$_{84}Po^{210} \xrightarrow{} _{82}Pb^{206} + _{2}He^{4}$. From the above equation,deduce the position of polonium in the periodic table (lead belongs to group $IV A$).

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?

The group displacement law was given by:

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

$_{92}U^{235}$ belongs to group $III-B$ of the periodic table. If it loses one $\alpha$-particle,the new element will belong to group: