# How does ""^238U undergo alpha-"decay" to give ""^234Th? Is this a chemical reaction? How does uranium metal react with fluorine?

Mar 20, 2017

Well, it is a nuclear reaction, rather than a chemical reaction........

#### Explanation:

And what we know for chemical reactions is that $\text{mass}$ and $\text{charge}$ are conserved. But for this $\text{NUCULAR}$ reaction, $\text{MASS}$ may not be conserved.

And thus we may write the $\alpha \text{-particle}$ decay of ""^(238)U (the $\text{92}$ subscript described in the question is superfluous, why?).

$\text{(i)}$ ${\text{^238U rarr ""^234Th +}}^{4} H e$

And for $\text{(ii)}$ we have a $\text{de jure}$ chemical reaction, when fluorine gas, the most reactive element on the Periodic Table, oxidizes uranium metal.

$\text{(ii)}$ $U \left(s\right) + 3 {F}_{2} \left(g\right) \rightarrow U {F}_{6} \left(s\right)$

Historically, reaction (ii) is significant, in that the product $U {F}_{6}$ is reasonably volatile. Thus this enables some means to separate the isotopomeric uranium products as their volatile fluorides. I do not think ""^238U is fissile, but for better info you should consult a text.