NCERT Solutions for Class 12 Physics Chapter 13 Nuclei
NCERT Solutions for Class 12 Physics Chapter 13 Nuclei provides complete explanations of nuclear composition, nuclear radius and density, mass defect, binding energy per nucleon curve, nuclear forces, radioactivity, nuclear fission, and nuclear fusion.
Overview of Topics & Subtopics (Click to Jump)
1. Atomic Mass Unit, Nuclear Composition & Size of Nucleus
Short Note:
• Atomic Mass Unit (u): Defined as 1/12th of the mass of a neutral Carbon-12 atom (1 u = 1.660539 × 10⁻²⁷ kg ≈ 931.5 MeV/c²).
• Composition of Nucleus: Contains protons ($Z$) and neutrons ($N$). Total mass number A = Z + N.
- Isotopes: Same atomic number $Z$, different mass number $A$ (e.g., ¹₁H, ²₁H, ³₁H).
- Isobars: Same mass number $A$, different atomic number $Z$ (e.g., ³₁H and ³₂He).
- Isotones: Same number of neutrons $N = A - Z$.
• Nuclear Size: Empirical formula for nuclear radius: R = R₀ A¹/³, where R₀ ≈ 1.2 × 10⁻¹⁵ m = 1.2 fm.
• Nuclear Density (ρ): Mass per unit volume of nucleus is nearly constant and independent of mass number A: ρ ≈ 2.3 × 10¹⁷ kg/m³.
2. Mass Defect, Binding Energy & Nuclear Forces
Short Note:
• Mass Defect (Δm): Difference between the total mass of individual constituent nucleons and the actual rest mass of the bound nucleus:
Δm = [Z m_p + (A - Z) m_n] - M_nucleus
• Nuclear Binding Energy (E_b): Energy released when nucleons come together to form a nucleus, or energy required to completely separate nucleons: E_b = Δm × c² = Δm (in u) × 931.5 MeV.
• Binding Energy per Nucleon (E_bn): Average binding energy per constituent particle: E_bn = E_b / A.
- Peak stability occurs around A = 56 (Fe-56, E_bn ≈ 8.75 MeV/nucleon).
- Average value is around 8.0 MeV/nucleon for $30 < A < 170$.
• Nuclear Force Properties: Strongest fundamental force in nature, extremely short-ranged (~ 1-2 fm), charge-independent, and exhibits saturation effects.
3. Radioactivity, Nuclear Fission & Fusion
Short Note:
• Nuclear Fission: Process in which a heavy nucleus ($A > 230$) splits into two lighter medium-mass fragments upon absorbing a neutron, releasing a large amount of energy (~200 MeV per fission event).
Example: ²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3 ¹₀n + Q
• Controlled Chain Reaction: Used in nuclear reactors with moderators (heavy water/graphite) and control rods (cadmium/boron).
• Nuclear Fusion: Process in which two or more lighter nuclei ($A \le 10$) combine under extreme temperature ($10⁷$ K) and pressure to form a heavier, more stable nucleus with energy release.
Example: ²₁H + ³₁H → ⁴₂He + ¹₀n + 17.6 MeV (Primary source of stellar energy in the Sun).
Intext & Conceptual Questions
• Features of the E_bn vs A Curve:
1. Intermediate Nuclei Stability: $E_{bn}$ is almost constant (around $8.0\text{ MeV/nucleon}$) for nuclei with mass numbers in the range $30 < A < 170$. It reaches a maximum peak of $8.75\text{ MeV/nucleon}$ at $A = 56$ (Fe-56).
2. Lighter Nuclei (A < 30): $E_{bn}$ is lower for light elements, but shows distinct local peaks at $A = 4, 8, 12, 16$ (such as ⁴₂He, ⁸₄Be, ¹²₆C, ¹⁶₈O), indicating special stability of even-even nuclei.
3. Heavier Nuclei (A > 170): $E_{bn}$ gradually decreases as mass number increases, dropping to about $7.6\text{ MeV/nucleon}$ for ²³⁸U.
• Key Inferences:
• Short-range & Saturation of Nuclear Forces: The constancy of $E_{bn}$ over a wide range implies a nucleon interacts only with its immediate neighbors.
• Energy Release in Fission & Fusion: Splitting a heavy nucleus ($A > 200$) or fusing light nuclei ($A < 10$) results in products with higher $E_{bn}$ (more tightly bound state), liberating massive quantities of nuclear energy.
| Feature | Nuclear Fission | Nuclear Fusion |
|---|---|---|
| Process | Splitting of a heavy unstable nucleus into lighter fragments. | Combining light nuclei to form a heavier stable nucleus. |
| Temperature Required | Can occur at normal room temperatures upon neutron absorption. | Requires ultra-high temperatures (~10⁷ K) to overcome Coulomb repulsion. |
| Energy per Unit Mass | High total energy per reaction (~200 MeV), lower energy yield per unit mass. | Significantly higher energy yield per unit mass than fission. |
| Radioactive Waste | Produces highly hazardous radioactive nuclear waste products. | Produces virtually non-radioactive or environmentally harmless byproducts (e.g. Helium). |
Textbook Chapter End Exercise Questions
For Nitrogen nucleus ¹⁴₇N:
• Number of protons ($Z$) = 7
• Number of neutrons ($N = A - Z$) = 14 - 7 = 7
Step 1: Calculate Total Mass of Constituent Nucleons
Mass of 7 protons = 7 × 1.007825 u = 7.054775 u
Mass of 7 neutrons = 7 × 1.008665 u = 7.060655 u
Total mass = 7.054775 + 7.060655 = 14.115430 u
Step 2: Calculate Mass Defect (Δm)
Δm = Total Mass - Rest Mass of ¹⁴₇N
Δm = 14.115430 u - 14.003070 u = 0.11236 u
Step 3: Calculate Binding Energy (E_b)
E_b = 0.11236 × 931.5 MeV ≈ 104.66 MeV
• Total Binding Energy = 104.66 MeV
• Binding Energy per Nucleon = 104.66 / 14 ≈ 7.48 MeV/nucleon
Formula for nuclear radius: R = R₀ A¹/³
Given:
• $A_1 = 27$, $R_1 = 3.6\text{ fm}$
• $A_2 = 125$, $R_2 = ?$
Step 1: Ratio of Radii
R₂ / R₁ = (A₂ / A₁)¹/³
R₂ / 3.6 = (125 / 27)¹/³ = 5 / 3
Step 2: Calculate R₂
R₂ = 3.6 × (5 / 3) = 1.2 × 5 = 6.0 fm
The radius of the nucleus with $A = 125$ is 6.0 fm (or 6.0 × 10⁻¹⁵ m).
Read More & Chapter Resources
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