Emission of a $\beta$-particle by an atom of an element results in the formation of its

The group displacement law was given by:

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:

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?

$A$ 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 $Group$ .......