The Value Of $\pi$

Copper in its metalic lattice is FCC structure, each atom has twelve closest neighbors,

coordinate number = 12

Each atom is shared by eight adjoining unit cells.  The atom at the face center also has twelve closest neighbors and is also shared by eight unit cells, when consider alone.

The frame of reference is displaced by half a unit cell, and the face center atom is now at the corner of a unit cell, without changing the lattice.  So, a corner atom is equivalent to to face center atom as far as the lattice structure is concerned.  Both atom share one electron with twelve neighbors, but

two electrons around one atom are shared by two neighboring copper atoms, in a chain of  two bonds, along one of the six directions which contains two bonds. Only one electron is associated with one $Cu$ atom freed by breaking two bonds.  Alternatively, we can consider that each electron is shared by two $Cu$ atom neighbors, then each $Cu$ atom has half an electron per bond.  It does not matter when the $Cu$ atoms are bonded this other way,

the removal of a pair of bonds releases one electron and leaves behind $2+$.

Unfortunately,

For a total of twelve bonds there are then six electrons $6e^{-}$, per $Cu^{12+}$ atom released.  The total positive charge left behind is $12+$.  This then requires a further four electrons $4e^{-}$, from the power source, to create one $Cu^{2+}$.

And since each atom is shared by eight unit cells, we have $\small{\cfrac{4}{8}}$ electron required per $Cu^{2+}$ released.

$n_e=\cfrac{1}{2}$

where $n_e$ is the number of electrons REQUIRED to creates one $Cu^{2+}$.

This is the charge required when a $Cu$ atom from a pure solid lattice goes into solution as $Cu^{2+}$.  Electrons have to flow into the electrode to break the bonds in the metal lattice.

This means, in an experiment where both electrode are PURE copper, the electrode connected to the negative terminal of the power source dissolve first, and $Cu^{2+}$ ions are released into the electrolyte.

At the cathode where $Cu^{2+}$ receives two electrons $2e^{-}$ but remains in solution as free $Cu$ atom. This $Cu$ atom will give $2e^{-}$ to the positive terminal and be oxidized to $Cu^{2+}$ again.  The measured current has two components in this case; one due to the free $Cu (atom) \rightarrow Cu^{2+}$, $Cu$ atoms already in solution after receiving $2e^{-}$; and the other in reverse due to $Cu$ atom in the lattice going into solution.  $n_e$ in this situation is,

$n_e=2-\cfrac{1}{2}=\cfrac{3}{2}$

In the experiment to measure Avogadro Constant,  A constant current is driven INTO a pure copper electrode, the mass lost is obtained by measuring the electrode before and after the current is passed for a time period, $t$.  The ions are removed and does not interact with the electrode again.

The measured current is due to $n_e=\cfrac{1}{2}$ only.

Since, the available charge per atom is proportional to current measured but inversely proportional to the number of atom oxidized, given a measured constant current.  The number of atoms given current is,

$n=\cfrac{A}{n_e}$

where $A=\cfrac{I}{e}t$, the constant current, $I$ divided by the unit charge, $e$.  $t$ is the time duration.

If we made a mistake and assume that $n_e=+2$...

$n_v=\cfrac{A}{2}$

 When the required charge is instead, $n_e=\cfrac{1}{2}$,

$n_D=2A$

then,

$\cfrac{n_D}{n_v}=4$

In which case we find that the factor $\cfrac{4}{3}\pi=4.188790$.

If however the constant current flows when $n_e=2-\cfrac{1}{2}=\cfrac{3}{2}$,  that is to say the constant current flows when $Cu^{2+}$ has returned to the negative terminal.

To find the correct answer, first we normalize the current, for $n_e=1$

$I_{1}=\cfrac{2}{3}I$

and the current due to $n_e=\cfrac{1}{2}$ is

$I_{1/2}=\cfrac{1}{2}I_{1}=\cfrac{1}{3}I$

$n_{v1}=\cfrac{A_n}{n_e}$

where $A_n=\cfrac{I_{1/2}}{e}t=\cfrac{1}{3}\cfrac{I}{e}t=\cfrac{1}{3}A$

So,  $n_{v1}=\cfrac{1}{3}\cfrac{A}{n_e}$

For $n_D$,

$n_e=\cfrac{1}{2}$

$n_{D}=\cfrac{1}{3}2.A$

But the answer assuming that all current creates $Cu^{2+}$ at a rate of $n_e=2e^{-}$ each.

$n_{v1}=\cfrac{A}{n_e}=\cfrac{A}{2}$

Which lead us to,

$\cfrac{n_D}{n_{v1}}=2.\cfrac{2}{3}=\cfrac{4}{3}=1.3333$

We compare this with the ratio of $\cfrac{9.029022}{6.022140}=1.4993$.

Second guessing what might have happened only result in lack of sleep.  Does the unit $kg$ includes the value of $\pi$?  Does it have chicken in it?