Nuclei — Class 12 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Physics chapter "Nuclei" — 5 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Nuclei" — 5 important questions with detailed answers for CBSE board exam preparation.
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 12 Physics — Nuclei. Step-by-step answers to all 5 textbook questions from this chapter, written for CBSE board preparation and free to use.
Nuclei — All 5 Questions Solved
Q1. What is nuclear binding energy? Derive the expression and explain its significance.
Nuclear Binding Energy: Energy released when nucleons combine to form nucleus; also the energy required to completely separate a nucleus into individual nucleons.
Mass Defect:
Mass of separated nucleons > mass of nucleus
Δm = (Z × m_p + N × m_n) - M_nucleus
Where:
Z = number of protons
N = number of neutrons (A - Z)
m_p = mass of proton = 1.007276 u
m_n = mass of neutron = 1.008665 u
M = mass of nucleus
u = atomic mass unit = 1.66054 × 10⁻²⁷ kg
Binding Energy:
BE = Δm × c² = (Z × m_p + N × m_n - M) × c²
Or in energy units (1 u = 931.5 MeV/c²):
BE = Δm (in u) × 931.5 MeV
Binding Energy per Nucleon:
BE/A = binding energy / mass number
Example: ₆C¹² nucleus
Δm = 6 × 1.007276 + 6 × 1.008665 - 12.000000
Δm = 6.043656 + 6.051990 - 12.000000 = 0.095646 u
BE = 0.095646 × 931.5 = 89.1 MeV
BE/A = 89.1/12 = 7.42 MeV/nucleon
Significance:
1. Measure of nuclear stability (higher BE/A → more stable)
2. Used in nuclear reactions (energy released)
3. Iron-56 most stable: BE/A ≈ 8.8 MeV
4. Light nuclei fuse → heavy nuclei (fusion energy)
5. Heavy nuclei split → medium nuclei (fission energy)
6. Shows conversion of mass to energy (E = mc²)
Curve of Binding Energy:
- Maximum near A = 56 (iron-56)
- Light nuclei: low BE/A (need fusion)
- Heavy nuclei: lower BE/A (can undergo fission)
- Explains stability trends
Q2. Define half-life and decay constant. Derive the radioactive decay law.
Half-life (T₁/₂): Time required for half of the nuclei in a sample to decay.
Decay Constant (λ): Probability per unit time that a nucleus will decay.
Relation: λ = 0.693/T₁/₂ = ln(2)/T₁/₂
Radioactive Decay Law Derivation:
Let N = number of nuclei at time t
Rate of decay proportional to number present:
dN/dt = -λN
Where negative sign indicates decrease
Separating variables:
dN/N = -λ dt
Integrating:
∫[N₀ to N] dN/N = -λ ∫[0 to t] dt
ln(N) - ln(N₀) = -λt
ln(N/N₀) = -λt
Taking exponential:
N = N₀e^(-λt)
Radioactive Decay Law: N(t) = N₀e^(-λt)
Half-life Derivation:
When t = T₁/₂: N = N₀/2
N₀/2 = N₀e^(-λT₁/₂)
1/2 = e^(-λT₁/₂)
ln(1/2) = -λT₁/₂
-ln(2) = -λT₁/₂
T₁/₂ = ln(2)/λ = 0.693/λ
Alternative forms:
N = N₀(1/2)^(t/T₁/₂)
N = N₀ × 2^(-t/T₁/₂)
Activity (Rate of decay):
A = -dN/dt = λN = λN₀e^(-λt)
A = A₀e^(-λt)
A₀ = λN₀ (initial activity)
Unit: 1 Becquerel (Bq) = 1 decay/second
1 Curie (Ci) = 3.7 × 10¹⁰ Bq
Mean Lifetime:
τ = 1/λ = T₁/₂/ln(2) = 1.44 × T₁/₂
Q3. Carbon-14 has half-life 5730 years. A sample has activity 100 Bq. Find the number of C-14 nuclei and activity after 2000 years.
Given:
Half-life: T₁/₂ = 5730 years
Initial activity: A₀ = 100 Bq
Time elapsed: t = 2000 years
Part 1: Initial Number of Nuclei
Decay constant: λ = ln(2)/T₁/₂ = 0.693/5730 years
λ = 1.21 × 10⁻⁴ year⁻¹
Converting to seconds:
T₁/₂ = 5730 × 365.25 × 24 × 3600 = 1.808 × 10¹¹ seconds
λ = 0.693/(1.808 × 10¹¹) = 3.83 × 10⁻¹² s⁻¹
From A₀ = λN₀:
N₀ = A₀/λ = 100/(3.83 × 10⁻¹²) = 2.61 × 10¹³ nuclei
Part 2: Activity After 2000 Years
Using decay law: A = A₀e^(-λt)
Calculating exponent:
λt = (1.21 × 10⁻⁴) × 2000 = 0.242
e^(-0.242) = 0.785
A = 100 × 0.785 = 78.5 Bq
Alternative using half-life:
A = A₀ × (1/2)^(t/T₁/₂)
A = 100 × (1/2)^(2000/5730)
A = 100 × (1/2)^0.349
A = 100 × 0.785 = 78.5 Bq
Number of nuclei after 2000 years:
N = N₀ × (1/2)^(t/T₁/₂)
N = 2.61 × 10¹³ × 0.785 = 2.05 × 10¹³ nuclei
Answers:
- Initial number of C-14 nuclei: 2.61 × 10¹³
- Activity after 2000 years: 78.5 Bq
- Number of nuclei after 2000 years: 2.05 × 10¹³
- About 21.5% decay in 2000 years (78.5% remains)
Q4. Explain nuclear fission and fusion. Compare energy release and conditions required.
Nuclear Fission: Splitting of heavy nucleus into two lighter nuclei, releasing energy.
Fission Process:
1. Heavy nucleus (U-235, Pu-239) absorbs slow neutron
2. Nucleus becomes unstable and splits
3. Two fission fragments produced (mass number ≈ A/2)
4. 2-3 neutrons released
5. Large energy released (~200 MeV)
Example: U-235 + n → (fission fragments) + 3n + 200 MeV
Typical: ₂₃₅U + n → ₉₂Kr + ₁₄₁Ba + 3n
Energy Release:
ΔE = (M_initial - M_final) × c²
About 200 MeV per fission
1 kg U-235: 8.2 × 10¹³ J
Comparable to 2000 tons of TNT
Conditions:
- Heavy nuclei only (A > 230)
- Requires neutron bombardment
- Can be controlled (nuclear reactor) or uncontrolled (bomb)
- Chain reaction possible (each fission produces neutrons for next)
Nuclear Fusion: Combining of light nuclei into heavier nucleus, releasing energy.
Fusion Process:
1. Two light nuclei approach with high velocity
2. Overcome Coulomb repulsion
3. Combine to form heavier nucleus
4. Energy released (~25 MeV or more)
Example: ₁H² + ₁H³ → ₂He⁴ + n + 17.6 MeV
Or: ₁H¹ + ₁H¹ → ₁H² + e⁺ + νₑ + 0.42 MeV
Energy Release:
1 kg deuterium: 3.3 × 10¹⁴ J
About 4× more than uranium fission
Conditions:
- Light nuclei only (A < 30)
- Extremely high temperature (10⁶ - 10⁸ K)
- High pressure and density
- Occurs in sun and stars (thermonuclear reactions)
Comparison:
Fission:
- Heavy nuclei split
- Chain reaction (self-sustaining)
- Can be controlled
- Radioactive waste problem
- Requires neutron source
- About 200 MeV/fission
Fusion:
- Light nuclei combine
- No chain reaction (hard to control)
- Difficult to maintain
- No long-lived waste
- Requires extreme conditions
- About 25-100 MeV per reaction
- More energy per unit mass
Practical Applications:
Fission: Nuclear power plants, nuclear weapons
Fusion: Sun and stars, experimental reactors (ITER), hydrogen bombs
Advantages Fusion over Fission:
- Abundant fuel (deuterium in seawater)
- No long-lived radioactive waste
- Higher energy output per reaction
- Inherently safer (no chain reaction)
- Challenge: temperature control
Q5. An alpha particle (He nucleus) is emitted from Ra-226. Write the decay equation and calculate the Q-value.
Alpha Decay Process:
Alpha particle = He-4 nucleus (2 protons + 2 neutrons)
Decay Equation:
₈₈Ra-226 → ₈₆Rn-222 + ₂He-4
Or: ₂₂₆Ra → ₂₂₂Rn + ⁴He
Mass number: 226 = 222 + 4 ✓
Atomic number: 88 = 86 + 2 ✓
Q-value Calculation:
Q = (M_parent - M_daughter - M_alpha) × c²
Using atomic mass units (u = 931.5 MeV/c²):
M_Ra-226 = 226.025406 u
M_Rn-222 = 222.017571 u
M_He-4 = 4.002603 u
Mass defect:
Δm = 226.025406 - 222.017571 - 4.002603
Δm = 226.025406 - 226.020174 = 0.005232 u
Q-value:
Q = 0.005232 × 931.5 MeV = 4.87 MeV
Interpretation:
- Positive Q means energy is released (exothermic)
- This energy is kinetic energy of products
- Divided between Rn nucleus and alpha particle
Kinetic Energy Distribution:
Using momentum conservation (initial momentum = 0):
p_Rn = p_alpha
m_Rn × v_Rn = m_alpha × v_alpha
From energy conservation:
Q = KE_Rn + KE_alpha
KE_alpha = Q × m_Rn/(m_Rn + m_alpha)
KE_alpha = 4.87 × 222/(222 + 4) ≈ 4.78 MeV
KE_Rn = Q × m_alpha/(m_Rn + m_alpha)
KE_Rn = 4.87 × 4/(222 + 4) ≈ 0.087 MeV
Notes:
- Most energy goes to lighter alpha particle (inverse mass ratio)
- Q-value is intrinsic property of decay
- Positive Q required for spontaneous decay
- Recoil nucleus has low velocity but still ionizes surroundings
Answers:
- Decay equation: ₈₈Ra-226 → ₈₆Rn-222 + ₂He-4
- Q-value: 4.87 MeV
- Kinetic energy of alpha: ~4.78 MeV
- Kinetic energy of recoil Rn: ~0.087 MeV
Nuclei — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. The nucleus of an atom consists of:
- A. Electrons and protons
- B. Protons and neutrons
- C. Electrons, protons, and neutrons
- D. Electrons and neutrons
Answer: B. Protons and neutrons — The nucleus of an atom is composed of positively charged protons and neutral neutrons. Electrons orbit the nucleus outside.
MCQ 2. One atomic mass unit (amu) is approximately equal to:
- A. 1.66 x 10⁻²⁴ kg
- B. 931.5 MeV/c²
- C. Mass of a proton
- D. 1/12th the mass of a carbon-12 atom
Answer: D. 1/12th the mass of a carbon-12 atom — One atomic mass unit (amu) is defined as exactly 1/12th the mass of an atom of carbon-12 in its ground state.
MCQ 3. The radius of a nucleus is proportional to the:
- A. Mass number (A)
- B. Cube root of the mass number (A¹/³)
- C. Atomic number (Z)
- D. Square of the mass number (A²)
Answer: B. Cube root of the mass number (A¹/³) — The empirical formula for the nuclear radius R is given by R = R₀A¹/³, where R₀ is a constant and A is the mass number.
MCQ 4. The nuclear density of all nuclei is:
- A. Different for different nuclei
- B. Dependent on the atomic number
- C. Approximately constant
- D. Dependent on the mass number
Answer: C. Approximately constant — Since the nuclear volume is proportional to the mass number (A) and the nuclear mass is also approximately proportional to A, the nuclear density is roughly constant for all nuclei.
MCQ 5. The mass defect of a nucleus is defined as the difference between:
- A. The mass of the nucleus and the sum of the masses of its constituent nucleons
- B. The sum of the masses of its constituent nucleons and the actual mass of the nucleus
- C. The mass of protons and the mass of neutrons in the nucleus
- D. The mass of the atom and the sum of the masses of electrons
Answer: B. The sum of the masses of its constituent nucleons and the actual mass of the nucleus — Mass defect (Δm) is the difference between the sum of the masses of the individual protons and neutrons (constituent nucleons) and the actual measured mass of the nucleus.
MCQ 6. What does a higher binding energy per nucleon indicate?
- A. The nucleus is less stable.
- B. The nucleus is more stable.
- C. The nucleus is radioactive.
- D. The nucleus has a larger mass defect.
Answer: B. The nucleus is more stable. — A higher binding energy per nucleon signifies that more energy is required to separate the nucleons, thus indicating greater stability of the nucleus.
MCQ 7. If 1 amu is equivalent to 931.5 MeV of energy, how much energy is released when 0.5 amu of mass is converted into energy?
- A. 931.5 MeV
- B. 465.75 MeV
- C. 1863 MeV
- D. 93.15 MeV
Answer: B. 465.75 MeV — If 1 amu corresponds to 931.5 MeV, then 0.5 amu will correspond to 0.5 * 931.5 MeV = 465.75 MeV.
MCQ 8. Which of the following is NOT a characteristic of the nuclear force?
- A. It is charge-independent.
- B. It is a short-range force.
- C. It is the strongest known fundamental force.
- D. It is a gravitational force.
Answer: D. It is a gravitational force. — The nuclear force (strong force) is much stronger than gravity, acts over very short distances, and is independent of the charge of the nucleons. It is not a gravitational force.
MCQ 9. In an alpha decay, what happens to the atomic number (Z) and mass number (A) of the parent nucleus?
- A. Z decreases by 2, A decreases by 4
- B. Z increases by 1, A remains unchanged
- C. Z decreases by 1, A decreases by 4
- D. Z remains unchanged, A decreases by 4
Answer: A. Z decreases by 2, A decreases by 4 — An alpha particle is a helium nucleus (⁴₂He). When a nucleus undergoes alpha decay, its atomic number (Z) decreases by 2, and its mass number (A) decreases by 4.
MCQ 10. During beta-minus (β⁻) decay, a neutron transforms into a proton. Which of the following is also emitted along with an electron?
- A. A positron
- B. An alpha particle
- C. An antineutrino
- D. A neutrino
Answer: C. An antineutrino — In beta-minus decay, a neutron converts into a proton, emitting an electron (beta particle) and an antineutrino (ν̅ₑ) to conserve lepton number and energy.
MCQ 11. Gamma decay involves the emission of:
- A. High-energy electrons
- B. High-energy photons
- C. Helium nuclei
- D. Neutrons
Answer: B. High-energy photons — Gamma decay occurs when an excited nucleus transitions to a lower energy state by emitting high-energy electromagnetic radiation, which are photons.
MCQ 12. If the half-life of a radioactive isotope is 10 days, what fraction of the original sample will remain after 30 days?
- A. 1/2
- B. 1/4
- C. 1/8
- D. 1/16
Answer: C. 1/8 — After 1 half-life (10 days), 1/2 remains. After 2 half-lives (20 days), 1/4 remains. After 3 half-lives (30 days), 1/8 remains.
MCQ 13. The decay constant (λ) of a radioactive substance is related to its half-life (T½) by the equation:
- A. λ = ln(2) / T½
- B. λ = T½ / ln(2)
- C. λ = 1 / T½
- D. λ = ln(2) * T½
Answer: A. λ = ln(2) / T½ — The half-life (T½) is the time required for half of the radioactive nuclei to decay, and it is related to the decay constant (λ) by the formula T½ = ln(2) / λ, hence λ = ln(2) / T½.
MCQ 14. Nuclear fission is the process where:
- A. Two light nuclei combine to form a heavier nucleus.
- B. A heavy nucleus splits into two or more lighter nuclei.
- C. A nucleus emits an alpha particle.
- D. A nucleus absorbs a neutron and becomes stable.
Answer: B. A heavy nucleus splits into two or more lighter nuclei. — Nuclear fission is the process in which a heavy nucleus (like Uranium-235) splits into two or more smaller nuclei when struck by a neutron, releasing a tremendous amount of energy.
MCQ 15. The energy released in nuclear fusion reactions is primarily due to:
- A. Conversion of mass into energy.
- B. Conversion of chemical energy into nuclear energy.
- C. Breaking of chemical bonds.
- D. Release of kinetic energy of initial particles.
Answer: A. Conversion of mass into energy. — In nuclear fusion, lighter nuclei combine to form a heavier nucleus. The mass of the resulting nucleus is slightly less than the sum of the masses of the original nuclei, and this mass difference (mass defect) is converted into a large amount of energy according to E=mc².
MCQ 16. The binding energy per nucleon curve peaks around which mass number (A)?
- A. A = 20-30
- B. A = 50-60
- C. A = 100-120
- D. A = 200-220
Answer: B. A = 50-60 — The binding energy per nucleon curve reaches its maximum value (around 8.7 MeV per nucleon) for nuclei with mass numbers (A) in the range of 50-60, typically around Iron-56, indicating maximum stability.
MCQ 17. In a nuclear reactor, what is the purpose of control rods?
- A. To speed up the fission process.
- B. To absorb excess neutrons and regulate the chain reaction.
- C. To moderate the speed of neutrons.
- D. To cool down the reactor core.
Answer: B. To absorb excess neutrons and regulate the chain reaction. — Control rods, typically made of cadmium or boron, are used in nuclear reactors to absorb excess neutrons, thereby regulating the rate of the chain reaction and preventing it from becoming supercritical.
MCQ 18. Which of the following is commonly used as a moderator in a nuclear reactor?
- A. Cadmium
- B. Heavy water (D₂O)
- C. Plutonium
- D. Uranium
Answer: B. Heavy water (D₂O) — A moderator, such as heavy water (D₂O) or graphite, is used to slow down fast neutrons produced during fission, making them more likely to cause further fission reactions.
MCQ 19. A radioactive sample has an activity of 2000 disintegrations per second (dps). After 20 minutes, its activity drops to 500 dps. What is the half-life of the sample?
- A. 5 minutes
- B. 10 minutes
- C. 15 minutes
- D. 20 minutes
Answer: B. 10 minutes — The activity dropped from 2000 dps to 500 dps (2000 → 1000 → 500), which represents two half-lives. Therefore, two half-lives occurred in 20 minutes, making one half-life equal to 10 minutes.
MCQ 20. Carbon dating is a technique used to determine the age of ancient artifacts by measuring the ratio of:
- A. ¹²C to ¹⁴C
- B. ¹⁶O to ¹⁸O
- C. ²³⁸U to ²⁰⁶Pb
- D. ¹H to ²H
Answer: A. ¹²C to ¹⁴C — Carbon dating uses the known half-life of radioactive carbon-14 (¹⁴C) and its decaying proportion relative to stable carbon-12 (¹²C) to determine the age of organic materials.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.