Mass-Energy, Nuclear Binding Energy, Nuclear Stability Questions in English

(A) The binding energy per nucleon is a measure of the stability of a nucleus. $A$ higher binding energy per nucleon indicates that the nucleus is more tightly bound and therefore more stable. Stable nuclei do not decay readily and are more likely to persist over long periods,which is why they are found to be abundantly available in nature.

(C) The binding energy per nucleon decreases for nuclei with $A > 100$ because the nuclear force is short-ranged,while the Coulomb repulsion between protons is long-ranged.
As the mass number $A$ increases,the size of the nucleus increases. Since the nuclear force acts only between nearest neighbors,the total nuclear binding energy increases roughly linearly with $A$.
However,the Coulomb repulsion acts between all pairs of protons,and the number of pairs increases as $A^2$. This causes the net binding energy per nucleon to decrease for heavier nuclei.
The Reason statement is technically incorrect because the nuclear force itself does not become 'weak' for heavier nuclei; rather,its short-range nature means it cannot compensate for the long-range Coulomb repulsion as the nucleus grows larger.

(A) In a nuclear decay process, energy is released because the total mass of the products is less than the mass of the parent nucleus. This mass difference, known as the mass defect $(\Delta m)$, is converted into energy according to Einstein's mass-energy equivalence principle, $E = \Delta m c^2$.
Since energy is released, the total rest mass energy of the products must be less than the rest mass energy of the parent nucleus to satisfy the law of conservation of energy.
Therefore, both the Assertion and the Reason are correct, and the Reason provides the correct explanation for the Assertion.

(A) The nucleus of a hydrogen atom $(_{1}^{1}H)$ consists of only a single proton.
Nuclear binding energy is defined as the energy required to disassemble a nucleus into its constituent protons and neutrons (nucleons).
Since a hydrogen nucleus contains only one nucleon (a proton),there are no other nucleons to bind with,and thus no nuclear force is acting to hold it together.
Therefore,the binding energy and mass defect of a hydrogen nucleus are zero.
Since the Assertion is true and the Reason correctly explains why the binding energy is zero (due to the presence of only one nucleon),the correct option is $A$.

(B) According to Einstein's mass-energy equivalence principle,the energy $E$ equivalent to a mass $m$ is given by the formula $E = mc^2$,where $c$ is the speed of light in a vacuum.
Given mass $m = 1\; g = 10^{-3}\; kg$.
The speed of light $c = 3 \times 10^{8}\; m/s$.
Substituting these values into the equation:
$E = 10^{-3} \times (3 \times 10^{8})^2\; J$
$E = 10^{-3} \times 9 \times 10^{16}\; J$
$E = 9 \times 10^{13}\; J$
Thus,the energy equivalent of $1\; g$ of substance is $9 \times 10^{13}\; J$.

(N/A) The mass of one atomic mass unit is $1 \, u = 1.6605 \times 10^{-27} \, kg$.
To convert this into energy units,we use Einstein's mass-energy equivalence relation $E = mc^2$:
$E = 1.6605 \times 10^{-27} \, kg \times (2.9979 \times 10^8 \, m/s)^2$
$E = 1.4924 \times 10^{-10} \, J$.
To convert this energy into $MeV$,we divide by $1.602 \times 10^{-13} \, J/MeV$:
$E = \frac{1.4924 \times 10^{-10}}{1.602 \times 10^{-13}} \, MeV \approx 931.5 \, MeV$.
Thus,$1 \, u = 931.5 \, MeV/c^2$.
For $_{8}^{16} O$,the mass defect $\Delta M = 0.13691 \, u$.
Converting this to $MeV/c^2$:
$\Delta M = 0.13691 \times 931.5 \, MeV/c^2 \approx 127.5 \, MeV/c^2$.

(N/A) The alpha decay of $^{238}_{92}U$ is represented by the equation: $^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He$. The energy released ($Q$-value) in this process is given by $Q = (M_{U} - M_{Th} - M_{He})c^{2}$.
Substituting the given atomic masses:
$Q = (238.05079 - 234.04363 - 4.00260) \; u \times c^{2}$
$Q = 0.00456 \; u \times c^{2}$
Using the conversion factor $1 \; u = 931.5 \; MeV/c^{2}$:
$Q = 0.00456 \times 931.5 \; MeV = 4.25 \; MeV$.
$(b)$ If $^{238}_{92}U$ were to spontaneously emit a proton,the decay process would be: $^{238}_{92}U \rightarrow ^{237}_{91}Pa + ^{1}_{1}H$.
The $Q$-value for this process is:
$Q = (M_{U} - M_{Pa} - M_{H})c^{2}$
$Q = (238.05079 - 237.05121 - 1.00783) \; u \times c^{2}$
$Q = -0.00825 \; u \times c^{2}$
$Q = -0.00825 \times 931.5 \; MeV = -7.68 \; MeV$.
Since the $Q$-value is negative,the process is endothermic and cannot proceed spontaneously. An external energy of $7.68 \; MeV$ must be supplied to the $^{238}_{92}U$ nucleus to trigger proton emission.

(N/A) chemical equation is balanced in the sense that the number of atoms of each element is the same on both sides. $A$ chemical reaction merely alters the combinations of atoms. In a nuclear reaction,elements may be transmuted,so the number of atoms of each element is not necessarily conserved. However,the total number of protons and the total number of neutrons are separately conserved in a nuclear reaction. Thus,nuclear reactions are balanced in terms of the total number of protons and neutrons.
$(b)$ The binding energy of a nucleus contributes negatively to its mass (mass defect). While the total number of protons and neutrons is conserved,the total binding energy of the nuclei on the reactant side may differ from that on the product side. This difference in binding energy manifests as energy released or absorbed. Since binding energy contributes to the mass,the difference in the total mass of the nuclei on both sides is converted into energy or vice-versa.
$(c)$ In principle,chemical reactions are similar to nuclear reactions regarding mass-energy interconversion. Energy released or absorbed in chemical reactions arises from the difference in chemical binding energies of atoms and molecules. Because chemical binding energy also contributes to the total mass (mass defect),the mass difference between reactants and products is converted into energy. However,these mass defects in chemical reactions are approximately a million times smaller than those in nuclear reactions,leading to the incorrect impression that mass-energy interconversion does not occur in chemical reactions.

(B) The atomic mass of nitrogen $\left(^{14}_{7} N \right)$ is $m = 14.00307 \; u$.
$A$ nucleus of $^{14}_{7} N$ contains $7$ protons and $7$ neutrons.
The mass defect $\Delta m$ is given by the formula:
$\Delta m = [Z m_{H} + (A - Z) m_{n}] - m_{nucleus}$
Where $Z = 7$ (number of protons) and $A - Z = 7$ (number of neutrons).
$\Delta m = 7 \times 1.007825 \; u + 7 \times 1.008665 \; u - 14.00307 \; u$
$\Delta m = 7.054775 \; u + 7.060655 \; u - 14.00307 \; u$
$\Delta m = 14.11543 \; u - 14.00307 \; u = 0.11236 \; u$
Since $1 \; u = 931.5 \; MeV/c^{2}$,the binding energy $E_{b}$ is:
$E_{b} = \Delta m \times 931.5 \; MeV/u$
$E_{b} = 0.11236 \times 931.5 \; MeV$
$E_{b} \approx 104.66 \; MeV$.
Thus,the binding energy of the nitrogen nucleus is $104.66 \; MeV$.

(C) For $^{56}_{26} Fe$:
Number of protons $Z = 26$,number of neutrons $N = 56 - 26 = 30$.
Mass defect $\Delta m = Z m_H + N m_n - m(^{56}_{26} Fe)$.
Using $m_H = 1.007825 \; u$ and $m_n = 1.008665 \; u$:
$\Delta m = 26(1.007825) + 30(1.008665) - 55.934939 = 0.528461 \; u$.
Binding Energy $E_{b1} = 0.528461 \times 931.5 \; MeV \approx 492.26 \; MeV$.
Binding energy per nucleon $= 492.26 / 56 \approx 8.79 \; MeV$.
For $^{209}_{83} Bi$:
Number of protons $Z = 83$,number of neutrons $N = 209 - 83 = 126$.
Mass defect $\Delta m' = 83 m_H + 126 m_n - m(^{209}_{83} Bi)$.
$\Delta m' = 83(1.007825) + 126(1.008665) - 208.980388 = 1.760877 \; u$.
Binding Energy $E_{b2} = 1.760877 \times 931.5 \; MeV \approx 1640.26 \; MeV$.
Binding energy per nucleon $= 1640.26 / 209 \approx 7.85 \; MeV$.

(N/A) Mass of a copper coin, $m' = 3\; g$.
Atomic mass of $_{29}^{63} Cu$ atom, $m = 62.92960\; u$.
The total number of $_{29}^{63} Cu$ atoms in the coin is $N = \frac{N_A \times m'}{\text{Mass number}}$.
Where $N_A = 6.023 \times 10^{23}\; \text{atoms/mol}$ and Mass number $= 63\; g/mol$.
$N = \frac{6.023 \times 10^{23} \times 3}{63} = 2.868 \times 10^{22}$ atoms.
$A$ $_{29}^{63} Cu$ nucleus has $29$ protons and $(63 - 29) = 34$ neutrons.
Mass defect of one nucleus, $\Delta m' = 29 \times m_H + 34 \times m_n - m$.
Using $m_H = 1.007825\; u$ and $m_n = 1.008665\; u$:
$\Delta m' = 29(1.007825) + 34(1.008665) - 62.9296 = 0.591935\; u$.
Total mass defect for all atoms, $\Delta m = 0.591935 \times 2.868 \times 10^{22} = 1.69767 \times 10^{22}\; u$.
Using $1\; u = 931.5\; MeV/c^2$, the binding energy $E_b = \Delta m \times 931.5\; MeV$.
$E_b = 1.69767 \times 10^{22} \times 931.5 = 1.581 \times 10^{25}\; MeV$.
Converting to Joules $(1\; MeV = 1.602 \times 10^{-13}\; J)$:
$E_b = 1.581 \times 10^{25} \times 1.602 \times 10^{-13} \approx 2.53 \times 10^{12}\; J$.

(N/A) The neutron separation energy $S_n$ is the energy required to remove a neutron from a nucleus,given by $S_n = [m(A-1, Z) + m_n - m(A, Z)] \times 931.5 \; MeV/u$.
For $_{20}^{41} Ca$:
Reaction: $_{20}^{41} Ca \rightarrow _{20}^{40} Ca + _{0}^{1} n$
Mass defect $\Delta m = m(_{20}^{40} Ca) + m_n - m(_{20}^{41} Ca)$
$\Delta m = 39.962591 + 1.008665 - 40.962278 = 0.008978 \; u$
$S_n = 0.008978 \times 931.5 \; MeV \approx 8.363 \; MeV$.
For $_{13}^{27} Al$:
Reaction: $_{13}^{27} Al \rightarrow _{13}^{26} Al + _{0}^{1} n$
Mass defect $\Delta m = m(_{13}^{26} Al) + m_n - m(_{13}^{27} Al)$
$\Delta m = 25.986895 + 1.008665 - 26.981541 = 0.014019 \; u$
$S_n = 0.014019 \times 931.5 \; MeV \approx 13.059 \; MeV$.

(N/A) The Equivalence of Mass and Energy: Albert Einstein proposed that mass and energy are interchangeable,related by the equation $E = mc^2$,where $E$ is energy,$m$ is mass,and $c$ is the speed of light in a vacuum $(3 \times 10^8 \ m/s)$. This implies that a small amount of mass can be converted into a large amount of energy.
$(b)$ Nuclear Energy: This is the energy released during nuclear reactions,such as nuclear fission (splitting of heavy nuclei) or nuclear fusion (combining of light nuclei). The energy released is due to the mass defect,where the mass of the products is slightly less than the mass of the reactants,with the difference converted into energy according to $E = \Delta mc^2$.
$(c)$ The Principle of Conservation of Energy: This principle states that energy can neither be created nor destroyed,only transformed from one form to another. In an isolated system,the total energy remains constant over time.

(B) Chemical energy is a form of potential energy stored in the bonds of chemical compounds.
During a chemical process,atoms are rearranged to form new substances.
This rearrangement involves the breaking of existing chemical bonds and the formation of new ones.
The energy change is primarily due to the change in the electrostatic potential energy of the electrons and nuclei as they move into new configurations.
Therefore,the correct reason is the change in the potential energy of the electrons and nuclei due to the rearrangement of atoms.

(N/A) The mass-energy equivalence principle,proposed by Albert Einstein,states that mass and energy are interchangeable. The equation is given by: $E = mc^2$,where $E$ is the energy,$m$ is the mass,and $c$ is the speed of light in a vacuum $(c \approx 3 \times 10^8 \ m/s)$.

(A) According to Einstein's mass-energy equivalence principle,the energy $E$ equivalent to a mass $m$ is given by the formula $E = mc^2$,where $c$ is the speed of light in a vacuum.
Given mass $m = 1 \ kg$.
The speed of light $c \approx 3 \times 10^8 \ m/s$.
Substituting these values into the equation:
$E = 1 \ kg \times (3 \times 10^8 \ m/s)^2$
$E = 1 \times 9 \times 10^{16} \ J$
$E = 9 \times 10^{16} \ J$.
Therefore,the energy equivalent to one kilogram of mass is $9 \times 10^{16} \ J$.

(N/A) When a proton and an electron combine to form a $H$-atom,the system moves to a state of lower potential energy to attain maximum stability. According to Einstein's mass-energy equivalence principle,$E = mc^2$,this reduction in potential energy corresponds to a decrease in the mass of the system. This mass defect is released as energy (binding energy) during the formation of the atom. Consequently,the mass of the $H$-atom is less than the sum of the masses of its constituent particles in their free state.

(N/A) According to Einstein's theory of special relativity,mass is considered another form of energy.
Before this theory,it was assumed that mass and energy were conserved separately in any physical reaction.
Einstein demonstrated that mass is equivalent to energy and can be converted into other forms of energy,such as kinetic energy,and vice versa.
This mass-energy equivalence is expressed by the equation $E = mc^2$,where $m$ is the mass and $c$ is the velocity of light in a vacuum,approximately $3 \times 10^8 \ m/s$.
Experimental verification of this relation has been achieved through the study of nuclear reactions involving nucleons,nuclei,electrons,and other subatomic particles.
In such reactions,the law of conservation of energy states that the total initial energy must equal the total final energy,accounting for the mass-energy conversion.
This concept is fundamental for understanding nuclear binding energy,nuclear masses,and the interactions between nuclei.

(N/A) The nucleus is made up of neutrons and protons. Therefore, it may be expected that the mass of the nucleus is equal to the total mass of its individual protons and neutrons.
However, the nuclear mass $M$ is found to be always less than the sum of the masses of its constituents (neutrons and protons) in the free state.
Suppose the mass of a nucleus $_{Z}^{A}X$ is $M$. If we indicate the masses of a proton and a neutron in the free state as $m_{p}$ and $m_{n}$ respectively, then $M < (Z m_{p} + N m_{n})$, where $N = A - Z$ is the neutron number. The difference between the total mass of the constituents of a nucleus and the actual mass of the nucleus is called the mass defect $(\Delta M)$.
$\therefore \Delta M = [Z m_{p} + (A - Z) m_{n}] - M$ is the formula for mass defect.
From Einstein's mass-energy equivalence, the energy equivalent to the mass defect is $E_{b} = \Delta M c^{2}$, where $c$ is the velocity of light in vacuum $(c \approx 3 \times 10^{8} \ m/s)$.
This energy $E_{b}$ is called the binding energy of the nucleus. It represents the energy that would be released when $Z$ protons and $N$ neutrons combine to form a nucleus, or the energy required to separate a nucleus into its individual nucleons.
By dividing the binding energy of the nucleus by the total number of nucleons $(A)$, the binding energy per nucleon $(E_{bn})$ is obtained.
$\therefore E_{bn} = \frac{E_{b}}{A}$
The binding energy per nucleon is a measure of the stability of a nucleus.

(N/A) The graph of binding energy per nucleon $(E_{bn})$ versus the atomic mass number $(A)$ is shown in the figure.
The main features of the plot are as follows:
$(i)$ The binding energy per nucleon,$E_{bn}$,is practically constant and independent of the atomic mass number for nuclei of intermediate mass number $(30 < A < 170)$. The maximum value of $E_{bn}$ is approximately $8.75 \text{ MeV/nucleon}$,which occurs for the iron $(^{56}\text{Fe})$ nucleus. For heavy nuclei like uranium $(^{238}\text{U})$,the value of $E_{bn}$ decreases to about $7.6 \text{ MeV/nucleon}$.
$(ii)$ $E_{bn}$ is lower for both light nuclei $(A < 30)$ and heavy nuclei $(A > 170)$,indicating that they are less stable compared to intermediate-mass nuclei. From the graph,it is evident that the binding energy per nucleon for the deuteron $(^{2}_{1}\text{H})$ is the lowest.
$(iii)$ For light nuclei,stability is generally achieved when the ratio of neutrons to protons is approximately $1:1$. However,for heavy nuclei,this ratio increases to about $3:2$ to compensate for the increased electrostatic repulsion between protons.

(N/A) The two main features of the binding energy per nucleon curve lead to the following conclusions:
$(i)$ The nuclear force is attractive and sufficiently strong to produce a binding energy per nucleon of approximately $8 \text{ MeV}$ for nuclei with $30 < A < 170$. This indicates that the nuclear force is short-ranged and exhibits the property of saturation.
$(ii)$ For very heavy nuclei $(A > 170)$,the binding energy per nucleon decreases. This occurs because the long-range Coulomb repulsion between protons becomes significant,reducing the stability. Consequently,if a heavy nucleus $(A = 240)$ splits into two lighter nuclei $(A = 120)$,the nucleons become more tightly bound,releasing energy. This process is known as nuclear fission.
$(iii)$ For very light nuclei $(A \leq 10)$,the binding energy per nucleon is low. When two light nuclei fuse to form a heavier nucleus,the resulting nucleus is more tightly bound,releasing energy. This process is known as nuclear fusion,which powers the Sun and hydrogen bombs.

(N/A) Einstein's special theory of relativity is based on two fundamental postulates:
$1$. The laws of physics are the same in all inertial frames of reference.
$2$. The speed of light in a vacuum is constant for all observers,regardless of the motion of the light source or the observer.
As a consequence of this theory,Einstein proposed the mass-energy equivalence formula,which states that mass and energy are interchangeable. The formula is given by:
$E = mc^2$
Where:
$E$ is the energy equivalent of the mass,
$m$ is the mass of the object,
$c$ is the speed of light in a vacuum (approximately $3 \times 10^8 \ m/s$).

(N/A) Mass defect is defined as the difference between the sum of the masses of individual nucleons (protons and neutrons) that constitute a nucleus and the actual observed mass of the nucleus.
It represents the mass that is converted into binding energy to hold the nucleus together.
The formula for mass defect $(\Delta m)$ is given by:
$\Delta m = [Z m_p + (A - Z) m_n] - M_{nucleus}$
Where:
$Z$ = Atomic number (number of protons)
$A$ = Mass number (total number of nucleons)
$m_p$ = Mass of a proton
$m_n$ = Mass of a neutron
$M_{nucleus}$ = Actual mass of the nucleus

(N/A) $1$. Binding Energy of the Nucleus $(BE)$: The binding energy of a nucleus is the energy required to separate the nucleons (protons and neutrons) of a nucleus to an infinite distance from each other. It is equivalent to the mass defect $(\Delta m)$ multiplied by the square of the speed of light $(c^2)$.
Formula: $BE = \Delta m \times c^2 = [Z m_p + (A - Z) m_n - M_{nucleus}] c^2$, where $Z$ is the atomic number, $A$ is the mass number, $m_p$ is the mass of a proton, $m_n$ is the mass of a neutron, and $M_{nucleus}$ is the actual mass of the nucleus.
$2$. Binding Energy per Nucleon $(BE/A)$: This is the average energy required to remove a single nucleon from the nucleus. It is calculated by dividing the total binding energy of the nucleus by the total number of nucleons (mass number $A$).
Formula: $BE/A = \frac{BE}{A}$.

(A) The binding energy per nucleon is defined as the total binding energy of a nucleus divided by its mass number $(A)$.
It is a direct measure of the stability of the nucleus.
$A$ higher value of binding energy per nucleon indicates that the nucleons are more tightly bound,making the nucleus more stable.
Therefore,the correct option is $A$.

(B) The binding energy per nucleon is a measure of the stability of a nucleus.
Experimental data shows that the binding energy per nucleon reaches a maximum for nuclei with mass numbers in the range of $A = 50$ to $A = 60$.
Specifically,the isotope Nickel-$62$ $(^{62}Ni)$ has the highest binding energy per nucleon,which is approximately $8.79 \ MeV$ to $8.8 \ MeV$.
While Iron-$56$ $(^{56}Fe)$ is often cited in textbooks due to its high abundance and stability,Nickel-$62$ holds the absolute maximum value.

(N/A) The energy released during a nuclear process is called nuclear energy.
The binding energy per nucleon is almost constant $(8.0 \text{ MeV})$ in the region between $A=30$ to $A=170$ in the figure.
For nuclei in the regions $A<30$ and $A>170$,the binding energy per nucleon is less than $8.0 \text{ MeV}$.
The greater the binding energy,the less is the total mass of a nucleus.
$[\text{Binding energy per nucleon} = \frac{E_{bn}}{A}]$
$\therefore$ Total binding energy of nucleus = (binding energy per nucleon) $\times A$.
If nuclei with less total binding energy transform into nuclei with greater binding energy,there will be a net energy release. As such,there are two types of nuclear processes for obtaining energy:
$1$. Nuclear Fission: When a heavy nucleus decays into two or more intermediate-mass fragments,energy is released. This is the principle of the atom bomb.
$2$. Nuclear Fusion: When two or more light nuclei fuse into a heavier nucleus,energy is also emitted. This is the principle of the hydrogen bomb.
Exothermic chemical reactions lie under conventional energy sources such as coal or petroleum. The energy associated with them is in the range of $\text{eV}$.
In a nuclear reaction,the energy release is of the order of $\text{MeV}$.
Thus,for the same quantity of matter,nuclear sources produce a million $(10^6)$ times more energy than a chemical source.
For example: $1 \text{ kg}$ of uranium generates $10^{14} \text{ J}$ of energy,compared to the burning of $1 \text{ kg}$ of coal,which gives $10^7 \text{ J}$.

(B) No,they do not have the same binding energy.
Binding energy depends on the number of protons and neutrons in the nucleus.
For $_{1}H^{3}$ (Tritium),the nucleus consists of $1$ proton and $2$ neutrons.
For $_{2}He^{3}$,the nucleus consists of $2$ protons and $1$ neutron.
The binding energy per nucleon is influenced by the nuclear force,which is attractive between nucleons.
Since $_{1}H^{3}$ has a higher neutron-to-proton ratio and different pairing effects,its total binding energy differs from that of $_{2}He^{3}$.
In general,$_{1}H^{3}$ has a binding energy of approximately $8.48 \text{ MeV}$,while $_{2}He^{3}$ has a binding energy of approximately $7.72 \text{ MeV}$.
Therefore,the binding energies are not the same.

(A) For the given nuclide $_{11}^{23}Na$,the atomic number $Z_1 = 11$ and the mass number $A = 23$. The number of neutrons is $N_1 = A - Z_1 = 23 - 11 = 12$.
For a mirror isobar,the new atomic number $Z_2 = N_1 = 12$ and the new number of neutrons $N_2 = Z_1 = 11$. The mass number remains $A = Z_2 + N_2 = 12 + 11 = 23$. The element with atomic number $12$ is Magnesium $(Mg)$. Thus,the mirror isobar is $_{12}^{23}Mg$.
$(b)$ The binding energy of a nucleus is influenced by the symmetry of protons and neutrons. In $_{11}^{23}Na$,there are $11$ protons and $12$ neutrons,while in $_{12}^{23}Mg$,there are $12$ protons and $11$ neutrons. The nucleus with more neutrons relative to protons generally experiences a slightly stronger net nuclear attraction due to the absence of additional Coulomb repulsion between protons. Therefore,$_{11}^{23}Na$ has a greater binding energy than $_{12}^{23}Mg$.

(D) Let the deuteron nucleus be at rest. When a $\gamma$-ray of energy $E$ is incident on it,the conservation of energy gives: $E = B + K_p + K_n$,where $K_p$ and $K_n$ are the kinetic energies of the proton and neutron.
From the conservation of linear momentum,the momentum of the photon must equal the sum of the momenta of the proton and neutron: $p_{\gamma} = p_p + p_n$. Since the photon's momentum is $p_{\gamma} = E/c$,we have $p_p + p_n = E/c$.
If $E = B$,then $K_p + K_n = 0$. Since kinetic energy cannot be negative,this implies $K_p = 0$ and $K_n = 0$,which means $p_p = 0$ and $p_n = 0$. However,this contradicts the momentum conservation equation $p_p + p_n = E/c$,as $E/c \neq 0$. Thus,$E = B$ is impossible.
For the process to occur,we must have $E > B$. Let $E = B + \Delta E$. The kinetic energies are $K_p = p_p^2 / 2m$ and $K_n = p_n^2 / 2m$. To minimize the energy required,we assume the particles move together with the same velocity,so $p_p = p_n = p/2$. Then $p = E/c$,so $p_p = p_n = E/2c$.
Substituting into the energy equation: $E - B = (E/2c)^2 / 2m + (E/2c)^2 / 2m = E^2 / 4mc^2$.
Since $E \approx B$,we have $\Delta E \approx B^2 / 4mc^2$.

(D) The binding energy of an $H$-atom is given by the formula:
$E = \frac{m_e e^4}{8 \epsilon_0^2 h^2} = 13.6 \text{ eV}$ ... $(1)$
Now,considering the deuteron nucleus as a binary system,we replace $m_e$ with the reduced mass $m'$ and $e$ with the effective charge $e'$. The binding energy of the deuteron is:
$E' = \frac{m' e'^4}{8 \epsilon_0^2 h^2} = 2.2 \times 10^6 \text{ eV}$ ... $(2)$
Taking the ratio of equation $(2)$ to equation $(1)$:
$\frac{m'}{m_e} \times \left(\frac{e'}{e}\right)^4 = \frac{2.2 \times 10^6}{13.6} \approx 1.617 \times 10^5$ ... $(3)$
The reduced mass $m'$ of the deuteron (proton-neutron system) is:
$m' = \frac{m_p m_n}{m_p + m_n} = \frac{m \times m}{2m} = \frac{m}{2}$
Given $m \approx 1836 m_e$,we have $m' = \frac{1836 m_e}{2} = 918 m_e$.
Thus,$\frac{m'}{m_e} = 918$.
Substituting this into equation $(3)$:
$918 \times \left(\frac{e'}{e}\right)^4 = 1.617 \times 10^5$
$\left(\frac{e'}{e}\right)^4 = \frac{1.617 \times 10^5}{918} \approx 176.14$
$\frac{e'}{e} = (176.14)^{1/4} \approx 3.65$.

(N/A) We know that $1\,u = 1.6605 \times 10^{-27}\,kg$ and $c = 3 \times 10^8\,m/s$.
Applying $E = mc^2$:
$E = (1.6605 \times 10^{-27}\,kg) \times (3 \times 10^8\,m/s)^2$
$E = 1.6605 \times 9 \times 10^{-11}\,J = 14.9445 \times 10^{-11}\,J$
To convert this into $MeV$,divide by $1.6 \times 10^{-13}\,J/MeV$:
$E = \frac{14.9445 \times 10^{-11}}{1.6 \times 10^{-13}}\,MeV \approx 934\,MeV$ (Using precise $1\,u = 1.660539 \times 10^{-27}\,kg$ yields $931.5\,MeV$).
$(b)$ The relation $1\,u = 931.5\,MeV$ is dimensionally incorrect because mass cannot be equal to energy. The correct relation is $1\,u \times c^2 = 931.5\,MeV$.

(A) The number of protons $Z = 50$ and the number of neutrons $N = A - Z = 120 - 50 = 70$.
The expected mass of the nucleus is $M_{expected} = Z m_{p} + N m_{n}$.
$M_{expected} = 50(1.00783) + 70(1.00867) = 50.3915 + 70.6069 = 120.9984 \, U$.
The mass defect $\Delta m = M_{expected} - m_{Sn} = 120.9984 - 119.902199 = 1.096201 \, U$.
The binding energy $B.E. = \Delta m \times 931 \, MeV/U = 1.096201 \times 931 \approx 1020.56 \, MeV$.
The binding energy per nucleon is $\frac{B.E.}{A} = \frac{1020.56}{120} \approx 8.5 \, MeV$.

(A) For a process to be allowed by energy conservation,the mass of the reactants must be greater than or equal to the mass of the products (considering the energy equivalent of mass).
Let us evaluate $A$: $n + p \rightarrow d + \gamma$.
Mass of reactants: $m_{n} + m_{p} = 1.0087 + 1.0072 = 2.0159 \ u$.
Mass of product: $m_{d} = 2.0141 \ u$.
Since $2.0159 \ u > 2.0141 \ u$,the mass difference is released as energy (photon $\gamma$),satisfying conservation laws.
Evaluating $B$: $e^{+} + e^{-} \rightarrow \gamma$ violates momentum conservation (a single photon cannot conserve both energy and momentum in the center-of-mass frame).
Evaluating $D$: $p \rightarrow n + e^{+} + \bar{v}$ is not possible for a free proton because $m_{p} < m_{n} + m_{e}$.
Thus,the correct process is $A$.

(C) According to Einstein's mass-energy equivalence principle,the energy $E$ is given by the formula $E = mc^2$.
Given mass $m = 0.5\, g = 0.5 \times 10^{-3}\, kg$.
The speed of light $c = 3 \times 10^8\, m/s$.
Substituting these values into the equation:
$E = (0.5 \times 10^{-3}\, kg) \times (3 \times 10^8\, m/s)^2$
$E = 0.5 \times 10^{-3} \times 9 \times 10^{16}\, J$
$E = 4.5 \times 10^{13}\, J$.

(B) The number of protons $Z = 26$ and the number of neutrons $N = 56 - 26 = 30$.
The mass defect $\Delta m$ is given by:
$\Delta m = [Z m_{p} + N m_{n} - m({^{56}Fe})]$
$\Delta m = [26 \times 1.00727 + 30 \times 1.00866 - 55.936] \ u$
$\Delta m = [26.18902 + 30.2598 - 55.936] \ u = 0.51282 \ u$.
The total binding energy $BE$ is:
$BE = \Delta m \times 931.5 \ MeV/u$
$BE = 0.51282 \times 931.5 \approx 477.7 \ MeV$.
The binding energy per nucleon is:
$BE/A = \frac{477.7}{56} \approx 8.53 \ MeV$.
Rounding to the nearest provided option,the answer is $8.52 \ MeV$.

(B) The energy of the $\gamma$-ray photon is $E_{\gamma} = h\nu$.
The momentum of the $\gamma$-ray photon is $p_{\gamma} = \frac{h\nu}{c}$.
By the law of conservation of momentum,the nucleus must recoil with an equal and opposite momentum $p_N = p_{\gamma} = \frac{h\nu}{c}$.
The kinetic energy of the recoiling nucleus is $K_N = \frac{p_N^2}{2M} = \frac{(h\nu/c)^2}{2M} = \frac{(h\nu)^2}{2Mc^2}$.
The total loss in internal energy of the nucleus is the sum of the energy of the emitted photon and the kinetic energy of the recoiling nucleus:
$\Delta E = E_{\gamma} + K_N = h\nu + \frac{(h\nu)^2}{2Mc^2} = h\nu \left[1 + \frac{h\nu}{2Mc^2}\right]$.

(C) The mass defect $\Delta m$ is calculated as: $\Delta m = (Z m_p + (A - Z) m_n) - M_{Al}$.
Here,$Z = 13$ (number of protons) and $A - Z = 14$ (number of neutrons).
$\Delta m = (13 \times 1.00726 + 14 \times 1.00866) - 26.98154$.
$\Delta m = (13.09438 + 14.12124) - 26.98154$.
$\Delta m = 27.21562 - 26.98154 = 0.23408 \, {u}$.
Given that $1 \, {u}$ corresponds to $1 \, {J}$ of energy,the binding energy $E = 0.23408 \, {J}$.
To express this in the form $x \times 10^{-3} \, {J}$,we have $E = 234.08 \times 10^{-3} \, {J}$.
Note: The provided mass of the aluminium nucleus in the prompt $(27.18846 \, {u})$ is physically incorrect as it exceeds the sum of its constituent nucleons. Using the standard mass of ${ }_{13}^{27} {Al}$ $(26.98154 \, {u})$,the result is approximately $234$. If we follow the prompt's specific (though incorrect) mass value,$\Delta m = 27.21562 - 27.18846 = 0.02716 \, {u}$,which gives $27.16 \times 10^{-3} \, {J}$. Rounding to the nearest integer,we get $27$.

(D) The energy released in a nuclear fission reaction is given by the difference between the total binding energy of the products and the total binding energy of the reactant.
Total binding energy of the reactant (Nucleus $A$):
$BE_A = 220 \times 5.6 \, MeV = 1232 \, MeV$.
Total binding energy of the products (Nuclei $B$ and $C$):
$BE_{B+C} = (105 \times 6.4) + (115 \times 6.4) \, MeV = (105 + 115) \times 6.4 \, MeV = 220 \times 6.4 \, MeV = 1408 \, MeV$.
The energy $Q$ released is:
$Q = BE_{products} - BE_{reactant}$
$Q = 1408 \, MeV - 1232 \, MeV = 176 \, MeV$.

(D) Consider a nuclear reaction of the form $x + p \rightarrow y + b$,where $x$ is the target nucleus,$p$ is the projectile,$y$ is the product nucleus,and $b$ is the emitted particle.
The $Q$-value of a nuclear reaction is defined as the difference between the initial kinetic energy and the final kinetic energy,or equivalently,the energy released due to the mass defect: $Q = K_{f} + K_{b} - K_{p}$.
Rearranging this equation,we get $Q + K_{p} = K_{f} + K_{b}$.
Since the kinetic energies of the product particles ($K_{f}$ and $K_{b}$) must be non-negative (i.e.,$K_{f} \geq 0$ and $K_{b} \geq 0$),their sum must be greater than or equal to zero.
Therefore,for the reaction to occur,the total energy available must satisfy the condition $(K_{p} + Q) > 0$.

(A) The reaction for removing a neutron from ${ }_5 B ^{11}$ is given by:
${ }_5 B ^{11} \longrightarrow { }_5 B ^{10} + { }_0 n ^1$
The total binding energy of a nucleus is calculated as:
$BE = (\text{Binding energy per nucleon}) \times (\text{Mass number } A)$
For ${ }_5 B ^{11}$:
$BE({ }_5 B ^{11}) = 7.5 \,MeV \times 11 = 82.5 \,MeV$
For ${ }_5 B ^{10}$:
$BE({ }_5 B ^{10}) = 8.0 \,MeV \times 10 = 80.0 \,MeV$
The energy required to remove the neutron is the difference in binding energies:
$E = BE({ }_5 B ^{11}) - BE({ }_5 B ^{10})$
$E = 82.5 \,MeV - 80.0 \,MeV = 2.5 \,MeV$

(B) The reaction for removing a proton from ${ }_8 O ^{16}$ is given by: ${ }_8 O ^{16} + \text{Energy} \longrightarrow { }_7 N ^{15} + { }_1 H ^1$.
The mass defect $\Delta m$ is calculated as: $\Delta m = (M_{N} + M_{H}) - M_{O}$.
Substituting the given values: $\Delta m = (15.000108 + 1.007825) - 15.994915 \text{ a.m.u.}$.
$\Delta m = 16.007933 - 15.994915 = 0.013018 \text{ a.m.u.}$.
The energy required is $E = \Delta m \times 931.5 \text{ MeV/a.m.u.}$.
$E = 0.013018 \times 931.5 \approx 12.126 \text{ MeV}$,which is approximately $12.13 \text{ MeV}$.

(B) The number of protons $Z = 6$ and the number of neutrons $N = 12 - 6 = 6$.
The mass defect $\Delta m$ is given by:
$\Delta m = [Z \cdot m_H + N \cdot m_n] - M_{nucleus}$
$\Delta m = [6 \times 1.007825 + 6 \times 1.008665] - 12.00000$
$\Delta m = [6.04695 + 6.05199] - 12.00000$
$\Delta m = 12.09894 - 12.00000 = 0.09894 \text{ a.m.u.}$
The binding energy $BE = \Delta m \times 931.5 \text{ MeV/a.m.u.}$
$BE = 0.09894 \times 931.5 \approx 92.162 \text{ MeV}$
The binding energy per nucleon is $BE/A = 92.162 / 12 \approx 7.68 \text{ MeV}$.
Comparing with the given options,the closest value is $7.675 \text{ MeV}$.

(B) The energy released in a nuclear reaction is given by the difference between the total binding energy of the products and the total binding energy of the reactants.
Total binding energy of the reactant $X = 200 \times 7.4 \, MeV = 1480 \, MeV$.
Total binding energy of the products $A$ and $B = (110 \times 8.2 \, MeV) + (90 \times 8.2 \, MeV) = (110 + 90) \times 8.2 \, MeV = 200 \times 8.2 \, MeV = 1640 \, MeV$.
Energy released $Q = \text{Total BE of products} - \text{Total BE of reactants}$.
$Q = 1640 \, MeV - 1480 \, MeV = 160 \, MeV$.

(D) The binding energy of the last proton (also known as the proton separation energy) is given by the mass defect of the reaction: ${ }_8 O ^{16} \rightarrow { }_7 N ^{15} + { }_1 H ^1$.
The mass defect $\Delta m$ is calculated as:
$\Delta m = [M({ }_7 N ^{15}) + m_p] - M({ }_8 O ^{16})$
Substituting the given values:
$\Delta m = 15.00011 \ amu + 1.00783 \ amu - 15.99492 \ amu$
$\Delta m = 16.00794 \ amu - 15.99492 \ amu = 0.01302 \ amu$
Since $1 \ amu = 931.5 \ MeV/c^2$,the binding energy $E$ is:
$E = 0.01302 \times 931.5 \ MeV \approx 12.13 \ MeV$.

(A) The stability of a nucleus is determined by its binding energy per nucleon $(BE/A)$.
For ${ }_1^2 H$: $BE/A = 2.22 / 2 = 1.11 \text{ MeV/nucleon}$.
For ${ }_2^4 He$: $BE/A = 28.3 / 4 = 7.075 \text{ MeV/nucleon}$.
For ${ }_{26}^{56} Fe$: $BE/A = 492 / 56 \approx 8.79 \text{ MeV/nucleon}$.
For ${ }_{92}^{235} U$: $BE/A = 1786 / 235 \approx 7.60 \text{ MeV/nucleon}$.
Since ${ }_{26}^{56} Fe$ has the highest binding energy per nucleon,it is the most stable nucleus among the given options.

(B) The helium nucleus $(_{2}^{4}He)$ consists of $2$ protons and $2$ neutrons.
The mass defect $(\Delta m)$ is given by: $\Delta m = (2 m_p + 2 m_n) - m_{He}$.
Substituting the given values: $\Delta m = [2(1.0073) + 2(1.0087)] - 4.0015$.
$\Delta m = [2.0146 + 2.0174] - 4.0015 = 4.0320 - 4.0015 = 0.0305\,u$.
Since $1\,u = 931.5\,MeV/c^2$, the binding energy $(B.E.)$ is:
$B.E. = \Delta m \times 931.5\,MeV$.
$B.E. = 0.0305 \times 931.5 \approx 28.4\,MeV$.

(D) For nucleus $A$, the atomic number $Z = 17$. Since the number of protons equals the number of neutrons, the number of neutrons $N = 17$. Therefore, the mass number $A_{mass} = Z + N = 17 + 17 = 34$.
The total binding energy of $A$ is given by: $BE_A = (\text{binding energy per nucleon}) \times (\text{mass number}) = 1.2 \, MeV \times 34 = 40.8 \, MeV$.
For nucleus $B$, the mass number $A_{mass} = 26$ and the binding energy per nucleon is $1.8 \, MeV$.
The total binding energy of $B$ is given by: $BE_B = 1.8 \, MeV \times 26 = 46.8 \, MeV$.
The difference in binding energy is: $\Delta BE = BE_B - BE_A = 46.8 \, MeV - 40.8 \, MeV = 6 \, MeV$.

(A) According to the semi-empirical mass formula:
$1$. The surface energy term is $E_s = -a_s A^{2/3}$. The surface energy per nucleon is $b_s = E_s/A = -a_s A^{-1/3}$. Thus,statement $A$ is incorrect.
$2$. The Coulomb energy term is $E_c = -a_c \frac{Z(Z-1)}{A^{1/3}}$. Thus,statement $B$ is incorrect as the denominator is $A^{1/3}$,not $A^{4/3}$.
$3$. The volume energy term is $E_v = a_v A$. The volume energy per nucleon is $b_v = E_v/A = a_v$. Statement $C$ is correct as volume energy is proportional to $A$.
$4$. The surface energy arises because nucleons at the surface have fewer neighbors than those in the interior. The number of surface nucleons is proportional to the surface area $(4\pi R^2 \propto A^{2/3})$. Thus,the decrease in binding energy is proportional to the surface area. Statement $D$ is correct.
$5$. In the liquid drop model,it is assumed that each nucleon interacts with a finite number of neighbors (typically $12$ in a close-packed structure),but the surface energy correction accounts for the missing interactions at the surface. Statement $E$ is a standard assumption in this model. Thus,$C, D, E$ are correct,but given the options,$C$ and $D$ are the most accurate descriptions.