$A$ nuclide of an alkaline earth metal undergoes radioactive decay by emission of two $\alpha$-particles in succession. The group of the periodic table to which the resulting daughter element would belong is $Group$ .......

$A$ 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?

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?

An isotope $_Y{A^X}$ undergoes a series of $m$ alpha and $n$ beta disintegrations to form a stable isotope $_{Y-10}{B^{X-32}}$. The values of $m$ and $n$ are respectively:

When a radioactive element emits an alpha particle,the daughter element is placed 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