Low Down on $T_{boom}$

Sodium, $Na$ $density=9680\,kgm^{-3}$  $M_m=22.9898\,g/mol$ and $Z=11$,

$v_{boom}=3.4354*\cfrac{9680}{26}=1279.0\,ms^{-1}$

$T_{boom}=1279.0^2*\cfrac{22.9898*10^{-3}}{3*8.3144}=1507.73\,K$

The boiling point of sodium being $1156.090\,K$

At its melting point ($370.944\,K$) its density is $9270\,kgm^{-3}$

$v_{boom}=3.4354*\cfrac{9270}{26}=1224.9\,ms^{-1}$

$T_{boom}=1224.9^2*\cfrac{22.9898*10^{-3}}{3*8.3144}=1382.77\,K$

Maybe there can be a $Na$ bomb, when it is completely vaporize at $1507.73\,K$ exactly (very high yield!), but its density remains at $9680\,kgm^{-3}$ because it is sealed completely ie, volume does not change and mass within the containment does not change.

Temperature is not $T_{boom}$ as density changes with temperature, because density decreases with temperature and so does $T_{boom}$.

Does $T_{boom}$ indicate that a substance can be use as a fuel?