Q:

Why does an atomic orbit/shell needs no 8 to be stable? Why not more or less than that??? I need a clear explanation about it because I know the VSPER and Quantum theories.

- Nikita (age 14)

Siliguri, W.B, India

- Nikita (age 14)

Siliguri, W.B, India

A:

This question amounts to asking why there are 8 electron quantum states with nearly the same energy in an atom, not more or less. It's only true in a limited range of atomic numbers, so I'll answer for the clearest case,, with 3 to 10 electrons.

The lowest energy waveform is spherically symmetrical, which corresponds to no rotation at all around the nucleus. There are just two states of this form, with the same waveform and opposite spins.

The next highest energy waveform can either be:

a) still spherical, but with a node as it wiggles away from the nucleus

b) non-spherical, so that it has some rotational energy.

That these two choices have the same energy (ignoring minor relativistic effects) is a special mathematical feature of the (1/r) electrical potential energy.

There's still just 1 spherical form of this type. But there are 3 of the rotating forms, which can be described as having -1, 0, or 1 unit of angular momentum around some chosen axis. The reasons why the angular momentum is quantized like that are hard to give in a short answer.

Anyway that leaves 4 different wave forms. For each, 2 different spins ar possible. That leaves 8 states.

If we were to go up another step in energy, then we could also have states with 2 units of angular momentum, -2,-1,0,1, or 2 along a given axis. That would add another 5x2 or 10 states, giving 18 in that "shell" when the states with 0 or 1 angular momentum are included. However, for larger numbers of electrons, the whole picture of starting with individual electron states, ignoring the interactions between the electrons, becomes less and less realistic.

Mike W.

*(published on 02/08/2017)*