An Overview of Antimatter
Normal matter exists all around us, everyday. As we all know, it is comprised of particles – protons, electrons, neutrons and other smaller particles. Antimatter is comprised of antiparticles, which are the exact opposite of normal particles. Just as there are protons, neutrons, and electrons, there are antiprotons, antineutrons and positrons. They have an opposite charge than their corresponding matter particles.
In 1928, Paul Dirac’s Dirac wave equation was an equation that combined elements of quantum theory and Einstein’s theory of special relativity. The equation had two solutions for an electron, and Dirac predicted that there was a negatively charged electron in addition to a positively charged electron. This equation was the beginning of research into antimatter.
Physicist Carl Anderson proved Dirac’s prediction of the positron when he found its existence in 1932. His discovery of this antimatter was fortuitous; when he first saw them he was photographing cosmic rays and looking at the electrons. He noticed that some particles had the exact same mass as an electron, and the same charge to mass ratio; but the charge was positive. He named this particle the positron.
An electron and positron can be bonded together in what is called Positronium. Positronium is considered to be an atom that is the opposite of a hydrogen atom. Instead of a proton in the nucleus, there is a positron. This atom is extremely unstable and decays in about 100 nanoseconds, producing two photons.
After the positron was discovered, it raised questions of whether there were other antiparticles than just the positron. This question was finally answered with the discovery of the antiproton. The opposite of the proton, the antiproton, was discovered in 1955 by Emilio Segrè and Owen Chamberlain along with two others of the University of California.
The Bevatron was a particle accelerator at the University of California at Berkeley. In the search for the elusive antiproton, it was proposed that the energy required to produce an antiproton would be 6 billion volts; therefore a machine was built specifically for this purpose. (BeV = billion volts, hence the name Bevatron)
They set up a device to detect the antiproton. It would record particles that moved at the velocity of an antiproton and block all the others that moved faster. In the course of seven hours, the team had counted 60 antiprotons. They received a Nobel Prize in 1959 for their discovery.
The antineutron is another antiparticle. It was discovered in 1956 by a team of physicists, who were once again from UC Berkeley, using the Bevatron. The team was led by Bruce Cork, and consisted of other physicists Oreste Piccioni, Glenn Lambertson, and William Wenzell. The scientists were shooting antiprotons towards protons, but they were just far enough as not to cause annihilation. This caused the antiproton’s charge to negate, turning it into an antineutron.
An antineutron has similar composition to a neutron, and has the same mass. But since an antielectron and an antiproton have opposite charges of their normal matter counterparts, this brings up the question of how there can be an opposite of a neutron, because the neutron is electrically neutral. The neutron consists of quarks (one up quark, two down quarks) and the antineutron consists of antiquarks. Antineutrons and neutrons both have a net charge of zero. But, quarks and antiquarks have very small electrical charges. In the case of the antineutron, it consists of one up antiquark (with a charge of -2/3) and two down antiquarks (both with charges of 1/3). This makes the antineutron’s charge a total of 0.
When matter and antimatter come into contact with one another, they annihilate (cancel each other out) and disappear, leaving behind a blast of energy. It is theorized that mass can be converted into energy; however, it is very difficult to convert all of something’s mass into pure energy. The only way that this can be achieved is by annihilation, which is the complete destruction of matter. This matter is then turned into pure energy.
According to Einstein’s mass energy equivalence equation, E=mc2 (where c2 is the speed of light, 300,000,000 meters per second), we could turn something into energy by having matter annihilate an equal mass of antimatter. For example, we could say we wanted to convert one atom of hydrogen into pure energy via contact with antihydrogen. We would take .000000000000000000000000000100794 (hydrogen’s weight in kilograms) times 90,000,000,000,000,000 (the speed of light squared) and get 9.07 x 1011 joules. Due to only the annihilation of a single hydrogen atom with its antimatter counterpart, the energy produced is enough to power 4 light bulbs if they were left on all year.
However, producing energy through matter-antimatter annihilation is not very efficient. It takes more energy to make antimatter than it puts out. You only get 10-10 of the energy back that you invested. There is no known way to contain the energy produced before it is lost, which is why antimatter creation is so inefficient. Antimatter has to be kept in a vacuum, and if the antimatter comes into contact with the walls of the container, it is lost. Also, antimatter is currently an extremely expensive substance. It is incredibly cost-inefficient at the current price of 0,000,000,000 per milligram. It is hard to produce antiparticles because only a few are made in particle accelerations. Also, the particle accelerators used to make antimatter are currently in use for other things. Only about billionth of a gram is produced every year.
People often wonder if the production of an antimatter weapon is possible; the short answer is no. We cannot make enough antimatter to produce one. So far, if you counted up all the antimatter made through particle accelerators throughout the years, it would amount to less than a gram. People look to these weapons as the next generation of weapons that will be used. They don’t have any radioactive fallout, but they are almost impossible to produce.
One of the greatest mysteries in physics is the question: Why is there more matter in the universe than antimatter? This brings up the other question of whether or not matter and antimatter are truly equal. If they were, they should have developed in equal occurrences. This challenges the Big Bang theory. Samuel Ting, an antimatter researcher, said:
“At the beginning, equal amounts of matter and antimatter were created [in the “big bang”]. Now there seems to be only matter. There have been theoretical speculations about the disappearance of antimatter, but no experimental support.” 1
However, there has to be some asymmetry between matter and antimatter because if so, they would have just annihilated each other during the formation of the universe and there would only be light. There is a theory called CP Violation. It states that almost all the matter and antimatter in the Big Bang was annihilated, except one in every billion particles of matter survived, but no antimatter. The ratio of matter to antimatter is 10,000,000,001:10,000,000,000. Current observations of galaxies support this, because they are made of matter and not antimatter.
Since annihilation happens in pairs (1 unit of matter to exactly 1 unit of antimatter) the 10,000,000,000 particle pairs of matter and antimatter annihilated and left behind the one particle of matter. These particles joined together and formed the universe we know now. This process of developing particles and their antiparticle counterparts is called baryogenesis.
It is often asked why there was an imbalance in the first place. This question has thus yet been unanswered and is crucial or our understanding of the fundamental laws of physics and the universe.
Another theory is that there are other galaxies, far away from us, composed entirely of antimatter. Antimatter is able to be produced naturally – nuclear fission produces antimatter all the time. In 2002, a solar flare created half a kilogram of antimatter. The antimatter then instantly annihilated when it collided with matter from the Sun. However, is antimatter able to exist in very large quantities with the abundance of matter in our universe?
There definitely aren’t any antimatter galaxies near us because we would have noticed the large gamma and x-ray emissions. If we managed to see an antimatter galaxy, it wouldn’t look any different from regular galaxies. The only way we could tell is by looking at the composition of the so called antimatter galaxy, by checking for large amounts of antihelium and antihydrogen. The Alpha Magnetic Spectrometer launched in 1998 was a device set up to detect antimatter galaxies. So far, results have been inconclusive. However, there is a possibility that there might be antimatter galaxies in the far reaches of our universe.
In our very own galaxy, there is a cloud of antimatter. It is proposed that it was caused by black holes spewing out antimatter after enveloping regular matter.
There you have it – antimatter in a nutshell. It’s one of the most baffling mysteries in physics today, and certainly all this information will change as our understanding of the universe progresses.
source: blogosphere
Question by Nerdy_Skittles_Lover: HELP! Nuclear Equations Fusion/Fission Reactions?
I can’t seem to figure out how to solve these problems.
1. Fission
1/0n + 235/92U–> 92/36Kr + 141/56Ba + ______ 1/0n
1/0n + 235/92U–>152/60Nd + _______+ 4 1/0n
1/0n+ 239/94Pu –> _______ + 97/40 Zr+2 1/0n
2. Fusion
2/1H + 2/1H –> 3/2 He+ _______
3/2He + 3/2He —> ______ +2 1/1H
3/2He +2/1H–> 4/2He+_______
These are all reactions, if that helps. Please explain your answer, (the process to get it) because I have no clue.
Best answer:
Answer by Jerry Lee
1 + 235 = 236 = 92 + 141 +*3*
236 – 152 – 4 = 80. 80 seems to be about the weight of Br (periodic table) so 80/35
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Fusion is easier because the numbers are smaller:
1/0n comes off the deuterium (2/1H) reaction
Related Nuclear Fission Equations Articles
ª/z X where
…… z = atomic number = number of protons = position of element in periodic table
…… a = mass number = number of protons (p) + number of neutrons (n)
…… X = kind of particle [ neutron or kind of nucleus of atom (element) ]
From the preceding, we find that
…… n = a – z
In nuclear reactions not involving the emission of negatively charged particles,
….. total number of protons before reaction = total number of that after reaction
….. total number of neutrons before reaction = total number of that after reaction
1) Fission
For the first reaction, 1/0n + 235/92U–> 92/36Kr + 141/56Ba + ______ 1/0n,
we have before the reaction,
neutron —–> z = p = 0 , a = p + n = 0 + n = 1
—————-> n = a – p = 1 – 0 = 1
U ————-> z = p = 92 , a = p + n = 92 + n = 235
—————-> n = 235 – 92 = 143
total p before = 0 + 92 = 92
total n before = 1 + 143 = 144
After the reaction,
Kr ————> z = p = 36 , a = p + n = 36 + n = 92
—————-> n = 92 – 36 = 56
Ba ————> z = p = 56 , a = p + n = 56 + n = 141
—————-> n = 141 – 56 = 85
neutron —–> z = p = 0 , a = p + n = 0 + n = 1
—————-> n = a – p = 1 – 0 = 1
total p after = 36 + 56 = 92
total n after = 56 + 85 + 1 = 142
Note that
….. n before (144) > n after (142)
….. n after is less by 2
The correct answer is therefore
1/0n + 235/92U–> 92/36Kr + 141/56Ba + 2 (1/0n )
For the second reaction, 1/0n + 235/92U –> 152/60Nd + _______+ 4 ( 1/0n ),
we have before the reaction,
neutron —–> z = p = 0 , a = p + n = 0 + n = 1
—————-> n = a – p = 1 – 0 = 1
235/92U —-> z = p = 92 , a = p + n = 92 + n = 235
—————–> n = 235 – 92 = 143
total p before = 0 + 92 = 92
total n before = 1 + 143 = 144
After the reaction,
152/60Nd —> z = p = 60 , a = p + n = 60 + n = 152
—————–> n = 152 – 60 = 92
4 ( 1/0n ) —-> z = p = 4 ( 0 ) = 0 , a = p + n = 0 + n = 4 (1)
—————–> n = 4 (1) – 0 = 4
total p after = 60 + 0 = 60
total n after = 92 + 4 = 96
Note that
….. p before (92) > p after (60)
….. p after is less by ( 92 – 60 ) = 32 —–> corresponds to Ge (z = 32)
and
….. n before (144) > n after (96)
….. n after is less by ( 144 – 96 ) = 48
That means Ge with mass number a = p + n = 32 + 48 = 80
The complete reaction for the second fission rection is then
1/0n + 235/92U –> 152/60Nd + 80/32Ge + 4 ( 1/0n )
For the third reaction, 1/0n+ 239/94Pu –> _______ + 97/40Zr + 2 ( 1/0n )
we have before the reaction,
neutron ——> z = p = 0 , a = p + n = 0 + n = 1
—————–> n = a – p = 1 – 0 = 1
239/94Pu —> z = p = 94 , a = p + n = 94 + n = 239
—————–> n = 239 – 94 = 145
total p before = 0 + 94 = 94
total n before = 1 + 145 = 146
After the reaction,
97/40Zr —–> z = p = 40 , a = p + n = 40 + n = 97
—————-> n = 97 – 40 = 57
2(1/0n) ——> z = p = 2 ( 0 ) = 0 , a = p + n = 0 + n = 2 (1)
—————–> n = 2 (1) – 0 = 2
total p after = 40 + 0 = 40
total n after = 57 + 2 = 59
Note that
….. p before (94) > p after (40)
….. p after is less by ( 94 – 40 ) = 54 —–> corresponds to Xe (z = 54)
and
….. n before (146) > n after (59)
….. n after is less by ( 146 – 59 ) = 87
That means Xe with mass number a = p + n = 54 + 87 = 141
The complete reaction for the third fission rection is then
1/0n+ 239/94Pu –> 141/54Xe + 97/40Zr + 2 ( 1/0n )
2) Fusion
For the first reaction, 2/1H + 2/1H –> 3/2 He + _______,
we have before the reaction
2/1 H ————> z = p = 1 , a = p + n = 1 + n = 2
——————–> n = 2 – 1 = 1
2/1 H ————> z = p = 1 , a = p + n = 1 + n = 2
——————–> n = 2 – 1 = 1
total p before = 1 + 1 = 2
total n before = 1 + 1 = 2
We have after
3/2 He ———-> z = p = 2 , a = p + n = 2 + n = 3
——————–> n = 3 – 2 = 1
total p after = 2
total n after = 1
Note that
….. n before (2) > n after (1)
….. n after is less by 1
The complete answer for the first reaction is therefore
2/1H + 2/1H –> 3/2 He + 1/0n
For the second reaction, 3/2He + 3/2He —> ______ +2 ( 1/1H ),
we have before the reaction
3/2 He ———-> z = p = 2 , a = p + n = 2 + n = 3
——————–> n = 3 – 2 = 1
3/2 He ———-> z = p = 2 , a = p + n = 2 + n = 3
——————–> n = 3 – 2 = 1
total p before = 2 + 2 = 4
total n before = 1 + 1 = 2
We have after
1/1 H ————> z = p = 1 , a = p + n = 1 + n = 1
—————> n = 1 – 1 = 0
1/1 H ————> z = p = 1 , a = p + n = 1 + n = 1
—————> n = 1 – 1 = 0
total p after = 1 + 1 = 2
total n after = 0
Note that
….. p before (4) > p after (2)
….. p after is less by 2
and
….. n before (2) > n after (0)
….. n after is less by 2
The complete reaction for the second fusion process is then
3/2He + 3/2He —> 4/2He + 2 ( 1/1H )
since
4/2He ————> z = p = 2 , a = p + n = 2 + n = 4
—————> n = 4 – 2 = 2
which gives
total p after = 1 + 1 + 2 = 4
total n after = 0 + 0 + 2 = 2
exactly the same as p and n before the reaction
For the third reaction, 3/2He +2/1H –> 4/2He + _______ ,
we have before the reaction
3/2 He ———-> z = p = 2 , a = p + n = 2 + n = 3
——————–> n = 3 – 2 = 1
2/1H ————-> z = p = 1 , a = p + n = 1 + n = 2
——————–> n = 2 – 1 = 1
total p before = 2 + 1 = 3
total n before = 1 + 1 = 2
We have afterwards
4/2 He ———-> z = p = 2 , a = p + n = 2 + n = 4
——————–> n = 4 – 2 = 2
total p after = 2
total n after = 2
Note that
….. p before (3) > p after (2)
….. p after is less by 1
That means the presence of 1/1H after the reaction.
The complete reaction for the third fusion process is then
3/2He +2/1H –> 4/2He + 1/1H