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Nuclei NCERT Solutions and Important Questions
If Chapter 13 Nuclei is giving you tension, don't worry. This guide provides the complete, Updated NCERT Solutions for Class 12 Physics Chapter 13. You will learn everything from nuclear binding energy to nuclear fission and fusion. This is a highly scoring chapter, perfect for boosting your CBSE Board Exam 2026 score and building a solid foundation for JEE and NEET. Bina tension ke, let's start!
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Chapter NameNuclei (Chapter 13)
SubjectPhysics
ClassClass 12
BoardCBSE (2026-27 Syllabus)
Important TopicsMass Defect, Binding Energy Curve, Nuclear Force, Fission & Fusion
Difficulty LevelEasy to Moderate (Calculation heavy)
Exam Weightage~3-4 Marks (Part of Modern Physics unit)
Learning Objectives
After completing this chapter, students will be able to:
- Understand the composition and size of a nucleus.
- Calculate mass defect and the corresponding nuclear binding energy.
- Analyze the Binding Energy per Nucleon curve and its implications for nuclear stability.
- Explain the characteristics of strong nuclear forces.
- Differentiate between nuclear fission and nuclear fusion processes.
- Apply Einstein's mass-energy equivalence principle to nuclear reactions.
Key Concepts, Definitions & Formulas
- Atomic Mass Unit (u): One atomic mass unit is defined as $1/12\text{th}$ of the mass of a carbon-12 atom.
$$1\text{ u} = 1.66 \times 10^{-27}\text{ kg}$$ - Einstein's Mass-Energy Equivalence: Mass and energy are interconvertible.
$$E = mc^2$$
Energy equivalent of 1 atomic mass unit: $1\text{ u} \approx 931.5\text{ MeV}$ - Nuclear Radius: The radius of a nucleus scales with its mass number ($A$).
$$R = R_0 A^{1/3}$$
Where $R_0 = 1.2 \times 10^{-15}\text{ m}$ (or $1.2\text{ fm}$). - Mass Defect ($\Delta m$): The difference between the sum of the masses of the individual nucleons forming a nucleus and the actual mass of the nucleus.
$$\Delta m = [Z m_p + (A - Z) m_n] - M$$ - Nuclear Binding Energy ($E_b$): The energy required to break a nucleus into its constituent protons and neutrons.
$$E_b = \Delta m \times c^2 \quad \text{(in Joules)}$$
$$E_b = \Delta m \times 931.5 \quad \text{(in MeV)}$$ - Nuclear Force: The strong, short-range attractive force that binds protons and neutrons together inside the nucleus, overcoming electrostatic repulsion.
- Nuclear Fission: The splitting of a heavy nucleus into two lighter nuclei of comparable masses, releasing a huge amount of energy.
- Nuclear Fusion: The fusing of two light nuclei to form a heavier, more stable nucleus, releasing energy (the process that powers the stars).
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Full NCERT Solutions
(Note: These solutions strictly follow the latest rationalized NCERT textbook for 2026-27 Board Exams. Questions related to radioactivity have been excluded as per the new syllabus).
Question 13.1: 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 u and 7.01600 u, respectively. Find the atomic mass of lithium.
Answer:
Step 1: Note down the given values.
Mass of first isotope ($m_1$) = $6.01512\text{ u}$
Abundance of first isotope ($p_1$) = $7.5\%$
Mass of second isotope ($m_2$) = $7.01600\text{ u}$
Abundance of second isotope ($p_2$) = $92.5\%$
Step 2: Apply the weighted average formula.
The atomic mass of an element is the weighted average of the masses of its isotopes:
$$\text{Average Mass} = \frac{m_1 p_1 + m_2 p_2}{100}$$
Step 3: Calculate the final value.
$$\text{Average Mass} = \frac{(6.01512 \times 7.5) + (7.01600 \times 92.5)}{100}$$
$$\text{Average Mass} = \frac{45.1134 + 648.98}{100}$$
$$\text{Average Mass} = \frac{694.0934}{100} = 6.941\text{ u}$$
Final Answer: The atomic mass of lithium is $6.941\text{ u}$.
Step 1: Note down the given values.
Mass of first isotope ($m_1$) = $6.01512\text{ u}$
Abundance of first isotope ($p_1$) = $7.5\%$
Mass of second isotope ($m_2$) = $7.01600\text{ u}$
Abundance of second isotope ($p_2$) = $92.5\%$
Step 2: Apply the weighted average formula.
The atomic mass of an element is the weighted average of the masses of its isotopes:
$$\text{Average Mass} = \frac{m_1 p_1 + m_2 p_2}{100}$$
Step 3: Calculate the final value.
$$\text{Average Mass} = \frac{(6.01512 \times 7.5) + (7.01600 \times 92.5)}{100}$$
$$\text{Average Mass} = \frac{45.1134 + 648.98}{100}$$
$$\text{Average Mass} = \frac{694.0934}{100} = 6.941\text{ u}$$
Final Answer: The atomic mass of lithium is $6.941\text{ u}$.
Question 13.2: Boron has two stable isotopes, $^{10}_{5}\text{B}$ and $^{11}_{5}\text{B}$. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of $^{10}_{5}\text{B}$ and $^{11}_{5}\text{B}$.
Answer:
Step 1: Define variables for abundances.
Let the abundance of $^{10}_{5}\text{B}$ be $x\%$.
Therefore, the abundance of $^{11}_{5}\text{B}$ will be $(100 - x)\%$.
Step 2: Set up the weighted average equation.
Given Average Mass = $10.811\text{ u}$
$$10.811 = \frac{10.01294 \times x + 11.00931 \times (100 - x)}{100}$$
Step 3: Solve for x.
$$1081.1 = 10.01294x + 1100.931 - 11.00931x$$
$$1081.1 - 1100.931 = x(10.01294 - 11.00931)$$
$$-19.831 = -0.99637x$$
$$x = \frac{19.831}{0.99637} \approx 19.9\%$$
Final Answer: The abundance of $^{10}_{5}\text{B}$ is roughly $19.9\%$, and the abundance of $^{11}_{5}\text{B}$ is $80.1\%$.
Step 1: Define variables for abundances.
Let the abundance of $^{10}_{5}\text{B}$ be $x\%$.
Therefore, the abundance of $^{11}_{5}\text{B}$ will be $(100 - x)\%$.
Step 2: Set up the weighted average equation.
Given Average Mass = $10.811\text{ u}$
$$10.811 = \frac{10.01294 \times x + 11.00931 \times (100 - x)}{100}$$
Step 3: Solve for x.
$$1081.1 = 10.01294x + 1100.931 - 11.00931x$$
$$1081.1 - 1100.931 = x(10.01294 - 11.00931)$$
$$-19.831 = -0.99637x$$
$$x = \frac{19.831}{0.99637} \approx 19.9\%$$
Final Answer: The abundance of $^{10}_{5}\text{B}$ is roughly $19.9\%$, and the abundance of $^{11}_{5}\text{B}$ is $80.1\%$.
Question 13.3: Obtain the binding energy (in MeV) of a nitrogen nucleus $^{14}_{7}\text{N}$, given $m_N = 14.00307\text{ u}$.
Answer:
Step 1: Identify the constituents of the Nitrogen-14 nucleus.
A $^{14}_{7}\text{N}$ nucleus has $Z = 7$ protons and $A - Z = 14 - 7 = 7$ neutrons.
Step 2: Use standard masses of proton and neutron (in u).
Mass of a proton ($m_p$) = $1.00783\text{ u}$
Mass of a neutron ($m_n$) = $1.00867\text{ u}$
Actual mass of Nitrogen nucleus ($M$) = $14.00307\text{ u}$
Step 3: Calculate the Mass Defect ($\Delta m$).
$$\Delta m = [7 \times m_p + 7 \times m_n] - M$$
$$\Delta m = [7 \times 1.00783 + 7 \times 1.00867] - 14.00307$$
$$\Delta m = [7.05481 + 7.06069] - 14.00307$$
$$\Delta m = 14.11550 - 14.00307 = 0.11243\text{ u}$$
Step 4: Calculate Binding Energy ($E_b$).
Since $1\text{ u} = 931.5\text{ MeV}$:
$$E_b = 0.11243 \times 931.5 = 104.7\text{ MeV}$$
Final Answer: The binding energy of the nitrogen nucleus is $104.7\text{ MeV}$.
Step 1: Identify the constituents of the Nitrogen-14 nucleus.
A $^{14}_{7}\text{N}$ nucleus has $Z = 7$ protons and $A - Z = 14 - 7 = 7$ neutrons.
Step 2: Use standard masses of proton and neutron (in u).
Mass of a proton ($m_p$) = $1.00783\text{ u}$
Mass of a neutron ($m_n$) = $1.00867\text{ u}$
Actual mass of Nitrogen nucleus ($M$) = $14.00307\text{ u}$
Step 3: Calculate the Mass Defect ($\Delta m$).
$$\Delta m = [7 \times m_p + 7 \times m_n] - M$$
$$\Delta m = [7 \times 1.00783 + 7 \times 1.00867] - 14.00307$$
$$\Delta m = [7.05481 + 7.06069] - 14.00307$$
$$\Delta m = 14.11550 - 14.00307 = 0.11243\text{ u}$$
Step 4: Calculate Binding Energy ($E_b$).
Since $1\text{ u} = 931.5\text{ MeV}$:
$$E_b = 0.11243 \times 931.5 = 104.7\text{ MeV}$$
Final Answer: The binding energy of the nitrogen nucleus is $104.7\text{ MeV}$.
Question 13.4: Obtain the binding energy of the nuclei $^{56}_{26}\text{Fe}$ and $^{209}_{83}\text{Bi}$ in units of MeV from the following data: $m(^{56}_{26}\text{Fe}) = 55.934939\text{ u}$, $m(^{209}_{83}\text{Bi}) = 208.980388\text{ u}$.
Answer:
Part 1: For Iron ($^{56}_{26}\text{Fe}$)
Protons ($Z$) = 26, Neutrons ($N$) = 30
$$\Delta m = [26(1.00783) + 30(1.00867)] - 55.934939$$
$$\Delta m = [26.20358 + 30.2601] - 55.934939$$
$$\Delta m = 56.46368 - 55.934939 = 0.528741\text{ u}$$
Binding Energy = $0.528741 \times 931.5 \approx 492.52\text{ MeV}$
Part 2: For Bismuth ($^{209}_{83}\text{Bi}$)
Protons ($Z$) = 83, Neutrons ($N$) = 126
$$\Delta m = [83(1.00783) + 126(1.00867)] - 208.980388$$
$$\Delta m = [83.64989 + 127.09242] - 208.980388$$
$$\Delta m = 210.74231 - 208.980388 = 1.761922\text{ u}$$
Binding Energy = $1.761922 \times 931.5 \approx 1641.23\text{ MeV}$
Part 1: For Iron ($^{56}_{26}\text{Fe}$)
Protons ($Z$) = 26, Neutrons ($N$) = 30
$$\Delta m = [26(1.00783) + 30(1.00867)] - 55.934939$$
$$\Delta m = [26.20358 + 30.2601] - 55.934939$$
$$\Delta m = 56.46368 - 55.934939 = 0.528741\text{ u}$$
Binding Energy = $0.528741 \times 931.5 \approx 492.52\text{ MeV}$
Part 2: For Bismuth ($^{209}_{83}\text{Bi}$)
Protons ($Z$) = 83, Neutrons ($N$) = 126
$$\Delta m = [83(1.00783) + 126(1.00867)] - 208.980388$$
$$\Delta m = [83.64989 + 127.09242] - 208.980388$$
$$\Delta m = 210.74231 - 208.980388 = 1.761922\text{ u}$$
Binding Energy = $1.761922 \times 931.5 \approx 1641.23\text{ MeV}$
Question 13.5: A given coin has a mass of 3.0 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 (of mass 62.92960 u).
Answer:
Step 1: Find the Binding Energy of a single Copper atom.
$Z = 29$, $N = 63 - 29 = 34$
$$\Delta m = [29(1.00783) + 34(1.00867)] - 62.92960$$
$$\Delta m = [29.22707 + 34.29478] - 62.92960$$
$$\Delta m = 63.52185 - 62.92960 = 0.59225\text{ u}$$
$E_b\text{ per atom} = 0.59225 \times 931.5 \approx 551.68\text{ MeV}$
Step 2: Find the total number of atoms in the 3.0 g coin.
Number of moles = Mass / Molar Mass = $3.0 / 63$
Number of atoms = $(3.0 / 63) \times 6.023 \times 10^{23} \approx 2.868 \times 10^{22}\text{ atoms}$
Step 3: Calculate total nuclear energy.
Total Energy = Number of atoms $\times E_b\text{ per atom}$
Total Energy = $2.868 \times 10^{22} \times 551.68\text{ MeV}$
Total Energy $\approx 1.58 \times 10^{25}\text{ MeV}$
Final Answer: The total nuclear energy required is approximately $1.58 \times 10^{25}\text{ MeV}$.
Step 1: Find the Binding Energy of a single Copper atom.
$Z = 29$, $N = 63 - 29 = 34$
$$\Delta m = [29(1.00783) + 34(1.00867)] - 62.92960$$
$$\Delta m = [29.22707 + 34.29478] - 62.92960$$
$$\Delta m = 63.52185 - 62.92960 = 0.59225\text{ u}$$
$E_b\text{ per atom} = 0.59225 \times 931.5 \approx 551.68\text{ MeV}$
Step 2: Find the total number of atoms in the 3.0 g coin.
Number of moles = Mass / Molar Mass = $3.0 / 63$
Number of atoms = $(3.0 / 63) \times 6.023 \times 10^{23} \approx 2.868 \times 10^{22}\text{ atoms}$
Step 3: Calculate total nuclear energy.
Total Energy = Number of atoms $\times E_b\text{ per atom}$
Total Energy = $2.868 \times 10^{22} \times 551.68\text{ MeV}$
Total Energy $\approx 1.58 \times 10^{25}\text{ MeV}$
Final Answer: The total nuclear energy required is approximately $1.58 \times 10^{25}\text{ MeV}$.
Extra Important Questions (Board Style 2026)
Multiple Choice Questions (MCQs)
1. The ratio of nuclear radii of elements with mass numbers 27 and 64 is:
A) 3:4
B) 27:64
C) 4:3
D) 9:16
Answer: A.
Explanation: $R \propto A^{1/3}$. Therefore, $(27)^{1/3} : (64)^{1/3} = 3 : 4$. (Difficulty: Easy)
Explanation: $R \propto A^{1/3}$. Therefore, $(27)^{1/3} : (64)^{1/3} = 3 : 4$. (Difficulty: Easy)
2. As the mass number $A$ increases, the binding energy per nucleon in a nucleus:
A) Increases continuously
B) Decreases continuously
C) First increases, reaches a maximum, and then decreases
D) Remains constant
Answer: C.
Explanation: It peaks around Iron ($A=56$) at $8.8\text{ MeV/nucleon}$ and drops for heavier nuclei. (Difficulty: Medium)
Explanation: It peaks around Iron ($A=56$) at $8.8\text{ MeV/nucleon}$ and drops for heavier nuclei. (Difficulty: Medium)
3. Nuclear forces are:
A) Charge-dependent
B) Long-range forces
C) Spin-dependent
D) Weaker than gravitational forces
Answer: C.
Explanation: Nuclear forces are charge-independent, short-range, and spin-dependent. (Difficulty: Easy)
Explanation: Nuclear forces are charge-independent, short-range, and spin-dependent. (Difficulty: Easy)
4. In nuclear fission, the energy released is primarily due to:
A) High binding energy per nucleon of the parent nucleus
B) High binding energy per nucleon of the product nuclei
C) Chemical reactions
D) Electromagnetic radiation
Answer: B.
Explanation: The products are more stable and have a higher binding energy per nucleon, releasing the mass difference as energy. (Difficulty: Medium)
Explanation: The products are more stable and have a higher binding energy per nucleon, releasing the mass difference as energy. (Difficulty: Medium)
5. 1 atomic mass unit (u) is equivalent to an energy of approximately:
A) 931.5 Joules
B) 931.5 MeV
C) 931.5 eV
D) $1.6 \times 10^{-19}\text{ J}$
Answer: B. (Difficulty: Easy)
Short Answer Questions (2-3 Marks)
6. Define mass defect. How is it related to binding energy?
Answer: Mass defect ($\Delta m$) is the difference between the sum of the masses of individual nucleons making up a nucleus and the rest mass of the actual nucleus. It is related to binding energy ($E_b$) by Einstein's equation: $E_b = \Delta m \cdot c^2$. (Difficulty: Easy)
7. Why is the density of all nuclei nearly constant?
Answer:
Nuclear volume is $V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.
Volume is directly proportional to mass number ($A$).
Density = Mass / Volume = $\frac{A \times \text{mass of nucleon}}{\frac{4}{3} \pi R_0^3 A}$.
The '$A$' cancels out, making nuclear density independent of mass number. (Difficulty: Medium)
Nuclear volume is $V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.
Volume is directly proportional to mass number ($A$).
Density = Mass / Volume = $\frac{A \times \text{mass of nucleon}}{\frac{4}{3} \pi R_0^3 A}$.
The '$A$' cancels out, making nuclear density independent of mass number. (Difficulty: Medium)
8. Draw a rough sketch showing the variation of binding energy per nucleon with mass number $A$.
Answer: [Students must practice drawing the standard BE/A vs A curve. Highlight the peak at Iron-56 (8.8 MeV) and the flat region for intermediate masses ($30 < A < 170$).] (Difficulty: Medium)
9. Explain why heavy nuclei undergo nuclear fission.
Answer: Heavy nuclei (like Uranium, $A > 170$) have a lower binding energy per nucleon compared to intermediate nuclei. To achieve greater stability (higher BE/nucleon), they split into lighter nuclei, releasing immense energy. (Difficulty: Easy)
10. What is nuclear fusion? Give one example.
Answer: The process where two light nuclei combine to form a single heavier, more stable nucleus is called nuclear fusion. Example: Two hydrogen isotopes fusing to form helium in the sun. (Difficulty: Easy)
Long Answer Questions (5 Marks)
11. (a) State the properties of nuclear force. (b) How does it differ from electrostatic force?
Answer:
(a) Properties: Strongest force in nature, very short-range (operative up to ~2-3 fm), charge-independent, and spin-dependent. It shows a saturation effect.
(b) Differences: Electrostatic force is long-range, charge-dependent (can be repulsive or attractive), and obeys the inverse-square law, whereas nuclear force does not obey the inverse-square law and is purely attractive beyond 0.7 fm. (Difficulty: Hard)
(a) Properties: Strongest force in nature, very short-range (operative up to ~2-3 fm), charge-independent, and spin-dependent. It shows a saturation effect.
(b) Differences: Electrostatic force is long-range, charge-dependent (can be repulsive or attractive), and obeys the inverse-square law, whereas nuclear force does not obey the inverse-square law and is purely attractive beyond 0.7 fm. (Difficulty: Hard)
12. Explain the terms: (i) Isotopes, (ii) Isobars, and (iii) Isotones with one example each.
Answer:
(i) Isotopes: Same atomic number ($Z$), different mass number ($A$). Example: $^{1}_{1}\text{H}, ^{2}_{1}\text{H}$.
(ii) Isobars: Same mass number ($A$), different atomic number ($Z$). Example: $^{40}_{18}\text{Ar}$ and $^{40}_{20}\text{Ca}$.
(iii) Isotones: Same number of neutrons ($A-Z$), different $Z$ and $A$. Example: $^{14}_{6}\text{C}$ and $^{16}_{8}\text{O}$. (Both have 8 neutrons). (Difficulty: Medium)
(i) Isotopes: Same atomic number ($Z$), different mass number ($A$). Example: $^{1}_{1}\text{H}, ^{2}_{1}\text{H}$.
(ii) Isobars: Same mass number ($A$), different atomic number ($Z$). Example: $^{40}_{18}\text{Ar}$ and $^{40}_{20}\text{Ca}$.
(iii) Isotones: Same number of neutrons ($A-Z$), different $Z$ and $A$. Example: $^{14}_{6}\text{C}$ and $^{16}_{8}\text{O}$. (Both have 8 neutrons). (Difficulty: Medium)
Case-Based Question
13. Case Study: Energy of the Stars
The sun and other stars shine due to nuclear fusion. In the sun, hydrogen nuclei fuse to form helium in a process called the proton-proton cycle. This process requires extremely high temperatures (around $10^7\text{ K}$) and high pressures to overcome the strong electrostatic repulsion between the positively charged protons.
Q1: Why are high temperatures required for nuclear fusion?
Answer: High temperature provides the required kinetic energy for protons to overcome their mutual electrostatic (Coulomb) repulsion and come close enough for the strong nuclear force to act.
Q2: What is the main product of the proton-proton cycle in the sun?
Answer: Helium ($^{4}_{2}\text{He}$). (Difficulty: Medium)
Answer: High temperature provides the required kinetic energy for protons to overcome their mutual electrostatic (Coulomb) repulsion and come close enough for the strong nuclear force to act.
Q2: What is the main product of the proton-proton cycle in the sun?
Answer: Helium ($^{4}_{2}\text{He}$). (Difficulty: Medium)
Assertion-Reason Questions
(A) Both A and R are true and R explains A.
(B) Both A and R are true but R does not explain A.
(C) A is true but R is false.
(D) A is false but R is true.
14. Assertion (A): The mass of a nucleus is always less than the sum of the masses of its constituent protons and neutrons.
Reason (R): During the formation of a nucleus, a certain mass is converted into binding energy.
Answer: A. The reason correctly explains the assertion via Einstein's mass-energy equivalence. (Difficulty: Easy)
15. Assertion (A): Nuclear forces are effective over long macroscopic distances.
Reason (R): Nuclear forces are responsible for binding protons and neutrons together.
Answer: D. The assertion is false because nuclear forces are short-range (only a few femtometers). The reason is a true statement. (Difficulty: Easy)
Common Mistakes Students Make
- Unit Confusion in Mass Defect: Always ensure that when you calculate mass defect ($\Delta m$), the unit is in u (atomic mass units). If asked for energy in MeV, directly multiply by 931.5. Don't convert to kg unless the answer is explicitly asked in Joules.
- Protons vs. Electrons: When calculating binding energy, use the mass of a proton ($m_p$), not an electron! Sometimes, atomic mass is given instead of nuclear mass. In such cases, the mass of electrons must be subtracted carefully.
- Volume vs. Density: Remember that nuclear density is constant for all elements, but nuclear volume scales directly with the mass number ($A$).
Exam Preparation Tips
- Practice the Binding Energy Curve: Be able to draw and label the BE/A vs. A curve perfectly. Explain fission and fusion directly from this graph.
- Formula Cheat Sheet: Write down $R = R_0 A^{1/3}$ and the formula for Mass Defect on a sticky note. Roz revise karo!
- Step-by-Step Numericals: CBSE gives marks for steps. Write down the given values, the formula used, and final answers with proper units (MeV or Joules).
- Time Management: Calculations in this chapter involve precise decimals (up to 5-6 places). Practice doing these quickly to save time during the board exam.
FAQ Section
Is Chapter 13 Nuclei important for CBSE Class 12 Boards?
Yes, it is very important. Along with Dual Nature and Atoms, it forms a highly scoring unit in Modern Physics.
Have radioactivity topics been removed from the CBSE syllabus?
Yes, for the 2026-27 rationalized syllabus, alpha, beta, gamma decay, and the radioactive decay law equations are generally excluded. However, always double-check the latest circular.
Which numericals are most frequently asked from this chapter?
Questions calculating the mass defect and the binding energy of an atom (like the Nitrogen or Iron examples in NCERT) are board favorites.
Where can I download NCERT PDF and chapter notes?
You can bookmark this page or print it directly to use it as a complete study guide and solution manual!
What is the value of $R_0$ in the nuclear radius formula?
The empirical constant $R_0$ is approximately $1.2 \times 10^{-15}\text{ meters}$ (or $1.2\text{ fm}$).
Conclusion: Mastering Chapter 13 Nuclei unlocks some of the most fundamental secrets of the universe, from atomic bombs to the energy of the sun. Because the syllabus is now rationalized, the focus is purely on mass-energy equivalence and nuclear stability. Keep practicing the numericals, memorize the binding energy graph, and you will easily secure full marks in this section. Download your notes, solve PYQs, and prepare confidently! Exam board phodenge!
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Ch 4: Moving Charges and Magnetism
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