Composition of Nucleus, Size of the Nucleus, Nuclear force Questions in English

(A) The radius of a nucleus is given by the relation $R = R_0 A^{1/3}$,where $R_0$ is a constant and $A$ is the mass number.
For nucleus $1$: $R_1 = R_0 A_1^{1/3}$.
For nucleus $2$: $R_2 = R_0 A_2^{1/3}$.
Taking the ratio: $\frac{R_2}{R_1} = \left( \frac{A_2}{A_1} \right)^{1/3}$.
Given that $A_2$ is $2\%$ larger than $A_1$,we have $A_2 = A_1(1 + 0.02) = 1.02 A_1$,so $\frac{A_2}{A_1} = 1.02$.
Using the binomial approximation $(1 + x)^n \approx 1 + nx$ for small $x$:
$\frac{R_2}{R_1} = (1.02)^{1/3} = (1 + 0.02)^{1/3} \approx 1 + \frac{1}{3}(0.02) = 1 + 0.00666...$
Thus,$\frac{R_2}{R_1} \approx 1 + \frac{2}{300} = 1 + \frac{2}{3} \%$.
Therefore,$R_2$ is larger than $R_1$ by $\frac{2}{3} \%$.

(B) There are four fundamental forces in nature: Gravitational force,Weak nuclear force,Electromagnetic force,and Strong nuclear force.
Among these,the strong nuclear force is the strongest force,which acts between nucleons (protons and neutrons) to hold the nucleus together.

(D) According to the fundamental forces in nature,the increasing order of strength of forces is given by:
Gravitational force $ < $ Weak nuclear force $ < $ Electromagnetic force $ < $ Strong nuclear force.
Therefore,the strong nuclear force is the strongest force in nature.

(A-(II), B-(IV), C-(I), D-(III)) The four fundamental forces of nature and their relative strengths are as follows:
$(i)$ Strong nuclear force: It is the strongest force,acting between nucleons (protons and neutrons) and quarks. Its relative strength is of the order of $1$.
(ii) Electromagnetic force: It acts between electrically charged particles. Its relative strength is of the order of $10^{-2}$.
(iii) Weak nuclear force: It acts between subatomic particles during certain radioactive decays. Its relative strength is of the order of $10^{-13}$.
(iv) Gravitational force: It is the force of attraction between masses. Its relative strength is the weakest,of the order of $10^{-39}$.
Comparing these with the given lists:
$(A)$ Strong nuclear force $\rightarrow$ (ii) $1$
$(B)$ Weak nuclear force $\rightarrow$ (iv) $10^{-13}$
$(C)$ Electromagnetic force $\rightarrow$ $(i)$ $10^{-2}$
$(D)$ Gravitational force $\rightarrow$ (iii) $10^{-39}$
Thus,the correct matching is $A-(ii), B-(iv), C-(i), D-(iii)$.

(B) The fundamental forces of nature in decreasing order of their strength are: Strong nuclear force $>$ Electromagnetic force $>$ Weak nuclear force $>$ Gravitational force.
Therefore,the gravitational force is the weakest force in nature,not the weak nuclear force.
Thus,the statement in option $B$ is incorrect.

(B) By definition,an anti-particle has the same mass and the same spin as its corresponding particle.
However,properties like electric charge,lepton number,and magnetic moment are opposite in sign to those of the particle.
Therefore,particles and their anti-particles have the same masses but opposite magnetic moments.

(C) The potential energy between two nucleons as a function of their separation distance is described by the nuclear force potential,often modeled by the Reid potential.
This potential curve shows that the nuclear force is strongly attractive at short distances and repulsive at very short distances.
The potential energy reaches its minimum value at a separation distance of approximately $0.8 \ fm$.
At this distance,the nucleons are in a stable configuration,which allows them to form a bound state with a negative binding energy.

(C) The nuclear force is the strong force that acts between nucleons (protons and neutrons) within the nucleus.
Experimental evidence shows that the nuclear force is charge-independent.
This means the force between two protons,two neutrons,or a proton and a neutron is essentially the same,provided the distance between them and their spin states are the same.
Therefore,$F_{pp} = F_{nn} = F_{np}$.

(D) The weak nuclear force is one of the four fundamental forces of nature.
It is responsible for processes like beta decay in nuclei.
Unlike the strong nuclear force, which has a range of about $10^{-15} \,m$, the weak nuclear force has an extremely short range.
The range of the weak nuclear force is approximately $10^{-16} \,m$ to $10^{-18} \,m$.
Therefore, the correct order of magnitude among the given options is $10^{-16} \,m$.

(B) The radius $R$ of a nucleus is related to its mass number $A$ by the formula $R = R_0 A^{1/3}$, where $R_0$ is a constant.
The surface area $S$ of a spherical nucleus is given by $S = 4 \pi R^2$.
Substituting the expression for $R$, we get $S = 4 \pi (R_0 A^{1/3})^2 = 4 \pi R_0^2 A^{2/3}$.
Thus, the surface area is proportional to $A^{2/3}$, i.e., $S \propto A^{2/3}$.
Given the ratio of mass numbers $A_1 : A_2 = 27 : 125$, the ratio of their surface areas is $S_1 : S_2 = (A_1 / A_2)^{2/3}$.
Substituting the values: $S_1 / S_2 = (27 / 125)^{2/3} = ((3^3) / (5^3))^{2/3} = (3/5)^2 = 9/25$.
Therefore, the ratio of their surface areas is $9 : 25$.

(A) The average mass of an element is calculated by taking the weighted average of the masses of its isotopes based on their relative abundances.
Given the ratio of abundances is $2:3:5$,the total parts are $2 + 3 + 5 = 10$.
The fractional abundances are:
For isotope $A$: $2/10 = 0.2$
For isotope $B$: $3/10 = 0.3$
For isotope $C$: $5/10 = 0.5$
The average mass is given by the formula: $\text{Average Mass} = (\text{fraction}_A \times m_1) + (\text{fraction}_B \times m_2) + (\text{fraction}_C \times m_3)$.
Substituting the values: $\text{Average Mass} = 0.2 m_1 + 0.3 m_2 + 0.5 m_3$.

(D) The radius $R$ of a nucleus with mass number $A$ is given by $R = R_0 A^{1/3}$,where $R_0$ is a constant.
Assuming the nucleus is spherical,its surface area $S$ is given by $S = 4 \pi R^2$.
Substituting the expression for $R$ into the formula for $S$:
$S = 4 \pi (R_0 A^{1/3})^2$
$S = 4 \pi R_0^2 A^{2/3}$
Since $4 \pi R_0^2$ is a constant,we have $S \propto A^{2/3}$.

(C) The radius of a nucleus is given by the relation $R = R_0 A^{1/3}$,where $A$ is the mass number.
Given $R_1 = R$ for $A_1 = 27$ and $R_2 = 2R$ for $A_2$.
Taking the ratio: $\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3}$.
Substituting the values: $\frac{R}{2R} = \left(\frac{27}{A_2}\right)^{1/3}$.
$\frac{1}{2} = \left(\frac{27}{A_2}\right)^{1/3}$.
Cubing both sides: $\frac{1}{8} = \frac{27}{A_2}$.
$A_2 = 27 \times 8 = 216$.
Nuclei with very high mass numbers (like $A=216$) are generally unstable and tend to undergo radioactive decay or fission to reach a more stable state.

(C) The radius of the atomic nucleus in terms of mass number $A$ is given by $R = R_0 A^{1/3}$.
As the mass number $A$ increases,the mass of the nucleus $(M \approx A \cdot m_p)$ and the volume of the nucleus $(V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi R_0^3 A)$ both increase.
Binding energy also changes as the number of nucleons increases.
However,the nuclear density $\rho$ is given by:
$\rho = \frac{\text{mass of nucleus}}{\text{volume of nucleus}} = \frac{m A}{\frac{4}{3} \pi R^3} = \frac{m A}{\frac{4}{3} \pi (R_0 A^{1/3})^3} = \frac{m A}{\frac{4}{3} \pi R_0^3 A} = \frac{3m}{4 \pi R_0^3}$.
Since $m$ (average nucleon mass) and $R_0$ are constants,the nuclear density is independent of the mass number $A$ and remains constant.

(D) The density of a nucleus is given by the formula: $\rho = \frac{\text{Mass}}{\text{Volume}}$.
Mass of the nucleus $M = A \times m_p$,where $A$ is the mass number and $m_p \approx 1.6 \times 10^{-27} \text{ kg}$.
Volume of the nucleus $V = \frac{4}{3} \pi R^3$,where $R = R_0 A^{1/3}$.
Substituting $V$: $V = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.
Density $\rho = \frac{A \times m_p}{\frac{4}{3} \pi R_0^3 A} = \frac{3 m_p}{4 \pi R_0^3}$.
Substituting the values: $\rho = \frac{3 \times 1.6 \times 10^{-27}}{4 \times 3.14 \times (1.2 \times 10^{-15})^3}$.
$\rho = \frac{4.8 \times 10^{-27}}{12.56 \times 1.728 \times 10^{-45}} \approx 2.2 \times 10^{17} \text{ kg m}^{-3}$.

(B) The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$,where $R_0$ is a constant and $A$ is the mass number.
Let $R_1$ be the radius of the nucleus with mass number $A_1 = 189$,so $R_1 = R_0 (189)^{1/3}$.
Let $R_2$ be the radius of the target nucleus with mass number $A_2$,so $R_2 = R_0 (A_2)^{1/3}$.
According to the problem,$R_2 = \frac{1}{3} R_1$.
Substituting the expressions,we get $R_0 (A_2)^{1/3} = \frac{1}{3} R_0 (189)^{1/3}$.
Canceling $R_0$ from both sides,we have $(A_2)^{1/3} = \frac{1}{3} (189)^{1/3}$.
Raising both sides to the power of $3$,we get $A_2 = (\frac{1}{3})^3 \times 189$.
$A_2 = \frac{1}{27} \times 189 = 7$.
Therefore,the mass number is $7$.

(C) The nuclear force is a short-range force that acts between nucleons (protons and neutrons) within the nucleus.
It is effective only when the distance between two nucleons is approximately $1$ fermi, which is equal to $10^{-15} \,m$.
Beyond this distance, the nuclear force decreases rapidly and becomes negligible.
Therefore, the range of the nuclear force is $10^{-15} \,m$.

(A) The radius of a nucleus is given by $R = R_0 A^{1/3}$.
The volume of a nucleus is $V = \frac{4}{3} \pi R^3$.
Substituting $R$,we get $V = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.
Given $R_0 \simeq 1.0 \times 10^{-15} \ m$,$\pi \simeq 3$,and $A = 27$:
$V = \frac{4}{3} \times 3 \times (1.0 \times 10^{-15} \ m)^3 \times 27$.
$V = 4 \times 10^{-45} \times 27 \ m^3 = 108 \times 10^{-45} \ m^3$.
Since $1 \ \text{Å} = 10^{-10} \ m$,then $1 \ m = 10^{10} \ \text{Å}$,so $1 \ m^3 = 10^{30} \ (\text{Å})^3$.
$V = 108 \times 10^{-45} \times 10^{30} \ (\text{Å})^3 = 108 \times 10^{-15} \ (\text{Å})^3 \approx 1 \times 10^{-13} \ (\text{Å})^3$.

(A) The radius $R$ of a nucleus with mass number $A$ is given by the formula $R = R_0 A^{1/3}$,where $R_0$ is a constant.
Given the mass numbers for iodine $(A_i = 125)$ and polonium $(A_p = 216)$.
The ratio of the radii is given by $\frac{R_i}{R_p} = \frac{R_0 A_i^{1/3}}{R_0 A_p^{1/3}} = \left( \frac{A_i}{A_p} \right)^{1/3}$.
Substituting the given values: $\frac{R_i}{R_p} = \left( \frac{125}{216} \right)^{1/3}$.
Since $125 = 5^3$ and $216 = 6^3$,we have $\frac{R_i}{R_p} = \frac{5}{6}$.
Therefore,the ratio is $5:6$.

(A) deuteron is the nucleus of deuterium,which is an isotope of hydrogen denoted as ${ }_1^2 H$.
It consists of exactly one proton and one neutron bound together by the strong nuclear force.
Therefore,the most accurate description among the given options is that it is composed of a proton and a neutron.

(D) The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$, where $R_0$ is a constant and $A$ is the mass number.
Given:
$R_1 = 1.5 \text{ fermi}$, $A_1 = 125$
$A_2 = 64$
We need to find $R_2$.
Taking the ratio of the radii:
$\frac{R_1}{R_2} = \left( \frac{A_1}{A_2} \right)^{1/3}$
$\frac{1.5}{R_2} = \left( \frac{125}{64} \right)^{1/3}$
$\frac{1.5}{R_2} = \frac{5}{4}$
$R_2 = \frac{1.5 \times 4}{5} = 0.3 \times 4 = 1.2 \text{ fermi}$.

(B) The relation between the radius $(R)$ and the mass number $(A)$ of a nucleus is given by $R = R_0 A^{1/3}$.
Therefore,the ratio of the radii of two nuclei is given by $\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3}$.
Given: $R_1 = 6 \text{ fermi}$,$A_1 = 125$,and $A_2 = 27$.
Substituting the values:
$\frac{6}{R_2} = \left(\frac{125}{27}\right)^{1/3} = \frac{5}{3}$.
Solving for $R_2$:
$R_2 = \frac{6 \times 3}{5} = \frac{18}{5} = 3.6 \text{ fermi}$.
Since $1 \text{ fermi} = 10^{-15} \text{ m}$,we have $R_2 = 3.6 \times 10^{-15} \text{ m}$.

(C) The nuclear density is independent of the mass number and is approximately constant for all nuclei,given by $\rho \approx 2.3 \times 10^{17} \ kg/m^3$. Thus,statement $A$ is true.
The relationship between the radius of the nucleus $R$ and the mass number $A$ is given by $R = R_0 A^{1/3}$,where $R_0$ is a constant. This implies $R^3 \propto A$,or $R \propto A^{1/3}$.
Given statement $B$ is $\sqrt{A} \propto R^{1/6}$. Squaring both sides gives $A \propto R^{1/3}$,which contradicts the established relation $R \propto A^{1/3}$ (or $A \propto R^3$). Therefore,statement $B$ is false.
Hence,$A$ is true and $B$ is false.

(A) According to the law of conservation of linear momentum,for a nucleus initially at rest,the magnitudes of momenta of the two fragments must be equal: $m_1 v_1 = m_2 v_2$.
This implies $\frac{m_1}{m_2} = \frac{v_2}{v_1}$.
Given the velocity ratio $\frac{v_1}{v_2} = \frac{3}{1}$,we have $\frac{m_1}{m_2} = \frac{1}{3}$.
Assuming the density $\rho$ of the nucleus is constant,the mass $m$ is proportional to the volume,so $m = \rho \cdot \frac{4}{3} \pi R^3$.
Therefore,$\frac{m_1}{m_2} = \frac{R_1^3}{R_2^3}$.
Equating the two expressions for the mass ratio: $\frac{R_1^3}{R_2^3} = \frac{v_2}{v_1} = \frac{1}{3}$.
Taking the cube root on both sides,we get $\frac{R_1}{R_2} = (\frac{1}{3})^{1/3} = 1 : 3^{1/3}$.

(C) The radius $R$ of a nucleus is related to its mass number $A$ by the formula $R = R_0 A^{1/3}$,where $R_0$ is a constant.
The volume $V$ of a nucleus is given by $V = \frac{4}{3} \pi R^3$.
Substituting the expression for $R$,we get $V = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.
This shows that the volume $V$ is directly proportional to the mass number $A$ $(V \propto A)$.
If the mass number is increased from $A$ to $2A$,the new volume $V'$ will be $V' \propto 2A$.
Therefore,$V' = 2V$.

(A) The relative strength of the strong nuclear force is approximately $1$.
The relative strength of the weak nuclear force is approximately $10^{-13}$.
Therefore,the ratio of the relative strengths of strong and weak nuclear forces is given by:
Ratio $= \frac{\text{Strong nuclear force}}{\text{Weak nuclear force}} = \frac{1}{10^{-13}} = 10^{13}$.

(C) Nuclear forces are the strongest forces in nature, acting only over very small distances within the nucleus (typically $r \approx 10^{-15} \,m$).
These forces are responsible for binding protons and neutrons together, overcoming the electrostatic repulsion between protons.
Therefore, nuclear forces are short-range attractive forces.

(A, C) The radius of a nucleus is given by the formula $r = R_0 A^{1/3}$,where $R_0$ is a constant and $A$ is the mass number.
Since $A_A = 64$ and $A_B = 125$,we have $r_A = R_0 (64)^{1/3} = 4R_0$ and $r_B = R_0 (125)^{1/3} = 5R_0$.
Thus,$r_A < r_B$.
Regarding binding energy per nucleon $(E_{bn})$,the value of $E_{bn}$ increases with mass number for light nuclei and reaches a maximum near $A = 56$ (Iron). For nuclei with $A > 60$,$E_{bn}$ gradually decreases.
Since $A_A = 64$ is closer to $56$ than $A_B = 125$,the binding energy per nucleon for nucleus $A$ is greater than that for nucleus $B$,i.e.,$E_{bnA} > E_{bnB}$.
Therefore,both options $A$ and $C$ are correct.

(D) In a nuclear reaction,both the total atomic number $(Z)$ and the total mass number $(A)$ must be conserved on both sides of the equation.
Let $X$ be represented as ${ }_{Z'}^{A'} X$.
The reaction is: ${ }_{7}^{14} N + { }_{Z'}^{A'} X \rightarrow { }_{6}^{14} C + { }_{1}^{1} H$.
Conservation of mass number $(A)$: $14 + A' = 14 + 1 \Rightarrow A' = 1$.
Conservation of atomic number $(Z)$: $7 + Z' = 6 + 1 \Rightarrow 7 + Z' = 7 \Rightarrow Z' = 0$.
$A$ particle with mass number $1$ and atomic number $0$ is a neutron,denoted as ${ }_{0}^{1} n$.

(D) In a nuclear reaction,both the total atomic number $(Z)$ and the total mass number $(A)$ must be conserved on both sides of the equation.
Given reaction: ${ }_4^9 Be + { }_2^4 He \rightarrow { }_6^{12} C + x$
Let the particle $x$ be represented by ${ }_Z^A X$.
Conservation of mass number $(A)$: $9 + 4 = 12 + A \Rightarrow 13 = 12 + A \Rightarrow A = 1$.
Conservation of atomic number $(Z)$: $4 + 2 = 6 + Z \Rightarrow 6 = 6 + Z \Rightarrow Z = 0$.
The particle with mass number $1$ and atomic number $0$ is a neutron,denoted as ${ }_0^1 n$.

(D) The radius of a nucleus is given by the formula $R = R_{0}A^{1/3}$,where $R_{0}$ is a constant and $A$ is the mass number.
For the first nucleus,$R_{\alpha} = R_{0}\alpha^{1/3}$.
For the second nucleus,$R_{\beta} = R_{0}\beta^{1/3}$.
Taking the ratio,we get $\frac{R_{\alpha}}{R_{\beta}} = \left(\frac{\alpha}{\beta}\right)^{1/3}$.
Given that $\beta = 8\alpha$,we substitute this into the ratio:
$\frac{R_{\alpha}}{R_{\beta}} = \left(\frac{\alpha}{8\alpha}\right)^{1/3} = \left(\frac{1}{8}\right)^{1/3} = \frac{1}{2} = 0.5$.

(C) The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $A$ is the mass number.
The surface area $S$ of a spherical nucleus is $S = 4 \pi R^2 = 4 \pi R_0^2 A^{2/3}$.
Thus,$S \propto A^{2/3}$.
Initial mass number $A_i = 8$.
Initial surface area $S_i \propto (8)^{2/3} = (2^3)^{2/3} = 2^2 = 4$.
After absorbing $10$ protons and $9$ neutrons (electrons do not contribute to the mass number),the new mass number $A_f = 8 + 10 + 9 = 27$.
Final surface area $S_f \propto (27)^{2/3} = (3^3)^{2/3} = 3^2 = 9$.
The percentage increase in surface area is given by $\frac{S_f - S_i}{S_i} \times 100$.
Percentage increase $= \frac{9 - 4}{4} \times 100 = \frac{5}{4} \times 100 = 125\%$.

(D) Isobars are defined as nuclei that have the same mass number $(A)$ but different atomic numbers $(Z)$.
In the pair ${}_{1}^{3}H$ and ${}_{2}^{3}He$:
$1$. For ${}_{1}^{3}H$,the mass number $A = 3$.
$2$. For ${}_{2}^{3}He$,the mass number $A = 3$.
Since both nuclei have the same mass number $(A = 3)$,they are isobars.

(D) The potential energy of the barrier at the point of contact is given by $U = k \frac{q_1 q_2}{r}$.
Here,the charge of each deuteron is $q_1 = q_2 = e = 1.6 \times 10^{-19}$ $C$.
The distance between the centers of the two deuterons at the point of contact is $r = R_1 + R_2 = 2 \text{ fm} + 2 \text{ fm} = 4 \text{ fm} = 4 \times 10^{-15}$ m.
Substituting the values into the formula:
$U = \frac{(9 \times 10^9 \text{ N m}^2/\text{C}^2) \times (1.6 \times 10^{-19} \text{ C})^2}{4 \times 10^{-15} \text{ m}}$
$U = \frac{9 \times 10^9 \times 2.56 \times 10^{-38}}{4 \times 10^{-15}}$
$U = \frac{23.04 \times 10^{-29}}{4 \times 10^{-15}}$
$U = 5.76 \times 10^{-14}$ $J$.

(A) Isotones are atoms of different elements that have the same number of neutrons.
To find the number of neutrons $(N)$,we use the formula $N = A - Z$,where $A$ is the mass number and $Z$ is the atomic number.
For $^{198}_{80}\text{Hg}$: $N = 198 - 80 = 118$.
For $^{197}_{79}\text{Au}$: $N = 197 - 79 = 118$.
Since both atoms have the same number of neutrons $(118)$,they are isotones.
For other options:
- $^3_1\text{H}$ $(N=2)$ and $^3_2\text{He}$ $(N=1)$ are not isotones.
- $^{214}_{82}\text{Pb}$ $(N=132)$ and $^{214}_{83}\text{Bi}$ $(N=131)$ are not isotones.
- $^{12}_6\text{C}$ $(N=6)$ and $^{14}_6\text{C}$ $(N=8)$ are isotopes,not isotones.

(B) To determine the relationship,we calculate the number of neutrons $(N)$ for each nuclide using the formula $N = A - Z$,where $A$ is the mass number and $Z$ is the atomic number.
For $^{197}_{79} \text{Au}$: $A = 197$,$Z = 79$. Thus,$N = 197 - 79 = 118$.
For $^{198}_{80} \text{Hg}$: $A = 198$,$Z = 80$. Thus,$N = 198 - 80 = 118$.
Since both nuclides have the same number of neutrons $(N = 118)$,they are classified as isotones.

(B) The nuclear radius $R$ is calculated using the formula $R = R_0 A^{1/3}$,where $R_0 \approx 1.2 \text{ fm}$ and $A$ is the mass number of the nucleus.
For a hydrogen atom,the nucleus consists of a single proton,so the mass number $A = 1$.
Substituting the value of $A$ into the formula:
$R = 1.2 \times (1)^{1/3} \text{ fm}$
$R = 1.2 \times 1 \text{ fm} = 1.2 \text{ fm}$.
Therefore,the nuclear radius of a hydrogen atom is $1.2 \text{ fm}$.
The correct option is $B$.

(B) The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $R_0$ is a constant and $A$ is the mass number.
The volume of a nucleus is $V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.
Since $\frac{4}{3} \pi R_0^3$ is a constant,the volume $V$ is directly proportional to the mass number $A$. Therefore,statement $A$ is false and statement $B$ is true.
Mass defect is defined as the difference between the sum of the masses of the individual nucleons (protons and neutrons) and the actual mass of the nucleus. Therefore,statement $C$ is false and statement $D$ is true.
Thus,statements $B$ and $D$ are true,while $A$ and $C$ are false.

(C) The mass of a nucleus is given by $M = A \times m_p$,where $A$ is the mass number and $m_p$ is the average mass of a nucleon (approximately $1.67 \times 10^{-27} \text{ kg}$).
Given the total mass of the nucleus $M = 19.926 \times 10^{-27} \text{ kg}$.
We can calculate the mass number $A$ as:
$A = \frac{M}{m_p} = \frac{19.926 \times 10^{-27} \text{ kg}}{1.67 \times 10^{-27} \text{ kg}} \approx 11.93$.
Rounding this value to the nearest integer,we get $A \approx 12$.
Thus,the mass number of the nucleus is $12$.