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Author Topic: why does n/p ratio inrease?  (Read 19002 times)

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sapta

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why does n/p ratio inrease?
« on: August 04, 2005, 06:38:48 AM »

why does n/p ratio inrease in the case of 92U238?it emits alpha particle.but why?
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sdekivit

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Re:why does n/p ratio inrease?
« Reply #1 on: August 04, 2005, 07:23:49 AM »

the Uranium is under the stabilitybelt. So this means it has less neutrons than it should have to be stable ( or: too many protons). What must a particle to do to gain stabiltiy when it's under the stabilitybelt? it must lose what?

--> there are more neutrons in the nucleus (check mass number). So what effect will be larger, the loss of two neutrons or two protons ?

See for a detailed explanation the foolowing text:

http://www.amscopub.com/images/file/File_36.pdf
« Last Edit: August 04, 2005, 07:38:11 AM by sdekivit »
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Mitch

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Re:why does n/p ratio inrease?
« Reply #2 on: August 04, 2005, 11:51:38 AM »

This is such a great question!!! But be warned great questions rarely have simple answers.

Your wondering why Uranium is unstable and why it emits alpha particles. But what is stabillity? The half life of U238 is ~109 years. Depending on who you talk to thats a very stable isotope. So, where do we draw the line on what is called "stable" or what isn't? When Uranium undergoes alpha decays it will emit helium nucleii and eventually become Lead 206. Lead 206 has a higher binding energy per nucleon. Meaning that the protons and neutrons are held more tightly in a  lead atom than an uranium atom. That being said, Uranium has a higher binding energy per nucleon than Helium! Helium is considered to be highly stable, why the contradiction? What would be a better way to measure nuclear stabillity? I'll leave the discussion open-ended so others can expand on it.
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Grejak

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Re:why does n/p ratio inrease?
« Reply #3 on: August 04, 2005, 02:05:26 PM »

why does n/p ratio inrease in the case of 92U238?it emits alpha particle.but why?

As the atomic number of a nucleus increases, the coulomb repulsion bewtween the protons becomes more and more important.  In light nuclei, this repulsion between the protons is not very important, so you get a proton to neutron number approximately equal to one.  With heavier nuclei the protons begin to repel each other and a stable nucleus needs to have more and more neutrons in it in order to balance out the increased distance between the protons.  Since uranium is a heavy nucleus, the protons are highly repelled from each other and more neutrons fit in-between them.  This will give you a large n/p number.  

When uranium emits an alpha particle, it emits an equal number of protons and neutrons.  At this point it becomes a simple math problem.  If the original nucleus is 238U, then the nucleus has 92 protons and 146 neutrons, giving it an n/p ratio of 146/92 or 1.59.  After it emits an alpha particle, the new nucleus is 234Th which has 90 protons and 144 neutrons.  For this nucleus the n/p ratio is 144/90 or 1.6, a slight increase.

As a side note (and one the Mitch has asked me to add), even though the n/p ratio has increased the stability of the new nucleus has decreased.  Something cool to think about :)
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sapta

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Re:why does n/p ratio inrease?
« Reply #4 on: August 04, 2005, 07:54:54 PM »

Quote
the Uranium is under the stabilitybelt.

but (238-92)/92=1.586. So,how can it be below the stability belt? ???

Quote
As a side note (and one the Mitch has asked me to add), even though the n/p ratio has increased the stability of the new nucleus has decreased.

why in the 1st place shud the ratio increase?wouldn't the daughter nuclide be farther away from the stability zone ???
« Last Edit: August 04, 2005, 07:58:16 PM by sapta »
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Mitch

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Re:why does n/p ratio inrease?
« Reply #5 on: August 04, 2005, 08:12:18 PM »

The more neutron rich the nuclei the more "stable" it is (to a limit).
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Borek

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Re:why does n/p ratio inrease?
« Reply #6 on: August 04, 2005, 10:46:32 PM »

but (238-92)/92=1.586. So,how can it be below the stability belt? ???why in the 1st place shud the ratio increase?wouldn't the daughter nuclide be farther away from the stability zone ???

Check Grejak's math - it is as simple as counting using fingers :)
« Last Edit: August 04, 2005, 10:53:11 PM by Borek »
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sapta

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Re:why does n/p ratio inrease?
« Reply #7 on: August 04, 2005, 11:33:22 PM »

Check Grejak's math - it is as simple as counting using fingers :)

still don't get it.where am i wrong?
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Borek

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Re:why does n/p ratio inrease?
« Reply #8 on: August 04, 2005, 11:54:07 PM »

Put aside all stability/unstability, binding energy per nucleon and similar problems. Look just for the quotient:

n/p

n is - at the moment of start - 146, p is 92

when the alpha particle is emitted you substract 2 from both numerator and denominator. As the denominator is smaller such substraction makes quotient larger. This is very simple math, nothing fancy, and nothing to do with the radiochemistry.

Now, if you want to know why alpha particle is emitted, you have to look for stabilities and so on, but that's completely different story an it requires knowledge. Mitch is much better here then me. I am using only common reasoning :)
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sapta

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Re:why does n/p ratio inrease?
« Reply #9 on: August 05, 2005, 09:10:47 AM »


when the alpha particle is emitted you substract 2 from both numerator and denominator. As the denominator is smaller such substraction makes quotient larger.

yes but shud it be that way?it shud have tried to bring down the ratio.
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Grejak

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Re:why does n/p ratio inrease?
« Reply #10 on: August 05, 2005, 09:32:24 AM »

yes but shud it be that way?it shud have tried to bring down the ratio.

I think your question it more specific than you want it to be.  You asked why alpha decay in specific increases the n/p ratio in a nucleus.  That is simple number manipulation and has been sufficiently answered multiple times.  No matter how many times you ask the question, alpha decay will always increase the n/p ratio in 238U.  Not due to the stability of the nucleus or any other factors, simply due to the definition of what alpha decay is.  That cannot be changed.

So what can be changed?  Well, the decay mode does not always have to be alpha, and different decay modes will change the n/p ratio in different ways.  Perhaps your question is why uranium undergoes alpha decay instead of one of the other decay modes.  That is a much more difficult question.
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sapta

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Re:why does n/p ratio inrease?
« Reply #11 on: August 05, 2005, 09:43:00 AM »



So what can be changed?  Well, the decay mode does not always have to be alpha, and different decay modes will change the n/p ratio in different ways.  Perhaps your question is why uranium undergoes alpha decay instead of one of the other decay modes.  That is a much more difficult question.


yes yes that's my question.never thought u were misunderstanding what i wanted ::)my poor English :-[
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Grejak

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Re:why does n/p ratio inrease?
« Reply #12 on: August 05, 2005, 04:09:50 PM »

yes yes that's my question.never thought u were misunderstanding what i wanted ::)my poor English :-[

There are several possible ways for radioactive nuclides to decay, most notably alpha decay, beta decay, electron capture and spontaneous fission:  Alpha decay involved the emission of a stripped 4He nucleus from the atom.  Beta decay involves the emission of an electron or a positron (along with neutrinos) from the nucleus.  During B+ decay a proton is converted into a neutron and the reverse is true for B- decay.  Electron capture is the process by which an orbital electron is absorbed in the nucleus, destroys a nuclear proton and creates a neutron.  Spontaneous fission involves the splitting of a nucleus into two or more smaller nuclei.

In order to simplify the 238U problem, let’s ignore the spontaneous fission decay mode and focus only on why 238U decays via alpha emission instead of beta decay or electron capture.

The Q value of a reaction determines whether or not a reaction is thermodynamically stable or not.  In the case of nuclear reactions, the Q value is simply the mass of the reactants minus the mass of the products (the reverse of most of chemistry).  A positive Q value implies that the reaction is exothermic and that over time the reaction will tend towards the products.  The reverse is true for a negative Q value.  In this case, the reaction is endothermic and over time will favor the reactants.

The values below are the Q values for each reaction:
Alpha decay: 4.27 MeV
B- decay: -0.147
B+ decay: -4.483
Electron Capture: -3.461

As you can see, the only decay mode that will release energy is alpha decay.  As the other modes all require energy, 238U will not decay by them.


On a fascinating side note, the Table of Radioactive Isotopes lists 238U as having a bb decay, which I assume means double beta decay. :)
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sapta

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Re:why does n/p ratio inrease?
« Reply #13 on: August 05, 2005, 04:49:23 PM »

i get it :)



On a fascinating side note, the Table of Radioactive Isotopes lists 238U as having a bb decay, which I assume means double beta decay. :)


interesting;what's double beta decay?(1 beta emission would create a new nuclide)
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Elgon

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Re:why does n/p ratio inrease?
« Reply #14 on: August 05, 2005, 06:48:24 PM »

Double beta decay is the simultaneous conversion of 2 neutrons into 2 protons under emission of 2 electrons and 2 anti neutrinos. The important part is that both conversions happen at the same time and not successively.
This type of decay has been postulated, and there are several large scale experiments planned to examine it. But as far as I know in the case of uranium there has only been a limit determined so far. One way of determining the rate for the double beta decay of uranium would be to measure the amount of plutonium 238 that is produced in a sample of uranium 238 over time. Unfortunately you have to make sure that your uranium does not contain any traces of plutonium from other sources. And this is very, very difficult.
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