When ${}_{92}^{238}U$ (Group $III B$) changes to ${}_{90}^{234}Th$ by the emission of an $\alpha$-particle,the daughter element will be in which group?

Whenever the parent nucleus emits a $\beta^{-}$ particle,the daughter element is shifted in the periodic table:

$A$ radioactive substance $_{88}X^{228}$ $(IIA)$ emits $3$ $\alpha$ and $3$ $\beta$ particles to form $Y$. To which group of the long form of the periodic table does $Y$ belong?

When a radium atom,which is placed in group $II$,loses an $\alpha$-particle,a new element is formed which should be placed in group ........

Group displacement law states that the emission of $\alpha$ or $\beta$ particles results in the daughter element occupying a position,in the periodic table,either to the left or right of that of the parent element. Which one of the following alternatives gives the correct position of the daughter element?

On emission of $\alpha$ particles

On emission of $\beta$ particles

$A$ radioactive substance ${}_{88}X^{228}$ (Group $IIA$) emits $3 \alpha$ and $3 \beta$ particles to form $Y$. To which group of the long form of the periodic table does $Y$ belong?