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

(C) The correct answer is $C$.
Free neutrons are unstable particles with a mean lifetime of approximately $880 \ s$ (often cited as $\approx 15 \ minutes$).
They undergo beta-minus decay $(n \rightarrow p + e^- + \bar{\nu}_e)$ to become a proton, emitting an electron and an antineutrino in the process.
In contrast, electrons and protons (in their free state) are stable particles.

(A) Isotones are nuclei that have the same number of neutrons $(N)$ but different atomic numbers $(Z)$.
Neutron number is calculated as $N = A - Z$,where $A$ is the mass number and $Z$ is the atomic number.
For option $A$: $_{34}Se^{74} \implies N = 74 - 34 = 40$ and $_{31}Ga^{71} \implies N = 71 - 31 = 40$.
Since both nuclei have $40$ neutrons,they are isotones.
For option $B$: $_{42}Mo^{92} \implies N = 92 - 42 = 50$ and $_{40}Zr^{92} \implies N = 92 - 40 = 52$. Not isotones.
For option $C$: These are isotopes (same $Z$,different $A$).
For option $D$: $_{20}Ca^{40} \implies N = 20$ and $_{16}S^{32} \implies N = 16$. Not isotones.
Therefore,the correct option is $A$.

(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.
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 for $_{13}^{27}Al$, $r_1 = 3.6 \, \text{Fermi}$ and $A_1 = 27$.
For $_{52}^{125}Te$, $A_2 = 125$.
Substituting the values: $\frac{3.6}{r_2} = \left( \frac{27}{125} \right)^{1/3}$.
$\frac{3.6}{r_2} = \frac{3}{5}$.
$r_2 = \frac{3.6 \times 5}{3} = 1.2 \times 5 = 6 \, \text{Fermi}$.

(B) The nuclear reaction is given by: $_{10}^{22}Ne \to 2(_2^4He) + _Z^A X$
Applying the law of conservation of mass number: $22 = 2(4) + A \implies 22 = 8 + A \implies A = 14$
Applying the law of conservation of atomic number: $10 = 2(2) + Z \implies 10 = 4 + Z \implies Z = 6$
The nucleus with atomic number $Z = 6$ is Carbon $(C)$.
Therefore,the unknown nucleus is Carbon.

(D) The initial nucleus is $^7_7X^{15}$.
An $\alpha$-particle is a helium nucleus,represented as $^2_2He^4$.
The reaction of the atom capturing an $\alpha$-particle is: $^7_7X^{15} + ^2_2He^4 \to ^{19}_9Y^*$.
Then,the resulting nucleus emits a proton $(^1_1H^1)$:
$^{19}_9Y^* \to ^1_1H^1 + ^{18}_8Z$.
Comparing the mass numbers: $15 + 4 = 1 + A \implies A = 18$.
Comparing the atomic numbers: $7 + 2 = 1 + Z \implies Z = 8$.
Thus,the resulting product has a mass number of $18$ and an atomic number of $8$.

(D) An $\alpha$-particle is identical to a helium nucleus $(^{4}_{2}He^{2+})$.
It consists of $2$ protons $(P)$ and $2$ neutrons $(N)$.
Therefore,the composition of an $\alpha$-particle is $2P + 2N$.

(C) Particles with half-integral spin are known as fermions.
Protons,neutrons,and electrons are examples of fermions,each possessing a spin of $1/2$.
Photons are bosons with an integer spin of $1$.
Pions and $K$-mesons are also bosons with a spin of $0$.

(A) By the law of conservation of momentum,for a nucleus at rest,the magnitudes of momenta of the two fragments must be equal:
$m_1 v_1 = m_2 v_2$
$\frac{m_1}{m_2} = \frac{v_2}{v_1} = \frac{1}{8}$
Since the mass of a nucleus is proportional to its mass number $A$ $(m \propto A)$,we have $\frac{A_1}{A_2} = \frac{m_1}{m_2} = \frac{1}{8}$.
The radius $r$ of a nucleus is related to its mass number $A$ by the relation $r = r_0 A^{1/3}$,where $r_0$ is a constant.
Therefore,the ratio of the radii is:
$\frac{r_1}{r_2} = \left( \frac{A_1}{A_2} \right)^{1/3} = \left( \frac{1}{8} \right)^{1/3} = \frac{1}{2}$.
Thus,the ratio of the radii is $1 : 2$.

(A) The density of a nucleus $(\rho)$ is defined as the ratio of its mass $(m)$ to its volume $(V)$, given by $\rho = m/V$.
Experimental observations show that the nuclear density is approximately constant for all nuclei, regardless of their mass number.
Since $\rho = \text{constant}$, we have $m/V = \text{constant}$.
Therefore, the mass of the nucleus is directly proportional to its volume, which can be expressed as $m \propto V$.

(C) The charge density inside a nucleus is approximately constant up to a certain radius,known as the nuclear radius $R$.
Beyond this radius,the density falls off rapidly to zero.
This behavior is characteristic of the distribution of nucleons within the nucleus.
Looking at the provided graphs,graph $C$ shows a constant value for $P$ near the center $(r=0)$ and a sharp decrease as $r$ approaches the nuclear surface.
Therefore,graph $C$ is the correct representation.

(A) The relationship between the nuclear radius $R$ and the mass number $A$ is given by the formula:
$R = R_0 A^{1/3}$
where $R_0$ is a constant $(R_0 \approx 1.2 \times 10^{-15} \text{ m})$.
Taking the natural logarithm on both sides:
$\ln R = \ln(R_0 A^{1/3})$
$\ln R = \ln R_0 + \ln(A^{1/3})$
$\ln R = \ln R_0 + \frac{1}{3} \ln A$
This equation is of the form $y = mx + c$,where $y = \ln R$,$x = \ln A$,the slope $m = 1/3$,and the intercept $c = \ln R_0$.
Since the slope is positive $(1/3)$,the graph of $\log R$ versus $\log A$ is a straight line with a positive slope.

(A) In a triclinic crystal system,the unit cell parameters are defined by the conditions where none of the edge lengths are equal $(a \ne b \ne c)$ and none of the interfacial angles are equal to each other or to $90^{\circ}$ $(\alpha \ne \beta \ne \gamma \ne 90^{\circ})$. This represents the least symmetric crystal system.

(D) $Cu$ (Copper) crystallizes in a face-centered cubic $(fcc)$ lattice structure.
In an $fcc$ crystal lattice,each atom is surrounded by $12$ nearest neighbors.
Therefore,the coordination number of $Cu$ is $12$.

(B) In a $bcc$ (body-centered cubic) lattice,the atoms touch each other along the body diagonal.
The length of the body diagonal is $\sqrt{3}a$,where $a$ is the edge length of the unit cell.
Since the body diagonal consists of two atomic radii $(r)$ from the corner atoms and the full diameter $(2r)$ of the central atom,we have $4r = \sqrt{3}a$,which gives the atomic radius $r = \frac{\sqrt{3}a}{4}$.
The nearest distance $(d)$ between two atoms is the distance between a corner atom and the central atom,which is $2r$.
Therefore,$d = 2 \times \left( \frac{\sqrt{3}a}{4} \right) = \frac{\sqrt{3}a}{2}$.

(D) According to the law of conservation of linear momentum,the initial momentum of the system is zero because the nucleus is at rest.
$\vec{P}_i = \vec{P}_f = 0$
Let $m_{\alpha}$ be the mass of the alpha particle and $m_R$ be the mass of the residual nucleus $(Th^{234})$.
$m_{\alpha} = 4$ units and $m_R = 234$ units.
$m_{\alpha}\vec{v}_{\alpha} + m_R\vec{v}_R = 0$
Given the speed of the alpha particle is $u$,so $\vec{v}_{\alpha} = u$.
$4u + 234\vec{v}_R = 0$
$\vec{v}_R = -\frac{4u}{234}$
The negative sign indicates that the residual nucleus moves in the opposite direction to the alpha particle.

(D) According to the law of conservation of linear momentum,the initial momentum is zero,so $m_1v_1 = m_2v_2$.
This implies $\frac{m_1}{m_2} = \frac{v_2}{v_1}$.
Given the ratio of velocities $\frac{v_1}{v_2} = \frac{8}{1}$,so $\frac{v_2}{v_1} = \frac{1}{8}$.
Assuming the density $\rho$ of the nucleus is constant,the mass $m$ is proportional to the volume,$m = \rho \cdot \frac{4}{3}\pi r^3$.
Therefore,$\frac{m_1}{m_2} = \frac{r_1^3}{r_2^3}$.
Equating the two expressions: $\frac{r_1^3}{r_2^3} = \frac{v_2}{v_1} = \frac{1}{8}$.
Taking the cube root on both sides: $\frac{r_1}{r_2} = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}$.

(D) The nuclear force is the strongest fundamental force in nature.
It is a short-range force,acting only within the dimensions of the nucleus (approximately $10^{-15} \ m$ to $10^{-14} \ m$).
It is charge-independent,meaning it acts equally between proton-proton,neutron-neutron,and proton-neutron pairs.
Therefore,all the given statements are correct.

(C) The nuclear force is a short-range,strong attractive force that acts between nucleons (protons and neutrons) within the nucleus.
It is independent of the charge of the nucleons.
It is not a long-range force,meaning it does not increase with the total number of nucleons; rather,it exhibits the property of saturation,where a nucleon only interacts with its nearest neighbors.
Therefore,the statement that the nuclear force increases as the number of nucleons increases is incorrect.

(C) Isotopes are atoms of the same element that have the same number of protons and electrons but a different number of neutrons.
For $U^{235}$: Atomic number $(Z)$ = $92$,Mass number $(A)$ = $235$. Number of neutrons $(N)$ = $A - Z = 235 - 92 = 143$.
For $U^{238}$: Atomic number $(Z)$ = $92$,Mass number $(A)$ = $238$. Number of neutrons $(N)$ = $A - Z = 238 - 92 = 146$.
Comparing the two,both have $92$ protons and $92$ electrons. The difference in the number of neutrons is $146 - 143 = 3$.
Therefore,$U^{238}$ has three more neutrons than $U^{235}$.

(C) The nuclear force is charge-independent.
This means that the strong nuclear force acting between two nucleons is independent of whether they are protons or neutrons.
Therefore,the magnitude of the nuclear force between two protons $(F_{pp})$,two neutrons $(F_{nn})$,and a proton and a neutron $(F_{pn})$ is approximately the same.
Thus,$F_{pp} = F_{pn} = F_{nn}$.

(A) Nuclear force is charge-independent.
This means the strong nuclear force between two nucleons is the same regardless of whether they are protons or neutrons.
Therefore,the magnitude of the nuclear force between two protons $(F_{pp})$,two neutrons $(F_{nn})$,and a proton and a neutron $(F_{pn})$ is approximately equal.
Thus,$F_{pp} \approx F_{nn} \approx F_{pn}$.

(A) The given nuclear reaction is $X + _0^1 n \rightarrow _2^4 He + _3^7 Li$.
Applying the law of conservation of mass number:
$A + 1 = 4 + 7$
$A + 1 = 11$
$A = 10$
Applying the law of conservation of atomic number (charge):
$Z + 0 = 2 + 3$
$Z = 5$
Thus,the element $X$ has mass number $10$ and atomic number $5$,which corresponds to Boron $(B)$.
Therefore,$X = _5^{10} B$.

(D) The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number.
Given that the radius of the unknown nucleus is half the radius of $Fe^{56}$,we have:
$R_{new} = \frac{1}{2} R_{Fe}$
$R_0 (A_{new})^{1/3} = \frac{1}{2} R_0 (A_{Fe})^{1/3}$
$(A_{new})^{1/3} = \frac{1}{2} (56)^{1/3}$
Taking the cube of both sides:
$A_{new} = (\frac{1}{2})^3 \times 56$
$A_{new} = \frac{1}{8} \times 56 = 7$
Therefore,the nucleus with mass number $A = 7$ is $Li^7$.

(A) The given nuclear reaction is $X + _0^1n \to _2^4He + _3^7Li$.
According to the law of conservation of mass number: $A_X + 1 = 4 + 7$,which gives $A_X = 10$.
According to the law of conservation of atomic number: $Z_X + 0 = 2 + 3$,which gives $Z_X = 5$.
Since the atomic number $Z = 5$ corresponds to Boron $(B)$,the element $X$ is $_5^{10}B$.

(A) The radius of a nucleus is given by the relation $R = R_0 A^{1/3}$,where $R_0 \approx 1.2 \times 10^{-15} \text{ m}$ and $A$ is the mass number.
The volume $V$ of a nucleus is given by the formula for the volume of a sphere:
$V = \frac{4}{3} \pi R^3$
Substituting the expression for $R$:
$V = \frac{4}{3} \pi (R_0 A^{1/3})^3$
$V = \frac{4}{3} \pi R_0^3 A$
Since $\frac{4}{3} \pi R_0^3$ is a constant,the volume $V$ is directly proportional to the mass number $A$ $(V \propto A)$.

(D) The radius of an atom is approximately $R_a = 10^{-10} \ m$.
The radius of a nucleus is approximately $R_n = 10^{-15} \ m$.
The volume of a sphere is given by $V = \frac{4}{3} \pi R^3$.
Therefore,the ratio of the volume of the atom to the volume of the nucleus is:
$\frac{V_a}{V_n} = \frac{\frac{4}{3} \pi (R_a)^3}{\frac{4}{3} \pi (R_n)^3} = \left( \frac{10^{-10}}{10^{-15}} \right)^3 = (10^5)^3 = 10^{15}$.
Thus,the volume of an atom is about $10^{15}$ times larger than the volume of its nucleus.

(C) The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number (number of nucleons).
Given that the radius of the Germanium nucleus $(R_{Ge})$ is twice the radius of the Beryllium nucleus $(R_{Be})$:
$R_{Ge} = 2 \times R_{Be}$
Substituting the formula $R = R_0 A^{1/3}$:
$R_0 (A_{Ge})^{1/3} = 2 \times R_0 (A_{Be})^{1/3}$
Given $A_{Be} = 9$,we have:
$(A_{Ge})^{1/3} = 2 \times (9)^{1/3}$
Cubing both sides:
$A_{Ge} = 2^3 \times 9$
$A_{Ge} = 8 \times 9 = 72$
Thus,the number of nucleons in Germanium is $72$.

(C) 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.
Therefore,the ratio of the radii of two nuclei is given by $\frac{R_1}{R_2} = \left( \frac{A_1}{A_2} \right)^{1/3}$.
For $_{13}Al^{27}$,the mass number $A_1 = 27$.
For $_{52}Te^{125}$,the mass number $A_2 = 125$.
Substituting these values into the ratio formula:
$\frac{R_{Al}}{R_{Te}} = \left( \frac{27}{125} \right)^{1/3} = \left( \frac{3^3}{5^3} \right)^{1/3} = \frac{3}{5}$.
Thus,the ratio is $3 : 5$.

(A) Isotones are nuclei that have the same number of neutrons $(N)$.
The number of neutrons is calculated as $N = A - Z$,where $A$ is the mass number and $Z$ is the atomic number.
For option $A$: $_{34}Se^{74} \implies N = 74 - 34 = 40$; $_{31}Ga^{71} \implies N = 71 - 31 = 40$.
Since both nuclei have $40$ neutrons,they are isotones.
For option $B$: $_{42}Mo^{92} \implies N = 92 - 42 = 50$; $_{40}Zr^{92} \implies N = 92 - 40 = 52$. (Not isotones)
For option $C$: $_{38}Sr^{84} \implies N = 84 - 38 = 46$; $_{38}Sr^{86} \implies N = 86 - 38 = 48$. (Not isotones)
For option $D$: $_{20}Ca^{40} \implies N = 40 - 20 = 20$; $_{16}S^{32} \implies N = 32 - 16 = 16$. (Not isotones)
Therefore,the correct option is $A$.

(B) The given nuclear reaction is $X(n, \alpha) \to _3^7Li$.
This can be written as: $X + _0^1n \to _3^7Li + _2^4He$.
Let the nucleus $X$ be represented as $_Z^AX$.
Applying the law of conservation of mass number: $A + 1 = 7 + 4$,which gives $A + 1 = 11$,so $A = 10$.
Applying the law of conservation of atomic number: $Z + 0 = 3 + 2$,which gives $Z = 5$.
Since the atomic number $Z = 5$ corresponds to Boron $(B)$ and the mass number $A = 10$,the nucleus is $_{5}^{10}B$. However,checking the provided options,the reaction $X(n, \alpha) \to _3^7Li$ implies $X + _0^1n \to _3^7Li + _2^4He$. Balancing the equation: $X = _3^7Li + _2^4He - _0^1n = _{3+2-0}^{7+4-1}X = _5^{10}X$. Given the options,there is a slight discrepancy in mass number,but $_5^{11}B$ is the standard isotope used in this reaction type $(_{5}^{10}B + _{0}^{1}n \to _{3}^{7}Li + _{2}^{4}He)$. Thus,the correct element is Boron.

(B) The mass number of $He$ is $A_{He} = 4$ and the mass number of $S$ is $A_S = 32$.
The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$,which implies $R \propto A^{1/3}$.
Therefore,the ratio of the radii is given by:
$\frac{R_S}{R_{He}} = \left( \frac{A_S}{A_{He}} \right)^{1/3}$
Substituting the values:
$\frac{R_S}{R_{He}} = \left( \frac{32}{4} \right)^{1/3} = (8)^{1/3} = 2$.
Thus,the radius of the $S$ nucleus is $2$ times the radius of the $He$ nucleus.

(B) The nuclear density $\rho$ is defined as the ratio of the mass of the nucleus to its volume.
$\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{A \cdot m_p}{\frac{4}{3}\pi R^3}$,where $A$ is the mass number,$m_p$ is the mass of a nucleon,and $R$ is the nuclear radius.
Since the nuclear radius $R$ is given by $R = R_0 A^{1/3}$,we have $R^3 = R_0^3 A$.
Substituting this into the density formula:
$\rho = \frac{A \cdot m_p}{\frac{4}{3}\pi (R_0^3 A)} = \frac{m_p}{\frac{4}{3}\pi R_0^3}$.
This shows that the nuclear density $\rho$ is independent of the mass number $A$.
Therefore,the ratio of the nuclear densities of two nuclei with any mass numbers is always $1:1$.

(D) 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: $m_1 v_1 = m_2 v_2$.
Therefore,the ratio of masses is inversely proportional to the ratio of velocities: $\frac{m_1}{m_2} = \frac{v_2}{v_1}$.
Given $\frac{v_1}{v_2} = \frac{2}{1}$,we have $\frac{m_1}{m_2} = \frac{1}{2}$.
The nuclear radius $R$ is related to the mass number $A$ (which is proportional to mass $m$) by the formula $R = R_0 A^{1/3}$.
Thus,the ratio of radii is $\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} = \left(\frac{m_1}{m_2}\right)^{1/3}$.
Substituting the mass ratio: $\frac{R_1}{R_2} = \left(\frac{1}{2}\right)^{1/3} = \frac{1}{2^{1/3}}$.
Hence,the ratio is $1:2^{1/3}$.

(A) The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number.
For $_{13}^{27}Al$,$A_1 = 27$ and $R_1 = 3.6 \ fm$.
So,$3.6 = R_0 (27)^{1/3} = R_0 (3) \implies R_0 = 1.2 \ fm$.
For $_{52}^{125}Te$,$A_2 = 125$.
Therefore,$R_2 = R_0 (A_2)^{1/3} = 1.2 \times (125)^{1/3} = 1.2 \times 5 = 6 \ fm$.

(A) At the distance of closest approach,the total initial kinetic energy of both deuterons is converted into electrostatic potential energy.
Let $K$ be the initial kinetic energy of each deuteron.
Total initial kinetic energy $= K + K = 2K$.
Electrostatic potential energy $U = \frac{k q_1 q_2}{r}$,where $q_1 = q_2 = e$ (charge of a deuteron is $+e$).
Given $r = 2 \, fm = 2 \times 10^{-15} \, m$.
$2K = \frac{k e^2}{r} = \frac{(9 \times 10^9) \times (1.6 \times 10^{-19})^2}{2 \times 10^{-15}} \, J$.
To convert to $eV$,we divide by $e = 1.6 \times 10^{-19} \, C$:
$2K = \frac{9 \times 10^9 \times 1.6 \times 10^{-19}}{2 \times 10^{-15}} \, eV = 7.2 \times 10^5 \, eV = 0.72 \, MeV$.
Therefore,$K = \frac{0.72}{2} = 0.36 \, MeV$.

(C) Inside the nucleus,protons and neutrons are held together by the strong nuclear force,which is an attractive force that acts between all nucleons (protons and protons,neutrons and neutrons,and protons and neutrons) at very short distances.
Additionally,there is an electrostatic (Coulombic) repulsive force between protons due to their like charges.
Therefore,both nuclear force and Coulombic force act between protons in the nucleus.

(B) The electric potential $V$ at the surface of a nucleus is given by the formula $V = \frac{1}{4\pi\epsilon_0} \frac{q}{R}$.
Here,the charge of the nucleus is $q = Ze$,where $Z = 50$ and $e = 1.6 \times 10^{-19} \ C$.
The radius $R = 9 \times 10^{-13} \ m$.
The constant $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \ N \ m^2/C^2$.
Substituting the values:
$V = (9 \times 10^9) \times \frac{50 \times 1.6 \times 10^{-19}}{9 \times 10^{-13}}$
$V = (9 \times 10^9) \times \frac{80 \times 10^{-19}}{9 \times 10^{-13}}$
$V = 80 \times 10^{9 - 19 + 13} \ V$
$V = 80 \times 10^3 \ V = 80 \ kV$.

(B) The electric potential $V$ at the surface of a charged sphere (nucleus) is given by the formula:
$V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r}$
Where $Q = Ze$ is the total charge of the nucleus.
Given:
$Z = 50$
$e = 1.6 \times 10^{-19} \ C$
$r = 9 \times 10^{-15} \ m$
$k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2$
Substituting the values:
$V = (9 \times 10^9) \times \frac{50 \times 1.6 \times 10^{-19}}{9 \times 10^{-15}}$
$V = (9 \times 10^9) \times \frac{80 \times 10^{-19}}{9 \times 10^{-15}}$
$V = 80 \times 10^{9-19+15}$
$V = 80 \times 10^5 \ V = 8 \times 10^6 \ V$

(C) The charge of an $\alpha$-particle is $q_{\alpha} = 2e$ and its mass is $m_{\alpha} = 4m_p$,where $e$ is the elementary charge and $m_p$ is the mass of a proton.
For a proton,the charge is $q_p = e$ and the mass is $m_p$.
The ratio of specific charge $\frac{q}{m}$ for an $\alpha$-particle is $(\frac{q}{m})_{\alpha} = \frac{2e}{4m_p} = \frac{1}{2} \frac{e}{m_p}$.
The ratio of specific charge for a proton is $(\frac{q}{m})_p = \frac{e}{m_p}$.
Therefore,the ratio of $\frac{q}{m}$ for an $\alpha$-particle to that of a proton is $\frac{(\frac{q}{m})_{\alpha}}{(\frac{q}{m})_p} = \frac{\frac{1}{2} \frac{e}{m_p}}{\frac{e}{m_p}} = \frac{1}{2}$.

(A) According to the law of conservation of linear momentum, the initial momentum of the nucleus is zero, so the final momenta of the two parts must be equal and opposite in magnitude: $m_1 v_1 = m_2 v_2$.
Given the ratio of velocities $\frac{v_1}{v_2} = \frac{8}{1}$, we have $\frac{m_1}{m_2} = \frac{v_2}{v_1} = \frac{1}{8}$.
Since the mass of a nucleus is proportional to its mass number $A$ $(m \propto A)$, we have $\frac{A_1}{A_2} = \frac{m_1}{m_2} = \frac{1}{8}$.
The radius $r$ of a nucleus is related to its mass number $A$ by the relation $r = R_0 A^{1/3}$, which implies $r \propto A^{1/3}$.
Therefore, the ratio of the radii is $\frac{r_1}{r_2} = \left( \frac{A_1}{A_2} \right)^{1/3}$.
Substituting the values: $\frac{r_1}{r_2} = \left( \frac{1}{8} \right)^{1/3} = \frac{1}{2}$.
Thus, the ratio of the radii is $1:2$.

(A) 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.
Therefore,the ratio of the radii of two nuclei is given by $\frac{R_1}{R_2} = \left( \frac{A_1}{A_2} \right)^{1/3}$.
For $_{13}^{27}Al$,the mass number $A_1 = 27$.
For $_{52}^{125}Te$,the mass number $A_2 = 125$.
Substituting these values into the ratio formula:
$\frac{R_1}{R_2} = \left( \frac{27}{125} \right)^{1/3} = \frac{(3^3)^{1/3}}{(5^3)^{1/3}} = \frac{3}{5}$.
Thus,the ratio of the radii is $3:5$.

(A) The atomic mass of $_6C^{14}$ is $14 \, g/mol$.
The number of moles in $14 \, g$ of $_6C^{14}$ is $n = \frac{14 \, g}{14 \, g/mol} = 1 \, mol$.
The number of atoms in $1 \, mol$ is equal to the Avogadro number,$N_A = 6 \times 10^{23}$.
In one atom of $_6C^{14}$:
Protons $(Z)$ = $6$
Neutrons $(N = A - Z)$ = $14 - 6 = 8$
Electrons = $6$ (since it is a neutral atom).
Total number of protons = $6 \times (6 \times 10^{23}) = 36 \times 10^{23}$.
Total number of neutrons = $8 \times (6 \times 10^{23}) = 48 \times 10^{23}$.
Total number of electrons = $6 \times (6 \times 10^{23}) = 36 \times 10^{23}$.

(C) The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $R_0 = 1.2 \, fm$ and $A = 64$.
$R = 1.2 \times (64)^{1/3} = 1.2 \times 4 = 4.8 \, fm$.
When two nuclei are in contact,the distance between their centers is $r = 2R = 2 \times 4.8 \, fm = 9.6 \times 10^{-15} \, m$.
The atomic number $Z$ for Copper $(Cu)$ is $29$. The charge on each nucleus is $q = Ze = 29e$.
The electrostatic potential energy is $U = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r} = \frac{k(Ze)^2}{r}$.
Substituting the values: $U = \frac{9 \times 10^9 \times (29 \times 1.6 \times 10^{-19})^2}{9.6 \times 10^{-15}} \, J$.
To convert to $MeV$,divide by $1.6 \times 10^{-13} \, J/MeV$:
$U = \frac{9 \times 10^9 \times 841 \times 2.56 \times 10^{-38}}{9.6 \times 10^{-15} \times 1.6 \times 10^{-13}} \approx 126.15 \, MeV$.

(D) The nuclear radius $R$ is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number and $R_0$ is a constant.
For the given nuclei:
For ${}_{13}^{27}Al$,$A_1 = 27$ and $R_1 = 3.6 \, fm$.
For ${}_{52}^{125}Te$,$A_2 = 125$ and we need to find $R_2$.
Taking the ratio:
$\frac{R_2}{R_1} = \left( \frac{A_2}{A_1} \right)^{1/3}$
Substituting the values:
$\frac{R_2}{3.6} = \left( \frac{125}{27} \right)^{1/3}$
$\frac{R_2}{3.6} = \frac{5}{3}$
$R_2 = \frac{5}{3} \times 3.6 = 5 \times 1.2 = 6 \, fm$.
Thus,the radius of ${}_{52}^{125}Te$ is $6 \, fm$.

(B) The momentum of the emitted photon is given by $p_{\text{photon}} = \frac{hf}{c}$.
According to the law of conservation of linear momentum,the magnitude of the momentum of the recoiling nucleus must be equal to the momentum of the photon:
$p_{\text{nucleus}} = p_{\text{photon}} = \frac{hf}{c}$.
Let $v$ be the recoil speed of the nucleus. Then,$Mv = \frac{hf}{c}$,which gives $v = \frac{hf}{Mc}$.
The recoil kinetic energy of the nucleus is given by $K = \frac{1}{2}Mv^2$.
Substituting the value of $v$:
$K = \frac{1}{2}M \left( \frac{hf}{Mc} \right)^2 = \frac{1}{2}M \left( \frac{h^2f^2}{M^2c^2} \right) = \frac{h^2f^2}{2Mc^2}$.

(C) The nuclear radius $R$ is given by the formula $R = R_0 A^{1/3}$,where $R_0$ is a constant and $A$ is the mass number.
For ${}_{13}^{27}Al$,$A_1 = 27$ and $R_1 = 3.6 \, fm$.
For ${}_{29}^{64}Cu$,$A_2 = 64$.
Taking the ratio: $\frac{R_2}{R_1} = \left( \frac{A_2}{A_1} \right)^{1/3}$.
Substituting the values: $\frac{R_2}{3.6} = \left( \frac{64}{27} \right)^{1/3}$.
$\frac{R_2}{3.6} = \frac{4}{3}$.
$R_2 = 3.6 \times \frac{4}{3} = 1.2 \times 4 = 4.8 \, fm$.

(B) The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number and $R_0$ is a constant.
For the $^{27}_{13}Al$ nucleus,the mass number $A_{Al} = 27$. Thus,$R_{Al} = R_0 (27)^{1/3} = 3 R_0$.
For the $^{125}_{53}Te$ nucleus,the mass number $A_{Te} = 125$. Thus,$R_{Te} = R_0 (125)^{1/3} = 5 R_0$.
Taking the ratio of the two radii:
$\frac{R_{Te}}{R_{Al}} = \frac{5 R_0}{3 R_0} = \frac{5}{3}$.
Therefore,$R_{Te} = \frac{5}{3} R_{Al}$.

(B) The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number.
Given the ratio of the radius of the target nucleus $(R)$ to the radius of the Helium nucleus $(R_{He})$ is $14^{1/3}$.
Using the relation $\frac{R}{R_{He}} = \left( \frac{A}{A_{He}} \right)^{1/3}$,we have $14^{1/3} = \left( \frac{A}{4} \right)^{1/3}$.
Cubing both sides,we get $14 = \frac{A}{4}$,which implies $A = 56$.
The mass number $A$ is the sum of protons $(Z)$ and neutrons $(N)$.
Given $N = 30$,we have $Z + 30 = 56$.
Therefore,$Z = 56 - 30 = 26$.