Why does an atom with too many neutrons become unstable and undergo nuclear decay?

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My teacher explained that when a large nucleus has too few neutrons, the binding energy is not enough to hold the nucleus together. But by adding neutrons the binding energy increases without increasing the repulsion between nucleons, and thus the nucleus can stay together. But atoms that have too many neutrons also undergo nuclear decay. If the previous logic is right, those atoms should be very stable, so why does too many neutrons make the atom unstable?
 
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do you understand why a single neutron is unstable?
 
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I have not heard of that. Can you explain?
 
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Cleonis
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atoms that have too many neutrons also undergo nuclear decay.[...] why does too many neutrons make the atom unstable?

If I recall correctly there are models for the the physics of the nucleon, where there are nested configurations, somewhat analogous to electron configuration as nested structures.

Recapitulating the case of electrons: elements where all atomic orbitals that are occupied by electrons are completely filled are chemically the most stable.

If I recall correctly: just as the electron configuration is reflected in the ordering of the elements in the periodic table, one can use nucleon configuration to order nucleons.

Presumably the instability of Technetium is explained along these lines. (For most elements there are several stable isotopes, but in the case of Technetium (42 protons) there is not even one stable isotope.)

It has also been hypothesized that above atomic number 100 there may be one or more 'islands of stability', for nucleon configurations where all the "shells" are perfectly filled.
 
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bcrowell
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FAQ: Why does the line of stability have the average over-all shape it does?

For light nuclei, the line of stability hugs the N=Z line, and this is because of the Pauli exclusion principle. If you have N=8 and Z=8 (16O), you can put the 8 neutrons in the 8 lowest energy states, and the 8 protons in the 8 lowest energy states. With N=10 and Z=6 (16C), the exclusion principle forces you to put those last few neutrons in high-energy states that weren't occupied in 16O.

For heavy nuclei, the mutual electrical repulsion of the protons breaks the symmetry in the way the strong nuclear force treats neutrons and protons. This effect favors higher N/Z ratios, so the line of stability bends away from N=Z.

The line of stability also has little wiggles superimposed on top of its broad over-all curve. These are caused by quantum mechanical shell effects, the nuclear analogs of the ones in atomic physics that make the noble gases so chemically stable. These shell effects have nothing to do with the over-all shape of the line of stability. For example, the nucleus 100Sn (N=50, Z=50) has two closed shells, but it is very far from the line of stability.
 
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Cleonis
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My teacher explained that when a large nucleus has too few neutrons, the binding energy is not enough to hold the nucleus together.

It dawned on me that the reply about nuclear configuration in shells, partly or completely filled, doesn't address the actual question.

Let me generalize, making the reasoning independent from how the nucleus is modeled (other models treat the nucleus as analogous to a drop of fluid)

The strong nuclear force acts between nucleons (protons and neutrons). The strong nuclear force is extremely short range. The Coulomb force between the protons falls off with distance too, (inversely proportional to the square of the distance) but the strong nuclear force is in effect only active between adjacent nucleons.

So you have the Coulomb force acting significantly over distances that span multiple adjacent nucleons, but the force that provides cohesion of the nucleus, the strong force, acts only between adjacent nucleons.
 
  • #9
Astronuc
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If I recall correctly there are models for the the physics of the nucleon, where there are nested configurations, somewhat analogous to electron configuration as nested structures.

Recapitulating the case of electrons: elements where all atomic orbitals that are occupied by electrons are completely filled are chemically the most stable.

If I recall correctly: just as the electron configuration is reflected in the ordering of the elements in the periodic table, one can use nucleon configuration to order nucleons.

Presumably the instability of Technetium is explained along these lines. (For most elements there are several stable isotopes, but in the case of Technetium (42 protons) there is not even one stable isotope.)

It has also been hypothesized that above atomic number 100 there may be one or more 'islands of stability', for nucleon configurations where all the "shells" are perfectly filled.
Slight correction, Tc has 43 protons, whereas Mo has Z=42. Tc is the lightest element without a stable isotope.

Promethium, Pm (Z=61) is another element lighter than Bi and without a stable isotope.

Bi-209 (Z=83) is the heaviest nuclide considered stable. All other nuclides of Z>83 are radioactive.

Nuclides with too few neutrons (or too many protons) may decay by positron emission (generally restricted to lighter radionuclides with neutron deficit) or electron capture.
 
  • #10
Vanadium 50
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Bi-209 is very, very slightly radioactive. About 10 years ago it was discovered that it has a half-life in the few x 1019 year ballpark. It had been calculated for quite some time that this was alpha-unstable to Ti-205 with a 3 MeV alpha, but it wasn't observed until then.

Of course, with a half-life this long (much longer than the age of the universe), it is for all intents and purposes stable. The most radioactive thing in a typical bottle of Pepto-Bismol is probably the glass. :wink:

The longest measured half-life that I am aware of is Te-128, at a few x 1024 years.
 

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