Artificial transmutation Questions in English

(A) In a nuclear reaction,both the mass number $(A)$ and the atomic number $(Z)$ must be conserved on both sides of the equation.
For the reaction $_7N^{14} + _2He^4 \to _ZX^A + _1H^1$:
Sum of mass numbers on the left side: $14 + 4 = 18$.
Sum of mass numbers on the right side: $A + 1 = 18$,which gives $A = 17$.
Sum of atomic numbers on the left side: $7 + 2 = 9$.
Sum of atomic numbers on the right side: $Z + 1 = 9$,which gives $Z = 8$.
Therefore,the product nucleus is $_8O^{17}$.

(B) To find the missing particle,we balance the mass numbers and atomic numbers on both sides of the equation.
Let the missing particle be $_Z^A X$.
The equation is $_3Li^6 + _Z^A X \to _2He^4 + _1H^3$.
Sum of mass numbers: $6 + A = 4 + 3 \implies 6 + A = 7 \implies A = 1$.
Sum of atomic numbers: $3 + Z = 2 + 1 \implies 3 + Z = 3 \implies Z = 0$.
$A$ particle with mass number $1$ and atomic number $0$ is a neutron,denoted as $_0n^1$.
Therefore,the missing particle is a neutron.

(C) The production of carbon-$14$ in the upper atmosphere occurs when cosmic ray neutrons collide with nitrogen-$14$ nuclei.
The nuclear reaction is given by: $_7N^{14} + _0n^1 \to _6C^{14} + _1H^1$.
Therefore,the correct option is $C$.

(C) The nuclear reaction is given by: $_{Z}X^{A} + _{1}H^{2} \to _{18}Ar^{38} + _{0}n^{1}$.
Applying the conservation of atomic number: $Z + 1 = 18 \implies Z = 17$.
Applying the conservation of mass number: $A + 2 = 38 + 1 = 39 \implies A = 37$.
Thus,the target nuclide is $_{17}Cl^{37}$.

(B) In a nuclear reaction,the sum of atomic numbers and the sum of mass numbers must be conserved on both sides of the equation.
Let the product nucleus be $_{Z}X^{A}$.
For atomic numbers: $12 + 2 = 0 + Z \implies Z = 14$.
For mass numbers: $24 + 4 = 1 + A \implies 28 = 1 + A \implies A = 27$.
Thus,the product nucleus is $_{14}Si^{27}$.

(B) The correct option is $(B)$.
In the upper atmosphere,cosmic rays produce neutrons which react with nitrogen-$14$ to form carbon-$14$.
The nuclear reaction is given by:
$_7N^{14} + _0n^1 \to _6C^{14} + _1H^1$
Thus,the fundamental particle involved is a neutron.

(A) For the given nuclear reaction: $_{92}U^{238} \to _{82}Pb^{206} + x\,_{2}He^{4} + y\,_{-1}\beta^{0}$
$1$. Balancing the mass number: $238 = 206 + 4x + 0y \implies 4x = 32 \implies x = 8$.
$2$. Balancing the atomic number: $92 = 82 + 2x - y \implies 92 = 82 + 2(8) - y \implies 92 = 82 + 16 - y \implies 92 = 98 - y \implies y = 6$.
Therefore,the values are $x = 8$ and $y = 6$.

(D) In an $(n, p)$ nuclear reaction,a neutron is captured by the target nucleus and a proton is emitted.
This is represented as $Target(n, p)Product$.
In option $D$,the reaction is $_{21}Sc^{45} + _{0}n^{1} \rightarrow _{20}Ca^{45} + _{1}H^{1}$.
Here,a neutron $(_{0}n^{1})$ is absorbed and a proton ($_{1}H^{1}$ or $p$) is released,which corresponds to the $(n, p)$ type transformation.

(B) In a nuclear reaction,both the total atomic number and the total mass number must be conserved on both sides of the equation.
Let the mass number of $Be$ be $A$.
The reaction is $_4Be^A + _2He^4 \to _6C^{12} + _0n^1$.
Equating the mass numbers on both sides:
$A + 4 = 12 + 1$
$A + 4 = 13$
$A = 13 - 4 = 9$.
Therefore,the mass number of the $Be$ atom is $9$.

(A) An $(n, p)$ reaction is a nuclear reaction where a neutron is captured by a nucleus and a proton is emitted.
In the reaction $_{13}Al^{27} + _{0}n^{1} \xrightarrow{\quad} _{12}Mg^{27} + _{1}H^{1}$,an aluminum-$27$ nucleus captures a neutron and emits a proton to form magnesium-$27$.
This is represented as $_{13}Al^{27}(n, p)_{12}Mg^{27}$.

(B) The given nuclear reaction sequence is: $_{92}X^{238}$ $\xrightarrow{-\alpha} Y$ $\xrightarrow{-\beta} Z$ $\xrightarrow{-\beta} L$ $\xrightarrow{-n\alpha} _{84}M^{218}$.
$1$. An $\alpha$-decay decreases the mass number by $4$ and atomic number by $2$.
$2$. $A$ $\beta$-decay does not change the mass number but increases the atomic number by $1$.
Let the initial nucleus be $_{92}X^{238}$ and the final nucleus be $_{84}M^{218}$.
The total change in mass number is $\Delta A = 238 - 218 = 20$.
Since $\beta$-decay does not change the mass number,the entire change in mass number is due to $\alpha$-decays.
Each $\alpha$-particle carries a mass number of $4$.
Total $\alpha$-particles emitted = $1$ (first step) + $n$ (last step) = $1 + n$.
Therefore,$(1 + n) \times 4 = 20$.
$1 + n = 5$.
$n = 4$.

(B) To identify $X$,we apply the law of conservation of mass number and atomic number.
Let $X$ be represented as $_{Z}A^{M}$.
For mass number: $32 + M = 30 + 4 \implies 32 + M = 34 \implies M = 2$.
For atomic number: $16 + Z = 15 + 2 \implies 16 + Z = 17 \implies Z = 1$.
Thus,$X$ is $_{1}X^{2}$,which corresponds to the deuteron $_{1}D^{2}$ (or $_{1}H^{2}$).

(B) In a nuclear reaction,both the mass number $(A)$ and the atomic number $(Z)$ must be conserved on both sides of the equation.
For option $(B)$: $_{11}Na^{23} + _{1}H^{1} \to _{10}Ne^{20} + _{2}He^{4}$.
Sum of $Z$: $11 + 1 = 12$ $(LHS)$ and $10 + 2 = 12$ $(RHS)$.
Sum of $A$: $23 + 1 = 24$ $(LHS)$ and $20 + 4 = 24$ $(RHS)$.
Since both are conserved,this reaction is correct.

(A) To balance the nuclear equation,we must conserve the mass number and the atomic number on both sides.
Left side: Total mass number = $23 + 1 = 24$. Total atomic number = $11 + 1 = 12$.
Right side: Mass number of $Mg = 23$. Atomic number of $Mg = 12$.
Let $x$ be $_{Z}A^{A}$. Then $23 + A = 24 \implies A = 1$ and $12 + Z = 12 \implies Z = 0$.
The particle with mass number $1$ and atomic number $0$ is a neutron,represented as $_{0}n^{1}$.

(A) In a nuclear reaction,both the total atomic number and the total mass number must be conserved on both sides of the equation.
Let the emitted particle be $_{z}X^{a}$.
The reaction is: $_{13}Al^{27} + _{2}He^{4} \to _{15}P^{30} + _{z}X^{a}$.
Equating the mass numbers: $27 + 4 = 30 + a \implies 31 = 30 + a \implies a = 1$.
Equating the atomic numbers: $13 + 2 = 15 + z \implies 15 = 15 + z \implies z = 0$.
The particle is $_{0}n^{1}$,which is a neutron.

(B) The given nuclear reaction is $_4^9Be + _1^1H \to X + _2^4He$.
Applying the law of conservation of mass number: $9 + 1 = A + 4$,which gives $A = 6$.
Applying the law of conservation of atomic number: $4 + 1 = Z + 2$,which gives $Z = 3$.
Therefore,the product $X$ is $_3^6Li$.

(A) In a nuclear reaction,the sum of atomic numbers $(Z)$ and the sum of mass numbers $(A)$ must be conserved on both sides of the equation.
For the given reaction: $_{12}^{24}Mg + _{1}^{2}D \to _{2}^{4}He + _{Z}^{A}X$
Sum of atomic numbers $(Z)$: $12 + 1 = 2 + Z \implies Z = 11$.
Sum of mass numbers $(A)$: $24 + 2 = 4 + A \implies A = 22$.
The element with atomic number $11$ is Sodium $(Na)$.
Therefore,the missing nuclide is $_{11}^{22}Na$.

(B) In a nuclear reaction,both the atomic number (sum of protons) and the mass number (sum of protons and neutrons) are conserved.
For the atomic number: $Z + 2 = 15 + 0$,which gives $Z = 15 - 2 = 13$.
For the mass number: $M + 4 = 30 + 1$,which gives $M = 31 - 4 = 27$.
Therefore,$Z = 13$ and $M = 27$.

(A) The nuclear reaction is given by: $_{96}X^{227} \to Y + 4(_{2}He^{4}) + 5(_{-1}e^{0})$.
To find the mass number $(A)$ of $Y$: $227 = A + 4 \times 4 + 5 \times 0 \implies 227 = A + 16 \implies A = 211$.
To find the atomic number $(Z)$ of $Y$: $96 = Z + 4 \times 2 + 5 \times (-1) \implies 96 = Z + 8 - 5 \implies 96 = Z + 3 \implies Z = 93$.
Thus,the atomic number is $93$ and the mass number is $211$.

(C) The nuclear reaction for the conversion of $_7^{14}N$ to $_8^{17}O$ is given by:
$_7^{14}N + _2^{4}He \to _8^{17}O + _1^{1}H$
Here,$_2^{4}He$ is an $\alpha$-particle.
Thus,an $\alpha$-particle is used to convert $_7^{14}N$ into $_8^{17}O$.

(B) Let the number of $\alpha$-particles emitted be $x$ and the number of $\beta$-particles emitted be $y$.
The change in mass number is given by $231 - 207 = 4x$.
$24 = 4x \implies x = 6$.
The change in atomic number is given by $89 - 82 = 2x - y$.
$7 = 2(6) - y$.
$7 = 12 - y \implies y = 5$.
Thus,the number of $\alpha$-particles is $6$ and the number of $\beta$-particles is $5$.

(A) To check the correctness of a nuclear reaction,we must balance the atomic number $(Z)$ and the mass number $(A)$ on both sides.
$A$. $_{96}^{242}Cm + _{2}^{4}He \rightarrow _{97}^{243}Bk + 2_{0}^{1}n$.
Left side: $Z = 96+2 = 98$,$A = 242+4 = 246$.
Right side: $Z = 97+0 = 97$,$A = 243+2 = 245$.
Since $98 \neq 97$ and $246 \neq 245$,this notation is incorrect.
$B$. $_{5}^{10}B + _{2}^{4}He \rightarrow _{7}^{13}N + _{0}^{1}n$.
Left side: $Z = 5+2 = 7$,$A = 10+4 = 14$.
Right side: $Z = 7+0 = 7$,$A = 13+1 = 14$.
This is correct.
$C$. $_{7}^{14}N + _{0}^{1}n \rightarrow _{6}^{14}C + _{1}^{1}H$.
Left side: $Z = 7+0 = 7$,$A = 14+1 = 15$.
Right side: $Z = 6+1 = 7$,$A = 14+1 = 15$.
This is correct.
$D$. $_{14}^{28}Si + _{1}^{2}H \rightarrow _{15}^{29}P + _{0}^{1}n$.
Left side: $Z = 14+1 = 15$,$A = 28+2 = 30$.
Right side: $Z = 15+0 = 15$,$A = 29+1 = 30$.
This is correct.

(A) The nuclear reaction for the bombardment of $_8^{16}O$ with deuterons $( _1^{2}H )$ is given by:
$_8^{16}O + _1^{2}H \to _9^{18}F$.
In this reaction,the sum of atomic numbers and mass numbers on both sides must be conserved.
Atomic number: $8 + 1 = 9$ (which corresponds to Fluorine,$F$).
Mass number: $16 + 2 = 18$.
Thus,the product formed is $_9^{18}F$.

(C) The nuclear reaction is given by: $_{12}^{24}Mg + \gamma \rightarrow {}_{11}^{23}Na + {}_{1}^{1}H$.
In this reaction,a gamma photon strikes the magnesium nucleus,causing the emission of a proton $(_{1}^{1}H)$.
By balancing the atomic numbers $(12 = 11 + 1)$ and mass numbers $(24 = 23 + 1)$,we identify the product as the sodium nuclide $_{11}^{23}Na$.

(B) For the decay of $_{92}U^{238}$ to $_{82}Pb^{206}$:
Number of $\alpha$-particles $(n_{\alpha})$ = $\frac{238 - 206}{4} = \frac{32}{4} = 8$.
Number of $\beta$-particles $(n_{\beta})$ = $2 \times n_{\alpha} - (Z_{initial} - Z_{final}) = 2 \times 8 - (92 - 82) = 16 - 10 = 6$.
Thus,the number of $\alpha$-particles is $8$ and $\beta$-particles is $6$.

(D) In a nuclear reaction,the sum of mass numbers and the sum of atomic numbers must be conserved on both sides of the equation.
For option $(D)$: $_{13}Al^{27} + _{2}He^{4} \to _{15}P^{30} + _{0}n^{1}$.
Mass balance: $27 + 4 = 31$ (reactants) and $30 + 1 = 31$ (products).
Atomic number balance: $13 + 2 = 15$ (reactants) and $15 + 0 = 15$ (products).
Since the product side includes $_{0}n^{1}$,this reaction results in the emission of a neutron.

(D) The nuclear reaction is represented as: $_{90}Th^{232} \to {}_{82}Pb^{208} + x({}_{2}He^{4}) + y({}_{-1}\beta^{0})$
Equating the mass number on both sides:
$232 = 208 + 4x + 0y$
$4x = 232 - 208 = 24$
$x = 6$
Equating the atomic number on both sides:
$90 = 82 + 2x - y$
$90 = 82 + 2(6) - y$
$90 = 82 + 12 - y$
$90 = 94 - y$
$y = 4$
Therefore,$6$ $\alpha$-particles and $4$ $\beta$-particles are emitted.

(A) Let the parent nucleus be $_Z^A X$.
Two successive $\beta$-particle emissions are represented as: $_Z^A X \xrightarrow{2 \beta} {}_7^{14} N$.
Each $\beta$-emission increases the atomic number by $1$ while the mass number remains constant.
Therefore,$Z + 2 = 7$,which gives $Z = 5$.
The mass number $A$ remains $14$.
The parent nucleus is $_5^{14} B$.
The number of neutrons $N = A - Z = 14 - 5 = 9$.

(D) The age of the most ancient geological formations is estimated using the $Uranium-Lead$ dating method.
$Carbon-14$ dating is primarily used for dating organic materials that are relatively young (up to $50,000$ years old).
$Uranium-Lead$ dating relies on the decay of $U^{238}$ to $Pb^{206}$,which has a very long half-life,making it suitable for dating rocks and minerals that are millions or billions of years old.

(C) The given reaction $_3Li^6 + _1H^2 \to 2_2He^4 + \Delta E$ involves the combination of two light nuclei (Lithium and Deuterium) to form a heavier nucleus (Helium).
This process of combining two lighter nuclei to form a heavier nucleus with the release of energy is known as a $Fusion$ reaction.

(D) The first artificial disintegration of an atomic nucleus was achieved by $Ernest \ Rutherford$ in $1919$.
He bombarded nitrogen gas with alpha particles $(^4_2He)$ to produce oxygen and a proton:
$^{14}_7N + ^4_2He \rightarrow ^{17}_8O + ^1_1H$.

(C) In a nuclear reaction,both the total mass number and the total atomic number must be conserved on both sides of the equation.
For the mass number conservation:
$249 + 15 = A + (4 \times 1)$
$264 = A + 4$
$A = 264 - 4 = 260$
Thus,the mass number of element $X$ is $260$.

(A) The given reaction is $_{29}^{63}Cu + _{2}^{4}He \to _{17}^{37}Cl + 14_{1}^{1}H + 16_{0}^{1}n$.
In this reaction,a high-energy alpha particle $(_{2}^{4}He)$ strikes a copper nucleus,causing it to break into several smaller fragments including a chlorine nucleus,protons,and neutrons.
This process,where a heavy nucleus is broken into many smaller fragments by high-energy bombardment,is known as a spallation reaction.
Therefore,the correct option is $A$.

(D) Neutrons are neutral particles,which allows them to penetrate the nucleus of an atom without being repelled by the electrostatic force of the positively charged nucleus. Therefore,they are widely used as projectiles in nuclear reactions,such as artificial transmutation,to induce nuclear changes.

(B) The given reaction involves the combination of two light nuclei ($^2_1H$ or deuterium) to form a heavier nucleus $(^3_2He)$ along with the release of a neutron $(^1_0n)$.
This process of combining light nuclei to form a heavier nucleus is known as nuclear fusion.

(C) The given reaction is $_{17}Cl^{37} + _1H^2 \to _{18}Ar^{38} + _0n^1$.
In this process,a stable nucleus of chlorine is bombarded with a projectile (deuteron,$_1H^2$) to form a new element,argon,and a neutron.
This is a classic example of artificial transmutation,where one element is converted into another by nuclear bombardment.

(C) The correct answer is $(C)$.
Artificial transmutation involves bombarding a target nucleus with high-energy particles.
Charged particles like protons,deuterons,and $\alpha$-particles experience electrostatic repulsion from the target nucleus.
However,$\alpha$-particles are relatively heavy and have a higher charge $(+2e)$,which makes them less effective at penetrating the nucleus compared to lighter particles or uncharged particles like neutrons.

(C) The nuclear reaction is given by:
$_{13}Al^{28} + _{z}X^{a} \to _{15}P^{31} + _{0}n^{1}$
By balancing the atomic number: $13 + z = 15 + 0 \implies z = 2$.
By balancing the mass number: $28 + a = 31 + 1 \implies a = 4$.
The projectile with atomic number $2$ and mass number $4$ is an alpha particle $(_{2}He^{4})$.

(D) . Radioactive iodine,specifically the isotope $I^{131}$,is used in medical applications to diagnose and treat diseases of the thyroid gland,such as hyperthyroidism and thyroid cancer,because the thyroid gland selectively absorbs iodine.

(D) Artificial transmutation involves the bombardment of a target nucleus with a projectile to induce a nuclear reaction.
Neutrons are the most effective projectiles for this purpose because they are electrically neutral.
Unlike protons,deuterons,or alpha particles (helium nuclei),neutrons do not experience electrostatic repulsion from the positively charged target nucleus.
Therefore,they can penetrate the nucleus even at low kinetic energies,making them highly effective for inducing nuclear transmutations.

(D) In a nuclear reaction,both the mass number and the atomic number must be conserved on both sides of the equation.
For the mass number: $235 + 1 = 140 + x + 2(1) \implies 236 = 142 + x \implies x = 94$.
For the atomic number: $92 + 0 = 56 + y + 2(0) \implies 92 = 56 + y \implies y = 36$.
Therefore,the values are $x = 94$ and $y = 36$.

(C) The first controlled artificial disintegration of an atomic nucleus was achieved by $John \ Cockcroft$ and $Ernest \ Walton$ in $1932$. They bombarded lithium nuclei with high-energy protons accelerated by a particle accelerator,resulting in the production of alpha particles. Therefore,the correct option is $C$.

(D) Artificial radioactivity was first discovered by $Irene \ Curie$ and $Frederic \ Joliot$ in $1934$. They observed that when light elements like $Al$ or $Mg$ were bombarded with $\alpha$-particles,they emitted positrons and neutrons,exhibiting radioactivity.

(B) Artificial transmutation,also known as artificial radioactivity or induced transmutation,was first achieved by $Ernest \ Rutherford$ in $1919$. He bombarded nitrogen gas with alpha particles to produce oxygen and a proton,represented by the nuclear reaction: $^ {14}_{7}N + ^ {4}_{2}He \rightarrow ^ {17}_{8}O + ^ {1}_{1}H$.

(A) Nuclear fusion is a process in which two light nuclei combine to form a heavier nucleus,releasing a large amount of energy.
In option $A$,two deuterium nuclei $(_{1}H^{2})$ fuse to form a helium nucleus $(_{2}He^{4})$,which is a classic example of nuclear fusion.
Option $B$ represents nuclear fission,and option $C$ represents an artificial transmutation reaction.

(A) An $(n, p)$ reaction involves the capture of a neutron and the emission of a proton.
For the production of $_{27}^{60}Co$ from a target nucleus $X$ via $(n, p)$ reaction,the equation is: $_{Z}^{A}X + _{0}^{1}n \rightarrow _{27}^{60}Co + _{1}^{1}p$.
By balancing the mass number: $A + 1 = 60 + 1$,so $A = 60$.
By balancing the atomic number: $Z + 0 = 27 + 1$,so $Z = 28$.
The target nucleus is $_{28}^{60}Ni$.

(C) hydrogen bomb operates on the principle of uncontrolled nuclear fusion.
In this process,light nuclei,typically isotopes of hydrogen like deuterium $(^2H)$ and tritium $(^3H)$,fuse together at extremely high temperatures to form a heavier nucleus,such as helium $(^4He)$,releasing a tremendous amount of energy.

(B) The nuclear reaction is given by:
$_7N^{14} + _2He^4 \to _8O^{17} + _1H^1$
Here,the nitrogen isotope $_7N^{14}$ is bombarded with $\alpha$-particles $(_2He^4)$ to produce oxygen-$17$ $(_8O^{17})$ and a proton $(_1H^1)$.

(C) The nuclear decay reaction is represented as: $_{92}U^{235} \to _{82}Pb^{207} + x(_{2}He^{4}) + y(_{-1}\beta^{0})$.
Number of $\alpha$-particles $(x)$ is calculated by the change in mass number: $x = \frac{235 - 207}{4} = \frac{28}{4} = 7$.
Number of $\beta$-particles $(y)$ is calculated by the change in atomic number: $92 = 82 + 2(x) - y \implies 92 = 82 + 2(7) - y \implies 92 = 82 + 14 - y \implies 92 = 96 - y \implies y = 4$.
Thus,the emitted particles are $7\alpha$ and $4\beta$.