Proof of maximum no. of electrons in a shell

[-ve]

how do you prove that the maximum no. of electrons in the nth shell of an atom is twice of n squared (2n^2)

 

[Borek]

Solving Schrödinger equation for a hydrogen atom.

 

[-ve]

thanks
^_^

 

[DrDu]

To be a bit more specific: The formula 2n^2 is based on the assumption that the shells are hydrogen like. There are deviations from this rule.
The factor 2 is due to the fact that each orbital can carry at most two electrons, one with spin up, the other with spin down.
So we have to explain why there are n^2 orbitals in each shell.
It is a peculiarity of the hydrogen atom that all orbitals having the same number of node surfaces have the same energy. There are radial nodes and spherical nodes. All orbitals in a given shell have n-1 nodes. The number of spherical nodes fixes whether we speak of an s, p, d, or f orbital. The number of spherical nodes is equal to the quantum number l with l=0 corresponding to s, l=1 to p etc. There are 2l+1 orbitals with the same value of l. So if e.g. n=4 the orbitals have 3 nodes. There are the following possibilities
# radial nodes #spherical nodes=l name multiplicity=2l+1
0 3 f 7
1 2 d 5
2 1 p 3
3 0 s 1

You can check that the sum of the multiplicities is 16=n^2.
In general $\sum_{l=0}^{n-1}(2l+1)=n^2$
as Kolmogorow, the father of modern statistics, realized as a 5 year old boy.

 

[dextercioby]

The only rigorous proof for 2n^2 would be to solve the SE for the atom which is mathematically impossible. Not even the helium atom admits a complete solution.

 

[DrDu]

Certainly. However, I think it is quite nice that in case of the H atom the energetic ordering of the orbitals depends only on the number of nodes (which can be traced back to the hidden SO(4) symmetry).
Is there a pedagogical way of making plausible that there are 2l+1independent spherical harmonics with given l?

 

[Einstein Mcfly]

Look at the experimental ionization potentials and note the pattern. Compare to the solution of the SE for the hydrogen atom.