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

(A) To find the missing particles,we apply the law of conservation of mass number and atomic number.
$(i)$ The reaction is $(n, p)$ on a target to produce ${}_{15}P^{32}$. The reaction is ${}_{16}S^{32} + {}_{0}n^{1} \to {}_{15}P^{32} + {}_{1}H^{1}$. Thus,the target is ${}_{16}S^{32}$.
$(ii)$ The reaction is $(p, \alpha)$ on a target to produce ${}_{8}O^{16}$. The reaction is ${}_{9}F^{19} + {}_{1}H^{1} \to {}_{2}He^{4} + {}_{8}O^{16}$. Thus,the target is ${}_{9}F^{19}$.
$(iii)$ The reaction is ${}_{7}N^{14} + ? \to {}_{6}C^{14} + {}_{1}H^{1}$. Balancing mass numbers: $14 + A = 14 + 1 \implies A = 1$. Balancing atomic numbers: $7 + Z = 6 + 1 \implies Z = 0$. The missing particle is ${}_{0}n^{1}$ (neutron).
Therefore,the missing components are ${}_{16}S^{32}, {}_{9}F^{19}, {}_{0}n^{1}$.

(A) The decay reaction is: $n \rightarrow p + e^- + \bar{\nu} + Q$.
The mass defect $\Delta m$ is given by: $\Delta m = m_n - (m_p + m_e)$.
Substituting the given values:
$\Delta m = 1.6725 \times 10^{-27} \ kg - (1.6725 \times 10^{-27} \ kg + 9 \times 10^{-31} \ kg)$.
$\Delta m = -9 \times 10^{-31} \ kg$.
The magnitude of mass defect is $9 \times 10^{-31} \ kg$.
The energy released $E$ is given by $E = \Delta m c^2$,where $c = 3 \times 10^8 \ m/s$.
$E = (9 \times 10^{-31} \ kg) \times (3 \times 10^8 \ m/s)^2 = 81 \times 10^{-15} \ J$.
To convert this into $MeV$,divide by $1.6 \times 10^{-13} \ J/MeV$:
$E = \frac{81 \times 10^{-15}}{1.6 \times 10^{-13}} \approx 0.51 \ MeV$.

(D) To determine the relationship between the nuclei ${ }_{6} C^{13}$ and ${ }_{7} N^{14}$,we calculate the number of neutrons $(N)$ in each.
For ${ }_{6} C^{13}$,the atomic number $Z = 6$ and the mass number $A = 13$. The number of neutrons is $N = A - Z = 13 - 6 = 7$.
For ${ }_{7} N^{14}$,the atomic number $Z = 7$ and the mass number $A = 14$. The number of neutrons is $N = A - Z = 14 - 7 = 7$.
Since both nuclei have the same number of neutrons $(N = 7)$,they are called isotones.

(A) To find the missing particle or nuclide,we apply the conservation of mass number $(A)$ and atomic number $(Z)$.
$(i) \, {_{Z}}X^{A} + {_{0}}n^{1} \rightarrow {_{15}}P^{32} + {_{1}}H^{1}$
Equating $Z$: $Z + 0 = 15 + 1 \Rightarrow Z = 16$.
Equating $A$: $A + 1 = 32 + 1 \Rightarrow A = 32$.
Thus,the nuclide is ${_{16}}S^{32}$.
$(ii) \, {_{Z}}X^{A} + {_{1}}H^{1} \rightarrow {_{8}}O^{16} + {_{2}}He^{4}$
Equating $Z$: $Z + 1 = 8 + 2 \Rightarrow Z = 9$.
Equating $A$: $A + 1 = 16 + 4 \Rightarrow A = 19$.
Thus,the nuclide is ${_{9}}F^{19}$.
$(iii) \, {_{7}}N^{14} + {_{Z}}X^{A} \rightarrow {_{6}}C^{14} + {_{1}}H^{1}$
Equating $Z$: $7 + Z = 6 + 1 \Rightarrow Z = 0$.
Equating $A$: $14 + A = 14 + 1 \Rightarrow A = 1$.
Thus,the particle is ${_{0}}n^{1}$ (neutron).
Therefore,the missing values are ${_{16}}S^{32}, {_{9}}F^{19}, {_{0}}n^{1}$.

(A) The number of electrons in a neutral atom is equal to the number of protons.
Summing the electrons from the configuration: $2 + 2 + 6 + 2 + 6 = 18$.
Thus,the number of protons $(Z)$ is $18$.
The mass number $(A)$ is given as $40$.
The number of neutrons $(N)$ is calculated as $N = A - Z$.
$N = 40 - 18 = 22$.
Therefore,the number of neutrons is $22$ and the number of protons is $18$.

(D) Nuclear density is defined as the ratio of the mass of the nucleus to its volume.
The volume $V$ of a nucleus with mass number $A$ is given by $V = \frac{4}{3} \pi R^3$,where $R = R_0 A^{1/3}$.
Thus,$V = \frac{4}{3} \pi R_0^3 A$.
The density $\rho$ is given by $\rho = \frac{M}{V} = \frac{A \cdot m_p}{\frac{4}{3} \pi R_0^3 A} = \frac{3 m_p}{4 \pi R_0^3}$,where $m_p$ is the mass of a nucleon.
Since $m_p$,$\pi$,and $R_0$ are constants,the nuclear density $\rho$ is independent of the mass number $A$.
Therefore,the density of the daughter nuclei remains the same as the parent nucleus,which is $\rho$.

(A) To check the correctness of a nuclear reaction,we must ensure the conservation of atomic number $(Z)$ and mass number $(A)$.
For option $A$: ${}_{96}Cm^{242} + {}_{2}He^{4} \to {}_{97}Bk^{243} + 2({}_{0}n^{1})$.
Checking atomic number: $96 + 2 = 98$ on the left,while $97 + 0 = 97$ on the right. Since $98 \neq 97$,the reaction is incorrectly balanced.
For option $B$: ${}_{5}B^{10} + {}_{2}He^{4} \to {}_{7}N^{13} + {}_{0}n^{1}$.
$Z: 5+2 = 7$ (Correct); $A: 10+4 = 14$ and $13+1 = 14$ (Correct).
For option $C$: ${}_{7}N^{14} + {}_{0}n^{1} \to {}_{6}C^{14} + {}_{1}H^{1}$.
$Z: 7+0 = 7$ and $6+1 = 7$ (Correct); $A: 14+1 = 15$ and $14+1 = 15$ (Correct).
For option $D$: ${}_{14}Si^{28} + {}_{1}H^{2} \to {}_{15}P^{29} + {}_{0}n^{1}$.
$Z: 14+1 = 15$ (Correct); $A: 28+2 = 30$ and $29+1 = 30$ (Correct).
Thus,option $A$ is the incorrectly matched reaction.

(C) The number of protons in the nucleus of an atom is equal to the atomic number $(Z)$.
The mass number $(A)$ of an atom is defined as the sum of the number of protons and the number of neutrons in the nucleus.
Therefore, the number of neutrons is given by the difference between the mass number $(A)$ and the atomic number $(Z)$.
$A = \text{Protons} + \text{Neutrons}$
$Z = \text{Protons}$
Substituting $Z$ for protons:
$A = Z + \text{Neutrons}$
$\text{Neutrons} = A - Z$
Thus, the nucleus has $Z$ protons and $A - Z$ neutrons.

(B) The radius of a nucleus is given by the relation $r = R_0 A^{1/3}$,where $A$ is the mass number.
Thus,the ratio of radii is $\frac{r_1}{r_2} = \left(\frac{A_1}{A_2}\right)^{1/3}$.
Given $\frac{r_1}{r_2} = \frac{3}{5}$ and $A_1 = 27$,we have $\frac{3}{5} = \left(\frac{27}{A_2}\right)^{1/3}$.
Cubing both sides,we get $\frac{27}{125} = \frac{27}{A_2}$,which implies $A_2 = 125$.
The number of neutrons $N$ is given by $N = A - Z$,where $Z$ is the atomic number.
For nucleus $X$,$A = 125$ and $Z = 52$.
Therefore,$N = 125 - 52 = 73$.

(C) Let the heavy nucleus break into two nuclei of masses $m_1$ and $m_2$ moving with velocities $V_1$ and $V_2$ respectively.
According to the law of conservation of linear momentum,the initial momentum is zero,so $m_1 V_1 = m_2 V_2$.
Given the ratio of velocities $\frac{V_1}{V_2} = \frac{8}{27}$,we have $\frac{m_1}{m_2} = \frac{V_2}{V_1} = \frac{27}{8}$.
Assuming constant nuclear density $\rho$,the mass of a nucleus is $m = \rho \times \frac{4}{3} \pi R^3$.
Therefore,$\frac{m_1}{m_2} = \frac{R_1^3}{R_2^3} = \frac{27}{8}$.
Taking the cube root on both sides,$\frac{R_1}{R_2} = \left( \frac{27}{8} \right)^{1/3} = \frac{3}{2}$.
Thus,the ratio of the radii is $3:2$.

(D) The nuclear mass density is given by the formula $\rho = \frac{M}{V} = \frac{A \cdot m_n}{\frac{4}{3} \pi R^3}$.
Since the nuclear radius $R$ is proportional to $A^{1/3}$ (i.e.,$R = R_0 A^{1/3}$),the volume $V$ is proportional to $A$.
Therefore,$\rho = \frac{A \cdot m_n}{\frac{4}{3} \pi R_0^3 A} = \frac{m_n}{\frac{4}{3} \pi R_0^3}$.
This shows that the nuclear mass density is independent of the mass number $A$.
Thus,the mass densities of all nuclei are approximately the same.
Consequently,the ratio of the mass densities of $^{40}Ca$ and $^{16}O$ is $1$.

(A) Nuclear forces are the strong forces that hold nucleons (protons and neutrons) together within the nucleus.
$1$. They are short-ranged,acting only over distances of the order of $10^{-15} \ m$ (femtometers).
$2$. They are strongly attractive,which overcomes the electrostatic repulsion between protons.
$3$. They are charge-independent,meaning the force between a proton-proton,neutron-neutron,or proton-neutron pair is essentially the same,provided the separation distance and spin states are identical.

(A) The notation ${ }_{Z} X^{A}$ represents an atom where $Z$ is the atomic number and $A$ is the mass number.
For ${ }_{11} Na^{23}$,the atomic number $Z = 11$ and the mass number $A = 23$.
The number of protons in the nucleus is equal to the atomic number $Z$,so protons = $11$.
The number of neutrons is given by $N = A - Z = 23 - 11 = 12$.
The nucleus of an atom contains only protons and neutrons; it does not contain electrons.
Therefore,the number of electrons in the nucleus is $0$.
Thus,the number of protons,neutrons,and electrons in the nucleus are $11, 12, 0$ respectively.

(C) The specific charge is defined as the ratio of charge to mass $(q/m)$.
For a proton $(p)$,the charge is $e$ and the mass is $m$. So,the specific charge is $(q/m)_p = e/m$.
For an $\alpha$-particle,the charge is $2e$ and the mass is $4m$ (since it consists of $2$ protons and $2$ neutrons).
So,the specific charge is $(q/m)_{\alpha} = 2e/4m = e/2m$.
The ratio of the specific charge of an $\alpha$-particle to that of a proton is:
$\frac{(q/m)_{\alpha}}{(q/m)_p} = \frac{e/2m}{e/m} = \frac{1}{2} = 1:2$.

(D) The radius of an atomic 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 $A_1 = 64$ and $A_2 = 125$.
The ratio of the radii is $\frac{r_1}{r_2} = \left(\frac{A_1}{A_2}\right)^{1/3}$.
Substituting the values,we get $\frac{r_1}{r_2} = \left(\frac{64}{125}\right)^{1/3}$.
Since $64 = 4^3$ and $125 = 5^3$,we have $\frac{r_1}{r_2} = \frac{4}{5}$.

(C) The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $R_0$ is a constant.
The volume of the nucleus is $V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.
The mass of the nucleus is approximately $M = A \times m_p$,where $m_p$ is the mass of a nucleon.
The mass density $\rho$ is defined as $\rho = \frac{M}{V} = \frac{A \times m_p}{\frac{4}{3} \pi R_0^3 A} = \frac{m_p}{\frac{4}{3} \pi R_0^3}$.
Since $m_p$,$R_0$,and $\pi$ are constants,the mass density $\rho$ is independent of the mass number $A$ and remains constant for all nuclei.

(A) According to the law of conservation of linear momentum,since the initial nucleus is at rest,the magnitudes of the momenta of the two fragments must be equal:
$m_1 v_1 = m_2 v_2$
Given the ratio of velocities is $v_1 / v_2 = 8 / 1$,we have:
$m_1 / m_2 = v_2 / v_1 = 1 / 8$
Since the mass of a nucleus is proportional to its mass number $A$ (i.e.,$m \propto A$),the ratio of mass numbers is $A_1 / A_2 = m_1 / m_2 = 1 / 8$.
The radius of a nucleus is given by $r = R_0 A^{1/3}$,so the ratio of the radii is:
$r_1 / r_2 = (A_1 / A_2)^{1/3}$
$r_1 / r_2 = (1 / 8)^{1/3} = 1 / 2$
Thus,the ratio of the radii is $1 : 2$.

(B) By definition,isobars are atoms of different elements that have the same mass number $(A)$ but different atomic numbers $(Z)$.
Therefore,the Assertion is correct.
Protons and neutrons are indeed present inside the nucleus,which is a correct statement regarding nuclear composition.
However,the fact that protons and neutrons exist in the nucleus does not explain why certain atoms are isobars.
Thus,the Reason is a correct statement but is not the correct explanation for the Assertion.

(D) Isotopes of an element have the same number of electrons and protons, but they differ in the number of neutrons, which leads to a difference in their atomic masses.
Because of this difference in atomic mass, isotopes can be separated using a mass spectrometer, which separates ions based on their mass-to-charge ratio $(m/q)$.
Since the Assertion claims that separation is due to the difference in electron numbers, the Assertion is incorrect.
However, the Reason stating that isotopes can be separated by a mass spectrometer is correct.
Therefore, the correct option is $D$.

(B) Neutrons penetrate matter more readily than protons because neutrons are electrically neutral,meaning they experience no electrostatic repulsion from the positively charged nucleus.
Protons,being positively charged,experience a strong electrostatic repulsive force from the nucleus,which hinders their penetration.
While it is true that the mass of a neutron $(m_n \approx 1.6749 \times 10^{-27} \ kg)$ is slightly greater than the mass of a proton $(m_p \approx 1.6726 \times 10^{-27} \ kg)$,this mass difference is not the reason for their higher penetrating power.
Therefore,both the Assertion and the Reason are correct,but the Reason is not the correct explanation for the Assertion.

(A) An $\alpha$-particle is a helium nucleus,denoted as $\alpha = {}_2^4 \text{He}^{2+}$.
It consists of $2$ protons and $2$ neutrons,having a total mass number of $4$ and an atomic number of $2$.

(A) The size of a nucleus is in the range of $10^{-15} \; m$ to $10^{-14} \; m$.
The tip of a sharp pin is in the range of $10^{-5} \; m$ to $10^{-4} \; m$.
To scale the nucleus to the size of a pin tip,we use a scaling factor of $10^{10}$ (since $10^{-15} \times 10^{10} = 10^{-5}$).
The size of an atom is approximately $10^{-10} \; m$.
Scaling the atom by the same factor of $10^{10}$,we get: $10^{-10} \; m \times 10^{10} = 1 \; m$.
Therefore,if the nucleus is scaled to the size of a pin tip,the atom would be roughly $1 \; m$ in size.

(N/A) The radius of a nucleus $r$ is given by the relation: $r = r_{0} A^{1/3}$ $(i)$,where $r_{0} = 1.2 \; f = 1.2 \times 10^{-15} \; m$.
The volume of the nucleus $V$ is given by: $V = \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi (r_{0} A^{1/3})^{3} = \frac{4}{3} \pi r_{0}^{3} A$.
The mass of a nucleus $M$ is approximately equal to its mass number $A$ in atomic mass units (amu): $M = A \times 1.66 \times 10^{-27} \; kg$.
The density of the nucleus $\rho$ is given by: $\rho = \frac{\text{Mass of nucleus}}{\text{Volume of nucleus}} = \frac{A \times 1.66 \times 10^{-27}}{\frac{4}{3} \pi r_{0}^{3} A} = \frac{3 \times 1.66 \times 10^{-27}}{4 \pi r_{0}^{3}} \; kg/m^{3}$.
This relation shows that the nuclear mass density depends only on the constant $r_{0}$ and is independent of $A$. Hence,the nuclear mass densities of all nuclei are nearly the same.
For a sodium nucleus,the density is: $\rho_{\text{sodium}} = \frac{3 \times 1.66 \times 10^{-27}}{4 \times 3.14 \times (1.2 \times 10^{-15})^{3}} \approx 2.29 \times 10^{17} \; kg/m^{3}$.

(N/A) Let $m, m_{1},$ and $m_{2}$ be the respective masses of the parent nucleus and the two daughter nuclei. The parent nucleus is at rest.
Initial momentum of the system (parent nucleus) $= 0$.
Let $v_{1}$ and $v_{2}$ be the respective velocities of the daughter nuclei having masses $m_{1}$ and $m_{2}$.
Total linear momentum of the system after disintegration $= m_{1}v_{1} + m_{2}v_{2}$.
According to the law of conservation of momentum:
Total initial momentum $=$ Total final momentum
$0 = m_{1}v_{1} + m_{2}v_{2}$
$v_{1} = -\frac{m_{2}v_{2}}{m_{1}}$
Here,the negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.

(N/A) proton has three quarks. Let there be $n$ up quarks in a proton,each having a charge of $+(2/3)e$.
Charge due to $n$ up quarks $= (2/3)e \cdot n$.
Number of down quarks in a proton $= 3-n$.
Each down quark has a charge of $-(1/3)e$.
Charge due to $(3-n)$ down quarks $= -(1/3)e \cdot (3-n)$.
Total charge on a proton $= +e$.
Therefore,$e = (2/3)en - (1/3)e(3-n)$.
$e = (2/3)en - e + (1/3)en$.
$2e = (3/3)en = en$.
Thus,$n = 2$.
Number of up quarks $= 2$,number of down quarks $= 3-2 = 1$. So,a proton is $uud$.
$A$ neutron has three quarks. Let there be $n$ up quarks in a neutron.
Charge due to $n$ up quarks $= (2/3)e \cdot n$.
Number of down quarks $= 3-n$,each with charge $-(1/3)e$.
Charge due to $(3-n)$ down quarks $= -(1/3)e \cdot (3-n)$.
Total charge on a neutron $= 0$.
Therefore,$0 = (2/3)en - (1/3)e(3-n)$.
$0 = (2/3)en - e + (1/3)en$.
$e = en$.
Thus,$n = 1$.
Number of up quarks $= 1$,number of down quarks $= 3-1 = 2$. So,a neutron is $udd$.

(B) In the Rutherford alpha-particle scattering experiment,the distance of closest approach is used to estimate the size of the nucleus.
For an alpha particle with kinetic energy $K$ directed towards the center of the nucleus with atomic number $Z$,the distance of closest approach $r_0$ is given by $r_0 = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{K}$.
Experimental results indicate that the dimension of the nucleus is of the order of $10^{-15} \ m$ (or $1 \ fm$).

(A) The mass of the iron nucleus is $M = 55.85 \, u$. Converting this to kilograms: $M = 55.85 \times 1.66 \times 10^{-27} \, kg \approx 9.27 \times 10^{-26} \, kg$.
The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $R_0 = 1.2 \times 10^{-15} \, m$.
The volume of the nucleus is $V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.
Nuclear density $\rho = \frac{M}{V} = \frac{M}{\frac{4}{3} \pi R_0^3 A}$.
Substituting the values: $\rho = \frac{55.85 \times 1.66 \times 10^{-27}}{\frac{4}{3} \pi (1.2 \times 10^{-15})^3 \times 56}$.
Since $M \approx A \times m_p$ (where $m_p$ is the mass of a nucleon),the density simplifies to $\rho = \frac{m_p}{\frac{4}{3} \pi R_0^3} \approx 2.29 \times 10^{17} \, kg \, m^{-3}$.

(N/A) Mass of $_{3}^{6} Li$ isotope,$m_{1} = 6.01512 \; u$
Mass of $_{3}^{7} Li$ isotope,$m_{2} = 7.01600 \; u$
Abundance of $_{3}^{6} Li$,$n_{1} = 7.5 \%$
Abundance of $_{3}^{7} Li$,$n_{2} = 92.5 \%$
The atomic mass of lithium is given by:
$m = \frac{m_{1} n_{1} + m_{2} n_{2}}{n_{1} + n_{2}}$
$m = \frac{6.01512 \times 7.5 + 7.01600 \times 92.5}{100} = 6.940934 \; u$
$(b)$ Mass of $_{5}^{10} B$ isotope,$m_{1} = 10.01294 \; u$
Mass of $_{5}^{11} B$ isotope,$m_{2} = 11.00931 \; u$
Let abundance of $_{5}^{10} B$ be $x \%$,then abundance of $_{5}^{11} B$ is $(100 - x) \%$.
Atomic mass of boron,$m = 10.811 \; u$
$10.811 = \frac{10.01294 \times x + 11.00931 \times (100 - x)}{100}$
$1081.1 = 10.01294 x + 1100.931 - 11.00931 x$
$0.99637 x = 19.831$
$x \approx 19.9 \%$
Abundance of $_{5}^{10} B \approx 19.9 \%$ and abundance of $_{5}^{11} B \approx 80.1 \%$.

(A) The average atomic mass is calculated using the weighted average formula:
$m_{avg} = \frac{m_1 \eta_1 + m_2 \eta_2 + m_3 \eta_3}{\eta_1 + \eta_2 + \eta_3}$
Given:
$m_1 = 19.99\;u, \eta_1 = 90.51\%$
$m_2 = 20.99\;u, \eta_2 = 0.27\%$
$m_3 = 21.99\;u, \eta_3 = 9.22\%$
Substituting the values:
$m_{avg} = \frac{(19.99 \times 90.51) + (20.99 \times 0.27) + (21.99 \times 9.22)}{90.51 + 0.27 + 9.22}$
$m_{avg} = \frac{1809.2949 + 5.6673 + 202.7478}{100}$
$m_{avg} = \frac{2017.71}{100} = 20.1771\;u$

(N/A) The nuclear 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.
For the gold isotope $^{197}_{79}Au$,the mass number $A_{Au} = 197$.
For the silver isotope $^{107}_{47}Ag$,the mass number $A_{Ag} = 107$.
The ratio of the nuclear radii is given by:
$\frac{R_{Au}}{R_{Ag}} = \left( \frac{A_{Au}}{A_{Ag}} \right)^{1/3} = \left( \frac{197}{107} \right)^{1/3}$.
Calculating the value:
$\frac{197}{107} \approx 1.841$.
$(1.841)^{1/3} \approx 1.2256$.
Rounding to two decimal places,the ratio is approximately $1.23$.

(N/A) The nuclear radius is given by the relation:
$R = R_{0} A^{1/3}$
where $R_{0}$ is a constant and $A$ is the mass number.
The nuclear matter density $\rho$ is defined as the ratio of the mass of the nucleus to its volume:
$\rho = \frac{\text{Mass of the nucleus}}{\text{Volume of the nucleus}}$
Let $m$ be the average mass of a nucleon. The total mass of the nucleus is approximately $m \times A$.
The volume of the nucleus,assuming it is spherical,is $V = \frac{4}{3} \pi R^{3}$.
Substituting the expression for $R$:
$\rho = \frac{mA}{\frac{4}{3} \pi (R_{0} A^{1/3})^{3}}$
$\rho = \frac{mA}{\frac{4}{3} \pi R_{0}^{3} A}$
Canceling $A$ from the numerator and denominator:
$\rho = \frac{3m}{4 \pi R_{0}^{3}}$
Since $m$,$\pi$,and $R_{0}$ are constants,the nuclear density $\rho$ is independent of the mass number $A$ and is nearly constant.

(A) Average atomic mass of magnesium,$m = 24.312 \; u$.
Mass of $_{12}^{24} Mg$ isotope,$m_1 = 23.98504 \; u$.
Mass of $_{12}^{25} Mg$ isotope,$m_2 = 24.98584 \; u$.
Mass of $_{12}^{26} Mg$ isotope,$m_3 = 25.98259 \; u$.
Abundance of $_{12}^{24} Mg$,$\eta_1 = 78.99 \%$.
Let the abundance of $_{12}^{25} Mg$ be $\eta_2 = x \%$.
Then,the abundance of $_{12}^{26} Mg$ is $\eta_3 = (100 - 78.99 - x) \% = (21.01 - x) \%$.
The average atomic mass is given by $m = \frac{m_1 \eta_1 + m_2 \eta_2 + m_3 \eta_3}{100}$.
Substituting the values: $24.312 = \frac{23.98504 \times 78.99 + 24.98584 \times x + 25.98259 \times (21.01 - x)}{100}$.
$2431.2 = 1894.5783 + 24.98584x + 545.8942 - 25.98259x$.
$2431.2 = 2440.4725 - 0.99675x$.
$0.99675x = 9.2725$.
$x \approx 9.30 \%$.
Thus,the abundance of $_{12}^{25} Mg$ is $9.30 \%$ and the abundance of $_{12}^{26} Mg$ is $21.01 - 9.30 = 11.71 \%$.

(N/A) In every atom,the positive charge and mass are densely concentrated at the center,which is called the nucleus of the atom.
The dimensions (radius) of a nucleus are much smaller than those of an atom.
Experiments on the scattering of $\alpha$-particles demonstrated that the radius of a nucleus is smaller than the radius of an atom by a factor of about $10^{4}$.
$\frac{\text{Volume of nucleus}}{\text{Volume of atom}} = \frac{4/3 \pi \times (10^{-15} \text{ m})^3}{4/3 \pi \times (10^{-10} \text{ m})^3} = 10^{-15} / 10^{-30} = 10^{-15}$ (approximate ratio based on standard nuclear radius $R = R_0 A^{1/3}$).
More simply,the volume ratio is approximately $10^{-12}$.
$\therefore$ The volume of a nucleus is about $10^{-12}$ times the volume of the atom.
Even so,the nucleus contains most (more than $99.9 \%$) of the mass of an atom. If we think of an atom enlarged to the size of a classroom,the nucleus would be the size of a pinhead.
Hence,the empty space in the atom is a very large area.

(A) The nucleus of an atom is a small,dense,and positively charged region located at the center of the atom.
It contains almost all of the atom's mass.
The nucleus is composed of protons (which are positively charged) and neutrons (which are electrically neutral),collectively known as nucleons.
These particles are held together by the strong nuclear force.

(B) In the $\alpha -$ scattering experiment conducted by Rutherford,the distance of closest approach was used to estimate the upper limit of the nuclear size.
For a gold nucleus $(Z = 79)$,the distance of closest approach $d$ is given by the formula $d = \frac{1}{4\pi\epsilon_0} \cdot \frac{2Ze^2}{K}$,where $K$ is the kinetic energy of the $\alpha -$ particle.
Experimental results indicated that the nucleus is extremely small compared to the atom,with a radius estimated to be in the order of $10^{-14} \ m$ to $10^{-15} \ m$.
Specifically,the radius of a nucleus is typically expressed as $R = R_0 A^{1/3}$,where $R_0 \approx 1.2 \times 10^{-15} \ m$.
Thus,the estimated order of magnitude for the radius of the nucleus from the scattering experiment is $10^{-14} \ m$.

(N/A) The radius of the nucleus $R$ is given by the relation $R = R_0 A^{1/3}$, where $R_0 \approx 1.2 \times 10^{-15} \ m$ and $A$ is the mass number.
The size of the nucleus is of the order of $10^{-15} \ m$ (femtometer).
The size of an atom is of the order of $10^{-10} \ m$ ($\mathring{A}$).
Therefore, the ratio of the size of the nucleus to the atomic size is approximately $10^{-15} / 10^{-10} = 10^{-5}$.
This indicates that the nucleus is about $10^5$ times smaller than the atom.

(A) The mass of an atom is primarily concentrated in its nucleus because protons and neutrons (nucleons) are much heavier than electrons.
The mass of a proton is approximately $1.67 \times 10^{-27} \text{ kg}$, while the mass of an electron is approximately $9.11 \times 10^{-31} \text{ kg}$.
Since the mass of a proton is about $1836$ times the mass of an electron, the electrons contribute very little to the total mass of the atom.
Therefore, more than $99.9\%$ of the total mass of an atom is concentrated in its nucleus.

(N/A) The measurement of atomic masses reveals the existence of different types of atoms of the same element which exhibit the same chemical properties but differ in mass. Such types of atoms are called isotopes.
Isotopes are atoms that have the same atomic number $Z$ but different mass number $A$ and neutron number $N$.
Isotopes occupy the same place in the periodic table of elements.
The relative abundance of different isotopes varies from element to element.
For example:
$(1)$ Chlorine has two isotopes having masses $34.98 \ u$ and $36.98 \ u$. The relative abundances of these isotopes are $75.4 \%$ and $24.6 \%$. Thus,the average mass of a chlorine atom is obtained by the weighted average of the masses of the two isotopes.
Average mass of chlorine $= \frac{75.4 \times 34.98 + 24.6 \times 36.98}{100} = 35.47 \ u$.
$(2)$ The lightest element,hydrogen,has three isotopes:
Mass of hydrogen atom $= 1.0078 \ u$
Mass of deuterium atom $= 2.0141 \ u$
Mass of tritium atom $= 3.0160 \ u$
The nucleus of hydrogen has a relative abundance of $99.985 \%$ and its nucleus is called a proton. In a hydrogen atom,there is no neutron,only one proton.
Mass of proton $m_{p} \approx 1.00727 \ u$.
$\therefore \text{Mass of proton } m_{p} = 1.00727 \times 1.660539 \times 10^{-27} \ kg = 1.67262 \times 10^{-27} \ kg$.
The mass of this proton is calculated as: Mass of hydrogen atom $-$ mass of electron $= 1.00783 \ u - 0.00055 \ u = 1.00728 \ u$.
Deuterium and tritium are unstable isotopes; they do not occur naturally and are produced artificially in laboratories. Hydrogen,deuterium,and tritium have only one proton in their nucleus,so their mass ratio is approximately $1:2:3$.

(N/A) In $1932$,James Chadwick verified the hypothesis that the atomic nucleus contains neutral particles in addition to protons.
Chadwick observed the emission of neutral radiation when beryllium nuclei were bombarded with $\alpha$-particles.
He further observed that this neutral radiation could knock out protons from light nuclei,such as those of helium,carbon,and nitrogen.
The only neutral radiation known at that time was photons (electromagnetic radiation).
Application of the principles of conservation of energy and momentum showed that if the neutral radiation consisted of photons,their energy would have to be much higher than what is available from the bombardment of beryllium nuclei with $\alpha$-particles.
Chadwick satisfactorily resolved this by assuming that the neutral radiation consists of a new type of neutral particle called neutrons.
Using the conservation of energy and momentum,he determined the mass of the new particle (neutron) to be very nearly the same as the mass of a proton.
Currently,the mass of a neutron is precisely $m_{n} = 1.00866 \ u = 1.6749 \times 10^{-27} \ kg$.
Chadwick was awarded the $1935$ Nobel Prize in Physics for his discovery of the neutron.

(N/A) The composition of a nucleus is described by the following terms:
$1$. Nucleon $(A)$: The total number of protons and neutrons present in the nucleus of an atom.
$2$. Atomic Number $(Z)$: The number of protons in the nucleus of an atom,which is equal to the number of electrons in a neutral atom.
$3$. Mass Number $(A)$: The total number of protons and neutrons in the nucleus,given by $A = Z + N$.
$4$. Neutron Number $(N)$: The number of neutrons present in the nucleus of an atom,calculated as $N = A - Z$.
$5$. Nuclide: $A$ specific type of nucleus characterized by its atomic number and mass number,represented by the symbol ${ }_{Z}^{A}X$,where $X$ is the chemical symbol of the element.
For example,the nuclide of gold is ${ }_{79}^{197}Au$,which contains $197$ nucleons,$79$ protons,and $118$ neutrons.

(N/A) Isotopes: Atoms having the same atomic number $Z$ but different mass numbers $A$ are called isotopes. Example: Hydrogen isotopes are ${ }_{1}^{1}H, { }_{1}^{2}H, { }_{1}^{3}H$. They have identical chemical properties.
Isobars: Atoms having the same mass number $A$ but different atomic numbers $Z$ are called isobars. Example: ${ }_{18}^{40}Ar$ and ${ }_{20}^{40}Ca$ are isobars.
Isotones: Atoms having the same number of neutrons $(N = A - Z)$ are called isotones. Example: ${ }_{6}^{14}C$ $(N=8)$ and ${ }_{8}^{16}O$ $(N=8)$ are isotones.
Isomers: Nuclei with the same atomic number $Z$ and same mass number $A$ but existing in different energy states are called nuclear isomers. Example: ${ }_{35}^{80}Br$ has a ground state and an excited state (metastable state denoted as ${ }_{35}^{80m}Br$).

(N/A) The unit of mass used in nuclear physics is the Atomic Mass Unit,denoted as $amu$ or $u$.
Definition: One atomic mass unit $(1 \ u)$ is defined as exactly $1/12$th of the mass of a carbon-$12$ atom.
Equivalence in kilograms: $1 \ u = 1.660539 \times 10^{-27} \ kg$.

(N/A) Mass is defined as the quantity of matter contained in a body.
In nuclear physics,the unit used to measure mass is the unified atomic mass unit,denoted by $u$ or $amu$.
$1$ unified atomic mass unit $(1 \ u)$ is defined as $\frac{1}{12}$ of the mass of a carbon-$12$ atom $(^{12}_{6}C)$.
The value of $1 \ u$ is approximately $1.66 \times 10^{-27} \ kg$.

(N/A) $1$ unified atomic mass unit $(1\,U) = \frac{1}{12}$ of the mass of a carbon-$12$ isotope $\left(^{12}_{6}C\right)$ including the mass of electrons $= 1.66 \times 10^{-27}\,\text{kg}$.
Generally,the mass of a body is measured by a common (physical) balance.
Large masses in the universe (like planets or stars) are determined based on Newton's universal law of gravitation $\left[m = \frac{Fr^2}{GM_e}\right]$.
To measure very small masses (like an atom),a mass spectrometer is used. In this instrument,the radius of the trajectory is proportional to the mass of the charged particle moving in uniform electric and magnetic fields.

(N/A) The mass of a proton is a fundamental physical constant.
In kilograms, the mass of a proton is approximately $m_p = 1.6726219 \times 10^{-27} \ kg$.
In atomic mass units $(u)$, the mass of a proton is approximately $m_p = 1.007276 \ u$.

(B) The neutron was discovered by the British physicist $James \ Chadwick$ in $1932$. He bombarded beryllium atoms with alpha particles,which resulted in the emission of neutral particles with a mass slightly greater than that of a proton. These particles were named neutrons.

(N/A) An atom consists of a central nucleus surrounded by electrons. The nucleus is composed of protons and neutrons.
$1$. $Z$ (Atomic Number): It is defined as the total number of protons present in the nucleus of an atom. It determines the identity of the element.
$2$. $A$ (Mass Number): It is defined as the total number of protons and neutrons (collectively called nucleons) present in the nucleus of an atom. $A = Z + N$.
$3$. $N$ (Neutron Number): It is defined as the total number of neutrons present in the nucleus of an atom. It is calculated as $N = A - Z$.

(N/A) $1$. Isotopes: Atoms of the same element that have the same atomic number $(Z)$ but different mass numbers $(A)$. They have the same number of protons but a different number of neutrons. Example: $^1_1H, ^2_1H, ^3_1H$.
$2$. Isobars: Atoms of different elements that have the same mass number $(A)$ but different atomic numbers $(Z)$. Example: $^{40}_{18}Ar$ and $^{40}_{20}Ca$.
$3$. Isotones: Atoms of different elements that have the same number of neutrons $(N = A - Z)$. Example: $^{14}_6C$ and $^{15}_7N$ both have $8$ neutrons.

(N/A) At Rutherford's suggestion,Geiger and Marsden performed their experiment on the scattering of $\alpha$-particles from thin gold foils. Their experiments revealed that the actual size of the nucleus of gold must be less than $4.0 \times 10^{-14} \ m$.
By performing scattering experiments in which fast electrons,instead of $\alpha$-particles,are projectiles that bombard targets made up of various elements,the size of the nuclei of various elements has been accurately measured,leading to the following formula:
$A$ nucleus of mass number $A$ has a radius $R = R_{0} A^{1/3}$,where $R_{0} = 1.2 \times 10^{-15} \ m = 1.2 \ fm$ and $1 \ fm = 10^{-15} \ m$.
The value of this constant is in the order of the range of the nuclear force. The volume of the nucleus is:
$V = \frac{4}{3} \pi R^{3} = \frac{4}{3} \pi (R_{0} A^{1/3})^{3} = \frac{4}{3} \pi R_{0}^{3} A$.
Therefore,$V \propto A$,which implies that the volume is directly proportional to the mass number. The density of the nucleus is:
$\rho = \frac{M}{V} = \frac{m A}{\frac{4}{3} \pi R_{0}^{3} A} = \frac{3m}{4 \pi R_{0}^{3}}$.
Hence,the density of the nucleus does not depend on the mass number $A$. Calculating the density:
$\rho = \frac{3 \times 1.66 \times 10^{-27}}{4 \times 3.14 \times (1.2 \times 10^{-15})^{3}} \approx 2.3 \times 10^{17} \ kg \ m^{-3}$.
This density is approximately $2.3 \times 10^{14}$ times that of water,indicating that the nucleus is extremely dense due to the large amount of empty space in an atom.

(B) In the alpha-particle scattering experiment,Rutherford observed that alpha particles were deflected by the nucleus.
By calculating the distance of closest approach,he estimated the size of the nucleus.
The distance of closest approach $r_0$ is given by the formula $r_0 = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{K}$,where $K$ is the kinetic energy of the alpha particle.
Based on his experimental results,Rutherford concluded that the radius of the nucleus is of the order of $10^{-15} \ m$ (or $1 \ fm$).