nuclear reaction

[vedmecum]

may be i am asking the foolish ques. but these ques. is really disturbing my hypothalamus .

 

we are all familiar with nuclear reaction . the most familiar one is the one with uranium . it changes to thorium, radon , polonium , radium or.......... depending upon the type of decay either its alpha, beta or any other .

 

now if we take not elemental uranium but its oxide which is widespread then how stiochiometry will be maintained i.e there may be formation of oxides of thorium, radon , polonium , radium or.......... , but from where oxygen comes ????????????:confused::confused::confused::confused::confused::confused::confused::confused::confused:

[timo]

1) Don't worry: It's not so many people familiar with nuclear reactions, certainly not all (I am not, for example).

2) Nuclear reactions, like the radioactive decay of uranium you mentioned, are a process happening in the nucleus, not in the electron shell. Chemistry is concerned with the electron shell of atoms (and the complex structures these shells can form).

3) As a rule of thumb the nucleus cares pretty little for the electron shell. An Uranium-Oxygen binding is a structure formed by the electron shell. So you can assume that at least as long you do not consider any form of complex interplay of a radioactive substance with its surrounding (like mediators in nuclear reactors) the radioactivity of Uranium is the same as the radioactivity of uranium oxide.

[SH3RL0CK]

1) Don't worry: It's not so many people familiar with nuclear reactions, certainly not all (I am not, for example).

2) Nuclear reactions, like the radioactive decay of uranium you mentioned, are a process happening in the nucleus, not in the electron shell. Chemistry is concerned with the electron shell of atoms (and the complex structures these shells can form).

 

This doesn't seem entirely correct to me.

 

Wouldn't a change in the number of protrons in the nucleus via alpha particle change the number of electrons in the shell? As such, couldn't Uranium oxide, for example, turn into another element which require a different chemical structure for the oxide? Or possibly the production of Radon (a noble gas) and free oxygen?

 

3) As a rule of thumb the nucleus cares pretty little for the electron shell. An Uranium-Oxygen binding is a structure formed by the electron shell. So you can assume that at least as long you do not consider any form of complex interplay of a radioactive substance with its surrounding (like mediators in nuclear reactors) the radioactivity of Uranium is the same as the radioactivity of uranium oxide.

 

Since the Uranium is diluted by the oxygen, and therefore in a given volume or mass there is less Uranium, I would suspect the radioactivity to be somewhat less.

[timo]

I was assuming the question was if the chemical binding of a radioactive atom has an influence on its radioactivity. The answer is basically "no". I might have misunderstood the question.

[SH3RL0CK]

You are correct, certainly there would be no influence as long as the atomic structure of the nucleous did not change. Most radioactive elements have a reasonably long half-life so that the vast majority of the atoms in a sample does not change appreciably.

 

However, the production of the radiation is associated with changes in the nuclear structure. So when an atom gives off, say an alpha particle (which is a helium nucleus), the atomic structure is changed and therefore the electon shells must change as well. This requires any chemical bonding associated with that atom to change as well.

[timo]

I think you are simply answering a different question than me, namely "does a radioactive decay influence a chemical binding?". In that case: I would the first significant influence expect to be the recoil on the (formerly-)Uranium nucleus. In the case of an alpha decay that is MeV kinetic energy against a binding energy in the eV. I wouldn't be too surprised if the binding ripped. Whether decay products recombine later then depend on the environment and of course also on what was left of the uranium nucleus.

 

EDIT: And I start to begin to understand why we did understand the OP differently. I think vedmecum did indeed ask what happens to the chemical binding, not whether the binding will influence the decay of the Uranium nucleus.

Edited April 2, 2009 by timo
inserted missing words ...

[insane_alien]

the decay product will become an anion which will affect chemical stability.

[Kaeroll]

I wonder if anyone's investigated this at all? Say... determining crystal structures of an oxide (or the like) of an unstable element before and after appreciable decay.

[insane_alien]

plenty of people will have. it is essential for nuclear reactor engineering and fuel reprocessing.

[amit]

it was a shock. i have never thought that. what happens to electrons in the shell of uranium after decay? i believe they must stay there but now the atom should be negatively charge? outer electrons will be less strongly bound?i cant undertstand it either. if we take elemental uranium and it decays then will the extra electrons just fly away in space? if not wont we have -ve charged particles repelling each other strongly?

[vedmecum]

 

EDIT: And I start to begin to understand why we did understand the OP differently. I think vedmecum did indeed ask what happens to the chemical binding, not whether the binding will influence the decay of the Uranium nucleus.

 

sorry because two of you have taken my ques. in a different sense but my exact ques. is what is written above . this is first time when i have no idea .

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it was a shock. i have never thought that. what happens to electrons in the shell of uranium after decay? i believe they must stay there but now the atom should be negatively charge? outer electrons will be less strongly bound?i cant undertstand it either. if we take elemental uranium and it decays then will the extra electrons just fly away in space? if not wont we have -ve charged particles repelling each other strongly?

 

no amit . log on http://en.wikipedia.org/wiki/Nuclear_decay

[swansont]

I think you are simply answering a different question than me, namely "does a radioactive decay influence a chemical binding?". In that case: I would the first significant influence expect to be the recoil on the (formerly-)Uranium nucleus. In the case of an alpha decay that is MeV kinetic energy against a binding energy in the eV. I wouldn't be too surprised if the binding ripped. Whether decay products recombine later then depend on the environment and of course also on what was left of the uranium nucleus.

 

Right.

 

After the decay you might expect the U to have 2 extra electrons, but it's more likely that the decay ionizes several electrons. When I was trapping radioactive K, studying the effects of decay, we found that the beta-plus decay of K-37 would only rarely leave you with Ar^- or Ar⁰. More often you'd end up with Arⁿ⁺, with n = 1, 2 or 3. i.e. very often the decay would strip off several electrons. This is only from a beta, i.e. a single charge and little nuclear recoil.

 

So after a U decay in an oxide, I don't think you can assume you have a molecule anymore.

[vedmecum]

Right.

 

After the decay you might expect the U to have 2 extra electrons, but it's more likely that the decay ionizes several electrons. When I was trapping radioactive K, studying the effects of decay, we found that the beta-plus decay of K-37 would only rarely leave you with Ar^- or Ar⁰. More often you'd end up with Arⁿ⁺, with n = 1, 2 or 3. i.e. very often the decay would strip off several electrons. This is only from a beta, i.e. a single charge and little nuclear recoil.

 

So after a U decay in an oxide, I don't think you can assume you have a molecule anymore.

 

i agree with you as per K-37 (that we get noble element in last ) is concerned but i think we can't apply it universally for all radioactive substance . for example log on

http://wiki.answers.com/Q/What_is_the_decay_products_of_uranium_238 and http://www.atral.com/U2381.html and there are many more links which indicates that the end product will be lead-206 . so now why not oxide of lead -206 form or it may form ?

[swansont]

i agree with you as per K-37 (that we get noble element in last ) is concerned but i think we can't apply it universally for all radioactive substance . for example log on

http://wiki.answers.com/Q/What_is_the_decay_products_of_uranium_238 and http://www.atral.com/U2381.html and there are many more links which indicates that the end product will be lead-206 . so now why not oxide of lead -206 form or it may form ?

 

Because the focus on those discussions is the nuclear aspect. The chemistry is being ignored, because it has no effect on the stability if the nucleus. The converse in not the case, though.

 

My point with the K decay is not that we get a noble gas (but one should also note that Radon is a noble gas and is in the decay chain of Uranium, so you do have this issue), it's that a singly-charged particle can cause multiple electrons to be stripped from the atom. That kind of disruption calls the viability of the chemical bond into question, and an alpha is going to be worse than a beta. More importantly, perhaps, is Atheist's point about the recoil, that will exceed the chemical binding energy — the recoiling nucleus will have more than 4% (4/92) of the decay energy, which means it will be a few hundreds of keV.

 

So there's simply no point in worrying about the chemistry. You will have to re-form bonds after the daughter comes to rest and neutralizes.

[John Cuthber]

One of the first syntheses of the perbromate ion (BrO4)- used beta decay of a radioactive version of (SeO4)2-

Most of the time you got all sorts of decomposition but sometimes the product was the perbromate. This was co precipitated with perchlorate and then monitored by looking for the decay of the radioisotope of bromine.

A staggeringly poor yield for a synthesis, but it did prove that the perbromate ion could be made.

http://pubs.acs.org/doi/abs/10.1021/ar50064a001

[vedmecum]

:-) finally we reach at that point where i think we need to discuss more of chemistry as mine first ques. still not completely unravel . from the above posts what i concluded is that reaction will take place but the major ques. which struck in my hypothalamus is how reaction will balance by itself ? because if , for ex , we take uranium oxide in which uranium is in x+ oxidation state and we end with oxide of different element in +y oxidation state then how stiochiometry will be maintained . there will be 2 or more situation possible:-

 

1. new comp. have lesser no. of oxygen , then where the remaining oxygen will go ? . will we get free oxygen (if we took uranium oxide) ? if so then isn't amazing ?

 

2. if new comp. have higher no. of oxygen then from where it comes ?

[John Cuthber]

Let's take an odd example, but one where the chemistry is relatively simple.

Potassium has a naturally ocurring isotope of mass 40 which decays (mainly- but lets ignore the other possibility for a minute) by beta decay to give argon. It has a half life of 1.25 billion years I think)

 

Imagine that we extract some of this isotope of potassium and turn it into the chloride.

 

It decays slowly but surely to argon, and of course, argon doesn't form a chloride.

 

So the reaction is

2 KCl --> 2Ar +Cl2.

The potassium was there as K+ ions and when they lost a beta particle they (briefly) became Ar++ ions.

Now Ar(II) is a much stronger oxidant that you get in any normal sort of chemistry. It's perfectly capable of oxidising chloride to chlorine.

 

In case you are wondering, the beta particle will hit something and stop. Then it's just an electron. Now tha electron (on it's own) is a very powerful reducing agent. It can certainly reduce Ar+ to Ar

[UC]

Let's take an odd example, but one where the chemistry is relatively simple.

Potassium has a naturally ocurring isotope of mass 40 which decays (mainly- but lets ignore the other possibility for a minute) by beta decay to give argon. It has a half life of 1.25 billion years I think)

 

Imagine that we extract some of this isotope of potassium and turn it into the chloride.

 

It decays slowly but surely to argon, and of course, argon doesn't form a chloride.

 

So the reaction is

2 KCl --> 2Ar +Cl2.

The potassium was there as K+ ions and when they lost a beta particle they (briefly) became Ar++ ions.

Now Ar(II) is a much stronger oxidant that you get in any normal sort of chemistry. It's perfectly capable of oxidising chloride to chlorine.

 

In case you are wondering, the beta particle will hit something and stop. Then it's just an electron. Now tha electron (on it's own) is a very powerful reducing agent. It can certainly reduce Ar+ to Ar

 

Actually, beta decay would convert potassium into calcium...

 

[ce] KCl -> CaCl^+ + e^- [/ce]

[vedmecum]

Let's take an odd example, but one where the chemistry is relatively simple.

Potassium has a naturally ocurring isotope of mass 40 which decays (mainly- but lets ignore the other possibility for a minute) by beta decay to give argon. It has a half life of 1.25 billion years I think)

 

Imagine that we extract some of this isotope of potassium and turn it into the chloride.

 

It decays slowly but surely to argon, and of course, argon doesn't form a chloride.

 

So the reaction is

2 KCl --> 2Ar +Cl2.

The potassium was there as K+ ions and when they lost a beta particle they (briefly) became Ar++ ions.

Now Ar(II) is a much stronger oxidant that you get in any normal sort of chemistry. It's perfectly capable of oxidising chloride to chlorine.

 

In case you are wondering, the beta particle will hit something and stop. Then it's just an electron. Now tha electron (on it's own) is a very powerful reducing agent. It can certainly reduce Ar+ to Ar

 

okay this unravel 1st situation and what about second because to balance UC reaction we need almost double amount of Cl- , upto my knowledge CaCl+ is less stable than CaCl2 .

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Actually, beta decay would convert potassium into calcium...

 

[ce] KCl -> CaCl^+ + e^- [/ce]

 

both u and john are correct . you are talking about beta- decay and john has given an example of beta+ decay .

[John Cuthber]

Very kind of you vedmecum, to try to bail me out but I screwed up.

I knew it did both forms of decay (which is why I said "mainly- but lets ignore the other possibility for a minute") but I got the decay mixed up.