Nuclear Fission, Fusion and Nuclear Reactor Questions in English

(A) The fusion reaction of deuterium is given by: $_1H^2 + _1H^2 \to _2He^3 + _0n^1 + 3.27\,MeV$.
In $1\, kg$ of deuterium $(^2H)$,the number of atoms is $N = \frac{1000}{2} \times 6.022 \times 10^{23} = 3.011 \times 10^{26}$ atoms.
Since each fusion reaction involves two deuterium nuclei,the number of fusion events is $N_{fusion} = \frac{N}{2} = 1.5055 \times 10^{26}$.
The energy released per reaction is $3.27\,MeV = 3.27 \times 1.602 \times 10^{-13}\,J \approx 5.238 \times 10^{-13}\,J$.
Total energy $E = N_{fusion} \times 5.238 \times 10^{-13}\,J = 1.5055 \times 10^{26} \times 5.238 \times 10^{-13} \approx 7.88 \times 10^{13}\,J$.
Rounding this gives approximately $8 \times 10^{13}\,J$.

(C) neutron moderator is a medium that reduces the speed of fast neutrons,typically emitted from nuclear fission,so they can sustain a chain reaction.
Heavy water $(D_2O)$ is considered the best neutron moderator because it has a very low neutron absorption cross-section compared to ordinary water,while its deuterium atoms are effective at slowing down neutrons through elastic collisions.
Therefore,heavy water is the most efficient moderator for nuclear reactors.

(A) Nuclear fission was discovered in $1938$ by the German chemists Otto Hahn and Fritz Strassmann. They observed that when uranium was bombarded with neutrons,it split into lighter elements like barium. This discovery was later explained by Lise Meitner and Otto Frisch. Therefore,the correct option is $A$.

(B) Control rods are used in nuclear reactors to control the fission rate of uranium and plutonium.
This is achieved by absorbing the excess neutrons from the reactor core.
The process relies on materials that possess a strong neutron absorption capacity.
Boron is one such element.
By regulating the insertion of boron rods into the reactor core during the fission process,the excess neutrons produced during the chain reaction are absorbed,thereby regulating the availability of neutrons and maintaining a steady reaction rate.

(D) Energy released in the fission of one nucleus $E = 200 \, MeV$.
Converting this into Joules: $E = 200 \times 10^6 \times 1.6 \times 10^{-19} \, J = 3.2 \times 10^{-11} \, J$.
The power of the plant is $P = 16 \, kW = 16 \times 10^3 \, W$.
The number of nuclei required per second $n$ is given by the ratio of power to the energy released per nucleus:
$n = \frac{P}{E} = \frac{16 \times 10^3}{3.2 \times 10^{-11}} = 5 \times 10^{14} \, \text{nuclei/s}$.

(A) Given: Power $P = 3.2 \text{ MW} = 3.2 \times 10^6 \text{ J/s}$.
Energy released per fission $E = 200 \text{ MeV} = 200 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-11} \text{ J}$.
Number of fissions per second $n = \frac{P}{E} = \frac{3.2 \times 10^6}{3.2 \times 10^{-11}} = 10^{17} \text{ fissions/s}$.
Number of fissions per minute $= n \times 60 = 10^{17} \times 60 = 6 \times 10^{18} \text{ fissions/min}$.

(C) In the process of nuclear fission of ${U^{235}}$ induced by thermal neutrons,the nucleus splits into two lighter fragments and releases energy along with several neutrons.
Experimental measurements show that the average number of neutrons produced per fission of ${U^{235}}$ is approximately $2.4$ to $2.5$.
Therefore,the value lies in the range of $2$ to $3$.
Thus,the correct option is $C$.

(A) In an atomic bomb,the energy is released due to the uncontrolled chain reaction of neutrons and $_{92}U^{235}$.
The nuclear fission reaction taking place is as follows:
$_{92}U^{235} + _{0}n^{1} \rightarrow _{56}Ba^{141} + _{36}Kr^{92} + 3_{0}n^{1} + Q \text{ (approx. } 200 \text{ MeV energy)}$.
The three neutrons generated in each fission event can react with three other $_{92}U^{235}$ nuclei,which leads to a self-sustaining chain reaction,releasing a massive amount of energy in a very short time.

(B) The nuclear fusion reaction is given by: ${}_1^2H + {}_1^2H \to {}_2^4He + Q$.
Total binding energy of the reactants: $2 \times (2 \times 1.1 \; MeV) = 4.4 \; MeV$.
Total binding energy of the product: $4 \times 7.1 \; MeV = 28.4 \; MeV$.
The energy released $Q$ is the difference between the total binding energy of the product and the reactants:
$Q = 28.4 \; MeV - 4.4 \; MeV = 24 \; MeV$.

(C) The number of nuclei in $1\, kg$ of $_{92}{U^{235}}$ is calculated using Avogadro's number $(N_A = 6.023 \times 10^{23}\, \text{atoms/mol})$.
Number of nuclei $N = \frac{\text{mass}}{\text{molar mass}} \times N_A = \frac{1000\, g}{235\, g/mol} \times 6.023 \times 10^{23}\, \text{nuclei/mol}$.
$N \approx 2.563 \times 10^{24}\, \text{nuclei}$.
Energy released per nucleus is $200\, MeV$.
Total energy $E = N \times 200\, MeV = 2.563 \times 10^{24} \times 200 \times 1.602 \times 10^{-13}\, J$.
$E \approx 8.21 \times 10^{13}\, J$.
Thus, the correct option is $C$.

(A) $^{235}U$ is a fissile isotope,meaning it can undergo nuclear fission when it captures a thermal (slow) neutron. This process releases energy and additional neutrons,which can sustain a chain reaction. In contrast,$^{238}U$ is fertile but not fissile with thermal neutrons; it requires fast neutrons to undergo fission,and even then,it is much less likely to fission compared to $^{235}U$. Spontaneous fission is a rare decay mode for $^{235}U$ compared to alpha decay. Therefore,the statement that $^{235}U$ is fissionable by thermal neutrons is correct.

(C) The binding energy per nucleon of fission products is approximately $8.5 \text{ MeV}$.
The binding energy per nucleon of the reactants (heavy nuclei like Uranium) is approximately $7.6 \text{ MeV}$.
The increase in binding energy per nucleon is $8.5 \text{ MeV} - 7.6 \text{ MeV} = 0.9 \text{ MeV}$.
The energy released per nucleon in fission is $0.9 \text{ MeV}$.
The fractional energy released is given by the ratio of energy released to the rest mass energy per nucleon,which is approximately $\frac{0.9 \text{ MeV}}{931 \text{ MeV}} \approx \frac{1}{1000} = 0.001$.
Therefore,the percentage of mass converted into energy during fission is $0.001 \times 100 = 0.1\%$.

(B) When a positron $(e^+)$ encounters an electron $(e^-)$,they annihilate each other because they are particle-antiparticle pairs.
This annihilation process results in the conversion of their mass-energy into two gamma-ray photons,typically emitted in opposite directions to conserve momentum.
This physical phenomenon is the fundamental principle behind Positron Emission Tomography,commonly abbreviated as $PET$ scan,which is used in medical diagnostics to image metabolic processes in the body.

(A) The reaction $^2H + ^2H \to ^4He$ involves the combination of two light nuclei to form a heavier nucleus,which is a nuclear fusion reaction.
Total mass of reactants $= 2 \times 2.0141 \,amu = 4.0282 \,amu$.
Total mass of products $= 4.0024 \,amu$.
Mass defect $\Delta m = \text{Mass of reactants} - \text{Mass of products} = 4.0282 \,amu - 4.0024 \,amu = 0.0258 \,amu$.
Energy released $E = \Delta m \times 931 \,MeV/amu = 0.0258 \times 931 \approx 24 \,MeV$.
Since energy is released,it is an exothermic fusion reaction.

(B) Since the $\pi^0$ meson is initially at rest,its initial linear momentum is $0$.
According to the law of conservation of linear momentum,the total final momentum of the system must also be $0$.
Let the momenta of the two $\gamma$ rays be $\vec{p}_1$ and $\vec{p}_2$. Then,$\vec{p}_1 + \vec{p}_2 = 0$,which implies $\vec{p}_1 = -\vec{p}_2$.
This indicates that the two $\gamma$ rays must move in exactly opposite directions to ensure the total momentum remains $0$.

(B) The nuclei of light elements have a lower binding energy per nucleon than that for the elements of intermediate mass. They are therefore less stable. Consequently,the fusion of light elements results in a more stable nucleus with a higher binding energy per nucleon,releasing a large amount of energy.

(A) In a nuclear reactor,fast neutrons produced during fission are difficult to capture by $U^{235}$ nuclei. $A$ moderator (such as heavy water or graphite) is used to slow down these fast neutrons to thermal energies (about $0.025 \ eV$),which increases the probability of further fission reactions. Therefore,the primary function of a moderator is to control the velocity of neutrons by slowing them down.

(B) The light energy emitted by stars is primarily due to the process of nuclear fusion.
In the core of stars,hydrogen nuclei (protons) undergo fusion to form helium nuclei.
This process of joining of nuclei releases a tremendous amount of energy in the form of light and heat,as the mass of the resulting nucleus is slightly less than the sum of the masses of the individual nuclei,with the mass defect being converted into energy according to Einstein's mass-energy equivalence principle,$E = \Delta mc^2$.

(B) The primary source of solar energy is the nuclear fusion process occurring in the core of the sun.
In this process,hydrogen nuclei (protons) fuse together to form heavier elements like helium.
This fusion reaction releases an enormous amount of energy in the form of heat and light due to the mass defect between the reactants and the products.

(C) The Kaiga Generating Station is a nuclear power generating station located at Kaiga,near the Kali river,in the Uttar Kannada district of Karnataka,India. It is primarily used for the generation of electricity,making it a power reactor.

(B) Heavy water is water that contains heavy hydrogen or deuterium.
Deuterium differs from the hydrogen usually found in water (protium) in that each atom of deuterium contains a proton and a neutron.
Heavy water is chemically represented as deuterium oxide,$D_2O$.
It occurs naturally in small quantities compared to regular water $(H_2O)$ and is commonly used as a moderator in nuclear reactors.

(C) In a nuclear fusion process, the mass of the resultant nucleus is always less than the sum of the masses of the initial nuclei.
This difference in mass is known as the mass defect $(\Delta m)$, which is converted into energy according to Einstein's mass-energy equivalence principle, $E = \Delta m c^2$.
Therefore, the relationship is ${m_3} < ({m_1} + {m_2})$.
Here, ${m_3}$ is the mass of the resultant nucleus, and ${m_1}$ and ${m_2}$ are the masses of the initial fusing nuclei.

(A) 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 nucleus be $_{Z}^{A}X$.
The reaction is: $_{0}^{1}n + _{92}^{235}U \to _{56}^{144}Ba + _{Z}^{A}X + 3(_{0}^{1}n)$.
Conservation of atomic number (charge): $0 + 92 = 56 + Z + 3(0) \implies 92 = 56 + Z \implies Z = 36$.
The element with atomic number $36$ is Krypton $(Kr)$.
Conservation of mass number: $1 + 235 = 144 + A + 3(1) \implies 236 = 147 + A \implies A = 89$.
Therefore, the missing nucleus is $_{36}^{89}Kr$.

(A) The fission reaction is given by: $_{92}U^{234} \rightarrow 2(_{Z}X^{A}) + 2(_{0}n^{1})$.
Applying the law of conservation of mass number: $234 = 2A + 2(1) \Rightarrow 234 = 2A + 2 \Rightarrow 2A = 232 \Rightarrow A = 116$.
Applying the law of conservation of atomic number: $92 = 2Z + 2(0) \Rightarrow 92 = 2Z \Rightarrow Z = 46$.
The element with atomic number $Z = 46$ is Palladium $(Pd)$.
Thus,the resultant nuclei are $_{46}Pd^{116}$.

(B) The power $P$ generated is given by the rate of energy release,$P = \frac{E}{t}$.
Using Einstein's mass-energy equivalence,$E = mc^2$,where $m$ is the mass consumed per unit time.
Given: $m/t = 1 \, mg/s = 1 \times 10^{-6} \, kg/s$ and $c = 3 \times 10^8 \, m/s$.
Substituting the values:
$P = (m/t) \times c^2$
$P = (1 \times 10^{-6} \, kg/s) \times (3 \times 10^8 \, m/s)^2$
$P = 1 \times 10^{-6} \times 9 \times 10^{16} \, W$
$P = 9 \times 10^{10} \, W$
Since $1 \, kW = 10^3 \, W$,we convert the power to kilowatts:
$P = \frac{9 \times 10^{10}}{10^3} \, kW = 9 \times 10^7 \, kW$.

(B) substance that is used in a nuclear reactor to slow down the fast-moving neutrons is called a moderator.
Moderators work by scattering neutrons,which reduces their kinetic energy through collisions.
Commonly used moderators include water,heavy water $(D_2O)$,beryllium,and graphite.
Cadmium is typically used as a control rod to absorb neutrons,not as a moderator.

(A) When a stable nucleus is bombarded with particles like neutrons,it can capture the neutron to become an unstable or radioactive isotope. This process of inducing radioactivity in a previously stable nucleus is known as artificial radioactivity or induced radioactivity. Therefore,the correct option is $A$.

(D) Nuclear fission is a process where a heavy nucleus splits into smaller nuclei after being struck by a particle. This process typically requires particles with sufficient energy and mass to overcome the Coulomb barrier or induce instability in the nucleus.
$1$. $\alpha$-particles,protons,and deuterons are charged particles that can interact with the nucleus through the strong nuclear force or by overcoming the electrostatic repulsion (Coulomb barrier) to induce fission.
$2$. Laser rays consist of photons. While high-energy gamma rays (photons) can cause photofission,standard laser rays (visible or infrared light) lack the energy required to overcome the binding energy of a heavy nucleus to induce fission.
Therefore,laser rays cannot cause fission in a heavy nucleus.

(A) The energy released in a nuclear process is given by Einstein's mass-energy equivalence formula: $E = \Delta m c^2$.
Given mass loss $\Delta m = 0.5\, g = 0.5 \times 10^{-3}\, kg$.
The speed of light $c = 3 \times 10^8\, m/s$.
Substituting the values: $E = (0.5 \times 10^{-3}) \times (3 \times 10^8)^2 = 0.5 \times 10^{-3} \times 9 \times 10^{16} = 4.5 \times 10^{13}\, J$.
To convert energy from Joules $(J)$ to kilowatt-hours $(kWh)$,we divide by $3.6 \times 10^6\, J/kWh$:
$E = \frac{4.5 \times 10^{13}}{3.6 \times 10^6} = 1.25 \times 10^7\, kWh$.

(B) The nuclear fission reaction is given by the conservation of mass number and atomic number.
The reaction is: $^1_0n + ^{235}_{92}U \rightarrow ^{94}_{36}Kr + X + 3(^1_0n)$.
Let the unknown product $X$ be $^{A}_{Z}Y$.
Equating the mass numbers: $1 + 235 = 94 + A + 3(1) \Rightarrow 236 = 97 + A \Rightarrow A = 139$.
Equating the atomic numbers: $0 + 92 = 36 + Z + 3(0) \Rightarrow 92 = 36 + Z \Rightarrow Z = 56$.
The element with atomic number $56$ is Barium $(Ba)$.
Thus,the product is $^{139}_{56}Ba$.

(A) In a nuclear fission process,a heavy nucleus splits into two or more lighter nuclei.
According to Einstein's mass-energy equivalence principle,$E = mc^2$,the energy released during the process is derived from the mass defect.
Since energy is released,the total mass of the fission products must be less than the mass of the parent nucleus.
Therefore,the ratio $\frac{\text{mass of fission products}}{\text{mass of parent nucleus}} < 1$.

(B) Nuclear fusion is a process in which two or more light nuclei combine to form a single heavier nucleus,releasing a large amount of energy.
In the Sun and stars,$4$ hydrogen nuclei (protons) fuse together to form a single helium nucleus $(He)$.
Therefore,the formation of $He$ from $H$ is a classic example of nuclear fusion.

(B) The correct option is $B$.
Cobalt therapy or cobalt-$60$ therapy is the medical use of gamma rays emitted from the radioisotope cobalt-$60$ $(Co^{60})$ to treat various conditions,most notably cancer. The high-energy gamma radiation is used to damage the $DNA$ of cancer cells,thereby inhibiting their growth and division.

(A) The term used to describe the catastrophic,large-scale destruction and loss of life resulting from the use of nuclear weapons is known as a $Nuclear$ $holocaust$. This event implies total devastation on a global or regional scale due to the immense energy release,radiation,and subsequent environmental impact of nuclear explosions.

(C) Nuclear fusion is a process where two light nuclei combine to form a heavier nucleus,releasing a large amount of energy.
In the Sun,hydrogen nuclei fuse to form helium,which is the primary source of its energy.
$A$ hydrogen bomb also operates on the principle of uncontrolled nuclear fusion of hydrogen isotopes.
Therefore,nuclear fusion is common to both the energy production in the Sun and the hydrogen bomb.

(C) The net reaction is: $3(_1H^2) \to _2He^4 + p + n$.
Mass defect $\Delta m = 3 \times M(H^2) - [M(He^4) + M(p) + M(n)]$
$\Delta m = 3(2.014) - [4.001 + 1.007 + 1.008] = 6.042 - 6.016 = 0.026 \, amu$.
Energy released per reaction $Q = 0.026 \times 931.5 \, MeV \approx 24.22 \, MeV$.
$Q = 24.22 \times 10^6 \times 1.6 \times 10^{-19} \, J \approx 3.875 \times 10^{-12} \, J$.
Since $3$ deuterons are used to produce $Q$ energy, energy per deuteron is $E_d = Q/3 = 1.29 \times 10^{-12} \, J$.
Total energy available $E_{total} = N \times E_d = 10^{40} \times 1.29 \times 10^{-12} = 1.29 \times 10^{28} \, J$.
Time taken $t = E_{total} / P = (1.29 \times 10^{28}) / 10^{16} = 1.29 \times 10^{12} \, s$.
Thus, the time is of the order of $10^{12} \, s$.

(A) The rest mass energy of an electron is $E_e = 0.51 \, MeV$ and that of a positron is $E_p = 0.51 \, MeV$.
When they annihilate, the total energy released is $E_{total} = E_e + E_p = 0.51 \, MeV + 0.51 \, MeV = 1.02 \, MeV$.
Converting this energy to Joules: $E_{total} = 1.02 \times 10^6 \times 1.6 \times 10^{-19} \, J = 1.632 \times 10^{-13} \, J$.
The wavelength $\lambda$ of the produced gamma-rays is given by $\lambda = \frac{hc}{E_{total}}$.
Substituting the values: $\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.632 \times 10^{-13}} \, m$.
$\lambda \approx 1.218 \times 10^{-12} \, m = 0.01218 \times 10^{-10} \, m = 0.012 \, \mathring{A}$.

(A) The average kinetic energy of the molecules of a gas at temperature $T$ is given by $E = \frac{3}{2}kT$.
To initiate the nuclear fusion reaction,the kinetic energy of the nuclei must be sufficient to overcome the repulsive potential energy barrier.
Equating the kinetic energy to the potential energy: $\frac{3}{2}kT = 7.7 \times 10^{-14} \ J$.
Substituting the value of Boltzmann's constant $k = 1.38 \times 10^{-23} \ J/K$:
$\frac{3}{2} \times (1.38 \times 10^{-23}) \times T = 7.7 \times 10^{-14}$.
$2.07 \times 10^{-23} \times T = 7.7 \times 10^{-14}$.
$T = \frac{7.7 \times 10^{-14}}{2.07 \times 10^{-23}} \approx 3.72 \times 10^9 \ K$.
Rounding to the nearest order of magnitude,the temperature is approximately $10^9 \ K$.

(C) The Sun's energy is generated through nuclear fusion reactions in its core.
In these reactions,hydrogen nuclei (protons) fuse together to form helium nuclei.
This process releases a tremendous amount of energy in the form of electromagnetic radiation,as the mass of the resulting helium nucleus is slightly less than the sum of the masses of the hydrogen nuclei that formed it,with the mass difference being converted into energy according to Einstein's mass-energy equivalence principle,$E = \Delta mc^2$.

(B) The Sun generates energy through the process of nuclear fusion. In its core,hydrogen nuclei (protons) undergo fusion to form helium nuclei. This process releases a tremendous amount of energy in the form of electromagnetic radiation,which allows the Sun to shine continuously and maintain its brightness.

(B) $1$. Isobars are nuclei with the same mass number $(A)$ but different atomic numbers $(Z)$. Nuclei with the same number of neutrons are called isotones. Thus,option $A$ is incorrect.
$2$. Radioactive decay is a spontaneous process that cannot be influenced by external factors like temperature or pressure. Nuclear fission in a reactor is a controlled chain reaction. Thus,option $B$ is correct.
$3$. Nuclear forces are short-range,strongly attractive,and charge-independent (they act equally between proton-proton,neutron-neutron,and proton-neutron). Thus,option $C$ is incorrect.
$4$. The de Broglie wavelength $\lambda = h/p$ is applicable to both matter particles and photons (where $p = E/c$). Thus,option $D$ is incorrect.

(B) In a nuclear explosion,the mass defect is converted into energy according to Einstein's mass-energy equivalence principle,$E = \Delta m c^2$.
Most of this energy is released as the kinetic energy of the fission fragments (the products of the reaction).
This kinetic energy is subsequently converted into heat and blast energy.

(A) The multiplication factor,denoted by $k$,is defined as the ratio of the number of neutrons present at a specific generation to the number of neutrons present in the preceding generation.
When $k = 1$,the chain reaction is said to be in a 'critical' state,meaning the reaction rate is steady and self-sustaining.
If $k < 1$,the reaction is sub-critical and will die out.
If $k > 1$,the reaction is super-critical and will grow exponentially.
Therefore,for a critical state,the value of the multiplication factor must be $1$.

(B) Control rods are used in nuclear reactors to control the fission rate of uranium and plutonium.
This is achieved by absorbing excess neutrons from the reactor core.
Boron is an element with a high neutron absorption capacity.
By regulating the insertion of boron rods into the reactor core,the number of neutrons available for the chain reaction is controlled,thereby regulating the fission rate.

(A) In a nuclear power station,uranium undergoes nuclear fission,which releases a large amount of heat energy. This heat is used to produce steam,which drives a turbine connected to a generator. The generator then converts this mechanical energy into electrical energy. Therefore,the primary purpose of using uranium in a nuclear power station is the production of electrical energy.

(B) In any nuclear process,if the binding energy per nucleon $(BE/A)$ of the product nuclei is greater than that of the reactant nuclei,energy is released.
From the given graph:
$1$. For $1 < A < 100$,$BE/A = 2 \text{ MeV}$.
$2$. For $100 < A < 200$,$BE/A = 8 \text{ MeV}$.
$3$. For $200 < A < 260$,$BE/A = 4 \text{ MeV}$.
Analyzing option $(B)$: If two nuclei with $A$ between $51$ and $100$ (where $BE/A = 2 \text{ MeV}$) fuse to form a nucleus with $A$ between $100$ and $200$ (where $BE/A = 8 \text{ MeV}$),the binding energy per nucleon increases. Therefore,energy is released.

(A) In a nuclear reactor,the chain reaction is controlled,meaning the multiplication factor $r$ is maintained at $r = 1$. This ensures a steady power output.
In an atomic bomb,the chain reaction is uncontrolled,meaning the multiplication factor $r$ is greater than $1$ $(r > 1)$,leading to an exponential release of energy.

(A) In a fast breeder reactor,fast neutrons are used to convert non-fissionable material (like $U^{238}$) into fissionable fuel (like $Pu^{239}$) while simultaneously releasing energy through the fission process. This process is highly efficient as it produces more fuel than it consumes.

(B) In the process of nuclear fusion, four hydrogen nuclei $(_1H^1)$ fuse to form one helium nucleus $(_2He^4)$. The mass of the resulting helium nucleus is less than the sum of the masses of the four hydrogen nuclei. This mass defect $(\Delta m)$ is converted into energy according to Einstein's mass-energy equivalence principle, $E = \Delta m c^2$. Since the binding energy per nucleon for helium is higher than that of hydrogen, the process is exothermic, meaning energy is released.

(D) The nuclear reaction is given by: $_3Li^7 + _1H^1 \to _4Be^8 + _ZX^A$.
Applying the law of conservation of charge (atomic number): $3 + 1 = 4 + Z$,which gives $Z = 0$.
Applying the law of conservation of mass number: $7 + 1 = 8 + A$,which gives $A = 0$.
$A$ particle with atomic number $Z = 0$ and mass number $A = 0$ is a $\gamma$-photon.