Where does the oxygen in yellowcake go?

originally posted to https://www.quora.com/What-happens-to-the-oxygen-in-for-example-uranium-oxide-after-the-uranium-undergoes-radiactive-decay/answer/Matt-Harbowy

This is an excellent question, but it is also difficult to answer since the vast majority of work on isotopes is not accessible in the public domain. However, there are some important conclusions that can be drawn from a purely theoretical standpoint, and there are practical research applications. In order to understand uranium oxide, it is best to study tritium (³H, or T) first.

By far, the most well researched effects of nuclear decay on chemistry are derived from a unique insight that I believe arose from the work of Kenneth Wilzbach, Fulvio Cacace, and Maurizio Speranza and has been well reviewed in the public domain at Radical formation via tritium to helium decay

(image credit: Measurement principle — KATRIN)

While this technique isn’t very fast, that’s probably a good thing, since there are a couple of elements of the decay of tritium that make it an astoundingly useful tool for chemical research. Despite its radioactive nature (which makes getting your hands on some an exercise in futility) it is a kind of goldilocks-style radioactivity.

  • Tritium releases only a relatively tiny amount of energy when it decays, and despite the boogeyman view of radiation, the experimenter is fairly well shielded since tritium’s beta rays only pass through a couple of inches of air. The biggest difficulty in dealing with tritium is that tritium gas (³H₂), like hydrogen (¹H₂), diffuses quite readily and exchanges with hydrogen in water, meaning that inadvertent releases are exceptionally difficult to contain.
  • Tritium decays fast enough that a small percentage decays every day, as opposed to studying uranium which has a much longer half life. Tritium decays on a time frame practical for chemical study, whereas uranium decays on a time frame practical for geologic study.
  • ³He is stable (non-radioactive) and “inert” so as a by-product it “doesn’t” participate in the reaction after that point nor does it create follow-on products that potentially confuse the analysis. The electron and antineutrino produced zips away, and you are left with a metastable helide product.
  • If the tritium is present on a carbon atom, that is, is part of an organic molecule, the helide steals an electron from the carbon to leave behind a pure carbocation and helium.
  • Generating carbocations in a way that they can be studied readily has never been easy, and garnered George Olah the Nobel for his work with Magic acid. There’s always going to be an abundance of “other chemistry” going on, so to find a reagent like tritium which forms carbocations without other significant chemically reactive species is (in my opinion) Nobel-worthy work.

OK, so that’s great, but what does that have to do with price of yellowcake in Nigeria?

First, notice how tritium forms helium. This means that tritiated water (T₂O, for the sake of discussion) forms the compound ³HeOT⁺, an electron, and an antineutrino when it decays. That compound likely further decays into ³He gas, tritium ion (T⁺, which probably goes on to form free tritonium with another tritiated water molecule, T₃O⁺), and a spectacularly reactive oxygen radical that likely forms tritiated peroxide (T₂O₂). Some of the research I looked at for this article (e.g. http://www.iaea.org/inis/collect...) assumes that the danger of decay of tritium to research animals comes from the radiation, but I suspect chemistry, not physics, to be the greater hazard, with free electrons, acid, and peroxide bouncing about. However, the study of tritium in aqueous solution is (irretrievably, I think) damaged by the irresponsible antics of Pons and Fleischmann, and no respectable chemist would want their name mentioned in the same breath as those two. Thankfully, I’m not respectable, but neither is Nicolau Saker Neto (major props to that person, whoever you are, and who is not me) and StackExchange is also helpful in this regard.

The point to be made, here, is that the major product of nuclear decay of tritium is reactive oxygen species, and noble gas. Therefore, by extension, I think this is incredibly illuminating as to what happens to uranium oxide over time.

image credit: File:U3O8lattice.jpg — Wikimedia Commons

OK, so let’s talk yellowcake.

Uranium is mined in the form U₃O₈, and is converted to UO₂ for use in reactors (see, Uranium: Its Uses and Hazards).

Like tritium decays to the noble gas helium, ²³⁸U decays through the intermediate noble gas radon.

image credit: File:Decay chain(4n+2, Uranium series).svg
image credit: File:Decay Chain of Actinium.svg

Thorium, another actinide akin to uranium, also decays similarly:

image credit: File:Decay Chain Thorium.svg

²³⁸U and ²³²Th were most likely formed during the Big Bang, and being the most stable, are the most abundant formed precursors still present. ²²⁰Rn and ²²²Rn, the most common forms of radon gas, are constantly being formed within the earth and bubble up through the foundations of buildings to deposit their radiation in unlucky home buyer’s wallets, but even this is only a temporary condition. Also consider- all of the intermediates between ²³⁸U/²³²Th and ²⁰⁶Pb/²⁰⁸Pb have lots of different oxidation states EXCEPT for radon, which is “inert”. The oxygen “left behind” doesn’t go anywhere- it might react with neighbors, absorb neutrons, dimerize to gas, or any other host of possibilities. It is most likely that it “sticks around” or bubbles up from the ground, too. For every 3 ²³⁸U atom in U₃O₈, there are 8 oxygen atoms, so by the time 3U becomes 3Pb, there are 5 “extra” oxygen atoms sitting around, since (despite the fact that lead is isoelectronic with carbon), lead oxide is principally PbO. The helium nuclei pick up electrons as they decelerate, the extra oxygens dimerize, so the best chemical equation to describe what goes on looks something like

over a period of hundreds of billions of years. Likewise, for thorium:

over a period of tens of trillions of years. There are dozens of intermediates that are formed: the intent of these chemical equations are to show the reaction if it goes to completion- no chemical reaction ever does, it just runs out of starting material.

This article originally appeared January 29, 2016 on Quora.

matt harbowy, a chemist of ill repute, is also a scientist, activist, and data management expert. He is one of the founders of the non-profit Counter Culture Labs, working to bring fairness and egalitarian ideals to people interested in learning about science and biotechnology. He is also a top writer on the question and answer site, Quora.



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