Uranium dioxide (UO_2), \(Z_m=235+8*2\), density \(10.97\,gmol^{-1}\), molar mass \(270.03\,gcm^{-3}\),
\(v_{boom}=3.4354*\cfrac{10.97*10^{3}}{235+8*2}=150.14\,ms^{-1}\)
\(T_{boom}=150.14^2*\cfrac{270.03*10^{-3}}{3*8.3144}=244.04\,K\) or \(-29.11^oC\)
\(T_{p}=150.14^2*\cfrac{270.03*10^{-3}}{2*8.3144}=366.05\,K\) or \(92.90\,^oC\)
\(T_{boom}\) closeness to room temperature is not indicative of radioactivity, the matter itself has to have an unstable nucleus in the first place.
If the most probable speed is \(v_{boom}\), ie \(v_p=v_{boom}\) and \(v_{boom}\) does induce nuclear disintegration then at \(T_{boom}\) radioactivity is most heightened. Putting off a meltdown is a different story, as at all temperature a finite portion of the particles are with \(v_{boom}\) velocity.
From,
\(v_{boom}=3.4354*\cfrac{density}{Z}\)
which was actually from,
\(\cfrac{1}{2}m_{\rho}v^{ 2 }_{ { max } }=E_{input/particle}=\cfrac{1}{2}c^{ 2 }*3.4354*\cfrac{density}{Z}\)
where \(Z\) denotes the number of basic particles of each type. In the case of elements, it is just the atomic number. In the case of molecules, it is the sum of the atomic numbers of the constituent elements.
The expression does not indicate a velocities at which \(v_{max}=0\), that no basic particles are emitted after collisions.
It is possible however, to find a non nuclear material with a stable nucleus that would absorb particles at \(v_{boom}\) velocity (also at resonance) without being radioactive. This material competes with the unstable radioactive material, and reduces the portion of colliding particles at \(v_{boom}\) velocity. Absorbed particles are confined within the volume of the stable material and can be removed safely.
Control rods; designed to readily absorbed into its interior, particles at \(v_{boom}\). This it does so by re-emissions into its interior particles absorbed. These are materials that share the same \(v_{boom}\) as the radioactive material.
Good evening.
