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Topic: electrons in "S" orbital  (Read 6262 times)

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Offline aeacfm

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electrons in "S" orbital
« on: November 09, 2011, 08:22:53 AM »
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

Offline Schrödinger

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Re: electrons in "S" orbital
« Reply #1 on: November 09, 2011, 10:06:14 AM »
Yeah, I think this is like one of those mathematical results that might not actually make sense. I think there is some sort of 'correction' for this..
"Destiny is not a matter of chance; but a matter of choice. It is not a thing to be waited for; it is a thing to be achieved."
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Offline aeacfm

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Re: electrons in "S" orbital
« Reply #2 on: November 09, 2011, 10:22:41 AM »
Yeah, I think this is like one of those mathematical results that might not actually make sense. I think there is some sort of 'correction' for this..

or you can say correct , but can't be imagined like other quantum theories ( thats what i think) !!!

Offline Enthalpy

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Re: electrons in "S" orbital
« Reply #3 on: November 11, 2011, 08:15:34 PM »
It is absolutely correct and sensible and imaginable.

S orbitals have their maximum probability density right at the nucleus, sure.
This is what makes radioactive decay by electron capture possible, for instance. It works only with S electrons because their probability density is nonzero at the nucleus, and usually with 1S electrons because these are more concentrated at the nucleus.

Non-S orbitals have a momentum, which means the phase of the wavefunction makes not zero turns on a path around the nucleus (when choosing the functions with an amplitude of circular symmetry). So at the nucleus, the wavefunction has all phases at the same time, and the only such number is zero - the probability density is zero at the nucleus for non-S orbitals.

"Pass through" may not be the best expression, as electrons on stationary states (orbitals) have no macroscopic movement. They have a probability density to be at any point in space, including at the nucleus.

To me, the best formulation is: why can't you imagine an electron right at the nucleus? And why do you imagine something special should happen then?

Offline fledarmus

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Re: electrons in "S" orbital
« Reply #4 on: November 12, 2011, 11:16:34 AM »

S orbitals have their maximum probability density right at the nucleus, sure.

As I understand it, the "probability density" is related to r2, not to r - http://en.wikipedia.org/wiki/Probability_amplitude and http://www.dartmouth.edu/~genchem/0405/spring/6belbruno/radial.pdf, a measure of the volume that you could expect to find an electron in. The volume of the nucleus is quite small related to the total volume occupied by the probability of finding an electron, and the actual highest probability of finding the electron is some distance away from the nucleus (for the hydrogen atom with one electron, this is the Bohr radius). However, there is still a finite possibility of finding the electron inside the nucleus, the penetration ability of an orbital.


Offline Jorriss

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Re: electrons in "S" orbital
« Reply #5 on: November 13, 2011, 12:19:52 AM »

S orbitals have their maximum probability density right at the nucleus, sure.

As I understand it, the "probability density" is related to r2, not to r - http://en.wikipedia.org/wiki/Probability_amplitude and http://www.dartmouth.edu/~genchem/0405/spring/6belbruno/radial.pdf, a measure of the volume that you could expect to find an electron in. The volume of the nucleus is quite small related to the total volume occupied by the probability of finding an electron, and the actual highest probability of finding the electron is some distance away from the nucleus (for the hydrogen atom with one electron, this is the Bohr radius). However, there is still a finite possibility of finding the electron inside the nucleus, the penetration ability of an orbital.


Probability density is related to psi*psi.

Offline aeacfm

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Re: electrons in "S" orbital
« Reply #6 on: November 14, 2011, 02:21:16 AM »

S orbitals have their maximum probability density right at the nucleus, sure.

As I understand it, the "probability density" is related to r2, not to r - http://en.wikipedia.org/wiki/Probability_amplitude and http://www.dartmouth.edu/~genchem/0405/spring/6belbruno/radial.pdf, a measure of the volume that you could expect to find an electron in. The volume of the nucleus is quite small related to the total volume occupied by the probability of finding an electron, and the actual highest probability of finding the electron is some distance away from the nucleus (for the hydrogen atom with one electron, this is the Bohr radius). However, there is still a finite possibility of finding the electron inside the nucleus, the penetration ability of an orbital.


yes it has a finite probability to be found in the nucleus , even electron capture is not happened every time in every element , while S electrons found in all elements . just  what i think

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