# Probability of finding electrons in nucleus (s orbitals)

1. Sep 4, 2013

### Enigman

Why is the probability of finding an electron of s orbital in the nucleus highest? Is Quantum Tunneling involved? If so, won't the electron need a large amount of energy to pass through the nucleus?

2. Sep 4, 2013

### Staff: Mentor

One type of wave function has the highest probability, and that type is called "s" for historical reasons. It has nothing to do with tunneling.
Neither the electron nor the nucleus are billard balls, they don't "collide" as classical objects would do. The electron does not need any energy to be (partially) in the nucleus, and it does not "pass" the nucleus.

3. Sep 4, 2013

### Enigman

That doesn't exactly answer my questions...my questions in are:
-Why is the amplitude of wave function reach an antinode at the nucleus, making the probability density highest at the nucleus?
-Does it have anything to do with angular momentum being zero?
-The probability density graphs of s orbitals show a highest density at the center of nucleus- so wouldn't an electron have to be fully in the nucleus? (-Thats probably why I got the weird idea of tunneling; mixed it up with penetration. ) Or is it just a schrodinger's cat scenario?
-Can an electron be 'detected' experimentally at the nucleus? (ie.Collapsing the wave form and fixing position at nucleus)
-Also a reference book suggestion would be great- the coursebook (Atkins- elements of phys. chem.) is rather vague about most of the part, I'm self studying by Shankar's book on Q.M. but it doesn't deal with atomic structure (as far as I've read).

Last edited: Sep 4, 2013
4. Sep 4, 2013

### Staff: Mentor

By symmetry (spherical symmetry of the potential), it has to reach a maximum or a minimum. All wave functions with a maximum there are called "s".
Yes, it is equivalent to that.
The electron "is" everywhere in its wave function "at the same time" - only a (very small) part of the wave function is in the nucleus.
In principle, this is possible. I don't know how an experimental realization of that would look like.

Books: no idea.

5. Sep 4, 2013

### 256bits

If you look at this site you can find some answers to your questions.
http://www.chemistry.mcmaster.ca/esam/Chapter_3/section_2.html

The electron in an atom is in a quantum state designated by n, l, m(l) and m(s).
For all s orbitals we have l=0, and m(l) =0, so subsequentially the angular momentum and magnetic moment are zero.

Orbital wave function can be broken up into a spherical polar coordinate part and a radial part.
What you are referring to is the probability of finding the electron at a certain radial distance from the nucleus.
There is another probability called the radial probability distribution, which is the radial probability of the wave function squared multiplied by the volume of a spherical shell of thickness dr at a distance r from the nucleuos.
The maximum probability is given as where the electron is most likely to be found. ( Can be considered the size of the orbital )

For determining the whereabouts of the electron, the site states:
I imagine if the wave function collapses in the nucleus, woulld not that mean it has been captured by a proton.

How the wave function for an electron is determined I cannot say but it has to do with Schroedinger, de Broglie, particle in a box, and all that stuff.
I kinda remember this stuff from old chemistry days and it is now somewhat vague.

6. Sep 4, 2013