# Probability of finding electrons in nucleus (s orbitals)

## Main Question or Discussion Point

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?

mfb
Mentor
Why is the probability of finding an electron of s orbital in the nucleus highest?
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.

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 billiard 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.
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).

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mfb
Mentor
-Why is the amplitude of wave function reach an antinode at the nucleus, making the probability density highest at the nucleus?
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".
-Does it have anything to do with angular momentum being zero?
Yes, it is equivalent to that.
-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?
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.
-Can an electron be 'detected' experimentally at the nucleus? (ie.Collapsing the wave form and fixing position at nucleus)
In principle, this is possible. I don't know how an experimental realization of that would look like.

Books: no idea.

256bits
Gold Member
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:
Two interpretations can again be given to the P1 curve. An experiment designed to detect the position of the electron with an uncertainty much less than the diameter of the atom itself (using light of short wavelength) will, if repeated a large number of times, result in Fig. 3-4 for P1. That is, the electron will be detected close to the nucleus most frequently and the probability of observing it at some distance from the nucleus will decrease rapidly with increasing r. The atom will be ionized in making each of these observations because the energy of the photons with a wavelength much less than 10-8 cm will be greater than K, the amount of energy required to ionize the hydrogen atom. If light with a wavelength comparable to the diameter of the atom is employed in the experiment, then the electron will not be excited but our knowledge of its position will be correspondingly less precise. In these experiments, in which the electron's energy is not changed, the electron will appear to be "smeared out" and we may interpret P1 as giving the fraction of the total electronic charge to be found in every small volume element of space. (Recall that the addition of the value of Pn for every small volume element over all space adds up to unity, i.e., one electron and one electronic charge.)
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.