What happens in electron shells?
- The disintegration series of Uranium to Pb 82 Z -

The orbital series as sums of the steps 5 - 4 - 3 - 2 - 1:

The point in this derivation of numbers is that we easier can see the potential dividing of electrons within the orbitals:

According to one illustration in a scientific article the d-orbital for example is divided into 1 circular formed electron around the origin of the coordinate system, 3 electrons divided along the 3 axes and 6 electrons (3 pairs), in the square quadrants of the 3 planes.

A multiplication operator acting in the electron shells?

An angle change from ⊕ to ⊗, from addition to multiplication relations in the e-shells, gives:

Through further division within orbitals the sum of the electrons gets 10 units higher.


The disintegration series of ²³⁸U from 92 Z to Pb 82 Z goes through
8 alpha-steps = - 16 Z: compare the change above in f-orbital from 64 → 48.
6 beta-steps   = +  6 Z: compare the change above in d-orbital from 18 → 24.

Compare from Physics:
f-orbital, from step 4→3: Mass identified as a property in this step. Mass reduction connected with alpha radiation.
 d-orbital from step 3 → 2: Charge identified as a property in this step. beta steps implying change of charge.


Surely the distribution of electrons on different orbitals above will not correspond to that which the physicists observe - if they have time to observe what really happens?!.

The same type of operations for Uranium 235 and Thorium 232, 90 Z?
Suggestions:

"Rearrangements" within the atoms:

²³⁵U: In this disintegration series we have - 28 A:
          - 7 α-steps,
          + 4 beta-steps = - 14 Z, + 4 Z
β-steps in this series occur in first steps in one(1) β+-radiation.

According to the suggestion above there should occur an inner β+-radiation.
(Possible to detection with its use as nuclear fuel?)

²³²Th, 90 Z: In the disintegration series to Pb 207 A we have - 26 A, - 8 Z:
           - 6 alfa-steps,
           + 4 beta-steps = - 12 Z, + 4 Z.

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