The nucleus and radioactivity...

[RyanJ]

Not shure if this is better here or in the physics section but I guess it kind of fits into both of them!

 

Why is it a certain ratio of neutrons to protons is stable and another one is unstable? What makes it unstable because the Neutrons contribute no charge so why would they be unstable in that way?

 

Cheers,

 

Ryan Jones

[ecoli]

I assume it's because of the mass, rather then charge.

[RyanJ]

I assume it's because of the mass, rather then charge.

 

Yea thats what I was thinking too but why

 

I mean bigger things seem to be more stable than small things for starters, a planet is quite stable for example.

 

What is it about the mass that makes it act like that if you know what I mean

 

Cheers,

 

Ryan Jones

[Daecon]

If it was purely mass then wouldn't an isotope of Helium be just as stable as Carbon?

[ecoli]

Yea thats what I was thinking too but why

 

I mean bigger things seem to be more stable than small things for starters' date=' a planet is quite stable for example.

 

What is it about the mass that makes it act like that if you know what I mean

 

Cheers,

 

Ryan Jones[/quote']

 

After doing a google search, almost all of the websites I cam up with just said something along the lines of, "the nucleus is too largeto stabilize itself"

 

One site compared it to "a pile of bricks that gets too large, topples over."

 

not very satisfactory answers, IMO.

[RyanJ]

If it was purely mass then wouldn't an isotope of Helium be just as stable as Carbon?

 

Maybe... perhaps it has something to do with the replusions of the subatomic particles in the nucleus too because it seems that mass alone is not the cause?

 

@ecoli: Wierd answers, a pile of bricks will not always fall over, a house is a pile of bricks and it likes to stay where it is without falling

 

Cheers,

 

Ryan Jones

[woelen]

Not shure if this is better here or in the physics section but I guess it kind of fits into both of them!

 

Why is it a certain ratio of neutrons to protons is stable and another one is unstable? What makes it unstable because the Neutrons contribute no charge so why would they be unstable in that way?

 

Cheers' date='

 

Ryan Jones[/quote']

There are more forces into play. There are also so-called weak nuclear forces, which have a very peculiar characteristic. Only at certain configurations, these lead to stable configurations. Unfortunately you cannot simply say that a certain configuration is stable and another is not. In the nucleus of an atom, there is a very delicate interaction between electrostatic forces and weak nuclear forces, which keeps a nucleus together. A slight unbalance makes the system as a whole unstable.

The mass (gravitational pull) plays no role at this scale. This force is negligible.

 

It is not a matter of continuous physics. Because of the small scale of the particles, only certain discrete configurations are stable and it is just as with the electronic configuration of atoms, certain nuclear configurations are more stable than others. Inside a nucleus, there also are orbitals, but now for quarks (or protons and neutrons, I'm not 100% sure about this) instead of electrons and certain configurations are stable and others are unstable, just as that certain molecular orbitals are unstable and lead to unstable molecules and other molecular orbitals are stable and lead to stable molecules.

 

For more detailed answers, I would like to refer to a physics expert.

[timo]

If I remember correctly, the Bethe-Weizsäcker Formula is pretty good at describing the energies of different nuclei. Most of the appearing terms should be rather easy to understand as it's mostly a classical model.

 

EDIT: I already had a strange feeling writing above because I remembered there was some trapdoor with Bethe. So I looked it up ... it´s the Bethe-Weizsäcker formula I meant, not the Bethe-Bloch formula. The first Wikipedia entry describing a bit about it was http://en.wikipedia.org/wiki/Semi-empirical_mass_formula

And in case the connection from a formula for the energy of a system to decays of nuclei isn´t obvious: You can use it to verify whether a nucleus would lose or gain energy when it was changing to another isotope by either absobing an electron or by releasing one (neutrinos left unconsidered).

[RyanJ]

I see, thanks.

 

So it is more related to forces than anything else... interesting and from your answer it also seems that one ratio will not fit all elements an isotopes, also very interesting More research too be done for me

 

Cheers,

 

Ryan Jones

[swansont]

The protons and neutrons in a nucleus have discrete energy levels. The neutron well is generally deeper (and more so for larger nuclei) because there is no electrostatic repulsion, so that's why there tend to be be more neutrons, and more so as you get larger nuclei - you want the wells filled to the same level. If there is a vacant energy level in the "other" well that is lower energy so that the resulting nucleus is at a lower energy (the proton, having a smaller mass, requires a small minimum difference), then the particle will undergo decay.

 

As has been said, you have to solve for the exact configuration to see if it would be stable or not, but there are some guidelines. Paired-up particles tend to be in a lower energy configuration, so a nucleus with an odd number of both neutrons and protons will likely be unstable (true for all nuclei larger than N-14). Even-even nuclei are more likely to be stable than even-odd nuclei. And nuclei with a filled shell of either or both is more likely to be stable (filled shells occurring at 2, 8, 20, 28, 50, 82 and 126 particles)

 

Large nuclei tend to decay because they're big - the nuclear force is finite-ranged, so the attractive force a nucleon feels doesn't keep growing with larger nuclei, but the repulsive force protons feel does (because that is an infinite-range force). So larger nuclei don't tend to be as tightly bound, and can get to a lower energy state by ejecting particles. Emitting alphas is a very efficient way of doing this, and a He-4 nucleus is very tightly bound (doubly magic)

[RyanJ]

Swansont, that makes sence - thanks!

 

And thanks everyone you've been a great help in explaining this too me, I understand now though I intend to do more research into the ratios.

 

Cheers,

 

Ryan Jones

[CanadaAotS]

what about "neutronium", as in what neutron stars are made of... apparently the neutrons are effected by degenerate neutron pressure

 

does this have anything to do with what RyanJ is talking about?

[woelen]

Neutron stars are very heavy and here the gravitational pull assures that the neutrons stay together. That is the dominant force for neutron stars, while gravitation does not play a role in nuclei, not even the largest ones.

[Goalfinder]

The protons and neutrons in a nucleus have discrete energy levels. The neutron well is generally deeper (and more so for larger nuclei) because there is no electrostatic repulsion' date=' so that's why there tend to be be more neutrons, and more so as you get larger nuclei - you want the wells filled to the same level. If there is a vacant energy level in the "other" well that is lower energy so that the resulting nucleus is at a lower energy (the proton, having a smaller mass, requires a small minimum difference), then the particle will undergo decay.

 

As has been said, you have to solve for the exact configuration to see if it would be stable or not, but there are some guidelines. Paired-up particles tend to be in a lower energy configuration, so a nucleus with an odd number of both neutrons and protons will likely be unstable (true for all nuclei larger than N-14). Even-even nuclei are more likely to be stable than even-odd nuclei. And nuclei with a filled shell of either or both is more likely to be stable (filled shells occurring at 2, 8, 20, 28, 50, 82 and 126 particles)

 

Large nuclei tend to decay because they're big - the nuclear force is finite-ranged, so the attractive force a nucleon feels doesn't keep growing with larger nuclei, but the repulsive force protons feel does (because that is an infinite-range force). So larger nuclei don't tend to be as tightly bound, and can get to a lower energy state by ejecting particles. Emitting alphas is a very efficient way of doing this, and a He-4 nucleus is very tightly bound (doubly magic)[/quote']

I agree it is the energy that keeps them together

refer nucleus by rutherford