## fredag 16 augusti 2013

### Quantum Contradictions 9: Shell Structure from Helium Scaling

To explain the observed sequence 2, 8, 18, 32,..., of the number of electrons in filled atomic shells of the periodic table, one may argue as follows (which is not the text book explanation) starting from the structure of Helium as described in a previous post:

Helium has 2 electrons filling a K-shell of width 1/2. The next element in the table is Lithium with 2 electrons filling a K-shell of width 1/3 and 1 electron in an outer L-shell of width 1 being attracted by a net charge of +1 from the kernel and the two inner shell electrons. In the model I am studying in this sequence of posts, the 3rd electron of Lithium of width 1 being attracted by a net charge of +1 finds a relative energy minimum in a second shell outside an inner shell of diameter 1/2 filled by two electrons.

More generally, seeking the number of electrons $Z$ required to fill a 2nd L-shell we compute the radius d of the 2nd shell to be
• $d_2 = \frac{1}{Z+2} + \frac{1}{Z}$
and thus the surface area $A$ of the 2nd shell to be
• $A = 4 \pi \times d_2^2$ .
In the Helium atom each electron can be viewed to fill a half-spherical surface and so $Z$ electrons of the 2nd shell of width $\frac{1}{Z}$ can be computed to require an area of
• $Z\times 2\pi\times\frac{1}{Z^2}$.
We thus obtain the following equation for a filled 2nd shell:
• $4\times\pi\times(\frac{1}{Z+2} + \frac{1}{Z})^2 = 2\pi\times\frac{1}{Z}$
with approximate solution
• $2\times (\frac{2}{Z})^2 = \frac{1}{Z}$
that is $Z = 8$.

The same argument for a 3rd M-shell gives the equation
• $2\times (\frac{3}{Z})^2 = \frac{1}{Z}$
with solution $Z = 18$. More generally we get for an Nth shell the following number $Z = Z(N)$ of electrons:
•  $Z(N) = 2 \times N^2$.
It is thus possible to derive the basic formula $Z(N) = 2\times N^2$ for the atomic shell structure of the periodic table by a scaling argument starting from the 2-valuedness (or Zweideutigkeit) of the Helium electron structure.

This is an entirely different argument from the standard one based on the sequence of of excited states of Hydrogen and the notion of 2-valued spin (combined with Pauli's exclusion principle) offering the Zweideutigkeit.

The above argument is another brick in support of an electron structure without Pauli's exclusion principle, a principle which Pauli himself did not consider to be a law of physics but rather an ad hoc invention constructed to prevent the outer electron of Lithium to enter into the K-shell.

PS1 The simple shell model under study gives the following ground state energy $E$ for Neon with $N = 2$ and $2 + 8 = 10$ electrons:
• E = - 2 x 10/(1/10) + 1/(1/10)^2 + 1/(2/10) - 8 x 8/d_e + 8/2 x 1/(1/8)^2 + E_e
• d_2 = 1/10 + 1/16
• E_e = 8 x 7/2 x 1/d_e
• d_e = 2 x 1/8
where we estimate the kernel distance d_2 of the 8 outer electrons to be 1/10 + 1/2 x 1/8 and their average distance d_e to be 2 x 1/8. This gives
• E = - 95 - 393 + 256 + 112 =  - 119
where - 95 as the inner shell energy is to be compared with the observed - 94 and the total energy E = -119 with the observed - 129. With a small perturbation of d_2 and d_e we can hit this target.

PS2 For Carbon with 2 + 4 = 6 electrons we get similarly
• E = [- 2 x 6/(1/6) + 1/(1/6)^2 + 1/(2/6)] + [- 4 x 4/(1/6 + 1/8) + 4/2 x 1/(1/4)^2 + 4 x 3/2 x 1/(1/2)]
• = [- 72 + 36 + 3] + [- 55 + 32 + 12] = - 33 - 11 = - 44
to be compared with - 32.4 - 5.4 = - 39. Again the target can be hit by perturbation of the kernel distance of the 4 outer electrons and their average distance.

PS3 For Oxygen O with 2 + 6 = 8 electrons, we get similarly:
• E = [- 2 x 8/(1/8) + 1/(1/8)^2 + 1/(2/8)] + [- 6 x 6/(1/8 + 1/12) + 6/2 x 1/(1/6)^2+6 x 5/2 x 1/(1/3)]
• = [- 128 + 64 + 4] + [- 178 + 108 + 45] = - 59 - 25 =  - 84
to be compared with the observed - 59 - 16 = - 75.