An Old Friend For New Reasons

When $v_{p}$ replaces $v_{rms}$,

$v_{p}=\sqrt{\cfrac{2RT_{boom}}{M_m}}$

$v^2_{boom}=v^2_{p}$         and

$M_m\rightarrow nM_m$

$v^2_{boom}=\cfrac{2RT}{nM_m}$

where $n$ is due to clustering, when $n$ particles coalesce and move as one,

$n.\cfrac{1}{2}{M_m}v^2_{boom}={\cfrac{1}{2}*2R.T_{boom}}=R.T_{boom}$

$\cfrac{\cfrac{1}{2}{M_m}v^2_{boom}}{R.T_{boom}}=\cfrac{1}{n}$

Let $M_m=N.m_a$,  $N$is the number in one mole and $m_a$ is particle mass,

$\cfrac{\cfrac{1}{2}{N.m_a}v^2_{boom}}{R.T_{boom}}=\cfrac{\cfrac{1}{2}{m_a}v^2_{boom}}{\cfrac{R}{N}T_{boom}}=\cfrac{1}{n}$

Let $E_a=\cfrac{1}{2}{m_a}v^2_{boom}$

$\cfrac{E_a}{kT_{boom}}=\cfrac{1}{n}$

and we are back to Maxwell-Boltzmann, but as in the post "$T_{boom}$ To The Power Of $T_{boom}$" dated 09 Nov 2017,

$P(energy\,state\,m)=e^{-\cfrac{E_a}{k.T_{boom}}}.\cfrac{\cfrac{1}{n^m}}{m!}$

Was Maxwell-Boltzmann measuring the most probable, and thus the maximum nuclear output due to $T_{boom}$ in the first place?

If so, $T_{boom}$ is $T$ in Maxwell-Boltzmann distribution.  We have been measuring $T_{boom}$ as temperature already, all the time.

But $T_{boom}$ is nuclear.  It is not just an increase in temperature.