Introduction
The splitting up of an atomic nucleus is known as nuclear fission . This is, in a sense, the opposite of nuclear fusion, which occurs inside the Sun and other stars. The very first atomic bombs, developed in the 1940s, made use of fission reactions to produce energy. More powerful weapons, created in the 1950s, used atomic fission bombs to produce the temperatures necessary to generate hydrogen fusion.
Human-caused And Natural Fission
The preceding problems involving oxygen and beryllium are given for illustrative purposes, but the actual breaking up of atomic nuclei is not such a simple business. A physicist can’t snap an atomic nucleus apart as if it were a toy. Nuclear reactions must take place under special conditions, and the results are not as straightforward as the foregoing problems suggest.
To split atomic nuclei in the laboratory, a particle accelerator is employed. This machine uses electric charges, magnetic fields, and other effects to hurl subatomic particles at extreme speeds at the nuclei of atoms to split them apart. The result is a fission reaction, often attended by the liberation of energy in various forms.
Some fission reactions occur spontaneously. Such a reaction can take place atom-by-atom over a long period of time, as is the case with the decay of radioactive minerals in the environment. The reaction can occur rapidly but under controlled conditions, as in a nuclear power plant. It can take place almost instantaneously and out of control, as in an atomic bomb when two sufficiently massive samples of certain radioactive materials are pressed together.
Matter And Antimatter
The proton, the neutron, and the electron each has its own nemesis particle that occurs in the form of antimatter . These particles are called antiparticles . The antiparticle for the proton is the antiproton; for the neutron it is the antineutron; for the electron it is the positron . The antiproton has the same mass as the proton, but in a negative sort of way, and it has a negative electric charge that is equal but opposite to the positive electric charge of the proton. The antineutron has the same mass as the neutron, but again in a negative sense. Neither the neutron nor the antineutron have any electric charge. The positron has same mass as the electron, but in a negative sense, and it is positively charged to an extent equal to the negative charge on an electron.
You might have read or seen in science-fiction novels and movies that when a particle of matter collides with its nemesis, they annihilate each other. This is true. What, exactly, does this mean? Actually, the particles don’t just vanish from the cosmos, but they change from matter into energy. The combined mass of the particle and the antiparticle is liberated completely according to the same Einstein formula that applies in nuclear reactions:
E = ( m + + m − ) c 2
where E is the energy in joules, m + is the mass of the particle in kilograms, m − is the mass of the antiparticle in kilograms, and c is the speed of light squared, which, as you recall, is approximately equal to 9 × 10 16 m 2 /s 2 .
Unimaginable Power
If equal masses of matter and antimatter are brought together, then in theory all the mass will be converted to energy. If there happens to be more matter than antimatter, there will be some matter left over after the encounter. Conversely, if there is more antimatter than matter, there will be some antimatter remaining.
In a nuclear reaction, only a tiny fraction of the mass of the constituents is liberated as energy; plenty of matter is always left over, although its form has changed. You might push together two chunks of 235 U, the isotope of uranium whose atomic mass is 235 amu, and if their combined mass is great enough, an atomic explosion will take place. However, there will still be a considerable amount of matter remaining. We might say that the matter-to-energy conversion efficiency of an atomic explosion is low.
In a matter-antimatter reaction, if the masses of the samples are equal, the conversion efficiency is 100 percent. As you can imagine, a matter-antimatter bomb would make a conventional nuclear weapon of the same total mass look like a firecracker by comparison. A single matter-antimatter weapon of modest size could easily wipe out all life on Earth.
Where Is All The Antimatter?
Why don’t we see antimatter floating around in the Universe? Why, for example, are the Earth, Moon, Venus, and Mars all made of matter and not antimatter? (If any celestial object were made of antimatter, then as soon as a spacecraft landed on it, the ship would vanish in a fantastic burst of energy.) This is an interesting question. We are not absolutely certain that all the distant stars and galaxies we see out there really are matter. However, we do know that if there were any antimatter in our immediate vicinity, it would have long ago combined with matter and been annihilated. If there were both matter and antimatter in the primordial solar system, the mass of the matter was greater, for it prevailed after the contest.
Most astronomers are skeptical of the idea that our galaxy contains roughly equal amounts of matter and antimatter. If this were the case, we should expect to see periodic explosions of unimaginable brilliance or else a continuous flow of energy that could not be explained in any way other than matter-antimatter encounters. However, no one really knows the answers to questions about what comprises the distant galaxies and, in particular, the processes that drive some of the more esoteric objects such as quasars.
Energy from Matter Practice Problems
Problem 1
Suppose that a 1.00-kg block of matter and a 1.00-kg block of antimatter are brought together. How much energy will be liberated? Will there be any matter or antimatter left over?
Solution 1
We can answer the second question first: There will be no matter or antimatter left over because the masses of the two blocks are equal (and, in a sense, opposite). As for the first part of the question, the total mass involved in this encounter is 2.00 kg, so we can use the famous Einstein formula. For simplicity, let’s round off the speed of light to c = 3.00 × 10 8 m/s. Then the energy E , in joules, is
E = mc 2
= 2.00 × (3.00 × 10 8 ) 2
= 2.00 × 9.00 × 10 16
= 1.80 × 10 17 J
This is a lot of joules. It is not easy to conceive how great a burst of energy this represents because the number 1.80 × 10 17 , or 180 quadrillion, is too large to envision. However, the quantity of energy represented by 1.80 × 10 17 J can be thought of in terms of another problem.
Problem 2
We know that 1 W = 1 J/s. How long would the energy produced in the preceding matter-antimatter reaction, if it could be controlled and harnessed, illuminate a 100-W light bulb?
Solution 2
Divide the amount of energy in joules by the wattage of the bulb in joules per second. We know this will work because, in terms of units,
J/W = J/(J/s) = J · (s/J) = s
The joules cancel out. Note also that the small dot (·) is used to represent multiplication when dealing with units, as opposed to the slanted cross (×) that is customarily used with numerals. Getting down to the actual numbers, let P be the power consumed by the bulb (100 W), let t s be the number of seconds the 100-W light bulb will burn, and let E be the total energy produced by the matter-antimatter reaction, 1.80 × 10 17 J. Thus
t s = E/P
= 1.80 × 10 17 /100
= 1.80 × 10 17 /10 2
= 1.80 × 10 15 s
This is a long time, but how long in terms of, say, years? There are 60.0 seconds in a minute, 60.0 minutes in an hour, 24.0 hours in a day, and, on average, 365.25 days in a year. This makes 31,557,600, or 3.15576 × 10 7 , seconds in a year. Let t yr be the time in years that the light bulb burns. Then
t yr = t s /(3.15576 × 10 7 )
= (1.80 × 10 15 )/(3.15576 × 10 7 )
= 0.570 × 10 8
= 5.70 × 10 7 yr
This is 57.0 million years, rounded to three significant figures (the nearest 100,000 years), which is all the accuracy to which we are entitled based on the input data.
Problem 3
Suppose that the amount of matter in the preceding two problems is doubled to 2.00 kg but the amount of antimatter remains 1.00 kg. How much energy will be liberated? Will there be any matter or antimatter left over?
Solution 3
The amount of liberated energy will be the same as in the examples shown by the preceding two problems: 1.80 × 10 17 J. There will be 1.00 kg of matter left over (the difference between the masses). However, assuming that the encounter produces an explosion, the matter won’t remain in the form of a brick. It will be scattered throughout millions of cubic kilometers of space.
Practice problems of these concepts can be found at: Particles of Matter Practice Test
Celebrate Memorial Day! Worksheets and Activities About American History 
May Workbooks are Here!
Get Outside! 10 Playful Activities
Add your own comment