\(2\cfrac{qq}{4\pi\varepsilon_o{r}^{2}_{ee}} sin(60^o) = \cfrac{q.3q}{4\pi\varepsilon_o{r}^2_{pe}}\)
\(\cfrac{2}{{r}^{2}_{ee}}\cfrac{\sqrt{3}}{2} = \cfrac{3}{{r}^2_{pe}}\)
\(\sqrt{3}{r}^2_{ee}={r}^2_{pe}\)
\(\sqrt[4]{3}{r}_{ee}={r}_{pe}\)
but from geometry,
\(2.{r}_{pe}cos(30^o)={r}_{ee}\)
\(\sqrt{3}{r}_{pe}={r}_{ee}\)
This means the configuration is not static, the electrons are in circular motion, instantaneously at least, in order not to fall into the positive charge. Let assume orbit speed is \(c\)
\(\cfrac{m_ec^2}{{r}_{pe}}= \cfrac{q.3q}{4\pi\varepsilon_o{r}^2_{pe}}-2\cfrac{qq}{4\pi\varepsilon_o{r}^{2}_{ee}} sin(60^o) \)
\(\cfrac{m_ec^2}{{r}_{pe}}= \cfrac{3q^2}{4\pi\varepsilon_o{r}^2_{pe}}-2\cfrac{q^2}{4\pi\varepsilon_o 3{r}^{2}_{pe}} .\cfrac{\sqrt{3}}{2} \)
\({{r}_{pe}}= \cfrac{q^2}{4\pi\varepsilon_o m_ec^2}(3-\cfrac{\sqrt{3}}{3}) \)
\({{r}_{pe}}\) = (1.60217657e-19)^2/(4*pi*8.8541878176e-12*9.10938291e-31*299792458^2)*(3-(3)^(0.5)/3)
But the atomic radius of Lithium. Li is 1.52e-10 m. There is one possibility, that the electrons are not orbiting a light speed \(c\). What speed are the electron orbiting at then? Then again electron mass has a relative uncertainty of 4.2e-10, so what of e-15. This value can not be more accurate than electron mass.
