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

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

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?

$Nd$ $(Z = 60)$ is a member of group $3$ in the periodic table. An isotope of it is $\beta$-active. The daughter nuclei will be a member of Group $...$ ...

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

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