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RRB CMA CBT-2 · Complete Study Notes · Day 3

NUCLEAR
PHYSICS

Structure · Radioactivity · Decay Laws · Fission · Fusion · Reactors

Nuclear Structure & Notation Alpha / Beta / Gamma Decay Half-Life Calculations Radioactive Decay Law Nuclear Binding Energy Mass Defect & Q-Value Nuclear Fission & Chain Reaction Nuclear Fusion Nuclear Reactor Components Radiation Dosimetry Applications & Hazards

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Numericals

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Section 01

Nuclear Structure

// atom → nucleus → nucleons → quarks | notation, isotopes, isobars, isotones

1.1 Atom & Nucleus

The nucleus is the dense central core of an atom containing protons and neutrons (collectively called nucleons). It contains 99.9% of the atom's mass in only 10^-15 m (1 femtometre) radius — compared to the atom's 10^-10 m radius.

Nuclear Notation

^A_ZX

• A = Mass number = Total nucleons = Protons + Neutrons
• Z = Atomic number = Number of protons = Number of electrons (neutral atom)
• N = Number of neutrons = A − Z
• Example: ²³⁸₉₂U → A=238, Z=92, N=238−92=146 neutrons

1.2 Nuclear Terminology

Term	Definition	Same	Different	Example
Isotopes	Same element, different mass numbers	Z (protons)	A and N	1H, 2H (deuterium), 3H (tritium)
Isobars	Different elements, same mass number	A	Z and N	40Ar and 40Ca (both A=40)
Isotones	Different elements, same number of neutrons	N (neutrons)	A and Z	14C(N=8) and 15N(N=8)
Isomers	Same Z and A, different energy states	Z and A	Energy state	99mTc (metastable technetium)
Nuclide	Any specific nucleus defined by Z and A	—		Specific Z, A, energy state

Memory Hook — ISO trick

Isotopes = Same Z (same proTon = same eLement)
Isobars = Same A (same mAss number)
Isotones = Same N (same Neutron count)

Hindi: "Top se Z, Middle se A, Neeche se N" — Iso-Z = Isotopes, Iso-A = Isobars, Iso-N = Isotones

1.3 Nuclear Radius & Density

R = R₀ × A^(1/3) where R₀ = 1.2 × 10⁻¹⁵ m = 1.2 fm Nuclear density ≈ 2.4 × 10¹⁷ kg/m³ (same for ALL nuclei!)

Nuclear density is independent of A — a key result. It is ~10¹⁴ times denser than water.

N-1.1 — Nuclear Radius Basic

Find the radius of nucleus of ⁶⁴Cu. Also compare with ⁸Be nucleus. (R₀ = 1.2 fm)

// Solution

1

For Cu-64: A = 64, so A^(1/3) = 64^(1/3) = 4
R = R₀ × A^(1/3) = 1.2 × 4 = 4.8 fm

2

For Be-8: A = 8, A^(1/3) = 8^(1/3) = 2
R = 1.2 × 2 = 2.4 fm

3

Ratio R_Cu / R_Be = 4.8/2.4 = 2 = (64/8)^(1/3) = 8^(1/3) = 2 ✓

Answer

R(Cu-64) = 4.8 fm. R(Be-8) = 2.4 fm. Cu nucleus is twice as large in radius as Be.

1.4 Nuclear Forces

Strong Nuclear Force Properties

• Strongest of all four fundamental forces at nuclear scale
• Short range: Acts only up to ~2–3 fm. Beyond this, it drops to zero.
• Charge independent: n-n = p-p = n-p attraction (same force regardless of charge)
• Saturates: Each nucleon interacts only with its nearest neighbours
• Non-central: Direction-dependent (unlike gravity and electrostatics)

Force	Relative Strength	Range	Mediator
Strong Nuclear	10³⁸ (strongest)	~2 fm	Gluons / Pions
Electromagnetic	10³⁶	Infinite	Photons
Weak Nuclear	10²⁵	~10⁻¹⁸ m	W, Z bosons
Gravitational	1 (weakest)	Infinite	Gravitons (theoretical)

Section 02

Binding Energy & Mass Defect

// E = mc² at work | the energy that holds the nucleus together

2.1 Mass Defect

Mass Defect Concept

The actual mass of a nucleus is less than the sum of masses of its constituent nucleons. This "missing mass" is the mass defect (Δm), converted to binding energy via E = mc².

Δm = [Z·m_p + (A−Z)·m_n] − M_nucleus BE = Δm × c² (in Joules) BE = Δm × 931.5 MeV (when Δm in atomic mass units, u) BE per nucleon = BE / A

1 u = 1.66054 × 10⁻²⁷ kg | 1 u × c² = 931.5 MeV | m_p = 1.007276 u | m_n = 1.008665 u | m_e = 0.000549 u

N-2.1 — Mass Defect of Helium-4 Classic

Calculate the mass defect and binding energy of He-4 nucleus. Given: m(He-4) = 4.002602 u, m_p = 1.007276 u, m_n = 1.008665 u.

// Solution

1

He-4 has Z=2 protons, N=2 neutrons
Sum of free nucleon masses = 2(1.007276) + 2(1.008665)
= 2.014552 + 2.017330 = 4.031882 u

2

Mass defect Δm = 4.031882 − 4.002602 = 0.029280 u

3

BE = Δm × 931.5 MeV = 0.029280 × 931.5 = 27.28 MeV

4

BE per nucleon = 27.28 / 4 = 6.82 MeV/nucleon

Answer

Δm = 0.02928 u | BE = 27.28 MeV | BE/nucleon = 6.82 MeV/nucleon

N-2.2 — Binding Energy of Fe-56 Important

Fe-56 has mass = 55.9349 u. Find mass defect, total BE, and BE/nucleon. (m_p = 1.007276 u, m_n = 1.008665 u)

// Solution

1

Fe-56: Z = 26, N = 30
Free mass = 26(1.007276) + 30(1.008665) = 26.18918 + 30.25995 = 56.44913 u

2

Δm = 56.44913 − 55.93490 = 0.51423 u

3

BE = 0.51423 × 931.5 = 479.0 MeV

4

BE/nucleon = 479.0 / 56 = 8.55 MeV/nucleon (HIGHEST — Fe-56 is most stable nucleus!)

Answer

Δm = 0.514 u | BE = 479 MeV | BE/nucleon = 8.55 MeV/nucleon (maximum — why Fe-56 is most stable)

2.2 BE/Nucleon Curve — The Most Important Graph in Nuclear Physics

Fig 2.1 — BE/nucleon curve: Fe-56 has maximum stability (8.55 MeV/nucleon). Energy released in both fusion (left→Fe) and fission (right→Fe).

Key Conclusions from BE Curve

• Fe-56 is most stable nucleus — maximum BE/nucleon (~8.55 MeV)
• Fusion: Light nuclei (H, He) have low BE/nucleon → merging them → products closer to Fe-56 → BE increases → energy released
• Fission: Heavy nuclei (U-235, Pu-239) have lower BE/nucleon than middle nuclei → splitting → products closer to Fe-56 → BE increases → energy released
• He-4 anomaly: BE/nucleon of He-4 is HIGHER than Li, Be, B — due to doubly magic configuration (Z=2, N=2)

Section 03

Radioactivity & Decay Types

// spontaneous nucleus disintegration | alpha, beta, gamma — properties & detection

3.1 What is Radioactivity?

Radioactivity is the spontaneous disintegration of unstable nuclei with emission of radiation (particles/waves). Discovered by Henri Becquerel (1896) in uranium. Named by Marie Curie.

Radioactivity is SPONTANEOUS — Key Features

• Cannot be controlled by temperature, pressure, chemical state, electric/magnetic field
• Depends ONLY on the nature of the nucleus (unstable N:Z ratio)
• Statistical in nature — we cannot predict which specific nucleus will decay next
• Accompanies nuclear reaction (chemical reactions do NOT produce radioactivity)

3.2 Alpha (α) Decay

Alpha Particle = Helium-4 Nucleus

Alpha particle = ⁴₂He (2 protons + 2 neutrons). It is identical to the helium-4 nucleus. High ionising power, low penetration.

Alpha decay: ᴬ_Z(X) → ᴬ⁻⁴_(Z-2)(Y) + ⁴_2(He) + energy

Mass number decreases by 4 | Atomic number decreases by 2

Fig 3.1 — Alpha decay of U-238: A decreases by 4, Z decreases by 2. Th-234 + He-4 (alpha particle)

3.3 Beta (β) Decay — Two Types

Beta-Minus (β⁻) Decay

A neutron converts to a proton inside the nucleus.

n → p + e⁻ + antineutrino (ν̄) ᴬ_Z(X) → ᴬ_(Z+1)(Y) + e⁻ + ν̄

A stays same | Z increases by 1 | Emits electron + antineutrino

Example: C-14 → N-14 + e⁻ + ν̄ (radiocarbon dating)

Beta-Plus (β⁺) Decay / Positron

A proton converts to a neutron inside the nucleus.

p → n + e⁺ + neutrino (ν) ᴬ_Z(X) → ᴬ_(Z-1)(Y) + e⁺ + ν

A stays same | Z decreases by 1 | Emits positron + neutrino

Example: Na-22 → Ne-22 + e⁺ + ν

3.4 Gamma (γ) Decay

Gamma Radiation

• Gamma rays are high-energy electromagnetic radiation (photons), NOT particles
• Emitted when excited nucleus transitions to lower energy state
• A and Z do NOT change in gamma decay — only energy changes
• No charge, no mass, travels at speed of light
• Example: ⁶⁰Co emits gamma radiation (used in cancer treatment)

ᴬ_Z(X*) → ᴬ_Z(X) + γ (* = excited state)

3.5 Comparison of Alpha, Beta, Gamma

Property	Alpha (α)	Beta (β⁻)	Gamma (γ)
Nature	Particle (He-4 nucleus)	Particle (electron)	EM radiation (photon)
Charge	+2e	−1e	0 (neutral)
Mass	4u	0.00055 u	0
Speed	~5–7% of c	Up to 99% of c	c (3×10⁸ m/s)
Ionising Power	Highest (~10,000×)	Medium (~100×)	Lowest (1×)
Penetrating Power	Lowest	Medium	Highest
Stopped by	Paper / skin / few cm air	Few mm aluminium	Several cm lead / concrete
A change	−4	0	0
Z change	−2	+1 (β⁻) or −1 (β⁺)	0
Deflected by fields	Yes (+ charge)	Yes (− charge)	No (neutral)

Memory Hook — Penetration vs Ionisation

"Alpha = Aam (common) = Stays in aam zindagi (paper stops it) but BITES HARD (highest ionisation)"
"Gamma = Goes everywhere (highest penetration) but TICKLES (lowest ionisation per cm)"

Inverse relationship: Higher penetration → Lower ionisation (per unit length)

Exam Trap Alerts — Radioactive Decay

• Trap 1: "Beta particle is emitted from electron shells" → FALSE. It comes from the NUCLEUS (neutron → proton + electron)
• Trap 2: "Gamma decay changes mass number" → FALSE. A and Z both unchanged in gamma decay
• Trap 3: "Alpha has high penetration because it's heavy" → FALSE. Alpha has LOWEST penetration precisely because it's heavy and charged (loses energy quickly)
• Trap 4: "Beta-minus emits only electron" → FALSE. It also emits an antineutrino (ν̄)
• Trap 5: "Gamma = beta⁺ (positron)" → FALSE. They are completely different. Positron (β⁺) is a particle; gamma is EM radiation

N-3.1 — Decay Equation Balancing Core Skill

Complete the decay equations: (a) U-238 undergoes alpha decay, (b) C-14 undergoes beta-minus decay, (c) Na-24 undergoes beta-minus decay followed by gamma emission.

// Solution

1

(a) Alpha decay: A−4=234, Z−2=90
²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He (Thorium-234)

2

(b) Beta-minus: A same=14, Z+1=7
¹⁴₆C → ¹⁴₇N + ⁰₋₁e + ν̄ (Nitrogen-14) — used in radiocarbon dating!

3

(c) Na-24 beta-minus: A=24, Z=11+1=12
²⁴₁₁Na → ²⁴₁₂Mg* + e⁻ + ν̄ Then gamma: ²⁴₁₂Mg* → ²⁴₁₂Mg + γ (A and Z unchanged)

Answer

(a) U-238 → Th-234 + He-4 | (b) C-14 → N-14 + e⁻ + ν̄ | (c) Na-24 → Mg-24* + e⁻ → Mg-24 + γ

N-3.2 — Decay Chain Problem Multi-step

A radioactive nucleus undergoes 3 alpha decays and 2 beta-minus decays. If starting nucleus is U-238 (Z=92, A=238), find the final nucleus.

// Solution

1

Each alpha decay: A−4, Z−2 3 alpha decays: A decreases by 3×4=12, Z decreases by 3×2=6

2

Each beta-minus decay: A unchanged, Z+1 2 beta decays: A unchanged, Z increases by 2×1=2

3

Final A = 238 − 12 = 226 Final Z = 92 − 6 + 2 = 88 (Radium, Ra)

Answer

Final nucleus: ²²⁶₈₈Ra (Radium-226). Formula: Final Z = Z₀ − 2α + β, Final A = A₀ − 4α

Section 04

Radioactive Decay Law & Half-Life

// N = N₀ e^(−λt) | the most-asked numerical topic in nuclear physics

4.1 Radioactive Decay Law

Rutherford-Soddy Law

The rate of disintegration of a radioactive substance at any instant is directly proportional to the number of radioactive atoms present at that instant.

dN/dt = −λN (negative because N is decreasing)

N(t) = N₀ × e^(−λt) A(t) = A₀ × e^(−λt) (Activity follows same law) T½ = ln2 / λ = 0.693 / λ (Half-life formula) λ = 0.693 / T½ (Decay constant from half-life) N(t) = N₀ × (1/2)^(t/T½) (Half-life form — MOST USEFUL) Activity A = λN (Activity = decay constant × number of atoms) Mean life τ = 1/λ = T½ / 0.693 = 1.443 × T½

N₀ = initial atoms | λ = decay constant (s⁻¹) | T½ = half-life | Activity unit: Becquerel (Bq) = 1 disintegration/second | Old unit: Curie (Ci) = 3.7×10¹⁰ Bq

4.2 Half-Life Intuition

Fig 4.1 — Exponential decay: N halves every half-life. After n half-lives: N = N₀/2ⁿ

N-4.1 — Basic Half-Life Easy

A radioactive sample has half-life of 5 years. Initially it has 8000 atoms. How many atoms remain after (a) 5 years, (b) 10 years, (c) 20 years?

// Solution

1

Use N = N₀ × (1/2)^(t/T½). Here N₀ = 8000, T½ = 5 years

2

(a) t = 5 years = 1 half-life: N = 8000 × (1/2)^1 = 4000 atoms

3

(b) t = 10 years = 2 half-lives: N = 8000 × (1/2)^2 = 8000/4 = 2000 atoms

4

(c) t = 20 years = 4 half-lives: N = 8000 × (1/2)^4 = 8000/16 = 500 atoms

Answer

(a) 4000 atoms | (b) 2000 atoms | (c) 500 atoms | After 4 half-lives = 1/16 remain

N-4.2 — Find Half-Life from Decay Medium

A radioactive substance reduces to 1/32 of its original amount in 25 days. Find its half-life.

// Solution

1

N/N₀ = 1/32 = 1/2⁵ → 5 half-lives have passed

2

Total time = 25 days, number of half-lives n = 5

3

T½ = Total time / n = 25 / 5 = 5 days

Answer

Half-life = 5 days. Quick trick: 1/32 = (1/2)⁵ → 5 half-lives → T½ = time/5

N-4.3 — Decay Constant Calculation Medium

Half-life of Iodine-131 is 8 days. (a) Find decay constant λ. (b) Find mean life. (c) How much of a 100 g sample remains after 24 days?

// Solution

1

(a) T½ = 8 days = 8×24×3600 = 691200 s λ = 0.693 / T½ = 0.693 / 691200 = 1.002 × 10⁻⁶ s⁻¹ Or simply: λ = 0.693/8 = 0.0866 per day

2

(b) Mean life τ = 1/λ = 1/0.0866 = 11.54 days = 1.443 × T½ = 1.443 × 8 = 11.54 days ✓

3

(c) t = 24 days, T½ = 8 days → n = 24/8 = 3 half-lives m = 100 × (1/2)³ = 100/8 = 12.5 g

Answer

(a) λ = 0.0866/day | (b) τ = 11.54 days | (c) 12.5 g remains after 24 days

N-4.4 — Exponential Decay Formula (Non-integer) Hard

A sample has 10⁶ radioactive atoms with T½ = 20 min. Find number of atoms after 50 minutes using the exponential formula.

// Solution

1

N = N₀ × e^(−λt) λ = 0.693 / T½ = 0.693 / 20 = 0.03465 min⁻¹

2

λt = 0.03465 × 50 = 1.7325

3

N = 10⁶ × e^(−1.7325) = 10⁶ × 0.1768 = 1.768 × 10⁵ atoms

4

Cross-check with half-life method: n = 50/20 = 2.5 half-lives N = 10⁶ / 2^2.5 = 10⁶ / 5.657 = 1.768 × 10⁵ ✓

Answer

N = 1.768 × 10⁵ atoms. Both methods give same answer. For non-integer n, use 2^n = 2^2.5 = 4√2 = 5.657

N-4.5 — Activity Calculation Medium

A radioactive source has initial activity 1200 Bq and half-life 30 seconds. Find: (a) Activity after 2 minutes, (b) Number of atoms initially present.

// Solution

1

(a) t = 2 min = 120 s, T½ = 30 s → n = 120/30 = 4 half-lives A = A₀ × (1/2)^4 = 1200/16 = 75 Bq

2

(b) Activity A₀ = λN₀ → N₀ = A₀/λ λ = 0.693/30 = 0.0231 s⁻¹ N₀ = 1200 / 0.0231 = 5.19 × 10⁴ atoms

Answer

(a) Activity after 2 min = 75 Bq | (b) Initial atoms = 5.19 × 10⁴

N-4.6 — Radiocarbon Dating Application

An ancient wooden artifact has C-14 activity of 3.1 disintegrations per minute per gram. Fresh wood has 15.3 dis/min/g. T½ of C-14 = 5730 years. Estimate the age of the artifact.

// Solution

1

Activity ratio = A/A₀ = 3.1/15.3 = 0.2026

2

A = A₀ × e^(−λt) → 0.2026 = e^(−λt) Taking ln: −λt = ln(0.2026) = −1.598

3

λ = 0.693 / 5730 = 1.209 × 10⁻⁴ per year t = 1.598 / (1.209 × 10⁻⁴) = 13,218 years ≈ 13,200 years

Answer

Age ≈ 13,200 years. Principle: Living organisms maintain C-14 equilibrium; after death, C-14 decays at known rate.

4.3 Half-Lives of Important Isotopes

Isotope	Half-Life	Application / Significance
U-238	4.47 × 10⁹ years	Uranium-lead geological dating; nuclear fuel
C-14	5730 years	Radiocarbon dating of organic materials up to ~50,000 years
Ra-226	1600 years	First isolated by Marie Curie; historical standard
Co-60	5.27 years	Gamma radiation in cancer radiotherapy (cobalt bomb)
I-131	8 days	Thyroid cancer treatment; nuclear reactor waste tracer
Tc-99m	6 hours	Most widely used medical imaging isotope (bone scans)
Rn-222	3.82 days	Indoor radon hazard (causes lung cancer)
Pu-239	24,100 years	Nuclear weapons; breeder reactor fuel
H-3 (Tritium)	12.3 years	Fusion weapon fuel; radioactive labelling
Sr-90	28.8 years	Radioactive fallout; mimics calcium in bones

Section 05

Nuclear Fission

// heavy nucleus splits → enormous energy release | chain reaction | critical mass

5.1 What is Nuclear Fission?

Nuclear Fission

When a heavy nucleus (Z > 90, typically U-235 or Pu-239) absorbs a slow (thermal) neutron, it becomes highly unstable and splits into two medium-mass fragments + 2–3 fast neutrons + enormous energy.

²³⁵₉₂U + ¹₀n → ²³⁶₉₂U* → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3¹₀n + ~200 MeV ²³⁵₉₂U + ¹₀n → ¹⁴⁰₅₄Xe + ⁹⁴₃₈Sr + 2¹₀n + energy (another common reaction)

~200 MeV released per fission | Compare: chemical combustion releases ~few eV per reaction | Fission releases ~10⁷ times more energy per atom

Fig 5.1 — U-235 fission chain: thermal neutron absorbed → U-236* excited → splits to Ba-141 + Kr-92 + 3 neutrons + ~200 MeV

5.2 Chain Reaction

Chain Reaction Concept

The 2–3 neutrons produced in each fission can trigger further fissions, creating a self-sustaining chain reaction. This is the basis of both nuclear reactors (controlled) and nuclear bombs (uncontrolled).

• Sub-critical: k < 1 → Chain reaction dies out (too many neutrons escape/absorb without fission)
• Critical: k = 1 → Steady chain reaction (nuclear reactor steady state)
• Super-critical: k > 1 → Exponential growth → nuclear explosion

k = multiplication factor = neutrons in generation n+1 / neutrons in generation n

Critical Mass

The minimum mass of fissile material required to sustain a chain reaction is the critical mass. Below critical mass, too many neutrons escape without causing further fission.

• Critical mass of U-235 ≈ 52 kg (bare sphere)
• Critical mass of Pu-239 ≈ 10 kg (much smaller due to higher fission cross-section)
• Reflectors (e.g., beryllium) can reduce critical mass by reflecting escaped neutrons back

N-5.1 — Energy from Fission Important

Calculate the energy released when 1 kg of U-235 undergoes complete fission. Each fission releases 200 MeV. (Avogadro's number = 6.022 × 10²³, 1 MeV = 1.6 × 10⁻¹³ J)

// Solution

1

Number of U-235 atoms in 1 kg: n = (1000 g / 235 g/mol) × 6.022 × 10²³ = 2.562 × 10²⁴ atoms

2

Energy per fission = 200 MeV = 200 × 1.6 × 10⁻¹³ = 3.2 × 10⁻¹¹ J

3

Total energy = 2.562 × 10²⁴ × 3.2 × 10⁻¹¹ = 8.2 × 10¹³ J = 8.2 × 10¹³ J

4

In TNT equivalent: 1 tonne TNT = 4.2 × 10⁹ J → 8.2 × 10¹³ / 4.2 × 10⁹ ≈ 19,500 tonnes TNT ≈ 20 kilotons

Answer

E = 8.2 × 10¹³ J ≈ 20 kilotons TNT equivalent. Hiroshima bomb ≈ 15 kilotons from ~64 kg U-235.

N-5.2 — Q-Value of Fission Advanced

Find the Q-value of the fission: U-235 + n → Ba-141 + Kr-92 + 3n. Given masses (in u): U-235 = 235.0439, n = 1.0087, Ba-141 = 140.9144, Kr-92 = 91.9262

// Solution

1

Mass of reactants = M(U-235) + M(n) = 235.0439 + 1.0087 = 236.0526 u

2

Mass of products = M(Ba-141) + M(Kr-92) + 3×M(n) = 140.9144 + 91.9262 + 3×1.0087 = 140.9144 + 91.9262 + 3.0261 = 235.8667 u

3

Mass defect Δm = 236.0526 − 235.8667 = 0.1859 u

4

Q = Δm × 931.5 MeV = 0.1859 × 931.5 = 173.2 MeV ≈ 173 MeV

Answer

Q-value = 173 MeV. This specific reaction releases 173 MeV (total energy including kinetic energy of fragments and neutrons).

Section 06

Nuclear Fusion

// light nuclei merge → heavier nucleus | sun's energy source | future of clean energy

6.1 What is Nuclear Fusion?

Nuclear Fusion

Two light nuclei combine at extremely high temperature (~10⁷ to 10⁸ K) to form a heavier nucleus, releasing enormous energy. This is the energy source of the sun and stars.

²₁H + ²₁H → ³₂He + ¹₀n + 3.27 MeV (D-D fusion) ²₁H + ³₁H → ⁴₂He + ¹₀n + 17.6 MeV (D-T fusion — most practical) 4¹₁H → ⁴₂He + 2e⁺ + 2ν + 26.7 MeV (Proton-proton chain in sun)

D = Deuterium (H-2) | T = Tritium (H-3) | Fusion releases ~3-4× more energy per unit mass than fission | Requires plasma at ~100 million °C

Fission vs Fusion Comparison

Feature	Fission	Fusion
Fuel	U-235, Pu-239	D, T (from seawater)
Energy/kg	~8×10¹³ J	~3×10¹⁴ J (higher!)
Trigger	Slow neutron	Extreme heat (~10⁸K)
Waste	Radioactive (thousands of years)	Minimal (He-4, short T½)
Status	Commercial use (reactors)	Experimental (ITER, NIF)
Weapons	Atomic bomb	Hydrogen (thermonuclear) bomb

Why Fusion Needs High Temperature

• Both nuclei are positively charged → strong Coulomb repulsion
• Need kinetic energy > Coulomb barrier to get close enough for strong force
• Temperature ~10⁸ K provides this kinetic energy (kT ~ MeV)
• At these temperatures, matter exists as plasma (fully ionised gas)
• Confinement methods: Tokamak (magnetic), ICF (laser)

N-6.1 — Energy from D-T Fusion Important

Calculate Q-value for D-T fusion: ²H + ³H → ⁴He + n. Masses: D = 2.01410 u, T = 3.01605 u, He-4 = 4.00260 u, n = 1.00867 u

// Solution

1

Mass of reactants = 2.01410 + 3.01605 = 5.03015 u

2

Mass of products = 4.00260 + 1.00867 = 5.01127 u

3

Δm = 5.03015 − 5.01127 = 0.01888 u

4

Q = 0.01888 × 931.5 = 17.59 MeV ≈ 17.6 MeV

Answer

Q = 17.6 MeV per D-T fusion event. Compare: U-235 fission gives ~200 MeV but per reaction, not per nucleon. Per nucleon: fusion gives MORE energy.

Section 07

Nuclear Reactor

// controlled fission chain reaction | 5 key components | thermal reactor design

7.1 Principle of Nuclear Reactor

A nuclear reactor maintains a controlled, self-sustaining fission chain reaction with multiplication factor k = 1 (exactly critical). The heat produced is used to generate steam, which drives turbines to produce electricity.

Fig 7.1 — Nuclear reactor: fuel rods + control rods + moderator inside shielded core; coolant carries heat to steam generator

7.2 Five Key Components of Nuclear Reactor

#	Component	Function	Material Used	Key Fact
1	Fuel	Source of fissile material for chain reaction	Enriched U-235 (3–5% in UO₂ pellets) or Pu-239	Natural uranium has only 0.72% U-235; needs enrichment
2	Moderator	Slows fast fission neutrons to thermal speeds (~0.025 eV) for efficient U-235 fission	Light water (H₂O), Heavy water (D₂O), Graphite	D₂O best moderator (low neutron absorption); Graphite used in Chernobyl; H₂O most common
3	Control Rods	Absorb excess neutrons to control reaction rate (k = 1)	Boron (B-10) or Cadmium (Cd)	Push in → absorb more neutrons → reaction slows; Pull out → reaction speeds up; Emergency shutdown = SCRAM
4	Coolant	Carries heat from reactor core to steam generator	H₂O (PWR, BWR), D₂O (CANDU), CO₂ (AGR), Liquid Na (fast breeder)	Must not absorb too many neutrons; high heat capacity; boiling point issues
5	Biological Shield	Absorbs radiation (n, γ) to protect workers and environment	Thick concrete + lead lining	Typical shield: 2–3 m thick concrete; containment building provides secondary barrier

Memory Hook — 5 Reactor Components

"Fuel Mango Chutney Cooks Slowly" = Fuel, Moderator, Control rods, Coolant, Shield
Or: "F-M-C-C-S" = "Five Men Control Cold Steel"

7.3 Types of Nuclear Reactors

Type	Full Name	Moderator	Coolant	Country/Example
PWR	Pressurised Water Reactor	H₂O	H₂O (pressurised)	Most common worldwide (USA, France)
BWR	Boiling Water Reactor	H₂O	H₂O (boiling)	Fukushima (Japan) was BWR
PHWR/CANDU	Pressurised Heavy Water	D₂O	D₂O	India (RAPS, MAPS, NAPS); uses natural uranium!
RBMK	Graphite-moderated	Graphite	H₂O	Chernobyl disaster reactor type
FBR	Fast Breeder Reactor	None (fast neutrons)	Liquid sodium	India (FBTR at Kalpakkam); breeds Pu-239 from U-238

Section 08

Radiation Dosimetry & Units

// measuring radiation exposure | units: Bq, Ci, Gy, Sv, rem | biological effects

8.1 Units of Radiation

Quantity	SI Unit	Old Unit	Conversion	What it measures
Activity	Becquerel (Bq)	Curie (Ci)	1 Ci = 3.7 × 10¹⁰ Bq	Disintegrations per second
Absorbed Dose	Gray (Gy)	rad	1 Gy = 100 rad = 1 J/kg	Energy deposited per kg of tissue
Equivalent Dose	Sievert (Sv)	rem	1 Sv = 100 rem	Biological effect (dose × quality factor)
Exposure	Coulomb/kg (C/kg)	Roentgen (R)	1 R = 2.58 × 10⁻⁴ C/kg	Ionisation in air (gamma/X-ray only)

Quality Factor (QF) / Radiation Weighting Factor

Radiation	Quality Factor (QF)	Implication
X-ray, Gamma, Beta	1	Equivalent dose = absorbed dose
Neutrons (fast)	10–20	10–20× more damaging than gamma
Alpha particles	20	20× more damaging than gamma (internally)
Heavy ions	20	Maximum biological damage per Gy

H (Sv) = D (Gy) × QF — The Sievert accounts for biological damage, not just energy deposition

8.2 Radiation Dose Limits & Effects

Dose (Sv)	Effect
0.001 (1 mSv)	Annual background radiation limit (public). Natural background ~2.4 mSv/year.
0.02 (20 mSv)	Annual occupational dose limit (radiation workers)
0.1–0.5	Temporary reduction in white blood cells
1–2	Radiation sickness begins: nausea, fatigue
4–5	LD₅₀/30 — lethal dose to 50% of population within 30 days without treatment
>6	Usually fatal. Bone marrow destroyed.
>10	Fatal within days (gastrointestinal syndrome)

N-8.1 — Equivalent Dose Calculation Practical

A radiation worker receives 0.05 Gy from gamma radiation and 0.002 Gy from alpha particles in one year. Calculate the total equivalent dose in Sv. Is this within safe limits?

// Solution

1

H = D × QF for each type Gamma: H_γ = 0.05 × 1 = 0.05 Sv

2

Alpha: QF = 20 H_α = 0.002 × 20 = 0.04 Sv

3

Total H = 0.05 + 0.04 = 0.09 Sv = 90 mSv

4

Occupational annual limit = 20 mSv/year (ICRP recommendation) 90 mSv >> 20 mSv → EXCEEDS SAFE LIMIT by 4.5×

Answer

Total dose = 90 mSv. Exceeds annual limit of 20 mSv. Note: tiny alpha dose (0.002 Gy) contributes 44% of total harm due to QF=20!

Section 09

Nuclear Applications

// medical, industrial, agricultural, environmental — all exam-important applications

Application	Isotope/Radiation	How it Works	Exam Angle
Cancer Radiotherapy	Co-60 (γ), I-131 (β)	High-energy radiation kills rapidly dividing cancer cells	Co-60 = cobalt therapy; I-131 = thyroid cancer treatment
Medical Imaging (PET)	F-18, Tc-99m	Beta+ emitter → positron annihilation → two 511 keV gamma photons detected	PET scan uses positron emitters; SPECT uses gamma emitters
Radiocarbon Dating	C-14 (T½=5730 yr)	Living organisms maintain C-14 equilibrium; after death it decays at known rate	Useful for organic material up to ~50,000 years
Uranium-Lead Dating	U-238 (T½=4.47×10⁹ yr)	U-238 decays to Pb-206; ratio gives age of rocks/Earth	Earth's age = 4.54 billion years by this method
Industrial Radiography	Ir-192 (γ), Co-60 (γ)	Like medical X-ray but for metal welds, pipes; detects cracks	Used in CMA/chemical plant equipment inspection
Tracer Studies	Na-24, P-32, I-131	Radioactive isotope follows same chemical path as stable isotope; tracked by Geiger counter	Blood flow: Na-24; Plant nutrient: P-32; Thyroid: I-131
Food Irradiation	Co-60 (γ), Cs-137 (γ)	Gamma rays kill bacteria/insects; extends shelf life; does NOT make food radioactive	Does NOT make food radioactive — gamma just ionises, no activation at these energies
Smoke Detectors	Am-241 (α)	Alpha particles ionise air between plates; smoke interrupts ionisation current → alarm	Alpha used (not gamma) because very short range — safe when contained
Nuclear Power	U-235, Pu-239	Controlled fission → heat → steam → turbine → electricity	~10% of world electricity from nuclear; India's 3-stage plan (Th-232 final goal)
Sterilisation	Co-60 (γ)	Gamma rays kill microorganisms in medical equipment, spices, cosmetics	Cold sterilisation — no heat needed

Application Trap Alerts

• Trap 1: "Irradiated food is radioactive" → FALSE. Gamma irradiation does not induce radioactivity in food at these energy levels
• Trap 2: "Alpha radiation is safe because it can't penetrate skin" → FALSE externally; but if INHALED or INGESTED, alpha is 20× more damaging (QF=20)
• Trap 3: "Radiocarbon dating works on rocks/minerals" → FALSE. C-14 dating only works on formerly living (organic) material. U-Pb dating for rocks.
• Trap 4: "PET scan uses gamma directly" → FALSE. It uses positron (β⁺) emitters; annihilation of positron with electron produces gamma

Section 10

Full Practice Set — 8 More Numericals

// mixed difficulty | covers all topics | exam-pattern questions

N-10.1 — Decay Series Identification Multi-step

Th-232 (Z=90) undergoes 6 alpha and 4 beta-minus decays before reaching a stable isotope. Find the final stable nucleus.

// Solution

1

6 alpha decays: ΔA = −24, ΔZ = −12 4 beta decays: ΔA = 0, ΔZ = +4

2

Final A = 232 − 24 = 208 Final Z = 90 − 12 + 4 = 82 (Lead, Pb)

Answer

Final nucleus: Pb-208 (Lead-208, Z=82). This is indeed the end product of Th-232 decay series!

N-10.2 — Fraction Remaining After Multiple Periods Easy

What fraction of a radioactive sample with T½ = 12 years remains after 48 years? What fraction has decayed?

// Solution

1

n = 48/12 = 4 half-lives Fraction remaining = (1/2)^4 = 1/16

2

Fraction decayed = 1 − 1/16 = 15/16 = 0.9375 = 93.75%

Answer

1/16 (6.25%) remains. 15/16 (93.75%) has decayed. Quick check: 1/16 = 0.0625 ✓

N-10.3 — Number of Atoms from Mass Conceptual

1 microgram (1 μg = 10⁻⁶ g) of Ra-226 (T½ = 1600 years). Find: (a) number of atoms, (b) initial activity in Bq.

// Solution

1

N = (mass in g / molar mass) × Avogadro's number N = (10⁻⁶ / 226) × 6.022 × 10²³ = 2.664 × 10¹⁵ atoms

2

T½ = 1600 yr = 1600 × 3.156 × 10⁷ s = 5.05 × 10¹⁰ s λ = 0.693 / T½ = 0.693 / 5.05 × 10¹⁰ = 1.372 × 10⁻¹¹ s⁻¹

3

Activity A = λN = 1.372 × 10⁻¹¹ × 2.664 × 10¹⁵ = 3.65 × 10⁴ Bq = 36,500 Bq

Answer

N = 2.664 × 10¹⁵ atoms | Activity = 3.65 × 10⁴ Bq ≈ 36.5 kBq

N-10.4 — Mean Life vs Half-Life Conceptual

A radioactive isotope has mean life τ = 14.4 days. Find: (a) decay constant, (b) half-life, (c) fraction remaining after 1 mean life.

// Solution

1

(a) λ = 1/τ = 1/14.4 = 0.0694 per day

2

(b) T½ = 0.693/λ = 0.693/0.0694 = 9.986 ≈ 10 days Or: T½ = τ × ln2 = 14.4 × 0.693 = 9.98 days

3

(c) At t = τ: N = N₀ × e^(−λτ) = N₀ × e^(−1) = N₀/e = 0.368 N₀ So 36.8% remains after 1 mean life (always 1/e = 36.8%)

Answer

λ = 0.0694/day | T½ = 10 days | After 1 mean life: 36.8% remains (always 1/e)

N-10.5 — Two Radioactive Sources Hard

Source A has T½ = 4 hours and initially 10⁸ atoms. Source B has T½ = 2 hours and initially N₀ atoms. At t = 8 hours, both sources have equal activity. Find N₀ for source B.

// Solution

1

At t = 8 hours: Source A: n_A = 8/4 = 2 half-lives → N_A = 10⁸/4 = 2.5×10⁷ Activity_A = λ_A × N_A = (0.693/4) × 2.5×10⁷ = 4.33×10⁶ per hour

2

Source B: n_B = 8/2 = 4 half-lives → N_B = N₀/16 Activity_B = λ_B × N_B = (0.693/2) × (N₀/16)

3

Setting Activity_A = Activity_B: 4.33×10⁶ = (0.3465/16) × N₀ N₀ = 4.33×10⁶ × 16 / 0.3465 = 2×10⁸

Answer

N₀ = 2 × 10⁸ atoms for source B. B starts with twice as many atoms, but decays faster, reaching same activity as A after 8 hours.

N-10.6 — Nuclear Reactor Fuel Burn Applied

A nuclear power plant operates at 1000 MW electrical power with 33% thermal efficiency. Each U-235 fission releases 200 MeV. How much U-235 is consumed per day?

// Solution

1

Thermal power = Electrical / efficiency = 1000 MW / 0.33 = 3030 MW = 3.03×10⁹ J/s

2

Energy per fission = 200 MeV = 200 × 1.6×10⁻¹³ = 3.2×10⁻¹¹ J

3

Fissions per second = 3.03×10⁹ / 3.2×10⁻¹¹ = 9.47×10¹⁹ fissions/s

4

Fissions per day = 9.47×10¹⁹ × 86400 = 8.18×10²⁴ fissions/day Mass of U-235 = (8.18×10²⁴ / 6.022×10²³) × 235 g = 3194 g ≈ 3.2 kg/day

Answer

~3.2 kg of U-235 consumed per day. Compare: a coal power plant of same output burns ~9000 TONNES of coal per day!

N-10.7 — Nuclear Stability Check Conceptual

For each nucleus, determine the likely decay mode: (a) C-14 (Z=6, N=8), (b) N-13 (Z=7, N=6), (c) U-238 (Z=92, N=146).

// Solution using N/Z ratio

1

(a) C-14: N/Z = 8/6 = 1.33 → too many neutrons (for light nuclei, stable N/Z ≈ 1) Excess neutrons → β⁻ decay (n→p) → N-14 ✓

2

(b) N-13: N/Z = 6/7 = 0.857 → too few neutrons (proton rich) Excess protons → β⁺ decay (p→n) → C-13 ✓

3

(c) U-238: Z=92 (very heavy). N/Z = 1.587 → Coulomb repulsion dominates Heavy nucleus → α decay preferred (reduces both Z and A) ✓

Answer

(a) C-14 → β⁻ | (b) N-13 → β⁺ | (c) U-238 → α decay. Rule: N/Z > stable line → β⁻; N/Z < stable line → β⁺; Z > 83 → α preferred

N-10.8 — Specific Activity Medium

Find the specific activity (activity per gram) of Co-60. T½ = 5.27 years.

// Solution

1

Number of atoms in 1 g of Co-60: N = (1/60) × 6.022×10²³ = 1.004×10²² atoms

2

T½ = 5.27 × 3.156×10⁷ = 1.663×10⁸ s λ = 0.693 / 1.663×10⁸ = 4.166×10⁻⁹ s⁻¹

3

Specific activity = λN = 4.166×10⁻⁹ × 1.004×10²² = 4.18×10¹³ Bq/g = 4.18×10¹³ / 3.7×10¹⁰ Ci/g = 1130 Ci/g

Answer

Specific activity of Co-60 = 4.18 × 10¹³ Bq/g ≈ 1130 Ci/g. Extremely high — why it's used in small quantities for radiotherapy.

Section 11

Formula Master Sheet

// every formula | revise in 2 minutes before exam

NUCLEAR STRUCTURE

Nuclear notationA_Z(X): A = Z+N, N = A−Z

Nuclear radiusR = R₀ × A^(1/3), R₀ = 1.2 fm

Nuclear density~2.4 × 10¹⁷ kg/m³ (same for all nuclei)

BINDING ENERGY

Mass defectΔm = [Z·m_p + N·m_n] − M_nucleus

Binding energyBE = Δm × 931.5 MeV (Δm in u)

BE per nucleonBE/A (max at Fe-56 = 8.55 MeV)

E = mc²1 u × c² = 931.5 MeV

RADIOACTIVE DECAY

Decay lawN = N₀ × e^(−λt)

Half-life formN = N₀ × (1/2)^(t/T½) = N₀ / 2^n

Half-life ↔ λT½ = 0.693/λ, λ = 0.693/T½

Mean lifeτ = 1/λ = 1.443 × T½. At t=τ: N = N₀/e

ActivityA = λN (Bq). 1 Ci = 3.7×10¹⁰ Bq

Alpha decayA−4, Z−2

Beta-minus decayA same, Z+1 (n→p+e⁻+ν̄)

Beta-plus decayA same, Z−1 (p→n+e⁺+ν)

Gamma decayA same, Z same (energy state only)

FISSION & FUSION

Q-valueQ = (mass of reactants − mass of products) × 931.5 MeV

U-235 fission energy~200 MeV per fission, ~8×10¹³ J/kg

D-T fusion energy17.6 MeV per reaction

Critical conditionk = 1 (reactor), k < 1 (sub-critical), k > 1 (bomb)

DOSIMETRY

Absorbed doseD (Gray, Gy) = energy/mass (J/kg)

Equivalent doseH (Sv) = D (Gy) × QF

QF valuesX/γ/β = 1, fast n = 10−20, α = 20

Annual limitsPublic: 1 mSv/yr, Workers: 20 mSv/yr

Unit conversions1 Gy = 100 rad, 1 Sv = 100 rem

Top 10 Nuclear Physics Exam Traps

• Isotopes = same Z (same element); Isobars = same A; Isotones = same N
• Gamma decay: A and Z BOTH unchanged — only energy changes
• Alpha has HIGHEST ionising power, LOWEST penetration
• Alpha is He-4 nucleus (+2 charge); Beta-minus is electron (−1 charge)
• Mean life τ = 1.443 × T½ (NOT equal to T½)
• After 1 mean life: N₀/e (36.8%) remains; after 1 half-life: N₀/2 (50%) remains
• BE/nucleon is MAXIMUM for Fe-56, not for heaviest nucleus
• Fission produces 2–3 neutrons (not 1); these can trigger chain reaction
• Control rods (B, Cd) ABSORB neutrons; moderator SLOWS neutrons
• Irradiated food is NOT radioactive — gamma ionises but doesn't activate food atoms at these energies