Class 12 Physics (Part 2) Chapter 26 Nuclear Structure

Key Concept: Quark Composition

c) Two down quarks and one up quark; total charge = $0$
[Solution Description]
A neutron consists of two down quarks (d) and one up quark (u). The charges are:
$\text{Up quark charge} = +\frac{2}{3}e \\ \\ \text{Down quark charge} = -\frac{1}{3}e \\ \\ \text{Total charge} = 2 \times \left(-\frac{1}{3}e\right) + \frac{2}{3}e = 0$
Thus, the neutron has zero net charge.

Key Concept: Energy equivalent calculation

c) 1863 MeV
[Solution Description]
The energy equivalent of 1 u is given as 931.5 MeV. To find the energy equivalent of 2 u, multiply 931.5 MeV by 2:
$2 \, \text{u} = 2 \times 931.5 \, \text{MeV} = 1863 \, \text{MeV}$

Key Concept: Radioactive Decay Series

b) 6
[Solution Description]
The total decrease in mass number is $238 - 206 = 32$. Each alpha decay reduces the mass number by 4, so the number of alpha decays is $\frac{32}{4} = 8$. The total decrease in atomic number is $92 - 82 = 10$. Each alpha decay reduces the atomic number by 2, contributing $8 \times 2 = 16$ to the atomic number decrease. Since beta decay increases the atomic number by 1, the number of beta decays must be $16 - 10 = 6$.

Key Concept: Nuclear size vs atomic size, Density independence

c) The nucleus is $10^5$ times smaller than the atom.
[Solution Description]
The nuclear diameter is $10^{-15}$ m, and the atomic diameter is $10^{-10}$ m. The ratio of sizes is $\frac{10^{-15}}{10^{-10}} = 10^{-5}$, meaning the nucleus is $10^5$ times smaller than the atom.
The nuclear density is independent of $A$ because the volume increases with $A$ as $R^3 \propto A$ (since $R = R_0 A^{1/3}$), and mass also increases linearly with $A$. Thus, density $\rho = \frac{mA}{\frac{4}{3}\pi R^3}$ remains constant for all nuclei.

Key Concept: Average Atomic Mass Calculation

b) 35.5
[Solution Description]
To find the average atomic mass, multiply each isotope's mass by its abundance and sum the results:
$\text{Average atomic mass} = (35 \times 0.75) + (37 \times 0.25)$
$\text{Average atomic mass} = 26.25 + 9.25 = 35.5$
The correct answer is 35.5.

Key Concept: \textbf{Isotopes, Isobars, Isotones}

c) $_{6}C^{13}$ and $_{7}N^{14}$
[Solution Description]
Isotones are nuclei with the same number of neutrons (A - Z). For example, $_{6}C^{13}$ has 6 protons and 7 neutrons (13 - 6 = 7), while $_{7}N^{14}$ has 7 protons and 7 neutrons (14 - 7 = 7). Thus, they are isotones. Now check the options to find an exactly matching pair.

Key Concept: Decay Series

c) 8
[Solution Description]
To find the number of alpha decays, compute the decrease in mass number:
$238 - 206 = 32$
Each alpha decay reduces the mass number by 4, so the number of alpha decays is:
$32 / 4 = 8$
Therefore, the correct answer is 8 alpha particles.

Key Concept: Shortcomings of Proton–Electron Hypothesis

c) Energy discrepancy in $\beta$-particle emissions
[Solution Description]
The uncertainty principle states that confining an electron within the nucleus ($\sim 10^{-15}$ m) would require energy around 100 MeV. However, $\beta$-particles (electrons emitted during nuclear decay) have energies only up to 2–3 MeV, which contradicts the hypothesis.

Key Concept: Nuclear Composition, Isobars

c) Its atomic number increases by 1, making it an isobar.
[Solution Description]
In $\beta^-$ decay, a neutron converts into a proton, emitting an electron and an antineutrino. The mass number ($A$) remains unchanged, but the atomic number ($Z$) increases by 1. Thus, the daughter nucleus is an isobar of the parent nucleus but belongs to a different element with different chemical properties.

Key Concept: Basic Nuclear Reaction

a) $_6C^{13}$
[Solution Description]
Balancing atomic numbers: 5 (Boron) + 2 (Helium) = Z (X) + 1 (Hydrogen). So, Z = 6.
Balancing mass numbers: 10 + 4 = A + 1. So, A = 13.
Therefore, $X$ is Carbon-13, written as $_6C^{13}$.

Key Concept: Isotones, Neutron Calculation

c) $_{5}B^{12}$
[Solution Description]
For $_{X}A^{Y}$, given neutrons = 8. So, $Y - X = 8$. To find isotones, we check $(A - Z)$ for other nuclei:
- $_{7}N^{15}$: $15 - 7 = 8$ (isotone).
- $_{8}O^{16}$: $16 - 8 = 8$ (isotone).
- $_{5}B^{12}$: $12 - 5 = 7$ (not isotone).
- $_{6}C^{14}$: $14 - 6 = 8$ (isotone).
Thus, $_{5}B^{12}$ is not an isotone.

Key Concept: Neutron Decay

c) Proton, electron, and antineutrino
[Solution Description]
A free neutron decays into a proton, an electron, and an antineutrino with the following reaction:
$\text{_0n^1 \rightarrow _1H^1 + _{-1}e^0 + \bar{\nu}}$

Key Concept: Nuclear Size and Density

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To analyze this assertion-reason pair:
1. The **Assertion** states that the nucleus is $10^5$ times smaller than the atom (size $\approx 10^{-15}$ m compared to $\approx 10^{-10}$ m) and has an extremely high density ($2 \times 10^{17}$ kg/m³ vs. water's density of $10^3$ kg/m³). This is correct based on experimental observations and nuclear physics principles.
2. The **Reason** explains the cause of the high nuclear density: nucleons (protons and neutrons) are tightly packed within a small volume. The formula $\rho = \frac{m}{\frac{4}{3}\pi R^3}$ calculates the nuclear density, and the radius dependence on mass number ($R = R_0 A^{1/3}$) ensures density remains nearly constant across nuclei. This reasoning logically and correctly explains the Assertion.
3. Thus, both Assertion and Reason are true, and the Reason provides the correct explanation for the Assertion.

Key Concept: Nucleus Density

c) $2 \times 10^{17}$
[Solution Description]
The density of the nucleus is given as:
$\rho = 2 \times 10^{17} \, \text{kg/m}^3$
This value is derived from the formula:
$\rho = \frac{mA}{\frac{4}{3}\pi R^3}$
Substituting the values, we get $\rho \approx 2 \times 10^{17} \, \text{kg/m}^3$.

Key Concept: Comparison with Ordinary Matter

d) $10^{15}$
[Solution Description]
The density of the nucleus is $2 \times 10^{17}$ kg/m$^3$ and the density of water is $1 \times 10^{3}$ kg/m$^3$. To find how many times denser the nucleus is compared to water, we divide the nuclear density by the density of water:
$\frac{2 \times 10^{17}}{1 \times 10^{3}} = 2 \times 10^{14}$
However, the syllabus states it is approximately $10^{15}$ times denser, which is a rounded estimate. Given the options, the closest correct answer is $10^{15}$.

Key Concept: Mass Number and Atomic Representation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the nucleus $_8^{16}O$ has 8 neutrons. To verify this, we use the reason provided: the formula $N = A - Z$. Here, $A = 16$ (mass number), and $Z = 8$ (atomic number). Substituting, we get $N = 16 - 8 = 8$ neutrons. Both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Nuclear Reactions, Conservation Laws

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that in a nuclear reaction, the total number of nucleons (protons + neutrons) is conserved. This is true as per the principle of nucleon number conservation in nuclear reactions. The reason states that nuclear reactions obey the law of conservation of mass-energy, which is also true. However, the reason does not explain why the total number of nucleons is conserved, as it refers to a different conservation law (mass-energy). Therefore, both are true, but the reason does not correctly explain the assertion.

Key Concept: Cross-sections, Charge independence

b) $2\sigma/3$
[Solution Description]
Charge independence predicts identical cross-sections for processes related by isospin symmetry. The initial states $pp$ (isospin $I=1$) and $np$ (isospin $I=0$ or $1$) must be decomposed into isospin states. For $I=1$, the matrix elements are identical. Using Clebsch-Gordan coefficients, the cross-section $\sigma(np \to \pi^0d)$ equals $\sigma(pp \to \pi^+d)$ multiplied by the squared ratio of Clebsch-Gordan coefficients: $\left(\sqrt{2/3}\right)^2 = 2/3$.

Key Concept: Energy Equivalent of Atomic Mass Unit

b) 1863 MeV
[Solution Description]
Given that $1 \, \text{u} = 931.5 \, \text{MeV}$, the energy equivalent of 2 u is calculated as:
$\text{Energy} = 2 \times 931.5 \, \text{MeV} = 1863 \, \text{MeV}$
The energy equivalent of 2 u is 1863 MeV.

Key Concept: Nuclide Notation and Modern Understanding

b) Number of protons in the nucleus.
[Solution Description]
In nuclide notation $\_Z^A X$, the subscript $Z$ represents the atomic number, which is the number of protons in the nucleus. This is a key identifier of the element and determines its chemical properties. The superscript $A$ represents the mass number (total nucleons, i.e., protons + neutrons).

Key Concept: Short-range nature of nuclear forces

c) Nucleons interact only with their nearest neighbors due to short-range forces.
[Solution Description]
Nuclear forces are short-range and saturate, meaning they act only over very small distances (~1-2 fm) and nucleons interact primarily with their immediate neighbors. This limited interaction range leads to the saturation effect, ensuring stable nuclear density and binding energy per nucleon.

Key Concept: Proton-Neutron Composition

b) 1 proton and 1 neutron
[Solution Description]
Deuterium has 1 proton and 1 neutron. Its atomic number ($Z$) is 1 (number of protons), and its mass number ($A$) is 2 (protons + neutrons). Hence, the correct representation is 1 proton and 1 neutron.

Key Concept: Nuclear Density

c) It is approximately the same for all nuclei.
[Solution Description]
Nuclear density is independent of the mass number and is approximately constant for all nuclei, with a value of around \$2 \times 10^{17} \, \text{kg/m}^3\$. Thus, it does not vary with the size or mass of the nucleus.

Key Concept: Mass number calculation, neutron count

c) 17
[Solution Description]
We know that mass number \$A = Z + N\$ where \$N\$ is number of neutrons.
Given \$A = 37\$ and \$N = 20\$.
To find \$Z\$, rearrange the formula:
\$Z = A - N = 37 - 20 = 17\$.

Key Concept: Nuclear Shape

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that most nuclei are spherical due to short-range nuclear forces. This is true as these forces tend to bind nucleons tightly, leading to a spherical shape for stability. The reason explains that the spherical shape minimizes surface area and thus reduces surface energy, which is indeed a key factor in nuclear binding energy. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Nuclear Reactions, Proton Discovery

d) Proton
[Solution Description]
Balancing the equation:
Atomic numbers: 7 + 2 = 8 + Z ⇒ Z = 1
Mass numbers: 14 + 4 = 17 + A ⇒ A = 1
Thus, X is $_1H^1$ which is a proton. This reaction led to proton discovery.

Key Concept: Nuclear Density

b) 4.4 \times 10^{16} \, \text{kg/m}^3
[Solution Description]
Nuclear density is given by $\rho = \frac{3m}{4\pi R^3}$. Substitute the given values:
$\rho = \frac{3 \times 1.67 \times 10^{-27} \, \text{kg}}{4\pi (3.07 \times 10^{-15} \, \text{m})^3}$
Calculate the denominator first:
$4\pi (3.07 \times 10^{-15})^3 \approx 1.14 \times 10^{-43} \, \text{m}^3$
Now compute the density:
$\rho = \frac{5.01 \times 10^{-27}}{1.14 \times 10^{-43}} \approx 4.39 \times 10^{16} \, \text{kg/m}^3$
The closest option is \$4.4 \times 10^{16}\$ kg/m$^3$.

Key Concept: Neutron Count

a) $M - Z$
[Solution Description]
The mass number ($M$) is the sum of protons ($Z$) and neutrons ($N$). Thus, the number of neutrons is calculated by subtracting the atomic number from the mass number: $N = M - Z$.

Key Concept: Neutron-to-proton ratio, radioactivity

c) $_{11}Na^{24}$ has an excess of neutrons, making it unstable.
[Solution Description]
The question compares $_{11}Na^{23}$ (stable) and $_{11}Na^{24}$ (radioactive). Stability depends on the neutron-to-proton ratio. $_{11}Na^{23}$ has 12 neutrons (23 - 11 = 12), giving a balanced ratio. $_{11}Na^{24}$ has 13 neutrons (24 - 11 = 13), creating an excess of neutrons, making it unstable and prone to beta decay to $_{12}Mg^{24}$.
Step 1: Calculate neutrons - $_{11}Na^{23}$ has 12 neutrons, $_{11}Na^{24}$ has 13 neutrons.
Step 2: Compare ratios - $_{11}Na^{23}$ has a balanced ratio; $_{11}Na^{24}$ has an excess of neutrons.
Step 3: Conclude - $_{11}Na^{24}$ undergoes beta decay due to the unstable neutron-to-proton ratio.

Key Concept: Discovery of Neutron

d) James Chadwick
[Solution Description]
James Chadwick discovered the neutron in 1932 by bombarding beryllium-9 with alpha particles, leading to the nuclear reaction:
$_{4}^{9}\text{Be} + _{2}^{4}\text{He} \rightarrow _{6}^{12}\text{C} + _{0}^{1}\text{n}$

Key Concept: Rutherford’s α-particle scattering experiment, Nuclear radius

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that backscattering confirms the negligible volume of the nucleus relative to the atom. This is true because most α-particles passed undeflected, showing the atom is mostly empty space, while only a few were backscattered, proving the nucleus is tiny. The reason provides the formula for nuclear radius, $R = R_0 A^{1/3}$, which mathematically explains why the nucleus is extremely small (order of $10^{-15}$ m). Thus, the reason correctly explains the assertion using the given mathematical relationship.

Key Concept: Average atomic mass calculation, Isotopic abundance

b) 35.453 u
[Solution Description]
The average atomic mass is calculated as a weighted average of the isotopic masses:
$\text{Average mass} = \left( \frac{75.77 \times 34.96885 + 24.23 \times 36.96590}{100} \right) \text{ u}$
$= \left( \frac{2649.4996 + 895.6918}{100} \right) \text{ u} = 35.453 \text{ u}$

Key Concept: Magnetic Moment: Predicted vs Observed

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the proton-electron hypothesis predicted a nuclear magnetic moment much larger than observed. This is true because electrons have a much larger magnetic moment compared to protons and neutrons. The Reason correctly explains that the observed nuclear magnetic moment is significantly smaller than $\mu_B$, which supports the conclusion that electrons are not present inside the nucleus. Thus, both statements are true, and the Reason correctly explains the Assertion.

Key Concept: Isotones

c) $_3Li^7$ and $_4Be^8$
[Solution Description]
Isotones are nuclei with the same number of neutrons ($A - Z$).
Checking the options:
- $_3Li^7$ and $_4Be^8$:
For $_3Li^7$, neutrons = 7 - 3 = 4.
For $_4Be^8$, neutrons = 8 - 4 = 4.
Both have the same neutron count.
Hence, the correct answer is the pair $_3Li^7$ and $_4Be^8$.

Key Concept: Angular Momentum

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Angular momentum ($L$) is given by $L = I\omega$, where $I$ is the moment of inertia and $\omega$ is the angular velocity. In an ideal closed system with no external torque, angular momentum is conserved. However, if external torques act on the system, they can alter the angular momentum, causing a discrepancy between the calculated value (assuming no external torque) and the experimental value (affected by real-world conditions).
Here:
- **(A)** is true because external torques can indeed lead to differences between calculated and experimental values.
- **(R)** is also true because external torques influence angular momentum as per $\tau = \frac{dL}{dt}$.
- **(R)** correctly explains **(A)** since external torques directly cause the difference in angular momentum.

Key Concept: Heisenberg's Uncertainty Principle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Heisenberg Uncertainty Principle states that the product of the uncertainties in position ($\Delta x$) and momentum ($\Delta p$) must be at least $\frac{\hbar}{2}$. If we confine an electron to a very small space, such as $10^{-15}$ m, then $\Delta x \approx 10^{-15}$ m. Using the principle:
$\Delta p \geq \frac{\hbar}{2 \Delta x}$
Substituting $\hbar \approx 1.05 \times 10^{-34}$ Js and $\Delta x = 10^{-15}$ m:
$\Delta p \geq \frac{1.05 \times 10^{-34}}{2 \times 10^{-15}} = 5.25 \times 10^{-20}\ \text{kg m/s}$
This means the uncertainty in momentum becomes extremely high, implying that the electron would require a very large kinetic energy, making such confinement practically impossible without violating quantum mechanical principles. Hence, both the assertion and reason are correct, and the reason explains the assertion.

Key Concept: Shortcomings of Proton-Electron Hypothesis

a) Heisenberg's uncertainty principle
[Solution Description]
The uncertainty principle states that confining an electron within the small size of the nucleus ($10^{-15}$ m) would require an energy of about 100 MeV. However, $\beta$-particles emitted from the nucleus have much lower energy (2-3 MeV), suggesting electrons cannot exist inside the nucleus.

Key Concept: Nucleus Size and Density

b) $10^{5}$
[Solution Description]
The size of the nucleus is about $10^{-15}$ meters, while the size of the atom is approximately $10^{-10}$ meters. To find how many times smaller the nucleus is compared to the atom, we divide the size of the atom by the size of the nucleus:
$\frac{10^{-10}}{10^{-15}} = 10^{5}$
So, the nucleus is $10^{5}$ times smaller than the atom.

Key Concept: Neutral atoms and ion formation

b) The atomic number remains 12, and the charge becomes $+2$
[Solution Description]
The atomic number is determined solely by the number of protons (12), which does not change when electrons are removed. Removing 2 electrons results in a net positive charge of $+2$ due to excess protons.

Key Concept: Nuclear Reactions - Conservation Laws

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that the total number of nucleons is conserved in a nuclear reaction, which is true according to the conservation laws. However, the reason states that the mass number (A) remains the same before and after the reaction, which is incorrect because the mass number can change if new particles are emitted or absorbed (e.g., an alpha particle emission changes the mass number by 4). Thus, Assertion is true, but Reason is false.

Key Concept: Alpha-proton reaction, Conservation laws

b) $_8O^{17}$
[Solution Description]
The given reaction is:
$_7N^{14} + _2He^4 \rightarrow Y + _1H^1$.
Apply conservation of atomic number ($Z$) and mass number ($A$):
For $Z$: $7 + 2 = Z + 1 \Rightarrow Z = 8$.
For $A$: $14 + 4 = A + 1 \Rightarrow A = 17$.
Thus, $Y$ is $_8O^{17}$ (Oxygen-17).

Key Concept: Typical Masses

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
From the given data, the mass of a neutron is $1.008665 \, \text{u}$, and the mass of a proton is $1.007276 \, \text{u}$. Therefore, the mass of a neutron is indeed greater than that of a proton, making the Assertion true. The Reason states that a neutron has no charge, while a proton does, which is also true. However, the mass difference arises due to the composition of quarks and binding energies, not directly due to the presence or absence of charge. Hence, the Reason does not correctly explain the Assertion.

Key Concept: Neutron Number Calculation

b) 5
[Solution Description]
Since $_{5}B^{10}$ is an isotone of $_{4}Be^{9}$, their neutron numbers must be equal.
Neutrons in $_{4}Be^{9}$:
$9 - 4 = 5$
Therefore, neutrons in $_{5}B^{10}$ must also be 5.

Key Concept: Nuclear Composition

b) 30
[Solution Description]
The mass number $A$ is 56 and the atomic number $Z$ is 26.
The number of neutrons $N$ is calculated as $N = A - Z$.
Therefore, $N = 56 - 26 = 30$.

Key Concept: Nuclear Structure and Composition

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The nuclear density $\rho$ is given by the formula $\rho = \frac{\text{mass}}{\text{volume}}$. The mass of the nucleus is approximately proportional to the mass number $A$ (since nucleons have nearly equal mass). The volume of the nucleus is calculated using the formula $V = \frac{4}{3}\pi R^3$, where $R = R_0 A^{1/3}$. Substituting, we get $V = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$. Thus, the volume is directly proportional to $A$. Now, $\rho = \frac{A m_p}{\frac{4}{3}\pi R_0^3 A} = \frac{m_p}{\frac{4}{3}\pi R_0^3}$, which is independent of $A$. Therefore, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Conservation of charge and nucleon number

b) $_{2}He^{4}$
[Solution Description]
Step 1: Check conservation of atomic number (charge).
Left side: $5$ (Boron) + $0$ (neutron) = $5$.
Right side: $3$ (Lithium) + atomic number of X. So, atomic number of X = $5 - 3 = 2$ (Helium).
Step 2: Check conservation of mass number.
Left side: $11$ (Boron) + $1$ (neutron) = $12$.
Right side: $8$ (Lithium) + mass number of X. So, mass number of X = $12 - 8 = 4$.
Therefore, X must be $_{2}He^{4}$ (alpha particle).

Key Concept: Nuclear Stability, Nuclear Forces

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For light nuclei (Z < 20), stability is typically achieved when the number of protons and neutrons is approximately equal (N ≈ Z). The nucleus $^{12}_6C$ has 6 protons and 6 neutrons (N/Z = 1), which is close to the stability line for light nuclei. On the other hand, $^{14}_6C$ has 6 protons and 8 neutrons (N/Z ≈ 1.33), which deviates from the ideal ratio, making it less stable. Thus, both Assertion and Reason are true, and the Reason correctly explains why $^{12}_6C$ is more stable than $^{14}_6C$.

Key Concept: Nuclear Radius, Nuclear Density

c) Assertion is true, but Reason is false.
[Solution Description]
The Assertion states that the nuclear density of gold is approximately equal to that of silver. This is true because nuclear density $\rho$ is given by:
$\rho = \frac{3m}{4\pi R_0^3}$
where $m$ is the average mass of a nucleon and $R_0$ is an empirical constant. Since $\rho$ does not depend on the mass number $A$, it remains nearly the same for all nuclei.
The Reason claims that nuclear density is inversely proportional to the mass number $A$. This is false because, as shown in the formula, $\rho$ is independent of $A$.
Therefore, the Assertion is true, but the Reason is false.

Key Concept: Key Observations from Rutherford’s Experiment

a) Passed straight through with no deflection
[Solution Description]
According to Rutherford’s experiment, most $\alpha$-particles passed straight through the gold foil without any deflection. This observation indicated that most of the atom is empty space because there was no dense or charged region to deflect the $\alpha$-particles.

Key Concept: Energy Equivalent

b) 931.5 MeV
[Solution Description]
According to Einstein's equation $\Delta E = (\Delta m) c^2$, the energy equivalent for 1 u is calculated as:
$\Delta E = (1.6605 \times 10^{-27} \, \text{kg}) (2.9979 \times 10^8 \, \text{ms}^{-1})^2$.
The result is $1.4924 \times 10^{-10}\, \text{J}$.
Since $1\, \text{MeV} = 1.6022 \times 10^{-13}\, \text{J}$, converting gives:
$\frac{1.4924 \times 10^{-10}}{1.6022 \times 10^{-13}} = 931.5\, \text{MeV}$.
Therefore, 1 u is equivalent to 931.5 MeV.

Key Concept: Saturation effect

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description] The assertion states that the binding energy per nucleon becomes nearly constant for heavy nuclei, which is true because nuclear forces exhibit saturation. The reason explains this by stating that each nucleon interacts only with its near neighbours due to the short-range nature of nuclear forces, which is also true and correctly explains the assertion. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Nuclear Density

c) $2 \times 10^{17} \, \text{kg/m}^3$
[Solution Description]
Nuclear density is given by:
$\rho = \frac{3m}{4\pi R_0^3}$
Substituting values:
$\rho = \frac{3 \times 1.67 \times 10^{-27}}{4\pi (1.2 \times 10^{-15})^3} \approx 2 \times 10^{17} \, \text{kg/m}^3$.

Key Concept: Nuclide Notation

b) $^{14}_6C$
[Solution Description]
The atomic number $Z$ represents the number of protons, which is 6 for carbon. The mass number $A$ is the sum of protons and neutrons, so $A = 6 + 8 = 14$. The nuclide notation is written as $^A_ZX$, where $X$ is the chemical symbol of the element. For carbon, the symbol is $C$. Therefore, the correct notation is $^{14}_6C$.

Key Concept: Atomic Number and Mass Number

c) 8
[Solution Description]
The number of neutrons is calculated using the formula:
Number of neutrons = Mass number (\$A\$) - Atomic number (\$Z\$).
Here, \$A = 14\$ and \$Z = 6\$, so:
Neutrons = \$14 - 6 = 8\$.
Therefore, there are 8 neutrons in the nucleus.

Key Concept: Nuclear Size

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the nuclear radius is of the order of $10^{-15}$ m, which is correct as nuclear radii typically range from $1-10$ fermi ($1 \text{ F} = 10^{-15} \text{ m}$). The Reason provides the formula for nuclear radius, $R = R_0 A^{1/3}$, which is also correct and explains why the nuclear radius is of the order of $10^{-15}$ m (since $R_0 \approx 1.2 \times 10^{-15}$ m). Hence, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Quark Composition, Charge Calculation

a) Before: 0, After: 0
[Solution Description]
Neutron ($_0^1 n$) has zero charge initially. After decay, it becomes a proton ($_1^1 H$), which has charge $+e$, an electron ($_{-1}^0 e$), which has charge $-e$, and an antineutrino ($\overline{\nu}$), which is neutral. The total final charge is $+e - e + 0 = 0$. Thus, the charge is conserved.

Key Concept: Nuclear Shape

c) Most nuclei are mostly spherical but may have minor deviations from perfect sphericity.
[Solution Description]
Most nuclei are mostly spherical, with minor deviations (about 10\%) from perfect sphericity in some cases.

Key Concept: Neutron Properties, Nuclear Reactions

d) 13
[Solution Description]
The reaction is balanced as: $_4^A Be + _2^4 He \rightarrow _8^{16} O + _0^1 n$. Conservation of mass number: $A + 4 = 16 + 1 \Rightarrow A = 13$. So, the initial beryllium isotope must have mass number 13.

Key Concept: Nuclear Composition

a) To balance the positive charge of the nucleus with negatively charged electrons.
[Solution Description]
The helium nucleus was thought to have 4 protons to account for its mass (since each proton has a mass approximately equal to 1 atomic mass unit). However, having 4 protons would make the charge +4e, but the observed charge was only +2e. Hence, it was hypothesized that 2 electrons were present inside the nucleus to neutralize 2 protons' charges, bringing the net charge to +2e.

Key Concept: Isotopes and Radioactivity

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. Since chemical properties depend on the electron configuration, which is determined by the number of protons (atomic number), isotopes share the same chemical properties. The Reason correctly explains the Assertion as it identifies the underlying cause (same number of protons) for the similarity in chemical properties. Thus, both Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Exchange nature of nuclear forces

d) Mesons possess mass, limiting the force range
[Solution Description]
The range of the nuclear force due to meson exchange is limited because mesons have mass. According to quantum field theory, massive exchange particles lead to a Yukawa potential with a finite range $r \approx \hbar / (mc)$. Pions, being the lightest mesons, determine the longest possible range, which is still short (about 1.4 fm).

Key Concept: Neutron Properties

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a neutron has a mass approximately equal to that of a proton, which is true as both have nearly the same mass. The reason provides the exact mass of a neutron, which is also correct and directly supports the assertion. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Nuclear Reactions and Nucleon Counting

b) Neutron, 1
[Solution Description]
First balance the atomic numbers (protons):
$4 (Be) + 2 (He) = 6 (C) + Z_X$
$6 = 6 + Z_X \Rightarrow Z_X = 0$
Then balance mass numbers (nucleons):
$9 + 4 = 12 + A_X$
$13 = 12 + A_X \Rightarrow A_X = 1$
Since Z=0 and A=1, the particle must be a neutron ($_0n^1$), which has 1 nucleon.

Key Concept: Atomic Number

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The atomic number (Z) represents the number of protons in an atom's nucleus. For carbon, this number is indeed 6, which means there are 6 protons in its nucleus. Therefore, both the Assertion and Reason are true. Additionally, the Reason correctly explains why the atomic number of carbon is 6, as it directly states the relation between atomic number and proton count.

Key Concept: Isotopes

a) $_{1}H^{1}$ and $_{1}H^{2}$
[Solution Description]
Isotopes are atoms of the same element with the same atomic number ($Z$) but different mass numbers ($A$). In the given options, only one pair has the same $Z$ but different $A$.
Option a) $_{1}H^{1}$ and $_{1}H^{2}$: Both are hydrogen atoms with $Z = 1$, but $A$ differs (1 and 2). They are isotopes.
Option b) $_{6}C^{12}$ and $_{7}N^{14}$: Different elements ($Z$ differs), not isotopes.
Option c) $_{3}Li^{7}$ and $_{4}Be^{8}$: Different elements ($Z$ differs), not isotopes.
Option d) $_{2}He^{4}$ and $_{1}H^{3}$: Different elements ($Z$ differs), not isotopes.

Key Concept: Isospin symmetry, Charge independence

a) Strong force between protons and neutrons
[Solution Description]
Charge independence implies that strong interactions are unaffected by the electric charge of particles. Isospin symmetry treats protons and neutrons as two states of the same particle (nucleon), with the strong force being invariant under isospin rotations. For example, the $\pi^+$, $\pi^-$, and $\pi^0$ mesons form an isospin triplet with nearly identical masses due to charge-independent strong interactions. Coulomb forces violate this symmetry.

Key Concept: Nuclear Size Calculation

b) 3.6 F
[Solution Description]
The radius $R$ of a nucleus is given by the formula:
$R = R_0 A^{1/3}$
Substituting $A = 27$ and $R_0 = 1.2 \times 10^{-15} \text{ m}$:
$R = 1.2 \times 10^{-15} \times (27)^{1/3}$
$27^{1/3} = 3 \quad (\text{since } 3^3 = 27)$
$R = 1.2 \times 10^{-15} \times 3 = 3.6 \times 10^{-15} \text{ m}$
Convert to fermi: $3.6 \text{ F}$.

Key Concept: Einstein's mass-energy equivalence

b) 465.75 MeV
[Solution Description]
Using Einstein's equation $\Delta E = (\Delta m)c^2$ and converting the mass loss to energy equivalent:
$\Delta E = 0.5 \, \text{u} \times 931.5 \, \text{MeV/u} = 465.75 \, \text{MeV}$

Key Concept: Assertion-Reason

c) Assertion is true, but Reason is false.
[Solution Description]
The Assertion states that the mass number ($A$) of an atom is always greater than its atomic number ($Z$). This is true because the mass number is the sum of protons and neutrons ($A = Z + N$), where $N$ represents the number of neutrons. For most atoms (except hydrogen-1, which has no neutrons), $N > 0$, so $A > Z$.
The Reason states that the mass number includes neutrons, which are more in number than protons for most atoms. While it's true that $A$ includes neutrons ($N$), the claim that neutrons are "more in number than protons" is not universally true—it depends on the element and isotope. For example, in carbon-12, $Z = 6$ and $N = 6$, so they are equal. Thus, the Reason is false.
Therefore, Assertion is true, but Reason is false.

Key Concept: Isobars

c) They have the same mass number
[Solution Description]
Isobars are nuclei with the same mass number (A) but different atomic numbers (Z). Hence, they belong to different elements and have different chemical properties. Let's verify the options:
- They have the same number of nucleons (since A is the same for both).
- They cannot have the same number of protons (different Z).
- They cannot have the same number of neutrons unless it's a coincidence.
- Their binding energies may or may not be equal.
So, the only consistent truth is that they have the same mass number.

Key Concept: Isotopes and Isobars

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that $_{6}C^{14}$ and $_{7}N^{14}$ are isotopes. However, isotopes are atoms of the same element with the same atomic number but different mass numbers. Here, $_{6}C^{14}$ has an atomic number of 6, while $_{7}N^{14}$ has an atomic number of 7, so they belong to different elements. Therefore, these are isobars (same mass number but different atomic numbers), not isotopes. The reason correctly defines isotopes but does not apply to the assertion. Hence, the assertion is false, but the reason is true.

Key Concept: Neutron interaction, slow vs fast neutrons

c) Slow neutrons spend more time near the nucleus, increasing capture probability.
[Solution Description]
Slow neutrons are more effective in causing nuclear reactions because their lower kinetic energy allows them to spend more time near the nucleus, increasing the probability of capture. Fast neutrons move quickly past the nucleus, reducing their chance of interaction. Additionally, slow neutrons are less likely to be scattered elastically due to their lower energy, making them more efficient for inducing fission.

Key Concept: Conservation Laws in Nuclear Reactions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify the assertion and reason:
1. Assertion checks if charge is conserved in the given nuclear reaction. Left side charge = $7 + 2 = 9$. Right side charge = $8 + 1 = 9$. Thus, charge is conserved.
2. Reason states that the sum of atomic numbers on both sides is equal, which is true ($9 = 9$).
3. The reason correctly explains why the assertion is true because charge conservation in nuclear reactions is mathematically expressed by the equality of the sum of atomic numbers.
Hence, both are true, and Reason is the correct explanation of Assertion.

Key Concept: Number of Neutrons Identification

c) The number of neutrons is equal to the mass number minus the atomic number.
[Solution Description]
The correct relationship amongst $A$, $Z$, and $N$ is given by the formula $N = A - Z$. This means the number of neutrons is obtained by subtracting the atomic number from the mass number. Hence, the statement that correctly describes this relationship is "The number of neutrons is equal to the mass number minus the atomic number."

Key Concept: Energy Equivalent, Unified Atomic Mass Unit

b) 217.0 MeV
[Solution Description]
The mass defect is given as $0.233 \, \text{u}$. Using the energy-mass equivalence formula:
$E = (\Delta m) c^2$
Since $1 \, \text{u} = 931.5 \, \text{MeV}$, the binding energy can be calculated as:
$E = 0.233 \, \text{u} \times 931.5 \, \text{MeV/u} = 217.04 \, \text{MeV}.$
Therefore, the binding energy is approximately $217.0 \, \text{MeV}$.

Key Concept: Multi-concept application

c) 75\%
[Solution Description]
Let \$x\$ be fraction of isotope-35, then fraction of isotope-37 is \$(1-x)\$.
Average atomic mass = (35 × \$x\$) + (37 × \$(1-x)\$) = 35.5
Solve: 35\$x\$ + 37 - 37\$x\$ = 35.5
37 - 2\$x\$ = 35.5
2\$x\$ = 1.5
\$x\$ = 0.75 or 75\%

Key Concept: Isobars, Nuclear Reactions

b) A proton with mass 1
[Solution Description]
Balancing the atomic and mass numbers:
For atomic number:
$7 + 2 = 8 + Z_X \implies Z_X = 1$
For mass number:
$14 + 4 = 17 + A_X \implies A_X = 1$
Thus, $X$ is a proton ($_1^1\text{H}$).

Key Concept: Nuclear Density, Nuclear Size

c) Becomes $1.26$ times the original
[Solution Description]
Nuclear density $\rho = \frac{3m}{4\pi R_0^3}$ is independent of mass number $A$.
Therefore, even if $A$ is doubled, the density remains unchanged.
But the radius $R = R_0 A^{1/3}$ depends on $A$.
Original radius: $R_1 = R_0 A^{1/3}$.
New radius: $R_2 = R_0 (2A)^{1/3} = 2^{1/3} R_0 A^{1/3} = 2^{1/3} R_1 \approx 1.26 R_1$.
Hence, the new radius is approximately $1.26$ times the original radius.

Key Concept: Implications of saturation effect

d) Because each nucleon interacts only with near neighbours
[Solution Description]
The saturation effect implies that each nucleon interacts only with its immediate neighbours. Hence, as the number of nucleons increases, the binding energy per nucleon does not significantly increase because additional nucleons do not contribute much to the total binding energy.

Key Concept: Nuclear vs Atomic Radius

c) $10^{15}$
[Solution Description]
The radius of an atom is $\approx 10^{-10}$ m, while the nuclear radius is $\approx 10^{-15}$ m. The volume ratio is calculated as:
$\left( \frac{10^{-10}}{10^{-15}} \right)^3 = 10^{15}$.

Key Concept: Nuclear Density

b) $2 \times 10^{14}$ times greater
[Solution Description]
Given that nuclear density is $\rho \approx 2 \times 10^{17} \, \text{kg/m}^3$ and the density of water is $1 \times 10^3 \, \text{kg/m}^3$, the ratio of nuclear density to water density is:
$\frac{\rho_{\text{nucleus}}}{\rho_{\text{water}}} = \frac{2 \times 10^{17}}{1 \times 10^3} = 2 \times 10^{14}$
Thus, nuclear density is $2 \times 10^{14}$ times greater than water density.

Key Concept: Example of Isobars

b) $_{1}H^{3}$ and $_{2}He^{3}$
[Solution Description]
Isobars have the same mass number ($A$) but different atomic numbers ($Z$). Among the given options, only $_{1}H^{3}$ and $_{2}He^{3}$ have the same mass number (3) but different atomic numbers (1 and 2 respectively).

Key Concept: Nuclear Notation

b) $_3^7 X$
[Solution Description]
The nuclear notation for an element is given as $_Z^A X$, where $Z$ is the atomic number and $A$ is the mass number. For this element, $Z = 3$ and $A = 7$. Thus, the correct notation is $_3^7 X$.

Key Concept: Exchange Nature of Nuclear Forces

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion (A) states that the exchange of mesons between nucleons is responsible for the strong attractive nuclear force, which is correct as per the syllabus. The reason (R) states that meson exchange explains the short-range and strongly attractive nature of nuclear forces, which is also correct and directly explains the assertion. Hence, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Nuclear magneton, proton-neutron model

b) $\mu_{Nn} = \frac{\mu_N}{1.0014}$
[Solution Description]
The nuclear magneton $\mu_N$ is given by:
$\mu_N = \frac{e\hbar}{2m_p}$
Similarly, the neutron magneton would be:
$\mu_{Nn} = \frac{e\hbar}{2m_n}$
Since $m_n \approx 1.0014 m_p$, we can write:
$\mu_{Nn} = \frac{e\hbar}{2 \times 1.0014 m_p} = \frac{\mu_N}{1.0014}$
Therefore, $\mu_{Nn}$ is slightly smaller than $\mu_N$ by a factor of approximately $1.0014$.

Key Concept: Nuclear Density

a) $\frac{1}{2}$
[Solution Description]
The nuclear radius is given by $R = R_0 A^{1/3}$. Therefore, the ratio of radii is:
$\frac{R_1}{R_2} = \frac{R_0 A_1^{1/3}}{R_0 A_2^{1/3}} = \left(\frac{A_1}{A_2}\right)^{1/3}$
Substitute $A_1 = 27$ and $A_2 = 216$:
$\frac{R_1}{R_2} = \left(\frac{27}{216}\right)^{1/3} = \left(\frac{1}{8}\right)^{1/3} = \frac{1}{2}$

Key Concept: Ratio of Nuclear Radii

b) $1.1 \, R_C$
[Solution Description]
The nuclear radius relation is:
$R = R_0 A^{1/3}$
For carbon, $R_C = R_0 (12)^{1/3}$. For oxygen, $R_O = R_0 (16)^{1/3}$. Taking the ratio:
$\frac{R_O}{R_C} = \left( \frac{16}{12} \right)^{1/3} = \left( \frac{4}{3} \right)^{1/3} \approx 1.100$
Thus, $R_O = 1.100 \, R_C$.

Key Concept: Isotopes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Isotopes of an element have the same atomic number (same number of protons and electrons), but different mass numbers (different number of neutrons). Since chemical properties are determined by the electron configuration, which is identical in isotopes due to the same number of electrons, their chemical properties are the same. Thus, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Key Concept: Nuclear Density, Nuclear Size

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The nuclear density $\rho$ is given by $$\rho = \frac{3m}{4\pi R_0^3}$$ where $m$ is the average mass of a nucleon and $R_0$ is the empirical constant. This formula shows that nuclear density does not depend on the mass number $A$, meaning all nuclei have approximately the same density regardless of their size. Thus, the Assertion is true as Uranium-238 and Carbon-12 will have the same nuclear density. The Reason correctly explains why the Assertion is true because it provides the formula showing the independence of $\rho$ from $A$.

Key Concept: Alpha Decay

a) $_{82} Pb^{214}$
[Solution Description]
An alpha particle is $_{2} He^{4}$. When $_{84} Po^{218}$ emits an alpha particle, its atomic number decreases by 2 ($84 - 2 = 82$) and its mass number decreases by 4 ($218 - 4 = 214$). Thus, the resulting nucleus is $_{82} Pb^{214}$.

Key Concept: Neutron Discovery

d) James Chadwick; $\text{_4Be^9 + _2He^4 \rightarrow _6C^{12} + _0n^1}$
[Solution Description]
James Chadwick discovered the neutron by bombarding beryllium with $\alpha$-particles. The nuclear reaction that took place was:
$\text{_4Be^9 + _2He^4 \rightarrow _6C^{12} + _0n^1}$

Key Concept: Proton–Neutron Hypothesis

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For the nuclide $^{23}_{11}Na$, the atomic number $Z = 11$ (number of protons) and the mass number $A = 23$. According to the Proton–Neutron Hypothesis, the number of neutrons ($N$) is calculated as $N = A - Z = 23 - 11 = 12$. Therefore, the Assertion is true as it correctly states the mass number and the number of neutrons. The Reason is also true because it correctly defines the relationship between mass number, atomic number, and number of neutrons. Moreover, the Reason provides the correct explanation for the Assertion.

Key Concept: Exchange nature $\to$ due to meson exchange

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that nuclear forces decrease significantly beyond 1-2 femtometers due to meson exchange having a limited range. The reason explains that mesons have finite mass and hence the range of the force they mediate is inversely proportional to their mass.
Step-by-step solution:
1. Nuclear forces are mediated by mesons (e.g., pions) as per quantum field theory.
2. According to Yukawa's theory, the potential of the nuclear force can be approximated as $V(r) \propto \frac{e^{-mr}}{r}$, where $m$ is the mass of the meson and $r$ is the distance between nucleons.
3. From this potential, the effective range of the force is given by $\lambda \approx \frac{\hbar}{mc}$, which shows that the range is indeed inversely proportional to the mass of the meson.
4. Since mesons like pions have a finite mass ($m_{\pi} \approx 140$ MeV/c$^2$), the predicted range is about $\lambda \approx 1.4$ fm, which matches the observed short-range nature of nuclear forces (~1-2 fm).
5. Thus, the assertion is true, and the reason correctly explains why meson exchange leads to short-range nuclear forces.
Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Nature of Nuclear Forces

c) Strongest and short-range
[Solution Description]
Nuclear forces are the strongest among fundamental forces but act only over extremely short ranges, unlike gravitational or electromagnetic forces, which have infinite range.

Key Concept: Nuclear Shape

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that most nuclei are spherical, which is correct as per the syllabus, with minor deviations (\~10\%). The reason provides the formula for nuclear radius, which is accurate but does not directly explain why nuclei are spherical. Therefore, both the assertion and reason are true, but the reason does not correctly explain the assertion.

Key Concept: Nuclear Reactions, Conservation Laws

d) Assertion is false, but Reason is true.
[Solution Description]
To verify the conservation laws in the given nuclear reaction, we analyze the atomic numbers (charge) and mass numbers on both sides.
Left side:
Atomic numbers = $5$ (Boron) + $1$ (Hydrogen) = $6$
Mass numbers = $10$ (Boron) + $1$ (Hydrogen) = $11$
Right side:
Atomic numbers = $4$ (Beryllium) + $2$ (Helium) = $6$
Mass numbers = $8$ (Beryllium) + $4$ (Helium) = $12$
Observation:
Charge is conserved (6 = 6), but the mass number is not conserved (11 $\neq$ 12). Thus, the Assertion is false. However, the Reason correctly states that nucleon conservation ensures mass number conservation in valid reactions.

Key Concept: Number of Neutrons Calculation

c) 12
[Solution Description]
To find the number of neutrons, use the formula $N = A - Z$, where $A$ is the mass number and $Z$ is the atomic number. Here, $A = 23$ and $Z = 11$. Substituting these values into the formula gives:
$N = 23 - 11 = 12$
Therefore, the number of neutrons is 12.

Key Concept: Nuclear Composition, Proton-Electron Hypothesis

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the proton-electron hypothesis cannot account for the observed magnetic moment of the nucleus. This is true because the proton-electron hypothesis suggests the presence of electrons in the nucleus, which have a large intrinsic magnetic moment ($\sim 10^3$ times that of a nucleon). However, experimentally, the nuclear magnetic moment is found to be much smaller, comparable to that of protons and neutrons. The Reason correctly explains why the Assertion is valid—it points out the discrepancy between the expected magnetic moment (if electrons were present) and the observed value. Therefore, both the Assertion and Reason are true, and the Reason provides the correct explanation for the Assertion.

Key Concept: Proton-electron Hypothesis

c) 4 protons and 2 electrons
[Solution Description]
The proton-electron hypothesis suggested that a helium nucleus contains 4 protons and 2 electrons to account for its mass (4 times the proton mass) and charge (+2e). This explanation was later disproven with the discovery of neutrons.

Key Concept: Proton-neutron reaction, Energy considerations

b) 17.32 MeV
[Solution Description]
First, identify $X$:
$Z$ conservation: $3 + 1 = 2 + Z \Rightarrow Z = 2$.
$A$ conservation: $7 + 1 = 4 + A \Rightarrow A = 4$.
Thus, $X$ is another $_2He^4$ (alpha particle).
Energy released $E$ is calculated using the mass defect $\Delta m$:
$E = \Delta m \times c^2 = 0.0186 \, \text{u} \times 931.5 \, \text{MeV/u} = 17.32 \, \text{MeV}$.

Key Concept: Radioactive Decay Sequences

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Given the decay of $_{92} U^{238}$ to $_{82} Pb^{206}$, we can calculate the number of alpha and beta decays as follows:
1. Calculate the number of alpha decays:
Mass number change = 238 - 206 = 32
Since each alpha decay reduces the mass number by 4, number of alpha decays = $\frac{32}{4} = 8$
2. After 8 alpha decays, the atomic number would be:
Initial atomic number = 92
Reduction due to alpha decays = 8 × 2 = 16
Resulting atomic number after alpha decays = 92 - 16 = 76
3. Final atomic number is 82, so the difference is compensated by beta decays:
Number of beta decays = 82 - 76 = 6
Therefore, both Assertion and Reason are true, and Reason correctly explains how the alpha and beta decays contribute to the final product.

Key Concept: Nuclear Composition

b) Proton-neutron hypothesis
[Solution Description]
The proton-electron hypothesis failed because it predicted a much larger nuclear magnetic moment. The proton-neutron hypothesis correctly explained the small observed magnetic moment by stating that neutrons (which are neutral) do not contribute significantly to the magnetic moment.

Key Concept: Physical properties, Mass number variation

d) Deuterium water has higher boiling point and density than protium water
[Solution Description]
Due to the difference in mass between protium and deuterium:
1. The boiling point of heavy water (D$_2$O) is higher than normal water (H$_2$O).
2. The density of D$_2$O is greater than H$_2$O.
3. The melting point of D$_2$O is higher than H$_2$O.
4. These differences occur because deuterium has a neutron while protium doesn't, making its nucleus and overall molecule heavier.

Key Concept: Mass Number, Nuclear Composition

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To solve this problem, let us break it down step by step:
1. Given the isotope notation $_3Li^7$, we know the atomic number ($Z$) is 3 (number of protons) and the mass number ($A$) is 7.
2. The mass number is defined as the sum of protons and neutrons in the nucleus: $A = Z + N$, where $N$ is the number of neutrons.
3. Rearranging the formula, we find the number of neutrons $N = A - Z = 7 - 3 = 4$.
4. Therefore, Assertion (A) stating that the number of neutrons is 4 is correct.
5. Reason (R) states the definition of mass number correctly and provides the correct atomic number ($Z = 3$), which directly supports the assertion.
6. Since both the assertion and reason are true, and the reason correctly explains why the assertion is true, the correct answer is option a).

Key Concept: Isotopes

c) 75\%
[Solution Description]
Let the natural abundance of Isotope A be $x\%$. Then, the natural abundance of Isotope B will be $(100 - x)\%$.
The average atomic mass is given by:
$35.5 = \frac{(35 \times x) + (37 \times (100 - x))}{100}$
Solving for $x$:
$35.5 \times 100 = 35x + 3700 - 37x$
$3550 = -2x + 3700$
$2x = 150 \implies x = 75\%$

Key Concept: Nuclear Size and Density

b) $R = R_0 A^{1/3}$
[Solution Description]
The nuclear radius $R$ is given by the empirical formula: $R = R_0 A^{1/3}$ where $R_0 = 1.2 \times 10^{-15}$ m. This formula shows how the radius depends on the mass number of the nucleus.

Key Concept: Nuclear Size and Density

c) $3.6 \times 10^{-15} \, \text{m}$
[Solution Description]
The radius of a nucleus is given by the formula $R = R_0 A^{1/3}$, where $R_0 \approx 1.2 \times 10^{-15} \, \text{m}$. For $A = 27$, calculate as follows:
$R = (1.2 \times 10^{-15}) \times (27)^{1/3}$
Since $27 = 3^3$, $27^{1/3} = 3$. Thus,
$R = 1.2 \times 10^{-15} \times 3 = 3.6 \times 10^{-15} \, \text{m}.$

Key Concept: Nuclear Reactions

b) Proton
[Solution Description]
The given nuclear reaction is:
$_7^{14}N + _2^4He \rightarrow _8^{17}O + _1^1H$
Here, a proton ($_1^1H$) is emitted, making this an ($\alpha, p$) reaction. The options include other particles, but only the proton matches the result.

Key Concept: Proton-electron Hypothesis

b) 4 protons and 2 electrons
[Solution Description]
The proton-electron hypothesis suggested that the nucleus of an atom is made up of protons and electrons. For a helium nucleus with mass roughly 4 times that of a proton and charge $+2e$, it was hypothesized to contain 4 protons (to account for the mass) and 2 electrons (to adjust the net charge to $+2e$).
Calculation: Total protons = 4, Total electrons = 2. Net charge = $(4 \times +e) + (2 \times -e) = +2e$.
Thus, the correct composition was 4 protons and 2 electrons.

Key Concept: Neutron Decay

d) Proton, electron, and antineutrino
[Solution Description]
In free state, a neutron is unstable and decays into a proton ($_{1}^{1}\text{H}$), an electron ($_{-1}^{0}\text{e}$), and an antineutrino ($\bar{\nu}$). The reaction is: $_{0}^{1}\text{n} \rightarrow _{1}^{1}\text{H} + _{-1}^{0}\text{e} + \bar{\nu}$.

Key Concept: Nuclear Composition and Stability

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion is true because nuclear stability is indeed influenced by the neutron-to-proton ratio (N/Z). For lighter nuclei (Z $<$ 20), stable nuclei typically have an N/Z ratio close to 1, while heavier nuclei require a higher N/Z ratio (up to $\sim$ 1.5) for stability due to increased proton-proton repulsion. The reason is also true, as additional neutrons contribute to the strong nuclear force without adding electrostatic repulsion, thereby enhancing stability. Moreover, the reason correctly explains why a higher N/Z ratio promotes stability in heavier nuclei.

Key Concept: Examples of Nuclear Reactions

d) Mass number, charge, and momentum
[Solution Description]
To verify conservation laws:
1. **Mass Number**: Left side: $14 + 4 = 18$, Right side: $17 + 1 = 18$ → Conserved.
2. **Charge**: Left side: $7 + 2 = 9$, Right side: $8 + 1 = 9$ → Conserved.
3. Momentum is inherently conserved in all nuclear reactions.
Thus, mass, charge, and momentum are conserved.

Key Concept: Alpha Decay

b) Atomic number decreases by 2; mass number decreases by 4
[Solution Description]
In alpha decay, an alpha-particle ($_{2}He^{4}$) is emitted from the nucleus. This causes:
- The atomic number ($Z$) to decrease by 2.
- The mass number ($A$) to decrease by 4.
Therefore, the correct option is the one that states the atomic number decreases by 2 and the mass number decreases by 4.

Key Concept: Nuclear Composition

b) 14
[Solution Description]
The mass number (\textit{A}) of a nucleus is the sum of the number of protons (\textit{Z}) and neutrons (\textit{N}) in it. Given: Number of protons = 6, Number of neutrons = 8. Therefore, $A = Z + N = 6 + 8 = 14$.

Key Concept: Nuclear Density, Mass Number

a) $\rho$
[Solution Description]
Nuclear density $\rho$ is independent of the mass number $A$ because the volume of the nucleus $\propto A$ while the mass $\propto A$. Thus,
$\rho = \frac{M}{V} = \frac{mA}{\frac{4}{3}\pi (R_0 A^{1/3})^3} = \frac{3m}{4\pi R_0^3},$
which is a constant. Therefore, the density remains unchanged even if the mass number changes.

Key Concept: Nuclear Density

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The nuclear density $\rho$ is given by:
$\rho = \frac{3m}{4 \pi R_0^3}$
where $m \approx 1.67 \times 10^{-27} \, \text{kg}$ is the mass of a nucleon, and $R_0 = 1.2 \times 10^{-15} \, \text{m}$ is an empirical constant. Substituting these values gives $\rho \approx 2 \times 10^{17} \, \text{kg/m}^3$, which is constant for all nuclei regardless of their mass number. Therefore, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Reaction Energy

b) 4.66 MeV
[Solution Description]
The energy released ($E$) in a nuclear reaction can be calculated using the mass defect ($\Delta m$) and the relation $E = \Delta m \times c^2$. Since 1 u $\approx$ 931 MeV, we have:
$E = 0.005 \, \text{u} \times 931 \, \text{MeV/u} = 4.655 \, \text{MeV}$
Rounding off, the energy released is approximately 4.66 MeV.

Key Concept: Evidence Against Proton-Electron Hypothesis

b) The magnetic moment was negligible compared to the electron's
[Solution Description]
The proton-electron hypothesis predicted the nuclear magnetic moment to be comparable to or larger than the electron's magnetic moment because electrons were assumed to be inside the nucleus. However, the observed nuclear magnetic moment was found to be much smaller (\$\sim\$ one-thousandth of an electron's magnetic moment). This discrepancy strongly suggested that electrons are not present in the nucleus, favoring the proton-neutron model.

Key Concept: Isotope Definition

b) Atoms with same proton numbers but different neutron numbers
[Solution Description]
Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons, resulting in different mass numbers.

Key Concept: Atomic number definition

c) Number of protons in the nucleus
[Solution Description]
The atomic number $Z$ of an atom is defined as the number of protons present in the nucleus of the atom. This uniquely identifies the element.

Key Concept: Quark Composition

d) $0e$
[Solution Description]
A neutron contains two down quarks, each with charge $-\frac{1}{3}e$, and one up quark with charge $+\frac{2}{3}e$. The net charge is calculated as:
$\text{Net Charge} = 2 \left( -\frac{1}{3}e \right) + \frac{2}{3}e = -\frac{2}{3}e + \frac{2}{3}e = 0e$

Key Concept: Nuclear Reactions, Conservation Laws

b) Proton ($_1^1 H$)
[Solution Description]
To identify $X$, we apply conservation laws:
- Conservation of charge: $7 (N) + 2 (He) = 8 (O) + Z_X \implies 9 = 8 + Z_X \implies Z_X = 1$.
- Conservation of nucleons: $14 + 4 = 17 + A_X \implies 18 = 17 + A_X \implies A_X = 1$.
Since $Z_X = 1$ and $A_X = 1$, $X$ must be a proton ($_1^1 H$).

Key Concept: Conservation Laws in Nuclear Reactions

d) All three (mass, charge, and momentum)
[Solution Description]
In nuclear reactions, mass (nucleon number), charge (atomic number), and momentum are always conserved. The problem asks for which quantity is conserved.

Key Concept: Nuclear Reactions, Conservation Laws

b) Proton ($^1_1H$); charge and nucleon number
[Solution Description]
Balance the mass numbers: $14 + 4 = 17 + A_X \Rightarrow A_X = 1$.
Balance the atomic numbers: $7 + 2 = 8 + Z_X \Rightarrow Z_X = 1$.
Hence, $X$ is a proton ($^1_1H$). The reaction conserves both atomic number (charge) and mass number (nucleons).

Key Concept: Alpha Decay

b) 2
[Solution Description]
An alpha particle ($_2He^4$) has 2 protons and 2 neutrons. When emitted, it reduces the parent nucleus’s atomic number by 2 (since 2 protons are lost) and mass number by 4 (2 protons + 2 neutrons).

Key Concept: Proton-electron Hypothesis, Magnetic Moment, Angular Momentum

c) Assertion is true, but Reason is false.
[Solution Description]
The Assertion correctly describes the proton-electron hypothesis for $_2He^4$, which suggested 4 protons and 2 electrons in the nucleus to explain its mass number (4) and atomic number (2). However, the Reason is false because the proton-electron hypothesis failed to match observed nuclear properties like magnetic moment and angular momentum, leading to its rejection. Thus, Assertion is true but Reason is false.

Key Concept: Proton-Electron Hypothesis, Nuclear Magnetic Moment

d) $\sim 10^{-3}$ times the electron’s magnetic moment
[Solution Description]
The proton-electron hypothesis suggested that nuclei contained both protons and electrons. Electrons have a magnetic moment ($\mu_e$). If the nucleus contained electrons, its magnetic moment should be comparable to $\mu_e$. However, experiments showed the nuclear magnetic moment to be $\sim 10^{-3} \mu_e$, which contradicted this hypothesis. This discrepancy was one reason the proton-electron model was abandoned.

Key Concept: Binding energy, Mirror nuclei

b) Coulomb repulsion in $^{11}$C
[Solution Description]
Mirror nuclei (like $^{11}$B and $^{11}$C) differ by swapping protons and neutrons. Charge independence predicts nearly identical nuclear binding energies, but the actual difference arises from Coulomb repulsion in $^{11}$C (extra proton) and the neutron-proton mass difference. The dominant effect is electromagnetic, violating charge independence.

Key Concept: Proton-Neutron Hypothesis, Quark Structure

c) $0$
[Solution Description]
A neutron has quark composition $udd$.
Step 1: Calculate the contribution from each quark:
$q_{\text{total}} = q_u + q_d + q_d = \left(+\frac{2}{3}e\right) + \left(-\frac{1}{3}e\right) + \left(-\frac{1}{3}e\right)$
Step 2: Simplify the expression:
$q_{\text{total}} = \frac{2}{3}e - \frac{1}{3}e - \frac{1}{3}e = 0$
The neutron is electrically neutral, consistent with experimental observations.

Key Concept: Nuclear Radius Formula

c) 7.4 F
[Solution Description]
The nuclear radius $R$ is given by:
$R = R_0 A^{1/3}$
where $R_0 = 1.2 \, \text{F}$. Substituting $A = 238$:
$R = 1.2 \times (238)^{1/3}$
Calculating $(238)^{1/3} \approx 6.20$, so:
$R \approx 1.2 \times 6.20 = 7.44 \, \text{F}$

Key Concept: Nuclear composition, isotopic notation

c) \${}_{27}Co^{60}\$
[Solution Description]
The standard notation is \${}_Z X^A\$ where:
\$Z\$ = atomic number (number of protons) = 27
\$X\$ = chemical symbol (from periodic table, element 27 is Cobalt, Co)
\$A\$ = mass number = 60
Therefore correct representation is \${}_{27}Co^{60}\$

Key Concept: Discovery of Neutrons, Importance of Neutron

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that neutrons are more effective than \$\alpha\$ particles for nuclear disintegration due to their uncharged nature, which is true because \$\alpha\$ particles are positively charged and get repelled by the nucleus.
The reason correctly explains why neutrons are effective: their lack of charge prevents repulsion, allowing them to interact with the nucleus more easily.
Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Angular Momentum Conservation

a) External torque was applied to the system during the experiment
[Solution Description]
The principle of conservation of angular momentum states that in the absence of external torques, angular momentum should remain constant. Here, $L_i \neq L_f$, indicating external torques may have acted on the system. Possible reasons include friction, air resistance, or other non-conservative forces.

Key Concept: Nuclear Density

b) $2 \times 10^{17} \, \text{kg/m}^3$
[Solution Description]
The nuclear density formula is:
$\rho = \frac{3m}{4 \pi R_0^3}.$
Substituting the given values:
$\rho = \frac{3 \times 1.67 \times 10^{-27}}{4 \pi (1.2 \times 10^{-15})^3} \approx 2 \times 10^{17} \, \text{kg/m}^3.$

Key Concept: Relation between mass number and atomic number

b) 10
[Solution Description]
The number of neutrons $N$ can be calculated using the formula $N = A - Z$. Given $A = 20$ and $Z = 10$:
$N = 20 - 10 = 10$

Key Concept: Atomic Number in Nuclear Notation

c) 26
[Solution Description]
In nuclear notation, the subscript before the element symbol indicates the atomic number (Z). For $_{26}Fe^{56}$, the atomic number is 26, representing the number of protons.

Key Concept: Nuclear Size

b) 3.6 fermi
[Solution Description]
The radius $R$ of a nucleus is given by the formula:
$R = R_0 A^{1/3}$.
Substituting $A = 27$ and $R_0 = 1.2 \times 10^{-15}$ m:
$R = (1.2 \times 10^{-15}) \times (27)^{1/3}$.
Since $3^3 = 27$, $(27)^{1/3} = 3$.
Thus, $R = 1.2 \times 10^{-15} \times 3 = 3.6 \times 10^{-15}$ m.
Therefore, the radius is approximately 3.6 fermi.

Key Concept: Isobars

b) The same mass number but different atomic numbers.
[Solution Description]
Isobars are nuclei with the same mass number ($A$) but different atomic numbers ($Z$). Since their proton and neutron counts differ, they represent different elements with distinct chemical and physical properties.

Key Concept: Nuclear forces characteristics

c) Saturation effect
[Solution Description]
The saturation effect of nuclear forces means that each nucleon interacts only with a limited number of other nucleons in its immediate vicinity. This property arises because the nuclear force is highly localized and does not extend over large distances within the nucleus, preventing long-range interactions.

Key Concept: Neutron Decay

d) Proton, electron, and antineutrino
[Solution Description]
A free neutron decays into a proton, an electron, and an antineutrino via beta decay. The reaction is:
$_{0}n^{1} \rightarrow _{1}H^{1} + _{-1}\beta^{0} + \bar{\nu}$
Here, $_{1}H^{1}$ is a proton ($p$), $_{-1}\beta^{0}$ is an electron ($e^-$), and $\bar{\nu}$ is an antineutrino.

Key Concept: Nuclear Composition, Isotopes

c) 3
[Solution Description]
For $_{92}U^{235}$, the number of neutrons is calculated as $N_1 = A - Z = 235 - 92 = 143$. Similarly for $_{92}U^{238}$, $N_2 = 238 - 92 = 146$. The difference is $N_2 - N_1 = 146 - 143 = 3$.

Key Concept: Nuclear Reactions

b) $_8O^{17}$
[Solution Description]
The given nuclear reaction is:
$_7N^{14} + _2He^4 \rightarrow Y + _1H^1$
To find the final nucleus $Y$, we apply conservation of nucleons and charge.
For nucleons: 14 (from N) + 4 (from He) = mass number of $Y$ + 1 (from H).
So, mass number of $Y = 17$.
For charge: 7 (from N) + 2 (from He) = atomic number of $Y$ + 1 (from H).
So, atomic number of $Y = 8$.
Therefore, the final nucleus is $_8O^{17}$.

Key Concept: Atomic Mass Unit Definition

b) One-twelfth the mass of a carbon-12 atom
[Solution Description]
1 unified atomic mass unit (u) is defined as one-twelfth the mass of a carbon-12 atom (${}^{12}_6C$). The mass of ${}^{12}_6C$ is exactly 12 u, making 1 u equal to $1.6605 \times 10^{-27}\, \text{kg}$.

Key Concept: Mass Number

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The mass number ($A$) of an atom is the sum of protons and neutrons in its nucleus. For lithium-7 ($_3Li^7$), the atomic number ($Z$) is 3 (number of protons). Given $A = 7$, the number of neutrons ($N$) can be calculated using the formula:
$N = A - Z = 7 - 3 = 4$
Therefore, the assertion that the mass number of lithium-7 is 7 is true, and the reason stating it has 3 protons and 4 neutrons is also true. The reason correctly explains the assertion because the mass number is indeed the sum of protons and neutrons.

Key Concept: Nuclear Composition, Neutron Decay

b) 0.782 MeV
[Solution Description]
The energy released ($E$) in the decay process is given by the mass difference between the initial and final particles multiplied by $c^2$. Here, the antineutrino mass is negligible.
Step 1: Find the total mass of the decay products:
Proton mass = 938.272 MeV/c$^2$
Electron mass = 0.511 MeV/c$^2$
Total final mass = 938.272 + 0.511 = 938.783 MeV/c$^2$
Step 2: Subtract the final mass from the initial neutron mass to find the mass defect:
Mass defect = 939.565 - 938.783 = 0.782 MeV/c$^2$
Step 3: Convert mass defect to energy using $E = \Delta m c^2$:
Energy released = 0.782 MeV

Key Concept: Unified Atomic Mass Unit, Isotopic Mass

b) 398.7 MeV
[Solution Description]
The mass defect is given as 0.428 u.
Energy equivalent of 1 u = 931.5 MeV.
Thus, binding energy = $0.428 \times 931.5 = 398.68 \, \text{MeV}$.

Key Concept: Quantum Confinement, Nuclear Size

b) The required confinement energy exceeds typical nuclear binding energies
[Solution Description]
According to Heisenberg's Uncertainty Principle, confining an electron to a nucleus would require such large momentum that its kinetic energy would be greater than its rest mass energy by orders of magnitude. This makes the electron highly relativistic and unstable in such small confines. Additionally, at these energy scales, other quantum field effects become dominant.

Key Concept: Number of Neutrons Calculation

c) 6
[Solution Description]
The formula to calculate the number of neutrons is $N = A - Z$. Given $A = 12$ and $Z = 6$, substituting these values into the formula gives:
$N = 12 - 6 = 6$
So, the number of neutrons is 6.

Key Concept: Proton–Neutron Hypothesis

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the mass number of lithium (\textsuperscript{7}\textsubscript{3}Li) is 7 due to its nucleus containing 3 protons and 4 neutrons, which aligns with the definition of mass number ($A = Z + N$, where $Z$ is the atomic number and $N$ is the number of neutrons). The reason correctly defines the mass number as the sum of protons and neutrons. Both statements are true, and the reason accurately explains the assertion.