Only The Light And Agile Applies

If we look at the formula for $T_{v\,boom}$,

$T_{v\,boom}=\left(\cfrac{3.4354*density}{Z}\right)^2*\cfrac{atomic\,mass*10^{-3}}{3*8.3144}$

where $Z$ is the atomic number, $R=8.3144\,J\,K^{-1}mol^{-1}$.
We have,

$T_{v\,boom}=M_m*\left(\cfrac{\rho}{Z}\right)^2*4.731*10^{-4}$

where $\rho$ is density, $M_m$ is molar mass and $Z$ is the atomic number.

The formula for $T_{v\,boom}$ is derived from kinetic energy expression for non interacting particles which is true when the material concerned is a sparse vapor at high temperature. This means $T_{v\,boom}$ should be beyond the boiling point of the element.

If such a $T_{v\,boom}$ can be achieved, at this temperature, all the material disintegrate at once.

For Osmium $Os$, the densest element,

$T_{v\,boom}=190.23*\left(\cfrac{22590}{76}\right)^2*4.731*10^{-4}=7951.3\,K$

which has a boiling point of $5285\,K$.

Maybe Uranium is not necessary.  Nuclear fission has nothing to do with radioactivity.


Note:  In general,

$T_{v\,boom}=\left(\cfrac{3.4354*density}{no.\,of\,particles}\right)^2*\cfrac{molar\,mass*10^{-3}}{3*8.3144}$

where molar mass (and atomic mass) is measured in $g\,mol^{-1}$.

$T_{v\,boom}$ is a resonance phenomenon, at any value other than $T_{v\,boom}$ disintegration does not occur.

Note:  Maybe just upon disintegration the material particles are freed from its lattice hold and are thus non-interacting particles (other than collisions) as the derivation of $v_{rms}$ requires.  If the material is already at a vapor state at $T_{v\,boom}$ then disintegration is instantaneous.  If not in a vapor state, then disintegration occurs only on the surface where material particles can break free; which make a sealed coal containment at $538.3^oC$ safe.