Atom Stability: Mass Number, Radioactive Decay & More

[Kyoma]

What's makes an atom stable or unstable?

From what I know, if an atom has a mass number Z that is more than 83, it is unstable and would undergo radioactive decay to make itself stable.

So, is there any criteria that classify an atom unstable?

 

[phyzguy]

Some nuclei are stable and others are not, and the ones that are unstable have a huge range of half-lives. for example, Be-8 has a half-life of ~10^-15 sec, while Bi-209 has a half-life of >10^19 years. Because the strong force (QCD) is so complex, we don't really have a good explanation as to which nuclei are stable and which are not, and nobody can predict th stability of a nucleus from first principles. There are some general rules, such as the fact that as the positive charge in the nucleus goes up, it takes more neutrons to make it stable. We don't even know for certain whether the "stable" nuclei are stable, or whether they just have extremely long half-lives. Here's a good resource on the different nuclei:

http://atom.kaeri.re.kr/

 

[Kyoma]

Thanks, but does the number of electrons affect the stability of an atom?

From what I know, atoms that have 8 valence electrons are considered stable (except for helium which has 2 valence electrons and is stable) and that all unstable atoms want to have 8 valence electrons to make itself stable. To do so, it needs to bond (covalently or other bonding methods).

 

[fatra2]

Hi there,

Thanks, but does the number of electrons affect the stability of an atom?

From what I know, atoms that have 8 valence electrons are considered stable (except for helium which has 2 valence electrons and is stable) and that all unstable atoms want to have 8 valence electrons to make itself stable. To do so, it needs to bond (covalently or other bonding methods).

Since the electrons are "spinning" around the nucleus, they don't affect its stability. If you want to understand the stability of nuclei, you need to look at the nuclear strong interaction. It is the only one that can explain the stability of nuclei.

In short, a nucleus is a soup of protons and neutrons. Protons have an electric charge, and repulse each other (Coulomb Law). When these particles get close enough to each other they start liking each other. Ok, so you have a nucleus, with so many protons and so many neutrons. It is easy to imagine "throwing" some neutrons in this soup. By adding neutrons, you are rearranging the interactions between the particles. Comes a point, when the interaction between them cannot compensate the repulsion. In this case, you have an unstable nucleus.

Hope this helps.

Cheers

 

[Bob S]

Generally, any isotope that has more mass than the sum of the masses of the decay products, it will eventually decay. Sometimes a proton-rich nucleus will capture an atomic electron and decay.

There is an island of stability beyond uranium where "doubly-magic" stable isotopes (Z = 114, 120, and 126) may someday be discovered (none yet). See

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

Bob S

 

[fatra2]

Hi there,

Well, very generally speaking, all the isotopes are fundamentally unstable. Their respective lifetime varies from a few ns to 10¹⁹ s. Therefore, a stable isotopes, per say, does not exist.

Cheers

 

[Bob S]

Well, very generally speaking, all the isotopes are fundamentally unstable. Their respective lifetime varies from a few ns to 10¹⁹ s. Therefore, a stable isotopes, per say, does not exist.

Helium has two protons and two neutrons. What would the helium nucleus decay into?

Bob S

 

[aabottom]

Hey gang.

I just joined PF. I'm interested in the topic in this thread.

Thank you for your useful replies.

 

[phyzguy]

Well, I just saw fatra2's reply that all nuclei are fundamentally unstable. While this may be true, this certainly is not known at present. All attempts to find decays of the proton have failed, so to the best of our current knowledge, the proton is absolutely stable, and if it is unstable, its lifetime is known to exceed 10^33 years. The same goes for other stable nuclei, like He-4 (alpha particle).

 

[fatra2]

Hi there,

the proton is absolutely stable, and if it is unstable, its lifetime is known to exceed 10^33 years. The same goes for other stable nuclei, like He-4 (alpha particle).

There seems to be a contradiction in your post. You say that protons are absolutely stable, but then you add that it's lifetime more than 10^33 years.

From my definition of stable means that the lifetime is infinite. We know that the proton has an extremely long lifetime, but we don't know (yet) if it is stable. So I stand behind what I said: from the knowledge we have today, we don't know any element that is completely stable, meaning that it will stay in this form and shape forever.

Cheers

 

[phyzguy]

fatra2:

There is a big difference between what you said in post #6, "all the isotopes are fundamentally unstable", and what you said in your most recent post, "we don't know any element that is completely stable". I agree with your second assertion, but not with the first, which says that we know that all elements will eventually decay. We don't know this.

 

[fatra2]

Hi thre,

I apologize for my first post. I guess I chose the wrong wording.

But then again, if even with the most stable isotopes/particles we know (proton) the lifetime is not infinite, then how can you call them stable? From my point of view, innocent until proven guilty applies here: unstable until proven stable. Even a particle with a very long lifetime, will eventually change = unstability.

 

[humanino]

It is not of much value to say "unstable until proven stable".

If the standard model were complete and the entire story, then the proton would be completely stable (up to electroweak non-perturbative corrections causing proton decay one time-scales many orders of magnitude longer than the universe lifetime). Still, the only thing we can get experimentally are lower bounds on its lifetime. Nothing is ever "proven stable". We never measure zero. We get a very large sample of protons, we wait for a very long time, and the only thing we get is an upper bound for the probability of decay.

However, some people come up with theories beyond the standard model, with new interactions which do induce proton decay. Now _this_ is what is interesting and has value, rather than general philosophical principles. As soon as you start having unifications between the strong and electroweak sector, you must get interactions allowing for transitions between the two, and thus some kinds of proton decay.

Independently, it is true that the notion of stability is not a sharp one. Most of the time, it is unambiguous, because the strong force acts at scales of a well-defined time and space span. The direct strong force extends up to about 1 fm, and that takes about 10^-23 s. If something lives much much longer than this, orders of magnitudes longer, then it is considered to be stable for the strong force. By the same token, the concept of a "resonance" is not always a sharp one. It is unambiguous only when the width of the resonance is negligible compared to its mass. If you were to consider all possible interactions and all possible scales when calling something "stable", then nothing would be stable at all : throw them in a black hole and they disappear !

 

[samtnz]

Hi there, please excuse any bad phrasing but is there a non zero probability that a nucleon could tunnel out of a Carbon-12 nucleus causing it to decay even in the absence of external force?

The quote below mentions the decay products would be heavier than the original nucleus but could energy be "borrowed" from the uncertainty principle to allow this to happen?

"Carbon-12 is stable because there's nothing it could decay to. Any combination of decay products would be heavier than the original nucleus. Law of conservation of energy prevents carbon-12 from decaying."

 

[lpetrich]

It's actually reasonably well-understood what gives atomic nuclei their binding energies. Though QCD is difficult to calculate with, it's been possible to make a fair amount of progress using semi-empirical approaches.

The two main ones are the liquid-drop model and the shell model.

In the liquid-drop model, one treats a nucleus as overlapping Fermi liquids of protons and neutrons that have constant overall nucleon density. The main effects:

• Volume -- each nucleon interacts with its neighbors
• Surface -- nucleons at the surface have no outside nucleons to interact with
• Electrostatic repulsion of protons
• Proton-neutron asymmetry -- whichever there is more of gets pushed to greater energies by its greater density
• Pairing effects -- protons like to form spin-0 pairs with other protons, and likewise with neutrons. Unpaired nucleons get extra energy.

Semi-empirical mass formula

In the shell model, one does shell-filling calculations like those for electrons in atoms. One starts with some overall potential and adds spin-orbit effects.
Nuclear shell model

More recently, it has been feasible to take nucleon-nucleon interaction potentials and calculate the structure of nuclei from them.
Nuclear structure

However, all these approaches have numerical parameters that are found by fitting to experimental results.

In theory, QCD should make it unnecessary to do so; one ought to be able to get nucleon-nucleon effective potentials from it, and from there, the liquid-drop and shell-model approximations. But such calculations are VERY difficult; it was only a half a decade ago that it became possible to calculate nucleon masses from QCD parameters to within a few percent.