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

(D) The size of an atomic nucleus is extremely small,typically on the order of $10^{-15} \ m$.
In physics,the unit $1 \ \text{fermi}$ (also known as a femtometre) is defined as $1 \ \text{fm} = 10^{-15} \ m$.
Since nuclear radii are measured in the range of $1$ to $10 \ \text{fm}$,the $Fermi$ is the standard and most appropriate unit for this measurement.
Therefore,the correct option is $(d)$.

(C) The barn is a non-$SI$ unit of area used in nuclear and particle physics to express the cross-section of scattering or absorption events.
By definition,$1 \text{ barn} = 10^{-28} \, m^2$.

(C) Neutrons are electrically neutral particles,so they do not experience electrostatic or electromagnetic forces.
$1$. Gravitational force: This is always attractive and acts between any two particles with mass. Since neutrons have mass,they experience an attractive gravitational force.
$2$. Nuclear force: This is the strong force that acts between nucleons (protons and neutrons). It is highly attractive at short distances (on the order of $10^{-15} \ m$).
Therefore,gravitational and nuclear forces are the two forces that provide an attractive force between two neutrons.

(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 $Q = Ze = 50 \times 1.6 \times 10^{-19} \, C$.
The radius $R = 9.0 \times 10^{-13} \, cm = 9.0 \times 10^{-15} \, m$.
The constant $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, N \cdot m^2/C^2$.
Substituting these values:
$V = (9 \times 10^9) \times \frac{50 \times 1.6 \times 10^{-19}}{9.0 \times 10^{-15}}$
$V = 10^9 \times 50 \times 1.6 \times 10^{-19} \times 10^{15}$
$V = 80 \times 10^5 \, V = 8 \times 10^6 \, V$.

(A) The electric potential $V$ at the surface of a nucleus is given by the formula: $V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Ze}{r}$.
Here,$Z = 47$,$e = 1.6 \times 10^{-19} \ C$,$r = 3.4 \times 10^{-14} \ m$,and $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \ N \ m^2/C^2$.
Substituting the values:
$V = (9 \times 10^9) \times \frac{47 \times 1.6 \times 10^{-19}}{3.4 \times 10^{-14}}$
$V = \frac{9 \times 47 \times 1.6}{3.4} \times 10^{9 - 19 + 14}$
$V = \frac{676.8}{3.4} \times 10^4$
$V \approx 199.05 \times 10^4 \ V = 1.99 \times 10^6 \ V$.

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

(A) The size of the nucleus is typically in the range of $10^{-15} \ m$ to $10^{-14} \ m$.
The Bohr radius of the hydrogen atom is given by $a_0 = \frac{4\pi\epsilon_0\hbar^2}{m_e e^2} \approx 0.529 \times 10^{-10} \ m$.
Therefore,the order of magnitude for the size of the nucleus is $10^{-14} \ m$ and for the Bohr radius is $10^{-10} \ m$.

(B) The correct option is $B$.
The nucleus of an atom is composed of protons and neutrons,which are collectively known as nucleons.
Protons are positively charged particles,and neutrons are electrically neutral particles.
Electrons are subatomic particles that revolve around the nucleus in extranuclear orbits and are not constituents of the nucleus.

(C) The chemical properties of an atom are determined by the number of electrons orbiting the nucleus,which is equal to the number of protons (atomic number $Z$) in the nucleus.
Adding protons would change the atomic number,thereby changing the element and its chemical properties.
Adding electrons would change the charge state (ion formation) or chemical reactivity.
Neutrons are electrically neutral particles. Adding neutrons to the nucleus changes the mass number $(A)$ of the atom,resulting in a different isotope of the same element. Since isotopes have the same atomic number,they exhibit identical chemical properties.

(C) 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 that he identified as neutrons.

(D) The mass number $(A)$ is the sum of the number of protons $(Z)$ and neutrons $(N)$ in a nucleus,so $A = Z + N$.
For the hydrogen nucleus $(_{1}^{1}H)$,the number of protons is $1$ and the number of neutrons is $0$,so the mass number is $1$,which is equal to the atomic number $(A = Z)$.
For all other nuclei,the number of neutrons is at least $1$ or more,making the mass number greater than the atomic number $(A > Z)$.
Therefore,the mass number is sometimes equal to and sometimes greater than the atomic number.

(B) The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$, where $R_0$ is a constant approximately equal to $1.2 \times 10^{-15} \ m$ to $1.5 \times 10^{-15} \ m$ and $A$ is the mass number.
Since the mass number $A$ typically ranges from $1$ to $238$, the value of $A^{1/3}$ is of the order of $1$ to $6$.
Therefore, the size of the nucleus is primarily determined by the order of $R_0$, which is $10^{-15} \ m$ (also known as $1 \ \text{fermi}$ or $1 \ \text{fm}$).
Thus, the correct order of magnitude for the size of the nucleus is $10^{-15} \ m$.

(C) Nuclear forces are short-range forces that act between nucleons (protons and neutrons) within the nucleus.
These forces become strongly attractive at a distance of approximately $1 \, \text{femtometre}$ $(1 \, \text{fm} = 10^{-15} \, m)$.
Beyond this range,the nuclear force decreases rapidly,and at distances greater than a few femtometres,it becomes negligible.

(D) Isobars are atoms of different chemical elements that have the same mass number $(A)$ but different atomic numbers $(Z)$.
In option $A$,$_1H^1$ and $_1H^2$ have the same atomic number $(Z=1)$ but different mass numbers,so they are isotopes.
In option $B$,$_1H^2$ and $_1H^3$ have the same atomic number $(Z=1)$ but different mass numbers,so they are isotopes.
In option $C$,$_6C^{12}$ and $_6C^{13}$ have the same atomic number $(Z=6)$ but different mass numbers,so they are isotopes.
In option $D$,$_{15}P^{30}$ and $_{14}Si^{30}$ have different atomic numbers ($Z=15$ and $Z=14$) but the same mass number $(A=30)$.
Therefore,$_{15}P^{30}$ and $_{14}Si^{30}$ are isobars.

(D) The mass number $(A)$ of an atom is defined as the total number of protons and neutrons present in its nucleus.
Protons and neutrons are collectively known as nucleons.
Therefore,the mass number is equal to the total number of nucleons in the nucleus of an atom.

(B) The notation for a nucleus is given by $_Z{X^A}$,where $Z$ is the atomic number (number of protons) and $A$ is the mass number (total number of protons and neutrons).
For the nucleus $_{88}Ra^{226}$:
Atomic number $Z = 88$,which represents the number of protons.
Mass number $A = 226$.
The number of neutrons $N$ is calculated as $N = A - Z$.
$N = 226 - 88 = 138$.
Therefore,there are $88$ protons and $138$ neutrons.

(B) The mass of a nucleus with mass number $A$ is approximately $m = A \times m_p$,where $m_p$ is the mass of a proton.
The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $R_0 \approx 1.2 \times 10^{-15} \ m$.
The volume of the nucleus is $V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.
The density $\rho$ is given by $\rho = \frac{m}{V} = \frac{A \times m_p}{\frac{4}{3} \pi R_0^3 A} = \frac{m_p}{\frac{4}{3} \pi R_0^3}$.
Substituting the values: $\rho = \frac{1.67 \times 10^{-27}}{\frac{4}{3} \times 3.14 \times (1.2 \times 10^{-15})^3} \approx \frac{1.67 \times 10^{-27}}{7.24 \times 10^{-45}} \approx 2.3 \times 10^{17} \ kg/m^3$.
Thus,the order of magnitude is $10^{17} \ kg/m^3$.

(C) The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number.
This implies that $R \propto A^{1/3}$.
Given for $_2He^4$,$R_1 = 3 \, fm$ and $A_1 = 4$.
For $_{82}Pb^{206}$,$A_2 = 206$.
Using the ratio: $\frac{R_2}{R_1} = \left( \frac{A_2}{A_1} \right)^{1/3}$.
$\frac{R_2}{3} = \left( \frac{206}{4} \right)^{1/3} = (51.5)^{1/3}$.
Calculating the cube root: $(51.5)^{1/3} \approx 3.72$.
Therefore,$R_2 = 3 \times 3.72 = 11.16 \, fm$.
Thus,the correct option is $C$.

(C) The atomic mass $(A)$ of the atom is $24$ and the atomic number $(Z)$ is $11$.
The atomic number $(Z)$ represents the number of protons in the nucleus.
Therefore,the number of protons = $11$.
The number of neutrons $(N)$ is given by the formula $N = A - Z$.
$N = 24 - 11 = 13$.
The nucleus of an atom consists only of protons and neutrons; it does not contain electrons.
Thus,the nucleus consists of $11$ protons and $13$ neutrons.

(A) Let the percentage of $_5B^{10}$ atoms be $x$. Then the percentage of $_5B^{11}$ atoms is $(100 - x)$.
The average atomic weight is given by the formula:
$\text{Average Atomic Weight} = \frac{(10 \times x) + (11 \times (100 - x))}{100} = 10.81$
Multiplying by $100$:
$10x + 1100 - 11x = 1081$
$-x = 1081 - 1100$
$-x = -19$
$x = 19$
So,the percentage of $_5B^{10}$ is $19\%$ and the percentage of $_5B^{11}$ is $81\%$.
Therefore,the ratio of $_5B^{10} : _5B^{11}$ is $19 : 81$.

(A) The mass of a neutron is approximately $1.6749 \times 10^{-27} \ kg$.
The mass of a proton is approximately $1.6726 \times 10^{-27} \ kg$.
Since these two values are very close to each other,the mass of a neutron is considered to be the same as that of a proton. Therefore,the correct option is $A$.

(A) Nuclear forces are the strong forces that hold nucleons (protons and neutrons) together inside the nucleus.
$1$. They are short-ranged,acting only over distances of the order of $10^{-15} \ m$ (femtometers).
$2$. They are primarily attractive in nature,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 approximately the same,provided the separation distance is the same.
Therefore,the correct option is $A$.

(D) $\pi$ mesons (pions) are subatomic particles that mediate the strong nuclear force. They exist in three charge states: positive $(\pi^+)$,negative $(\pi^-)$,and neutral $(\pi^0)$. These particles are produced in high-energy particle interactions,such as:
$1$. $p \to n + \pi^+$
$2$. $n \to p + \pi^-$
$3$. $p \to p + \pi^0$
Therefore,$\pi$ mesons can be $\pi^+, \pi^-,$ or $\pi^0$.

(C) The helium nucleus is represented as $_2He^4$.
In any nucleus,the atomic number $Z$ represents the number of protons.
Here,$Z = 2$,so there are $2$ protons.
The mass number $A$ is the sum of protons and neutrons,$A = 4$.
The number of neutrons $N = A - Z = 4 - 2 = 2$.
Therefore,a helium nucleus consists of $2$ protons and $2$ neutrons.

(A) Isotopes are defined as atoms of the same element that have the same atomic number $(Z)$ but different mass numbers $(A)$.
Since the atomic number $(Z)$ represents the number of protons,isotopes have the same number of protons.
The mass number $(A)$ is the sum of protons $(Z)$ and neutrons $(N)$,so $A = Z + N$.
Because the mass number $(A)$ differs while the atomic number $(Z)$ remains the same,the number of neutrons $(N = A - Z)$ must be different for isotopes.
Therefore,option $A$ is correct.

(C) The radius $R$ of a nucleus with mass number $A$ is given by the relation $R = R_0 A^{1/3}$,where $R_0$ is a constant.
Therefore,the ratio of the radii of two nuclei is given by $\frac{R_S}{R_{He}} = \left( \frac{A_S}{A_{He}} \right)^{1/3}$.
Given $A_S = 32$ and $A_{He} = 4$,we have:
$\frac{R_S}{R_{He}} = \left( \frac{32}{4} \right)^{1/3} = (8)^{1/3} = 2$.
Thus,the radius of the sulphur nucleus is $2$ times larger than that of the helium nucleus.

(B) 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: $m_1 v_1 = m_2 v_2$.
Given the velocity ratio $\frac{v_1}{v_2} = \frac{2}{1}$,we have $\frac{m_2}{m_1} = \frac{v_1}{v_2} = \frac{2}{1}$.
Since the density $\rho$ of nuclear matter is constant,the mass $m$ is proportional to the volume $V$,where $V = \frac{4}{3} \pi r^3$. Thus,$\frac{m_2}{m_1} = \frac{r_2^3}{r_1^3}$.
Equating the ratios: $\frac{r_2^3}{r_1^3} = \frac{2}{1}$.
Taking the cube root on both sides: $\frac{r_2}{r_1} = 2^{1/3}$.
Therefore,the ratio of their radii $r_1 : r_2 = 1 : 2^{1/3}$.

(A) The atomic mass number $M$ (often denoted as $A$) represents the total number of nucleons,which is the sum of the number of protons $(Z)$ and the number of neutrons $(N)$.
Mathematically,$M = Z + N$.
Therefore,the number of neutrons is given by $N = M - Z$.

(C) Inside the nucleus,protons are subject to two primary forces:
$1$. The electrostatic (Coulombic) force of repulsion due to their positive charges.
$2$. The strong nuclear force,which is an attractive force acting between nucleons (protons and neutrons) at short ranges.
Since both forces act between protons within the nucleus,the correct answer is $C$.

(D) For light nuclei,stability is achieved when the number of neutrons is approximately equal to the number of protons,i.e.,$N \approx Z$ or $\frac{N}{Z} = 1$.
For heavier nuclei,the electrostatic repulsion between protons increases,requiring more neutrons to provide additional strong nuclear force to maintain stability. Thus,for heavier nuclei,$N > Z$,which implies $\frac{N}{Z} > 1$.
Combining these two conditions,the general relation for a stable nucleus is $N \ge Z$.

(B) The nuclear force is the strongest force in nature and acts between nucleons (protons and neutrons) at short distances (approximately $1 \times 10^{-15} \ m$).
One of the fundamental properties of the nuclear force is that it is charge-independent.
This means that the strong nuclear force between two neutrons $(n-n)$,two protons $(p-p)$,or a proton and a neutron $(p-n)$ is identical in magnitude,provided the separation distance is the same.
Therefore,$F_1 = F_2 = F_3$.

(D) The radius $R$ of a nucleus is related to its mass number $A$ by the empirical formula:
$R = R_0 A^{1/3}$
where $R_0$ is a constant (approximately $1.2 \times 10^{-15} \ m$).
From this relation,it is clear that the radius $R$ is directly proportional to the cube root of the mass number $A$,i.e.,$R \propto A^{1/3}$.

(D) The nucleus of an atom contains only protons and neutrons.
For $_{11}^{23}\text{Na}$,the atomic number $Z = 11$ represents the number of protons.
The mass number $A = 23$ represents the total number of protons and neutrons.
The number of neutrons is calculated as $N = A - Z = 23 - 11 = 12$.
Electrons are located outside the nucleus,not inside it.
Therefore,the sodium nucleus contains $12$ neutrons.

(C) The atomic number $Z$ represents the number of protons and electrons in a neutral atom. For both $^{12}C$ and $^{14}C$,$Z = 6$,so both have $6$ protons and $6$ electrons.
The mass number $A$ is the sum of protons and neutrons $(A = Z + N)$.
For $^{12}C$: $A = 12$,so $N = 12 - 6 = 6$ neutrons.
For $^{14}C$: $A = 14$,so $N = 14 - 6 = 8$ neutrons.
Comparing the two,$^{14}C$ has $8 - 6 = 2$ extra neutrons and the same number of protons and electrons. Thus,option $C$ is correct.

(A) The nuclear force is a short-range force that acts between nucleons (protons and neutrons) within the nucleus.
It is effective only at distances on the order of $10^{-15} \, m$ (or $1 \, \text{fermi}$).
Beyond this range, the nuclear force becomes negligible compared to the electrostatic repulsion between protons.
Among the given options, $10^{-15} \, m$ is the characteristic range of the nuclear force. Since the original option A was $10^{-14} \, m$, which is the closest order of magnitude to the nuclear range, it is the correct choice.

(C) The nuclear reaction for the bombardment of a neutron $(_{0}n^{1})$ on Boron $(_{5}B^{10})$ is given by:
$_{5}B^{10} + _{0}n^{1} \to _{3}Li^{7} + _{2}He^{4}$
Here,$_{2}He^{4}$ represents the $\alpha$-particle emitted.
By balancing the atomic numbers $(5 + 0 = 3 + 2)$ and mass numbers $(10 + 1 = 7 + 4)$,we confirm the product nucleus is Lithium-$7$ $(_{3}Li^{7})$.

(D) The given nuclear reaction is $_2He^4 + _zX^A \to _{z+2}Y^{A+3} + A$.
Applying the law of conservation of mass number:
$4 + A = A + 3 + m$,where $m$ is the mass number of $A$.
$4 + A = A + 3 + m \Rightarrow m = 1$.
Applying the law of conservation of atomic number (charge):
$2 + z = z + 2 + q$,where $q$ is the atomic number of $A$.
$2 + z = z + 2 + q \Rightarrow q = 0$.
$A$ particle with mass number $1$ and atomic number $0$ is a neutron $(_{0}n^1)$.
Therefore,$A$ denotes a neutron.

(A) The nuclear reaction is given by: $_8^{18}O + _1^1H \rightarrow _9^{18}F + X$.
To balance the atomic number (charge): $8 + 1 = 9 + Z \Rightarrow Z = 0$.
To balance the mass number: $18 + 1 = 18 + A \Rightarrow A = 1$.
Thus,the particle $X$ is a neutron,represented as $_0^1n$.

(B) In a nuclear reaction,both the total mass number and the total atomic number must be conserved.
For the mass number (the superscript):
$24 + 4 = X + 1$
$28 = X + 1$
$X = 28 - 1$
$X = 27$
Therefore,the value of $X$ is $27$.

(C) Isotopes are atoms of the same element that have the same number of protons and electrons but a different number of neutrons.
For Uranium $(U)$,the atomic number $(Z)$ is $92$,which means both $^{235}U$ and $^{238}U$ have $92$ protons and $92$ electrons.
The number of neutrons $(N)$ is calculated as $N = A - Z$,where $A$ is the mass number.
For $^{238}U$: $N = 238 - 92 = 146$ neutrons.
For $^{235}U$: $N = 235 - 92 = 143$ neutrons.
Comparing the two,$^{238}U$ has $146 - 143 = 3$ more neutrons than $^{235}U$.

(C) neutrino is an elementary subatomic particle that interacts only via the weak subatomic force and gravity. It has no electric charge (it is chargeless) and possesses an intrinsic angular momentum (spin) of $1/2$. Therefore,the correct description is that it is chargeless and has spin.

(A) The given nuclear reaction is $X(n, \alpha) {_3Li^7}$.
This can be written as: $_ZX^A + _0n^1 \to _3Li^7 + _2He^4$.
According to the law of conservation of mass number and atomic number:
For atomic number $(Z)$: $Z + 0 = 3 + 2 \implies Z = 5$.
For mass number $(A)$: $A + 1 = 7 + 4 \implies A + 1 = 11 \implies A = 10$.
Thus,the nucleus $X$ is $_5X^{10}$,which corresponds to Boron-$10$ $(_{5}B^{10})$.

(B) 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.
For the given reaction: $_7N^{14} + _2He^4 \to X + _1H^1$
Let $X$ be represented as $_Z^AX$.
Conservation of atomic number $(Z)$: $7 + 2 = Z + 1 \implies 9 = Z + 1 \implies Z = 8$.
Conservation of mass number $(A)$: $14 + 4 = A + 1 \implies 18 = A + 1 \implies A = 17$.
Since the element with atomic number $8$ is Oxygen $(O)$,the product $X$ is $_8O^{17}$.

(D) The nuclear reaction can be represented as: $_8O^{16} + _1H^2 \to _Z^AX + _2He^4$.
Applying the law of conservation of mass number: $16 + 2 = A + 4$,which gives $A = 14$.
Applying the law of conservation of atomic number: $8 + 1 = Z + 2$,which gives $Z = 7$.
Thus,the product nucleus is $_7N^{14}$.

(A) The correct statement is that nuclei of different elements can have the same number of neutrons. These are known as isotones.
For example,consider $_4Be^9$ and $_5B^{10}$.
For $_4Be^9$: Number of neutrons $N = A - Z = 9 - 4 = 5$.
For $_5B^{10}$: Number of neutrons $N = A - Z = 10 - 5 = 5$.
Since both have $5$ neutrons,they are isotones.

(C) By definition, $1 \, \text{atomic mass unit (amu)}$ or $1 \, \text{u}$ is defined as exactly one-twelfth $(1/12)$ of the mass of an unbound neutral atom of carbon-$12$ at rest and in its ground state.
Therefore, $1 \, \text{u} = \frac{1}{12} \times (\text{mass of one } C-12 \text{ atom})$.
Thus, option $C$ is correct.

(B) Isotopes are atoms of the same element that have the same atomic number (number of protons) but different mass numbers due to a different number of neutrons in the nucleus.
Since chemical properties are primarily determined by the electronic configuration of an atom,and the electronic configuration depends on the number of protons (atomic number),isotopes exhibit identical chemical properties.