F3 Examples of Mass-Energy Conservation
The Fusion of Hydrogen into Helium
If a cloud of hydrogen gas is sufficiently hot and dense, then there will no longer be H2 molecules, nor even H atoms, but only a plasma of free protons and electrons. In this plasma direct collisions of protons occur very frequently. At lower temperatures they can be slowed to zero by the coulomb repulsion (there is always a coordinate system, in which the total momentum of two particles is zero…) and then propelled back to where they came from. If the collision is not completely direct, then they whiz past each other on hyperbolic orbits. If the temperature is however sufficiently high, then they impact in a direct collision in such a manner that the short ranged strong force between the two nuclear particles begins to act and unites them as a deuterium nucleus consisting of a proton and a neutron. This reaction emits a positron e+ and a neutrino ν. The neutrino is needed only to satisfy further conservation laws (here the lepton number) of particle physics. The positron will soon encounter an electron e-, whereby the two particles 'annihilate' each other, i.e. dematerialize into two energy quanta (aka photons) (see F5).
Two deuterium nuclei could then fuse directly into a He nucleus, consisting of two protons and two neutrons. More frequently however a further proton will merge with the deuterium into a He-3 nucleus and two such He-3 nuclei will fuse to a normal He-4 nucleus while emitting two protons. Yet other fusions are possible – but in the long run it is always the case that from 4 protons and 2 electrons a He-4 nucleus is created while emitting two neutrinos.
The rest masses of all these particles are known with high precision (keyword mass spectroscopy). We set up a balance sheet for the mass:
|before||4 protons||4 • 1.007,825 u|
|2 electrons||2 • 0.000,056 u|
|after||1 He-4 nucleus||1 • 4.002,603 u|
|2 neutrinos||2 • 0.000,000 u|
|'missing' mass||0.029,817 u
The fusion of only one He nucleus from protons frees an amount of energy corresponding to 0.029,817 atomic mass units. If we fuse a whole mol of helium, then we can multiply this amount by Avogadro's number and obtain about 2.6 • 1012 J. With this fusion 0,029,817 / 4.032,420 ≈ 0.74% of the original mass ‘disappears’.
This fusion process occurs as we said only under extreme conditions (hydrogen bombs must therefore be ignited with an ‘ordinary’ uranium bomb…). No material container could include such a plasma. Research reactors are currently being built however, in which this process can be run in a controlled environment. Fusion would have the tremendous advantage over nuclear fission that it does not produce long-lived radioactive substances.
By the way the fusion reaction we described also illustrates the conservation of electrical charge!
The Squandering of Energy by Our Sun
The sun has been radiating tremendous amounts of energy for millions of years. Still around 1900 one did not have the slightest idea where it got its energy. One could figure out that a sun made from pure coal (considered apart from the oxygen needed to burn) would burn out after a few 1000 years. Today one would naturally do the calculation with oil…
The total energy output of the sun can be computed quite simply: In the alps one measures an energy flow of approximately 1380 W/m2, the so-called ‘solar constant’. If one assumes that the sun delivers its radiation symmetrically in all directions of a sphere, then one can multiply this value by the surface area, whose radius is the average radius of the Earth's orbit. In this way one obtains the 3.85 • 1026 W, with which to label the light bulb ‘sun’.
This energy is essentially produced (as with all ‘main sequence stars’) by the fusion of hydrogen into helium. 3.85 • 1026 Js per second are radiated away. We obtain the appropriate mass loss, if we divide this number by c2: Per second the sun loses about 4.28 • 109 kg of matter - that is 4.28 million tons! In one year that amounts to 1.35 • 1017 kg, and in 10 billion years 1.35 • 1027 kg. Considering this value in relationship to the total mass of the sun: 1.35 • 1027 / 1.99 • 1030 ≈ 0.000,678, we see that in 10 billion years the sun loses less than 1 part per thousand of its entire mass!
which shows up intensively as UV light of ionized helium at 304 angstroms
http://soho.esac.esa.int/gallery/images/superprom.html (© ESA and NASA)
Today we assume that the sun and the planet system formed approximately 5 billion years ago from the ‘waste’ of an earlier star generation (otherwise there would be no heavy elements such as carbon, oxygen, iron and uranium on earth). The sun will continue radiating quite stably and with the same intensity for about 5 billion years. Then another phase will begin…
Today astrophysics can model in great detail the birth, life and death of different types of stars. Here I hoped only to give you a little taste.
Radioactive Decay and the Splitting of Heavy Atomic Nuclei
The two protons and the two neutrons in helium are held together by the strong force. This binding energy corresponds to the energy that is released during fusion. It is now possible to determine the middle binding energy per nuclear particle for all atoms, or even better, for all isotopes giving the following diagram:
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The energy gain is particularly large by the fusion of protons to helium. In addition, one can win energy, if one splits heavy nuclei. In nuclei that are heavier than iron (Fe-57), the nuclear particles are again on the average less strongly bounded together. Thus energy is released, if one splits one heavy nucleus into two moderately heavy nuclei.
In particular the uranium isotope U-235 needs only to be bombarded with neutrons of suitable kinetic energy, in order to elicit its decay into Kr-89 and Ba-144, for example. In the process three other fast neutrons are produced, making for a suitable nuclear chain reaction:
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If we know the rest masses of the nuclei involved, we can once again do a balance sheet as we did with the fusion of hydrogen to helium. The two fission products are however extremely unstable (by far too many neutrons in the nucleus) and therefore their rest masses are not specified in the tables. Thus we make another calculation (which in the long run likewise depends on the precise measurements of the rest masses): In Uranium-235 the middle binding energy per nucleus is about 7.6 MeV, with Krypton-89 this amounts to the appropriate value of 8.6 MeV and with the Barium-144 it is about 8.4 MeV (see table above). The result is that the splitting of a single U-235 nucleus releases an energy of
89 • 8.6 MeV + 144 • 8.4 MeV – 235 • 7.6 MeV ≈ 198 MeV
Since both products practically immediately decay further (beta decay), there are some additional MeV set free, with which one comes to a total energy of 210 MeV for splitting one U-235 nucleus. Let’s extrapolate for a mol of Uranium-235: The complete splitting of 235 grams of U-235 releases the energy of 6.02 • 1023 • 210 MeV ≈ 2.0•1013 Joules ≈ 20 TJ. The corresponding ‘mass loss’ is 20 TJ / c2 ≈ 0.225 gram or somewhat less than one part per thousand.
Excellent information about the fundamentals of fission technology and the different reactor types in use is offered in the publication , which was published by the German nuclear power station operators and also from which the three illustrations in this section were taken. Only the section “Disposal of Highly Radioactive Wastes” is - given the present state of the project work – rather tersely handled …
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Energy is also released during radioactive decay: The helium nucleus spontaneously released during α-decay of a radium atom contains a lot of kinetic energy. This topic is also excellently covered by . I would like to thank the Vattenfall Europe AG and the Informationskreis KernEnergie in Berlin for permission to reproduce the three illustrations shown above.