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

(A) The atomic number $Z = 14$ and the mass number $A = 29$.
The number of protons is $Z = 14$.
The number of neutrons is $N = A - Z = 29 - 14 = 15$.
The mass defect $\Delta m$ is given by $\Delta m = [Z m_p + N m_n] - M_{nucleus}$.
$\Delta m = [14 \times 1.007276 u + 15 \times 1.008664 u] - 28.976495 u$.
$\Delta m = [14.101864 u + 15.129960 u] - 28.976495 u$.
$\Delta m = 29.231824 u - 28.976495 u = 0.255329 u$.
The binding energy $B.E. = \Delta m \times 931.5 MeV/u$.
$B.E. = 0.255329 \times 931.5 MeV \approx 237.84 MeV$.
Rounding to the nearest provided option,the correct answer is $237.86 MeV$.

(B) Statement $(i)$ is false because the mass of a nucleus is always less than the sum of the masses of its individual nucleons. This mass difference,known as the mass defect,corresponds to the binding energy released when the nucleus is formed. Therefore,the mass energy of a nucleus is less than the total mass energy of its individual protons and neutrons.
Statement $(ii)$ is true. To separate a nucleus into its constituent nucleons,work must be done against the strong nuclear force,which requires an energy input equal to the binding energy.
Statement $(iii)$ is true. Binding energy per nucleon is a standard measure of nuclear stability; higher values indicate that nucleons are more tightly bound.
Statement $(iv)$ is true. In nuclear fission,a heavy nucleus splits into lighter nuclei with higher binding energy per nucleon,resulting in the release of energy.
Thus,statements $(ii)$,$(iii)$,and $(iv)$ are correct.

(A) The alpha particle $\left({ }^{4} He\right)$ consists of $2$ protons and $2$ neutrons.
The mass of the constituents is $m_{c} = 2(m_{p} + m_{n}) = 2(1.00783 + 1.00867) = 2(2.01650) = 4.03300 \ amu$.
The mass defect $\Delta m$ is given by $\Delta m = m_{c} - m_{He} = 4.03300 - 4.00300 = 0.0300 \ amu$.
The binding energy $E$ is calculated using the conversion factor $1 \ amu = 931 \ MeV$.
$E = \Delta m \times 931 \ MeV/amu = 0.0300 \times 931 = 27.93 \ MeV$.
Thus,the binding energy is approximately $27.9 \ MeV$.

(A) The stability of a nucleus is directly related to its binding energy per nucleon. However,in these reactions,we can compare the binding energy $(BE)$ directly to determine relative stability.
From the first reaction: ${ }_2 He ^3 + { }_0 n ^1 \rightarrow { }_2 He ^4 + 20 \ MeV$. The energy released $(Q = 20 \ MeV)$ implies that $BE({ }_2 He ^4) - BE({ }_2 He ^3) = 20 \ MeV$. Thus,$BE({ }_2 He ^4) > BE({ }_2 He ^3)$.
From the second reaction: ${ }_2 He ^4 + { }_0 n ^1 \rightarrow { }_2 He ^5 - 0.9 \ MeV$. The energy absorbed $(Q = -0.9 \ MeV)$ implies that $BE({ }_2 He ^5) - BE({ }_2 He ^4) = -0.9 \ MeV$. Thus,$BE({ }_2 He ^4) > BE({ }_2 He ^5)$.
Comparing the two,we find $BE({ }_2 He ^4) > BE({ }_2 He ^3)$ and $BE({ }_2 He ^4) > BE({ }_2 He ^5)$.
Since ${ }_2 He ^4$ is a very stable alpha particle (magic number $Z=2, N=2$),it has the highest stability. Comparing ${ }_2 He ^3$ and ${ }_2 He ^5$,${ }_2 He ^3$ is more stable than ${ }_2 He ^5$ because ${ }_2 He ^5$ is highly unstable and decays rapidly.
Therefore,the order of stability is $X_4 > X_3 > X_5$.

(D) Statement $I$ is false because the binding energy per nucleon is not a monotonically increasing function of the mass number. It increases initially,reaches a maximum for iron $(Fe)$,and then decreases for heavier nuclei.
Statement $II$ is true because nuclei with lower binding energy per nucleon are less stable. Through processes like nuclear fission (for heavy nuclei) or nuclear fusion (for light nuclei),they tend to transform into more stable nuclei with higher binding energy per nucleon to achieve a lower energy state.

(C) According to Einstein's mass-energy equivalence principle,the energy $E$ is given by the formula $E = mc^2$.
Here,$m$ is the mass of the substance and $c$ is the speed of light in a vacuum.
Given:
Mass $m = 1.0 \text{ kg}$
Speed of light $c = 3 \times 10^8 \text{ m/s}$
Substituting these values into the formula:
$E = 1.0 \text{ kg} \times (3 \times 10^8 \text{ m/s})^2$
$E = 1.0 \times 9 \times 10^{16} \text{ J}$
$E = 9 \times 10^{16} \text{ J}$
Therefore,the energy equivalent of $1.0 \text{ kg}$ of substance is $9 \times 10^{16} \text{ J}$.
The correct option is $C$.

(B) The number of protons $Z = 83$,and the number of neutrons $N = 209 - 83 = 126$.
The mass defect $\Delta m$ is calculated as: $\Delta m = [Z m_p + N m_n - M_{nucleus}]$.
$\Delta m = [83 \times 1.007825 + 126 \times 1.008665 - 208.980388] \text{ u}$.
$\Delta m = [83.649475 + 127.09179 - 208.980388] \text{ u} = 1.760877 \text{ u}$.
The total binding energy is $BE = \Delta m \times 931 \text{ MeV/u} = 1.760877 \times 931 \approx 1639.376 \text{ MeV}$.
The binding energy per nucleon is $\frac{BE}{A} = \frac{1639.376}{209} \approx 7.84 \text{ MeV}$.

(C) The nuclear reaction is given by: $2 \times X(A=3) + Y(A=4) \to Z(A=10)$.
Total binding energy of reactants $= (2 \times 3 \times 5.6 \text{ MeV}) + (4 \times 7.4 \text{ MeV}) = 33.6 \text{ MeV} + 29.6 \text{ MeV} = 63.2 \text{ MeV}$.
Total binding energy of the product $= 10 \times 6.1 \text{ MeV} = 61.0 \text{ MeV}$.
The energy change in the process is $\Delta E = E_{\text{product}} - E_{\text{reactants}} = 61.0 \text{ MeV} - 63.2 \text{ MeV} = -2.2 \text{ MeV}$.
The magnitude of the energy change involved in the process is $2.2 \text{ MeV}$.
Therefore,$\Delta Mc^2 = 2.2 \text{ MeV}$.

(B) The nucleus of $^{12}_{6}C$ contains $6$ protons and $6$ neutrons.
The mass of the constituent nucleons is calculated as:
$M_{nucleons} = 6 \times m_p + 6 \times m_n$
$M_{nucleons} = 6 \times 1.00727 \text{ u} + 6 \times 1.00866 \text{ u}$
$M_{nucleons} = 6.04362 \text{ u} + 6.05196 \text{ u} = 12.09558 \text{ u}$.
The mass defect $\Delta m$ is the difference between the sum of the masses of the nucleons and the actual mass of the nucleus:
$\Delta m = M_{nucleons} - M_{nucleus}$
$\Delta m = 12.09558 \text{ u} - 12.00000 \text{ u} = 0.09558 \text{ u}$.