L-26: Nuclei and Radioactivity
Physics — Class 12 · NIOS Code 312 · Module 7 · Source: 312_Physics_Eng_Lesson26.pdf
Nuclei and Radioactivity
The nucleus contains protons and neutrons held by strong nuclear forces. Unstable nuclei disintegrate spontaneously — radioactivity. This lesson covers nuclear structure, binding energy, nuclear forces, decay laws, and applications. Module 7: Atoms and Nuclei.
26.1 The Atomic Nucleus
Discovered by Rutherford (1911); neutron discovered by Chadwick (1932). Nucleus built from nucleons (protons + neutrons).
- Atomic number Z = number of protons (= electrons in neutral atom)
- Mass number A = protons + neutrons = nucleon count
- Neutrons N = A − Z (usually N > Z for heavier nuclei)
- Nuclear charge = Ze; notation: ᴬ_Z X (e.g. ³⁵₁₇Cl)
Isotopes, Isobars, Isotones
- Isotopes: same Z, different A (same element, different neutrons) — identical chemistry
- Isobars: same A, different Z (different elements) — e.g. ⁴⁰₁₈Ar and ⁴⁰₂₀Ca
- Isotones: same N = A − Z — e.g. ²³₁₁Na and ²⁴₁₂Mg (both N = 12)
H (A=1): R ≈ 1.2 fm · U-238: R ≈ 7.5 fm
Volume ∝ A · nuclear density ~ 10¹⁷ kg/m³
Atom volume ~ 10⁵ times nuclear volume. Nuclear matter is extraordinarily dense — Earth at nuclear density would be a sphere of radius ~184 m.
Unified Atomic Mass (u)
1 u = (1/12) mass of ¹²C atom = 1.66 × 10⁻²⁷ kg. Proton: 1.00727 u; neutron: 1.00865 u.
Used to express nuclear binding energies in MeV
26.1.5 Mass Defect and Binding Energy
BE = binding energy — energy to break nucleus apart
Deuterium: Δm ≈ 3.96×10⁻³⁰ kg; BE ≈ 2.223 MeV
Decreases to ~7.6 MeV for U-238
Light nuclei (A<20) less stable · even-even nuclei (⁴He, ¹²C, ¹⁶O) extra stable
26.2 Nuclear Force
Cannot be electromagnetic (proton-proton repulsion) or gravitational (~10⁻³⁹ relative strength). New strong nuclear force binds nucleons.
- Short range: ~10⁻¹⁵ m (operates between neighbours only)
- Charge independent: p–p, p–n, n–n forces equivalent in binding
- Saturation: each nucleon interacts with neighbours only, not entire nucleus
- Repulsive core: attraction for r > ~0.4 fm; repulsion below critical distance
26.3 Radioactivity
Discovered by Becquerel (1896) — uranium fogged sealed photographic plates. Marie & Pierre Curie isolated radium and polonium. Rutherford identified α and β; Villard found γ-rays.
Nature of Radiations
| Property | α | β | γ |
|---|---|---|---|
| Nature | ⁴₂He nucleus | Fast e⁻ or e⁺ | High-energy EM wave |
| Charge | +2e | ±e | 0 |
| Ionizing power | Highest | Medium | Lowest |
| Penetration | ~0.02 mm Al | ~mm Al | cm of Pb/Fe |
| Deflection in E/B | Yes | Yes | No |
26.3.3 Radioactive Decay Equations
Z and A conserved in nuclear reactions:
- α-decay: ᴬ_Z X → ᴬ⁻⁴_Z₋₂ Y + ⁴₂He (Z−2, A−4)
- β⁻-decay: ᴬ_Z X → ᴬ_Z₋₁ Y + ⁰₋₁e (Z+1, A unchanged)
- γ-decay: excited nucleus → ground state + γ (Z, A unchanged)
26.3.4–26.3.5 Law of Radioactive Decay
Rate proportional to atoms present at that instant
Independent of temperature, pressure — law of chance
¹⁴C: T₁/₂ = 5730 years (carbon dating)
After n half-lives: N = N₀/2ⁿ
Activity units: 1 becquerel (Bq) = 1 disintegration/s; 1 curie (Ci) = 3.7×10¹⁰ dps; 1 rutherford (rd) = 10⁶ dps.
Applications
- Medicine: radiotherapy (Co-60 γ-rays), tracer technique (²⁴Na for ulcers)
- Agriculture: γ-irradiation of seeds for better yield; preservation of food
- Geology: carbon-14 dating (~15 dpm/g living → decays after death); U-Pb dating (Earth ~4 billion years)
- Industry: γ-rays detect internal flaws in machinery (air bubbles transmit more)
NUCLEI & RADIOACTIVITY — KEY POINTS
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Notation : ᴬ_Z X ; N = A − Z
Radius : R = r₀A^(1/3) ; r₀ = 1.2 fm
1 u : 931.3 MeV
Mass defect : Δm = Zmₚ + (A−Z)mₙ − M
Binding energy : BE = Δmc² ; B = BE/A
Nuclear force : short-range, charge-independent, saturated
α, β, γ : ionizing α>β>γ ; penetration α<β<γ
Decay law : N = N₀e^(−λt)
Half-life : T₁/₂ = 0.693/λ
Carbon dating : ¹⁴C T₁/₂ = 5730 y
Quick Revision
- Atomic number cannot differ for atoms of the same element.
- Neutron slightly heavier than proton; almost all atomic mass in nucleus.
- BE/A curve explains why fission (heavy) and fusion (light) release energy.
- Radioactivity is nuclear — changes Z and/or A in α and β decay.
- 10 g with T₁/₂ = 5 y → 2.5 g in 10 y (two half-lives).
Q1. The number of neutrons in ²³⁸₉₂U is:
Q2. Atoms with the same mass number A but different atomic number Z are called:
Q3. Nuclear radius is proportional to:
Q4. 1 atomic mass unit (u) is equivalent to approximately:
Q5. Binding energy per nucleon is maximum near:
Q6. In α-decay, the daughter nucleus has atomic number:
Q7. Which radiation has the greatest penetrating power?
Q8. The law of radioactive decay is:
Q9. Half-life T₁/₂ is related to decay constant λ by:
Q10. Nuclear force is:
PYQ — Previous Year Questions
Extracted from NIOS Physics (312) board exam papers in your PDF. Chapter L26 — Nuclei and Radioactivity only. Use Model Answer for marking points; Explanation for concept clarity.
Chapter 26 — Nuclei and Radioactivity (L26)
8 questions · Sections A & B · Sources: 312/TUS/104A, 312/MAY/204A, 68/ESS/1-312-A
Section A — Objective (1 mark)
PYQ1. Equal numbers of nuclei of radioactive elements A and B are given. Half-life of A is 20 minutes and of B is 40 minutes. After 80 minutes the ratio of nuclei in A and B will be: (A) 1 : 1 (B) 2 : 1 (C) 4 : 1 (D) 1 : 4
Model Answer
After 80 min: A undergoes 80/20 = 4 half-lives → NA = N₀/16
B undergoes 80/40 = 2 half-lives → NB = N₀/4
NA : NB = 1 : 4 → (D)
Explanation
N = N₀(1/2)n with n = t/T1/2 (L26 §26.3). Same initial N₀ cancels in ratio.
PYQ2. Read the passage on nuclear reactions. Which particle is the best projectile to trigger a nuclear reaction? (A) ⁴₂He (B) ¹₁H (C) ²₁H (D) ¹₀n
Model Answer
(D) neutron (¹₀n)
Neutrons have no Coulomb barrier → easily penetrate nucleus and initiate nuclear reactions.
Explanation
Charged projectiles (α, p, d) need high energy to overcome electrostatic repulsion. Slow neutrons are ideal for capture/transmutation (L26 §26.1).
PYQ3. Read the passage: beta-decay of radioactive elements showed an apparent breach of energy conservation, leading to discovery of: (A) electron (B) proton (C) neutron (D) neutrino
Model Answer
(D) neutrino (Pauli, 1930; experimentally confirmed later)
Explanation
β-spectrum is continuous → missing energy carried by antineutrino ν̄. Restores conservation in nuclear decay (L26 §26.3).
Section A — Short Answer (2 marks)
PYQ4. Write ‘True’ for correct statement and ‘False’ for incorrect: (a) The decay rate of a radioactive element increases with increase in temperature. (b) Using techniques of artificial radioactivity, it is now theoretically viable to convert iron into gold.
Model Answer
(a) False — radioactive decay rate depends on decay constant λ only; unaffected by temperature/pressure.
(b) False — transmutation is possible in nuclear reactions but not practically/theoretically viable as alchemy to make gold from iron.
Explanation
Decay is a nuclear spontaneous process (law of chance), not chemical/thermal (L26 §26.3).
PYQ5. Fill in the blanks (attempt any two; options: electron, nucleus, neutron, proton, α-particles): (a) In chemical reactions, _____ is not affected. (b) The best projectile for triggering a nuclear reaction is _____. (c) When ²³⁸₉₂U is bombarded with a neutron, along with ²³⁹₉₃Np, energy and radiation are released. (d) When nitrogen was bombarded with high-energy _____, it transformed into oxygen.
Model Answer
(a) nucleus — chemical reactions involve electrons only
(b) neutron
(c) γ-ray emitted (capture reaction); nucleus transmuted to Np
(d) α-particles — Rutherford's ¹⁴₇N + ⁴₂He → ¹⁷₈O + ¹₁H
Explanation
Distinguishes chemical (electron shell) vs nuclear (nucleus) changes — core L26 theme (L26 §26.1, §26.3).
PYQ6. Write TRUE or FALSE: β-particles have the highest ionizing power.
Model Answer
FALSE
α-particles have the highest ionizing power; β is medium; γ is lowest.
Explanation
Heavy +2e α ionizes strongly over short range; γ penetrates most (L26 §26.3 table).
PYQ7. Write TRUE or FALSE: The radius R of a nucleus is proportional to the cube root of its mass number.
Model Answer
TRUE
R = r₀ A1/3 with r₀ ≈ 1.2 fm
Explanation
Nuclear volume ∝ A implies R ∝ A1/3 — empirical nuclear radius formula (L26 §26.1).
Section B — Short Answer (2 marks)
PYQ8. A radioactive substance decays to 1/32 of its activity in 25 days. Calculate its half-life.
Model Answer
N/N₀ = 1/32 = (1/2)n → n = 5 half-lives in 25 days
T1/2 = 25/5 = 5 days
Explanation
Activity ∝ N; after n half-lives N = N₀/2n. Solve n then T1/2 = t/n (L26 §26.3).
Problem Solving — L26 Nuclei and Radioactivity
Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). Explanations open by default.
Define mass number A and atomic number Z. What is N for ¹⁴₆C?
Solution — step by step with formulas
- A = nucleons; Z = protons.
- N = A − Z = 8.
Final answer: N = 8 for ¹⁴C
Formulas used in this problem
Textbook formal language
Nucleus contains Z protons and N = A−Z neutrons.
Working formula set for this problem: A = Z + N. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Carbon-14: 6 protons, 8 neutrons.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Nuclear composition
Isotopes: same Z, different N.
Link to chapter notes (L26 — Nuclear composition): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: A = Z + N. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write A = Z + N before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
What is mass defect and how is binding energy obtained?
Solution — step by step with formulas
- Δm = (Zm_p + Nm_n − M_nucleus).
- BE = Δm c².
Final answer: BE = mass defect × c²
Formulas used in this problem
Textbook formal language
Bound nucleus has less mass than free nucleons; deficit appears as binding energy.
Working formula set for this problem: BE = Δm c². In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Glued nucleus weighs less; missing mass is the “glue energy.”
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Binding energy
BE per nucleon peaks near Fe—explains fusion/fission energy release.
Link to chapter notes (L26 — Binding energy): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: BE = Δm c². In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write BE = Δm c² before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
A sample’s activity halves every 8 days. Find decay constant λ (in day⁻¹).
Solution — step by step with formulas
- T½ = 8 d ⇒ λ = ln2 / 8 ≈ 0.0866 day⁻¹.
Final answer: λ ≈ 0.0866 d⁻¹
Formulas used in this problem
Textbook formal language
Exponential law follows constant decay probability per nucleus per time.
Working formula set for this problem: N = N₀ e^(−λt); T½ = ln2/λ. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Half-life 8 days means λ = 0.693/8 per day.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Decay law
Activity A = λN.
Link to chapter notes (L26 — Decay law): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: N = N₀ e^(−λt); T½ = ln2/λ. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write N = N₀ e^(−λt); T½ = ln2/λ before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Compare α, β, γ in charge and ionising/penetrating power qualitatively.
Solution — step by step with formulas
- α: +2e, high ionisation, low penetration.
- β: ±e, medium.
- γ: 0 charge, low ionisation, high penetration.
Final answer: α least penetrating; γ most penetrating
Textbook formal language
Emission types differ in mass, charge, and interaction with matter.
Working formula set for this problem: (see solution steps). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Alpha is a helium nucleus bullet—stopped by paper; gamma is pure energy ray—needs lead.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Decay modes
β is electron/positron from weak decay in nucleus.
Link to chapter notes (L26 — Decay modes): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: (see solution steps). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write (see solution steps) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Why is Q-value positive for spontaneous radioactive decay?
Solution — step by step with formulas
- Mass of products less than parent ⇒ energy released; Q > 0 allows spontaneous decay.
Final answer: Q > 0 for spontaneous decay
Formulas used in this problem
Textbook formal language
Kinematics and mass-energy allow decay only if final rest mass energy is lower.
Working formula set for this problem: Q = (m_i − m_f)c². In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Nature likes lower mass-energy; leftover becomes KE/photons of products.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Energy in decay
Neutrino shares energy in β decay (continuous β spectrum).
Link to chapter notes (L26 — Energy in decay): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: Q = (m_i − m_f)c². In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write Q = (m_i − m_f)c² before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Why do heavy nuclei have N > Z?
Solution — step by step with formulas
- More neutrons dilute proton repulsion; n–p force saturates; heavy nuclei need N>Z.
Final answer: Extra neutrons stabilise against Coulomb repulsion
Textbook formal language
Coulomb repulsion grows with Z²; neutron excess helps stability until fission/instability sets in.
Working formula set for this problem: (see solution steps). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Too many protons push each other apart; extra neutrons act as spacers and strong-force partners.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Nuclear stability
Magic numbers mark especially stable shells.
Link to chapter notes (L26 — Nuclear stability): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: (see solution steps). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write (see solution steps) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).