg Į Fg
Mi = F/a
Different masses have different accelerations when the same net force acts on them.
mg Į Fg
Different masses have different gravitational forces acting on them them.
All experiments to measure each type of mass for an object have shown that, within the experimental uncertainty, an object’s gravitational mass is numerically equal to its inertial mass.
(LQVWHLQ¶V3ULQFLSOHRI(TXLYDOHQFH: a postulate that states there is no difference between an accelerating frame of reference and a gravitational field
Einstein’s elevator “thought experiments”
I.
A ball dropped in an elevator at rest on the Earth’s surface accelerates to the floor due to gravity.
A ball dropped in an elevator accelerating upward in a gravity free region (deep space) will act the same as the floor accelerates up to meet it.
II.
A person in an nonaccelerating elevator (rocket) in a gravity free region (deep space) feels no forces.
A person in an elevator
freely falling due to
gravity feels no forces.
19
IB 12
Conclusion of elevator thought experiments:
An observer inside the elevator interprets the situations as identical. That is, there is no way to distinguish between being in an accelerating reference frame and being in a gravitational field.
Based on the principle of equivalence, Einstein predicted that . . . .
light rays will bend in gravitational fields. (gravity bends light)
A photon emitted from the right side of the accelerating rocket ship will appear to trace a curved path as the rocket accelerates beneath it.
By the equivalence principle, a beam of light in an
elevator at rest in a gravitational field should also bend.
([SHULPHQWDOHYLGHQFHIRUWKHEHQGLQJRIOLJKWE\DJUDYLWDWLRQDOILHOG (GGLQJWRQ¶VVRODUHFOLSVHPHDVXUHPHQWV The positions of several stars were measured against a background of fixed stars. Six months later, those stars were hidden behind the Sun due to Earth’s new position in its yearly revolution. It was predicted by General Relativity that these stars should still be visible if the gravitational field of the Sun bent the light rays around it and deflected the light rays toward Earth. However, these “hidden” stars would still not be visible due to the glare of the Sun. But an expedition led by Sir Arthur Eddington sent to the island of Principe sought to measure the deflection of these light rays during a total eclipse of the Sun in 1919 when the stars would be briefly visible. He measured the new positions of the stars against the background of fixed stars and found that they had apparently shifted position. This was experimental evidence that gravitational fields do deflect light rays.
20
*UDYLWDWLRQDOOHQVLQJ Massive galaxies can deflect the light from quasars or other very distant sources of light so that the rays bend around the galaxy. The galaxy acts like a lens so that observers on Earth can see multiple images of the quasar.
IB 12
(LQVWHLQFURVV: four images of the same astronomical object, produced by a gravitational lens A TXDVDU (contraction of 48$6LVWHOO$5UDGLRVRXUFH) is an extremely powerful and distant active galactic nucleus. They were first identified as being high redshift sources of electromagnetic energy, including radio waves and visible light that were point-like, similar to stars, rather than extended sources similar to galaxies. While there was initially some controversy over the nature of these objects, there is now a scientific consensus that a quasar is a compact region 10-10,000 Schwarzschild radii across surrounding the central supermassive black hole of a galaxy.
*UDYLWDWLRQDOUHGVKLIW Based on the principle of equivalence, Einstein predicted that . . . . time slows down near a massive body. (gravity slows time)
Clocks near the surface of the Earth . . . run more slowly than clocks at higher altitudes
([SODQDWLRQ: The frequency of vibration of an object or of an electromagnetic wave (a photon) is essentially a measurement of time. Slowing the frequency of vibration means that time is running slower. The frequency of vibration in a gravitational field is slowed due to something like the Doppler Effect. Consider a rocket with a light source at the bottom (1) and a detector at the top (2), as shown in (a). If the rocket accelerates upward, the detector will be accelerating away from the light source. Thus, the light waves from 1 will reach the detector at 2 less frequently, hence the received frequency will now be less than the emitted frequency, that is, the frequency will be shifted to a lower (redder) frequency. Since, by the principle of equivalence, an observer cannot distinguish between an accelerating reference frame and a gravitational field, the same effect will be noted for a stationary rocket on the surface of Earth, as shown in (b). As a photon moves from 1 to 2, its frequency will be shifted lower. Similarly, a clock at 1 will have a lower frequency (run slower) than a clock at 2. This effect is known as the JUDYLWDWLRQDOUHGVKLIW
21
IB 12
(YLGHQFHWRVXSSRUWJUDYLWDWLRQDOUHGVKLIWHIIHFW 1. Experiment: Pound-Rebka experiment In the early 60's physicists Pound, Rebka,and Snyder at the Jefferson Physical Laboratory at Harvard measured the shift in gamma rays emitted from iron-57 by placing a source at the base of Harvard Tower and a detector at its top, a distance of 22.6 m higher. They were able to measure the shift in frequency of the photons and the results agreed with the predicted value to within 1%.
*UDYLWDWLRQDO5HG6KLIW )UHTXHQF\)RUPXOD
'I I
J 'K
F2
Assumption: relatively constant “g” between two points
Calculate the shift in frequency for gamma photon radiation whose wavelength is 8.62 x 10-11 m.
Schematic representation of the gravitational redshift of a light wave escaping from the surface of a massive body
2. Experiment: Hafele-Keating experiment Frequency shift in atomic clocks aboard commercial jets circling Earth. Clocks in planes ran faster due to gravitational red-shift, and faster or slower due to time dilation.
3. Experiment: Shapiro time delay experiment Delay in time taken for a radar pulse to travel to a nearby planet (Venus or Mercury) and return due to gravitational field of the Sun was measured in 1960s. Results agreed with general relativity predictions. 22
6SDFHWLPH
IB 12
6SDFHWLPH four dimensional coordinates used to describe any event (three spatial dimensions and time)
NOTE: Moving objects follow the shortest path between two points in spacetime.
1HZWRQ¶VH[SODQDWLRQRIJUDYLWDWLRQDODWWUDFWLRQ Two masses exert a force on each other, pulling each other closer. (LQVWHLQ¶VH[SODQDWLRQRIJUDYLWDWLRQDODWWUDFWLRQ Any mass warps (distorts) spacetime - the greater the mass, the greater the warping. Particles, such as planets, moving in spacetime follow the shortest path. The path becomes more curved as the object approaches the central mass. Following this curvature of spacetime is interpreted as a force. “Mass tells space how to curve – space tells mass how to move.”
%ODFN+ROH a region of spacetime with extreme curvature due to the presence of a mass &HQWHURID%ODFN+ROH VLQJXODULW\ ± the single point to which all mass would collapse 6XUIDFHRID%ODFN+ROH HYHQWKRUL]RQ ± where the escape speed is equal to F and within this surface, mass has “disappeared” from the universe 6FKZDU]VFKLOG5DGLXV 56 ± a particular distance from the center of black hole where the escape velocity is equal to the speed of light
23
6FKZDU]FKLOG UDGLXVIRUPXOD
2*0 YHVF U OHW vesc F DQG U 56
F
2*0 56
56
2*0 F2
IB 12
1. Calculate the size of a black hole that has the same mass as our Sun (m = 1.99 x 1030 kg).
2.9 x 103 m
The closer one gets to a black hole, the slower time runs to an outside observer. At the event horizon . . .
*UDYLWDWLRQDOWLPHGLODWLRQ QHDUDEODFNKROH JUDYLWDWLRQDOUHGVKLIW
time appears to stop
'W
'W0 5 1 V U
2. A person who is a distance 3RS from the event horizon of a black hole measures an event to last 4.0 s. Calculate how long the event would appear to last for a person very far from the black hole.
24
Note that both gravitational lensing and gravitational red-shift can be explained by the curvature of spacetime.
Light gains potential energy as it climbs the curvature of spacetime and so must lose some of its energy. Losing energy results in a lower frequency.
IB 12
The picture shows a simulated black hole of ten solar masses as seen from a distance of 600 kilometers with the Milky Way in the background. Notice the gravitational lensing effect.
Just for fun . . .
Do you think the math we’re doing is hard? Just look at what Einstein really wrote.
The (LQVWHLQILHOGHTXDWLRQV (()() or (LQVWHLQ VHTXDWLRQV are a set of ten equations in Einstein's theory of general relativity in which the fundamental force of gravitation is described as a curved spacetime caused by matter and energy. They were first published in 1915. The Einstein field equations (EFE) may be written in the form:
where 5ȝȞ is the Ricci curvature tensor, 5 the scalar curvature, JȝȞ the metric tensor, is the cosmological constant, * is the gravitational constant, F the speed of light, and 7ȝȞ the stress-energy tensor.
25
7KHUPDO3K\VLFV Internal Energy: total potential energy and random kinetic energy of the molecules of a substance
IB 12 Symbol: U
Units: J
Internal Kinetic Energy: arises from random translational, vibrational, and rotational motion
Internal Potential Energy: arises from forces between the molecules
Temperature (Definition #1): a measure of the average random kinetic energy of all the particles of a system
Symbol: T Units: oC, K
Thermal Energy (Heat): the transfer of energy between two substances by nonmechanical means – conduction, convection and radiation
Symbol: Q
Units: J
Temperature (Definition #2): a property that determines the direction of thermal energy transfer between two objects Thermal Equilibrium: at same temperature – no thermal energy transfer – independent of mass, etc.
Thermal Capacity: amount of energy required to raise the temperature of a substance by 1 K Formula: C = Q/ǻT
Q = CǻT
Symbol: C
Units: J/K
Symbol: c
Units: J/(kg K)
Specific Heat Capacity: amount of energy required per unit mass to raise the temperature of a substance by 1 K Formula: c = Q/mǻT
Q = mcǻT
1
IB 12 1. Compare the thermal capacities and specific heat capacities of these samples.
Why do different amounts of the same substances have different thermal capacities? more molecules to store internal potential and kinetic energy
$
lower C same c
higher C
same c
higher C higher c
lower C lower c
%
Why do the same amounts of different substances have different specific heat capacities? substances contain different numbers of molecules with different molecular masses
2. The thermal capacity of a sample of lead is 3.2 x 103 J K-1. How much thermal energy will be released if it cools from 610 C to 250 C?
Q = CǻT
Q = 1.2 x 105 J
3. How much thermal energy is needed to raise the temperature of 2.50 g of water from its freezing point
to its boiling point?
Slope
mercury
water
VORSH
Q = mcǻT Q = (2.50 x 10-3) (4.186 x 103) (100 – 0) Q = 1.05 x 103J
Compare your answer to the amount of thermal energy needed to raise the temperature of liquid mercury the same amount.
more Q needed for water since higher c 2
'7 1 '4 PF
4. A hole is drilled in an 800g iron block and an electric heater is placed inside. The heater provides thermal energy at a constant rate of 600 W.
IB 12
a) Assuming no thermal energy is lost to the surrounding environment, calculate how long it will take the iron block to increase its temperature by 150 C. 9.0 s
b) The temperature of the iron block is recorded as it varies with time and is shown at right. Comment on reasons for the shape of the graph. begins at room temp increases linearly as Q = cmǻT as gets hotter, more energy lost to environment levels out when heat gained by heater = heat lost to room
c) Calculate the initial rate of increase in temperature. 1.7 0C/s
5. An active solar heater is used to heat 50 kg of water initially at 120 C. If the average rate that the thermal energy is absorbed in a one hour period is 920 J min-1, determine the equilibrium temperature after one hour. 120 C
3
&DORULPHWU\
IB 12
&DORULPHWU\: determining the specific heat capacity (or latent heat capacity) of a substance
Conservation of Energy
4F 4K PF FF '7F
PK FK' 7K
Assumption: no thermal energy lost to environment, container, thermometer 1. A 0.10 kg sample of an unknown metal is heated to 1000 C by placing it in boiling water for a few minutes. Then it is quickly transferred to a calorimeter containing 0.40 kg of water at 100 C. After thermal equilibrium is reached, the temperature of the water is 150 C. a) What is the specific heat capacity of the metal sample?
b) What is the thermal capacity of the metal sample?
0HWKRGRI0L[WXUHV
983 J/(kg 0C)
98.3 J/ 0C
2. A 3.0 kg block of copper at 900 C is transferred to a calorimeter containing 2.00 kg of water at 200 C. The mass of the calorimeter cup, also made of copper, is 0.210 kg. Determine the final temperature of the water. 28.30 C
4
3KDVHVRI0DWWHU
IB 12
.LQHWLFWKHRU\ says that: 1. All matter is made up of atoms, and 2. the atoms are in continuous random motion at a variety of speeds. 3. Whether a substance is a solid, liquid, or gas basically depends on how close together its molecules are and how strong the bonds are that hold them together.
6ROLG 0DFURVFRSLF GHVFULSWLRQ
Definite volume Definite shape
0LFURVFRSLF GHVFULSWLRQ
Molecules are held in fixed positions relative to each other by strong bonds and vibrate about a fixed point in the lattice
&RPSDUDWLYH GHQVLW\
High
.LQHWLFHQHUJ\
Vibrational
High 3RWHQWLDOHQHUJ\ $YHUDJHPROHFXODU Atomic radius (10-10 m) VHSDUDWLRQ
/LTXLG
*DV
Definite volume Variable shape Molecules are closely packed with strong bonds but are not held as rigidly in place and can move relative to each other as bonds break and reform
Variable volume Variable shape Molecules are widely spaced apart without bonds, moving in random motion, and intermolecular forces are negligible except during collisions
High
Low
Vibrational Rotational Some translational Higher
Mostly translational Higher rotational Higher vibrational Highest
Atomic radius (10-10 m)
10 x atomic radius (10-9)
0ROHFXOHVSHUP
1028
1028
1025
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IB 12
1. Describe and explain the process of phase changes in terms of molecular behavior.
When thermal energy is added to a solid, the molecules gain kinetic energy as they vibrate at an increased rate. This is seen macroscopically as an increase in temperature. At the melting point, a temperature is reached at which the kinetic energy of the molecules is so great that they begin to break the permanent bonds that hold them fixed in place and begin to move about relative to each other. As the solid continues to melt, more and more molecules gain sufficient energy to overcome the intermolecular forces and move about so that in time the entire solid becomes a liquid. As heating continues, the temperature of the liquid increases due to an increase in the vibrational, translational and rotational kinetic energy of the molecules. At the boiling point, a temperature is reached at which the molecules gain sufficient energy to overcome the intermolecular forces that hold them together and escape from the liquid as a gas. Continued heating provides enough energy for all the molecules to break their bonds and the liquid turns entirely into a gas. Further heating increases the translational kinetic energy of the gas and thus its temperature increases.
2. Explain in terms of molecular behavior why temperature does not change during a phase change. The making or breaking of intermolecular bonds involves energy. When bonds are broken (melting and vaporizing), the potential energy of the molecules is increased and this requires input energy. When bonds are formed (freezing and condensing), the potential energy of the molecules is decreased as energy is released. The forming or breaking of bonds happens independently of the kinetic energy of the molecules. During a phase change, all energy added or removed from the substance is used to make or break bonds rather than used to increase or decrease the kinetic energy of the molecules. Thus, the temperature of the substance remains constant during a phase change.
3. Explain in terms of molecular behavior the process of evaporation. Evaporation is a process by which molecules leave the surface of a liquid, resulting in the cooling of the liquid. Molecules with high enough kinetic energy break the intermolecular bonds that hold them in the liquid and leave the surface of the substance. The molecules that are left behind thus have a lower average kinetic energy and the substance therefore has a lower temperature.
Factors affecting the rate of evaporation: a) surface area
b) drafts
c) temperature
d) pressure
e) latent heat of vaporization
4. Distinguish between evaporation and boiling. Evaporation – process whereby liquid turns to gas, as explained above - occurs at any temperature below the boiling temperature - occurs only at surface of liquid as molecules escape - causes cooling of liquid Boiling –
process whereby liquid turns to gas when the vapor pressure of the liquid equals the atmospheric pressure of its surroundings
-
occurs at one fixed temperature, dependent on substance and pressure
-
occurs throughout liquid as bubbles form, rise to surface and are released
-
temperature of substance remains constant throughout process
6
6SHFLILF/DWHQW+HDW 6SHFLILF/DWHQW+HDW amount of energy per unit mass required to change phase of a substance at constant temperature and pressure
IB 12
Symbol: L Units: J/kg
Formula: L = Q/m
Q = mL
Specific latent heat of fusion: melting and freezing
Lf
Specific latent heat of vaporization: Lv
boiling and condensing
1. How much energy is needed to change 500 grams of ice into water? a) Assume the ice is already at its melting point.
b) Assume the ice is at -150 C.
2. Thermal energy is supplied to a pan containing 0.30 kg of water at 200 C at a rate of 400 W for 10 minutes. Estimate the mass of water turned into steam as a result of this heating process. 0.060 kg
7
7KH.LQHWLF0RGHORIDQ,GHDO*DV
IB 12
.LQHWLFWKHRU\ views all matter as consisting of individual particles in continuous motion in an attempt to relate the macroscopic behaviors of the substance to the behavior of its microscopic particles. Certain microscopic assumptions need to be made in order to deduce the behavior of an ideal gas, that is, to build the .LQHWLF0RGHORIDQ,GHDO*DV. Assumptions: 1. A gas consists of an extremely large number of very tiny particles (atoms or molecules) that are in continuous random motion with a variety of speeds. 2. The volume of the particles is negligible compared to the volume occupied by the entire gas. 3. The size of the particles is negligible compared to the distance between them. 4. Collisions between particles and collisions between particles and the walls of the container are assumed to be perfectly elastic and take a negligible amount of time. 5. No forces act between the particles except when they collide (no intermolecular forces). As a consequence, the internal energy of an ideal gas consists solely of random kinetic energy – no potential energy. 6. In between collisions, the particles obey Newton’s laws of motion and travel in straight lines at a constant speed.
([SODLQLQJ0DFURVFRSLF%HKDYLRULQWHUPVRIWKH.LQHWLF0RGHO 3UHVVXUH Macroscopic definition: force per unit area acting on a surface
Formula:
P = F/A
$WPRVSKHULF3UHVVXUH
Units: N/m2 = Pa (Pascals)
1. A cylinder with diameter 3.00 cm is open to the air. What is the pressure on the gas in this open cylinder?
Weight per unit area of all air above
2. What is the pressure on the gas after a 500. gram piston and a 5.00 kg block are placed on top?
Atmospheric pressure at sea level 1.01x 105 N/m2 = 1.01 x 105 Pa = 101 kPa = 760 mm Hg
8
IB 12
3UHVVXUH Microscopic definition: total force per unit area from the collisions of gas particles with walls of container Explanation: 1) A particle collides with the wall of container and changes momentum. By Newton’s second law, a change in momentum means there must have been a force by the wall on the particle.
2) By Newton’s third law, there must have been an equal and opposite force by the particle on the wall. 3) In a short interval of time, there will be a certain number of collisions so the average result of all these collisions is a constant force on the container wall. 4) The value of this constant force per unit area is the pressure that the gas exerts on the container walls.
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1. 0DFURVFRSLFEHKDYLRU: Ideal gases increase in pressure when more gas is added to the container. 0LFURVFRSLFH[SODQDWLRQ: More gas means more gas particles in the container so there will be an increase in the number of collisions with the walls in a given interval of time. The force from each particle remains the same but an increased number of collisions in a given time means the pressure increases. 2. 0DFURVFRSLFEHKDYLRU: Ideal gases increase in temperature when their volume is decreased.
animation: Serway: chap 12: TDM06AN1 0LFURVFRSLFH[SODQDWLRQ: As the volume is reduced, the walls of the container move inward. Since the particles are now colliding with a moving wall, the wall transfers momentum (and kinetic energy) to the particles, making them rebound faster from the moving wall. Thus the kinetic energy of the particles increases and this means an increase in the temperature of the gas. 3. 0DFURVFRSLFEHKDYLRU: At a constant temperature, ideal gases increase in pressure when their volume decreases. 0LFURVFRSLFH[SODQDWLRQ: The decrease in volume means the particles hit a given area of the wall more often. The force from each particle remains the same but an increased number of collisions in a given time means the pressure increases. 5HODWLRQVKLS pressure is inversely related to volume (Boyle’s Law)
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4. 0DFURVFRSLFEHKDYLRU: At a constant volume, ideal gases increase in pressure when their temperature increases. 0LFURVFRSLFH[SODQDWLRQ: The increased temperature means the particles have, on average, more kinetic energy and are thus moving faster. This means that the particles hit the walls more often and, when they do, they exert a greater force on the walls during the collision. For both these reasons, the total force on the wall in a given time increases which means that the pressure increases. 5HODWLRQVKLS pressure is directly related to temperature (Pressure Law – Admonton Law)
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5. 0DFURVFRSLFEHKDYLRU: At a constant pressure, ideal gases increase in volume when their temperature increases. 0LFURVFRSLFH[SODQDWLRQ: A higher temperature means faster moving particles that collide with the walls more often and
with greater force. However, if the volume of the gas is allowed to increase, the rate at which these particles hit the walls
will decrease and thus the average force exerted on the walls by the particles, that is, the pressure can remain the same.
5HODWLRQVKLS volume is directly related to temperature (Charles Law – Gay-Lussac Law)
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,GHDO*DV/DZV
IB 12
Hot air balloon
volume
Heat can of soup pressure
pressure
volume
Squeeze a balloon
temperature
temperature
Control = temperature
Control = Pressure
Control = volume
P Į 1/V PV = k
VĮT
PĮT
P = k/V
V = kT
P = kT
pressure
K = C + 273
temperature (0 C)
temperature (K)
$EVROXWH=HUR: temperature at which gas would exert no pressure .HOYLQVFDOHRI7HPSHUDWXUH: an absolute scale of temperature in which 0 K is the absolute zero of temperature
MROH an amount of a substance that contains as many particles as there are atoms in 12 grams of carbon-12. $YRJDGUR¶VFRQVWDQW the number of atoms in 12 g of carbon 12. NA = 6.02 x 1023 particles/mole
NA = 6.02 x 1023 mol-1
0RODUPDVV: the mass of one mole of a substance. As a general rule, the molar mass in grams of a substance is numerically equal to its mass number. a) 1 mole of
7 3
/L
b) 2 moles of 27 13
$O
has a mass of 7 g
has a mass of 54 g
c) How atoms are in 8 grams of helium (mass number = 4)?
8 g/ 4 = 2 moles 2 moles x NA = 1.20 x 1024 atoms
10
,GHDO*DV(TXDWLRQRI6WDWH Derivation:
IB 12
P Į 1/V
PĮT
PV Į nT
VĮT
VĮn
PV = nRT
Equation of State: PV = nRT The “state” of a fixed amount of a gas is described by the values of its pressure, volume, and temperature. Gas constant: R = 8.31 J/(mol K)
,GHDO*DV a gas that follows the ideal gas equation of state PV = nRT for all values of p, V, and T
Compare real gases to an ideal gas: a) real gases only behave like ideal gases only at low pressures and high temperatures
b) ideal gases cannot be liquefied but real gases can
Combined Gas Law derivation: State 1: P1V1 = nRT1
P1V1/T1 = nR
State 2: P2V2 = nRT2
P2V2/T2 = nR
1. What is the volume occupied by 16 g of oxygen (mass number = 8) at room temperature and atmospheric pressure? 200 C 0.049 m3
P1V1/T1 = P2V2/T2 T must be in K
2. A weather balloon with a volume of 1.0 m3 contains helium (mass number = 4) at atmospheric pressure and a temperature of 350 C. What is the mass of the helium in the balloon? 0.160 kg
11
7KHUPRG\QDPLFV
IB 12
7KHUPRG\QDPLFV is the branch of physics that deals with the way in which a system interacts with its surroundings. 7KHUPRG\QDPLF6\VWHPsubstance - usually an ideal gas 6XUURXQGLQJV everything else – walls of container, outside environment 6WDWHRIWKHV\VWHPIRUDJDVDSDUWLFXODUVHWRIYDOXHVRI39QDQG7 ,QWHUQDOHQHUJ\ total potential energy and random kinetic energy of the molecules of a substance
Symbol: U Units: J
:RUN product of force and displacement in the direction of the force
Symbol: W Units: J
7KHUPDO(QHUJ\ +HDW the transfer of energy between two substances by non-mechanical means – conduction, convection and radiation
Symbol: Q Units: J
7KHLQWHUQDOHQHUJ\RIDV\VWHPFDQFKDQJHE\
Heating Q = +400 J
Cooling Q = -400 J
Expansion W = +100 J
Compression W = -100 J
Qin = +400 J
Qin = -400 J
Wby = +100 J
Wby = -100 J
Qout = -400 J
Qout = +400 J
Won = -100 J
Won = +100 J
Definitions: Q = thermal energy added to system W = work done by the system ǻU = change in internal energy of the system 12
IB 12 1. A sample of gas is heated with a Bunsen burner and allowed to expand. If 400 J of thermal energy are transferred to the gas during heating and the gas does 100 J of work by expanding, what is the resulting change in the internal energy of the gas?
2. A sample of gas is warmed by placing it in a bath of hot water, adding 400 J of thermal energy. At the same time, 100 J of work is done compressing the gas manually. What is the resulting change in the internal energy of the gas?
ǻU = 400 J -100 J
ǻU = 400 J – (-100 J)
ǻU = 300 J
ǻU = 500 J
ǻU = Q - W
ǻU = Q - W
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4 ǻ8 :
)LUVW/DZRI7KHUPRG\QDPLFV The thermal energy transferred to a system from its surroundings is equal to the work done by the system plus the change in internal energy of the system.
NOTE: The First Law is a statement of . . . the principle of conservation of energy.
In each case, determine the change in the internal energy of the gas. a) A gas gains 1500 J of heat from its surroundings, and expands, doing 2200 J of work on the surroundings. 4 ǻ8:
b) A gas gains 1500 J of heat at the same time as an external force compresses it, doing 2200 J of work on it.
Heating a hot air balloon and let canvas expand
1500 J= ǻU + (-2200 J)
1500 J= ǻU + (+2200 J)
Heating and stirring water
ǻU = +3700 J
ǻU = -700 J
Internal energy of many substances depends on . . .
4 ǻ8:
temperature and intermolecular bonds
Internal energy of an ideal gas depends on . . . 1. only on temperature since there are no intermolecular bonds U Į T so ǻU Į ǻT
U increases if T increases, if +ǻT then +ǻU
2. the change in internal energy of ideal gas is path independent
13
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IB 12
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The gas in the cylinder is expanding isobarically because the pressure is held constant by the external atmosphere and the weight of the piston and the block. Heat can enter or leave through the non-insulating walls.
,VRFKRULF ,VRYROXPHWULF 3URFHVV
The gas in the cylinder is being heated isochorically since the volume of the cylinder is held fixed by the rigid walls. Heat can enter or leave through the non-insulating walls.
,VRWKHUPDO3URFHVV
The gas in the cylinder is being allowed to expand isothermally since it is in contact with a water bath (heat reservoir) that keeps the temperature constant. Heat can enter or leave through the non-insulating walls.
$GLDEDWLF 3URFHVV
The gas in the cylinder is being compressed adiabatically since the cylinder is surrounded by an insulating material.
,VREDULF3URFHVV :RUN,QYROYHGLQD9ROXPH&KDQJHDW&RQVWDQW3UHVVXUH How is pressure held constant? weight of brick, piston, and atmosphere constant How is work done by the gas? Molecules strike piston and transfer momentum and KE to it causing it to move upward/outward - as KE decreases, so does internal energy and T
How much work is done by the gas if it expands at constant pressure?
W = F s cos ș W = p A s cos 00
animation: Serway: chap 12: TDM06AN1
W = p ǻV
14
IB 12 What does an isobaric process look like on a diagram of pressure vs. volume (P-V diagram)?
Expansion of gas
Compression of gas
How can the amount of work done by a gas during a process be determined from a P-V diagram?
work done by the gas = area underneath curve
arrow to right = positive work done by the gas = expansion
arrow to left = negative work done by the gas = compression
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VW ODZ
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If gas is ideal, U increases when T increases
U Į T
So + ǻT means + ǻU
PV/ T = PV/ T P1 = P2
ǻQ = ǻU + ǻW
So V1/T1 = V2/T2 If V increases, so does T
Example: A gas is allowed to expand isobarically by adding 1000 J of thermal energy, causing the gas to increase its internal energy by 200 J. How much work is done by the gas in expanding? &RPSUHVVLRQDW FRQVWDQWSUHVVXUH
(+) = (+) + (+)
More heat is added than work done if isobaric
800 J
*DVODZV
PV/ T = PV/ T P1 = P2 So V1/T1 = V2/T2 If V decreases, so does T
VW ODZ If gas is ideal, U decreases when T decreases U Į T
So - ǻT means - ǻU
ǻQ = ǻU + ǻW
(-) = (-) + (-)
More heat is removed than work done if
isobaric - Heat leaves system 15
,VRFKRULF ,VRYROXPHWULF 3URFHVV :RUN Means ǻV = 0
IB 12
*DVODZ PV/T = PV/T V=V
VW ODZ
Q = ǻU + W
ǻ8 4 (since W = 0)
If ǻV = 0 then W = 0
So P/T = P/T
No area = no work
If T increases, P increases
If Q+, then ǻU+ If ideal gas, ǻU Į ǻT so ǻT+
1. One mole of an ideal gas is heated at a constant volume of 2.0 x 10-3 m3 from an initial pressure of 1.0 x 105 Pa to a final pressure of 5.0 x 105 Pa. a) Determine the initial and final temperatures of the gas.
b) Does the internal energy of the gas increase or decrease? Justify your answer.
c) Determine the work done by the gas during this process.
c) If the change in internal energy of the gas is 1200J, determine the amount of thermal energy added to the gas.
16
2. In each case shown below, an ideal gas at 5.0 x 105 Pa and 1.0 x 10-3 m3 expands to 4.0 x 10-3 m3 at a pressure of 1.0 x 105 Pa by a different process or series of processes.
,
,,
IB 12
,,,
a) Compare the change in internal energy of the gas as a result of each process. Justify your answer.
b) Compare the work done by (or on) the case during each process. Justify your answer.
c) Compare the thermal energy added to or removed from the gas during each process. Justify your answer.
d) If the change in internal energy in each case is 500 J, calculate the work done and thermal energy exchanged in each case.
Conclusions: 1) Change in internal energy does not depend on the path taken – only on the change in temperature – path independent.
2) Work done and thermal energy transferred depend on the path taken between the initial and final states.
17
,VRWKHUPDO3URFHVV
IB 12
Heat reservoir: hot or cold water bath that maintains constant temperature of gas by supplying or removing thermal energy
VW /DZ
*DV/DZ
Q = ǻU + W
PV/T = PV/T
If ideal gas, ǻU Į ǻT so ǻT= 0 means ǻU = 0
T=T PV = PV
so
4 :
If V increases, P decreases
Expansion: thermal energy flows in at same rate as work is done by gas Compression: thermal energy flows out at same rate as work done on the gas
Ideal Gas Equation of State P1V1/T1 = P2V2/T2
PV = nRT P = nRT/V - hyperbola for fixed T
P1V1 = P2V2 on one isotherm
,VRWKHUP hyperbola of constant temperature
Conclusions: 1) all states on one isotherm have same U since have same T – ǻT and ǻU = 0 moving along same isotherm
2) isotherms further from origin – higher T so higher U 3) ǻU between two isotherms is path independent – same ǻU since same ǻT and ǻU Į ǻT
Expansion arrow to right work done by + Q added +
Compression arrow to left work done on Q removed -
18
animation: Serway: chap 10: TDA06AN3
$GLDEDWLF3URFHVV
IB 12
Adiabatic walls: insulating walls so no thermal energy can enter or leave system
NOTE: Rapid expansion or compression of gas is approximately adiabatic
VW /DZ
Q = ǻU + W
Q = 0
VR ǻ8 :
If ideal gas, ǻU Į ǻT so W Į -ǻT
Expansion: work done by gas cools gas down as it loses internal energy
PV/T = PV/T P decreases and V increases and T decreases
W Į -ǻT so +W means temperature goes down = jumps to lower isotherm = gas cools down
Compression: work done on gas heats gas up as it gains internal energy
PV/T = PV/T P increases and V decreases and T increases
W Į -ǻT so -W means temperature goes up = jumps to higher isotherm = gas gets hotter
1. If 410 J of heat energy are added to an ideal gas causing it expand at constant temperature,
2. If an ideal gas is allowed to expand adiabatically, the internal energy of the gas changes by2500 J.
a) what is the change in internal energy of the gas?
a) Does the internal energy of the gas increase or decrease? Justify your answer.
b) how much work is done by the gas? Determine: b) the thermal energy added or removed from the gas. c) how much work is done on the gas?
c) the work done by the gas.
19
IB 12
&\FOHV &\FOH: a series of processes that returns a gas to its initial state
The cycle shown below represents processes performed on an ideal gas initially at P0 = 1.0 x 105 Pa and V0 = 2.0 x 10-3 m3.
4
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:
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1. Compare the temperatures at each state A, B, C, and D.
2. During process AĺB, 600 J of thermal energy were added to the gas. Complete the chart.
20
IB 12 3URSHUWLHVRIWKHLQGLYLGXDOWKHUPDOSURFHVVHV Q in
$ĺ%
Q in
%ĺ&
Q out
&ĺ' Q out
'ĺ$
isochoric, temperature increase, U increase, W = 0, Q + = Q in isobaric expansion, +W, temperature increase, U increase, Q in (more than W by) isochoric, temperature decrease, U decrease, W = 0, Q - = Q out isobaric compression, - W, temperature decrease, U decrease, Q out (more than Won)
1HW:RUNIRUD&\FOH
3URSHUWLHVRIWKHHQWLUHF\FOH 1) gas returns to same P, V, and T 2) ǻT = 0 so ǻU = 0 (for all ideal gases) 3) ǻU = 0 so net Q = netW
4) net W = area enclosed by figure so positive area enclosed means positive net work = work done by gas = net work out
5) net Q = W so Q+ so more heat added than removed during cycle = net heat in
21
IB 12
An ideal gas is confined in a cylinder with a movable piston. The gas starts at 300 K in state A and proceeds through the cycle shown in the diagram. a) Find the temperatures at B and at C. 900 K isothermal
b) State whether ǻU, W and Q are +, - , or 0 for each of the three processes and for the entire cycle.
4
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:
A to B: W = 0, Q +, ǻU+ B to C: ǻU = 0, W+ and Q+
$ĺ%
C to A: Q -, W -, ǻU –
%ĺ&
Cycle: ǻU = 0, Q+, W+
&ĺ$ &\FOH
c) The internal energy of the gas changes by 1520 J during process A to B. 1700 J of heat are added to the gas during process B to C. Find ǻU, W, and Q for each process and for the entire cycle.
4 $ĺ% %ĺ&
ǻ8
: A to B: Q = ǻU = 1520 J, W = 0 B to C: ǻU = 0, Q = +1700J, W = +1700J C to A: ǻU = -1520 J, W = - 1000 J, Q = -2520 J Cycle: ǻU = 0, Q = 700 J, W = 700 J
&ĺ$ &\FOH
22
7KH6HFRQG/DZRI7KHUPRG\QDPLFVDQG(QWURS\
IB 12
The Second Law of Thermodynamics implies that . . . thermal energy cannot spontaneously transfer from a region of low temperature to a region of high temperature.
(QWURS\ a system property that expresses the degree of disorder in the system.
6HFRQG/DZRI7KHUPRG\QDPLFV:
1) the overall entropy of the universe is increasing
2) all natural processes increase the entropy of the universe
$OWKRXJKORFDOHQWURS\FDQGHFUHDVHDQ\SURFHVVZLOOLQFUHDVHWKHWRWDOHQWURS\RIDV\VWHPDQGLWVVXUURXQGLQJV WKHXQLYHUVH 1. Discuss this statement for the case of a puddle of water freezing into a block of ice.
2. A block of ice is placed in a thermally insulated room initially at room temperature. Discuss any changes in the total energy, total entropy, and temperature of the room.
3. An operating refrigerator with its door open is placed in a thermally insulated room. The refrigerator operates for a long period of time. Discuss any changes in the total energy, total entropy, and temperature of the room.
23
:DYH3KHQRPHQD
IB 12
3KDVH The phase of any particle is its position in its cycle of oscillation. ,QSKDVH: (A,E,I) (B,F) (D,H) (C,G) The SKDVHGLIIHUHQFH between any points LQSKDVH is 0. &RPSOHWHO\RXWRISKDVH: (A,C) (B,D) (A,G) (B,H) The SKDVHGLIIHUHQFH between any points FRPSOHWHO\RXWRISKDVH is ʌ or 1800. 5HIOHFWLRQDWD%RXQGDU\EHWZHHQ7ZR0HGLD , )L[HG(QG5HIOHFWLRQ +DUG5HIOHFWLRQ
3KDVH'LIIHUHQFH ʌ or 1800.
,, )UHH(QG5HIOHFWLRQ 6RIW5HIOHFWLRQ
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Reason: Rope applies upward force on support (incident pulse). By Newtons 3rd law, support applies downward force on rope (reflected pulse).
6XSHUSRVLWLRQDQG,QWHUIHUHQFH 3ULQFLSOHRI/LQHDU6XSHUSRVLWLRQ: When two or more waves (pulses) meet, the resultant displacement is the vector sum of the individual displacements.
&RQVWUXFWLYH,QWHUIHUHQFH: superposition of two or more pulses or waves in phase
'HVWUXFWLYH,QWHUIHUHQFH: superposition of two or more pulses or waves out of phase
Equal Amplitudes
Equal Amplitudes – Complete destructive interference
Unequal Amplitudes
Unequal Amplitudes
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IB 12
How are standing waves formed? 1. A traveling wave moving in one direction in a medium is reflected off the end of the medium. 2. This sends a reflected wave traveling in the opposite direction in the medium. This second wave is (nearly) identical with the first traveling wave. (same frequency, same wavelength, almost same amplitude) 3. The two identical waves traveling in opposite directions interfere with each other creating the standing wave whose amplitude at any point is the superposition of the components’ amplitudes.
When the two component waves meet in phase, constructive interference takes place.
When the two component waves meet out of phase, destructive interference takes place.
2QHSRVVLEOHVWDQGLQJZDYHRQDVWULQJ Node: location of constant complete destructive interference
Anti-Node: location of maximum constructive interference
6WDQGLQJ 6WDWLRQDU\ :DYH Energy is not transferred by the wave, but it does have energy stored in it. All points on the wave have a different amplitude (= variable $PSOLWXGH amplitude). The maximum amplitude is 2A at the antinodes and 0 at the nodes. Same as the wavelength of the component waves. :DYHOHQJWK Equal to twice the distance between any two consecutive nodes (or antinodes). All particles oscillate in SHM with the )UHTXHQF\ same frequency as the component waves. All points in one section between two consecutive nodes (or antinodes) are moving in phase. All points in the next section are 1800 out of phase with 3KDVH all points in this section. Thus, the only possible phase differences are 00 and 1800. (QHUJ\
7UDYHOLQJ:DYH Energy is transferred by the wave. All points on the wave have the same amplitude (= fixed amplitude) – provided energy is not dissipated. Equal to the shortest distance along the wave between any two points that are in phase. All particles oscillate in SHM with the same frequency. All points within one wavelength have a different phase. Thus, all phase differences are possible.
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IB 12
v = 1200 m/s L = ½ Ȝ1 Ȝ1 = 2 L = 12 m VW +DUPRQLF IXQGDPHQWDO
f1 = v / Ȝ1 = 1200 / 12 = 100 Hz
v = 1200 m/s L = Ȝ2 Ȝ2 = L = 6 m QG +DUPRQLF VW RYHUWRQH
f2 = v / Ȝ2 = 1200 / 6 = 200 Hz = 2 f1
v = 1200 m/s L = 3/2 Ȝ3 Ȝ3 = 2/3 L = 4 m UG +DUPRQLF QG RYHUWRQH
f3 = v / Ȝ3 = 1200 / 4 = 300 Hz = 3 f1
Boundary conditions for transverse standing waves on a string: 2 fixed ends – node at each end
Fundamental wavelength and frequency:
L = ½ Ȝ1
so
Ȝ1 = 2L
f1 = v/Ȝ1 = v / 2L Other natural frequencies (Resonant modes):
fn = n f1 = n [ v/ (2 L)]
where n = 1,2,3,4,… 3
/RQJLWXGLQDO6WDQGLQJ:DYHV 3LSHRSHQDWERWKHQGV
IB 12
v = 340 m/s L = ½ Ȝ1 Ȝ1 = 2 L = 1.2 m VW +DUPRQLF IXQGDPHQWDO
f1 = v / Ȝ1 = 340 / 1.2 = 283 Hz
v = 340 m/s L = Ȝ2 Ȝ2 = L = 0.6 m QG +DUPRQLF VW RYHUWRQH
f2 = v / Ȝ2 = 340 / 0.6 = 567 Hz = 2 f1
v = 340 m/s L = 3/2 Ȝ3 Ȝ3 = 2/3 L = 0.4 m UG +DUPRQLF QG RYHUWRQH
f3 = v / Ȝ3 = 340 / 0.4 = 850 Hz = 3 f1
Boundary conditions for a pipe open at both ends: 2 free ends – antinode at each end
Fundamental wavelength and frequency:
L = ½ Ȝ1
so
Ȝ1 = 2L
f1 = v/Ȝ1 = v / 2L Other natural frequencies (Resonant modes):
fn = n f1 = n [ v/ (2 L)]
where n = 1,2,3,4,…
4
/RQJLWXGLQDO6WDQGLQJ:DYHV 3LSHFORVHGDWRQHHQG
IB 12
v = 340 m/s L = ¼ Ȝ1 Ȝ1 = 4 L = 2.4 m f1 = v / Ȝ1 = 340 / 2.4 = 142 Hz
VW +DUPRQLF IXQGDPHQWDO
v = 340 m/s L = ¾ Ȝ3 Ȝ3 = 4/3 L = 0.8 m f3 = v / Ȝ3 = 340 / 0.8 = 425 Hz = 3 f1
UG +DUPRQLF
v = 340 m/s L = 5/4 Ȝ5 Ȝ5 = 4/5 L = 0.48 m f5 = v / Ȝ5 = 340 / 0.48 = 708 Hz = 5 f1
WK +DUPRQLF
Boundary conditions for a pipe closed at one end: 1 fixed and one free end – one node and one antinode
Fundamental wavelength and frequency:
L = ¼ Ȝ1
so
Ȝ1 = 4L
f1 = v/Ȝ1 = v / 4L Other natural frequencies (Resonant modes):
fn = n f1 = n [ v/ (4 L)]
where n= 1,3,5…
5
IB 12 1. A violin string that is 40.0 cm long has a fundamental frequency of 440 Hz. What is the speed of the waves on this string?
2. A flute is essentially a pipe open at both ends. What are the first two harmonics of a 66.0 cm flute with all of its keys closed (making the vibrating column of air approximately equal to the length of the flute)? Assume the flute is at room temperature.
3. What is the fundamental frequency and wavelength of a 0.50 m organ pipe that is closed at one end, when the speed of sound in the pipe is 352 m/s?
5HVRQDQFH7XEH 4. A 256 Hz tuning fork is used in a resonance tube and the first two resonances of the fundamental frequency are found at 0.32 m and 0.98 m. What is the speed of sound under the present laboratory conditions?
L1 = ¼ Ȝ L2 = ¾ Ȝ L2 – L1 = ½ Ȝ = 0.66 m v = f Ȝ = (256 Hz) (1.32 m) = 338 m/s 6
7KH'RSSOHU(IIHFW
IB 12
'RSSOHU(IIHFW The change in frequency of a wave detected by an observer because the wave source and the observer have different velocities with respect to the medium of the wave propagation. Sound – change in pitch Light – change in frequency (color)
1RWH: In general, the velocities of the source and/or detector are specified with respect to the medium of propagation. However, light is unique in that there is no medium of propagation so it is the relative velocity of the source and detector that is relevant.
6WDWLRQDU\VRXUFHDQGVWDWLRQDU\REVHUYHUV The number of compressions reaching each observer’s ear per second is the same so each hears a sound of the same frequency. This frequency is identical to the frequency of the source so there is no Doppler shift. 0RYLQJVRXUFHDQGVWDWLRQDU\REVHUYHUV Source moving away from observer #1:
Source moving toward observer #2: More waves detected per second Higher frequency Higher pitch heard Shorter wavelength
For truck moving at constant velocity: one constant high pitch when
moving toward and one constant low pitch heard when moving away
For truck speeding up: pitch increases and then decreases
'RSSOHU)RUPXOD PRYLQJVRXUFH
§ Y · I ' I ¨
¸ ©Y r XV ¹ f = original frequency f’ = shifted frequency v = speed of sound in medium us = speed of source relative to medium STS – source toward subtract
frequency
Fewer waves detected per second Lower frequency Lower pitch heard Longer wavelength
time
EXAMPLE - A high-speed train is traveling at a speed of 44.7 m/s (100 mi/h) when the engineer sounds the 415-Hz warning horn. The speed of sound in air is 343 m/s. What are the frequency and wavelength of the sound, as perceived by a person standing at a crossing, when the train is approaching?
§ 343 · I ' 415 ¨ ¸ ©343 44.7 ¹ I ' 477 Hz
O'
Y 343 0.72 P I ' 477
7
'RSSOHUVKLIWIRUPRYLQJREVHUYHUDQG
VWDWLRQDU\VRXUFH
IB 12 Observer moving toward source: Encounters more waves per second Higher frequency = higher pitch
Observer moving away from source: Encounters fewer waves per second Lower frequency = lower pitch
Is a moving observer equivalent to a moving source? No. since velocity of each with respect to medium is not the same. Wind but no motion of either source or observer = no change in pitch
'RSSOHU)RUPXOD PRYLQJREVHUYHU
I'
§ Y rXR · I ¨ ¸ © Y ¹
uo = speed of observer relative to medium
EXAMPLE - The security alarm on a parked car goes off and produces a frequency of 960 Hz. The speed of sound in air is 343 m/s. What is the frequency you perceive as you drive toward this parked car at 20. m/s?
§ 343 20 · I ' 960 ¨ ¸ © 343 ¹ I ' 1016 Hz
OTA – observer toward add 'RSSOHU6KLIWIRU(0UDGLDWLRQ Blue shift: source and observer moving towards each other Red shift: source and observer moving away from each other
'RSSOHU)RUPXOD (0UDGLDWLRQ
Y
'I | I
F when Y F v = relative speed of source and observer ǻf = frequency shift
EXAMPLE – A star is moving away from Earth at a speed of 3.0 x 105 m/s. One of the elements in the star emits light with a frequency of 6.0 x 1014 Hz. By how much is the frequency shifted when it is received by a telescope on Earth?
3.0 [105 14 6.0 [
10 'I | 3.0 [ 108 'I 6.0 [1011 +] 8
IB 12
([DPSOHVRIWKH'RSSOHU(IIHFW 1. Measuring the speed of vehicles Police use radar to measure the speed of moving vehicles to see if they are
breaking the speed limit. A pulse of radio waves of known frequency is
emitted, reflected off the moving vehicle and reflected back to the source.
The change between the frequency emitted and the frequency received is
used to calculate the speed of the car.
2. Blood-flow measurements Doctors use a ‘Doppler flow meter’ to measure the speed of blood flow. Transmitting and
receiving elements are placed directly on the skin and an ultrasound signal (sound whose
frequency is around 5 MHz) is emitted, reflected off moving red blood cells and then
received. The difference in transmitted and received frequencies is then used to calculate
how fast blood is flowing which can help doctors identify constricted arteries.
3. Determining rotation rates When an object such as a cyclone or a distant star rotates, one side is moving toward the observer and one side is moving away. For a cyclone, radio pulses known as radar are transmitted and reflected from each side of the rotating air mass and the difference in Doppler shift from each side can be used to calculate the rate at which it is rotating. Similarly, the atomic absorption spectrum of the light from a star will be blue-shifted from one side and redshifted from the other so this difference can be used to determine its rotational rate.
'LIIUDFWLRQ 'LIIUDFWLRQ: the bending or spreading of a wave when it passes through a small opening (aperture) or around a barrier 1. The wave diffracts around the barrier at the edges, leaving a “shadow” region behind it where the wave does not reach. The smaller the object is compared to the wavelength, the smaller the shadow region and the more the wave reaches behind the obstacle. If the obstacle is small compared to the wavelength, no noticeable diffraction occurs.
2. A wave also spreads out (diffracts) when it passes through a small opening (aperture). In this way, a plane wave can act like a point source.
3. The larger the aperture size (slit width) compared to the
wavelength of the wave, the less the wave diffracts.
E VOLWZLGWK±DSHUWXUHVL]H
&RQGLWLRQIRUQRWLFHDEOHGLIIUDFWLRQ
Size of opening should be approximately equal to the wavelength (b§Ȝ)
9
,QWHUIHUHQFHRI:DYHVLQ7ZR'LPHQVLRQV
IB 12
3DWK/HQJWK Ɛ – distance traveled by a wave from source to a location 3DWK'LIIHUHQFH ǻƐ – difference in path lengths between two waves = | Ɛ1 - Ɛ2 |
10
$QWLQRGDO/LQH line of maximum constructive interference
IB 12 1RGDO/LQH line of constant complete destructive interference
&RQGLWLRQVIRU$QWLQRGDO/LQH
&RQGLWLRQVIRU1RGDO/LQH
Phase difference: 0
Phase difference: 1800 or ʌ
Path difference: ǻƐ = n Ȝ where n = 0, 1, 2, 3, …
Path difference: ǻƐ = (n + ½ ) Ȝ where n = 0, 1, 2, 3, …
&RQGLWLRQVIRUDVWDEOHLQWHUIHUHQFHSDWWHUQ 1) waves have approximately same amplitude/intensity 2) sources are coherent Coherent waves: waves that have a constant phase relationship – not necessarily in phase
1. A square is 3.5 m on a side, and point A is the midpoint of one of its sides. On the side opposite this spot, two in-phase loudspeakers are located at adjacent corners. Standing at point A, you hear a loud sound and as you walk along the side of the square toward either empty corner, the loudness diminishes gradually but does not entirely disappear until you reach either empty corner, where you hear no sound at all. Find the wavelength of the sound waves.
ǻƐ = ½ Ȝ = 1.449 Ȝ = 2.9 m
2. The same set-up as above is used but now the frequency of sound emitted by both speakers is increased to 700 Hz. This time, as you walk along the side of the square from A toward an empty corner, you hear the loud sound at A alternately diminish to no sound and then increase to a maximum again. By the time you arrive at the corner, you have noticed the sound disappear three times. Use this information to estimate a speed for sound.
ǻƐ = 3 Ȝ = 1.449 Ȝ = 0.483 m v=f Ȝ = (700)(0.483) = 388 m/s
11
,QWHUIHUHQFHRI/LJKW
IB 12
6LQJOH6OLW'LIIUDFWLRQDQG,QWHUIHUHQFH Why is there an interference pattern when light travels through a single slit? different parts of the wave interfere with itself as it spreads
7KHLQWHQVLW\SDWWHUQIRUVLQJOHVOLWGLIIUDFWLRQ
Features of the Single Slit Intensity Pattern
a) Wide central maximum – wider than other fringes
b) Bright central maximum – much brighter than other fringes
c) secondary maxima – equal width and spacing
3RVLWLRQRIILUVWPLQLPXP
T
O
E
ș must be in radians Narrow slit
Wide slit 12
'HULYDWLRQRI6LQJOH6OLW)RUPXOD 1. Break wavefront into separate point sources. Assume screen is very far from the slit so that the rays are parallel when they interfere destructively at the first dark fringe (first minimum).
2. Match points in pairs that are ½ slit width apart: 1 and 3, 2 and 4 (ignore 5)
IB 12
3. Draw small right triangle perpendicular to rays. Wavelet pairs interfere destructively if the path difference is ½ Ȝ.
4. Calculate:
+DOIZLGWKRIFHQWUDOPD[LPXP location of first dark fringe ș Ȝ E IRUșLQUDGLDQV draw small triangle to show ș = Ȝ/b = d/D
(;$03/( Light passes through a single slit and shines on a flat screen that is located 0.40 m away. The width of the slit is 4.0×10-6 m. Determine the width of the central bright fringe when the wavelength of the light in a vacuum is Ȝ = 690 nm (red).
Half width: ș = 0.1725 rad = 9. 9o s = 0.069 m width: 2ș = 0.35 rad = 19.8o 2s = 0.14 m
13
5HVROXWLRQ
IB 12
5HVROXWLRQ the ability to distinguish between two sources of light
The ability toUHVROYH two sources of light depends on . . . distance away from aperture distance between two sources wavelength size of aperture 5HVROXWLRQRIWZRVRXUFHV
WKURXJKDVLQJOHVOLW
5HVROXWLRQRIWZRVRXUFHV WKURXJKDFLUFXODUDSHUWXUH Examples: pupil of eye, telescope, microscope
,QWHQVLW\3DWWHUQV :HOOUHVROYHG
-XVWUHVROYHG
1RWUHVROYHG
5HVROYLQJ3RZHU the minimum angle between sources for them to be “just resolved” = ½ width of one diffraction pattern
6LQJOH6OLW
&LUFXODU $SHUWXUH
5D\OHLJK&ULWHULRQ For two sources to be “just resolved,” the first minimum of one diffraction pattern is located on top of the central maximum of the other diffraction pattern. 14
IB 12 'LVWDQFH5HODWLRQVKLS
1. The brightest star in the winter sky in the Northern Hemisphere is Sirius. In reality, Sirius is a system of two stars that orbit each other. The Hubble Space Telescope (diameter 2.4 m) is pointed at the Sirius system, which is 7.98 x 1016 meters from Earth. a) What is the minimum separation needed between the stars in order for the telescope to just resolve them? Assumption: average light coming in from the stars has a wavelength of 500 nm.
2.2 x 1010 m
b) What is the resolving power of the telescope?
2. a) What is the resolving power of your eye? Assumptions:
2 x 10-4 rad
Diameter of pupil approx = 3 mm
b) How far away can a car be for you to just distinguish between the two headlights at night? Assumptions:
R = 10 km
Distance between headlights = 2 m
15
IB 12
6LJQLILFDQFHRI5HVROXWLRQ Due to diffraction effects, all devices have a limit on their ability to perceive and to resolve between sources of light. For instance, our eyes can never see atoms since atoms are smaller than the wavelength of visible light so light waves will just diffract around them. Here are some cases where diffraction and resolution are important. 1. CDs and DVDs CDs and DVDs store digital information as “bumps” and “pits” etched into a plastic surface. Music CDs have data tracks approximately 5 x 10-7 m wide with the bumps and pits just over 1 x 10-7 m high. The bumps and pits on a DVD are much smaller so that more data can be stored. The data is read by reflecting a laser beam off the surface. The wavelength of laser light used to read the data and the size of the aperture of the lens used to receive the laser light places a limit on how close together the bumps and pits can be placed, that is, places a limit on the resolution of the data.
2. Electron Microscopes In order to resolve objects beyond the limits imposed by the wavelength of
visible light, the wave properties of electrons are used in electron microscopes.
The de Broglie wavelength of an electron is much smaller than the wavelength
of a photon of visible light so a microscope using an electron beam can resolve
objects that are much smaller than those of a light microscope.
3. Radio Telescopes Astronomers often wish to detect the radio waves emitted by very distant objects like quasars and galaxies. However, since the wavelength of radio waves is much larger than visible light, the ability of a radio telescope to resolve sources is more limited than that of light telescopes. To get around this limitation, astronomers use two or more radio telescopes separated by a large distance, called a Very Large Array (VLA). For instance, in New Mexico, there is a VLA consisting of 27 parabolic dishes each of diameter 25 m arranged in a Y-shape that covers an area of 570 km2. EXAMPLE: The Galaxy Cygnus A can be resolved optically as an elliptically shaped galaxy. However, it is also a strong emitter of radio waves of wavelength 0.15 m. The Galaxy is estimated to be 5.0 x 1024 m from Earth. Use of a radio telescope shows that the radio emission is from two sources separated by a distance of 3.0 x 1021 m. Estimate the diameter of the dish required to just resolve the sources.
ș = 6.0 x 10-4 rad d = 3000 m (much too big for one dish)
16
3RODUL]DWLRQRI/LJKW
IB 12
Nature of EM Waves: 1. produced by oscillation of electric charge 2. creates varying electric and magnetic fields that are perpendicular 3. transverse wave 4. plane of vibration is arbitrarily plane of E field vibration
3RODUL]HG/LJKW±light in which the electric field vector vibrates in one plane only
8QSRODUL]HG/LJKW±light in which the electric field vectors vibrates in random directions
Polarization of sunlight: 1. IURPVFDWWHULQJE\PROHFXOHVLQDWPRVSKHUH: Sunlight is polarized in a direction perpendicular to direction of wave velocity. 2. E\UHIOHFWLRQIURPQRQPHWDOOLFVXUIDFH Incident sunlight is unpolarized. Reflected and refracted rays are partially polarized by the surface. The plane of polarization of the reflected light is horizontal (parallel to surface). If the reflected and refracted rays are perpendicular, the reflected light is completely polarized. The angle of incidence for this complete polarization depends on relative indices of refraction of the two substances and is known as Brewster’s angle ( I ).
%UHZVWHU¶V/DZ± When light is incident on a surface at such an angle that the reflected and transmitted rays are perpendicular and the reflected ray is totally plane polarized, then the index of refraction of the substance is equal to the tangent of the angle of incidence. (Q WDQ I ) 'HULYDWLRQ
QL sin TL QU sin T U if QL 1.00 Q
sin TL sin T U
([DPSOH What is Brewster’s angle for sunlight reflected off a lake?
Q tan I
L U 90 0
1.33 tan I
sin TL Q sin(900 TL)
530
Q tan TL Q tan I
I
17
IB 12 3RODUL]HU±device that produces plane polarized light from an unpolarized beam 7UDQVPLVVLRQD[LV±direction of polarization that a polarizer allows through $VLPSOHPRGHORIDSRODUL]HUXVLQJDZDYHRQDURSH
Transmission axis of polarizer is parallel to the plane of polarization of the wave.
Transmission axis of polarizer is perpendicular to the plane of polarization of the wave.
NOTE: only transverse waves can be polarized – not longitudinal waves – sound cannot be polarized. $PRUHVRSKLVWLFDWHGPRGHORIDSRODUL]HUXVLQJOLJKW Polarizer allows ½ original intensity through since ½ of components of all waves are parallel to transmission axis
How do polarized sunglasses reduce glare? Transmission axis is vertical – does not allow glare to pass through since glare is light that has been horizontally polarized by reflection from non-metallic surface
$QDO\]HU±polarizer used to detect polarized light
When the transmission axis of the analyzer is parallel to that of the polarizer . . .polarized light passes through
When the transmission axis of the analyzer is perpendicular to that of the polarizer . . .no light passes through
18
IB 12 What happens when the analyzer is neither parallel nor perpendicular to the polarizer? The component of the polarized
light parallel to the transmission
axis of the analyzer is allowed
to pass through
'HULYDWLRQ
,D$2 ,D( ( cos T) 2 ,D( 2 cos 2 T
ș = angle between transmission axis of polarizer and analyzer
, , 0 cos 2T 0DOXV¶/DZ± the transmitted intensity of polarized light is equal to the product of the incident intensity times the square of the cosine of the angle between the direction of the analyzer and the direction of the electric field vibration of the polarized light (, ,R FRV ș ) 1. Natural, unpolarized light of intensity 6.0 W m-2 is incident on two polaroids oriented at 600 to each other. Find the intensity of the light transmitted through both of them. ½ through first one Malus law = 0.75 W m-2 =0.75 cd How can light be transmitted through “crossed polarizers?” Insert a third polarizer between the original polarizer and the analyzer. Some component of light from the first will make it through the second and some component of the second will make it through the third The intermediate polarizer “rotates the plane of polarization” at the cost of lost intensity.
2SWLFDOO\$FWLYH6XEVWDQFH± 1) one that rotates the plane of polarization of the light that passes through it
2) one that changes the plane in which the electric field vector of the light vibrates
19
Applications:
IB 12
1. Determining the concentration of solutions Eg – sugar solutions are optically active – intensity of light passing through is inversely proportional to concentration and distance through solution Using a polarizer and analyzer to find angle of rotation for maximum intensity allows calculation of concentration
2. Stress analysis Some materials are optically active under stress and allow different colors to pass through at different angles. Engineers can build models out of plastic and subject them to stress and view through polarizers and analyzers to determine points of probable mechanical failure due to high stress.
3. Liquid crystal displays (LCD) Liquid crystal rotates plane of polarization to block light when activated – seen as a dark spot – number, letter, etc.
20
