How Long can Atom live ?

[Solve]

If an Atom is kept away from any form of external influence then how long it will live ? forever ?

[Klaynos]

What do you mean by live?

 

Have a read of:

 

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

[ajb]

Think about the hydrogen atom in particular. This consists of a proton and an electron. We can have the reaction

 

[math]p^{+} + e^{-} \rightarrow n + \nu_{e}[/math],

 

which is known as electron capture or inverse beta decay.

 

The question: Is the ground state of the hydrogen atom stable?

[swansont]

It also depends on which atom.

[CaptainPanic]

Since the reverse reactions also exists, there is an equilibrium situation.

 

You can also ask yourself if water is stable.

If you know that in any glass of water, billions of molecules of water dissociate into a hydronium and a hydroxide ion every microsecond, then it doesn't sound very stable... only the reverse reaction where water is formed again proceeds just as fast. Does that mean that water is stable?

 

Call me pedantic, but in this case it's worth the effort to define the exact question first... and that might mean you even have to define "stable".

[timo]

I don't really see what you're missing CaptainPanic? "If an Atom is kept away from any form of external influence then how long it will live ?" seems to be a well-defined question to me (and I am pedantic). The answer is, as Tom said, that it depends on the atom. While some atoms decay very fast (->radioactivity), there are atoms that will live forever - at least as far as I know.

[swansont]

What it basically boils down to is whether there is a lower-energy configuration available to the system. If not, the system should be stable.

[CaptainPanic]

I don't really see what you're missing CaptainPanic? "If an Atom is kept away from any form of external influence then how long it will live ?" seems to be a well-defined question to me (and I am pedantic). The answer is, as Tom said, that it depends on the atom. While some atoms decay very fast (->radioactivity), there are atoms that will live forever - at least as far as I know.

I responded to this post:

 

Think about the hydrogen atom in particular. This consists of a proton and an electron. We can have the reaction

 

[math]p^{+} + e^{-} \rightarrow n + \nu_{e}[/math],

 

which is known as electron capture or inverse beta decay.

 

The question: Is the ground state of the hydrogen atom stable?

I definitely stepped outside of my own field of expertise here... but I'll describe what I meant to say:

The post of ajb describes a reaction, and a reverse reaction. That always means that there is an equilibrium - which might however be completely at one side, completely at the other. Note that this says nothing about the rate (or frequency) of this reaction. It only deals with the relative frequencies of the two reactions.

 

If the beta decay and inverse beta decay can both occur, then we have an equilibrium situation. That means that the hydrogen atom will exist for some time as a hydrogen atom, and for some time as a neutron. And it can theoretically go back and forth (although it might not happen too often).

 

The question is then, do you call this "stable" if the hydrogen atom spends 99% of its lifetime as a hydrogen atom, but jumps back and forth several times... or is it unstable then?

 

I realize that this is more relevant in chemisty than in nuclear physics.

[timo]

I see. I though you were referring to the original question. Ajb's statement was a bit weird - while it's true that in principle excited states can contain enough energy to undergo a transition that wouldn't be available to the ground state, this doesn't apply to the example given, where a few hundred keV would be needed and only up to 13.6 eV are available from excited states.

 

In particle physics, lifetime usually refers to a scenario where the object is at rest and not influenced from the outside. That includes no thermal fluctuations, i.e. a constant-energy and not a temperature-based scenario. Incidently, that's pretty much what the OP was asking for. Since energy is conserved and no kinetic energy is available, this directly means that anything can only decay into decay products that have the same or less mass - the decay product may have momentum, so a energy difference in mass (E=mc²) can be compensated by kinetic energy of the final-state products. This directly means that objects for which no final state with a lower (summed up) mass than the current one exists cannot decay into anything and are stable. Such a case is the Hydrogen atom.

 

The question to what extent that scenario is realistic is indeed a different question. In principle, the question about anything having an infinite lifetime has a very simple answer once you allow for interference from the outside: anything can be destroyed if you kick it hard enough. That's not a very useful statement in practice.

What you presumably are very used to is thermal equilibrium situations. In principle, you can get arbitrary energy kicks there, too (except for T=0). Thermal equilibrium is a very important and powerful concept (as a matter of fact I recently published a paper where I propose that structures expected in living systems can be understood from the static equilibrium case). It is, however, not always appropriate. Particularly in physics, you often have processes that happen on entirely different time scales (from chemistry you might know the Born-Oppenheimer approximation). In such a case, thermal equilibrium might not be the correct concept for at least part of the process.

Take for instance a bit of Uranium. The motion of the atoms is nicely equilibrated with the surrounding - average kinetic energy is constant. However, the ratio of radioactive (A) and already-decayed (D) Uranium atoms is not equilibrated with the surroundings. Instead of maintaining a constant number of As, you have a constantly-decreasing number of them. With respect to the A/D ratio, the system is far from thermal equilibrium and instead relaxing towards it. On a time-scale that a Chemist would most likely call "very slowly", to say the least. To close the circle of my statement: this system of decaying Uranium is an example where the particle physicist's understanding of lifetime seems like a more appropriate concept to use than thermal equilibrium. The decay of the Uranium is (afaik) largely unaffected by putting it in a fridge or in some solvent - on human timescales, that is.

 

In short: Thanks for clarification. While equilibrium is a powerful concept, it's not what was asked for.

Edited February 3, 2011 by timo

[ajb]

Ajb's statement was a bit weird - while it's true that in principle excited states can contain enough energy to undergo a transition that wouldn't be available to the ground state, this doesn't apply to the example given, where a few hundred keV would be needed and only up to 13.6 eV are available from excited states.

 

Ok, I was thinking out loud.

 

Is it clear from the Schrödinger equation that the ground state of hydrogen is stable? I honestly am not really sure about this. (This is more a question about analysis than physics I expect.)

[swansont]

A lone proton will not spontaneously change into a neutron via beta decay, since the neutron has more mass. Even the electron capture, which has a smaller mass discrepancy, is not energetically favored.

[timo]

My assumption that the Hydrogen is stable is indeed partly intuitive. In particular: I cannot think of any lower-energetic system. I'm not fully sure what you mean with being clear from the Schrödinger eq. Today's mainstream physics (the SM) and a proper mathematical treatment of QM do not cooperate very well, as far as I know.

 

- The energy levels of the standard Hamiltonian (the one always being asked for in diploma exams) have a minimum, which is the ground state. I faintly remember that the solutions for a relativistic electron (i.e. based on the Dirac equation of motion) can't be written down that easy anymore and must be approximated graphically/numerically, but still preserve the structure of discrete levels with a minimum. I don't know enough QM to say something about a full-fledged QED Hydrogen or even an SM hydrogen.

- If you're assuming BSM physics in which the proton decays, say P -> e+ pi+, then the hydrogen can decay rather trivially (from the perspective of a physicist who just ignores the presence of the electron for the decay of the nucleus).

- In SM physics, it boils down to the fact that I cannot think of a lower-energetic state. The first idea would indeed be moving the electron's charge to the proton. But the resulting neutron+neutrino state is higher in energy. For the proton, there is not that much you can do with it, either. All Feynman vertices I can think of at the moment conserve the total number of quarks. So your final state would in some way have to have a total charge of zero, three net quarks, and one net lepton. From the physicists perspective, you could just go through all combinations of known particles that satisfy this criterion (I have to admit didn't do that). That of course heavily relies on the (somewhat reasonable) assumption that the particles in the sought-for final state are already known. For a rigorous treatment I think you're already stuck because of not being able to handle QCD bound states.

[Solve]

<br />What do you mean by live?<br /><br />Have a read of:<br /><br /><a href='http://en.wikipedia.org/wiki/Stable_isotope' class='bbc_url' title='External link' rel='nofollow external'>http://en.wikipedia..../Stable_isotope</a><br />

<br /><br /><br />

 

There are confusions in my mind.

 

Because i have two thoughts.

 

1. There are several phenomenon in the universe which externally maintain the integrity called "Atom" . Outta those phenomenons only very few are within the grasp of Humans and their technology such as Energy and Mass. And If the integrity called "Atom" is deprived from all those external phenomenons (known and unknown) then it will collapse.

 

2. OR, The Integrity called "Atom" is isolated can live forever and doesn't need any external influence to sustain.