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(D) The Group Displacement Law describes the change in the position of an element in the periodic table upon radioactive decay.$1$. On emission of an

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$2$ groups to the left,$1$ group to the right

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(D) The Group Displacement Law describes the change in the position of an element in the periodic table upon radioactive decay.$1$. On emission of an $\alpha$-particle $(_{2}^{4}He^{2+})$,the atomic number of the parent element decreases by $2$,causing the daughter element to shift $2$ groups to the left in the periodic table.$2$. On emission of a $\beta$-particle $(_{-1}^{0}e)$,the atomic number of the parent element increases by $1$,causing the daughter element to shift $1$ group to the right in the periodic table.Therefore,the correct option is $D$.

Similar Questions

The group displacement law was given by:

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

$A$ 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?

$_{92}U^{235}$ 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:

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