\(\cfrac{1}{2}m_{\rho}v^{ 2 }_{ { max } }=E_{input/proton}=\cfrac{1}{2}c^{ 2 }*3.4354*density*\cfrac{1}{particle\,count\,per\,type}\)
where we obtain the boom velocity,
\(v_{boom}=3.4354*density*\cfrac{1}{particle\,count\,per\,type}\)
Obviously the higher this velocity is the greater the stability of the material is with regard to both temperature and collisions. The factor,
\(S_v=density*\cfrac{1}{particle\,count\,per\,type}\)
is then indicative of the relative stability of the materials concerned. In the case of the Elements,
\(S_v=\cfrac{element\,\,density}{atomic\,\,number}\)
where the density of the elements is at room temperature.
\(S_v\) is indicative of the relative stability of the elements. \(S_v\) is nuclear stability as we are concern here with the interactions of \(\psi\) particles. A \(S_v\) vs atomic number plot is given below,
This is not as expected; it is not in the profile of nuclear binding energy vs number of nucleons. In this plot, data was not available to \(Z=85\) and \(Z=87\), a break occurs along the atomic number axis. Elements that are gases under normal conditions are low in the plot. The most stable element is \(^5B\), Boron, followed by \(^4Be\), Beryllium.
Note: Nuclear binding energy, confined to the nucleus of an atomic, has nothing to do with material density, a property concerning the compactness between atoms. Maybe, nuclear binding energy is the same for all nucleus, but the element's density (an attribute derived from the space between atoms) that endows the element with strength against disintegration. Stability is a group thing.
