Sciencemadness Discussion Board

Why is Hydrogen-1 stable?

Parabombastus - 31-12-2007 at 09:28

I know this is more in the realm of Physics but it does relate to Chemistry...

What holds the Hydrogen-1 nucleus "in place"? That is, why doesn't it move towards the electrons (I understand that the e- are restricted to a certain distance from the nucleus but what kind of force imposes this spearation? it must be working both ways) and why doesn't it leave the atom leaving it to disintergrate?

12AX7 - 31-12-2007 at 09:37

Well, they want to be together, because charges attract.

The reason the electron doesn't fall into a lower state is, there are no lower states. The electron, a fermion, excludes itself from reaching a lower state.

The atom could take a more compact form (p+ + e- --(W)--> n), but it's an extremely slow decay (due to the heavy weak-coupling particle and the small probability of the electron being in the nucleus), and not even favored, as free neutrons have a half-life of some 11 minutes, reverting back to p + e-.

Tim

I am a fish - 1-1-2008 at 06:52

The electron isn't trapped on the nucleus due to Heisenberg's Uncertainty principle. If the electron was trapped, then its position would be "known" to a very high degree of accuracy. By the uncertainty principle, this would lead to a great uncertainty in its momentum. This momentum would then allow the electron to escape.

This is a huge oversimplification, however, as when Heisenberg Uncertainty becomes significant, it becomes meaningless to speak of an electron being at a fixed location. It is a common misconception that Heisenberg's Uncertainty principle merely makes it impossible to measure both the position and momentum of a particle with complete accuracy. This implicitly assumes the so called "hidden variable" formulation of quantum mechanics (meaning that the numbers somehow exist, but just can't be read), which has been shown to be logically inconsistent (see Bell's Theorem). Therefore the electron can only be thought of in terms of a so-called wavefunction, which roughly speaking, gives the probability that electron is at a given point in space. The electron wavefunction in a hydrogen atom is centered around the proton, and is as tightly localized as the uncertainty principle allows it to be.

Interestingly, the peak of the wavefunction occurs within the nucleus. If this seems odd, it is because you are applying everyday perceptions to fundamental physics. It may seem obvious that two objects can't occupy the same space, but ultimately this is only due to electrical repulsion (if you try to push two objects into one another, the electrons on their surface will repel each other). On a fundamental level, there is nothing to say that particles can't overlap, and hence the electron will happily go "through" the proton.

Furthermore, it is possible to replace electrons in atoms with muons (particles which are very much like electrons, but about 200 times heavier). These orbit so tightly, that they do so almost entirely within the nucleus.

maniacscientist - 1-1-2008 at 11:43

The paradigma of the nucleus having not the chance of "0" by schrödinger is explained by the principle that the heisenberg principle has to be related to schrödingers equation which is the psi² , which is the relation of the
diameter of the ring(the cloud) to Heisenbergs uncertainty, given the nucleus being the zero in a co-ordinate system, the chance of the electron being in the nucleus is 0 even when factorised with the Heisenberg.


Theoretically, the electron could be anywhere at any place in the universe within a certain probability, the maximum being caculated by a "probabilitycloud" which also relates to the wavestructure of the electron, but as a particle, the e- could be anywhere in space, but the nucleus.(it´s a bit of an uncertain maybe wrong explanation, cause I don´t have the literature at hand right now, but I hope it makes it clear, that there´s two factors that have to be looked at, one which doesn´t explain the statistically chance of the electron by looking at the co-ordinate system but has to be related to another equation, as it doesn´t refer to the real world of atoms as a stand-alone theoreme)

I am a fish - 1-1-2008 at 13:44

Quote:
Originally posted by maniacscientist
The paradigma of the nucleus having not the chance of "0" by schrödinger is explained by the principle that the heisenberg principle has to be related to schrödingers equation which is the psi² , which is the relation of the
diameter of the ring(the cloud) to Heisenbergs uncertainty, given the nucleus being the zero in a co-ordinate system, the chance of the electron being in the nucleus is 0 even when factorised with the Heisenberg.


Theoretically, the electron could be anywhere at any place in the universe within a certain probability, the maximum being caculated by a "probabilitycloud" which also relates to the wavestructure of the electron, but as a particle, the e- could be anywhere in space, but the nucleus.(it´s a bit of an uncertain maybe wrong explanation, cause I don´t have the literature at hand right now, but I hope it makes it clear, that there´s two factors that have to be looked at, one which doesn´t explain the statistically chance of the electron by looking at the co-ordinate system but has to be related to another equation, as it doesn´t refer to the real world of atoms as a stand-alone theoreme)


This is complete and utter drivel.


The ground state wavefunction of the electron has a maximum at the nucleus. Integrating psi² over the nucleus's spatial extent gives a small but non-zero value. Therefore, the electron can be within the proton. Heisenburg's uncertainty principle doesn't prevent this from happening; instead, it explains why the electron doesn't get stuck there.

Co-ordinate systems have nothing to do with it. The laws of physics are completely independent of the type of co-ordinates that may or may not be used to describe them.

Furthermore, the Heisenburg uncertainty principle isn't something that has to be explicitly applied to quantum mechanical problems. Instead, it is an intrinsic and unavoidable part of wave-mechanics, which will follow from any derivation using (for example) the Schrödinger equation.

maniacscientist - 4-1-2008 at 13:18

Quote:
Originally posted by Parabombastus
I know this is more in the realm of Physics but it does relate to Chemistry...

What holds the Hydrogen-1 nucleus "in place"? That is, why doesn't it move towards the electrons (I understand that the e- are restricted to a certain distance from the nucleus but what kind of force imposes this spearation? it must be working both ways) and why doesn't it leave the atom leaving it to disintergrate?


The forcesa re: 1. the positive charge of the nucleus (one proton) attracts the negative charge of the electron.

The electron has a pulse, which is p=mv (m=mass= the energy of the electron in mV; v=velocity)

This pulse can be related to the frequency of the e- and the energy(0,5 mv²; m=eo), which can´t be lower than it is within the hydrogen.
these factors determine the chance of finding the electron in a certain space within the atom as time goes by which is determined by the Schrödinger equation psi², going from the wave function of the electron psi, which defines for every point of the function a position of the electron, to psi² which makes it a cloud, by the exponent.This cloud described the "chance-density" for the ring around the nucleus.

Now the probability density is _not_ the probability.
the real-world probability, in which one is most likely to find the electron, has a Volume-factor, which is described by another equation, having the radius of the electron to the nucleus as an exponent and the equation also includes the main count n of the element and all other (here n=1) and all other possible energy states of the electron.
Now for the volume element (n= 1,2,3) which makes a potential of the electron within the integrated radial form of the orbital, there is a zero at the nucleus.

But the nucleus is not a fixed, static point, with a defined middle. it has also a frequency, a wave function, by which it can be charachterised...


As I´m a fish states, it´s just a model and the nucleus has much more to it than just a positive charge or just a proton.



[Edit:] Once again, this is complete gibberish. Please refrain from pontificating about things you manifestly don't understand.

[Edited on 4-1-2008 by I am a fish]

MagicJigPipe - 4-1-2008 at 14:42

I am a fish, please explain why this is complete gibberish. Just saying so isn't enough.

Plus, more correct information can never hurt.

len1 - 5-1-2008 at 03:43

Thats indeed an interesting question. Before the advent of quantum mechanics in the 20s the stability of the atom was an enigma. If one assumes the electron orbits around the nucleus by classical mechanics, much as the earth around the sun, then being charged it should radiate, and eventually lose all its energy and fall into the nucleus.

Quantum mechanics produced a new set of rules which, if you accept the rules, explain why the atom is stable.

I agree with Fish in that the main reason - the centrifugal force if you like - why the electron doesnt 'fall in' is the uncertainty principle, although the question itself is not well posed, in the context of the atom 'falling in is meaningless'. However the way Fish used the uncertainty principle also doesnt correspond to the usual interpretation.

I shall explain in order.

The reason that the electron can not spend more time closer to the nucleus than it does, is that by the uncertainty principle it would be more localised, and therefore have more energy than it actually has - it would be a contradiction to the known energy of an electron in an atom, several electron volts. If you put this energy into the uncertainty principle delP del x > h you'll see the electron can not be localised tighter than an Angstrom - a factor 10^5 greater than the average nuclear radius.

Lets put it another way assuming you have an elementary knowledge of physics

an electron localised to radius r has P > h/r

it then has energy E > (h/r)^2/2m

its electrostatic energy pulling it towards the nucleus is e^/r

The confinement energy grows as 1/r^2 while the electrostatic goes as 1/r so eventually the uncertainty principle overtakes the electrostatic attraction, at the minimum allowable average radius these are equal

(h/r)^2/2m = e^2/r

solving for r gives whats known as the bohr radius. This is in fact what the Schrodinger equation does (adjusting for a small factor). Note all these numbers are averages - thats all QM can predict - there is a non zero probability for an electron to be as close to the origin as you like - though the prob for it to be at a point is zero.

The second point is regarding 'falling in'. The nucleus is not an apple, it has no surface. When one starts to talk about its size one necessarily becomes concerned with its structure. It turns out its made of elementary pointlike constituents (quarks) which have a probability density distribution with a mean radius that of the proton. But thats just a mean radius. In light of this talking about 'falling in' is meaningless. The quarks are point particles and the likelyhood of them and the electron occupyng the same point in space time is zero. All that would happen if the electron had a mean density distribution comparable to the proton radius is its interaction energy with the nucleus would increase 10^5 fold, for instance if it and the proton charge were upped 300 fold. This would of course make the fine structure constant about 1000 and quantum electrodynamics would become strong.

The muon is 200 times heavier than the electron and so its average radius in an atom would be root(200) closer than the electrons, unfortunately it lives only 10^-5 secs.

A propos what that other gentleman wrote, it is incoherent, there is no sense in interpreting it. Len

[Edited on 5-1-2008 by len1]

[Edited on 5-1-2008 by len1]

I am a fish - 5-1-2008 at 04:24

Quote:
Originally posted by len1
...the main reason - the centrifugal force if you like - why the electron doesnt 'fall in' is the uncertainty principle...


Centrifugal force is completely different. In fact the ground-state electron doesnt "orbit" at all, as it has zero angular momentum.
Quote:

...the way Fish used the uncertainty principle also doesnt correspond to the usual interpretation.


Quantum mechanics has no "usual interpretation". :) In what way am I wrong?
Quote:
The second point is regarding 'falling in'. The nucleus is not an apple, it has no surface. When one starts to talk about its size one necessarily becomes concerned with its structure. It turns out its made of elementary pointlike constituents (quarks) which have a probability density distribution with a mean radius that of the proton. But thats just a mean radius. In light of this talking about 'falling in' is meaningless. The quarks are point particles and the likelyhood of them and the electron occupyng the same point in space time is zero.


Even if you accept the (IMHO dubious) notion of point particles, there is no reason in principle that they could not occupy the same point in space. The probability of this happening is zero, but the happenstance would violate no physical rule. (Of course, currently speculative physical theories such as string theory, may eventually provide a reason why particles can't overlap.)
Quote:
Originally posted by MagicJigPipe
I am a fish, please explain why this is complete gibberish. Just saying so isn't enough.

Plus, more correct information can never hurt.


I explained my reasons in response to his first post. His second post seems to be little more than a rehash of his first. (Then again, it is so incoherent that it is difficult to tell.)

len1 - 5-1-2008 at 04:34

Quote:
Centrifugal force is completely different. In fact the ground-state electron doesnt "orbit" at all, as it has zero angular momentum.


Did you notice I used the term 'if you like'. What that means is we are not talking about the centripetal force, but something that takes its place in balancing the electrostatic energy. This is demosntrated in the calculation I give.
Quote:
in what way am I wrong


You mentioned the electron 'escaping'. That is non-standard parlance since it suggests the elctron in an atom has a trajectory, it doesnt.

This is the first time I meet a person who thinks the notion of point particles is 'dubious'. Youd be in line for the Nobel prise if ou can justify that.

I am a fish - 5-1-2008 at 09:50

Quote:
Originally posted by len1
You mentioned the electron 'escaping'. That is non-standard parlance since it suggests the elctron in an atom has a trajectory, it doesnt.


My very next words were "This is a huge oversimplification, however...". I think, however, that we're talking cross-purposes. When you said "...the way Fish used the uncertainty principle also doesnt correspond to the usual interpretation", I assumed that you were refering to my comments about hidden variables and Bell's theorem, when in fact you were referring to my earlier comments. I freely concede that what I said wasn't physically accurate, but I never claimed it was!

Furthermore, the explaination you give is also a toy model, as it implicity assumes the existence of other energy levels in which the electrostatic attraction and the Heisenberg "force" aren't balanced. Ultimately, quantum mechanics is beyond human intuition, and can only be adequately explained with mathematics.
Quote:
This is the first time I meet a person who thinks the notion of point particles is 'dubious'. Youd be in line for the Nobel prise if ou can justify that.


And you'd be in line for a Nobel prize if you could justify the existence of point particles. I'm not saying they're impossible, I just think that reality is probably more complicated. Despite your amazement, such an opinion is hardly novel (for a start, it is axiomatic to string theory).

len1 - 5-1-2008 at 15:51

Quote:
Furthermore, the explaination you give is also a toy model, as it implicity assumes the existence of other energy levels in which the electrostatic attraction and the Heisenberg "force" aren't balanced.


Bingo! Indeed 'my toy model' predicts other higher energy levels, because the H uncertainty principle only restricts the energy from below. These energy levels are labelled by the principle quantum number n, and are responsible for the rows of the periodic table, on which all the elements and properties of chemicals discussed on this forum are based. In the mathematics n comes out of the differential operator h^2/2m del^2 phi, with the del^2 acting on the wavefunction bringing out n^2. The lowest value n=1 gives the ground level I described in the previous post where the inequality of H is essentially an equality. I thank you for attributing the schrodinger eqn of the hydrogen atom to me, alas Im 80 years too late
Quote:
And you'd be in line for a Nobel prize if you could justify the existence of point particles.


Thanks for your kindd words. Alas Dirac has already beaten me to that


[Edited on 6-1-2008 by len1]

I am a fish - 6-1-2008 at 05:00

Quote:
Originally posted by len1
Bingo! Indeed 'my toy model' predicts other higher energy levels, because the H uncertainty principle only restricts the energy from below. These energy levels are labelled by the principle quantum number n, and are responsible for the rows of the periodic table, on which all the elements and properties of chemicals discussed on this forum are based. In the mathematics n comes out of the differential operator h^2/2m del^2 phi, with the del^2 acting on the wavefunction bringing out n^2. The lowest value n=1 gives the ground level I described in the previous post where the inequality of H is essentially an equality. I thank you for attributing the schrodinger eqn of the hydrogen atom to me, alas Im 80 years too late.


I wasn't refering to the other solutions of the Schrodinger equation. I was refering to the fact that in your explaination, you implicitly assume a continuous set of wavefunctions with varying degrees of localisation, and that then the true solution arises when the electrostatic attraction and Heisenberg uncertainty cancel out. However, these unbalanced (for want of a better word) states simply do not exist, and so the properties of the true wavefunction can't be derived by considering perturbations to them.

The excited states have nothing to do with it.
Quote:

Quote:
And you'd be in line for a Nobel prize if you could justify the existence of point particles.


Thanks for your kindd words. Alas Dirac has already beaten me to that


Please cite a reference. (And whilst you're at it, please forward it to all those stupid string theorists who have presumably overlooked it also.) Yes, the electron being a point particle is assumed, but there is no fundamental justification for this. Furthermore, it is not even known for certain that the electron lacks internal structure and is not itself composed of more particles.

len1 - 6-1-2008 at 05:34

Im sorry, I assume no such thing as you say. What I wrote is a standard interpretation of the schrodinger equation which in my extensive teaching of the subject I dont recall having problems getting understood. Lacking further detail of what you are thinking I cant help.

Regarding Dirac his seminal paper has constributed more in the way of verified experimental predictions than string theory, which has had no experimental success. Certainly all the verified predictions of quantum theory have been on the basis of point particle theories. I think its a bit loud to call these theories 'dubious' unless you can improve on what these theories have achieved.

I am a fish - 6-1-2008 at 10:39

Quote:
Originally posted by len1
Im sorry, I assume no such thing as you say. What I wrote is a standard interpretation of the schrodinger equation which in my extensive teaching of the subject I dont recall having problems getting understood. Lacking further detail of what you are thinking I cant help.


In your original post, you equate the confinement force (due to the uncertainty principle) to the Coulomb attraction, and then solve for r. Therefore, r must be free to move (otherwise, you will have an overdetermined system). What do the out-of-equilibrium states (in which the forces aren't balanced, and r differs from the Bohr radius) correspond to?

If, for example, you wish to find the equilibrium position of a mass hanging from a spring (using classical physics), you will proceed by finding expressions for the the spring's tension and the mass's weight. By equating them, and then solving for position, you can find the equilibrium position. This procedure is only valid, because the mass is free to move over a continuum of positions, and hence the position is a free parameter.
Quote:
Regarding Dirac his seminal paper has constributed more in the way of verified experimental predictions than string theory, which has had no experimental success. Certainly all the verified predictions of quantum theory have been on the basis of point particle theories. I think its a bit loud to call these theories 'dubious' unless you can improve on what these theories have achieved.


The theories have undoubtedly achieved great experimental success, but they also have a fundamental flaw, namely the incompatibility with general relativity. String theory (which as you rightly point out is highly speculative, and has little empirical backing) is (nevertheless) one of the most promising ways of resolving this problem.

I am not for a moment bashing the great achievements of Dirac and others. I am merely pointing out (the completely noncontroversial fact) that reality could well be more complicated. Also, please try to appreciate that by "dubious", I am not trying to say "badly thought out". Instead, I simply mean "may not be true", and nothing more. I think that point particles are possible, but I also think other possibilities are just as likely.

Also, there is a big gulf between saying that an assumption may not hold true at a fundamental level (which I did), and denying that Dirac provided a profound insight into nature (which I most certainly did not). Similarly, if I pointed out the limits to Newtonian gravity, I would be in no way questioning the greatness of Newton. I have nothing but admiration for Dirac and others, but that is no reason to turn their word into unimpeachable law.

len1 - 6-1-2008 at 14:23

I understand what you had in mind now, but that is not how I wanted what I was saying to be understood. By ballancing out of opposing forces I meant a way to converge to a solution of the qm equation. Your variation of r in the mass problem then corresponds to a variation of the wavefunction of the electron in the atom. By varying that function you are not assuming that the real wavefunction can really take all the forms you are using in your variation - these are completely arbitrary. But the end result - the minimum - is not, its the actual ground state of the electron

Lets look at the details - the famous schrodinger eqn of qm

coulomb energy + kinetic energy (constrained by uncertainty principle) = total energy

e^2/r psi - h^2/2m del^2 psi = E psi

We need a solution for psi. Can we say something about it without solving the equation (which introduces mathematical difficulties being 3-dimensional)? Yes. Spherically symmetric psi (called the s wavefunction) will have some mean radius R. For the ground state E is not large, we can negect it without a large error. So

e^2/r psi(r) ~ h^2/2m del^2 psi(r)

At this stage you can substitute normalised trial wavefunctions of characteristic radius R, say a exp(-r/R), take averages by multiplying by psi*(r) integrating, and solving for R. But you can gauge the result even without doing that.

e^2/R ~ h^2/2mR^2 this gives us R = h^2/2me^2

This is how what I was saying relates to the actual math. The electron can not be constrained around the mean radius of the proton because the RHS which comes from the uncertainty principle is much larger than the LHS, the coulomb attraction, at these radii. However the H uncertainty energy drops off more rapidly than coulomb energy with increasing R and theres an optimum solution, the one you get from solving for R above.

[Edited on 6-1-2008 by len1]