Nuclear Physics – Determining Half-Life of Observationally Stable Nuclei

elementshalf-lifenuclear-physicsradioactivity

For example, a reaction

$${}^{132} \mathrm{Ba} \rightarrow {}^{128}\mathrm{Xe} + \alpha + 22.19\mathrm{keV}$$

would be energetically favorable, contrary that ${}^{132} \mathrm{Ba}$ is observationally stable. However, energetically favorable reactions happen, only not very quickly. There are also a lot of similar examples.

Does any estimation exist, what could be its half-life? How could it be calculated? I think, it might depend on some potential gate calculation.

Best Answer

There is the empirically found Geiger-Nuttall law relating the energy of the $\alpha$ decay to its half-life:

$$\log T_{1/2} = \frac{A(Z)}{\sqrt{E}}+B(Z)$$ where $T_{1/2}$ is the half-life, $E$ the total kinetic energy (of the alpha particle and the daughter nucleus), and $A$ and $B$ are coefficients that depend on the isotope's atomic number $Z$.

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(image from this question)

You see, for decay energy less than $4$ MeV the half-life is longer than $10^{18}$ s ($\approx 10^{11}$ years), so that such a decay is hard to observe because it would require a very large number of atoms.