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:

$Nd$ $(Z = 60)$ 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 $...$ ...

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

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

$A$ 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?

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