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APEX CLASS ESTD 2018

IITJEE: Binding Energy Curve (BE/A), Fission/Fusion & Radioactive Decay

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Ashish Kumar Bahekar

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APEX CLASS • IITJEE Nuclear Physics & Radioactivity Lab

Binding Energy Curve (BE/A), Fission/Fusion & Radioactive Decay

Master Fe-56 Stability Peak, Q-Value Numericals, Constant Nuclear Density, and Half-Life Ratios!

🔬 1 amu (u) = 931.5 MeV/c². Select a Nuclear Physics Concept:
8.75 8.0 6.0 4.0 2.0 BE / A (MeV/nucleon) 0 56 (Fe) 120 238 (U) Mass Number (A) → ²H (1.1 MeV) ⁴He ★ Peak: ⁵⁶Fe (8.75 MeV/A) ²³⁸U (7.6 MeV/A) ↗ Fusion (A < 20) ↖ Fission (A > 170) 🏆 4 NCERT BE/A Curve Conclusions 1. Flat Plateau (30 < A < 170): BE/A ≈ 8 MeV is constant ⇒ Nuclear Force is SHORT-RANGED & SATURATED! 2. Light Peaks (⁴He, ¹²C, ¹⁶O): More stable than neighbors (Even-Even) 3. Both Fission & Fusion Release Energy: Because both move TOWARDhigher BE/A! 🎯 Fundamental Rule: Higher Binding Energy per Nucleon (BE/A) = Greater Stability = Lower Rest Mass per Nucleon! Heavy X (A = 240) BE/A = 7.6 MeV Y₁ (120) Y₂ (120) BE/A = 8.5 MeV! 🏆 Most Repeated JEE Main Fission Q-Value Numerical • Golden Rule: Q = Total BE(Products) – Total BE(Reactants) 1. Initial BE of Heavy Nucleus (A = 240): BE_initial = 240 × 7.6 MeV = 1824 MeV 2. Final BE of Two Fragments (120 + 120 = 240): BE_final = 240 × 8.5 MeV = 2040 MeV 3. Energy Released: Q = 240 × (8.5 – 7.6) = 216 MeV! (≈ 200 MeV/U-235) ⚠️ Sign Trap: For Binding Energy, Q = BE_products – BE_reactants; for Rest Mass, Q = (Mass_reactants – Mass_products) × 931.5 MeV! 1. Thermonuclear Fusion in the Sun (T ≈ 10⁷ K) 4 ¹₁H → ⁴₂He + 2 e⁺ + 2 ν + 26.7 MeV • Deuteron-Deuteron Fusion: ²₁H + ²₁H → ⁴₂He + 24 MeV • Why High Temperature (10⁷ K) is Required? To overcome strong Coulomb repulsion between +Ze nuclei! • Energy per Nucleon: 26.7 / 4 ≈ 6.7 MeV/nucleon! 2. Fission vs. Fusion Comparison (JEE Trap) • Energy per Reaction: Fission (~200 MeV) > Fusion (~26.7 MeV) • Energy per Unit Mass (per Nucleon): Fusion (6.7 MeV/u) >> Fission (0.85 MeV/u)! (1 kg of Hydrogen fusion releases ~8× more energy than 1 kg U-235!) ☀️ Reactor Moderator Fact: Heavy Water (D₂O) and Graphite slow down fast neutrons (2 MeV → 0.025 eV thermal neutrons); Cadmium rods absorb them! ²⁷Al R₁ = 3 R₀ ¹²⁵Te R₂ = 5 R₀ Radius Ratio: R₁/R₂ = (27/125)¹/³ = 3 / 5 Nuclear Size & Constant Density Proof 1. Nuclear Radius: R = R₀ A¹/³ (R₀ = 1.2 × 10⁻¹⁵ m) 2. Nuclear Volume: V = (4/3)π R³ = (4/3)π R₀³ · A (V ∝ A!) 3. Nuclear Matter Density (Mass / Volume): ρ = (m_p · A) / [(4/3)π R₀³ · A] = 3 m_p / (4π R₀³) 🏆 A cancels out! ρ ≈ 2.3 × 10¹⁷ kg/m³ = CONSTANT! 🎯 Classic 5-Second JEE Question: “Ratio of nuclear densities of ²⁷Al and ¹²⁵Te is…” → Answer is ALWAYS 1 : 1 (Independent of A)!
Mass Defect (Δm)
Z m_p + (A – Z)m_n – M
BE = Δm × 931.5 MeV
Nuclear Radius Law
R = 1.2 A¹/³ fermi (fm)
1 fermi = 10⁻¹⁵ meters
Nuclear Density Ratio
ρ₁ : ρ₂ = 1 : 1 (Always!)
ρ ≈ 2.3 × 10¹⁷ kg/m³
Strong Nuclear Force
Charge Independent
F_pp = F_nn = F_np (At ~1.5 fm)
☢️ Rutherford-Soddy Law: N(t) = N₀ e⁻λᵗ = N₀ (½)ⁿ where n = t / T_½. Select Topic:
N₀ N₀/2 N₀/4 Parent N_P = N₀ / 2ⁿ Daughter N_D = N₀(1 – 1/2ⁿ) T_½ (1:1) 2 T_½ (1:3) 3 T_½ (1:7) 🏆 Parent : Daughter Ratio Cheat Table • Formula: N_P / N_D = 1 / (2ⁿ – 1) • n = 1 (t = 1 T_½): N_P : N_D = 1 : 1 (50% : 50%) • n = 2 (t = 2 T_½): N_P : N_D = 1 : 3 (25% : 75%) • n = 3 (t = 3 T_½): N_P : N_D = 1 : 7 (12.5% : 87.5%) • n = 4 (t = 4 T_½): N_P : N_D = 1 : 15 (6.25% Left) 🎯 Rock Dating Question: “Ratio of Parent X to Stable Daughter Y in a rock is 1 : 7. Find age!” → 1/(2ⁿ – 1) = 1/7 ⇒ 2ⁿ = 8 ⇒ t = 3 T_½! 1. Half-Life (T_½) vs. Mean Life (τ) T_½ = ln(2) / λ = 0.693 / λ = 0.693 τ Mean Life: τ = 1 / λ = 1.44 × T_½! • At t = T_½: Remaining N = 50% N₀, Decayed = 50% • At t = τ (1/λ): Remaining N = N₀/e = 36.8% N₀! 2. Activity Law & Probability of Survival • Activity: A(t) = |dN/dt| = λ N(t) = A₀ e⁻λᵗ • SI Unit: 1 Becquerel (Bq) = 1 decay / sec • 1 Curie (Ci) = 3.7 × 10¹⁰ Bq | 1 Rutherford = 10⁶ Bq • Survival Probability for time t: P_survive = e⁻λᵗ • Decay Probability in time t: P_decay = 1 – e⁻λᵗ ⚠️ Notice that Mean Life (τ = 1.44 T_½) is ALWAYS longer than Half-Life (T_½)! Parent X (N nuclei) Mode 1 (λ₁, T₁) Y₁ (α) Mode 2 (λ₂, T₂) Y₂ (β) Simultaneous (Parallel) Decay Formulas λ_eff = λ₁ + λ₂ ⇒ 1 / τ_eff = 1 / τ₁ + 1 / τ₂ T_eff = (T₁ × T₂) / (T₁ + T₂)! • Example: T₁ = 6 hours, T₂ = 12 hours: T_eff = (6 × 12) / (6 + 12) = 72 / 18 = 4 hours! • Branching Fraction for Mode 1: f₁ = λ₁ / (λ₁ + λ₂) 🔀 Parallel Resistor Analogy: Simultaneous Half-Lives combine just like two resistors in parallel: T_eff = T₁T₂ / (T₁ + T₂)! 1. Soddy-Fajans α, β⁻, β⁺, γ Shift Rules • Alpha (α = ⁴₂He): A → A – 4, Z → Z – 2 • Beta-Minus (β⁻ = e⁻): A → Same, Z → Z + 1 (n → p + e⁻ + ν̄) • Beta-Plus (β⁺ = e⁺): A → Same, Z → Z – 1 (p → n + e⁺ + ν) • Gamma (γ Photon): A → Same, Z → Same (Excited → Ground) • Trick: Count ΔA / 4 FIRST to find number of α-particles! 2. Alpha Decay Recoil Energy Split (p_α = p_D) Parent (Mass A at rest) → Daughter (A – 4) + α (Mass 4) Since K = p² / (2m), K_α / K_D = (A – 4) / 4 K_α = [ (A – 4) / A ] × Q_total (Gets ~98% of Q!) K_Daughter = (4 / A) × Q_total (Gets ~2% Recoil) (α spectrum is discrete; β spectrum is continuous due to ν!) 🎯 2-Step α/β Counting Formula: Always find N_α = (A₁ – A₂) / 4 FIRST; then find N_β⁻ = Z₂ – (Z₁ – 2 N_α)!
Remaining Nuclei (N)
N = N₀ (½)ⁿ = N₀ e⁻λᵗ
Where n = t / T_½ (Half-lives)
Half-Life vs Mean Life
T_½ = 0.693 τ = 0.693 / λ
Mean Life τ = 1 / λ = 1.44 T_½
Parallel Decay Half-Life
T_eff = (T₁ T₂) / (T₁ + T₂)
Because λ_eff = λ₁ + λ₂
Alpha Kinetic Energy
K_α = [(A – 4) / A] × Q
Conservation of Linear Momentum
🔥 2 Final JEE Nuclear Shortcuts:
1. Isotopes, Isobars & Isotones: Isotopes = Same Protons (Z); Isobars = Same Mass Number (A); Isotones = Same Neutrons (A – Z).
2. Penetrating vs Ionizing Power: Ionizing Power: α >> β >> γ  |  Penetrating Power: γ >> β >> α!
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