Isotopes-Isotones and Nuclear isomers Questions in English

(A) Let the mass number and atomic number of $X$ be $A$ and $Z_{num}$ respectively.
$X(A, Z_{num}) \xrightarrow{-\alpha} Y(A-4, Z_{num}-2)$
$Y(A-4, Z_{num}-2) \xrightarrow{-\beta} Z(A-4, Z_{num}-1)$
$Z(A-4, Z_{num}-1) \xrightarrow{-\beta} W(A-4, Z_{num})$
Isotopes have the same atomic number but different mass numbers.
Comparing $X(A, Z_{num})$ and $W(A-4, Z_{num})$,they have the same atomic number $Z_{num}$ but different mass numbers $A$ and $A-4$.
Therefore,$X$ and $W$ are isotopes.

(B) The reaction sequence is: $_{92}U^{235}$ $\xrightarrow{-\alpha } (A)$ $\xrightarrow{-\beta } (B)$ $\xrightarrow{-\beta } (C)$
$1$. Alpha decay of $_{92}U^{235}$ results in $(A)$:
$_{92}U^{235} \xrightarrow{-\alpha } _{90}A^{231}$
$2$. Beta decay of $(A)$ results in $(B)$:
$_{90}A^{231} \xrightarrow{-\beta } _{91}B^{231}$
$3$. Beta decay of $(B)$ results in $(C)$:
$_{91}B^{231} \xrightarrow{-\beta } _{92}C^{231}$
Isotopes are atoms of the same element having the same atomic number but different mass numbers. Here,$_{92}U^{235}$ and $_{92}C^{231}$ both have an atomic number of $92$. Therefore,the isotopes are $_{92}U^{235}$ and $C$.

(A) The radioactive isotope of carbon,$_{6}C^{14}$,is used as a tracer to study the metabolic pathways and the mechanism of photosynthesis in plants.
Therefore,the correct option is $A$.

(C) Let the initial atom be $^m_Z X$.
After the emission of an $\alpha$-particle $(^4_2 He)$,the nucleus becomes $^{m-4}_{Z-2} Y$.
Then,after the emission of two $\beta$-particles $(^0_{-1} e)$,the mass number remains $m-4$ and the atomic number becomes $(Z-2) + 2 = Z$.
Since the final element has the same atomic number $Z$ as the initial atom,it is an isotope.

(C) The correct option is $(C)$.
$Co^{60}$ (Cobalt-$60$) is a radioactive isotope that emits gamma rays,which are used in radiotherapy to destroy cancer cells.

(A) Let the nucleus $R$ be represented as $_Z A^m$.
$1$. After $\alpha$-decay: $_Z A^m \xrightarrow{\alpha} _{Z-2} X^{m-4}$.
$2$. After first $\beta$-decay: $_{Z-2} X^{m-4} \xrightarrow{\beta} _{Z-1} Y^{m-4}$.
$3$. After second $\beta$-decay: $_{Z-1} Y^{m-4} \xrightarrow{\beta} _Z Z^{m-4}$.
Thus,$R$ $(_{Z} A^m)$ and $Z$ $(_{Z} Z^{m-4})$ have the same atomic number $(Z)$ but different mass numbers ($m$ and $m-4$).
Therefore,$R$ and $Z$ are isotopes.

(B) Isotopes are atoms of the same element that have the same atomic number $(Z)$ but different mass numbers $(A)$.
Since the atomic number is defined by the number of protons,isotopes have the same number of protons.
However,because the mass number is the sum of protons and neutrons $(A = Z + N)$,a difference in mass number implies a difference in the number of neutrons $(N)$.
Therefore,isotopes differ in the number of neutrons.

(C) An isotope has the same atomic number $(Z)$ but a different mass number $(A)$.
When a nucleus loses one $\alpha$-particle $(_{2}He^{4})$,the atomic number decreases by $2$ and the mass number decreases by $4$.
When a nucleus loses one $\beta$-particle $(_{-1}e^{0})$,the atomic number increases by $1$ and the mass number remains unchanged.
To maintain the same atomic number $Z$,the net change in $Z$ must be zero.
Loss of one $\alpha$-particle $(Z-2)$ followed by two $\beta$-particles $(Z-2+1+1 = Z)$ results in the same atomic number.
The mass number changes by $-4$ (from the $\alpha$-particle),thus producing an isotope.
Therefore,the correct option is $(C)$.

(D) Isotones are atoms of different elements that have the same number of neutrons but different atomic numbers $(Z)$ and different mass numbers $(A)$.
For example,$_{18}^{39}Ar$ and $_{19}^{40}K$ both have $21$ neutrons ($39-18 = 21$ and $40-19 = 21$).

(B) Atoms of different elements having different atomic numbers but the same mass number are called isobars.
In the given pair,$_6C^{11}$ and $_5B^{11}$,both have the same mass number $(A = 11)$ but different atomic numbers ($Z = 6$ and $Z = 5$).
Therefore,they are isobars.

(D) When an element $A$ emits an $\alpha$-particle $(^4_2He)$,its mass number decreases by $4$ and its atomic number decreases by $2$.
Let $A$ be represented as $^Z_A X^M$,where $M$ is the mass number and $Z$ is the atomic number.
After $\alpha$-emission,the new element $B$ is $^Z_{A-2} Y^{M-4}$.
Since the mass number changes by $4$ and the atomic number changes by $2$,the difference between the number of neutrons $(N = M - Z)$ for $A$ and $B$ is:
$N_A = M - Z$
$N_B = (M - 4) - (Z - 2) = M - Z - 2$
The difference in the number of neutrons is $(N_A - Z_A) - (N_B - Z_B) = 0$,which defines isodiaspheres (elements having the same $(N - Z)$ value).

(B) Atoms of different elements having different atomic numbers but the same mass number are called isobars.

(D) Natural uranium consists of three primary isotopes: $_{92}U^{238}$ (approx. $99.27\%$),$_{92}U^{235}$ (approx. $0.72\%$),and $_{92}U^{234}$ (approx. $0.0055\%$).
$_{92}U^{239}$ is a synthetic isotope produced in nuclear reactors by neutron capture of $_{92}U^{238}$ and is not found in significant quantities in natural uranium deposits.

(A) Isotones are nuclei that have the same number of neutrons $(n = A - Z)$.
For $^{14}_{6}C$: $n = 14 - 6 = 8$.
For $^{15}_{7}N$: $n = 15 - 7 = 8$.
For $^{17}_{9}F$: $n = 17 - 9 = 8$.
Since all three nuclei have $8$ neutrons,they are isotonic.

(B) Isotopes are atoms of the same element having the same atomic number but different atomic masses.
Neutron has an atomic number of $0$ and an atomic mass of $1$.
Therefore,the loss or gain of a neutron changes the mass number while keeping the atomic number constant,thus generating an isotope.
For example: $_{92}U^{238} + _{0}n^{1} \rightarrow _{92}U^{239}$

(C) The decay process is $_{35}X^{88}$ $\xrightarrow{I} Y$ $\xrightarrow{II} _{32}Z^{84}$.
Step $I$: $_{35}X^{88} \rightarrow Y + _{1}e^{0}$ (Positron emission) gives $Y = _{34}Y^{88}$.
Step $II$: $_{34}Y^{88} \rightarrow _{32}Z^{84} + _{2}He^{4}$ ($\alpha$ decay).
Alternatively,if $I$ is $\alpha$ decay: $_{35}X^{88} \rightarrow _{33}Y^{84} + _{2}He^{4}$,then $II$ is $\beta^-$ decay: $_{33}Y^{84} \rightarrow _{32}Z^{84} + _{-1}e^{0}$.
Checking isodiaphers: Isodiaphers have the same $(N-Z)$ value.
For $Y$ $(_{34}Y^{88})$: $N-Z = (88-34) - 34 = 54 - 34 = 20$.
For $Z$ $(_{32}Z^{84})$: $N-Z = (84-32) - 32 = 52 - 32 = 20$.
Since both have $N-Z = 20$,$Y$ and $Z$ are isodiaphers.

(B) Isotones are atoms that have the same number of neutrons.
For $_{32}Ge^{76}$,the number of neutrons $= 76 - 32 = 44$.
For $_{33}As^{77}$,the number of neutrons $= 77 - 33 = 44$.
Since both have $44$ neutrons,$_{33}As^{77}$ is an isotone of $_{32}Ge^{76}$.

(A) Isotones are atoms or nuclei that contain the same number of neutrons $(n = A - Z)$.
For $_{6}^{14}C$: $n = 14 - 6 = 8$.
For $_{7}^{15}N$: $n = 15 - 7 = 8$.
For $_{9}^{17}F$: $n = 17 - 9 = 8$.
Since all three nuclei have $8$ neutrons,they are isotones.

(D) . $A$ nucleus $X$ captures a $\beta$-particle and then emits a neutron and $\gamma$-ray to form $Y$. The nuclear reaction is as follows:
${}_{Z}X^{A} + {}_{-1}e^{0}$ $\rightarrow {}_{Z-1}X^{A}$ $\xrightarrow{-{}_{0}n^{1}} {}_{Z-1}Y^{A-1} + \gamma$
Here,$A$ is the mass number of $X$ and $Z$ is the atomic number of $X$.
Number of neutrons in $X = A - Z$.
Number of neutrons in $Y = (A - 1) - (Z - 1) = A - Z$.
Since both $X$ and $Y$ have the same number of neutrons,they are isotones.

(B) Isotones are atoms of different elements that have the same number of neutrons in their nuclei.
To find the number of neutrons,we use the formula: $n = A - Z$,where $A$ is the mass number and $Z$ is the atomic number.
For ${ }_{6}^{12}C$: $n = 12 - 6 = 6$.
For ${ }_{5}^{11}B$: $n = 11 - 5 = 6$.
Since both have $6$ neutrons,they are isotones.

(D) Isotones are atoms of different elements that have the same number of neutrons in their nuclei.
To find the number of neutrons,we use the formula: $n = A - Z$,where $A$ is the mass number and $Z$ is the atomic number.
For $_{6}^{12}C$: $n = 12 - 6 = 6$.
For $_{5}^{11}B$: $n = 11 - 5 = 6$.
Since both have $6$ neutrons,they are isotones.

(D) Isotones are defined as atoms or nuclides that possess the same number of neutrons $(N)$ but different numbers of protons $(Z)$,which consequently leads to different mass numbers $(A = Z + N)$.
Since they have different atomic numbers,they exhibit different chemical properties and occupy different positions in the periodic table.

(B) $S^{35}$ is a radioactive isotope of sulphur.
When it comes in contact with the skin,it can cause skin irritation or dermatitis.

(C) Isotones are atoms of different elements that have the same number of neutrons.
For ${ }_{19}K^{39}$,number of neutrons = $39 - 19 = 20$.
For ${ }_{20}Ca^{40}$,number of neutrons = $40 - 20 = 20$.
Since both have $20$ neutrons,they are isotones.

(B) The relative abundance of ${}^{14}C$ isotope in nature is extremely low,approximately $10^{-10} \%$.
Among the given options,$2 \times 10^{-10}$ is the correct value.

(B) Isotones are species having the same number of neutrons.
In $_{20}Ca^{40}$,the number of neutrons is $40 - 20 = 20$.
In $_{18}Ar^{40}$,the number of neutrons is $40 - 18 = 22$.
Since the number of neutrons is different,$_{20}Ca^{40}$ and $_{18}Ar^{40}$ are not isotones.
Therefore,the statement in option $B$ is incorrect.

(C) Let the initial element be ${}_{Z}E^{M}$,where $Z$ is the atomic number and $M$ is the atomic mass.
When an $\alpha$-particle $({}_{2}He^{4})$ is emitted,the atomic number decreases by $2$ and the atomic mass decreases by $4$: ${}_{Z}E^{M} \rightarrow {}_{Z-2}E'^{M-4} + {}_{2}He^{4}$.
When two $\beta$-particles $({}_{-1}e^{0})$ are emitted,the atomic number increases by $1$ for each $\beta$-particle (total increase of $2$),while the atomic mass remains unchanged: ${}_{Z-2}E'^{M-4} \rightarrow {}_{Z}E''^{M-4} + 2({}_{-1}e^{0})$.
The final element is ${}_{Z}E''^{M-4}$.
Since the final element has the same atomic number $Z$ as the initial element $E$ but a different atomic mass $M-4$,it is an isotope of $E$.