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L-27: Nuclear Fission and Fusion

Physics — Class 12 · NIOS Code 312 · Module 7 · Source: 312_Physics_Eng_Lesson27.pdf

Nuclear Fission and Fusion

Nuclear fission releases enormous energy in reactors and weapons; nuclear fusion powers stars and promises clean future energy. This lesson compares chemical and nuclear reactions, explains fission/fusion mechanisms, reactor design, and peaceful applications. Module 7: Atoms and Nuclei.

27.1 Chemical and Nuclear Reactions

27.1.1 Chemical Reactions

Valence electrons rearrange; nucleus unaffected. Energy ~eV (e.g. C + O₂ → CO₂ + 4.08 eV). Mass change ~10⁻³⁵ kg — practically conserved. Atom count of each element balanced on both sides.

27.1.2 Nuclear Reactions

Nuclei interact → new elements (transmutation). Energy ~MeV. Coulomb barrier (~3 MeV for C, ~20 MeV for Pb) must be overcome by projectile. Neutrons are ideal projectiles — no Coulomb repulsion; even thermal neutrons (0.0253 eV) can induce reactions.

Rutherford (1919): ⁴He + ¹⁴N → ¹⁷O + ¹H + 6.5 MeV (endothermic overall needs 1.2 MeV supplied). Exothermic example: ²⁷Al + ⁴He → ³⁰Si + ¹H + 10.7 MeV (~700,000× energy per event vs burning one C atom).

Fig 27.1 — Coulomb Barrier Near Nucleus Coulomb barrier nucleus projectile Charged projectiles need KE > barrier · neutrons penetrate easily
Fig 27.1 — Electrostatic repulsion prevents low-energy charged particles from reaching nucleus

27.1.3 Conservation Laws

  • Sum of mass numbers A conserved
  • Sum of atomic numbers Z conserved
  • Total energy (including mass-energy) conserved
  • Momentum conserved — kinetic energy shared among products

27.2 Nuclear Fission

Discovery (1938): Hahn & Strassmann — slow neutrons on uranium produced barium (not transuranics) + ~200 MeV. Meitner & Frisch explained via liquid-drop model; named "fission."

²³⁵U + ¹n → ¹⁴¹Ba + ⁹²Kr + 3¹n + Q ≈ 200 MeV
One of 40+ fission modes for ²³⁵U
Mass defect Δm ≈ 0.215 u → ~200 MeV
Average 2.54 neutrons per fission · event in ~10⁻¹⁷ s

Liquid-drop model (Bohr & Wheeler): neutron capture adds ~6.8 MeV/nucleon excitation; nucleus oscillates spherical ↔ dumbbell; Coulomb repulsion between fragments overcomes surface tension → split. ²³⁵U more fissile than ²³⁸U.

Fig 27.2–27.3 — Nuclear Fission (Liquid Drop Model) ²³⁵U n fragment 2–3 n Fissile: ²³³Th, ²³³U, ²³⁵U, ²³⁹Pu (odd A, even Z)
Fig 27.2–27.3 — Thermal neutron induces oscillation; nucleus splits with neutron emission

27.2.2 Nuclear Chain Reaction

Each fission releases 2–3 neutrons → can trigger more fissions. Self-sustained: neutron production rate = loss rate.

  • Controlled chain reaction: nuclear reactor (control rods absorb excess neutrons)
  • Uncontrolled chain reaction: atom bomb (Hiroshima, Aug 6 1945 — ~20,000 ton TNT equivalent)
Fig 27.4 — Nuclear Chain Reaction 1 fission → 2–3 neutrons → exponential growth if uncontrolled
Fig 27.4 — Each fission neutron can cause further fissions in fissile material

One fission event releases ~7×10⁵ times energy of burning one carbon atom.

27.3 Nuclear Reactor

First reactor: Fermi, Chicago (1942). Components:

  • Core: fuel rods (²³⁵U), moderator (slows neutrons in thermal reactors), control rods (Cd/B — absorb neutrons)
  • Coolant: removes fission heat (heavy water or ordinary water)
  • Reflector: reduces neutron leakage
  • Pressure vessel, shielding (concrete), airtight reactor building

Heat → steam → turbine → electricity. Research reactors discharge heat to sea/river.

Fig 27.5 — Nuclear Power Reactor (Schematic) reactor core heat exch. turbine generator Control rods · moderator · shielding · fuel runs ~6 months per load
Fig 27.5 — Fission heat drives steam turbine; control rods maintain steady chain reaction

27.4 Nuclear Fusion

Two light nuclei combine → heavier nucleus + energy. From BE/A curve: fusion of H→He releases more energy per nucleon than fission.

²H + ²H → ⁴He + Q ≈ 24 MeV
~6 MeV per nucleon in fusion vs ~0.84 MeV/nucleon in fission
Requires ~10 million K (Sun centre ~20 million K)
Overcome Coulomb repulsion between positive nuclei
4(¹H) → ⁴He + 2e⁺ + Q ≈ 26.8 MeV
Bethe's proton-proton cycle — solar energy source
Sun consumes ~400×10⁶ ton hydrogen per second
1 g deuterium → ~100,000 kWh · ocean deuterium nearly inexhaustible

Fusion harder to achieve on Earth than fission (extreme temperature + confinement). Controlled thermonuclear fusion is active research area. Hydrogen bomb uses uncontrolled fusion.

27.5 Nuclear Energy

India (Bhabha's 3-stage plan): (1) PHWR with natural uranium → electricity + Pu; (2) fast breeder reactors → breed U-233 from thorium; (3) thorium-based surplus fissile material. Reactors: Tarapur, Kota, Kaiga, Narora, Kalpakkam, Kakrapar.

27.5.2 Hazards and RBE

Radiation causes ionisation, cancer, genetic damage, sterility. RBE (relative biological effectiveness): γ/X/β = 1; thermal neutrons = 2–5; fast neutrons = 10; α = 10–20. Safety: avoid nuclear tests, careful waste disposal (salt mines), minimal medical radiation doses.

         FISSION & FUSION — KEY POINTS
         =================================
    Chemical reaction :  ~eV ; nucleus unchanged
    Nuclear reaction  :  ~MeV ; transmutation
    Conservation      :  A, Z, energy, momentum
    Fission           :  ²³⁵U + n → fragments + 2–3n + 200 MeV
    Chain reaction    :  controlled = reactor ; uncontrolled = bomb
    Reactor parts     :  fuel, moderator, control rods, coolant, shield
    Fusion            :  light nuclei → heavier ; needs ~10⁷ K
    Sun               :  4¹H → ⁴He + 26.8 MeV
    India reactors    :  PHWR (pressurised heavy water)

Quick Revision

  • Neutrons best for inducing nuclear reactions — no Coulomb barrier.
  • ²³⁸U becomes β-active after neutron capture (high n/p ratio).
  • Fissile materials have odd mass number, even atomic number.
  • Fusion releases more energy per unit mass than fission (~6.7 vs ~0.84 MeV/u).
  • Spent reactor fuel is highly radioactive; wastes stored in sealed salt mines.
20 cards · click any card to flip
Chemical vs nuclear reaction
Chemical: valence electrons, ~eV, nucleus unchanged. Nuclear: nuclei interact, ~MeV, transmutation (Rutherford 1919: ⁴He + ¹⁴N → ¹⁷O + ¹H).
Coulomb barrier
Electrostatic repulsion around nucleus. ~3 MeV for carbon, ~20 MeV for lead. Charged projectiles need sufficient KE. Neutrons (neutral) penetrate easily — even thermal neutrons work.
Conservation laws in nuclear reactions
Mass number A conserved. Atomic number Z conserved. Total energy (including mass) conserved. Momentum conserved — KE distributed among products.
Discovery of nuclear fission
1938: Hahn & Strassmann — uranium + slow neutrons → barium + ~200 MeV (unexpected). Meitner & Frisch: liquid-drop model, named fission. Fermi: first chain reaction (1942).
Liquid-drop model of fission
Bohr & Wheeler: nucleus like charged liquid drop. Neutron capture excites nucleus; oscillates to dumbbell shape; Coulomb repulsion splits it when separation exceeds critical value.
Energy released in ²³⁵U fission
~200 MeV per fission event. Mass defect ~0.215 u. Average 2.54 neutrons emitted. 40+ fission modes → ~80 different fragment nuclei. Event in ~10⁻¹⁷ s.
Fissile materials
Undergo fission by thermal (slow) neutrons: ²³³Th, ²³³U, ²³⁵U, ²³⁹Pu. All have odd mass number and even atomic number. ²³⁵U more fissile than ²³⁸U.
Nuclear chain reaction
Each fission releases 2–3 neutrons that can cause more fissions. Self-sustained when production = loss. Controlled: reactor. Uncontrolled: atom bomb (~20,000 ton TNT at Hiroshima).
Nuclear reactor components
Core (fuel + moderator + control rods), coolant, reflector, pressure vessel, concrete shield, heat exchanger, turbine-generator. Control rods: Cd or B. Moderator/coolant: heavy or ordinary water.
Moderator and control rods
Moderator slows fast fission neutrons to thermal speeds for further fission. Control rods (Cd/B) absorb neutrons to regulate chain reaction rate. Absorber prevents runaway reaction.
Nuclear fusion
Two light nuclei fuse into heavier nucleus + energy. Example: ²H + ²H → ⁴He + 24 MeV. Needs ~10 million K to overcome Coulomb repulsion. More energy per nucleon than fission (~6 vs ~0.84 MeV/u).
Energy source of the Sun
Thermonuclear fusion at ~20 million K core. 4¹H → ⁴He + 2e⁺ + 26.8 MeV (Bethe). Sun is H/He — not fission. Consumes ~400×10⁶ ton H per second. Enough H for ~8 billion more years.
Fusion vs fission energy per mass
Fusion: ~6.7 MeV per nucleon. Fission: ~0.84 MeV per nucleon (200/238). Fusion more efficient per unit mass but harder to initiate on Earth.
Deuterium as fusion fuel
Abundant in oceans. 1 g deuterium → ~100,000 kWh. Once controlled fusion achieved, potentially endless clean electricity without greenhouse gases.
India's nuclear power programme
Bhabha's 3-stage plan: PHWR (natural U) → fast breeder (Pu, breed U-233 from Th) → thorium utilisation. Sites: Tarapur, Kota, Kaiga, Narora, Kalpakkam, Kakrapar. ~3% of India's electricity from nuclear.
Why ²³⁸U becomes β-active after neutron capture?
Neutron capture increases n/p ratio above stable value. Nucleus emits β⁻ (neutron effectively becomes proton) to restore stability — not fission for ²³⁸U with slow neutrons.
RBE factors
Relative biological effectiveness: γ, X-rays, β = 1. Thermal neutrons = 2–5. Fast neutrons = 10. α-particles = 10–20. α most harmful per unit energy due to high ionisation.
Advantages of nuclear power
Fuel loaded for ~6 months continuously. Low greenhouse gas emission vs coal. One fission ≈ 7×10⁵× energy of burning one C atom. Powers submarines and ships.
Spent fuel and waste disposal
Spent fuel highly radioactive (many isotopes). India reprocesses for useful isotopes. Wastes embedded in steel cases in deep salt mines. Major environmental and safety concern.
Thermal vs fast reactors
Thermal reactors use moderated (slow) neutrons — PHWR in India. Fast reactors use fast neutrons without moderator — breeder reactors at Kalpakkam (India developing).

Q1. In a chemical reaction, the energy involved is typically of the order of:

Q2. The best projectile for inducing nuclear reactions is:

Q3. Nuclear fission of ²³⁵U releases approximately:

Q4. A substance that undergoes fission by thermal neutrons is called:

Q5. In a controlled nuclear chain reaction, excess neutrons are absorbed by:

Q6. The moderator in a thermal reactor is used to:

Q7. Nuclear fusion requires temperature of approximately:

Q8. The energy source of the Sun is:

Q9. India primarily uses which type of reactor for power generation?

Q10. Which radiation has the highest RBE (relative biological effectiveness)?

Chemical ~eV · Nuclear ~MeV
ΣA, ΣZ, E, p conserved
²³⁵U + ¹n → ¹⁴¹Ba + ⁹²Kr + 3¹n + Q
Δm ≈ 0.215 u → Q ≈ 200 MeV
Fission: ~0.84 MeV/nucleon
²H + ²H → ⁴He + Q ≈ 24 MeV
4(¹H) → ⁴He + 2e⁺ + Q ≈ 26.8 MeV
Fusion: ~6 MeV/nucleon
1 fission ≈ 7×10⁵ × 1 C atom
⁴He + ¹⁴N → ¹⁷O + ¹H + 6.5 MeV
²⁷Al + ⁴He → ³⁰Si + ¹H + 10.7 MeV
Chain: 1 fission → 2–3 neutrons

1. Formulas & Definitions

Full Ch 27 study guide — nuclear fission, fusion, reactors, and energy comparisons.

Chemical ~eV · Nuclear ~MeV

Definition: Energy scale comparison — chemical vs nuclear reactions.

Derivation

Chemical: valence electrons rearrange; nuclear: nuclei transmute. Mass change negligible in chemistry (~10⁻³⁵ kg).

Variables

Chemical: ~few eV (e.g. C+O₂ → 4.08 eV) · Nuclear: ~MeV per event

Why it works

Explains why nuclear energy is millions of times more powerful per atom.

Historical context

Rutherford (1919) first artificial transmutation.

Deep understanding

Nucleus unchanged in chemical reactions; new elements possible in nuclear reactions.

2. Diagrams & Visuals

Chemical: eV Nuclear: MeV

Color-coded visual · step-by-step breakdown below

  1. Identify if electrons or nuclei change
  2. Chemical → eV scale
  3. Nuclear → MeV scale
  4. Compare energy per atom

3. Solved Examples

Basic

Q: CO₂ formation energy?

Solution: ~4 eV

Answer: ~4 eV

Intermediate

Q: ²³⁵U fission?

Solution: ~200 MeV

Answer: ~200 MeV

Advanced

Q: Ratio?

Solution: ~10⁶×

Answer: Nuclear >> chemical

Exam

Q: Nuclear energy unit?

Solution: MeV

Answer: Sec 27.1

ΣA, ΣZ, E, p conserved

Definition: Conservation laws in all nuclear reactions.

Derivation

Mass number and atomic number balanced; total energy (incl. mass-energy) and momentum conserved.

Variables

A = mass number · Z = atomic number · products share kinetic energy

Why it works

Used to balance every nuclear equation and predict missing particles.

Historical context

Same principles as chemical reactions, extended to mass-energy.

Deep understanding

Momentum conservation explains why fission fragments fly apart with shared KE.

2. Diagrams & Visuals

ΣA_in = ΣA_out · ΣZ_in = ΣZ_out

Color-coded visual · step-by-step breakdown below

  1. Write reactants and products
  2. Balance total A on both sides
  3. Balance total Z on both sides
  4. Account for energy Q released/absorbed

3. Solved Examples

Basic

Q: α decay A change?

Solution: A−4

Answer: −4

Intermediate

Q: Missing particle?

Solution: balance Z

Answer: Use conservation

Advanced

Q: Endothermic?

Solution: KE supplied

Answer: Needs input energy

Exam

Q: Conserved quantities?

Solution: A, Z, E, p

Answer: Sec 27.1

²³⁵U + ¹n → ¹⁴¹Ba + ⁹²Kr + 3¹n + Q

Definition: Typical thermal-neutron fission of uranium-235.

Derivation

Slow neutron captured; nucleus splits into two medium-mass fragments + 2–3 neutrons + ~200 MeV.

Variables

Q ≈ 200 MeV · one of 40+ fission modes · event ~10⁻¹⁷ s

Why it works

Foundation of nuclear reactors and weapons; releases enormous energy per event.

Historical context

Hahn & Strassmann (1938); Meitner & Frisch explained fission.

Deep understanding

²³⁵U more fissile than ²³⁸U; liquid-drop model explains oscillation and split.

2. Diagrams & Visuals

²³⁵U + n → fragments + neutrons

Color-coded visual · step-by-step breakdown below

  1. Thermal neutron hits ²³⁵U
  2. Nucleus captures neutron
  3. Splits into two fragments
  4. Emits 2–3 neutrons + Q ≈ 200 MeV

3. Solved Examples

Basic

Q: Neutrons released?

Solution: 2–3

Answer: 2–3

Intermediate

Q: Energy Q?

Solution: ~200 MeV

Answer: ~200 MeV

Advanced

Q: Fragment count?

Solution: ~80 types

Answer: 40+ modes

Exam

Q: ²³⁵U fission equation?

Solution: Ba+Kr+3n

Answer: Sec 27.2

Δm ≈ 0.215 u → Q ≈ 200 MeV

Definition: Mass defect converted to fission energy for ²³⁵U.

Derivation

Q = Δm c²; using 1 u = 931.3 MeV gives ~200 MeV from ~0.215 u mass loss.

Variables

Δm in unified atomic mass units · Q in MeV

Why it works

Links measurable mass difference to released fission energy.

Historical context

Einstein E = mc²; average 2.54 neutrons per ²³⁵U fission.

Deep understanding

Energy comes from higher BE/A of fission fragments vs parent nucleus.

2. Diagrams & Visuals

0.215 u × 931.3 ≈ 200 MeV

Color-coded visual · step-by-step breakdown below

  1. Find mass of reactants
  2. Find mass of products
  3. Δm = m_react − m_prod
  4. Q = Δm × 931.3 MeV/u

3. Solved Examples

Basic

Q: Δm=0.215 u.

Solution: Q≈200 MeV

Answer: ~200 MeV

Intermediate

Q: 1 u energy?

Solution: 931.3 MeV

Answer: 931.3 MeV

Advanced

Q: Why energy released?

Solution: higher BE/A

Answer: More stable fragments

Exam

Q: Fission energy from Δm?

Solution: Δmc²

Answer: Sec 27.2

Fission: ~0.84 MeV/nucleon

Definition: Average energy per nucleon in ²³⁵U fission.

Derivation

Q ≈ 200 MeV divided by A = 238 nucleons in uranium → ~0.84 MeV/nucleon.

Variables

Compare with fusion ~6 MeV/nucleon

Why it works

Benchmark for fission efficiency; fusion releases more per nucleon.

Historical context

Heavy nuclei (U, Pu) undergo fission; light nuclei fuse.

Deep understanding

BE/A curve: fission profitable for A > ~200; fusion for light H isotopes.

2. Diagrams & Visuals

200 MeV / 238 ≈ 0.84 MeV/n

Color-coded visual · step-by-step breakdown below

  1. Total fission energy Q
  2. Divide by mass number A
  3. Compare to fusion value
  4. Note per-nucleon efficiency

3. Solved Examples

Basic

Q: 200/238?

Solution: ~0.84 MeV

Answer: 0.84 MeV/n

Intermediate

Q: Fusion value?

Solution: ~6 MeV/n

Answer: ~6 MeV/n

Advanced

Q: Which more per nucleon?

Solution: fusion

Answer: Fusion wins

Exam

Q: Fission MeV/nucleon?

Solution: ~0.84

Answer: Sec 27.2

²H + ²H → ⁴He + Q ≈ 24 MeV

Definition: Deuterium-deuterium fusion — laboratory example.

Derivation

Two deuterium nuclei fuse to helium-4; Coulomb barrier overcome at ~10⁷ K.

Variables

Q ≈ 24 MeV · ~6 MeV per nucleon · needs extreme temperature

Why it works

Model for controlled fusion research; deuterium abundant in oceans.

Historical context

Hydrogen bomb uses uncontrolled D-D or D-T fusion.

Deep understanding

1 g deuterium → ~100,000 kWh if fusion achieved.

2. Diagrams & Visuals

→ ⁴He + 24 MeV

Color-coded visual · step-by-step breakdown below

  1. Heat fuel to ~10 million K
  2. Overcome Coulomb repulsion
  3. Light nuclei fuse
  4. Energy Q released as KE of products

3. Solved Examples

Basic

Q: Products?

Solution: ⁴He

Answer: Helium-4

Intermediate

Q: Q value?

Solution: ~24 MeV

Answer: 24 MeV

Advanced

Q: Temperature needed?

Solution: ~10⁷ K

Answer: 10 million K

Exam

Q: D-D fusion equation?

Solution: ²H+²H→⁴He

Answer: Sec 27.4

4(¹H) → ⁴He + 2e⁺ + Q ≈ 26.8 MeV

Definition: Solar proton-proton (Bethe) cycle — Sun's energy source.

Derivation

Four hydrogen nuclei fuse to helium at Sun's core (~20 million K); positrons emitted.

Variables

Q ≈ 26.8 MeV · Sun consumes ~400×10⁶ ton H/s

Why it works

Stars shine by fusion, not fission; explains solar lifetime (~8 billion years H left).

Historical context

Hans Bethe explained stellar nucleosynthesis.

Deep understanding

Sun is H/He — thermonuclear fusion at core, not uranium fission.

2. Diagrams & Visuals

→ ⁴He + 26.8 MeV

Color-coded visual · step-by-step breakdown below

  1. Four ¹H at stellar core temperature
  2. Fusion builds ⁴He
  3. Positrons (and neutrinos) emitted
  4. 26.8 MeV carried away as radiation/KE

3. Solved Examples

Basic

Q: Sun's reaction type?

Solution: fusion

Answer: Fusion

Intermediate

Q: Energy per cycle?

Solution: 26.8 MeV

Answer: 26.8 MeV

Advanced

Q: Core temperature?

Solution: ~2×10⁷ K

Answer: 20 million K

Exam

Q: Solar energy equation?

Solution: 4¹H→⁴He

Answer: Sec 27.4

Fusion: ~6 MeV/nucleon

Definition: Energy per nucleon in hydrogen fusion — higher than fission.

Derivation

From BE/A curve: light nuclei gain stability fusing to He; ~6–6.7 MeV/n vs fission ~0.84.

Variables

Fusion ~6.7 MeV/u · Fission ~0.84 MeV/u

Why it works

Fusion more efficient per unit mass but harder to initiate on Earth.

Historical context

Controlled thermonuclear fusion still active research (ITER, etc.).

Deep understanding

Coulomb barrier + confinement make Earth fusion much harder than fission.

2. Diagrams & Visuals

Fusion ~6 Fission ~0.84

Color-coded visual · step-by-step breakdown below

  1. Calculate total fusion Q
  2. Divide by total nucleons involved
  3. Compare to fission per nucleon
  4. Note initiation difficulty

3. Solved Examples

Basic

Q: Which higher per nucleon?

Solution: fusion

Answer: Fusion

Intermediate

Q: Fusion value?

Solution: ~6 MeV/n

Answer: ~6 MeV/n

Advanced

Q: Why hard on Earth?

Solution: 10⁷ K needed

Answer: Extreme T

Exam

Q: Fusion vs fission efficiency?

Solution: ~6 vs ~0.84

Answer: Sec 27.4

1 fission ≈ 7×10⁵ × 1 C atom

Definition: Energy comparison — one fission vs burning one carbon atom.

Derivation

C + O₂ releases ~4 eV; ²³⁵U fission ~200 MeV ≈ 7×10⁵ times larger per event.

Variables

Also: ²⁷Al + ⁴He exothermic ~700,000× vs one C atom

Why it works

Dramatic illustration of nuclear vs chemical energy density.

Historical context

Explains submarine and ship nuclear propulsion advantage.

Deep understanding

Nuclear fuel loaded for ~6 months continuous reactor operation.

2. Diagrams & Visuals

Fission : Chemical ≈ 7×10⁵ : 1

Color-coded visual · step-by-step breakdown below

  1. Chemical energy ~eV per atom
  2. Fission energy ~200 MeV
  3. Convert to same units
  4. Form ratio ~7×10⁵

3. Solved Examples

Basic

Q: Fission vs coal atom?

Solution: fission wins

Answer: ~10⁵–10⁶×

Intermediate

Q: Order of magnitude?

Solution: 7×10⁵

Answer: 700,000×

Advanced

Q: Al+He reaction?

Solution: ~7×10⁵×

Answer: Similar ratio

Exam

Q: Energy ratio fission/C?

Solution: 7×10⁵

Answer: Sec 27.2

⁴He + ¹⁴N → ¹⁷O + ¹H + 6.5 MeV

Definition: Rutherford's first artificial nuclear transmutation (1919).

Derivation

Alpha particle bombards nitrogen; proton (¹H) knocked out; oxygen-17 formed.

Variables

Overall endothermic — needs 1.2 MeV supplied to α particle

Why it works

Proved nucleus can be altered; birth of nuclear physics.

Historical context

First human-made nuclear reaction; not fission or fusion.

Deep understanding

Charged α must overcome Coulomb barrier (~few MeV for light targets).

2. Diagrams & Visuals

⁴He + ¹⁴N → ¹⁷O + ¹H (Rutherford 1919)

Color-coded visual · step-by-step breakdown below

  1. Fast ⁴He approaches ¹⁴N
  2. Overcomes Coulomb barrier
  3. Transmutation to ¹⁷O
  4. Proton ejected with kinetic energy

3. Solved Examples

Basic

Q: Who discovered?

Solution: Rutherford

Answer: 1919

Intermediate

Q: New element formed?

Solution: ¹⁷O

Answer: Oxygen-17

Advanced

Q: Projectile type?

Solution: alpha (⁴He)

Answer: Charged

Exam

Q: First transmutation?

Solution: N→O

Answer: Sec 27.1

²⁷Al + ⁴He → ³⁰Si + ¹H + 10.7 MeV

Definition: Exothermic nuclear reaction example from lesson.

Derivation

Alpha captures on aluminium; silicon and proton produced with 10.7 MeV released.

Variables

Q = 10.7 MeV exothermic · ~700,000× one C atom burn

Why it works

Shows nuclear reactions can release large energy without fission.

Historical context

Typical (α, p) reaction studied in accelerators.

Deep understanding

Balance: A: 27+4=31=30+1; Z: 13+2=15=14+1.

2. Diagrams & Visuals

²⁷Al + ⁴He → ³⁰Si + ¹H + 10.7 MeV

Color-coded visual · step-by-step breakdown below

  1. Write reactants ²⁷Al + ⁴He
  2. Products ³⁰Si + ¹H
  3. Check A and Z balance
  4. Energy Q = 10.7 MeV released

3. Solved Examples

Basic

Q: Exothermic or endo?

Solution: exothermic

Answer: Releases energy

Intermediate

Q: Q value?

Solution: 10.7 MeV

Answer: 10.7 MeV

Advanced

Q: Balance Z?

Solution: 13+2=14+1

Answer: Conserved

Exam

Q: Exothermic example?

Solution: Al+He→Si+H

Answer: Sec 27.1

Chain: 1 fission → 2–3 neutrons

Definition: Nuclear chain reaction — neutrons from one fission trigger more.

Derivation

Self-sustained when neutron production rate equals loss rate; average ~2.54 for ²³⁵U.

Variables

Controlled: reactor (Cd/B rods) · Uncontrolled: atom bomb

Why it works

Makes sustained power (reactor) or explosive release (bomb) possible.

Historical context

Fermi: first chain reaction, Chicago 1942.

Deep understanding

Moderator slows fast neutrons to thermal energy for further ²³⁵U fission.

2. Diagrams & Visuals

2–3 n per fission

Color-coded visual · step-by-step breakdown below

  1. One nucleus fissions
  2. 2–3 neutrons released
  3. Neutrons hit more ²³⁵U
  4. Control or absorb excess neutrons

3. Solved Examples

Basic

Q: Neutrons per fission?

Solution: 2–3

Answer: 2–3

Intermediate

Q: Reactor control?

Solution: control rods

Answer: Absorb neutrons

Advanced

Q: Self-sustained condition?

Solution: prod = loss

Answer: Criticality

Exam

Q: Chain reaction neutrons?

Solution: 2–3

Answer: Sec 27.2

5. Special Features & Extras

Complete study guide for Nuclear Fission and Fusion.

Exam Tips & Tricks

  • Energy scale: chemical ~eV · nuclear ~MeV.
  • Conservation: balance ΣA, ΣZ, energy, momentum in every equation.
  • Fissile materials: ²³³Th, ²³³U, ²³⁵U, ²³⁹Pu (odd A, even Z).
  • ²³⁵U fission: Q ≈ 200 MeV · Δm ≈ 0.215 u · 2–3 neutrons.
  • Chain reaction: controlled = reactor · uncontrolled = atom bomb.
  • Reactor parts: fuel, moderator, control rods (Cd/B), coolant, reflector, shield.
  • Fusion: needs ~10⁷ K · Sun = 4¹H → ⁴He + 26.8 MeV.
  • India: Bhabha 3-stage plan · PHWR → breeder → thorium.
  • RBE: γ/β = 1 · thermal n = 2–5 · fast n = 10 · α = 10–20.

Common Student Mistakes

  • Thinking the Sun runs on fission (it is fusion)
  • Confusing fissile (²³⁵U) with fertile (²³⁸U — absorbs n, β⁻ decay)
  • Forgetting to balance A and Z in nuclear equations
  • Using fast neutrons for ²³⁵U fission in reactors (need thermal ~0.025 eV)
  • Mixing up control rods (absorb) and moderator (slow down)
  • Assuming fusion is easier than fission on Earth

Memory Aids & Mnemonics

Fission vs Fusion: "Fission splits heavy U; Fusion fuses light H"
Reactor parts: "Fuel Moderate Control Cool Shield — FMCCS"
Chain reaction: "2–3 neutrons out — one fission in"
Sun energy: "Four H make one He — Bethe's PP cycle"

Which Formula When?

  • Chemical vs nuclear energy? → eV vs MeV
  • Balance reaction? → ΣA, ΣZ conserved
  • ²³⁵U fission energy? → Q ≈ 200 MeV, Δm ≈ 0.215 u
  • Per nucleon comparison? → fission ~0.84 · fusion ~6 MeV/n
  • Solar fusion? → 4¹H → ⁴He + 26.8 MeV
  • Lab fusion example? → ²H + ²H → ⁴He + 24 MeV
  • First transmutation? → Rutherford ⁴He + ¹⁴N
  • Chain reaction? → 2–3 neutrons per fission

QUICK REFERENCE — Ch 27 Nuclear Fission & Fusion

Chemical ~eV · Nuclear ~MeVΣA, ΣZ, E, p conserved²³⁵U + ¹n → ¹⁴¹Ba + ⁹²Kr + 3¹n + QΔm ≈ 0.215 u → Q ≈ 200 MeVFission: ~0.84 MeV/nucleon²H + ²H → ⁴He + Q ≈ 24 MeV4(¹H) → ⁴He + 2e⁺ + Q ≈ 26.8 MeVFusion: ~6 MeV/nucleon1 fission ≈ 7×10⁵ × 1 C atom⁴He + ¹⁴N → ¹⁷O + ¹H + 6.5 MeV²⁷Al + ⁴He → ³⁰Si + ¹H + 10.7 MeVChain: 1 fission → 2–3 neutrons

Key reactions: ²³⁵U+n→fragments+2–3n+200 MeV · ²H+²H→⁴He+24 MeV · 4¹H→⁴He+26.8 MeV

Reactor: fuel · moderator · control rods · coolant · reflector · shield

Tip: Thermal neutrons (<0.1 eV) sustain ²³⁵U fission in reactors — use moderator.

PYQ — Previous Year Questions

Extracted from NIOS Physics (312) board exam papers in your PDF. Chapter L27 — Nuclear Fission and Fusion only. Use Model Answer for marking points; Explanation for concept clarity.

Chapter 27 — Nuclear Fission and Fusion (L27)

3 questions · Sections A & B · Sources: 312/MAY/204A–C, 68/ESS/1-312-A

Section A — Objective (1 mark)

PYQ1. Read the passage: “In nuclear reactions, the nuclei of the reactants interact with each other and result in the formation of new elements… nuclear reactions can be endothermic or exothermic.” (b) Which of the following energy values of neutron can cause fission chain reaction in ²³⁵₉₂U? (A) Less than 0·1 eV (B) Greater than 1·0 eV (C) Greater than 7·7 MeV (D) Less than 7·0 eV but greater than 1·0 eV

1 mark · Section A Q19 (b) · 312/MAY/204A (also 204B/C)

Model Answer

(A) Less than 0·1 eV

Thermal (slow) neutrons (~0.025 eV at room temperature) efficiently induce fission in ²³⁵U and sustain a controlled chain reaction in a reactor.

Explanation

²³⁵U is fissile by thermal neutrons; fast neutrons are less likely to cause fission capture. Moderator in reactors slows neutrons to this energy range (L27 §27.2).

Section A — Short Answer (2 marks)

PYQ2. Fill in the blanks: (i) ___________ is a device in which a sustained, nuclear chain reaction is carried out in a controlled manner. (ii) The enormous energy produced by the sun is generated as a result of __________ reaction.

2 marks · Section A Q20 · 68/ESS/1-312-A

Model Answer

(i) Nuclear reactor — controlled chain reaction (control rods absorb excess neutrons)

(ii) nuclear fusion — Sun fuses hydrogen into helium at ~10⁷ K core temperature

Explanation

Reactor = peaceful controlled fission; Sun’s power = thermonuclear fusion, not fission (L27 §27.2–§27.3).

Section B — Short Answer (2 marks)

PYQ3. What is nuclear fusion? Write an equation of nuclear fusion to support your answer.

2 marks · Section B Q31 · 68/ESS/1-312-A (Marking Scheme)

Model Answer

Nuclear fusion is the process in which two nuclei of lighter elements (such as hydrogen) fuse to form a heavier nucleus (such as helium), and a neutron may be emitted; a large amount of energy is released.

Example: ²₁H + ²₁H → ³₂He + ¹₀n + energy

(Sun: 4¹H → ⁴He + 2e⁺ + 2ν + 26.8 MeV also acceptable as fusion equation.)

Explanation

Fusion releases more energy per nucleon than fission but needs ~10⁷ K to overcome Coulomb repulsion between positive nuclei (L27 §27.3).

Problem Solving — L27 Nuclear Fission and Fusion

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.

Question 1 of 6Fission

Define nuclear fission and name one fissile isotope used in reactors.

Solution — step by step with formulas

  1. Heavy nucleus splits into medium-mass fragments with energy release.
  2. e.g. ²³⁵U.

Final answer: ²³⁵U fission example

Textbook formal language

Fission liberates energy because BE/nucleon is higher for medium mass products.

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)

Big nucleus cracks into mid-size pieces that are more tightly bound—energy spills out.

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 fission

Neutrons emitted can sustain a chain reaction.

Link to chapter notes (L27 — Nuclear fission): 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).
Question 2 of 6Chain

Distinguish controlled and uncontrolled chain reactions.

Solution — step by step with formulas

  1. Controlled: reactor, neutron population steady (moderator/control rods).
  2. Uncontrolled: explosion, rapid multiplication.

Final answer: Reactor vs bomb: neutron economy control

Textbook formal language

Multiplication factor k ≈ 1 critical; k>1 supercritical.

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)

If each fission’s neutrons cause one more fission, steady power; if more than one, runaway.

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 — Chain reaction

Control rods absorb neutrons (Cd, B).

Link to chapter notes (L27 — Chain reaction): 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).
Question 3 of 6Fusion

Define fusion and state why high temperature is required.

Solution — step by step with formulas

  1. Light nuclei combine to heavier; need high KE to overcome Coulomb barrier.

Final answer: High T for Coulomb barrier penetration

Textbook formal language

Fusion of light nuclei increases BE/nucleon up to Fe region.

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)

Nuclei repel; must smash together very fast (hot plasma) to stick.

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 fusion

Powers stars; hydrogen isotopes fuse in stages.

Link to chapter notes (L27 — Nuclear fusion): 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).
Question 4 of 6Energy

Why does fission of ²³⁵U release energy in terms of binding energy per nucleon?

Q = Δm c²

Solution — step by step with formulas

  1. Products have higher BE per nucleon than ²³⁵U ⇒ total mass decreases ⇒ energy released.

Final answer: Higher BE/A of fragments ⇒ energy out

Formulas used in this problem

Q = Δm c²

Textbook formal language

Mass defect of reaction appears as KE of fragments and neutrons plus radiation.

Working formula set for this problem: Q = 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)

Pieces are ‘more glued’ than original—leftover mass becomes 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 — Energy release

Typical ~200 MeV per ²³⁵U fission.

Link to chapter notes (L27 — Energy release): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: Q = 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 Q = 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).
Question 5 of 6Moderator

Why are moderators used in thermal reactors?

Solution — step by step with formulas

  1. Slow fast fission neutrons to thermal energies where fission cross-section of ²³⁵U is high.

Final answer: Slow neutrons for efficient ²³⁵U fission

Textbook formal language

Elastic scattering on light nuclei reduces neutron energy without much absorption ideally.

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)

Fast neutrons zoom past fuel; slowed (water/graphite) they are more easily captured for fission.

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 — Moderator role

H₂O, D₂O, graphite common moderators.

Link to chapter notes (L27 — Moderator role): 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).
Question 6 of 6Fusion vs fission

Give one advantage of fusion over fission for power (in principle).

Solution — step by step with formulas

  1. Abundant fuel (isotopes of H), less long-lived radioactive waste (in ideal scenarios).

Final answer: Cleaner fuel cycle / abundant fuel (principle)

Textbook formal language

Engineering confinement (magnetic/inertial) remains challenging on Earth.

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)

Fusion aims for star-like power with lighter radioactive burden, but hard to contain.

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 — Comparison

Both release energy via BE curve.

Link to chapter notes (L27 — Comparison): 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).