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

(B) The prefix 'mega' $(M)$ in the $SI$ system represents a factor of $10^6$.
Therefore,$1\,MeV$ is equivalent to $10^6\,eV$.

(C) The binding energy per nucleon in a nucleus represents the average energy required to remove a single nucleon from the nucleus. Experimental observations show that this energy is typically in the range of $7$ to $9 \ MeV$ for most stable nuclei. Therefore,the binding energy of a nucleon is of the order of a few $MeV$.

(C) The mass of $1 \text{ a.m.u.}$ is $1.66 \times 10^{-27} \text{ kg}$.
Using Einstein's mass-energy equivalence formula,$E = mc^2$,where $c = 3 \times 10^8 \text{ m/s}$ is the speed of light.
$E = (1.66 \times 10^{-27} \text{ kg}) \times (3 \times 10^8 \text{ m/s})^2$
$E = 1.66 \times 10^{-27} \times 9 \times 10^{16} \text{ J}$
$E \approx 14.94 \times 10^{-11} \text{ J} \approx 1.5 \times 10^{-10} \text{ J}$.
Therefore,the correct option is $C$.

(B) According to Einstein's mass-energy equivalence principle,$E = mc^2$,energy and mass are interchangeable. When water is cooled to form ice,it releases latent heat to the surroundings. Since the system loses energy,there must be a corresponding decrease in its total mass. Therefore,the mass of the water decreases when it turns into ice.

(D) According to Einstein's mass-energy equivalence principle,the energy $E$ is given by the formula $E = mc^2$.
Here,$m = 1 \text{ kg}$ and $c \approx 3 \times 10^8 \text{ m/s}$ (speed of light).
Substituting these values into the equation:
$E = 1 \times (3 \times 10^8)^2$
$E = 1 \times 9 \times 10^{16}$
$E = 9 \times 10^{16} \text{ J}$
This value is approximately equal to $10^{17} \text{ J}$.
Therefore,the correct option is $D$.

(D) The nuclear binding energy $(B.E.)$ is the energy required to disassemble a nucleus into its constituent protons and neutrons.
According to Einstein's mass-energy equivalence principle, this energy is equivalent to the mass defect $(\Delta m)$ of the nucleus.
The relationship is given by the formula: $B.E. = \Delta m \times c^2$.
In terms of atomic mass units, it is expressed as: $B.E. = \Delta m \times 931 \, MeV$.

(A) The binding energy $(BE)$ of a nucleus is related to the mass defect $(\Delta m)$ by the equation: $BE = \Delta m \times 931.5 \, MeV/a.m.u.$
Given, $BE = 2.23 \, MeV$.
Therefore, $\Delta m = \frac{BE}{931.5} = \frac{2.23}{931.5} \approx 0.00239 \, a.m.u.$
Rounding this value, we get $\Delta m \approx 0.0024 \, a.m.u.$

(A) The helium nucleus ${ }_2^4 \text{He}$ consists of $2$ protons and $2$ neutrons.
Mass of $2$ protons $= 2 \times 1.0073 \; \text{u} = 2.0146 \; \text{u}$.
Mass of $2$ neutrons $= 2 \times 1.0087 \; \text{u} = 2.0174 \; \text{u}$.
Total mass of nucleons $= 2.0146 + 2.0174 = 4.0320 \; \text{u}$.
Mass defect $\Delta m = (\text{Total mass of nucleons}) - (\text{Mass of nucleus}) = 4.0320 \; \text{u} - 4.0015 \; \text{u} = 0.0305 \; \text{u}$.
Binding energy $\text{B.E.} = \Delta m \times 931.5 \; \text{MeV/u} = 0.0305 \times 931.5 \approx 28.4 \; \text{MeV}$.

(B) The mass defect $(\Delta m)$ for the helium nucleus is $0.0303 \, a.m.u.$
We know that $1 \, a.m.u. = 931 \, MeV$ of energy.
Total binding energy $(BE)$ = $\Delta m \times 931 \, MeV = 0.0303 \times 931 \approx 28.21 \, MeV$.
Helium nucleus $(_{2}^{4}He)$ has $4$ nucleons ($2$ protons and $2$ neutrons).
Binding energy per nucleon = $\frac{BE}{\text{Number of nucleons}} = \frac{28.21}{4} \approx 7.05 \, MeV$.
Thus, the binding energy per nucleon is approximately $7 \, MeV$.

(C) The mass-energy equivalence is given by Einstein's equation: $E = \Delta m c^2$.
Given mass $\Delta m = 1 \ \mu g = 10^{-6} \ g = 10^{-9} \ kg$.
The speed of light $c = 3 \times 10^8 \ m/s$.
Substituting these values into the equation:
$E = 10^{-9} \ kg \times (3 \times 10^8 \ m/s)^2$
$E = 10^{-9} \times 9 \times 10^{16} \ J$
$E = 9 \times 10^7 \ J$.
Therefore,the correct option is $C$.

(C) The binding energy per nucleon of a nucleus is defined as the total binding energy of the nucleus divided by its mass number $(A)$,which represents the total number of nucleons.
Experimental observations of the binding energy per nucleon versus mass number graph show that for most nuclei (excluding very light nuclei),the value remains relatively constant.
The average binding energy per nucleon for stable nuclei is approximately $8 \, MeV$ per nucleon.

(C) The binding energy of a nucleus is defined as the energy equivalent of the mass defect.
When nucleons (protons and neutrons) come together to form a stable nucleus, the mass of the nucleus is found to be less than the sum of the individual masses of the constituent nucleons.
This difference in mass, known as the mass defect $(\Delta m)$, is released as energy during the formation of the nucleus.
Therefore, the binding energy is the energy released from the nucleus during its formation, which is equivalent to the work required to disassemble the nucleus into its constituent nucleons.
Thus, option $(C)$ is the correct description.

(C) The energy required to remove a nucleon from a nucleus is related to the nuclear binding energy,which is governed by the strong nuclear force.
The energy required to remove an electron from an atom is related to the ionization energy,which is governed by the electromagnetic (Coulomb) force.
Since the strong nuclear force is significantly stronger than the electromagnetic force,the energy required to remove a nucleon $({E_n})$ is much greater than the energy required to remove an electron $({E_e})$.
Therefore,${E_n} > {E_e}$.

(C) $1 \, a.m.u. = 1.66 \times 10^{-27} \, kg$.
According to Einstein's mass-energy equivalence,$E = mc^2$,where $c$ is the speed of light $(c \approx 3 \times 10^8 \, m/s)$.
Substituting the values:
$E = 1.66 \times 10^{-27} \times (3 \times 10^8)^2 \, J$
$E = 1.66 \times 10^{-27} \times 9 \times 10^{16} \, J$
$E = 14.94 \times 10^{-11} \, J$.
To convert this energy into electron-volts $(eV)$,divide by $1.6 \times 10^{-19} \, J/eV$:
$E = \frac{14.94 \times 10^{-11}}{1.6 \times 10^{-19}} \, eV$
$E \approx 9.3375 \times 10^8 \, eV \approx 931 \times 10^6 \, eV$.
Since $10^6 \, eV = 1 \, MeV$,the equivalent energy is $931 \, MeV$.

(B) The mass defect $(\Delta m)$ is the difference between the sum of the masses of individual nucleons and the actual mass of the nucleus.
Packing fraction $(f)$ is defined as the ratio of the mass defect $(\Delta m)$ to the mass number $(A)$ of the nucleus.
Mathematically, $f = \frac{\Delta m}{A}$.
Therefore, the mass defect per nucleon is known as the packing fraction.

(B) The binding energy per nucleon $(B.E./A)$ is a measure of the stability of a nucleus.
According to the binding energy curve,the value of $B.E./A$ increases with the mass number $A$ for light nuclei,reaches a maximum value of approximately $8.8 \text{ MeV}$ for iron $(_{26}^{56}Fe)$,and then gradually decreases for heavier nuclei.
Therefore,the binding energy per nucleon is maximum for $_{26}^{56}Fe$.

(A) The mass of an $\alpha$-particle is less than the sum of the masses of its constituent particles (two protons and two neutrons).
This difference in mass is known as the mass defect $(\Delta m)$.
The mass defect is converted into binding energy,which is required to hold the nucleons together within the nucleus of the $\alpha$-particle.
Therefore,the correct option is $A$.

(D) The binding energy $(B.E.)$ of a nucleus is calculated as: $B.E. = (\text{number of nucleons}) \times (\text{binding energy per nucleon})$.
For $Li^7$, $B.E. = 7 \times 5.60 \ MeV = 39.20 \ MeV$.
For $He^4$, $B.E. = 4 \times 7.06 \ MeV = 28.24 \ MeV$.
The reaction is $Li^7 + p \to 2He^4$.
The binding energy of the proton $(p)$ is $0 \ MeV$ as it is a single nucleon.
The total binding energy of the reactants is $39.20 \ MeV + 0 \ MeV = 39.20 \ MeV$.
The total binding energy of the products is $2 \times 28.24 \ MeV = 56.48 \ MeV$.
The energy released in the reaction ($Q$-value) is the difference between the total binding energy of the products and the reactants: $Q = 56.48 \ MeV - 39.20 \ MeV = 17.28 \ MeV$.
Rounding to one decimal place, we get $17.3 \ MeV$.

(A) According to Einstein's mass-energy equivalence principle,the energy $E$ is given by the formula $E = mc^2$.
Here,the mass $m = 1 \, g = 1 \times 10^{-3} \, kg$.
The speed of light $c = 3 \times 10^8 \, m/s$.
Substituting these values into the equation:
$E = (1 \times 10^{-3} \, kg) \times (3 \times 10^8 \, m/s)^2$
$E = 10^{-3} \times 9 \times 10^{16} \, J$
$E = 9.0 \times 10^{13} \, J$.
Therefore,the correct option is $A$.

(A) The actual mass of a nucleus $(M)$ is always less than the sum of the individual masses of its constituent nucleons (protons and neutrons).
This difference in mass is known as the mass defect $(\Delta m)$, which is converted into binding energy according to Einstein's mass-energy equivalence principle $(E = \Delta m c^2)$.
Therefore, the mass of the nucleus $M$ is related to the sum of the masses of $N$ neutrons and $Z$ protons as:
$M < (N M_n + Z M_p)$.

(B) The mass of a $H_2$ nucleus (deuteron) is approximately $2.014 \, amu$.
However,in the context of standard physics problems regarding energy equivalence of nucleons,the mass of a single nucleon (proton or neutron) is approximately $1 \, amu$.
Since $1 \, amu$ is equivalent to $931 \, MeV$ of energy,and a deuteron consists of $2$ nucleons,the energy would be approximately $2 \times 931 \, MeV = 1862 \, MeV$.
Given the provided options,the question likely refers to the energy equivalent of $1 \, amu$ (the mass of a single proton),which is $931 \, MeV$.
Thus,the correct option is $(b)$.

(B) $1\, amu$ is defined as one-twelfth of the mass of an unbound neutral atom of carbon-$12$ in its nuclear and electronic ground state.
Mass of $1$ mole of $C$ atoms $= 12\, g = 0.012\, kg$.
$1\, amu = \frac{0.012\, kg}{6.023 \times 10^{23}} \approx 1.66 \times 10^{-27}\, kg$.
Using Einstein's mass-energy equivalence formula,$E = mc^2$:
$E = (1.66 \times 10^{-27}\, kg) \times (3 \times 10^8\, m/s)^2 = 1.494 \times 10^{-10}\, J$.
Converting Joules to $MeV$ $(1\, eV = 1.602 \times 10^{-19}\, J)$:
$E = \frac{1.494 \times 10^{-10}}{1.602 \times 10^{-13}}\, MeV \approx 931.5\, MeV \approx 930\, MeV$.
Thus,the correct option is $B$.

(A) The mass of a nucleus is always less than the sum of the masses of its individual constituent nucleons (protons and neutrons). This difference in mass is known as the mass defect $(\Delta m)$,which is converted into binding energy according to Einstein's mass-energy equivalence principle $(E = \Delta mc^2)$.
Given that the nucleus $_Z{X^A}$ contains $Z$ protons and $(A - Z)$ neutrons,the sum of the masses of the individual nucleons is $(A - Z)m_n + Zm_p$.
Since the mass of the nucleus $m$ is less than the sum of the masses of its constituents due to the mass defect,we have:
$m < (A - Z)m_n + Zm_p$.

(B) The energy released in an atomic explosion is given by Einstein's mass-energy equivalence relation: $E = \Delta m C^2$.
Let the original speed of light be $C$ and the new speed of light be $C' = \frac{2}{3} C$.
The new energy released $E'$ is given by: $E' = \Delta m (C')^2$.
Substituting the value of $C'$: $E' = \Delta m \left(\frac{2}{3} C\right)^2 = \Delta m \left(\frac{4}{9} C^2\right) = \frac{4}{9} E$.
The fraction by which the energy is decreased is given by: $\text{Fraction} = \frac{E - E'}{E}$.
Substituting $E' = \frac{4}{9} E$: $\text{Fraction} = \frac{E - \frac{4}{9} E}{E} = \frac{\frac{5}{9} E}{E} = \frac{5}{9}$.

(D) In nuclear reactions,the total energy (including mass-energy equivalence),linear momentum,angular momentum,and charge are conserved. According to the law of conservation of mass-energy,the total mass-energy of the system remains constant. Since no external forces act on the system,the total linear momentum is also conserved. Therefore,mass (in terms of mass-energy),energy,and momentum are all conserved.

(B) In a nuclear fission process,a heavy nucleus splits into two lighter nuclei.
For the process to be energetically favorable and release energy,the total binding energy of the product nuclei must be greater than the binding energy of the parent nucleus.
This is because the binding energy per nucleon increases as we move from a heavy nucleus towards the middle of the periodic table (near iron).
Therefore,the sum of the binding energies of the daughter nuclei $B$ and $C$ is greater than the binding energy of the parent nucleus $A$.
Thus,${E_b} + {E_c} > {E_a}$.

(A) According to Einstein's mass-energy equivalence relation,$E = mc^2$,where $E$ is energy,$m$ is mass,and $c$ is the speed of light $(c = 3 \times 10^8\,m/s)$.
To find the mass $m$,we use the formula $m = \frac{E}{c^2}$.
Given $E = 931\, MeV = 931 \times 10^6 \times 1.602 \times 10^{-19}\,J \approx 931 \times 1.6 \times 10^{-13}\,J$.
Substituting the values: $m = \frac{931 \times 1.6 \times 10^{-13}}{(3 \times 10^8)^2} = \frac{1489.6 \times 10^{-13}}{9 \times 10^{16}} \approx 1.66 \times 10^{-27}\,kg$.

(D) The process of pair production involves the conversion of a high-energy photon into an electron-positron pair.
The rest mass energy of an electron is $m_e c^2 \approx 0.511 \, MeV$.
The rest mass energy of a positron is also $m_p c^2 \approx 0.511 \, MeV$.
To create this pair,the total energy of the $\gamma$-ray photon must be at least equal to the sum of the rest mass energies of the electron and the positron.
Therefore,the minimum energy required is $E_{min} = 0.511 \, MeV + 0.511 \, MeV = 1.02 \, MeV$.

(C) In any nuclear reaction,several fundamental quantities are conserved. These include:
$1$. Conservation of mass number $(A)$: The total number of nucleons (protons + neutrons) remains constant.
$2$. Conservation of atomic number $(Z)$: The total charge of the system remains constant.
$3$. Conservation of energy: The total energy,including the rest mass energy $(E = mc^2)$,is conserved.
$4$. Conservation of linear momentum and angular momentum.
Therefore,atomic number,mass number,and energy are all conserved in a nuclear reaction. The correct option is $C$.

(B) When a proton and an anti-proton annihilate, their total mass is converted into energy according to Einstein's mass-energy equivalence principle, $E = mc^2$.
The mass of a proton is approximately $1.67 \times 10^{-27} \; kg$, which is equivalent to $1 \; amu$.
The mass of an anti-proton is the same as that of a proton, i.e., $1 \; amu$.
Total mass annihilated = $1 \; amu + 1 \; amu = 2 \; amu$.
We know that $1 \; amu$ is equivalent to $931.5 \; MeV$.
Therefore, total energy released = $2 \times 931.5 \; MeV = 1863 \; MeV$.
Converting this energy into Joules:
$E = 1863 \times 10^6 \times 1.602 \times 10^{-19} \; J \approx 2.98 \times 10^{-10} \; J$.
Rounding this value, we get approximately $3 \times 10^{-10} \; J$.

(B) The mass of an electron $(m_e)$ is $9.1 \times 10^{-31} \ kg$.
Since a positron has the same mass as an electron,the total mass involved in the annihilation is $2m_e$.
The energy released $(E)$ is given by Einstein's mass-energy equivalence formula: $E = (2m_e)c^2$.
Substituting the values: $E = 2 \times (9.1 \times 10^{-31} \ kg) \times (3 \times 10^8 \ m/s)^2$.
$E = 18.2 \times 10^{-31} \times 9 \times 10^{16} \ J$.
$E = 163.8 \times 10^{-15} \ J = 1.638 \times 10^{-13} \ J$.
Rounding to the nearest provided option,the energy released is approximately $1.6 \times 10^{-13} \ J$.

(D) The binding energy per nucleon is a direct measure of the stability of a nucleus. $A$ higher binding energy per nucleon indicates that the nucleons are more tightly bound,making the nucleus more stable. Therefore,the binding energy is a measure of nuclear stability.

(D) The packing fraction $(f)$ is defined as the difference between the actual atomic mass $(M)$ and the mass number $(A)$,divided by the mass number $(A)$.
Mathematically,it is expressed as:
$f = \frac{M - A}{A}$
This quantity provides a measure of the stability of a nucleus.

(C) The binding energy $B$ of a nucleus is defined as the energy equivalent of the mass defect $\Delta m$.
The mass defect is given by $\Delta m = (Z M_p + N M_n) - M(N, Z)$.
According to Einstein's mass-energy equivalence relation,$B = \Delta m c^2$.
Substituting the expression for $\Delta m$,we get $B = [Z M_p + N M_n - M(N, Z)] c^2$.
Rearranging the terms to solve for $M(N, Z)$:
$B / c^2 = Z M_p + N M_n - M(N, Z)$
$M(N, Z) = Z M_p + N M_n - B / c^2$.

(B) The binding energy per nucleon $(B.E./A)$ is a measure of the stability of a nucleus.
$A$ higher value of $B.E./A$ indicates that the nucleons are more tightly bound together,making the nucleus more stable.
The curve of $B.E./A$ versus mass number $(A)$ shows a sharp peak for the helium nucleus $(^4_2He)$,which indicates that it has a very high binding energy per nucleon compared to its neighbors.
Therefore,the helium nucleus is exceptionally stable.

(D) Using Einstein's mass-energy equivalence principle,$E = \Delta m c^2$.
Given,$\Delta m = 1\, kg$ and $c = 3 \times 10^8\, m/s$.
$E = 1 \times (3 \times 10^8)^2 = 9 \times 10^{16}\, J$.
To convert energy from Joules $(J)$ to electron-volts $(eV)$,divide by $1.6 \times 10^{-19}\, J/eV$:
$E = \frac{9 \times 10^{16}}{1.6 \times 10^{-19}} = 5.625 \times 10^{35}\, eV$.
Since $1\, MeV = 10^6\, eV$,we have:
$E = \frac{5.625 \times 10^{35}}{10^6} = 5.625 \times 10^{29}\, MeV$.

(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.
In the given reaction: $_1^2H + _1^3H \to _2^4He + _0^1n$.
The binding energy of the neutron $_0^1n$ is $0 \ MeV$.
The total binding energy of the reactants is $a + b$.
The total binding energy of the products is $c + 0 = c$.
Therefore, the energy released $Q = (\text{Total Binding Energy of Products}) - (\text{Total Binding Energy of Reactants})$.
$Q = c - (a + b) = c - a - b \ MeV$.

(D) The intensity of solar radiation at the earth's surface is $I = 1.4 \; kW/m^2 = 1.4 \times 10^3 \; J/(s \cdot m^2)$.
The total power radiated by the sun is $P = I \times A$, where $A$ is the surface area of a sphere with radius $r = 1.5 \times 10^{11} \; m$.
$A = 4 \pi r^2 = 4 \times 3.14 \times (1.5 \times 10^{11})^2 \approx 2.827 \times 10^{23} \; m^2$.
Total energy radiated per day $(E)$ is $P \times t$, where $t = 86400 \; s$.
$E = (1.4 \times 10^3 \; J/s \cdot m^2) \times (2.827 \times 10^{23} \; m^2) \times (86400 \; s) \approx 3.42 \times 10^{31} \; J$.
Using Einstein's mass-energy equivalence, $E = mc^2$, where $c = 3 \times 10^8 \; m/s$:
$m = \frac{E}{c^2} = \frac{3.42 \times 10^{31}}{(3 \times 10^8)^2} = \frac{3.42 \times 10^{31}}{9 \times 10^{16}} \approx 3.8 \times 10^{14} \; kg$.

(C) The nuclear reaction for removing a neutron from ${O^{17}}$ is given by: ${O^{17}} \to {O^{16}} + {n^1}$.
The total binding energy $(B.E.)$ of a nucleus is calculated as: $B.E. = (\text{Number of nucleons}) \times (\text{Binding energy per nucleon})$.
Total $B.E.$ of ${O^{17}} = 17 \times 7.75 \,MeV = 131.75 \,MeV$.
Total $B.E.$ of ${O^{16}} = 16 \times 7.97 \,MeV = 127.52 \,MeV$.
The energy required to remove the neutron is the difference between the total binding energies of the parent and daughter nuclei:
$\text{Energy required} = B.E.({O^{17}}) - B.E.({O^{16}})$
$\text{Energy required} = 131.75 \,MeV - 127.52 \,MeV = 4.23 \,MeV$.

(D) The mass of a nucleus is always less than the sum of the masses of its constituent nucleons due to the mass defect,which accounts for the binding energy of the nucleus.
For the $_{10}^{20}Ne$ nucleus,which consists of $10$ protons and $10$ neutrons,the mass ${M_1}$ is given by ${M_1} < 10({m_p} + {m_n})$. This confirms option $(a)$.
For the $_{20}^{40}Ca$ nucleus,which consists of $20$ protons and $20$ neutrons,the binding energy per nucleon is higher than that of $_{10}^{20}Ne$. Since the binding energy per nucleon increases with the mass number for lighter nuclei,the mass defect per nucleon is greater for $_{20}^{40}Ca$ than for $_{10}^{20}Ne$.
Therefore,the mass of the $_{20}^{40}Ca$ nucleus ${M_2}$ is less than twice the mass of the $_{10}^{20}Ne$ nucleus ${M_1}$,i.e.,${M_2} < 2{M_1}$. This confirms option $(c)$.
Thus,both $(a)$ and $(c)$ are correct.

(A) The binding energy per nucleon $(B_N)$ is a measure of the stability of a nucleus.
Experimental observations show that $B_N$ increases rapidly for light nuclei and reaches a maximum value of approximately $8.8 \text{ MeV}$ for iron $(Fe^{56})$,which corresponds to a mass number $A = 56$.
For nuclei with $A > 56$,the binding energy per nucleon gradually decreases as the mass number increases.
Therefore,the graph that correctly represents this dependence shows a peak at $A = 56$.

(C) The binding energy per nucleon $(BE/A)$ is a measure of the stability of a nucleus.
For light nuclei,the $BE/A$ increases rapidly with the mass number $(A)$.
It reaches a maximum value of about $8.8 \text{ MeV}$ for nuclei with mass numbers between $30$ and $170$ (e.g.,$^{56}Fe$).
For heavier nuclei,the $BE/A$ decreases gradually as the mass number increases due to the increased Coulomb repulsion between protons.
Curve $C$ correctly represents this behavior,showing an initial increase,a broad maximum,and a subsequent gradual decrease.
Therefore,the correct option is $C$.

(C) Energy is released in a nuclear process when the total binding energy $(B.E.)$ of the products is greater than the total binding energy of the reactants.
From the graph, we can identify the mass number $(A)$ and binding energy per nucleon $(B.E./A)$ for each nucleus:
For $W$: $A = 120, B.E./A = 7.5 \, MeV \implies \text{Total } B.E. = 120 \times 7.5 = 900 \, MeV$
For $X$: $A = 90, B.E./A = 8.0 \, MeV \implies \text{Total } B.E. = 90 \times 8.0 = 720 \, MeV$
For $Y$: $A = 60, B.E./A = 8.5 \, MeV \implies \text{Total } B.E. = 60 \times 8.5 = 510 \, MeV$
For $Z$: $A = 30, B.E./A = 5.0 \, MeV \implies \text{Total } B.E. = 30 \times 5.0 = 150 \, MeV$
Now, checking option $(c)$: $W \to 2Y$
Total $B.E.$ of reactants $= 900 \, MeV$
Total $B.E.$ of products $= 2 \times (510) = 1020 \, MeV$
Since the total $B.E.$ of products $(1020 \, MeV)$ is greater than the total $B.E.$ of reactants $(900 \, MeV)$, energy is released in this process.

(D) The atomic mass unit $(\text{a.m.u.})$ is defined as $1/12$ of the mass of a carbon-$12$ atom.
Using the mass-energy equivalence principle $E = mc^2$, the energy equivalent of $1 \, \text{a.m.u.}$ is calculated as follows:
$1 \, \text{a.m.u.} = 1.660539 \times 10^{-27} \, \text{kg}$.
$E = (1.660539 \times 10^{-27} \, \text{kg}) \times (2.9979 \times 10^8 \, \text{m/s})^2$.
$E \approx 1.4924 \times 10^{-10} \, \text{J}$.
Converting Joules to $MeV$ $(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J})$:
$E \approx (1.4924 \times 10^{-10}) / (1.602 \times 10^{-13} \, \text{J/MeV}) \approx 931.5 \, \text{MeV}$.
Therefore, $1 \, \text{a.m.u.} \approx 931 \, \text{MeV}/c^2$.

(D) In a nuclear decay process,energy is released because the system moves towards a more stable state.
Stability is directly related to the binding energy per nucleon.
$A$ nucleus with higher binding energy per nucleon is more stable.
Since the parent nucleus decays into daughter nuclei,the daughter nuclei must be more stable than the parent nucleus.
Therefore,the binding energy per nucleon of the daughter nuclei $(E_2)$ must be greater than the binding energy per nucleon of the parent nucleus $(E_1)$.
Thus,$E_2 > E_1$.

(B) The binding energy per nucleon $(BE/A)$ is not a monotonic function of the mass number $(A)$.
For light nuclei $(A < 30)$, $BE/A$ generally increases with $A$.
For intermediate nuclei $(30 < A < 170)$, $BE/A$ is relatively constant (around $8 \text{ MeV}$).
For heavy nuclei $(A > 170)$, $BE/A$ decreases as $A$ increases.
Since the question asks for the general trend as mass number increases across the entire range, it is observed that for heavy nuclei, the binding energy per nucleon decreases. However, in the context of standard physics curriculum questions, if the question implies the trend for heavy nuclei or the general behavior of stability, the most appropriate answer is that it decreases for heavy nuclei.

(B) The binding energy per nucleon $(BE/A)$ curve shows that for light nuclei $(A < 30)$,the binding energy per nucleon is low.
As the mass number increases,$BE/A$ increases rapidly and reaches a maximum value of approximately $8.8 \text{ MeV}$ for nuclei with mass numbers between $50$ and $80$.
Specifically,for iron $(^{56}_{26}Fe)$,the binding energy per nucleon is maximum.
Therefore,the range of mass numbers where the binding energy per nucleon is maximum is between $50$ and $100$.

(D) The binding energy per nucleon curve shows that for mass numbers $A > 56$,the binding energy per nucleon gradually decreases as the mass number increases.
Because heavy nuclei (high mass number) have lower binding energy per nucleon compared to intermediate-mass nuclei,they become more stable by splitting into smaller fragments,which is the process of nuclear fission.

(A) The binding energy per nucleon is a measure of the stability of a nucleus. For most stable nuclei,the binding energy per nucleon is found to be approximately $8 \, MeV$. This value is relatively constant for nuclei with mass numbers between $30$ and $170$.