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[[Category:Pan American Games medalists in boxing]] |
[[Category:Pan American Games medalists in boxing]] |
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[[Category:Medalists at the 2011 Pan American Games]] |
[[Category:Medalists at the 2011 Pan American Games]] |
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[[Category:21st-century Canadian sportsmen]] |
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Canadian boxer
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Brody Blair |
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|---|---|
| Born | December 27, 1991 |
| Statistics | |
| Weight(s) | Super Middleweight |
| Height | 170 cm (5 ft 7 in) |
| Boxing record | |
| Total fights | 2 |
| Wins | 2 |
| Wins by KO | 2 |
Brody Blair (born December 27, 1991) is a Canadian professional boxer who has represented Canada in multiple international amateur competitions as a middleweight.
[edit]
Born in New Glasgow, Nova Scotia, Blair won a bronze medal at the 2011 Pan American Games in the middleweight division. After receiving a bye in the first round, Blair defeated Mario Bernal of El Salvador in the quarterfinals 24–12 to advance to the medal round. In the semifinals Blair lost to eventual gold medalist Emilio Correa of Cuba 30–7.[1]
[edit]
At the 2012 American Boxing Olympic Qualification Tournament, Blair won two bouts to advance to the quarterfinals as a middleweight. In the quarterfinals he lost to Junior Castillo of Dominican Republic. Blair needed Castillo to win the tournament to advance to the Olympics, but Castillo lost in the finals.[2]
[edit]
At the 2013 AIBA World Boxing Championships in the middleweight division Blair defeated Sanjin-Pol Vrgoč of Croatia and Mahmoud Shabab of the Palestinian territories in the first two rounds 3–0 each before losing to Zoltán Harcsa of Hungary 3–0 in the third round.[3]
Blair went on to represent Canada at the 2014 Commonwealth Games.
[edit]
Blair turned professional on May 27, 2017 in Fredericton, New Brunswick and won by TKO in the 2nd round. He won his second professional fight in November.[5]
| 2 fights | 2 wins | losses |
|---|---|---|
| By knockout | 2 | 0 |
==Outline of proposed changes==
==Outline of proposed changes==
{{Dashboard.wikiedu.org_bibliography/outline}}I think two most important sections I could add is History and Collections. Besides, I could also add a reputation section.
{{Dashboard.wikiedu.org_bibliography/outline}}I think two most important sections I could add is History and Collections. Besides, I could also add .
History
History
There are a few sources that detail the history of the archives including Bin’s and Bartlett’s articles.
Bin’s
Collections
Collections
Mao’s article and Lai’s article provide information on the collections of the archives. The archive’s official website is helpful in this regard. I could also add some photos of the galleries and collections in the archives.
I could also add some photos of the gallery
”’Reputation”’
I need to find some news reports about the archives to collect information on the reviews the archives receive.
”’Governance and Departments”’
I can find related information on the official website.
You will be compiling your bibliography and creating an outline of the changes you will make in this sandbox.
 |
Bibliography
As you gather the sources for your Wikipedia contribution, think about the following:
If you’re not sure whether a source is reliable, ask a librarian! If you have questions about Wikipedia’s sourcing rules, you can use the Get Help button below to contact your Wikipedia Expert. |
Moss, W. W. (1996). Dang’an: contemporary Chinese archives. The China Quarterly, 145, 112-129.
Moss, W. W. (1982). Archives in the People’s Republic of China. The American Archivist, 45(4), 385-409.
Mao, L., & Ma, Z. (2012). “Writing History in the Digital Age”: The New Qing History Project and the Digitization of Qing Archives. History Compass, 10(5), 367-374.
Lai, C. K. (2017). Enterprise history: studies and archives. In Chinese Business History (pp. 169-188). Routledge.
Bartlett, B. S. (2007). A world-class archival achievement: the People’s Republic of China archivists’ success in opening the Ming-Qing central-government archives, 1949–1998. Archival Science, 7, 369-390.
Bartlett, B. S. (1981). An archival revival: the Qing central government archives in Peking today. Ch’ing-shih wen-t’i, 4(6), 81-110.
Bin. G. (2024). Description of documentation relating to the history of Macao held in the First Historical Archive of China. http://www.icm.gov.mo/rc/viewer/20019/1023#’LAB2001900180001′)
The First Historical Archives of China https://fhac.com.cn/index.html
Edit this section to compile the bibliography for your Wikipedia assignment. Add the name and/or notes about what each source covers, then use the “Cite” button to generate the citation for that source.
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Examples:
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[edit]
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Now that you have compiled a bibliography, it’s time to plan out how you’ll improve your assigned article.
In this section, write up a concise outline of how the sources you’ve identified will add relevant information to your chosen article. Be sure to discuss what content gap your additions tackle and how these additions will improve the article’s quality. Consider other changes you’ll make to the article, including possible deletions of irrelevant, outdated, or incorrect information, restructuring of the article to improve its readability or any other change you plan on making. This is your chance to really think about how your proposed additions will improve your chosen article and to vet your sources even further. Note: This is not a draft. This is an outline/plan where you can think about how the sources you’ve identified will fill in a content gap. |
I think two most important sections I could add is History and Collections. Besides, I could also add sections like Reputation including both positive and negative reviews if those information is available, Governance, and Departments.
History
There are a few sources that detail the history of the archives including Bin’s and Bartlett’s articles.
Collections
Mao’s article and Lai’s article provide information on the collections of the archives. The archive’s official website is helpful in this regard. I could also add some photos of the galleries and collections in the archives.
Reputation
I need to find some news reports about the archives to collect information on the reviews the archives receive.
Governance and Departments
I can find related information on the official website.
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Copyright © 2024 by IOP Publishing Ltd and individual contributors
From Wikipedia, the free encyclopedia
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==Licensing== |
==Licensing== |
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{{Non-free |
{{Non-free poster}} |
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| Description | Poster for the German film Women You Rarely Greet |
|---|---|
| Author or copyright owner |
Unknown |
| Source (WP:NFCC#4) | https://letterboxd.com/film/frauen-die-man-oft-nicht-grut/releases/ |
| Use in article (WP:NFCC#7) | Women You Rarely Greet |
| Purpose of use in article (WP:NFCC#8) | to serve as the primary means of visual identification at the top of the article dedicated to the work in question. |
| Not replaceable with free media because (WP:NFCC#1) |
Any derivative work based upon the cover art would be a copyright violation, so creation of a free image is not possible. |
| Minimal use (WP:NFCC#3) | For use in the relevant article infobox only |
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The use of a low resolution image of a work’s cover will not impact the commercial viability of the work. |
| Fair useFair use of copyrighted material in the context of Women You Rarely Greet//en.wikipedia.org/wiki/File:Women_You_Rarely_Greet.jpgtrue | |
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This image is of a film poster, and the copyright for it is most likely owned by either the publisher or the creator of the work depicted. It is believed that the use of scaled-down, low-resolution images of film posters
qualifies as fair use under the copyright law of the United States. Any other uses of this image, on Wikipedia or elsewhere, may be copyright infringement. See Wikipedia:Non-free content for more information. |
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Copyright © 2024 by IOP Publishing Ltd and individual contributors
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Copyright © 2024 by IOP Publishing Ltd and individual contributors
Canadian equestrian (b. 1968)
| Eric Lamaze | |
|---|---|
 |
|
| Full name | Eric Lamaze |
| Nationality |  Canada |
| Discipline | Show jumping |
| Born | April 17, 1968 Montreal, Quebec, Canada |
| Height | 5 ft 8 in (1.73 m) |
Eric Lamaze (born April 17, 1968) is a Canadian showjumper and Olympic champion.[1] He won individual gold and team silver at the 2008 Beijing Olympics, riding Hickstead. Lamaze has won three Olympic medals, as well as four Pan American Games medals and one World Equestrian Games bronze. He is considered one of Canada’s best showjumpers. He is currently banned from participating in equestrian activities until 2027.
Lamaze was born in Montreal, Quebec.[1][2] He started riding at age twelve and worked in exchange for time in the saddle.[3] He was considered a promising junior rider,[1] and trained under Roger Deslauriers, George Morris, Jay Hayes and Hugh Graham.[2]
Lamaze began competing at the grand prix level in 1991[2] or 1992.[3] A year later, he was named to the Canadian equestrian team.[2] His first major competition as a national team member was the 1994 World Equestrian Games.[1]
Lamaze was named to the Canadian team for the 1996 Summer Olympics in Atlanta, Georgia, but lost his place and received a four-year suspension after testing positive for cocaine.[4] Arbitrator Ed Ratushny overturned the suspension, although Lamaze had already missed the Atlanta Games when the ruling was delivered.[5]
Lamaze rebuilt his career and ascended the rankings, being again regarded as a key member of the Canadian team for the Sydney Games. However he tested positive for a banned stimulant, which resulted in his removal from the team and facing a lifetime ban.[4] Right afterwards, a despondent Lamaze contemplated suicide and while drunk he smoked a cigarette laced with cocaine. Forty-eight hours later, the test for the banned stimulant was reversed on appeal, however Lamaze then tested positive for cocaine which would also have meant a lifetime ban. Arbitrator Ed Ratushny overturned the cocaine test, but the Canadian Olympic Committee refused to reinstate Lamaze on the Canadian team.[5]
[edit]
In 2007, Lamaze became the first Canadian jumping rider in 20 years to make the top ten in the world rankings. He was also the first North American jumping rider to exceed one million in prize money a year, a third of these earnings the result of winning the CN International Grand Prix at Spruce Meadows.[5] The CN International Grand Prix was Lamaze’s first major win with Hickstead.
Lamaze competed in the Beijing Olympics, riding the stallion Hickstead.[6] He was awarded a silver medal after a strong performance in the team event.[7] Lamaze went on to win a gold medal in the individual show jumping event of the 2008 Beijing Olympics at the Shatin Equestrian Venue in Hong Kong as a result of a jump off between himself riding Hickstead and the Swedish rider Rolf-Göran Bengtsson, riding Ninja.[6]
In the January 2009 Rolex World Rankings for show jumping by the International Equestrian Federation, Lamaze was named to the top spot for the first time.[8][9] In October 2009, Lamaze won the €120,000 Equita Masters in Lyon, France, riding Hickstead.[10]
Lamaze returned to first place in the Rolex Rankings for July 2010. In July that year, he had two major wins with Hickstead, at the Aachen World Equestrian Festival[11] and the Spruce Meadows Queen Elizabeth II Cup.
In 2011, Lamaze and Hickstead won the €200,000 Rome Grand Prix, the €200,000 La Baule Grand Prix, the Spruce Meadows Queen Elizabeth II Cup, the €23,000 1.55m in Rotterdam,[12] the $1 million CN International Grand Prix, and the €100,000 Barcelona Grand Prix.[13]
Hickstead died during an event for the FEI Show Jumping World Cup in Verona, Italy. Lamaze was distraught and considered retirement.[14]
After the death of Hickstead in 2011,[15] Lamaze selected the nine-year-old mare Derly Chin De Muze to ride at the 2012 London Olympics.[16]
In July 2016, he was again named to Canada’s Olympic team, serving as the leader following Ian Millar’s decision to not compete again. Lamaze rode the Hanoverian mare, Fine Lady 5.[17] As a member of Canada’s jumping team, he competed in a climactic jump-off for the bronze medal, which was ultimately won by the German team. Later, he won a bronze medal in the individual jumping event, a single knocked rail preventing him from earning a second gold medal.[18]
In 2017, Lamaze claimed he was diagnosed with brain cancer, which he revealed to the public in 2019.[19] He continued competing for some time, winning a gold medal at the Spruce Meadows Masters tournament in June 2019.[20][21]
In 2021, he was selected for random drug testing at an event in the Netherlands, which he failed to comply.[22] Also that year he announced that he would not seek to be part of the Canadian Olympic team for the 2020 Summer Olympics in Tokyo, saying that while his health was stable he felt there were too many risks.[23] Lamaze’s battle with brain cancer continued, and on March 31, 2022 he announced that he would be retiring from competition in order to focus on his health. He planned to remain as the Canadian showjumping team’s chef d’équipe. After announcing his retirement, he said: “I’ve always said that I will retire under my own terms when the time is right. The situation with my health has forced me to make the decision earlier than I had envisioned, but the silver lining is that I still have the will to win and can contribute to the Canadian team and the sport I love through my new role.”[24]
During a lawsuit in 2023, Lamaze testified his brain cancer, a glioblastoma, had shrunk by 2021, which did free him of brain cancer, leaving him only with throat cancer, despite his claims of having brain cancer at the time.[25]
Citing Lamaze’s refusal to undertake drug testing and his forgery of medical documents, the FEI banned Lamaze from competition until September 11, 2027.[26][27]
In 2010, Iron Horse Farm brought a suit against Lamaze for misrepresenting horses he sold to the farm. As the case slowly made its way through the court system, evidence emerged suggesting Lamaze had forged his medical history.
[edit]
In August 2023, it was found that Lamaze had forged medical documents submitted to the Ontario Supreme Court.[28][4][29]
As a result of these findings, Lamaze’s lifelong attorney, Tim Danson, stepped down as counsel.
It is alleged that Lamaze’s ongoing cancer and alleged treatments during his career were faked by Lamaze in order to delay and avoid several ongoing court cases brought against Lamaze in relation to horse sale disputes.
In order to dismiss this allegation that his cancer and health issues are entirely faked, Lamaze would need to produce reliable medical evidence to prove his claims. Lamaze has publicly declined to do so, stating in an 11 September 2023 article:
“They want a doctor’s note; have they seen me? People were telling me to go home, every day for years, as I was so sick. People who have never seen me and don’t know, believe what they see online. I’ve always spoken the truth and will continue to.”
Lamaze has said he was unaware of letters and documents submitted to the court and cited issues such as data protection relating to his medical records, and the need to protect people who have treated him.[30]
In a Florida lawsuit from November 2023, the judge ruled that Lamaze’s conduct to obfuscate his medical condition was fraudulent and deliberate.[31] Lamaze was found liable for $1.4 million in damages for fraud, breach of contract, and breaking Florida law, related to the purchase and sale of two horses.[32]
After nearly 15 years in the court system, in August 2024, a Canadian court found in support of Iron Horse Farm, who brought suit against Lamaze for misrepresenting horses he later sold to the farm. The court ordered Lamaze to return the money for the horse sales, as well as to cover damages and court costs, totaling $786,000.[33] Iron Horse Farm said they did not expect Lamaze would be in a position to pay back the farm for their damages.[34]
[edit]
{{cite web}}: CS1 maint: multiple names: authors list (link)
While basking in the warmth of the summer sun, a student reads of the latest breakthrough in achieving sustained thermonuclear power and vaguely recalls hearing about the cold fusion controversy. The three are connected. The Sun’s energy is produced by nuclear fusion (see Figure 1). Thermonuclear power is the name given to the use of controlled nuclear fusion as an energy source. While research in the area of thermonuclear power is progressing, high temperatures and containment difficulties remain. The cold fusion controversy centered around unsubstantiated claims of practical fusion power at room temperatures.
Nuclear fusion is a reaction in which two nuclei are combined, or fused, to form a larger nucleus. We know that all nuclei have less mass than the sum of the masses of the protons and neutrons that form them. The missing mass times [latex]{c^2}[/latex] equals the binding energy of the nucleus—the greater the binding energy, the greater the missing mass. We also know that [latex]{\text{BE/}A}[/latex] , the binding energy per nucleon, is greater for medium-mass nuclei and has a maximum at Fe (iron). This means that if two low-mass nuclei can be fused together to form a larger nucleus, energy can be released. The larger nucleus has a greater binding energy and less mass per nucleon than the two that combined. Thus mass is destroyed in the fusion reaction, and energy is released (see Figure 2). On average, fusion of low-mass nuclei releases energy, but the details depend on the actual nuclides involved.
The major obstruction to fusion is the Coulomb repulsion between nuclei. Since the attractive nuclear force that can fuse nuclei together is short ranged, the repulsion of like positive charges must be overcome to get nuclei close enough to induce fusion. Figure 3 shows an approximate graph of the potential energy between two nuclei as a function of the distance between their centers. The graph is analogous to a hill with a well in its center. A ball rolled from the right must have enough kinetic energy to get over the hump before it falls into the deeper well with a net gain in energy. So it is with fusion. If the nuclei are given enough kinetic energy to overcome the electric potential energy due to repulsion, then they can combine, release energy, and fall into a deep well. One way to accomplish this is to heat fusion fuel to high temperatures so that the kinetic energy of thermal motion is sufficient to get the nuclei together.
You might think that, in the core of our Sun, nuclei are coming into contact and fusing. However, in fact, temperatures on the order of [latex]{^{108} \text{K}}[/latex] are needed to actually get the nuclei in contact, exceeding the core temperature of the Sun. Quantum mechanical tunneling is what makes fusion in the Sun possible, and tunneling is an important process in most other practical applications of fusion, too. Since the probability of tunneling is extremely sensitive to barrier height and width, increasing the temperature greatly increases the rate of fusion. The closer reactants get to one another, the more likely they are to fuse (see Figure 4). Thus most fusion in the Sun and other stars takes place at their centers, where temperatures are highest. Moreover, high temperature is needed for thermonuclear power to be a practical source of energy.
The Sun produces energy by fusing protons or hydrogen nuclei [latex]{^1 \textbf{H}}[/latex] (by far the Sun’s most abundant nuclide) into helium nuclei [latex]{^4 \text{He}}[/latex]. The principal sequence of fusion reactions forms what is called the proton-proton cycle:
[latex]\begin{array} {r @{{} \rightarrow{}} l @{{} \;\;\; {}} l} {^1 \textbf{H} + {^1 \textbf{H}}} & {{^2 \textbf{H}} + e ^+ + v_e} & {(0.42 \;\text{MeV})} \\[1em] {{^1 \textbf{H}} + {^2 \textbf{H}}} & {{^3 \text{He}} + \gamma} & {(5.49 \;\text{MeV})} \\[1em] {{^3 \text{He}} + {^3 \text{He}}} & {{^4 \text{He}} + {^1 \textbf{H}} + {^1 \textbf{H}}} & {(12.86 \text{MeV})} \end{array}[/latex]
where [latex]{e ^+}[/latex] stands for a positron and [latex]{v_e}[/latex] is an electron neutrino. (The energy in parentheses is released by the reaction.) Note that the first two reactions must occur twice for the third to be possible, so that the cycle consumes six protons ( [latex]{^1 \textbf{H}}[/latex] ) but gives back two. Furthermore, the two positrons produced will find two electrons and annihilate to form four more [latex]{\gamma}[/latex] rays, for a total of six. The overall effect of the cycle is thus
[latex]\begin{array} {r @{{} \rightarrow {}}l @{{} \;\;\; {}} l} {{2e ^-} + 4 {^1 \textbf{H}}} & {{^4 \text{He}} + {2v_e} + {6 \gamma}} & {(26.7 \;\text{MeV})} \end{array}[/latex]
Theories of the proton-proton cycle (and other energy-producing cycles in stars) were pioneered by the German-born, American physicist Hans Bethe (1906–2005), starting in 1938. He was awarded the 1967 Nobel Prize in physics for this work, and he has made many other contributions to physics and society. Neutrinos produced in these cycles escape so readily that they provide us an excellent means to test these theories and study stellar interiors. Detectors have been constructed and operated for more than four decades now to measure solar neutrinos (see Figure 6). Although solar neutrinos are detected and neutrinos were observed from Supernova 1987A (Figure 7), too few solar neutrinos were observed to be consistent with predictions of solar energy production. After many years, this solar neutrino problem was resolved with a blend of theory and experiment that showed that the neutrino does indeed have mass. It was also found that there are three types of neutrinos, each associated with a different type of nuclear decay.
The proton-proton cycle is not a practical source of energy on Earth, in spite of the great abundance of hydrogen ([latex]{^1 \textbf{H}}[/latex]). The reaction [latex]{{^1 \textbf{H}} + {^1 \textbf{H}} \rightarrow {^2 \textbf{H}} + e^+ + v_e}[/latex] has a very low probability of occurring. (This is why our Sun will last for about ten billion years.) However, a number of other fusion reactions are easier to induce. Among them are:
[latex]\begin{array} {r @{{} \rightarrow{}} l @{{} \;\;\; {}} l} {^2 \textbf{H} + {^2 \textbf{H}}} & {{^3 \textbf{H}} + {^1 \textbf{H}}} & {(4.03 \;\text{MeV})} \\[1em] {{^2 \textbf{H}} + {^2 \textbf{H}}} & {{^3 \text{He}} + n} & {(3.27 \;\text{MeV})} \\[1em] {{^2 \textbf{H}} + {^3 \textbf{H}}} & {{^4 \text{He}} + n} & {(17.59 \text{MeV})} \\[1em] {{^2 \textbf{H}} + {^2 \textbf{H}}} & {{^4 \text{He}} + \gamma} & {(23.85 \;\text{MeV})} \end{array}[/latex]
Deuterium ([latex]{^2 \textbf{H}}[/latex]) is about 0.015% of natural hydrogen, so there is an immense amount of it in sea water alone. In addition to an abundance of deuterium fuel, these fusion reactions produce large energies per reaction (in parentheses), but they do not produce much radioactive waste. Tritium ([latex]{^3 \textbf{H}}[/latex]) is radioactive, but it is consumed as a fuel (the reaction [latex]{{^2 \textbf{H}} + {^3 \textbf{H}} \rightarrow {^4 \text{He}} + n}[/latex]), and the neutrons and [latex]{\gamma}[/latex] s can be shielded. The neutrons produced can also be used to create more energy and fuel in reactions like
[latex]\begin{array} {r @{{} \rightarrow {}}l @{{} \;\;\; {}} l} {{n} + {^1 \textbf{H}}} & {{^2 \textbf{H}} + {\gamma}} & {(20.68 \;\text{MeV})} \end{array}[/latex]
and
[latex]\begin{array} {r @{{} \rightarrow {}}l @{{} \;\;\; {}} l} {{n} + {^1 \textbf{H}}} & {{^2 \textbf{H}} + {\gamma}} & {(2.22 \;\text{MeV})} \end{array}[/latex]
Note that these last two reactions, and [latex]{{^2 \textbf{H}} + {^2 \textbf{H}} \rightarrow {^4 \text{He}} + \gamma}[/latex], put most of their energy output into the [latex]{\gamma}[/latex] ray, and such energy is difficult to utilize.
The three keys to practical fusion energy generation are to achieve the temperatures necessary to make the reactions likely, to raise the density of the fuel, and to confine it long enough to produce large amounts of energy. These three factors—temperature, density, and time—complement one another, and so a deficiency in one can be compensated for by the others. Ignition is defined to occur when the reactions produce enough energy to be self-sustaining after external energy input is cut off. This goal, which must be reached before commercial plants can be a reality, has not been achieved. Another milestone, called break-even, occurs when the fusion power produced equals the heating power input. Break-even has nearly been reached and gives hope that ignition and commercial plants may become a reality in a few decades.
Two techniques have shown considerable promise. The first of these is called magnetic confinement and uses the property that charged particles have difficulty crossing magnetic field lines. The tokamak, shown in Figure 8, has shown particular promise. The tokamak’s toroidal coil confines charged particles into a circular path with a helical twist due to the circulating ions themselves. In 1995, the Tokamak Fusion Test Reactor at Princeton in the US achieved world-record plasma temperatures as high as 500 million degrees Celsius. This facility operated between 1982 and 1997. A joint international effort is underway in France to build a tokamak-type reactor that will be the stepping stone to commercial power. ITER, as it is called, will be a full-scale device that aims to demonstrate the feasibility of fusion energy. It will generate 500 MW of power for extended periods of time and will achieve break-even conditions. It will study plasmas in conditions similar to those expected in a fusion power plant. Completion is scheduled for 2018.
The second promising technique aims multiple lasers at tiny fuel pellets filled with a mixture of deuterium and tritium. Huge power input heats the fuel, evaporating the confining pellet and crushing the fuel to high density with the expanding hot plasma produced. This technique is called inertial confinement, because the fuel’s inertia prevents it from escaping before significant fusion can take place. Higher densities have been reached than with tokamaks, but with smaller confinement times. In 2009, the Lawrence Livermore Laboratory (CA) completed a laser fusion device with 192 ultraviolet laser beams that are focused upon a D-T pellet (see Figure 9).
(a) Calculate the energy released by the fusion of a 1.00-kg mixture of deuterium and tritium, which produces helium. There are equal numbers of deuterium and tritium nuclei in the mixture.
(b) If this takes place continuously over a period of a year, what is the average power output?
Strategy
According to [latex]{{^2 \textbf{H}} + {^3 \textbf{H}} \rightarrow {^4 \text{He}} + n}[/latex], the energy per reaction is 17.59 MeV. To find the total energy released, we must find the number of deuterium and tritium atoms in a kilogram. Deuterium has an atomic mass of about 2 and tritium has an atomic mass of about 3, for a total of about 5 g per mole of reactants or about 200 mol in 1.00 kg. To get a more precise figure, we will use the atomic masses from Appendix A. The power output is best expressed in watts, and so the energy output needs to be calculated in joules and then divided by the number of seconds in a year.
Solution for (a)
The atomic mass of deuterium ([latex]{^2 \textbf{H}}[/latex]) is 2.014102 u, while that of tritium ([latex]{^3 \textbf{H}}[/latex]) is 3.016049 u, for a total of 5.032151 u per reaction. So a mole of reactants has a mass of 5.03 g, and in 1.00 kg there are [latex]{(1000 \;\text{g})/(5.03 \;\text{g/mol}) = 198.8 \;\text{mol of reactants}}[/latex]. The number of reactions that take place is therefore
[latex]{(198.8 \;\text{mol})(6.02 \times 10^{23} \;\text{mol}^{-1}) = 1.20 \times 10^{26} \;\text{reactions}}[/latex]
The total energy output is the number of reactions times the energy per reaction:
[latex]{E = (1.20 \times 10^{26} \;\text{reactions})(17.59 \;\text{MeV/reaction}) (1.602 \times 10^{-13} \;\text{J/MeV}) = 3.37 \times 10^{14} \;\text{J}}[/latex]
Solution for (b)
Power is energy per unit time. One year has [latex]{3.16 \times 10^7 \;\text{s}}[/latex], so
[latex]\begin{array}{ r @{{}={}} l} {P} & {\frac{E}{t} = \frac{3.37 \times 10^{14} \;\text{J}}{3.16 \times 10^7 \;\text{s}}} \\[1em] & {1.07 \times 10^7 \;\text{W} = 10.7 \;\text{MW}} \end{array}[/latex]
Discussion
By now we expect nuclear processes to yield large amounts of energy, and we are not disappointed here. The energy output of [latex]{3.37 \times 10^{14} \;\text{J}}[/latex] from fusing 1.00 kg of deuterium and tritium is equivalent to 2.6 million gallons of gasoline and about eight times the energy output of the bomb that destroyed Hiroshima. Yet the average backyard swimming pool has about 6 kg of deuterium in it, so that fuel is plentiful if it can be utilized in a controlled manner. The average power output over a year is more than 10 MW, impressive but a bit small for a commercial power plant. About 32 times this power output would allow generation of 100 MW of electricity, assuming an efficiency of one-third in converting the fusion energy to electrical energy.
[latex]\begin{array} {r @{{} \rightarrow{}} l @{{} \;\;\; {}} l} {^1 \textbf{H} + {^1 \textbf{H}}} & {{^2 \textbf{H}} + e ^+ + v_e} & {(0.42 \;\text{MeV})} \\[1em] {{^1 \textbf{H}} + {^2 \textbf{H}}} & {{^3 \text{He}} + \gamma} & {(5.49 \;\text{MeV})} \\[1em] {{^3 \text{He}} + {^3 \text{He}}} & {{^4 \text{He}} + {^1 \textbf{H}} + {^1 \textbf{H}}} & {(12.86 \text{MeV})} \end{array}[/latex]
being the principal sequence of energy-producing reactions in our Sun.
[latex]\begin{array} {r @{{} \rightarrow {}}l @{{} \;\;\; {}} l} {{2e ^-} + 4 {^1 \textbf{H}}} & {{^4 \text{He}} + {2v_e} + {6 \gamma}} & {(26.7 \;\text{MeV})} \end{array}[/latex]
where the 26.7 MeV includes the energy of the positrons emitted and annihilated.
1: Why does the fusion of light nuclei into heavier nuclei release energy?
2: Energy input is required to fuse medium-mass nuclei, such as iron or cobalt, into more massive nuclei. Explain why.
3: In considering potential fusion reactions, what is the advantage of the reaction [latex]{{^2 \textbf{H}} + {^3 \textbf{H}} \rightarrow {^4 \text{He}} + n}[/latex] over the reaction [latex]{{^2 \textbf{H}}+{^2 \textbf{H}} \rightarrow {^3 \text{He}} +n}[/latex] ?
4: Give reasons justifying the contention made in the text that energy from the fusion reaction [latex]{{^2 \textbf{H}} + {^2 \textbf{H}} \rightarrow {^4 \text{He}} + \gamma}[/latex] is relatively difficult to capture and utilize.
1: Verify that the total number of nucleons, total charge, and electron family number are conserved for each of the fusion reactions in the proton-proton cycle in
[latex]{{^1 \textbf{H}} + {^1 \textbf{H}} \rightarrow {^2 \textbf{H}} + e^+ + v_e ,}[/latex]
[latex]{{^1 \textbf{H}} + {^2 \textbf{H}} \rightarrow {^3 \text{He}} + \gamma ,}[/latex]
and
[latex]{{^3 \text{He}} + {^3 \text{He}} \rightarrow {^4 \text{He}} + {^1 \textbf{H}} + {^1 \textbf{H}} .}[/latex]
(List the value of each of the conserved quantities before and after each of the reactions.)
2: Calculate the energy output in each of the fusion reactions in the proton-proton cycle, and verify the values given in the above summary.
3: Show that the total energy released in the proton-proton cycle is 26.7 MeV, considering the overall effect in [latex]{{^1 \textbf{H}} + {^1 \textbf{H}} \rightarrow {^2 \textbf{H}} + e^+ + v_e}[/latex], [latex]{^1 \textbf{H} + {^2 \textbf{H}} \rightarrow {^3 \text{He} + \gamma} }[/latex] , and [latex]{{^3 \text{He}} + {^3 \text{He}} \rightarrow {^4 \text{He}} + {^1 \textbf{H}} + {^1 \textbf{H}}}[/latex] and being certain to include the annihilation energy.
4: Verify by listing the number of nucleons, total charge, and electron family number before and after the cycle that these quantities are conserved in the overall proton-proton cycle in [latex]{2e-+ 4 {^1 \textbf{H}} \rightarrow {^4 \text{He}} + 2v_{\textbf{e}} + 6 \gamma}[/latex].
5: The energy produced by the fusion of a 1.00-kg mixture of deuterium and tritium was found in Example Example 1 – Calculating Energy and Power from Fusion. Approximately how many kilograms would be required to supply the annual energy use in the United States?
6: Tritium is naturally rare, but can be produced by the reaction [latex]{n + {^2 \textbf{H}} \rightarrow {^3 \textbf{H}} + \gamma}[/latex]. How much energy in MeV is released in this neutron capture?
7: Two fusion reactions mentioned in the text are
[latex]{n + {^3 \text{He}} \rightarrow {^4 \text{He}} + \gamma}[/latex]
and
[latex]{n+ {^1 \textbf{H}} \rightarrow {^2 \textbf{H}}+ \gamma}[/latex].
Both reactions release energy, but the second also creates more fuel. Confirm that the energies produced in the reactions are 20.58 and 2.22 MeV, respectively. Comment on which product nuclide is most tightly bound, [latex]{^4 \text{He}}[/latex] or [latex]{^2 \textbf{H}}[/latex].
8: (a) Calculate the number of grams of deuterium in an 80,000-L swimming pool, given deuterium is 0.0150% of natural hydrogen.
(b) Find the energy released in joules if this deuterium is fused via the reaction [latex]{{^2 \textbf{H}} + {^2 \textbf{H}} \rightarrow {^3 \text{He}} + n}[/latex].
(c) Could the neutrons be used to create more energy?
(d) Discuss the amount of this type of energy in a swimming pool as compared to that in, say, a gallon of gasoline, also taking into consideration that water is far more abundant.
9: How many kilograms of water are needed to obtain the 198.8 mol of deuterium, assuming that deuterium is 0.01500% (by number) of natural hydrogen?
10: The power output of the Sun is [latex]{4 \times 10^{26} \;\text{W}}[/latex].
(a) If 90% of this is supplied by the proton-proton cycle, how many protons are consumed per second?
(b) How many neutrinos per second should there be per square meter at the Earth from this process? This huge number is indicative of how rarely a neutrino interacts, since large detectors observe very few per day.
Another set of reactions that result in the fusing of hydrogen into helium in the Sun and especially in hotter stars is called the carbon cycle. It is
[latex]\begin{array}{l @{{}\rightarrow{}}l} {^{12} \text{C} + ^1 \textbf{H}} & {^{13} \textbf{N} + \gamma ,} \\[1em] {^{13} \textbf{N}} & {^{13} \text{C} + e^+ + v_{e} ,} \\[1em] {^{13} \text{C} + ^1 \textbf{H}} & {^{14} \textbf{N} + \gamma ,} \\[1em] {^{14} \textbf{N} + ^{1} \textbf{H}} & {^{15} \textbf{O}+ \gamma ,} \\[1em] {^{15} \textbf{O}} & {^{15} \textbf{N} + e^+ + v_e,} \\[1em] {^{15} \textbf{N} + ^1 \textbf{H}} & {^{12} \text{C} + ^4 \text{He} .} \end{array}[/latex]
11: Write down the overall effect of the carbon cycle (as was done for the proton-proton cycle in [latex]{2e^- + ^{41} \textbf{H} \rightarrow ^4 \text{He} + 2v_e + 6 \gamma}[/latex] ). Note the number of protons ( [latex]{^1 \textbf{H}}[/latex]) required and assume that the positrons (
[latex]{e^+}[/latex]) annihilate electrons to form more [latex]{\gamma}[/latex] rays.
12: (a) Find the total energy released in MeV in each carbon cycle (elaborated in the above problem) including the annihilation energy.
(b) How does this compare with the proton-proton cycle output?
13: Verify that the total number of nucleons, total charge, and electron family number are conserved for each of the fusion reactions in the carbon cycle given in the above problem. (List the value of each of the conserved quantities before and after each of the reactions.)
14: Integrated Concepts
The laser system tested for inertial confinement can produce a 100-kJ pulse only 1.00 ns in duration. (a) What is the power output of the laser system during the brief pulse?
(b) How many photons are in the pulse, given their wavelength is [latex]{1.06 \;\mu \text{m}}[/latex] ?
(c) What is the total momentum of all these photons?
(d) How does the total photon momentum compare with that of a single 1.00 MeV deuterium nucleus?
15: Integrated Concepts
Find the amount of energy given to the [latex]{^4 \text{He}}[/latex] nucleus and to the [latex]{\gamma}[/latex] ray in the reaction [latex]{n + {^{3} \text{He}} \rightarrow {^{4} \text{He}} + \gamma}[/latex] , using the conservation of momentum principle and taking the reactants to be initially at rest. This should confirm the contention that most of the energy goes to the [latex]{\gamma}[/latex] ray.
16: Integrated Concepts
(a) What temperature gas would have atoms moving fast enough to bring two [latex]{^3 \text{He}}[/latex] nuclei into contact? Note that, because both are moving, the average kinetic energy only needs to be half the electric potential energy of these doubly charged nuclei when just in contact with one another.
(b) Does this high temperature imply practical difficulties for doing this in controlled fusion?
17: Integrated Concepts
(a) Estimate the years that the deuterium fuel in the oceans could supply the energy needs of the world. Assume world energy consumption to be ten times that of the United States which is [latex]{8 \times 10^{19}}[/latex] J/y and that the deuterium in the oceans could be converted to energy with an efficiency of 32%. You must estimate or look up the amount of water in the oceans and take the deuterium content to be 0.015% of natural hydrogen to find the mass of deuterium available. Note that approximate energy yield of deuterium is [latex]{3.37 \times 10^{14}}[/latex] J/kg.
(b) Comment on how much time this is by any human measure. (It is not an unreasonable result, only an impressive one.)
Problems & Exercises
1: (a) [latex]{A = 1+1=2}[/latex] , [latex]{Z=1+1=1+1}[/latex] , [latex]{\text{efn}=0= -1+1}[/latex]
(b) [latex]{A=1+2=3}[/latex] , [latex]{Z=1+1=2}[/latex] , [latex]{\text{efn}=0=0}[/latex]
(c) [latex]{A = 3+3=4+1+1}[/latex] , [latex]{Z=2+2=2+1+1}[/latex] , [latex]{\text{efn}=0=0}[/latex]
3: [latex]\begin{array}{r @{{}={}} l} {E} & {(m_{\textbf{i}} – m_{\textbf{f}})c^2} \\[1em] & {[4m (1\textbf{H}) – m(4 \text{He})]c^2} \\[1em] & {[4(1.007825) – 4.002603](931.5 \;\text{MeV})} \\[1em] & {26.73 \;\text{MeV}} \end{array}[/latex]
5: [latex]{3.12 \times 10^5 \;\text{kg}}[/latex] (about 200 tons)
7:
[latex]\begin{array}{r @{{}={}} l} {E} & {(m_{\textbf{i}} – m_{\textbf{f}})c^2} \\[1em] {E_1} & { (1.008665 + 3.016030 – 4.002603)(931.5 \;\text{MeV})} \\[1em] & {20.58 \;\text{MeV}} \\[1em] {E_2} & {(1.008665 + 1.007825 – 2.014102)(931.5 \;\text{MeV})} \\[1em] & {2.224 \;\text{MeV}} \end{array}[/latex]
[latex]{^4 \text{He}}[/latex] is more tightly bound, since this reaction gives off more energy per nucleon.
9: [latex]{1.19 \times 10^4 \;\text{kg}}[/latex]
[latex]{2e^- + ^{41} \textbf{H} \rightarrow ^4 \text{He} + 7 \gamma +2v_e}[/latex]
13: (a) [latex]{A=12+1=13}[/latex] , [latex]{Z=6+1=7}[/latex] , [latex]{\text{efn}=0=0}[/latex]
(b) [latex]{A=13=13}[/latex] , [latex]{Z=7=6+1}[/latex] , [latex]{\text{efn}=0=-1+1}[/latex]
(c) [latex]{A=13+1=14}[/latex] , [latex]{Z=6+1=7}[/latex] , [latex]{\text{efn}=0=0}[/latex]
(d) [latex]{A=14+1=15}[/latex] , [latex]{Z=7+1=8}[/latex] , [latex]{\text{efn}=0=0}[/latex]
(e) [latex]{A=15=15}[/latex] , [latex]{Z=8=7+1}[/latex] , [latex]{\text{efn}=0=-1+1}[/latex]
(f) [latex]{A=15+1=12+4}[/latex] , [latex]{Z=7+1=6+2}[/latex] , [latex]{\text{efn} = 0 = 0}[/latex]
15: [latex]{E_{\gamma} = 20.6 \;\text{MeV}}[/latex]
[latex]{E_{^4 \text{He}} = 5.68 \times 10^{-2} \;\text{MeV}}[/latex]
(a) [latex]{3 \times 10^9 \;\textbf{y}}[/latex]
(b) This is approximately half the lifetime of the Earth.
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