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.
