Question

In a series of three steps in radioactivity disintegration sequence starting with ${}_{88}^{228}Ra$, the particles emitted are, successively, $\beta -,\beta -$ and $\alpha -$particles. The resulting product is an isotope of :

• A. ${}_{92}U$
• B. ${}_{90}Th$
• C. ${}_{88}Ra$
• D. ${}_{86}Rn$

Answer 1

The correct option is C ${}_{88}Ra$