Class 12 Physics Chapter 13 Nuclei

 Key Concept: Neutron-Proton Ratio and Stability, Binding Energy Curve Analysis

c) Assertion is true, but Reason is false.

[Solution Description]

Assertion is true because the neutron-to-proton ratio influences nuclear stability; heavier nuclei generally require more neutrons to offset proton repulsion due to increased Coulomb forces. This helps achieve nuclear stability. However, Reason is false because while binding energy per nucleon for heavy nuclei like lead-208 shows relatively high values, it does not imply higher stability compared to all lighter nuclei. Lighter nuclei often have a higher binding energy per nucleon, making them more stable on average. Therefore, the reason provided is incorrect.

 Key Concept: Characteristics of Nuclear Force

c) It is a short-range force.

[Solution Description]

Nuclear forces are known to have specific characteristics:

1. They are short-range forces, meaning they operate only at a very small distance (on the order of a few femtometers).

2. They are very strong compared to electromagnetic forces.

3. They act independently of nucleon type, meaning they affect protons and neutrons equally.

Thus, the correct characteristic from the given options is that it is a short-range force.

Key Concept: Decay Constant

b) s$$^{-1}$$
[Solution Description] The decay constant $$\lambda$$ has units of inverse time because it represents the probability of decay per unit time. Hence, the unit for the decay constant is $$\text{s}^{-1}$$ (per second).

 Key Concept: Advanced Calculations, Complex Reaction Chains

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] To calculate the Q-value in nuclear reactions such as the fusion of deuterium and tritium, it involves determining the difference between the sum of initial masses and the sum of final masses, then multiplying by $$c^2$$. The reaction:

$$ ^2_1H + ^3_1H \rightarrow ^4_2He + n $$

The binding energy per nucleon increases for products compared to reactants in this reaction, leading to a significant mass defect. This results in a positive Q-value, indicating an exothermic process with energy release. In fusion reactions involving light nuclei like deuterium and tritium, the substantial increase in binding energy when forming helium leads to a high Q-value. Thus, both assertion and reason are true, and reason correctly explains the assertion.

 Key Concept: Size Comparison

d) A small stadium

[Solution Description]

Given that the nucleus is much smaller than the atom, specifically with a volume about $$10^{-12}$$ times the volume of the atom, if a nucleus were the size of a marble, the atom would be extremely large possibly comparable to a stadium or larger structure.

Key Concept: Types of Decay

d) Delta decay
[Solution Description] Natural radioactive decay processes include alpha decay, beta decay, and gamma decay. Delta decay is not recognized as a natural decay process.

Key Concept: Nuclear Stability, Neutron-to-Proton Ratio

b) Stable, as the ratio is exactly 1.5 which is ideal for heavy nuclei
[Solution Description]
The neutron-to-proton ratio is given as $$\frac{150}{100} = 1.5$$. For heavy nuclei, stability often requires a neutron-to-proton ratio greater than 1 but typically closer to about 1.5 to 1.6 due to increasing repulsion between protons. A ratio of 1.5 is within the stable range for heavy elements, suggesting relative stability.

 Key Concept: Saturation Property

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion is about the range of the nuclear force, which acts only over a short distance and affects nucleons that are very close to each other. This results in saturation, meaning that adding more nucleons does not significantly increase the interaction energy beyond a certain point. The reason provided explains this concept further; the short-range nature of nuclear forces means that medium and large nuclei have nearly constant binding energy per nucleon because these forces saturate within the nucleus. Thus, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

 Key Concept: Nuclear Radius

c) $$4.8 \times 10^{-15} \, m$$

[Solution Description] To find the nuclear radius, substitute the given values into the formula:

$$R = R_0 A^{1/3} = 1.2 \times 10^{-15} \times 64^{1/3}$$

Calculate $$64^{1/3} = 4$$:

$$R = 1.2 \times 10^{-15} \times 4 = 4.8 \times 10^{-15} \, m$$

Therefore, the radius of the nucleus is $$4.8 \times 10^{-15} \, m$$.

Key Concept: Fission and Fusion

c) Fusion
[Solution Description] In nuclear fusion, lighter nuclei combine to form a heavier nucleus, resulting in the release of energy due to an increase in binding energy per nucleon.

Key Concept: Energy Release

c) 200 MeV
[Solution Description] The nuclear fission of $$^{235}_{92}U$$ releases a large amount of energy, primarily in the form of kinetic energy of the fragments and neutrons. It is known that each fission event releases approximately 200 MeV of energy.

Key Concept: Charge Independence

d) None; all are affected
[Solution Description] Nuclear force is independent of electric charge, which means it affects all nucleon-nucleon interactions equally. Therefore, neutron-neutron, proton-neutron, and proton-proton interactions are all influenced by this characteristic. Hence, none of the interactions are unaffected.

Key Concept: Types of Decay

a) Alpha decay
[Solution Description]
Alpha decay involves the release of an alpha particle from the nucleus, which consists of two protons and two neutrons. This reduces the atomic number by 2 and the mass number by 4.

Key Concept: Basic Fusion Reaction

a) $$^2_1H + ^2_1H \rightarrow ^3_2He + n$$
[Solution Description] The reaction involving two deuterons ($$^2_1H$$) combining to form a helium isotope is represented as: $$^2_1H + ^2_1H \rightarrow ^3_2He + n$$. This depicts two deuterium nuclei fusing to produce a helium-3 nucleus and a neutron.

Key Concept: Nuclear Force

d) They are the same for neutron-neutron, proton-neutron, and proton-proton.
[Solution Description]
Nuclear forces are fundamental forces that act between nucleons in the nucleus. They are strong enough to keep the nucleus together despite the repulsion between protons due to their positive charge. Importantly, these forces do not depend on electric charge, meaning they are equally effective among neutron-neutron, proton-neutron, and proton-proton pairs. Therefore, the correct description of nuclear forces from the given options is that they are the same for neutron-neutron, proton-neutron, and proton-proton interactions.

Key Concept: Mass Spectrometer Use

a) Mass spectrometer
[Solution Description]
The mass spectrometer is an analytical device used to measure the mass-to-charge ratio of ions. It is widely used for determining the precise masses of isotopes and elements.

Key Concept: Advanced Fusion Processes, Energy Conversion

a) $$Helium \rightarrow Carbon \rightarrow Oxygen$$

[Solution Description]

After hydrogen in a star is depleted, the core contracts and heats up, allowing for helium fusion into carbon through the triple-alpha process. As this progresses, further fusion can create heavier elements like oxygen. The sequence of these reactions results in energy release as each step binds nucleons more tightly than before, converting mass into energy according to Einstein’s equation $$E=mc^2$$.

Key Concept: Fission Product Identification

c) $$^{144}_{56}Ba$$ and $$^{89}_{36}Kr$$
[Solution Description] In neutron-induced fission of $$^{235}_{92}U$$, one possibility for the fragments produced is that their atomic numbers and mass numbers add up to those of uranium plus the neutron, i.e., $$236_{92}$$U. For example:

– If $$^{144}_{56}Ba$$ and $$^{89}_{36}Kr$$ are formed, we find:
Atomic number: $$56 + 36 = 92$$

Mass number: $$144 + 89 + 3 \times 1 = 236$$ (considering 3 neutrons are emitted)

The balance checks out, so this possibility can occur.

Thus, option C is the correct answer.

 Key Concept: Advanced Calculations, In-depth Analysis

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

To solve this problem, we need to understand the relationships between the mean life $$(\tau),$$ the half-life $$T_{1/2}$$, and the decay constant $$\lambda$$.

The mean life $$\tau$$ is given by:

$$\tau = \frac{1}{\lambda}$$

The half-life $$T_{1/2}$$ is related to the decay constant by:

$$T_{1/2} = \frac{\ln(2)}{\lambda}$$

Comparing these two equations:

$$\tau = \frac{1}{\lambda}, \quad T_{1/2} = \frac{\ln(2)}{\lambda}$$

It can be observed that:

$$\tau > T_{1/2} \text{ because } 1 > \ln(2)$$

So, the assertion is true.

Now, let’s consider the reason. As $$\lambda$$ increases, $$T_{1/2}$$ decreases because it is inversely proportional to $$\lambda$$. Similarly, $$\tau$$ also decreases for the same reason. However, the relationship described in the reason does not specifically explain why $$\tau$$ is always greater than $$T_{1/2}$$. It merely states that both decrease as $$\lambda$$ increases without explaining their comparative magnitudes.

Therefore, while both statements are true, the reason provided doesn’t adequately explain the assertion.

Key Concept: Energy Release

c) The combined mass of the fragments is less than the original nucleus.
[Solution Description] According to mass-energy equivalence ($$E = mc^2$$), when a heavy nucleus splits into smaller fragments during fission, the total mass of the resulting fragments is less than the original nucleus. This mass difference is converted into energy, which is released.

Key Concept: Energy Units

a) 931.5 MeV
[Solution Description] One atomic mass unit (u) is equivalent to 931.5 MeV/c² in terms of energy. This conversion factor is crucial in calculating energies involved in nuclear reactions and processes.

Key Concept: Basic Reaction

a) $$^{144}_{56}Ba$$
[Solution Description] When $$^{235}_{92}U$$ absorbs a neutron, it becomes $$^{236}_{92}U$$, which is unstable and subsequently splits into smaller nuclei. Among typical fission products are $$^{144}_{56}Ba$$ and $$^{89}_{36}Kr$$. The $$^{144}_{56}Ba$$ fragment is a known fission product.

 Key Concept: Nuclear Force Characteristics

d) They are strong, short-ranged, and saturating.

[Solution Description] Nuclear forces are short-range forces that hold nucleons together within the nucleus. They are significantly stronger than Coulomb forces which are responsible for electrical interaction but act over shorter ranges, leading to saturation where each nucleon interacts effectively only with its nearest neighbors.

 Key Concept: Critical Mass, Energy Calculation

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion discusses the concept of critical mass, which is crucial in maintaining a continuous chain reaction, as it allows the emitted neutrons to be absorbed by other fissile nuclei, thereby sustaining the reaction.

The reason involves the calculation of energy released during a fission process based on binding energy differences. While both statements are true individually, the reason does not directly explain why a critical mass is necessary. Critical mass relates more to the availability of sufficient neutrons within a system rather than the calculation of energy release.

Thus, while both the Assertion and Reason are true, the Reason is not the correct explanation of the Assertion.

Key Concept: Fusion Conditions, Isotopes and Stability

b) High energy collisions allow overcoming electrostatic repulsion.
[Solution Description] Fusion reactions require extremely high temperatures because the positively charged nuclei must overcome the electrostatic repulsion force between them, known as the Coulomb barrier. At high temperatures, the kinetic energy of the deuterium nuclei increases, allowing them to come close enough for the strong nuclear force to act, facilitating the fusion process. Moreover, higher temperatures ensure that a significant fraction of particles have sufficiently high velocities to surmount the potential barrier.

Key Concept: Decay Types

a) Alpha decay
[Solution Description] In alpha decay, a helium nucleus ($$ ^4_2He $$) is emitted from the unstable nucleus.

Key Concept: Technical Specifications, Comparative Analysis

b) Cloud chambers allow direct viewing of alpha particle paths, unlike Geiger-Müller counters
[Solution Description]
Geiger-Müller counters detect ionizing radiation by measuring the electrical pulses generated by the ionization of gas within the chamber, but they generally do not distinguish between types of particles. Cloud chambers visualize charged particle tracks through vapor condensation, which allows detailed analysis of the paths and characteristics of different particles like alpha particles. Therefore, while both devices can detect alpha particles, the cloud chamber provides visualization advantages that facilitate analysis.

Key Concept: Nuclear Density

c) $$2.3 \times 10^{17}$$ kg/m³
[Solution Description]
The density of nuclear matter is considered to be almost constant and independent of the mass number $$ A $$. The approximate value is often stated as $$ 2.3 \times 10^{17} $$ kg/m³, which allows us to assume uniformity in how closely packed nucleons are within different nuclei.

Key Concept: Magic Numbers and Nuclear Models

b) $$^{40}_{20}Ca$$
[Solution Description]
The concept here involves identifying which nucleus has either proton or neutron counts that match one of the known magic numbers: 2, 8, 20, 28, 50, 82, or 126. The presence of a magic number indicates a stable configuration due to filled nuclear shells.

– Looking at $$^{40}_{20}Ca,$$ it has 20 protons (a magic number).
– Therefore, this configuration is associated with increased stability according to the nuclear shell model.

Answer: $$^{40}_{20}Ca$$

Key Concept: Saturation Property

b) A nucleon influences only a few neighboring nucleons
[Solution Description] The saturation property of nuclear forces indicates that a nucleon within a nucleus interacts strongly only with a limited number of nearby nucleons, rather than all nucleons in the nucleus. This occurs because nuclear forces are short-ranged, and they do not extend beyond a certain distance within the nucleus.

Key Concept: Energy Release

c) 200 MeV
[Solution Description] Each fission event of uranium-235 releases about 200 MeV of energy. This number is derived from the difference in binding energy before and after the fission.

Key Concept: Energy Release

d) 200 MeV
[Solution Description] A single fission event of a $$^{235}_{92}U$$ nucleus typically releases about 200 MeV of energy, as calculated from the products and the binding energies.

 Key Concept: Cause and Effect

b) Energy is released

[Solution Description] When a heavy nucleus undergoes fission, it splits into smaller nuclei which usually have higher binding energy per nucleon. Since energy is released during this process, because the resulting nuclei are more stable than the original nucleus. Thus, energy is released.

Key Concept: Neutron Count

c) 12
[Solution Description]
The neutron number $$N$$ can be found using the formula $$N = A – Z$$, where $$A$$ is the mass number and $$Z$$ is the atomic number. For $$^{23}_{11}\text{Na}$$, $$A = 23$$ and $$Z = 11$$. Thus, $$N = 23 – 11 = 12$$.

Key Concept: Basic Definition

b) The process of a nucleus becoming stable through particle emission
[Solution Description] Radioactive decay is a nuclear phenomenon where an unstable nucleus undergoes decay by emitting particles or radiation to become more stable.

Key Concept: Binding Energy

a) 8.45 MeV/nucleon
[Solution Description]
The total binding energy of the original nucleus can be calculated by multiplying the number of nucleons by the average binding energy per nucleon: $$236 \times 7.6 \text{ MeV}$$. This gives:
$$
236 \times 7.6 = 1793.6 \text{ MeV}
$$
After fission, the energy released is 200 MeV, indicating that the total binding energy of the resulting fragments must be increased by this amount:
$$
1793.6 + 200 = 1993.6 \text{ MeV}
$$
Assuming the mass number remains approximately the same (236), the new average binding energy per nucleon is:
$$
\frac{1993.6}{236} \approx 8.45 \text{ MeV/nucleon}
$$

Key Concept: Basic Fission

b) Splitting of a heavy nucleus into smaller ones with release of energy
[Solution Description] Nuclear fission is a process where a heavy nucleus splits into smaller nuclei, releasing energy. This is a fundamental concept in understanding how energy is derived from nuclear reactions.

Key Concept: Nuclear Reaction Energy, Binding Energy

d) Because fusion results in a product with a higher binding energy per nucleon.
[Solution Description] The binding energy per nucleon increases as we move from lighter nuclei like hydrogen towards iron on the binding energy curve. This increase indicates that products formed in fusion have higher binding energy per nucleon, meaning they are more stable than the reactants. Therefore, the excess energy due to this increase in stability is released, which is substantial for light nuclei.

Key Concept: Proton Count

a) 9
[Solution Description] The number of protons in an isotope is given by its atomic number ($$Z$$). For fluorine ($$^{19}\text{F}$$), the atomic number is 9, which means it has 9 protons.

Key Concept: Basic Decay Types

a) Electron
[Solution Description] In beta decay, either an electron or a positron is emitted from the nucleus.

 Key Concept: Fusion Reactor Development, Mass-Energy Conversion

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description] The assertion states that the energy release during nuclear fusion is due to mass-energy conversion, which is true because the mass difference between reactants and products results in energy release as per Einstein’s equation $$E=mc^2$$. The reason explains why high temperatures are necessary in fusion reactors: to provide charged particles enough kinetic energy to overcome their electrostatic repulsion (Coulomb barrier). While both statements are true, they address different aspects of fusion reactions—one related to energy mechanisms and the other concerning operational requirements.