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.

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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 Jul 23, 2026

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Key Questions Covered:

What is nuclear binding energy? Derive the expression and explain its signifi…
Define half-life and decay constant. Derive the radioactive decay law.
Carbon-14 has half-life 5730 years. A sample has activity 100 Bq. Find the nu…
Explain nuclear fission and fusion. Compare energy release and conditions req…
An alpha particle (He nucleus) is emitted from Ra-226. Write the decay equati…

Solutions Summary:

Question	Status
What is nuclear binding energy? Derive the expression and…	✓ Solved
Define half-life and decay constant. Derive the radioacti…	✓ Solved
Carbon-14 has half-life 5730 years. A sample has activity…	✓ Solved
Explain nuclear fission and fusion. Compare energy releas…	✓ Solved
An alpha particle (He nucleus) is emitted from Ra-226. Wr…	✓ Solved

Showing 5 of 5 questions

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 ×...

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...

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...

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 ×...

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...

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