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Topic: Instability of nuclei  (Read 8340 times)

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Memy

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Instability of nuclei
« on: October 24, 2005, 08:03:24 AM »
Hello,

This is a question about radioactivity. Basically, nuclei decay since they are unstable in order to achieve the proton to neutron ratio that stable nuclei have.

I was just wondering why nuclei that lie above the band of stability ( neutron rich ) are unstable....

I know that the nuclei that lie under the band of stability ( proton rich) are unstable since the repulsion between the protons is greater than the strong force and therefore decay by ejecting protons, but in the case of the neutron rich nuclei what makes them unstable?

cheers
« Last Edit: October 25, 2005, 09:23:07 AM by Memy »

Offline Grejak

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Re:Instability of nuclei
« Reply #1 on: October 27, 2005, 12:30:52 AM »
Unfortunately your question does not have a quick answer. First of all, proton rich nuclei normally decay via beta decay or electron capture.  It is only in the region near the proton drip line that you will find proton rich nuclei that decay by proton emission or any of its variants.

The stability of a nucleus is determined largely/mainly by its binding energy.  There are five terms that contribute to binding energy: volume, surface area, Coulomb, symmetry and pairing.  Of these, only the volume term is consistently positive and increases proportional the A.  The pairing term is dependent on whether your nucleus is even-even, even-odd, odd-even or odd-odd (for the number of protons-neutrons).  Almost every book has a different value for the pairing energy, and it contributes less than an MeV of energy for elements up to and beyond uranium, so lets ignore it :).

That leaves three other terms to deal with.  The Coulomb term is negative and increases proportional to Z2/A1/3.  So as you increase the number of neutrons the term becomes ‘less’ negative.  So, up to this point, everything states that the more neutrons you add, the more stable your nucleus.

This is where the two remaining terms come into play.  The surface energy term is also negative and increases proportional to A2/3.  The symmetry term is also negative for neutron deficient isotopes and increases as (Z-N)2/A.  So, in order to determine the most stable nuclei for a given element or a give mass, you need to balance these five terms.  Anything that is to one side or the other has potential for decay.



Offline Grejak

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Re:Instability of nuclei
« Reply #2 on: October 27, 2005, 12:47:37 AM »
Another answer to your question requires some working knowledge of the shell model and is more conceptual in nature.  Anyway, the shell model for the nucleus is based on similar concepts as that of the electrons.  There are multiple levels corresponding to different energies inside of a potential well.  Wikipedia does a better job at explaining it: http://en.wikipedia.org/wiki/Shell_model

There are similar, yet separate shells for the protons and for the neutrons.  For example, it is possible to have a proton in the s1/2 shell with an identical spin as a neutron in the s1/2 shell without violating any laws.  However, due to the Coulomb repulsion of the protons, there is a slight increase in the spacing of the proton shell levels compared to those of neutrons.
  With a low number of nucleons, this spacing is not very important and you end up with equal numbers of protons and neutrons.  But as you increase the number of nucleons, then you have more neutron shells within the same energy region, leading to stable nuclei with more neutrons than protons.

For example: The 2d5/2 for protons will be higher in energy than the corresponding 2d5/2 level for neutrons.  As such, if there is one proton in the 2d5/2, and a free hole in the neutron 2d5/2, then the proton will decay with a Q value proportional to the difference in the level energies.

How does this relate to the question in the original post?  Well, the same is true for neutron rich nuclei.  If you have a neutron that is in a level that has a higher energy than an unfilled proton level, then the nucleus will decay via beta decay.

Probably not the best explanation, but interesting none the less :).

Memy

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Re:Instability of nuclei
« Reply #3 on: November 09, 2005, 11:26:38 AM »
Can i simply say that neutrons are larger in mass than protons and therefore according to einstein equation E=mc2 neutrons have more energy and when they can they decay into protons since these have less energy?

( i am a biochemist so i don't have a lot of knowledge about particle physics)    

Offline Grejak

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Re:Instability of nuclei
« Reply #4 on: November 11, 2005, 11:26:15 AM »
The reason why a lone neutron will decay into a proton is due to the differences in masses, where a lone proton is more energetically favorable then a lone neutron.  Once you start dealing with two or more nucleons then you need to factor in the binding energy of the last nucleon as well as the mass defect of the nucleus.  If you simply state that the neutron is heavier in mass and will decay, then the logical conclusion is that nuclei should be composed only of protons.

Memy

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Re:Instability of nuclei
« Reply #5 on: November 14, 2005, 07:34:13 AM »
Can I say then that in neutron-rich nuclei, there are unoccupied proton states avalaible, so the system can achieve a lower energy by having only a certain number of neutrons decaying?

thank you very much

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