Causes of radioactivity and Group displacement law Questions in English

(D) According to the group displacement law:
$1$. Emission of an $\alpha$-particle decreases the atomic number by $2$,shifting the element two groups to the left.
$2$. Emission of a $\beta$-particle increases the atomic number by $1$,shifting the element one group to the right.
Let the group of $Po$ be $G$.
After $\alpha$-emission,the group of $Pb$ is $(G - 2)$.
After $\beta$-emission from $Pb$,the group of $Bi$ is $(G - 2 + 1) = (G - 1)$.
Given that $Bi$ belongs to group $15$,we have $G - 1 = 15$,which implies $G = 16$.
Therefore,$Po$ belongs to group $16$.

(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 $D$.

(B) The radioactive decay of a nucleus by $\beta$-emission involves the conversion of a neutron into a proton,which increases the atomic number $(Z)$ by $1$.
According to the Group Displacement Law,when an element undergoes $\beta$-decay,its atomic number increases by $1$,causing the element to shift one group to the right in the periodic table.
Since $Nd$ $(Z = 60)$ belongs to Group $3$,the daughter nucleus will have an atomic number of $61$ $(Pm)$ and will belong to Group $4$.

(D) The nuclear reaction is given by: $_Y{A^X} \to {}_{Y-10}{B^{X-32}} + m({}_2He^4) + n({}_{-1}e^0)$.
For mass number conservation: $X = (X - 32) + 4m$,which gives $4m = 32$,so $m = 8$.
For atomic number conservation: $Y = (Y - 10) + 2m - n$. Substituting $m = 8$: $Y = Y - 10 + 16 - n$,which simplifies to $0 = 6 - n$,so $n = 6$.
Therefore,the values of $m$ and $n$ are $8$ and $6$ respectively.

(B) Alkaline earth metals belong to $Group$ $2$ of the periodic table.
When a radioactive element emits one $\alpha$-particle $(_{2}He^{4})$,its atomic number decreases by $2$,shifting its position in the periodic table by two groups to the left.
Starting from $Group$ $2$:
After the first $\alpha$-emission: $2 - 2 = 0$ (which corresponds to $Group$ $18$ in the periodic table).
After the second $\alpha$-emission: $18 - 2 = 16$.
Therefore,the resulting daughter element belongs to $Group$ $16$.

(C) The group displacement law,also known as the $Soddy-Fajan$ rule,was proposed by $Frederick \ Soddy$ and $Kasimir \ Fajan$ in $1913$. This law describes the change in the position of an element in the periodic table when it undergoes radioactive decay.

(D) According to the Group Displacement Law,when an element emits an $\alpha$-particle $(_{2}He^{4})$,its atomic number decreases by $2$ and it moves two groups to the left in the periodic table.
Since $Pb$ (lead) belongs to group $IV A$,the parent element $Po$ (polonium) must be located two groups to the right of $Pb$.
Therefore,the group of $Po = IV A + 2 = VI A$.

(A) When a parent nucleus emits a $\beta^{-}$ particle (an electron),the atomic number $(Z)$ of the nucleus increases by $1$ while the mass number $(A)$ remains unchanged.
According to the group displacement law,an increase in the atomic number by $1$ shifts the daughter element one place to the right in the periodic table.

(D) According to the group displacement law,when an element emits an $\alpha$-particle,its atomic number decreases by $2$ and its mass number decreases by $4$.
As a result,the element shifts two positions to the left in the periodic table.
Since radium is in group $II$,the new element formed will be in group $II - 2 = 0$ (the noble gas group).

(A) According to the Group Displacement Law,when a radioactive element emits an $\alpha$-particle $(^4_2He^{2+})$,its atomic number decreases by $2$ and its mass number decreases by $4$.
As the atomic number decreases by $2$,the daughter element shifts two groups to the left of the parent element in the periodic table.

(D) When an atom emits a $\beta$-particle $(_{-1}^{0}e)$,its atomic number increases by $1$ while the mass number remains unchanged.
For example,the decay of tritium is represented as: $_{1}^{3}H \to _{2}^{3}He + _{-1}^{0}e$.
Since $_{1}^{3}H$ and $_{2}^{3}He$ have the same mass number $(A=3)$ but different atomic numbers,they are classified as isobars.

(C) The emission of an $\alpha$-particle $(_{2}He^{4})$ results in the decrease of the atomic number by $2$ units.
Since the original element is Uranium $(Z = 92)$,the new element will have an atomic number of $92 - 2 = 90$.
The element with atomic number $90$ is Thorium $(Th)$.
Thorium belongs to the Actinide series,which are all placed in group $III-B$ of the periodic table.

(A) The emission of an $\alpha$-particle $(_{2}^{4}He^{2+})$ results in the atomic number of the parent element decreasing by $2$ and the mass number decreasing by $4$.
Given the parent element is ${}_{92}^{238}U$ in group $III B$.
After the emission of an $\alpha$-particle,the atomic number becomes $92 - 2 = 90$.
The element with atomic number $90$ is Thorium $(Th)$.
According to the periodic table,Thorium $(Th)$ belongs to the Actinide series,which is placed in group $III B$ (or group $3$ in modern $IUPAC$ notation).

(D) The initial radioactive substance is $_{88}^{228}X$.
Each $\alpha$ emission decreases the atomic number by $2$ and the mass number by $4$.
Each $\beta$ emission increases the atomic number by $1$ and does not change the mass number.
After emitting $3$ $\alpha$ and $3$ $\beta$ particles,the atomic number of $Y$ becomes: $88 - (3 \times 2) + (3 \times 1) = 88 - 6 + 3 = 85$.
The element with atomic number $85$ is Astatine $(At)$,which belongs to group $VIIA$ (Halogens) of the periodic table.

(D) The initial element is ${}_{88}X^{228}$ belonging to group $IIA$ (atomic number $Z = 88$).
When an element emits one $\alpha$-particle,its atomic number decreases by $2$ and mass number by $4$.
When an element emits one $\beta$-particle,its atomic number increases by $1$ and mass number remains unchanged.
For the emission of $3 \alpha$ and $3 \beta$ particles:
Change in atomic number $(Z)$ = $(3 \times -2) + (3 \times 1) = -6 + 3 = -3$.
New atomic number of $Y$ = $88 - 3 = 85$.
An element with atomic number $88$ is Radium $(Ra)$,which is in group $IIA$ (alkaline earth metals).
Group $IIA$ elements have a valence shell configuration of $ns^2$.
Atomic number $88$ is in the $7^{th}$ period. The element with atomic number $85$ is Astatine $(At)$.
Astatine $(Z=85)$ belongs to the halogen family,which is group $VIIA$ (or group $17$).