Sciencemadness Discussion Board

Uranium compounds

copperastic - 22-3-2014 at 17:30

Hi, I wasn't sure were this should go. What happens when a uranium compound decays? Would it turn into another element (down its decay chain) compound or separate?
Thanks.

TheChemiKid - 22-3-2014 at 18:46

What isotrope of Uranium are you talking about? This is easily accessible information, and you should find it with no problem. This is very basic and should be posted in beginnings, if posted at all.

Texium - 22-3-2014 at 20:02

Quote: Originally posted by TheChemiKid  
This is easily accessible information, and you should find it with no problem. This is very basic and should be posted in beginnings, if posted at all.


Is it really? I just did some Googling out of curiosity and couldn't find much very clear or general information on the subject. How about you share some helpful information regarding the topic since you know enough about it to dismiss it as being so basic?

IrC - 22-3-2014 at 20:35

Is there something preventing either of you from going to google and entering "Uranium radioactive decay" in the search terms?

This thread belongs in the subset of beginnings titled "I insist you explain it I refuse to search". I'll even add two links from said search.

http://en.wikipedia.org/wiki/Uranium-238

http://en.wikipedia.org/wiki/Radioactive_decay

Texium - 22-3-2014 at 20:54

Well that information is very helpful if the question was simply about uranium decay, but that wasn't the question. Copperastic's question, was regarding the effects of the decay on uranium containing compounds, not elemental uranium. Did you actually read the entire original post?
I did happen to find this, after additional search queries though. Not sure about the reliability of it, but it's closer to the question at hand.
http://physics.stackexchange.com/questions/8081/what-happens...

Zephyr - 22-3-2014 at 21:12

I do not think that there is a general rule for the decay of radioactive compounds, for instance, if the compound is compatible with the next element on the decay chain, the compound may remain intact. However would the energy created by the radioactive decay, in some cases, split the compound?

Texium - 22-3-2014 at 22:26

That's what that thing that I found said:
"In general, radioactive decay is so energetic, that any chemical bonds/lattice forces are broken. What happens then is very complicated and not to be answered by a simple scheme."

Zephyr - 22-3-2014 at 22:28

Where Is this information from?
Could you please cite your source?

Texium - 22-3-2014 at 22:47

Quote: Originally posted by zts16  
I did happen to find this, after additional search queries though. Not sure about the reliability of it, but it's closer to the question at hand.
http://physics.stackexchange.com/questions/8081/what-happens...


It was this same thing which I posted earlier, sorry for the lack of clarity.

Chemosynthesis - 23-3-2014 at 00:05

My understanding is that this is not a simple matter at all; each case would have to be specific to the nuclide and ligand. Secondary complexes can form due either to lability or degradation of parent to daughter nuclides, changing the LCAO's. Some decays, such as those of 99m Tc to Tc in radiopharmaceuticals, change the activity of a radiopharmaceutical without affecting the chemistry.

IrC - 23-3-2014 at 05:05

"Did you actually read the entire original post?"

Not closely enough obviously I see your point. What he is asking is so very complex I doubt it is discussed often even in many otherwise good texts on the subject of radioactive decay. Sorry copperastic you are actually asking a very good question which I think has a complex answer. One with many possible outcomes.

I would think one starts with the decay chain and step by step works it out. Time would be important since it may take longer for the new element to decay than it does for the newly 'homeless' elements to react in some way with elements in their surroundings. If you had an NO3 left over and Radon floating away what would it do? What is around it with which it could react? How 'locked in' to the surrounding lattice are the leftover Oxygen and pair of NO3's? This would require carefully going step by step from initial conditions, considering the chemical properties of every element in question, including new arrivals. Bond energies of many elements needs to be considered as well as crystal lattice considerations.

Say you have Uranyl nitrate, UO2(NO3)2. I think trying to discover what is happening chemically as the uranium decays down the chain is very complex and difficult. In many years of study I cannot recall any book I have read which covers this in depth. I get the feeling this is one of those questions one could spend years studying (with great difficulty). Say doing chemical analysis of samples where even all gasses released while crushing up the sample are carefully collected, measured and analyzed. A work both physically and theoretically so tedious it is no wonder I cannot even suggest a good book on the subject.

copperastic - 23-3-2014 at 05:14

Maybe the uranium (lets say UO2) decays into lets say thallium then it becomes ThO2.

Texium - 23-3-2014 at 09:23

Well, uranium-238 first decays into thorium-234 (Th), not thallium (Tl), although thallium-210 is farther down the chain

copperastic - 23-3-2014 at 09:26

oh ok zts16.

Zyklon-A - 23-3-2014 at 10:19

Would this happen?
40KF -->beta radiation ~50%40Ca+ ~50%40CaF +electron capture ~10% 40Ar + gamma ray.
It's hard to balance this equation as 40K decays into different things in different proportions.

[Edited on 23-3-2014 by Zyklonb]

Subcomputer - 30-3-2014 at 19:39

It's amazing how complicated this can get, even with thought experiment pure conditions. A pure sample of Uranium-238 as UO2, degrading into Thorium-234 as ThO2. Can I calcine it hard enough to draw that into the simulation? If I manage a crystal structure, what would be optimal to provide strength to the molecules as the atoms split? If I allow other elements for secondary support elements, could they react before the Th and O radicals had a chance to reintegrate?

Answering this would require (as IrC said) gathering incredibly detailed data, and LOTS of simulations.

Pyrovus - 31-3-2014 at 00:57

Perbromate was first prepared via the decay of radioactive selenate:

83SeO42- -> 83BrO4- + e-

Presumably the low energy of the decay allowed the ion to remain intact, though I don't know what the yield was.

Steam - 31-3-2014 at 05:42

Remember that chemical bonds occur because of electrons, not because of the nucleus. During decay, the number of electrons won't change (except for electron capture). Theirfore the way a radioisotope decays will (for the most part) not effect the way it bonds.

That being said, exceptions can be made because the oxidation number of the species might change during decay, so decay compounds must be looked at in a case by case basis.

neptunium - 31-3-2014 at 08:48

its a fascinating question that i have thought about years ago as well....When heavy elements from spent nuclear reactor are being discarded in containers , a valve is added to release the helium gas due to alpha decay so it doesnt built preassure over time. Fluorine gas can be found trapped in bubbles inside crystals due to slow natural radioactivity . so clearly electron capture allows elements to be naturaly separated (even F2!!!) it is not un heard of to find many ions and free radicals (like NO3-) inside radioactive chemicals.
thermoluminescence dating is based on the fact that over time (milleniums..) electrical charges are building up in any given sample (mostly it is used for clay and rock dating where the ions and free radicals dont have a long free travel track and have been heated ) unless the temperature gets high enough (arround 500C) and things get shaffled and reorganized these ions will continue to build up due to natural radioactivity both from passing ionizing radiations and left over from decayed atoms.
thats the general answer . on the case by case decay left over molecules and ions it is much more complicated .
lets not forget that for 1atom being observed we are talking about rare event in a rare situation over thousands of years (if not millions and billions for U238) of stability in which nothing at all is happening...

annaandherdad - 31-3-2014 at 09:04

It might help to look at a simpler example of decay, say tritium->He3 plus electron (plus neutrino, but that escapes). Tritium is a gas. The electron is emitted with some Kev of energy, much more than the binding energy, so it goes some distance away in the decayed atom, which is now a He3+. Over time the gas will attain some degree of ionization, which is determined by the balance between ionization (due to decay) and recombination. The former goes at a steady rate, while the rate of the latter increases with density, so a low density gas of tritium will achieve a higher fraction of ionization in equilibrium than a dense gas.

If the tritium atom is part of a molecule, say HTO or T2O, then the molecule will behave as if one of the nuclei of hydrogen suddenly changed its charge from +1 to +2. That's because the electron emitted by the beta decay escapes the atom at a velocity much greater than the orbital velocities of the electrons left in the molecule. So, to find out what happens to the molecule itself, you have to solve the time-dependent Schr"odinger equation for the electrons with an initial condition which is the wave function of the original (pre-decay) molecule, but with the charge of one nucleus changed. In general, this is a linear combination of energy eigenstates of the new Hamiltonian, so there will be different outcomes with different probabilities (branching ratios). For example, starting with HTO, you can end up with HO+ plus He3, or HO plus He3+, or other possibilities.

Decay of radioactive nuclei in a solid is more complicated, of course, because, depending on the conductivity of the solid, the electrons (speaking of beta decay) will not be able to move from their final locations after they lose all their kinetic energy.

alpha decay is similar, but the alpha particle won't travel as far as the electron in beta decay. gamma decay doesn't change the nucleus, but the recoil of the nucleus could cause a dislocation of the atomic lattice site in a crystal.

annaandherdad - 31-3-2014 at 09:12

Sorry, neptunium, you posted your reply while I was writing mine, and I didn't see it.

neptunium - 31-3-2014 at 11:37

no problem,
which makes the whole study much more interesting to me !!

neptunium - 2-4-2014 at 17:38

not many people interested it seemed...

annaandherdad - 3-4-2014 at 09:25

Maybe your post was so definitive that no one has anything to add.

I'll add a few thoughts that this discussion makes me think about. You know in an insulator the charges can't move. If you take a block of insulator, say glass, and expose it to a beam of charged particles that are able to penetrate into the material, you can deposit charges at random locations distributed throughout the material. In this way the material picks up a net charge. I'm not sure, but it's possible this charged object will collect electrons or ions on its surface in an attempt to remain overall neutral. In any case, in the interior the electric field builds up and can get quite strong. If you keep adding charges, you get to a point where the material breaks down, and you get effectively sparks jumping inside the material. These can make complex looking fractures inside the glass (supposing it's glass). I've seen pictures of this, but I don't remember where, unfortunately.

If the insulator contains a radioactive substance, then the alpha or beta particles emitted will travel some distance from the decaying atoms and then get stuck. This is different from the previous situation because charges of both signs are imbedded in the insulator. So the electric fields will be much less than in the previous situation. However, it's still not an equilibrium situation, that is, if the charges were free to move they would do so, so there is energy stored in the configuration. I don't know what this does to the insulator, if the amount of separated charge is allowed to grow, but I know there has been talk of storing waste from nuclear reactors fused into some kind of glass. So if this is done, then such separated charges must develop over time. If you something about this, why don't you tell me.

annaandherdad - 3-4-2014 at 09:28

Maybe your post was so definitive that no one has anything to add.

I'll add a few thoughts that this discussion makes me think about. You know in an insulator the charges can't move. If you take a block of insulator, say glass, and expose it to a beam of charged particles that are able to penetrate into the material, you can deposit charges at random locations distributed throughout the material. In this way the material picks up a net charge. I'm not sure, but it's possible this charged object will collect electrons or ions on its surface in an attempt to remain overall neutral. In any case, in the interior the electric field builds up and can get quite strong. If you keep adding charges, you get to a point where the material breaks down, and you get effectively sparks jumping inside the material. These can make complex looking fractures inside the glass (supposing it's glass). I've seen pictures of this, but I don't remember where, unfortunately.

If the insulator contains a radioactive substance, then the alpha or beta particles emitted will travel some distance from the decaying atoms and then get stuck. This is different from the previous situation because charges of both signs are imbedded in the insulator. So the electric fields will be much less than in the previous situation. However, it's still not an equilibrium situation, that is, if the charges were free to move they would do so, so there is energy stored in the configuration. I don't know what this does to the insulator, if the amount of separated charge is allowed to grow, but I know there has been talk of storing waste from nuclear reactors fused into some kind of glass. So if this is done, then such separated charges must develop over time. If you something about this, why don't you tell me.

neptunium - 3-4-2014 at 11:24

i remember seen a video (cant remember where..) they put a certain type of transparen glass bottle into a strong source of Co 60 or inside a running nuclear reactor and left it there a few hours .
When it came out it was completly opake and to get it transparent again it was put into an oven to heat up...
this was described as the charged particles generated inside of it were trapped (as you pointed out) and were absorbing photons whereas the clear looking glass had no (or very little) isolated charges..
if i find the link to the video i`ll post it here...
but it goes along what you were just saying..
So to go back to the original question, isolated charged and ions or even whole group (like SO4--) are likely inside radioactive chemicals and even common.

MrHomeScientist - 3-4-2014 at 11:57

Quote: Originally posted by annaandherdad  
If you take a block of insulator, say glass, and expose it to a beam of charged particles that are able to penetrate into the material, you can deposit charges at random locations distributed throughout the material. In this way the material picks up a net charge.
...
If you keep adding charges, you get to a point where the material breaks down, and you get effectively sparks jumping inside the material. These can make complex looking fractures inside the glass (supposing it's glass). I've seen pictures of this, but I don't remember where, unfortunately.

You're probably thinking of Lichtenburg Figures (the first image on that page). I have one on my desk right now! I bought mine from these guys - they have lots of different styles and all are beautiful.

annaandherdad - 4-4-2014 at 10:59

Yes, that's it exactly! Thanks for the link, and sorry about the double post.

phlogiston - 13-9-2014 at 15:13

I suspect in many alpha and beta decays the recoil on the nucleus from emitting a very energetic particle is enough to break a chemical bond, so it is then not only a matter of working out what happens electronically.

careysub - 13-9-2014 at 18:00

Quote: Originally posted by phlogiston  
I suspect in many alpha and beta decays the recoil on the nucleus from emitting a very energetic particle is enough to break a chemical bond, so it is then not only a matter of working out what happens electronically.


A simple physics BOTE calculation. The mass ratio between and alpha particle and the Th-234 emitting atom is (formerly a U-238 pre-emission atom) is about 1:60 (1:58.5 more precisely).

Conservation of momentum requires that mv (alpha) = mv (atom). Since the alpha is 60 times lighter, it must travel 60 times faster than the atom post recoil or momenta to equal.

Kinetic energy is KE= (mv^2)/2 so the atom/alpha kinetic energy ratio is:
(60*1^2)/(1*60^2) = 1/60

Thus if the total decay energy is 4269.7 keV, then about 1/60 of that would end up as kinetic energy of the atom, or about 70 keV.

Since molecular bond energies are several eV the recoil energy of the atom will not only disrupt those bonds, it will cause the atom to the ejected into another part of the lattice structure. Additionally the alpha particle will zip along disrupting chemical bonds as it goes until stopped., whereupon it become an alien embedded helium atom.

So alpha decay at least completely destroys any chemical bonding that the parent atom participated in, and then some.

Nuclear decay is a process that drives the evolution of uranium minerals in noticeable ways.

For beta decay the ratio is more like 400,000:1; that Th-234 decays into U-234 with a decay energy of 270 keV, so the kinetic energy is only ~0.6 eV. The electron shell configuration changes though (different element) so does to associated ion states available. U has a +6 state, Th does not, so the chemical compound can still change especially with the excitation energy of 0.6 eV to help things along.