Checking Discharge Voltage

From the post "Neon Light Boom" dated 22 Dec 2017,

$V_{min}=A\left(\cfrac{P}{T}\right)^2$

and $A=\cfrac{3.4354^2}{2}\cfrac{m_i^3}{qZ^2k^2}$

but

$\cfrac{P}{T}=\cfrac{n}{V_{ol}}R=\cfrac{M}{m_iN_AV_{ol}}R=\cfrac{density}{m_i}k$

$V_{min}=\cfrac{3.4354^2}{2}\cfrac{m_i^3}{qZ^2k^2}\left(\cfrac{density}{m_i}k\right)^2$

$V_{min}=\cfrac{3.4354^2}{2}\cfrac{m_i}{qZ^2}density^2$

$V_{min}=\cfrac{1}{2}\cfrac{m_i}{q}\left(3.4354*\cfrac{density}{Z}\right)^2$ ---(*)

as $v_{boom}=3.4354*\cfrac{density}{Z}$

$qV_{min}=\cfrac{1}{2}{m_i}v_{boom}^2$

which just says the voltage applied converts to $KE$ at $v_{boom}$, provided that the gap length between the electrodes are sufficiently wide.  This implies that the expression for $V_{min}$ is dimensional-ly consistent.

However, if $q\rightarrow \cfrac{1}{4}q$ then,

$\cfrac{1}{4}qV_{min}=\cfrac{1}{2}{m_c}v_{boom}^2$

and (*) becomes,

$V_{min\,c}=2\cfrac{m_c}{q}\left(3.4354*\cfrac{density}{Z}\right)^2$ ---(**)

$m_c$, however, is not available.

How does this tally with experimental values of $V_{min}$?