# Homework Help: Orbitals around the nucleus problem

1. Aug 13, 2007

### ksssudhanva

Rsepected sir
Please read my doubts in the attachment and kindly answer it.
Thankyou.

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2. Aug 13, 2007

### Staff: Mentor

From your first question:

Why do you believe that? It is not true. If you are thinking of the Bohr-Sommerfeld model in which electrons travel in classical planet-like circular or elliptical orbits, that model has not been considered valid for about eighty years now. It was superseded by the quantum mechanics of Schrödinger, Heisenberg, et al.

The probability per unit volume for finding a 1s electron (in a hydrogen atom) at a given location is given by the square of the wave function $\psi$ for n = 1, as given near the bottom of this page:

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html

Notice that this function has its maximum value at r = 0!

3. Aug 13, 2007

### chemisttree

4. Aug 13, 2007

### Staff: Mentor

It depends on which probability you're talking about. The probability per unit of volume is given by $|\psi|^2$ which for the 1s orbital is maximum at r = 0, but the probability per unit of radius goes like $r^2|\psi|^2$ which goes to zero as r goes to zero.

If a particle is equally likely to be found anywhere within the volume of a sphere (uniform $|\psi|^2$), it is less likely to have a small r than a large r, because (loosely speaking) there are fewer points with small r than with large r. I consider this variation to be purely a geometrical artifact.

5. Aug 14, 2007

### chemisttree

Yes, that is very counterintuitive. When I read 'probability per unit volume' I immediately think it is the probability of finding a certain electron within a volume element (thin shell) centered on the nucleus and a function of r and r + dr and, of course, that probability is per unit of radius and has a maximum value at [TEX]r=a_0[/TEX].

6. Aug 14, 2007

### chemisttree

ignore this one.

Last edited: Aug 14, 2007
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