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

1 Answer
Mar 20, 2017

Answer:

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

Explanation:

And what we know for chemical reactions is that #"mass"# and #"charge"# are conserved. But for this #"NUCULAR"# reaction, #"MASS"# may not be conserved.

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

#"(i)"# #""^238U rarr ""^234Th +""^4He#

And for #"(ii)"# we have a #"de jure"# chemical reaction, when fluorine gas, the most reactive element on the Periodic Table, oxidizes uranium metal.

#"(ii)"# #U(s) +3F_2(g) rarr UF_6(s)#

Historically, reaction (ii) is significant, in that the product #UF_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.