A Party of Four

Obviously,

That is why in a true vacuum, devoid of space itself, where charges loses their electrostatic property and uncoils from the nucleus of the atom, we see geometrical shape such as this one.  So if Beryllium is placed inside a true vacuum, take a picture with the cap still on the lens, the film will capture a tetrahedron such as the one shown atop.  Basically, the electrons that hits the metal containment generates X-rays.  Does it work with a digital camera?  May be not.

Since this is a tetrahedron,

$\cfrac{{r}_{ee}}{{r}_{pe}}=\cfrac{2\sqrt{2}}{\sqrt{3}}$

${r}^2_{ee}=\cfrac{8}{3}{r}^2_{pe}$

$sin(\theta)=\cfrac{{r}_{pe}}{{r}_{ee}}.\cfrac{3}{2}=\cfrac{3\sqrt{3}}{4\sqrt{2}}$

Consider centripetal force,

$\cfrac{{m}_{e}c^2}{{r}_{pe}}=\cfrac{q.4q}{4\pi\varepsilon_o{r}^2_{pe}}-\cfrac{3q.q}{4\pi\varepsilon_o{r}^2_{ee}}.sin(\theta)$

$\cfrac{{m}_{e}c^2}{{r}_{pe}}=\cfrac{q^2}{4\pi\varepsilon_o {r}^2_{pe}}.(4-\cfrac{3}{\frac{8}{3}}).\cfrac{\sqrt{3}}{2\sqrt{2}}\cfrac{3}{2}$

${{r}_{pe}}= \cfrac{q^2}{4\pi\varepsilon_o m_ec^2}(4-\cfrac{ 9 }{8}).\cfrac{3\sqrt{3}}{4\sqrt{2}}$

${{r}_{pe}}$ = (1.60217657e-19)^2/(4*pi*8.8541878176e-12*9.10938291e-31*299792458^2)*(4-9/8)*(3)^(0.5)/(2)^(0.5)*(3/4))

${r}_{pe}$ = 7.4418e-15 m

And the published atomic radius of Be, Beryllium is 112e-12 m.