Artificial transmutation Questions in English

(C) In a nuclear fission reaction,both the mass number and the atomic number must be conserved on both sides of the equation.
Let the missing particle be $_{Z}^{A}X$.
The equation is: $_{92}^{235}U + _{0}^{1}n \to _{56}^{146}Ba + _{Z}^{A}X + 3_{0}^{1}n$.
Conservation of mass number: $235 + 1 = 146 + A + 3(1) \implies 236 = 149 + A \implies A = 87$.
Conservation of atomic number: $92 + 0 = 56 + Z + 3(0) \implies 92 = 56 + Z \implies Z = 36$.
The element with atomic number $36$ is Krypton $(Kr)$.
Thus,the missing particle is $_{36}^{87}Kr$.

(D) The nuclear reaction is given by: $_{92}U^{235} + _{0}n^{1} \to _{54}Xe^{139} + _{38}Sr^{94} + x(_{0}n^{1})$.
Applying the law of conservation of mass number: $235 + 1 = 139 + 94 + x$.
$236 = 233 + x$.
$x = 3$.
Therefore,the product $X$ is $3$ neutrons.

(B) The nuclear reaction is given by:
$_3Li^7 + _1H^1 \to _4Be^8 + \gamma$
In this reaction,a lithium-$7$ nucleus captures a proton $(H^+)$ to form a beryllium-$8$ nucleus,accompanied by the emission of $\gamma$-rays.

(A) For the first reaction: $_7N^{14} + _2He^4 \to _8O^{17} + X_1$.
Sum of atomic numbers: $7 + 2 = 8 + Z_1 \implies Z_1 = 1$.
Sum of mass numbers: $14 + 4 = 17 + A_1 \implies A_1 = 1$.
Thus,$X_1$ is $_1H^1$ (a proton).
For the second reaction: $_{13}Al^{27} + _1D^2 \to _{14}Si^{28} + X_2$.
Sum of atomic numbers: $13 + 1 = 14 + Z_2 \implies Z_2 = 0$.
Sum of mass numbers: $27 + 2 = 28 + A_2 \implies A_2 = 1$.
Thus,$X_2$ is $_0n^1$ (a neutron).
Therefore,$X_1$ and $X_2$ are $_1H^1$ and $_0n^1$ respectively.

(B) The nuclear reaction is given by: $_{13}Al^{27} + _{2}He^{4} \to _{15}P^{30} + _{0}n^{1}$.
Here,the particle used to bombard the aluminum nucleus is the $\alpha-$ particle $(_{2}He^{4})$.

(C) The nuclear transformation is ${}_{92}^{238}U \rightarrow {}_{92}^{234}U + \text{emissions}$.
First,the emission of one $\alpha$-particle $({}_{2}^{4}He)$ results in: ${}_{92}^{238}U \rightarrow {}_{90}^{234}Th + {}_{2}^{4}He$.
To reach the final product ${}_{92}^{234}U$,the ${}_{90}^{234}Th$ must undergo beta decay: ${}_{90}^{234}Th \rightarrow {}_{91}^{234}Pa + {}_{-1}^{0}e$ and ${}_{91}^{234}Pa \rightarrow {}_{92}^{234}U + {}_{-1}^{0}e$.
Thus,the total emissions are one $\alpha$-particle and two $\beta^-$-particles.

(D) The nuclear fission reaction is represented as: ${}_{92}^{235} U + {}_{0}^{1} n \rightarrow {}_{54}^{142} Xe + {}_{38}^{90} Sr + x({}_{0}^{1} n)$.
By balancing the mass number: $235 + 1 = 142 + 90 + x$.
$236 = 232 + x$.
$x = 236 - 232 = 4$.
Therefore,the number of neutrons emitted is $4$.

(A) The reaction for path $(i)$ is: ${ }_{13}^{27} Al +{ }_2^4 \alpha \rightarrow{ }_{14}^{30} Si +{ }_1^1 p (X)$.
Here,$X$ is a proton $({ }_1^1 p)$ because the mass number and atomic number are conserved ($27+4 = 30+1$ and $13+2 = 14+1$).
The reaction for path $(ii)$ is: ${ }_{13}^{27} Al +{ }_2^4 \alpha \rightarrow{ }_{15}^{30} P +{ }_0^1 n (Y)$.
Here,$Y$ is a neutron $({ }_0^1 n)$ because the mass number and atomic number are conserved ($27+4 = 30+1$ and $13+2 = 15+0$).
The subsequent decay of phosphorus is: ${ }_{15}^{30} P \rightarrow{ }_{14}^{30} Si +{ }_{+1}^{0} e (Z)$.
Here,$Z$ is a positron $({ }_{+1}^{0} e)$ because the atomic number decreases by $1$.
Therefore,$X$ is proton,$Y$ is neutron,and $Z$ is positron.
Thus,the correct option is $A$.

(A) The given nuclear reaction is $_4^9 Be + X \longrightarrow _4^8 Be + Y$.
For the reaction to be balanced,the sum of atomic numbers and mass numbers must be conserved on both sides.
Case $1$: If $X = \gamma$ (photon,$0^0\gamma$),then $_4^9 Be + _0^0\gamma \longrightarrow _4^8 Be + _0^1 n$. Here $Y = n$ (neutron).
Case $2$: If $X = p$ (proton,$_1^1 p$),then $_4^9 Be + _1^1 p \longrightarrow _4^8 Be + _1^2 D$. Here $Y = D$ (deuteron).
Thus,both $(A)$ and $(B)$ are valid pairs for $(X, Y)$.

(D) The nuclear decay reaction for ${ }_{92}^{238} U$ to ${ }_{82}^{206} Pb$ is given by: ${ }_{92}^{238} U \rightarrow { }_{82}^{206} Pb + 8 { }_{2}^{4} He + 6 { }_{-1}^{0} e$.
Initially, the system contains $1 \ mol$ of air (gaseous) and $1 \ mol$ of solid ${ }_{92}^{238} U$.
Since the vessel is rigid and the temperature is constant at $298 \ K$, the pressure is directly proportional to the number of moles of gas present $(P \propto n_{gas})$.
Initial moles of gas $(n_i)$ = $1 \ mol$ (air).
After complete decay, the solid ${ }_{92}^{238} U$ is converted into $1 \ mol$ of solid ${ }_{82}^{206} Pb$ and $8 \ mol$ of gaseous ${ }_{2}^{4} He$ ($\alpha$-particles).
The final moles of gas $(n_f)$ = $1 \ mol$ (air) + $8 \ mol$ (He) = $9 \ mol$.
The ratio of the final pressure to the initial pressure is given by $P_f / P_i = n_f / n_i = 9 / 1 = 9$.

(B) The nuclear reaction is given by: ${ }_{x} A^{y} \longrightarrow { }_{82} B^{207} + 5 { }_{2} \alpha^{4} + 4 { }_{-1} \beta^{0}$
By balancing the atomic number $(x)$: $x = 82 + (5 \times 2) + (4 \times -1) = 82 + 10 - 4 = 88$
By balancing the mass number $(y)$: $y = 207 + (5 \times 4) + (4 \times 0) = 207 + 20 = 227$
Thus,the element $A$ is ${ }_{88} A^{227}$.
Number of protons = $88$
Number of neutrons = $227 - 88 = 139$

(A) The balanced nuclear reaction is:
$_7N^{14} + _0n^1 \longrightarrow _6C^{12} + _1T^3$
Here,$_1T^3$ (Tritium) is a radioactive isotope of hydrogen,which is produced in the upper atmosphere by the interaction of cosmic ray neutrons with nitrogen-$14$ nuclei.

(C) In a nuclear reaction,the sum of atomic numbers and the sum of mass numbers must be conserved on both sides.
For atomic numbers: $12 + 1 = 12 + Z$,which gives $Z = 1$.
For mass numbers: $26 + 2 = 27 + A$,which gives $A = 1$.
Since the atomic number is $1$ and the mass number is $1$,the particle $X$ is a proton,represented as ${}_{1}H^{1}$.

(C) Let the unknown substance be $_{Z}X^{A}$.
The nuclear reaction is: $_{Z}X^{A} + _{6}C^{12} \rightarrow _{98}Cf^{246} + _{0}n^{1}$.
According to the law of conservation of mass number: $A + 12 = 246 + 1$,which gives $A = 235$.
According to the law of conservation of atomic number: $Z + 6 = 98$,which gives $Z = 92$.
The element with atomic number $92$ is Uranium $(U)$.
Therefore,the unknown substance is $_{92}U^{235}$.

(B) The given nuclear reaction is: $_{5}B^{10} + _{2}He^{4} \rightarrow _{b}^{a}X + _{0}n^{1}$
By balancing the atomic numbers (sum of protons):
$5 + 2 = b + 0$
$b = 7$
By balancing the mass numbers (sum of protons and neutrons):
$10 + 4 = a + 1$
$14 = a + 1$
$a = 13$
Therefore,the product $X$ is $_{7}N^{13}$.

(B) The reaction is $^{14}N + {}_{2}^{4}He \rightarrow {}^{17}O + {}_{1}^{1}H$.
Since energy is absorbed,the mass of the products is greater than the mass of the reactants.
Mass defect $(\Delta m) = 0.00124 \ amu$.
$(m_{{}^{17}O} + m_{p}) - (m_{reactants}) = \Delta m$.
$m_{{}^{17}O} + 1.00782 \ amu - 18.00567 \ amu = 0.00124 \ amu$.
$m_{{}^{17}O} = 18.00567 + 0.00124 - 1.00782 = 16.99909 \ amu$.
Rounding to four decimal places,we get $16.9991 \ amu$.