I Physical Significance of 2s Orbital Peaks?

the physical significance is that one has to throw out the Bohr's model of atom and as the probability density cloud is interpenetrating the electrons can not be thought of as confined to some circular orbital regions around the nucleus.
therefore those effects which are due to exchange of electrons from one orbital to another or sharing can be accounted for.
e.g. in He calculations for energy states we encounter overlap integrals or exchange integrals.
though we are looking at s-states here - if one goes to p and other orbitals there shapes get further modified by the angular momentum considerations .

 

well, i could not follow the relation of the thread with black body radiation.
i know that there are very many limitation of Bohr's theory of atomic structure.

 

As I understand it, an electron in the H atom in the excited n = 2, state is most likely at the r corresponding to the peak in the square of the wavefunction. This will not be at an integral multiple of the Bohr radius. This idea is a total non - sequitur.

 

The fact that it's proportional to the Bohr radius is not coincidence - that's required to get the right dimensionality and behavior. The fact that the constant of proportionality of is 1 and not 3/2 is because you chose to look at the distribution predicted by QM and ask "what is the mode of this distribution?" and not "what is the mean of this distribution?"

 

The expected value is related to the Bohr radii. The Bohr radii scale as ##n^2##. So, for the n=2 state, ##r=4a_0##, where ##a_0## is the Bohr radius.

If you calculate an expected value for ##\frac{1}{r}##, it does give you ##\frac{1}{n^2 a_0}##. This follows from the virial theorem.

You calculate the expected value by
##\int \frac{1}{r} (r R_{nl})^2##
where R is the radial wavefunction.

If you calculate the expected value for ##r## instead of ##1/r##, you get ##\frac{3}{2} a_0 n^2##

 

You keep asking this question in different forms and not accepting the answer you get. See message #7.

 

As I said, the functional form is driven by the physics. The number "1" is your choice, and corresponds to the mode of the distribution. If you picked another reasonable choice, like the mean, you'd have a different number, like 3/2.

 

In Old Quantum Theory, there were no 2s or 3s states. There were just the n=2 and 3 states.

 

I don't understand this debate. The physical meaning of the wave function is that its modulus squared gives the probability distribution for position of the electron bound to a proton. Case closed, question answered!

 

I don't understand again. #17 is my own posting, which I read carefully again. So?