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The largest known elements, particularly synthetics such as Unilquadium and Unilenium, are known to have spontaneously fissile isotopes; That is, they break up outside of a geometric reaction. The decay products consist mainly of small, fully stable atoms, and stable large isotopes, many of which are
candidates for fission in geometric reaction.
Why not use the spontaneous fission in a reactor? It would have to be refueled fairly often, but it would produce minimal amounts of waste, potentially churning out good quality fuel for traditional reactors. An autofissile reactor would also be easier to control, since a complex process of particle absorbtion would not be necessary.
To scram or shut down the reactor, the fuel could be exposed to a regular high neutron source, such as weapons grade plutonium. this would quickly degrade the fuel. Graphite control rods would then be dropped into the plutonium to prevent a geometric reaction from occurring.
Table of nuclides
http://atom.kaeri.re.kr/ton/
Looks pretty up-to-date and complete [Vernon, Oct 04 2004, last modified Oct 21 2004]
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//particularly synthetics//
I should think the energy you used to make these synthetic elements in the first place would exceed the output of this reactor, making it a losing proposition. |
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Most synthetics have exceedingly short half-lifes, the most stable isotope of Meitnerium (aka unnilennium) is only 3.4 ms. |
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There are several spontaneously fissile materials available in large quantities. |
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I think now the best fuel would probably be Californium-254. Very little energy is needed to produce this isotope, and his has a half-life of about 60 days. |
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That's right, clothist. Californium-254 is a laid-back isotope. |
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Don't you realize that nuclear reactor design is the gayest activity on the planet? |
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Besides, Uranium is the highest Z-number naturally occuring element (maybe not the heaviest). Anything that spontaneously fizzes much has already done so over the last 4.5 to 15 billion years, so there ain't much left. Where's that supernova when you need it? |
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//Don't you realize that nuclear reactor design is the gayest activity on the planet?// |
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is it really that much fun? do they deck the halls with boughs of holly and sing falalalala lala la la, too? the facilities look so serious. |
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That's the idea, captain_entropy. Use synthetics. SF Californium and Curium isotopes are produced en masse in standard and fast flux reactors. In the process of refueling and recycling, extract these isotopes and make them into radioactive foils (This part is baked, Cf foil is popular as a lightweight and portable radiation source for research; I'm proposing a new usage for the foils) and use them to fuel the Autofissile Reactor. Then place the fuel produced in the AF reactor back into the standard reactor, generate more AF fuel, and keep going. Although the fuel output would eventually become so low that replacement would be necessary, you would be able to make use of a lot more of each fuel rod's potential. |
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//Don't you realize that nuclear reactor design is the gayest activity on the planet?// |
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It is an interesting, challenging line of work with good pay and excellent prospects for career advancement. What's wrong with that? |
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//The decay products consist mainly of small, fully stable atoms//
I strongly doubt the validity of that assertion. There is a fact about neutron-to-proton ratios in nuclei that is a definite factor here. This ratio inexoribly increases (more neutrons per proton) as you climb the table of elements. This is a consequence of the role that neutrons play in stabilizing a nucleus against the mutual electrostatic repulsion of all the protons. To be more specific, here is a list of some STABLE (and nearly stable) elements, (atomic number) with least-massive and most-massive nuclei:
(1) Hydrogen-1: 0 neutron/proton
(1) Hydrogen-2: 1 neutron/proton
For Hydrogen-1, its single proton doesn't have the problem of all other elements (fighting electrostatic repulsion of other protons in the nucleus), so of course no neutrons are needed for it to be stable. Hydrogen-3/Tritium is not being listed because it is radioactive with a moderately short half-life, 12 years.
(2) Helium-3: .5 neutron/proton
(2) Helium-4: 1 neutron/proton
(Note Hydrogen-2 is first and Calcium-40 is last having 1:1 ratio)
(20) Calcium-40: 1 neutron/proton
(20) Calcium-48: 1.4 neutrons/proton
(My source indicates Calcium-48 to decay after several quintillion years -- that's nearly stable in my book!)
(50) Tin-112: 1.24 neutrons/proton
(50) Tin-124: 1.48 neutrons/proton
(Tin has ten stable isotopes, more than any other element)
(82) Lead-204: 1.4878 neutrons/proton
(82) Lead-208: 1.5366 neutrons/proton
(element 83, Bismuth, is the very last element to have any stable isotopes --one!)
(92) Uranium-233: 1.5326 neutrons/proton
(92) Uranium-238: 1.587 neutrons/proton
Above I included U-233 because although its half-life is only 159,000 years, that's stable enough to be useful (we can make it from Thorium-232, and it is a good fission-fuel, like U-235 or Plutonium-239). Oh, and U-238 has a half-life of 4.5 billion years, which is pretty stable (and this is the stuff being mentioned in "depleted Uranium munitions"). |
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Anyway, the point is that for elements such as unilquadium, the neutron/proton ratio is even higher, and in fact a major part of the difficulty of making those elements has to do with getting enough neutrons into them, to give them any semblance of stability at all (say, a microsecond). |
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OK, so THIS is the problem with fissioning such an element: The fission products will basically have the same neutron/proton ratio as the parent nucleus (slightly reduced because fission tends to yield a small number of loose neutrons --average of 2.5 in Uranium or Plutonium fission). Well, when the number of protons is small, such a high ratio of neutrons ALWAYS means that the nuclei will be radioactive! And therefore I shall desire the evidence for the statement I quoted at the start of this annotation. |
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Surely it's going to take a huge wodge of energy to make these trans-Uranic elements? So far they've only been made for fleeting fractions of seconds and in minute quantities, so I can't see them being produced en masse for spontaneous fission. |
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[hazel], yes, making those elements is also not energy-efficient. But I gathered that the major rationale for this Idea was the claim regarding nonradioactive fission products, so that claim needed to be refuted more. |
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..but you could use a firecracker-hot isotope in a small hohlrahm cavity with LiD fuel. it would set it off like a mini H-bomb and the neutrons could strike something that makes MORE isotope. that is the answer to fusion! |
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\\I should think the energy you used to make these synthetic elements in the first place would exceed the output of this reactor, making it a losing proposition.\\ |
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hmm. but if the resulting power source is energy dense, it has its uses. Like hydrogen. it takes more energy to split water into hydrogen than is gotten back by burning the hydrogen, BUT hydrogen is used like a battery, to store energy |
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\\I should think the energy you used to make these synthetic elements in the first place would exceed the output of this reactor, making it a losing proposition.\\ |
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hmm. but if the resulting power source is energy dense, it has its uses. Like hydrogen. it takes more energy to split water into hydrogen than is gotten back by burning the hydrogen, BUT hydrogen is used like a battery, to store energy. A possible fuel source for vehicles |
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\\ OK, so THIS is the problem with fissioning such an element: The fission products will basically have the same neutron/proton ratio as the parent nucleus (slightly reduced because fission tends to yield a small number of loose neutrons --average of 2.5 in Uranium or Plutonium fission). Well, when the number of protons is small, such a high ratio of neutrons ALWAYS means that the nuclei will be radioactive! \\ |
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but if its less radioactive than uranium fission products then its still an improvement |
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