NCERT Solutions Class 12 Physics Chapter 13 Nuclei
Get comprehensive NCERT Solutions Class 12 Physics Chapter 13 Nuclei to score top marks in your CBSE Class 12 Board examinations and competitive tests like JEE and NEET. This dedicated guide delivers step-by-step solutions for every textbook exercise, in-depth derivations of nuclear physics laws, solved previous-year board questions, and master tips to avoid common numerical pitfalls. Download our structured explanations to build an unshakeable conceptual foundation in nuclear structure, mass defect, binding energy, and nuclear reactions.
Chapter Overview: Class 12 Physics Chapter 13 Nuclei
In the CBSE Class 12 Physics syllabus, Chapter 13 Nuclei belongs to Unit VIII (Atoms and Nuclei), which collectively carries a weightage of 6 to 8 marks in the annual board examination. While Chapter 12 explores the atomic electron orbitals based on Rutherford and Bohr models, Chapter 13 delves directly into the core of the atom: the atomic nucleus.
The study of the nucleus is governed by strong nuclear forces, mass-energy equivalence, and nuclear transformations. The central themes covered in this chapter include:
- Composition and Size of the Nucleus: Protons, neutrons, atomic mass unit (u), isotopes, isobars, isotones, and nuclear radius proportionality ($R = R_0 A^{1/3}$).
- Nuclear Density: The constant density of nuclear matter across all elements ($\approx 2.3 \times 10^{17}\text{ kg/m}^3$).
- Mass Defect and Binding Energy: Einstein’s mass-energy equation ($E = \Delta m \cdot c^2$), total binding energy, and binding energy per nucleon ($E_{bn}$).
- Binding Energy Curve: Stability of nuclei, saturation of nuclear forces, and explanation of nuclear fission and fusion based on the curve.
- Nuclear Forces: Characteristics of the fundamental strong interaction (short-range, non-central, charge-independent, saturated).
- Nuclear Energy: Energy released in nuclear fission of $^{235}_{92}\text{U}$ and nuclear fusion in stellar cores (proton-proton cycle).
Important Formulas, Laws, and Key Concepts
Before solving the NCERT exercises, review the core mathematical equations and conceptual definitions summarized in the table below:
| Concept / Physical Quantity | Mathematical Formula | Key Constants & SI Units |
|---|---|---|
| Unified Atomic Mass Unit | $1\text{ u} = \frac{1}{12} \times \text{Mass of } ^{12}_{6}\text{C}\text{ atom}$ | $1\text{ u} = 1.660539 \times 10^{-27}\text{ kg} \approx 931.5\text{ MeV}/c^2$ |
| Nuclear Radius | $R = R_0 A^{1/3}$ | $R_0 \approx 1.2 \times 10^{-15}\text{ m} = 1.2\text{ fm}$, $A$ = Mass Number |
| Nuclear Density | $\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{A \cdot m}{\frac{4}{3}\pi R_0^3 A} = \frac{3m}{4\pi R_0^3}$ | $\rho \approx 2.3 \times 10^{17}\text{ kg/m}^3$ (Independent of $A$) |
| Mass Defect ($\Delta m$) | $\Delta m = [Z m_p + (A - Z) m_n] - M_{\text{nucleus}}$ | $m_p = 1.007276\text{ u}$, $m_n = 1.008665\text{ u}$ |
| Total Binding Energy ($E_b$) | $E_b = \Delta m \times 931.5\text{ MeV}$ | Expressed in $\text{MeV}$ or Joules ($1\text{ MeV} = 1.602 \times 10^{-13}\text{ J}$) |
| Binding Energy per Nucleon ($E_{bn}$) | $E_{bn} = \frac{E_b}{A}$ | Measures nuclear stability; maximum $\approx 8.75\text{ MeV}$ for $^{56}_{26}\text{Fe}$ |
| $Q$-value of Nuclear Reaction | $Q = (M_{\text{reactants}} - M_{\text{products}}) \times c^2$ | Exothermic if $Q > 0$; Endothermic if $Q < 0$ |
Properties of the Nuclear Force
- Strongest Interaction: It is approximately $100$ times stronger than the electrostatic repulsive force and $10^{38}$ times stronger than gravitational force at subatomic distances.
- Short-Range Force: Operates effectively only over distances of the order of a few femtometers ($\approx 1\text{ to } 2\text{ fm}$). Beyond $2\text{ to }3\text{ fm}$, it drops sharply to zero.
- Charge Independence: The nuclear force acts identically between proton-proton ($p-p$), neutron-neutron ($n-n$), and proton-neutron ($p-n$).
- Non-Central & Spin-Dependent: The magnitude of the force depends on the relative spin orientations of the nucleons.
- Saturation Property: A nucleon interacts only with its immediate neighboring nucleons, leading to a constant binding energy per nucleon for medium-sized nuclei ($30 < A < 170$).
NCERT Solutions Class 12 Physics Chapter 13 Nuclei: Step-by-Step Exercise Solutions
Here are detailed, whiteboard-style solutions for the exercise problems in NCERT Class 12 Physics Chapter 13.
Exercise 13.1
Question: Two stable isotopes of lithium $^{6}_{3}\text{Li}$ and $^{7}_{3}\text{Li}$ have respective abundances of $7.5\%$ and $92.5\%$. These isotopes have masses $6.01512\text{ u}$ and $7.01600\text{ u}$, respectively. Find the atomic mass of lithium.
Solution:
- Given:
- Isotope 1 ($^{6}_{3}\text{Li}$): Mass $m_1 = 6.01512\text{ u}$, Relative abundance $a_1 = 7.5\%$
- Isotope 2 ($^{7}_{3}\text{Li}$): Mass $m_2 = 7.01600\text{ u}$, Relative abundance $a_2 = 92.5\%$
- To Find: Average atomic mass of lithium ($m_{\text{avg}}$).
- Formula:
$$m_{\text{avg}} = \frac{a_1 m_1 + a_2 m_2}{a_1 + a_2} = \frac{(7.5 \times m_1) + (92.5 \times m_2)}{100}$$
- Substitution & Calculation:
$$m_{\text{avg}} = \frac{(7.5 \times 6.01512) + (92.5 \times 7.01600)}{100}$$
$$7.5 \times 6.01512 = 45.1134\text{ u}$$
$$92.5 \times 7.01600 = 648.9800\text{ u}$$
$$m_{\text{avg}} = \frac{45.1134 + 648.9800}{100} = \frac{694.0934}{100} = 6.940934\text{ u} \approx 6.941\text{ u}$$
- Final Answer: The average atomic mass of lithium is 6.941 u.
Exercise 13.2
Question: Boron has two stable isotopes, $^{10}_{5}\text{B}$ of mass $10.01294\text{ u}$ and $^{11}_{5}\text{B}$ of mass $11.00931\text{ u}$. The average atomic mass of naturally occurring boron is $10.811\text{ u}$. Obtain the relative abundances of $^{10}_{5}\text{B}$ and $^{11}_{5}\text{B}$.
Solution:
- Given:
- Mass of $^{10}_{5}\text{B}$ ($m_1$) = $10.01294\text{ u}$
- Mass of $^{11}_{5}\text{B}$ ($m_2$) = $11.00931\text{ u}$
- Average atomic mass ($m_{\text{avg}}$) = $10.811\text{ u}$
- To Find: Abundance of $^{10}_{5}\text{B}$ ($x\%$) and abundance of $^{11}_{5}\text{B}$ ($(100 - x)\%$).
- Formula:
$$m_{\text{avg}} = \frac{x \cdot m_1 + (100 - x) \cdot m_2}{100}$$
- Substitution & Calculation:
$$10.811 = \frac{x(10.01294) + (100 - x)(11.00931)}{100}$$
$$1081.1 = 10.01294 x + 1100.931 - 11.00931 x$$
$$1081.1 - 1100.931 = (10.01294 - 11.00931) x$$
$$-19.831 = -0.99637 x$$
$$x = \frac{19.831}{0.99637} \approx 19.90\%$$
$$\text{Abundance of } ^{11}_{5}\text{B} = 100 - 19.90 = 80.10\%$$
- Final Answer: Abundance of $^{10}_{5}\text{B}$ is 19.9% and abundance of $^{11}_{5}\text{B}$ is 80.1%.
Exercise 13.3
Question: Calculate the binding energy per nucleon of a $^{4}_{2}\text{He}$ nucleus. Given: Mass of $^{4}_{2}\text{He}$ nucleus = $4.00150\text{ u}$, Mass of proton $m_p = 1.007276\text{ u}$, Mass of neutron $m_n = 1.008665\text{ u}$.
Solution:
- Given:
- Helium nucleus $^{4}_{2}\text{He}$: Atomic number $Z = 2$, Mass number $A = 4$, Number of neutrons $N = A - Z = 2$
- Nuclear mass $M_{\text{He}} = 4.00150\text{ u}$
- Proton mass $m_p = 1.007276\text{ u}$
- Neutron mass $m_n = 1.008665\text{ u}$
- To Find: Binding energy per nucleon ($E_{bn}$).
- Formulas:
- Total mass of individual constituents: $M' = Z m_p + N m_n$
- Mass defect: $\Delta m = M' - M_{\text{He}}$
- Total Binding Energy: $E_b = \Delta m \times 931.5\text{ MeV}$
- Binding Energy per Nucleon: $E_{bn} = \frac{E_b}{A}$
- Substitution & Calculation:
$$M' = (2 \times 1.007276\text{ u}) + (2 \times 1.008665\text{ u}) = 2.014552\text{ u} + 2.017330\text{ u} = 4.031882\text{ u}$$
$$\Delta m = 4.031882\text{ u} - 4.001500\text{ u} = 0.030382\text{ u}$$
$$E_b = 0.030382 \times 931.5\text{ MeV} = 28.300833\text{ MeV} \approx 28.30\text{ MeV}$$
$$E_{bn} = \frac{28.300833\text{ MeV}}{4} = 7.0752\text{ MeV/nucleon}$$
- Final Answer: Binding energy per nucleon of $^{4}_{2}\text{He}$ is 7.075 MeV/nucleon.
Exercise 13.4
Question: A given coin has a mass of $3.0\text{ g}$. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity, assume that the coin is entirely made of $^{63}_{29}\text{Cu}$ atoms (mass of $^{63}_{29}\text{Cu} = 62.92960\text{ u}$, $m_p = 1.007276\text{ u}$, $m_n = 1.008665\text{ u}$, $m_e = 0.000548\text{ u}$).
Solution:
- Given:
- Mass of copper coin $m = 3.0\text{ g}$
- Molar mass of copper $M_{\text{molar}} \approx 63\text{ g/mol}$
- Avogadro number $N_A = 6.022 \times 10^{23}\text{ atoms/mol}$
- Copper atom $^{63}_{29}\text{Cu}$: $Z = 29$, $N = 63 - 29 = 34$
- Atomic mass of $^{63}_{29}\text{Cu} = 62.92960\text{ u}$
- Mass of 1 hydrogen atom ($m_H \approx m_p + m_e$) = $1.007825\text{ u}$
- Mass of neutron $m_n = 1.008665\text{ u}$
- Step 1: Calculate the number of atoms in the coin ($N_{\text{total}}$):
$$N_{\text{total}} = \frac{m}{M_{\text{molar}}} \times N_A = \frac{3.0}{63} \times 6.022 \times 10^{23} \approx 2.8676 \times 10^{22}\text{ atoms}$$
- Step 2: Calculate mass defect ($\Delta m$) of a single $^{63}_{29}\text{Cu}$ nucleus:
$$\text{Mass of constituents} = 29 \times m_H + 34 \times m_n$$
$$= (29 \times 1.007825\text{ u}) + (34 \times 1.008665\text{ u}) = 29.226925\text{ u} + 34.294610\text{ u} = 63.521535\text{ u}$$
$$\Delta m = 63.521535\text{ u} - 62.929600\text{ u} = 0.591935\text{ u}$$
- Step 3: Calculate the binding energy per nucleus:
$$E_b = 0.591935 \times 931.5\text{ MeV} \approx 551.387\text{ MeV}$$
- Step 4: Calculate the total nuclear energy required:
$$E_{\text{total}} = N_{\text{total}} \times E_b = (2.8676 \times 10^{22}) \times 551.387\text{ MeV} \approx 1.581 \times 10^{25}\text{ MeV}$$
$$\text{In Joules: } E_{\text{total}} = 1.581 \times 10^{25} \times 1.602 \times 10^{-13}\text{ J} \approx 2.53 \times 10^{12}\text{ J}$$
- Final Answer: The total energy required to separate all nucleons is 2.53 × 10¹² J (or 1.58 × 10²&sup5; MeV).
Exercise 13.5
Question: Calculate the energy released in the fission of $1.0\text{ kg}$ of $^{235}_{92}\text{U}$, assuming that an average of $200\text{ MeV}$ is released per fission event.
Solution:
- Given:
- Mass of $^{235}_{92}\text{U}$ sample $m = 1.0\text{ kg} = 1000\text{ g}$
- Molar mass of $^{235}_{92}\text{U} = 235\text{ g/mol}$
- Energy released per fission $E_{\text{fission}} = 200\text{ MeV} = 200 \times 10^6 \times 1.602 \times 10^{-19}\text{ J} = 3.204 \times 10^{-11}\text{ J}$
- Avogadro number $N_A = 6.023 \times 10^{23}\text{ nuclei/mol}$
- To Find: Total energy released ($E_{\text{total}}$).
- Step 1: Number of nuclei in 1 kg of $^{235}\text{U}$ ($N$):
$$N = \frac{\text{Mass}}{\text{Molar Mass}} \times N_A = \frac{1000}{235} \times 6.023 \times 10^{23} \approx 2.563 \times 10^{24}\text{ nuclei}$$
- Step 2: Total energy released:
$$E_{\text{total}} = N \times E_{\text{fission}} = 2.563 \times 10^{24} \times 3.204 \times 10^{-11}\text{ J} = 8.212 \times 10^{13}\text{ J}$$
$$\text{In kilowatt-hours (kWh): } E_{\text{total}} = \frac{8.212 \times 10^{13}}{3.6 \times 10^6\text{ J/kWh}} \approx 2.28 \times 10^7\text{ kWh}$$
- Final Answer: Energy released is 8.21 × 10¹³ J (approximately 2.28 × 10⁷ kWh).
Exercise 13.6
Question: Show that the nuclear density of any nucleus is independent of its mass number $A$. Estimate the order of magnitude of nuclear matter density.
Solution:
- Derivation:
- Let $A$ be the mass number of a nucleus and $m$ be the average mass of a nucleon ($m \approx 1.66 \times 10^{-27}\text{ kg}$).
- Total mass of the nucleus: $M = A \cdot m$
- Assuming a spherical nucleus of radius $R = R_0 A^{1/3}$, where $R_0 = 1.2 \times 10^{-15}\text{ m}$:
$$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$$
- Nuclear density $\rho$ is the ratio of mass to volume:
$$\rho = \frac{M}{V} = \frac{A \cdot m}{\frac{4}{3}\pi R_0^3 A} = \frac{3m}{4\pi R_0^3}$$
- Since the mass number $A$ cancels out completely in the expression, $\rho$ is a constant, independent of $A$.
- Order of Magnitude Estimation:
$$\rho = \frac{3 \times 1.66 \times 10^{-27}\text{ kg}}{4 \times 3.1416 \times (1.2 \times 10^{-15}\text{ m})^3} = \frac{4.98 \times 10^{-27}}{4 \times 3.1416 \times 1.728 \times 10^{-45}} \approx 2.29 \times 10^{17}\text{ kg/m}^3$$
- Final Answer: Nuclear density is independent of mass number A and has an extremely high order of magnitude of 10¹ͬ kg/m³.
Official CBSE Board Important Questions & Solutions
Mastering fundamental definitions and high-probability board questions from the official CBSE Question Bank is vital for securing a centum in Class 12 Physics. Below are essential questions with rigorous answers:
CBSE Official Question Bank Key Set
Q1: Define electric flux. Write its SI unit.
Answer: Electric flux ($\Phi$) through a given surface placed inside an electric field represents the total number of electric field lines passing normally through that surface. Mathematically, it is defined as the surface integral of the electric field vector $\vec{E}$ over the area vector $\vec{A}$:
$$\Phi = \vec{E} \cdot \vec{A} = E A \cos \theta$$
where $\theta$ is the angle between the electric field $\vec{E}$ and the normal to the surface area $\vec{A}$.
SI Unit: $\text{N}\cdot\text{m}^2/\text{C}$ (Newton meter squared per Coulomb) or $\text{V}\cdot\text{m}$ (Volt meter).
Q2: State Coulomb's Law. Write its vector form.
Answer: Coulomb's Law states that the magnitude of the electrostatic force of attraction or repulsion between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The force acts along the line joining the centers of the two charges:
$$F = \frac{1}{4\pi \varepsilon_0}\frac{|q_1 q_2|}{r^2}$$
Vector Form: The force $\vec{F}_{12}$ exerted on charge $q_1$ by charge $q_2$ separated by position vector $\vec{r}_{12}$ is given by:
$$\vec{F}_{12} = \frac{1}{4\pi \varepsilon_0}\frac{q_1 q_2}{r^2}\hat{r}_{21} = \frac{1}{4\pi \varepsilon_0}\frac{q_1 q_2}{|\vec{r}_1 - \vec{r}_2|^3}(\vec{r}_1 - \vec{r}_2)$$
Q3: What is an equipotential surface? List its properties.
Answer: An equipotential surface is any geometric surface over which the electric potential has the same constant value at every point ($V_A = V_B = \text{constant}$).
Key Properties:
- Zero Work Done: No work is done in moving an electric test charge from one point to another on an equipotential surface ($W = q \Delta V = 0$).
- Perpendicular Electric Field: The electric field lines are always normal to the equipotential surface at every point ($\vec{E} \perp d\vec{r}$).
- No Intersection: Two equipotential surfaces can never intersect each other, as two different values of electric potential cannot exist at the point of intersection.
- Field Strength Indicator: Equipotential surfaces are closer together in regions of strong electric fields and farther apart in regions of weak electric fields ($E = -dV/dr$).
Q4: Define drift velocity. How is it related to current?
Answer: Drift velocity ($v_d$) is defined as the average velocity with which free electrons in a metallic conductor drift towards the positive terminal under the influence of an applied external electric field ($v_d = \frac{e E \tau}{m}$, where $\tau$ is relaxation time).
Relation to Electric Current:
$$I = n A e v_d$$
where $I$ is the electric current, $n$ is the number density of free electrons, $A$ is the cross-sectional area of the conductor, and $e$ is the elementary charge ($1.6 \times 10^{-19}\text{ C}$).
Q5: State Kirchhoff's laws for electrical circuits.
Answer:
- Kirchhoff's First Law (Junction Rule / KCL): In any electrical network, the algebraic sum of currents meeting at any junction is zero. That is, total current entering a junction equals total current leaving it ($\sum I = 0$). This rule is a direct consequence of the law of conservation of electric charge.
- Kirchhoff's Second Law (Loop Rule / KVL): In any closed loop of an electrical circuit, the algebraic sum of changes in potential (including EMFs and potential drops across resistors) around the loop is equal to zero ($\sum \Delta V = \sum \mathcal{E} - \sum I R = 0$). This rule is a direct consequence of the law of conservation of energy.
Q6: What is total internal reflection? State the conditions for its occurrence.
Answer: Total Internal Reflection (TIR) is the optical phenomenon in which a ray of light traveling from an optically denser medium to an optically rarer medium is completely reflected back into the denser medium at the interface without any refraction into the rarer medium.
Essential Conditions:
- The light ray must travel from an optically denser medium towards an optically rarer medium.
- The angle of incidence ($i$) in the denser medium must be strictly greater than the critical angle ($i_c$) for the given pair of media ($\sin i_c = 1/\mu$).
Q7: What is the electric field at the surface of a charged spherical conductor of radius $R$ and charge $Q$?
Answer: For a spherical conductor of radius $R$ carrying a total charge $Q$ distributed uniformly over its surface, the electric field at any surface point ($r = R$) is:
$$E = \frac{1}{4\pi \varepsilon_0}\frac{Q}{R^2} = \frac{\sigma}{\varepsilon_0}$$
where $\sigma = \frac{Q}{4\pi R^2}$ is the surface charge density. The field is directed radially outwards if $Q > 0$ and radially inwards if $Q < 0$. Inside the conductor ($r < R$), the electrostatic field is strictly zero ($E = 0$).
Q8: Two point charges $+2\,\mu\text{C}$ and $-2\,\mu\text{C}$ are placed $5\text{ cm}$ apart. What is the electric dipole moment?
Answer:
- Given: Magnitude of charge $q = 2\,\mu\text{C} = 2 \times 10^{-6}\text{ C}$, Separation distance $2a = 5\text{ cm} = 0.05\text{ m}$.
- Formula: Electric dipole moment $\vec{p} = q \times 2\vec{a}$
- Calculation:
$$p = 2 \times 10^{-6}\text{ C} \times 0.05\text{ m} = 1.0 \times 10^{-7}\text{ C}\cdot\text{m}$$
- Direction: Directed along the dipole axis from the negative charge ($-2\,\mu\text{C}$) to the positive charge ($+2\,\mu\text{C}$).
- Final Value: p = 1.0 × 10⁻⁷ C·m
High-Yield Chapter 13 Nuclei Board Exam Questions
Q9: Draw the curve showing the variation of binding energy per nucleon ($E_{bn}$) with mass number $A$. State two main conclusions drawn from this curve.
Answer:
The binding energy per nucleon curve increases rapidly for light nuclei ($A < 20$), exhibits sharp local peaks for exceptionally stable configurations ($^{4}\text{He}$, $^{12}\text{C}$, $^{16}\text{O}$), reaches a broad maximum of $\approx 8.75\text{ MeV/nucleon}$ around $A = 56$ ($^{56}\text{Fe}$), and drops steadily to $\approx 7.6\text{ MeV/nucleon}$ for heavy nuclei like $^{238}\text{U}$.
Key Conclusions:
- Nuclear Fission of Heavy Nuclei: For heavy nuclei ($A > 200$), $E_{bn}$ is lower ($\approx 7.6\text{ MeV}$). When a heavy nucleus breaks into two intermediate nuclei ($A \approx 100\text{ to }120$), the binding energy per nucleon increases to $\approx 8.5\text{ MeV}$. The nucleons become more tightly bound, releasing massive energy ($Q \approx 200\text{ MeV}$).
- Nuclear Fusion of Light Nuclei: For very light nuclei ($A \le 10$), $E_{bn}$ is low. When two very light nuclei fuse together to form a heavier, more tightly bound nucleus (such as $^{4}_{2}\text{He}$), the binding energy per nucleon increases drastically, releasing immense energy.
Common Mistakes and Tips
Board examiners frequently penalize small conceptual and arithmetic errors in Chapter 13. Pay close attention to these common pitfalls:
- Confusing Atomic Masses with Nuclear Masses: Textbook tables list atomic masses (which include atomic electrons), not purely nuclear masses. When computing the mass defect $\Delta m$, always express the masses in terms of hydrogen atoms $m_H = m_p + m_e$ so that the electron masses cancel out accurately on both sides.
- Direct Conversion of ‘u’ to ‘kg’ Unnecessarily: Do not convert unified atomic mass units ($\text{u}$) into kilograms first and then apply $E = m c^2$. Instead, multiply the mass defect in $\text{u}$ directly by $931.5\text{ MeV}$ to save time and prevent calculation mistakes.
- Neglecting to Divide by Mass Number $A$: Nuclear stability is determined strictly by binding energy per nucleon ($E_{bn} = E_b / A$), NOT total binding energy ($E_b$). A heavy nucleus like $^{238}\text{U}$ has a much higher total binding energy than $^{56}\text{Fe}$, but $^{56}\text{Fe}$ is far more stable because its $E_{bn}$ ($8.75\text{ MeV}$) exceeds that of $^{238}\text{U}$ ($7.6\text{ MeV}$).
- Radius Ratio Errors: In problems asking for the ratio of radii of two nuclei, students often forget the cube root ($R_1 / R_2 = (A_1 / A_2)^{1/3}$). For instance, the radius ratio of $^{27}_{13}\text{Al}$ to $^{125}_{53}\text{I}$ is $(27/125)^{1/3} = 3/5$, not $27/125$.
- Assuming Nuclear Density Changes with Mass: Nuclear density is completely constant across all nuclides. Whether you calculate density for Hydrogen or Lead, it remains approximately $2.3 \times 10^{17}\text{ kg/m}^3$.
Board Exam Relevance and Marking Scheme
In the CBSE Class 12 Physics paper, Chapter 13 questions typically appear across different sections:
- Section A (1 Mark - MCQs & Assertion-Reason): Questions on nuclear radius ratios, properties of nuclear force, and identifying isotopes/isobars/isotones.
- Section B (2 Marks - Short Answer): Conceptual questions on nuclear density independence from $A$, reasons for the constancy of binding energy in the range $30 < A < 170$, and characteristics of nuclear force.
- Section C (3 Marks - Short Answer Numerical): Quantitative problems on mass defect, binding energy per nucleon, and energy released in fission or fusion reactions.
- Section E (5 Marks - Case-Based / Long Answer): Comprehensive questions combining binding energy curve analysis with fission/fusion numerical derivations.
More NCERT Solutions and Practice
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