Sciencemadness Discussion Board - electrons in "S" orbital

electrons in "S" orbital

aeacfm - 9-11-2011 at 05:32

this part from wikipedia

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

Quote:

In a similar way, all s electrons have a finite probability of being found inside the nucleus, and this allows s electrons to occasionally participate in strictly nuclear-electron interaction processes, such as electron capture and internal conversion.

did the meaning of this that "S" electrons pass through the nucleus ?
if yes , isn't this mean the end of atomic system ?

thanks in advance

ScienceSquirrel - 9-11-2011 at 05:38

The probability that the electron is close to the nucleus is vanishingly small. Most of the charge of the electron is held within a small spherical shell around the nucleus.

kmno4 - 9-11-2011 at 06:07

Quote: Originally posted by aeacfm

(...)
did the meaning of this that "S" electrons pass through the nucleus ?
if yes , isn't this mean the end of atomic system ?

Yes, they "pass". You can find electron even in the center of nucleus. It is pure quantum effect and chemists do not have to bother with it.

No. Atoms exist with or without our chemistry, mathematics or quantum mechanics. Whatever model you choose - it is only model. We cannot imagine what exactly is an atom and what happens there.

BTW.
Electron density (for s shell) rises to nucleus - the closer, the larger concentration of electronic charge.
But it is something different than radial distribution function.

aeacfm - 9-11-2011 at 06:20

Quote: Originally posted by ScienceSquirrel

The probability that the electron is close to the nucleus is vanishingly small. Most of the charge of the electron is held within a small spherical shell around the nucleus.

thats what i was thinking first , but digging in the wiki i dont think so

aeacfm - 9-11-2011 at 06:27

Quote: Originally posted by kmno4

i know it is not maths or imagination which control all this , but trying to find out general concept of what is going there ....

so you mean this is a special property for S orbital and quantum mechanics

turd - 9-11-2011 at 06:55

Quote: Originally posted by ScienceSquirrel

The probability that the electron is close to the nucleus is vanishingly small. Most of the charge of the electron is held within a small spherical shell around the nucleus.

The way you put it is not correct.
The nucleus is the most likely place to find an s-electron - the density function has a maximum there. Just look at any of those crystal structure models - all of them have the maximum electron density at the centre of the atoms.

What you are probably thinking about is the radial density distribution. Of course it is 0 for d=0, because you multiply with the surface of a sphere, which is 0. Of course a zero-volume has a zero-probability of containing the electron. That is true for every point in space. Your sentence is therefore tautological and uninteresting.

Think of electrons not as marbles but as fields with interaction probabilities and the mental problems will resolve. I mean think of neutrons: they have mass and go right through a lot of materials that would simply stop photons.

PS: There is already a thread on this and it is terrible. Lots of morons with at most superficial knowledge acting like Nobel price quantum chemists (me included).

ScienceSquirrel - 9-11-2011 at 07:13

Sorry, I should have said probability density. The probability that the electron will interact with the nucleus is very small because it is very seldom 'there'.
What we have is models that sort of describe something that we cannot see and is outside our direct experience.
I can imagine what it is like to be a cat but I cannot step inside my cat's mind and body and really experience it. However I know that it would be bloody weird!

aeacfm - 9-11-2011 at 07:31

Quote: Originally posted by turd

The nucleus is the most likely place to find an s-electron - the density function has a maximum there. Just look at any of those crystal structure models - all of them have the maximum electron density at the centre of the atoms.

yeah thats supported by the following text in wiki here

Quote:

A mental "planetary orbit" picture closest to the behavior of electrons in s orbitals, all of which have no angular momentum, might perhaps be that of the path of an atomic-sized black hole, or some other imaginary particle which is able to fall with increasing velocity from space directly through the Earth, without stopping or being affected by any force but gravity, and in this way falls through the core and out the other side in a straight line, and off again into space, while slowing from the backwards gravitational tug. If such a particle were gravitationally bound to the Earth it would not escape, but would pursue a series of passes in which it always slowed at some maximal distance into space, but had its maximal velocity at the Earth's center (this "orbit" would have an orbital eccentricity of 1.0). If such a particle also had a wave nature, it would have the highest probability of being located where its velocity and momentum were highest, which would be at the Earth's core. In addition, rather than be confined to an infinitely narrow "orbit" which is a straight line, it would pass through the Earth from all directions, and not have a preferred one. Thus, a "long exposure" photograph of its motion over a very long period of time, would show a sphere.

i can't get the idea here

Quote:

What you are probably thinking about is the radial density distribution. Of course it is 0 for d=0, because you multiply with the surface of a sphere, which is 0. Of course a zero-volume has a zero-probability of containing the electron. That is true for every point in space. Your sentence is therefore tautological and uninteresting. Think of electrons not as marbles but as fields with interaction probabilities and the mental problems will resolve. I mean think of neutrons: they have mass and go right through a lot of materials that would simply stop photons.

annaandherdad - 9-11-2011 at 11:47

In quantum mechanics the state of the electron is described by a complex wave function psi(x), where x is the 3-dimensional position of the electron. The square of the wave function, |psi(x)|^2, is the (nonnegative) probability density for finding the electron at various locations in space, when measurements are made on an ensemble of systems that are all in the same quantum state.

When the atom is in a state of definite energy, the probability density |psi(x)|^2 is independent of time. In the case that the electron is moving in a central force field, a wave function psi(x) that has a definite energy also has a definite angular momentum, designated by the symbols s,p,d,f, etc, which correspond to l (lower case L, the quantum number of angular momentum) equal to 0,1,2,3, etc. These are s-waves, p-waves, etc.

The wave function psi(x) in these cases is a product of a radial wave function R(r) times a spherical harmonic Y_{lm}(theta,phi), which gives the angular dependence. The radial wave function behaves as a constant times r^l near r=0, that is, the Taylor series expansion of R(r) about r=0 begins with the term r^l. Thus s-waves go like r^0=1, p-waves like r^1=r, d-waves like r^2 etc near r=0.

Someone said s-waves are maximum at r=0; that is not necessarily true, it depends on the potential and the value of l, but s-waves do have a *local* maximum at r=0 (this is just because the derivative of the constant is 0, so the slope of R(r) is 0 at r=0 in the case of s-waves).

Someone also said that the electron is most likely to be found at r=0 (or inside the nucleus) for s-waves; that is certainly not true, because the probability of finding the electron inside the nucleus is the *integral* of the probability density over the volume of the nucleus, which is very small compared to the total probability in all cases, due to the small size of the nucleus compared to the size of the atom.

What is true is that the probability of finding an s-wave electron inside the nucleus is much larger than the probability of finding a p-wave or any other value of l>0 inside the nucleus. All waves with l>0 go to zero at r=0, more and more rapidly as l increases, so the integral of the probability density over the volume of the nucleus is very small for such waves. Usually as a practical matter one only needs to consider s-waves when interactions with the nucleus are concerned.

The original question seemed to ask how it is possible for an electron to be inside a nucleus. There is no problem with this. It is not even possible to assign a size to an electron, which as far as anyone knows is a point particle. The electron *wave function* has a size, but not the electron itself. (I know we can argue about the "classical radius of the electron").

Nuclei are not hard spheres, they are made up of protons and neutrons. And protons and neutrons are not hard spheres, either, they are made up of quarks. And quarks, like electrons, are, as far as anyone knows, point particles.

turd - 9-11-2011 at 12:59

Yes, we're proud of you - you've taken an introduction to quantum chemistry.

Quote: Originally posted by annaandherdad

Someone also said that the electron is most likely to be found at r=0 (or inside the nucleus) for s-waves; that is certainly not true, because the probability of finding the electron inside the nucleus is the *integral* of the probability density over the volume of the nucleus, which is very small compared to the total probability in all cases, due to the small size of the nucleus compared to the size of the atom.

Jeeze.

Since the probability density is at a maximum at the nucleus (let's talk 1s for simplicity) - it is the volume element with the highest probability of finding an electron there among all volume elements with the same volume. Ergo the nucleus is the most likely place to find an electron. Duh.

So yes, finding an electron in dV at r=0 is more likely than finding it in a (one of the many) dV at r<>0.

Of course the radial probability density must be 0 at r=0, because you're giving the density of a surface at a point.

kmno4 - 9-11-2011 at 14:32

http://www.everyscience.com/Chemistry/Inorganic/Atomic_Struc...

turd - 9-11-2011 at 14:57

Quote: Originally posted by kmno4

http://www.everyscience.com/Chemistry/Inorganic/Atomic_Struc...

Yes, this site confirms what I say - the probability density is highest at the nucleus, therefore this is the place you will most likely find an electron.

But ... in the friggin' first sentence I read on that page there was a glaring mistake:

Quote:

Another important function we need to consider if the Radial Distribution Function, Pnl(r). This is defined as the probability that an electron [...] will be found at a distance r from the nucleus.

W-T-F? The probability of finding an electron at a fixed distance r from the nucleus is ... zero! The points with a fixed distance from the center build a sphere - an object with 0 volume, therefore the integral over the probability density function is 0.

The internet sucks. How are people supposed to learn from with such fundamental mistakes - I'm going to cry now. Good bye.

aeacfm - 9-11-2011 at 15:10

Quote: Originally posted by annaandherdad

Nuclei are not hard spheres, they are made up of protons and neutrons. And protons and neutrons are not hard spheres, either, they are made up of quarks. And quarks, like electrons, are, as far as anyone knows, point particles.

logic

aeacfm - 9-11-2011 at 15:12

Quote: Originally posted by turd

Jeeze.

great shoot
really fabulous

annaandherdad - 9-11-2011 at 16:29

Hey, turd, you're right, if your statement, "The nucleus is the most likely place to find an s-electron" means that the probability density is highest at the nucleus (probability per unit volume).

But it certainly can be interpreted as meaning that if you measure the position of the electron, you're most likely to find it inside the nucleus. That is not true.

By the way, I've never taken a course in introductory quantum chemistry.

watson.fawkes - 9-11-2011 at 20:19

Quote: Originally posted by turd

Quote:

Another important function we need to consider if the Radial Distribution Function, Pnl(r). This is defined as the probability that an electron [...] will be found at a distance r from the nucleus.

This isn't a mistake, although they use the standard imprecise language physicists are notorious for. P<sub>n,l</sub> is a probability density function. Physicist insist on calling probability densities "distributions", a habit that goes back to the 19th century and the early work in statistical mechanics. The flaw you pointed out is another standard bit of imprecise language, saying that a density represents "probability at". Strictly speaking, probability is for events, which are integrals of density. The right was to say this is that it is the "probability density that an electron will be a distance r", but I don't see much of that precision usage in the physics world.

aeacfm - 10-11-2011 at 01:00

i 've read a sentence may make me understand more

http://www.uwgb.edu/dutchs/Petrology/WhatAtomsLookLike.HTM

Quote:

the electron occupies the whole orbital at once. It does not travel a circular path.

so electron found every where in the orbital at same time , when "S" orbital spherical around nucleus it could found in the nucleus also at the same time

may be i am wrong ???still confused

turd - 10-11-2011 at 02:13

Quote: Originally posted by annaandherdad

By the way, I've never taken a course in introductory quantum chemistry.

Heh... sorry. I was a bit cynical yesterday, because these orbital/schmorbital threads always end up the same and never progress.

Quote:

This isn't a mistake, although they use the standard imprecise language physicists are notorious for.

Since I'm not part of the physics community, I take your word for it. I just know that in my field many referees would give you a swift kick in the nuts (or other place that hurts) for such a fuzzy usage of mathematical language.

watson.fawkes - 10-11-2011 at 04:03

Quote: Originally posted by turd

Well, I'm with you. Unfortunately, there's barely a physicist out there that doesn't consider this ordinary language, and the referees for physics papers are physicists. There's positive scorn for mathematical rigor in the physics world. Decades of that attitude led to string theory, but that rant is getting too off-topic.

The real point I want to make is that if you're going to read the writings of physicists, for completely practical purposes you've got to put up with their bullshit lack of rigor and hazy language.

phlogiston - 10-11-2011 at 05:58

You are all ignoring the fact that the probability density functions as discussed in nearly every quantumchemistry textbook is based only on the electrostatic forces between the nucleus and the atom. The actual probability distribution in the vicinity and inside the nucleus will be heavily influenced by other forces acting on the electron, but this is of little consequence to chemistry because the electronic interactions between atoms (ie the field of chemistry) is dictated mostly by overlap of the orbitals quite distant from the nucleus.
It may well be that the actual probability of an electron being found inside the volume of the nucleus is minuscule, if all forces are taken into account.

turd - 10-11-2011 at 07:35

Quote: Originally posted by phlogiston

You are all ignoring the fact that the probability density functions as discussed in nearly every quantumchemistry textbook is based only on the electrostatic forces between the nucleus and the atom. The actual probability distribution in the vicinity and inside the nucleus will be heavily influenced by other forces acting on the electron [...]
It may well be that the actual probability of an electron being found inside the volume of the nucleus is minuscule, if all forces are taken into account.

Please be more precise. What forces are you talking about?

I think what people are missing is that the whole discussion on electron density at the core is a red herring. This is not the reason electrons don't "react" with protons. According to a quick google search, the actual reason is that you need lots of energy to make a neutron from an electron and a proton. Energy which core electrons usually do not have. Only in special cases like certain radioactive elements electron capture is observed: http://en.wikipedia.org/wiki/Electron_capture

I hope this answers the original question even though I have no idea of what I'm talking about.

Edit: typo

[Edited on 10-11-2011 by turd]

annaandherdad - 10-11-2011 at 08:17

Quote: Originally posted by aeacfm

i 've read a sentence may make me understand more

http://www.uwgb.edu/dutchs/Petrology/WhatAtomsLookLike.HTM

Quote:

the electron occupies the whole orbital at once. It does not travel a circular path.

so electron found every where in the orbital at same time , when "S" orbital spherical around nucleus it could found in the nucleus also at the same time

may be i am wrong ???still confused

If you make a measurement of the electron's position, you find it at one place. But if you make a series of measurements on an ensemble of atoms all in the same quantum state, you get a distribution of positions, that is described by the wave function. The predictions of quantum mechanics are fundamentally statistical. See http://bohr.physics.berkeley.edu/classes/221/1112/notes/post... for some notes on this subject.

In the quote you cite, where it says that the electron is found everywhere in the orbital at the same time, is rough language for the fact that the distribution of electron positions, measured in an ensemble of atoms in the same state, is described by the orbital. Here "orbital" means the same as "wave function", whose square is the probability density.

Although a measurement of the position of the electron produces a single value, it does not make sense in quantum mechanics to talk about the trajectory or orbit of the electron.

One can, however, take the energy and other quantum numbers of a quantum state, and ask what the trajectory would be for the same value of energy and other observables, if classical mechanics were valid. If you do that, the classical orbit corresponding to an s-wave would be a needle-like orbit passing through the nucleus (the electron makes a hair-pin turn right at the nucleus). For a given energy, the orbit with maximum angular momentum (for example, a d-wave for n=3 in hydrogen) corresponds to a circular orbit in classical mechanics.

The needle-like classical orbit (for an s-wave) does not look at all like the wave function, which is spherically symmetric. The reason is that the s-wave wave function is a precise state of angular momentum, so the angles (ie the orientation of the orbit) are completely undetermined. This is an example of the uncertainty relations. Effectively, the needle-orbit is averaged over all angles (all directions) to produce a spherically symmetric distribution.

annaandherdad - 10-11-2011 at 08:36

Quote: Originally posted by turd

Quote: Originally posted by phlogiston

I explained above that the principal interactions of the electron inside the nucleus are electrostatic, and that the weak interactions are much much smaller and usually negligible. The reaction you refer to (inverse beta decay) e+p -> n + nu, requires an electron energy of roughly an MeV for free protons (it is the p-n mass difference, times c^2). But for protons inside a nucleus it may require less energy, which is what makes electron capture possible in some atoms (with some isotopes of some nuclei). In any case, this is a weak interaction.

But even if this or other reactions with nucleons are impossible because of insufficient energy, still the weak interactions with the nucleons change the electron wave function, by a very small amount. For example, it causes the electron wave function not to be an eigenstate of parity, at a very small level. There is an active experimental program going on at various places measuring weak interactions in atomic physics, and for these experiments knowing the value of the electron wave function inside the nucleus is important.

It is almost never important for chemistry, although there have been researches on parity-violating effects in molecules as a possible explanation for the observed asymmetry in chirality in biological systems. (It seems that the weak interactions do not explain the asymmetry.)