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State Board · Class 11 · Chemistry

Lesson 13: Nuclear Chemistry and Radioactivity

7 concepts · 56 practice questions with answers · Free to read

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The nucleus: composition, size and classes of nuclides

From daily life

The americium-241 in a smoke detector and sunlight are both nuclear energy at work.

Simple definition

A nucleus = protons (Z) + neutrons (N); mass number A = Z + N. An atom with a specific Z and A is a nuclide.

NuclidesClasses of nuclidesX (Z, A)A = Z + NIsotopes: same Z¹²C, ¹³C, ¹⁴CIsobars: same A¹⁴C, ¹⁴NIsotones: same N¹³C, ¹⁴NIsomers: same A and Z⁶⁰ᵐCo, ⁶⁰CoRadius r = r₀ A^(1/3), r₀ = 1.33 fm (10⁻¹⁵ m)A = 27 ⇒ r ≈ 4.0 fm; A = 64 ⇒ 5.32 fmDensity ≈ 1.7 × 10¹⁷ kg/m³ (independent of A)

In the figure : Classes of nuclides.

Rule 1: Classification — Isotopes: same Z; isobars: same A; isotones: same N; isomers: same A and Z but different energy state.

  • Easy : ¹²C and ¹⁴C are isotopes.
  • Medium : ¹⁴C and ¹⁴N are isobars.
  • Exam-type : ¹³C and ¹⁴N are isotones (N = 7).

Rule 2: Size of the nucleus — r = r₀A^(1/3), r₀ = 1.33 fm (1.33 × 10⁻¹⁵ m in the textbook; some books use 1.2 fm). Volume ∝ A ⇒ nearly constant density ≈ 1.7 × 10¹⁷ kg/m³.

  • Easy : A = 8 ⇒ r = 1.33 × 2 = 2.66 fm.
  • Medium : A = 27 ⇒ r = 1.33 × 3 ≈ 4.0 fm.
  • Exam-type : A = 64 ⇒ r = 1.33 × 4 = 5.32 fm.

Rule 3: Particles — Proton (+1, 1.0073 u), neutron (0, 1.0087 u), electron (−1, 0.00055 u). The nucleus holds nearly all the mass.

  • Easy : ²³⁵U: Z = 92, N = 143.
  • Medium : ¹⁴C: N = 8.
  • Exam-type : ⁵⁶Fe: Z = 26, N = 30.

Calculation: radius of a nucleus with A = 64

  • Step 1: r = r₀A^(1/3) = 1.33 × 64^(1/3).
  • Step 2: 64^(1/3) = 4.
  • Step 3: r = 1.33 × 4 = 5.32 fm.

CET level (MCQ)

  • CET 1) ¹⁴C and ¹⁴N are? 1) isotopes 2) isobars 3) isotones 4) isomers — Answer: 2).
  • CET 2) Nuclear density is? 1) depends on A 2) nearly constant 3) Z² 4) 1/A — Answer: 2).

Real-life use: Smoke detectors, medical imaging, energy.

Remember

A = Z + N; r = 1.33 A^(1/3) fm; density ≈ 1.7×10¹⁷.

Common mistake

Swapping isotopes and isobars.

CET trap

Electrons are not counted in A.

Shortcut

Memory clue: "iso-tope = same place Z, iso-bar = same weight A".

Spot the difference

ClassSameExample
IsotopesZ¹²C, ¹⁴C
IsobarsA¹⁴C, ¹⁴N
IsotonesN¹³C, ¹⁴N

Practice questions

Easy
  1. 1What is a nuclide?
  2. 2What are isobars?
  3. 3Radius for A = 27?
Medium
  1. 4N in ²³⁵U?
  2. 5Nuclear density?
  3. 6Why constant density?
Challenging
  1. 7What are isomers?
  2. 8What does A count?
Show answers (try them yourself first)
  1. 1An atom of specific Z and A.
  2. 2Same A.
  3. 3≈ 4.0 fm.
  4. 4143.
  5. 5≈ 1.7 × 10¹⁷ kg/m³.
  6. 6Volume ∝ A.
  7. 7Same A and Z, different energy.
  8. 8Protons + neutrons.

Nuclear stability, N/Z ratio and magic numbers

From daily life

Lead and gold are stable while uranium slowly changes — stability differs from nucleus to nucleus.

Simple definition

About 270 nuclides are stable; they lie in the band of stability on an N vs Z plot. N/Z ≈ 1 for light, ≈ 1.5 for heavy nuclei.

Stability bandBand of stability: N vs ZN = ZUnstable: N-rich ⇒ β⁻Unstable: N-poor ⇒ β⁺0Z = 80Magic numbers: 2, 8, 20, 28, 50, 82, 126 · even-even most stable

In the figure : Band of stability.

Rule 1: Even–odd rule — Even-Z even-N most stable (about 160); even-odd/odd-even (about 100); odd-odd only 4 stable: ²H, ⁶Li, ¹⁰B, ¹⁴N.

  • Easy : ⁴He, ¹²C, ¹⁶O are even-even.
  • Medium : ⁹Be is even-odd.
  • Exam-type : ¹⁴N is odd-odd (exception).

Rule 2: N/Z ratio — Too many neutrons ⇒ β⁻ decay (n → p). Too few ⇒ β⁺ or K-capture (p → n). Z > 83 ⇒ α decay.

  • Easy : ¹⁴C (N/Z = 1.33) ⇒ β⁻.
  • Medium : ¹¹C (N/Z = 0.83) ⇒ β⁺.
  • Exam-type : ²³⁸U ⇒ α.

Rule 3: Magic numbers and forces — 2, 8, 20, 28, 50, 82, 126 protons or neutrons ⇒ extra stable (closed shell). Nuclear force: strong, short-range (~2 fm), charge-independent.

  • Easy : ⁴He: Z = N = 2 doubly magic.
  • Medium : ²⁰⁸Pb: 82 and 126.
  • Exam-type : ¹⁶O: 8 and 8.

Reasoning: decay mode of ¹⁴C

  • Step 1: Z = 6, N = 8 ⇒ N/Z = 1.33.
  • Step 2: Stable ≈ 1 for light nuclei ⇒ neutron-rich.
  • Step 3: n → p + e⁻ + ν̄ ⇒ β⁻ decay.

CET level (MCQ)

  • CET 1) A magic number? 1) 6 2) 20 3) 30 4) 100 — Answer: 2).
  • CET 2) A stable odd-odd nucleus? 1) ¹²C 2) ¹⁴N 3) ¹⁶O 4) ⁴He — Answer: 2).

Real-life use: Reactor design, isotope research.

Remember

Magic: 2, 8, 20, 28, 50, 82, 126; N-rich ⇒ β⁻; N-poor ⇒ β⁺.

Common mistake

Assuming N/Z is always 1; it is 1.5 in heavy nuclei.

CET trap

The four stable odd-odd nuclei: ²H, ⁶Li, ¹⁰B, ¹⁴N.

Shortcut

Memory clue: "2-8-20-28-50-82-126".

Spot the difference

ConditionDecay
too many nβ⁻
too few nβ⁺ / K
Z > 83α

Practice questions

Easy
  1. 1What is the band of stability?
  2. 2List the magic numbers.
  3. 3Decay of ¹⁴C?
Medium
  1. 4Decay of ¹¹C?
  2. 5Four stable odd-odd nuclei.
  3. 6Properties of nuclear force.
Challenging
  1. 7Why is ²⁰⁸Pb special?
  2. 8Z > 83 ⇒ ?
Show answers (try them yourself first)
  1. 1N–Z band of stable nuclides.
  2. 22, 8, 20, 28, 50, 82, 126.
  3. 3β⁻.
  4. 4β⁺ or K-capture.
  5. 5²H, ⁶Li, ¹⁰B, ¹⁴N.
  6. 6Strong, short-range.
  7. 7Doubly magic (82, 126).
  8. 8α decay.

Mass defect and binding energy

From daily life

A built wall is more stable than loose bricks; assembling nucleons releases energy.

Simple definition

Mass defect Δm = (Zmₚ + Nmₙ) − m(nucleus). Binding energy = Δm × 931.5 MeV/u (textbook: 931.4). BE per nucleon is the stability measure.

Binding energyBinding energy per nucleon (MeV)²H⁴He¹²C¹⁶O⁵⁶Fe¹⁰⁷Ag²³⁸Ufusion releases energyfission releases energyA (log)⁴He: Δm = 0.030376 u ⇒ 28.3 MeV ⇒ 7.07 MeV/nucleon

In the figure : BE/A curve.

Rule 1: Δm and E — E = Δm c². 1 u = 931.5 MeV; 1 g = 9 × 10¹³ J. mₚ = 1.007276 u, mₙ = 1.008665 u.

  • Easy : ⁴He: Δm = 0.030376 u.
  • Medium : BE = 28.3 MeV.
  • Exam-type : BE/A = 7.07 MeV.

Rule 2: BE/A curve — Peak near ⁵⁶Fe (≈ 8.8 MeV). Joining light nuclei (fusion) or splitting heavy ones (fission) releases energy.

  • Easy : ²H: 1.1; ⁴He: 7.07.
  • Medium : ⁵⁶Fe: 8.79, most stable.
  • Exam-type : ²³⁸U: 7.57.

Rule 3: Stability and the curve — Higher BE/A ⇒ more stable. Peaks at ⁴He, ¹²C, ¹⁶O ⇒ magic/α-particle structure.

  • Easy : ⁴He >> ²H.
  • Medium : ⁵⁶Fe > ²³⁸U.
  • Exam-type : Light ⇒ fusion pays.

Calculation: BE/A of ⁴He

  • Step 1: Δm = 2(1.007276) + 2(1.008665) − 4.001506 = 0.030376 u.
  • Step 2: BE = 0.030376 × 931.5 = 28.3 MeV.
  • Step 3: BE/A = 28.3 ÷ 4 = 7.07 MeV.

CET level (MCQ)

  • CET 1) Highest BE/A? 1) ²H 2) ⁵⁶Fe 3) ²³⁸U 4) ⁴He — Answer: 2).
  • CET 2) 1 u = ? MeV 1) 93.15 2) 931.5 3) 9315 4) 0.931 — Answer: 2).

Real-life use: Nuclear power, stars, energy sources.

Remember

1 u = 931.5 MeV; ⁴He 7.07; ⁵⁶Fe 8.79.

Common mistake

Confusing total BE with BE/A.

CET trap

Use the nuclear mass (or account for electrons) when finding Δm.

Shortcut

Memory clue: "iron = the peak".

Spot the difference

ItemFissionFusion
Nucleiheavylight
BE/Arisesrises

Practice questions

Easy
  1. 1What is mass defect?
  2. 21 u = ? MeV
  3. 3BE of ⁴He?
Medium
  1. 4BE/A of ⁴He?
  2. 5Most stable nucleus?
  3. 6Why does fusion release energy?
Challenging
  1. 7Energy from 1 g mass?
  2. 8Why does BE/A matter?
Show answers (try them yourself first)
  1. 1Sum of nucleon masses − nuclear mass.
  2. 2931.5 MeV.
  3. 328.3 MeV.
  4. 47.07 MeV.
  5. 5⁵⁶Fe.
  6. 6BE/A rises.
  7. 79 × 10¹³ J.
  8. 8It measures stability.

Radioactivity and modes of decay

From daily life

In a smoke alarm α particles ionise air; smoke blocks them and triggers the alarm.

Simple definition

Radioactivity: an unstable nucleus emits α, β or γ and becomes stable (Becquerel, 1896).

Radioactive decayModes of radioactive decayα²³⁸U → ²³⁴Th + ⁴HeA − 4, Z − 2β⁻¹⁴C → ¹⁴N + e⁻ + ν̄n → p; Z + 1β⁺²²Na → ²²Ne + e⁺ + νp → n; Z − 1γ⁶⁰Co* → ⁶⁰Co + γA, Z unchangedPenetration: α < β < γαβγpaperAl sheetlead

In the figure : Decay modes.

Rule 1: α and β decay — α: ⁴He, A − 4, Z − 2. β⁻: n → p + e⁻ + ν̄, Z + 1. β⁺: p → n + e⁺ + ν, Z − 1. A is unchanged in β.

  • Easy : ²³⁸U → ²³⁴Th + α.
  • Medium : ¹⁴C → ¹⁴N + e⁻.
  • Exam-type : ²²Na → ²²Ne + e⁺.

Rule 2: γ and K-capture — γ: electromagnetic photon, no change in A or Z. K-capture: the nucleus takes a K-shell electron ⇒ Z − 1.

  • Easy : ⁶⁰Co* → ⁶⁰Co + γ.
  • Medium : ⁷Be + e⁻ → ⁷Li.
  • Exam-type : γ energy in MeV.

Rule 3: Penetration and ionisation — Ionising power: α > β > γ; penetration: γ > β > α. α is stopped by paper, β by aluminium, γ by thick lead.

  • Easy : α: helium nucleus, +2.
  • Medium : β: e⁻ or e⁺.
  • Exam-type : γ: no charge.

Calculation: ²³⁸U (Z = 92) ⇒ ²⁰⁶Pb (Z = 82)

  • Step 1: ΔA = 238 − 206 = 32 ⇒ α = 32 ÷ 4 = 8.
  • Step 2: 8α gives Z = 92 − 16 = 76; need 82 ⇒ gain of 6.
  • Step 3: 6 β⁻ decays.

CET level (MCQ)

  • CET 1) After α emission A, Z change by? 1) −2, −4 2) −4, −2 3) −4, +2 4) 0, −2 — Answer: 2).
  • CET 2) Most penetrating? 1) α 2) β 3) γ 4) proton — Answer: 3).

Real-life use: Smoke alarms, cancer therapy, industrial thickness gauging.

Remember

α: A−4, Z−2; β⁻: Z+1; β⁺: Z−1; γ: no change.

Common mistake

Thinking A changes in β decay.

CET trap

²³⁸U ⇒ ²⁰⁶Pb: 8α + 6β⁻.

Shortcut

Memory clue: "α: 4-2; β⁻: +1".

Spot the difference

RayChargePenetration
α+2low
β−1medium
γ0high

Practice questions

Easy
  1. 1Change in A, Z in α decay?
  2. 2What happens in β⁻?
  3. 3What are γ rays?
Medium
  1. 4²³⁸U ⇒ ²⁰⁶Pb: α and β?
  2. 5What is K-capture?
  3. 6Order of penetration?
Challenging
  1. 7What stops α?
  2. 8Z in β⁺?
Show answers (try them yourself first)
  1. 1−4, −2.
  2. 2n → p + e⁻ + ν̄.
  3. 3High-energy photons.
  4. 48 α and 6 β⁻.
  5. 5Nucleus takes a K electron.
  6. 6γ > β > α.
  7. 7Paper.
  8. 8Z − 1.

Rate of decay, decay constant and units

From daily life

Like ice melting a fixed fraction per day, a radioactive sample decays by a fixed fraction.

Simple definition

Decay rate: A = −dN/dt = λN. Decay constant λ: unit s⁻¹. N = N₀e^(−λt). Mean life τ = 1/λ.

Decay curveDecay curve: N = N₀ e^(−λt)0N₀1 t½N₀/22 t½N₀/43 t½N₀/84 t½N₀/16t½ = 0.693 / λ · mean life τ = 1/λ = 1.44 t½1 Bq = 1 decay/s · 1 Ci = 3.7 × 10¹⁰ Bq

In the figure : Decay curve.

Rule 1: Law of decay — Decay rate is proportional to the nuclei present; N = N₀e^(−λt); ln N is linear in t.

  • Easy : λ = 0.0693 d⁻¹ ⇒ t½ = 10 d.
  • Medium : τ = 14.4 d.
  • Exam-type : Smaller N ⇒ lower rate.

Rule 2: Units — 1 becquerel (Bq) = 1 decay/s; 1 curie (Ci) = 3.7 × 10¹⁰ Bq; 1 rutherford = 10⁶ Bq.

  • Easy : 1 Ci = 3.7 × 10¹⁰ decays/s.
  • Medium : 5 mCi = 1.85 × 10⁸ Bq.
  • Exam-type : 1 μCi = 3.7 × 10⁴ Bq.

Rule 3: t½ and λ — t½ = 0.693/λ = 0.693τ ⇒ τ = 1.44 t½. Larger λ ⇒ shorter t½.

  • Easy : t½ = 5730 y ⇒ λ = 1.21 × 10⁻⁴ y⁻¹.
  • Medium : τ = 1.44 × 5730 ≈ 8250 y.
  • Exam-type : Doubling λ halves t½.

Calculation: λ and τ for t½ = 10 d

  • Step 1: λ = 0.693 ÷ 10 = 0.0693 d⁻¹.
  • Step 2: τ = 1/λ = 14.4 d.
  • Step 3: τ = 1.44 × 10 = 14.4 d ✔.

CET level (MCQ)

  • CET 1) 1 Ci = ? Bq 1) 3.7×10⁷ 2) 3.7×10¹⁰ 3) 10⁶ 4) 1 — Answer: 2).
  • CET 2) Unit of λ? 1) s 2) s⁻¹ 3) Bq 4) Ci — Answer: 2).

Real-life use: Medical doses, waste management.

Remember

A = λN; t½ = 0.693/λ; τ = 1.44 t½; 1 Ci = 3.7×10¹⁰ Bq.

Common mistake

Writing the unit of λ as s.

CET trap

Rate in Bq; decay constant in s⁻¹.

Shortcut

Memory clue: "0.693 = ln 2".

Spot the difference

QuantityUnit
λs⁻¹
ActivityBq, Ci
t½, τs

Practice questions

Easy
  1. 1State the decay law.
  2. 2Unit of λ?
  3. 31 Ci = ? Bq
Medium
  1. 4t½ = 10 d ⇒ λ?
  2. 5τ = ? (t½ = 10 d)
  3. 65 mCi = ? Bq
Challenging
  1. 7Shape of ln N vs t?
  2. 8Meaning of larger λ?
Show answers (try them yourself first)
  1. 1Rate ∝ N.
  2. 2s⁻¹.
  3. 33.7 × 10¹⁰.
  4. 40.0693 d⁻¹.
  5. 514.4 d.
  6. 61.85 × 10⁸ Bq.
  7. 7Straight line, slope −λ.
  8. 8Shorter half-life.

Half-life, decay calculations and radiocarbon dating

From daily life

The age of ancient wood is known because its ¹⁴C halves every 5730 years.

Simple definition

Half-life t½: the time for half the nuclei to decay. N = N₀(½)^(t/t½).

Carbon datingRadiocarbon dating¹⁴N + n → ¹⁴C + pupper atmosphere¹⁴CO₂plants, animalsdeath: ¹⁴C decayst½ = 5730 yt = (2.303 / λ) log (N₀ / N)λ = 0.693 / 5730 = 1.21 × 10⁻⁴ y⁻¹25 % ¹⁴C left ⇒ 2 half-lives = 11 460 years50 % left ⇒ 5730 yearsLimit: roughly 50 000 yearsAssumes constant ¹⁴C in living tissue

In the figure : Radiocarbon.

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Half-life, decay calculations and radiocarbon dating

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Rule 1: Half-life calculations — N = N₀(½)ⁿ with n = t ÷ t½. Fraction left = (½)ⁿ.

  • Easy : 80 g, 3 t½ ⇒ 10 g.
  • Medium : t½ = 10 d, 25 d ⇒ (½)^2.5 = 17.7 %.
  • Exam-type : 2 t½ ⇒ 25 %.

Rule 2: Formula method — t = (2.303/λ) log(N₀/N); λ = 0.693/t½.

  • Easy : N₀/N = 4 ⇒ t = 2 t½.
  • Medium : t½ = 10 d, N₀/N = 4 ⇒ 20 d.
  • Exam-type : ln 2 = 0.693.

Rule 3: Radiocarbon dating — ¹⁴N + n → ¹⁴C + p in the atmosphere; living tissue keeps a constant ¹⁴C level; after death it decays. t½ = 5730 y; usable to about 50 000 y.

  • Easy : 25 % left ⇒ 11 460 y.
  • Medium : 50 % left ⇒ 5730 y.
  • Exam-type : 12.5 % ⇒ 17 190 y.

Calculation: % left after 25 d with t½ = 10 d

  • Step 1: n = 25 ÷ 10 = 2.5.
  • Step 2: (½)^2.5 = 0.1768.
  • Step 3: ≈ 17.7 %.

CET level (MCQ)

  • CET 1) 80 g after 3 half-lives? 1) 40 2) 20 3) 10 4) 5 — Answer: 3).
  • CET 2) 25 % ¹⁴C left (t½ = 5730)? 1) 5730 2) 11460 3) 17190 4) 2865 — Answer: 2).

Real-life use: Archaeology and geology.

Remember

(½)ⁿ; 25 % = 2 t½; 12.5 % = 3 t½; ¹⁴C: 5730 y.

Common mistake

Thinking nothing is left after two half-lives; 25 % remains.

CET trap

n can be a fraction (2.5).

Shortcut

Memory clue: "100-50-25-12.5".

Spot the difference

n of t½Fraction left
150 %
225 %
312.5 %

Practice questions

Easy
  1. 1What is half-life?
  2. 280 g after 3 t½?
  3. 3t½ = 10 d, 25 d ⇒ %?
Medium
  1. 4Age if 25 % ¹⁴C left?
  2. 5How is ¹⁴C formed?
  3. 6Limit of the method?
Challenging
  1. 7Age for 12.5 %?
  2. 8Find λ for t½ = 5730 y.
Show answers (try them yourself first)
  1. 1Time for half to decay.
  2. 210 g.
  3. 317.7 %.
  4. 411 460 y.
  5. 5¹⁴N + n → ¹⁴C + p.
  6. 6≈ 50 000 y.
  7. 717 190 y.
  8. 81.21 × 10⁻⁴ y⁻¹.

Nuclear reactions, reactors and uses of radioisotopes

From daily life

A nuclear power reactor and a hospital PET scan both rely on nuclear reactions.

Simple definition

Transmutation: one element becoming another. Fission: a heavy nucleus splitting. Fusion: light nuclei joining.

Nuclear reactionsFission, fusion and the reactorFission²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n+ ≈ 200 MeVchain reactionFusion²H + ³H → ⁴He + n+ 17.6 MeVSun, > 10⁷ KReactor: fuel (U) · moderator (heavy water/graphite)control rods (Cd, B) · coolant · shieldingModerator slows neutrons; Cd absorbs neutronsRadioisotopes: ⁶⁰Co (cancer) · ¹³¹I (thyroid) · ³²P · ¹⁴CTransmutation: ¹⁴N + ⁴He → ¹⁷O + ¹H (Rutherford)

In the figure : Fission, fusion, reactor.

Rule 1: Transmutation — ¹⁴N + ⁴He → ¹⁷O + ¹H (Rutherford, 1919). A and Z balance on both sides.

  • Easy : ²⁷Al + ⁴He → ³⁰P + n.
  • Medium : ⁹Be + ⁴He → ¹²C + n.
  • Exam-type : A: 14 + 4 = 17 + 1.

Rule 2: Fission and the reactor — ²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n + ≈ 200 MeV. Chain reaction; control rods (Cd, B), moderator (heavy water/graphite), coolant. 1 g of ²³⁵U ⇒ ≈ 8.2 × 10¹⁰ J.

  • Easy : A: 235 + 1 = 141 + 92 + 3.
  • Medium : The moderator slows neutrons.
  • Exam-type : A Cd rod absorbs neutrons.

Rule 3: Fusion and radioisotopes — ²H + ³H → ⁴He + n + 17.6 MeV (> 10⁷ K). Uses: ⁶⁰Co (cancer), ¹³¹I (thyroid), ³²P (plant studies), ¹⁴C (dating).

  • Easy : The Sun runs on fusion.
  • Medium : ¹³¹I collects in the thyroid.
  • Exam-type : ⁶⁰Co emits γ rays.

Check: balance the fission equation

  • Step 1: A on the left 235 + 1 = 236; on the right 141 + 92 + 3 = 236.
  • Step 2: Z on the left 92; on the right 56 + 36 = 92.
  • Step 3: Both match ⇒ the equation is balanced.

CET level (MCQ)

  • CET 1) A reactor moderator? 1) Cd 2) heavy water 3) U 4) Pb — Answer: 2).
  • CET 2) Temperature needed for fusion? 1) 10³ K 2) 10⁵ K 3) 10⁷ K 4) 300 K — Answer: 3).

Real-life use: Power, cancer therapy, agricultural research.

Remember

Fission ≈ 200 MeV; fusion 17.6 MeV; Cd = controller; heavy water = moderator.

Common mistake

Confusing control rods with the moderator.

CET trap

Check both A and Z in the equation.

Shortcut

Memory clue: "moderator slows, cadmium captures".

Spot the difference

ItemFissionFusion
Nucleiheavylight
Energy≈ 200 MeV17.6 MeV
ExamplereactorSun

Practice questions

Easy
  1. 1What is transmutation?
  2. 2Rutherford's reaction.
  3. 3Write a fission equation.
Medium
  1. 4Role of a moderator?
  2. 5What are control rods made of?
  3. 6Condition for fusion?
Challenging
  1. 7Use of ¹³¹I?
  2. 8Energy from 1 g ²³⁵U?
Show answers (try them yourself first)
  1. 1One element converted to another.
  2. 2¹⁴N + ⁴He → ¹⁷O + ¹H.
  3. 3²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n.
  4. 4Slow down neutrons.
  5. 5Cd or B.
  6. 6> 10⁷ K.
  7. 7Thyroid treatment.
  8. 8≈ 8.2 × 10¹⁰ J.

Quick revision of the whole lesson

ConceptKey formula / fact
r1.33 A^(1/3) fm
Magic2, 8, 20, 28, 50, 82, 126
BEΔm × 931.5 MeV
α / β⁻ / β⁺A−4,Z−2 / Z+1 / Z−1
DecayA = λN
t½0.693 / λ
¹⁴C5730 y
Fission / fusion≈ 200 / 17.6 MeV
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🚀 Quick revision of the whole lesson

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🎯 Most important points for the exam

  • Nucleus and nuclides.
  • Stability.
  • Mass defect.
  • Decay modes.
  • Decay rate.
  • Half-life, ¹⁴C.
  • Nuclear reactions.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.