*(Fig.1) ↓ "de Broglie wave interference" influences valence electrons' number*

The number of valence electrons is **fixed** in each atom. Helium has two, and Neon has eight valence electrons.

What determines this number of valence electrons ? Electron spin is *unrealistic* and too weak to affect valence electrons and molecular bonds.

In fact, ( destructive ) interference among electrons' de Broglie **waves** determines the maximum number of valence electrons in each atom

*(Fig.2) ↓ One de Broglie wavelength orbit and its phases*

Electron de Broglie wave has been proved in experiments such as Davisson-Germer. One de Broglie wavelength consists of two **opposite** phases ( crest+ and trough- ).

To **avoid** destructive interference between these two opposite wave phases, each orbital length has to be an integer times de Broglie wavelength.

*(Fig.3) "Opposite" wave phases interfere with each other, destructively. → "perpendicular".*

Helium atoms contains two **1 ×** de Broglie wavelength orbits ( **n = 1** ).

Two overlapping **opposite** wave phases cause "destructive" interference and *instability*.

To avoid this destructive interference, two orbits of helium have to cross each other "**perpendicularly**", because "perpendicular" means each wave phase can be **independent** from another.

*(Fig.4) Helium consists of two 1 × de Broglie wavelength orbits.*

Helium consists of *two* **1** × de Broglie wavelengt orbits. Fig.4 shows how two electrons are **moving** in Helium orbits.

To avoid the destructive interference between two de Broglie waves, these electrons' orbits are perpendicular to each other.

*(Fig.5) ↓Combining four 1 × de Broglie wavelength orbits. ←Impossible*

Helium consists of **two** *1* × de Broglie wavelength orbits. The maximum orbital number in 1 × de Broglie wavelength is **two** orbits.

Combining **four** *1* × de Broglie wavelength orbits in one atom (= Be ? ) is **impossible** due to destructive interference among electron de Broglie waves.

*(Fig.6) ↓ e3 electron is surrounded by opposite wave phases of e1 and e2.*

The maxmum orbital number in 1 × de Broglie wavelength is **two** orbits as seen in Helium. Then, why we **cannot** mix **four** *1* × de Broglie wavelength orbits in a single atom ?

In Fig.6, the **e3** electron is completely **surrounded** by the *opposite* phases of *e1*,*e2* de Broglie waves.

So in this case, this e3 electron is **kicked out** from the atom due to destructive interference among de Broglie waves.

This is the reason why **four** orbits in *1* × de Broglie wavelength is **impossible**.

*(Fig.7) ↓ Four 2 × de Broglie wavelength orbits is possible = Neon.*

**Opposite** phases of de Broglie wave have to cross *perpendicularly* to avoid destructive interference. Neon has **four** 2 × de Broglie wavelength orbits.

Why is four 2 × de Broglie wavelength orbits is possible, different from 1 × de Broglie wavelength ?

As seen in Fig.7, the e3 electron is on *midlines* (= the *mid*point between two opposite phases ), when it's just between two other orbits.

Midline (= mid-*phase* ) does **not** cause destructive interference in electrons. So **four** *2* × de Broglie wavelength is **possible**.

*(Fig.8) Opposite phase is longer in 1 × de Broglie wavelength (= 1/2 of orbit )*

As an atom contains **more** orbits, the *distance* between two orbits becomes **shorter**.

When an electron *U-turn*s between orbits ( see e3 in Fig.7 ), it has to be in other electron's **mid**line (= mid-point ) to avoid destructive interference.

The length of each opposite phase is **longer** (= 1/2 ) in 1 × de Broglie wavelength than 2 × de Broglie wavelength (= 1/4 ).

So when **four** *1* × de Broglie wavelength orbits are mixed, each electron are surrounded in opposite phases of other electrons.

As the number of de Broglie wavelength is larger ( 1 → 2 de Broglie wavelength ), it can contain **more** orbits (= distance between neighboring orbits is **shorter** ).

2016/11/11 updated. Feel free to link to this site.