NILFiR

Jumpboy11j — Wed, 01 Sep 2010 17:22:59 GMT

←Older revision		Revision as of 17:22, 1 September 2010
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-	NILFiR stands for Neutron Induced Lithium ~~FIssion~~ Reaction	+	NILFiR stands for Neutron Induced Lithium Fission Reaction. So far as I know, I am the first to think of the potential that this reaction has in power generation (though I will retract this claim if anyone finds a citation for something earlier than March 3, 2008 on this topic). The reaction itself releases 16 MeV. Though not yet tested, this reaction looks very promising for power generation the world over, not least because it cannot possibly be used in a nuclear bomb. The reaction would most likely be in the liquid phase, with a mixture of pure elemental Lithium and Beryllium in a carefully controlled ratio. The reactor would involve fairly similar technology to a Molten Salt reactor.

-	~~This is the reaction:~~

-	~~Li-7 + N => Li-8~~

-	~~Li-8~~ =~~> Be-8 + 16 MeV~~	+	==The Reaction==

-	Be-8 => 2 He-4 ~~+ 90 KeV.~~	+	Li-7+n->Li-8->Be-8->2 He-4

-	The ~~total energy gain~~ is ~~~16~~.~~1 MeV~~. ~~The energy density~~ is ~~about~~ 2 ~~MeV/AMU (193 trillion joules/ Kg) more than twice Uranium~~.	+	The reaction is caused by Lithium-7 absorbing a Neutron. This creates Lithium-8, which has a half life of about 838 milliseconds. This is one out of 2 reasons why this reaction could not be used in a nuclear bomb: 838 seconds after a uranium nuclear bomb goes off, it has already long exploded. This would mean that a nuclear bomb based on NILFiR would "fizzle" and prevent any serious detonation. The other reason is that the Beryllium involved in neutron production would take a significant amount of time, on the scales of a nuclear explosion, to generate neutrons.

-	~~Now~~ all of this ~~so far~~ is ~~great~~, ~~but you would need an expensive and power hungry external neutron source because this reaction doesn't produce any neutrons~~.	+	The energy is released primarily in the reaction Li-8->Be-8. This is a beta decay reaction. It releases the 16 MeV in 2 forms: 13 MeV of it in a very high energy electron, and 3 MeV of it in a neutrino. The neutrino will in essentially all cases be lost to the rest of the universe without interacting with anything else. That leaves 13 MeV (technically, 12.9645 MeV. Close enough.). 90 further KeV are released in the reaction Be-8->2 He-4. However, in comparison to the amount of energy released through other means, this is very much negligible. It is enough energy to make the 13 MeV figure accurate, though.

-	~~This is how I would alleviate this:~~	+	===Radioisotopes===

-	~~Deuterium contains a proton~~ and ~~a neutron~~. The ~~binding energy per nucleon~~ is ~~about 1~~.~~25 MeV. That meant that~~ to ~~split it~~, ~~and create~~ a ~~proton~~ and a ~~neutron~~, ~~you need 2.5 MeV (actually 2.45, but close enough). The fission products (2 alpha particles) have 8 MeV each, so a collision with deuterium will be enough to split~~ it. ~~If bound up~~ in ~~LiD (Lithium Deuteride)~~, I ~~would think~~ that about 3 neutrons per fission would be generated, although I have no real idea. Another possible way to generate neutrons is Beryllium. An alloy of lithium and beryllium would stay solid at higher temperatures and need less energy per neutron.	+	In terms of radioactive isotopes, none are inherently created in this reaction. All light elements are involved, and any radioisotopes released in this reaction will decay to negligible levels literally within one minute of being in the reactor. The reactor vessel, however, is another matter. A neutron flux will inevitably lead to some level of radioactivity in the reactor vessel. However, this will be radiation of a more low level and would not pose a very serious risk to public health, provided that it is disposed of carefully. There is one instance in which the production of radioisotopes becomes possible, but I will expand upon that later.

-	~~The cross section for~~ the ~~reaction is~~ small, .~~05 Barns, but~~ the ~~high~~ neutron ~~yield per mass~~, ~~and the fact that the lithium atom is small to begin with, should keep the critical mass small~~. It ~~will~~ also ~~help~~ that ~~D is a~~ neutron ~~moderator. Another positive is that~~ the ~~most radioactive thing this reaction can POSSIBLY create is tritium~~, which ~~isn't that bad~~.	+	It is possible that Beryllium-10 could be produced as a result of Beryllium-9 in the reactor absorbing a neutron. This will happen in a fairly small fraction of cases. At the optimal Be:Li ratio for neutron production of 1.68, 22% of neutrons will be absorbed by Beryllium instead of Lithium. It is also possible that fast neutrons, which are an intermediate in neutron production, could be absorbed by the Lithium nucleus, which would cause the release of a tritium nucleus.

-	~~Sources:~~	+	Neither of these two isotopes are extremely dangerous. They both decay via beta decay. Beryllium-10 has a half life of 1.5 million years, and Tritium 12 years. A relatively small amount of shielding will keep both well contained.

-	Li-8 decay: http://nucleardata.nuclear.lu.se/~~Nuc...~~.asp?iZA=30008	+	==Neutron Production==
-	Be-8 decay: http://nucleardata.nuclear.lu.se/~~Nuc...~~.asp?iZA=40008	+
-	~~Reaction~~ Cross-Section: http://www.ncnr.nist.gov/resources/n...~~ements~~/li.html	+	Next up is the reaction for neutron production. Pure lithium, when bombarded with low energy neutron, will generate 1 high-energy electron and 2 moderate-energy Helium-4 nuclei. Pure Lithium will not generate neutrons to make this a self-sustaining chain reaction unless some mechanism is added to create the neutrons. However (and this is important, and it seems, fairly often confused), by the addition of Beryllium to the Lithium, the high energy particles can cause the Beryllium nucleus to split. The reaction is:
		+
		+	Be-9-> 2He-4+n
		+
		+	This reaction absorbs, by my calculations, 1.67 MeV. I got this by subtracting the mass of a Beryllium atom from the mass of 2 Helium-4 atoms and a neutron. This is in approximate agreement with the number given on Wikipedia of 1.6 MeV per reaction. Since my number is higher and more precise, I am inclined to go with it. instead. This energy is obtained by a collision with a fission fragment with a high energy. These collisions can happen until the fission fragments have energies below 1.67 MeV, at which point neutron production from that particular reaction stops. By my calculations, when the Be:Li ratio (by number of atoms, not by mass) is 2:1, 3.64 neutrons will be produced per fission. However, a certain
		+
		+
		+	==Sources==
		+
		+	*Li-8 decay: [http://nucleardata.nuclear.lu.se/nucleardata/toi/nuclide.asp?iZA=30008]
		+	*Be-8 decay: [http://nucleardata.nuclear.lu.se/nucleardata/toi/nuclide.asp?iZA=40008]
		+	*Lithium Neutron Cross Section: [http://www.ncnr.nist.gov/resources/n-lengths/elements/li.html]
		+	*Beryllium Neutron Cross Section: [http://www.ncnr.nist.gov/resources/n-lengths/elements/be.html]

Nilfir

Jumpboy11j — Wed, 01 Sep 2010 17:00:42 GMT

[[Nilfir]] moved to [[NILFiR]] capitalization in order to conform with acronym- jumpboy11j

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