Radioactivity: Nature’s Nuclear Clock
The Marie Curie Legacy 🎬
Before Oppenheimer harnessed nuclear energy, Marie Curie discovered radioactivity - winning two Nobel Prizes and revolutionizing physics! Her work with polonium and radium revealed that atoms aren’t immutable - they can spontaneously transform, ticking away like cosmic clocks.
“Nothing in life is to be feared, it is only to be understood. Now is the time to understand more, so that we may fear less.” - Marie Curie
Discovery of Radioactivity
Historical Timeline
1896: Henri Becquerel
- Discovered uranium emits “rays”
- Accidentally left uranium salt on photographic plate
- Plate darkened even wrapped in black paper!
- Not phosphorescence - something new! ⚛️
1898: Marie & Pierre Curie
- Discovered polonium (Po) and radium (Ra)
- Coined term “radioactivity”
- Isolated radioactive elements
1899: Rutherford
- Identified α and β radiation
- Different penetrating powers
1900: Villard
- Discovered γ radiation
- Most penetrating type
Types of Radioactive Decay
α-Decay (Alpha Decay)
Emission of helium nucleus ($^4_2He$ or α)
:::box
$$^A_ZX → ^{A-4}_{Z-2}Y + ^4_2He + Q$$Changes:
- Mass number: A → A - 4
- Atomic number: Z → Z - 2
- New element (2 places left in periodic table) :::
Example:
$$^{238}_{92}U → ^{234}_{90}Th + ^4_2He + 4.27 \text{ MeV}$$Characteristics:
- Heavy nuclei (A > 200) typically
- Low penetration (~few cm in air)
- Stopped by paper
- High ionization
- Energy: 4-9 MeV typically
β⁻-Decay (Beta-minus Decay)
Neutron converts to proton
$$n → p + e^- + \bar{ν}_e$$:::box
$$^A_ZX → ^{A}_{Z+1}Y + e^- + \bar{ν}_e + Q$$Changes:
- Mass number: A unchanged
- Atomic number: Z → Z + 1
- New element (1 place right in periodic table)
- Electron and antineutrino emitted :::
Example:
$$^{14}_6C → ^{14}_7N + e^- + \bar{ν}_e + 0.156 \text{ MeV}$$Characteristics:
- Neutron-rich nuclei
- Moderate penetration (~few mm in Al)
- Continuous energy spectrum (due to neutrino)
- Medium ionization
β⁺-Decay (Beta-plus/Positron Decay)
Proton converts to neutron
$$p → n + e^+ + ν_e$$:::box
$$^A_ZX → ^{A}_{Z-1}Y + e^+ + ν_e + Q$$Changes:
- Mass number: A unchanged
- Atomic number: Z → Z - 1
- New element (1 place left)
- Positron and neutrino emitted :::
Example:
$$^{22}_{11}Na → ^{22}_{10}Ne + e^+ + ν_e + 0.545 \text{ MeV}$$Note: Requires M(parent) > M(daughter) + 2m_e (threshold!)
Electron Capture (EC)
Nucleus captures inner shell electron
$$p + e^- → n + ν_e$$:::box
$$^A_ZX + e^- → ^{A}_{Z-1}Y + ν_e + Q$$Changes:
- Mass number: A unchanged
- Atomic number: Z → Z - 1
- X-rays emitted (electron vacancy filled)
- Competes with β⁺ decay :::
Example:
$$^{7}_4Be + e^- → ^{7}_3Li + ν_e$$γ-Decay (Gamma Decay)
Excited nucleus releases energy
:::box
$$^A_ZX^* → ^A_ZX + γ$$Changes:
- No change in A or Z!
- Same element, lower energy state
- Pure electromagnetic radiation
- Often follows α or β decay :::
Example:
$$^{60}_{27}Co^* → ^{60}_{27}Co + γ (1.17 \text{ MeV} + 1.33 \text{ MeV})$$Characteristics:
- Very high penetration (many cm of Pb)
- No mass or charge
- Electromagnetic wave (photon)
- Low ionization
- Energy: keV to MeV
Comparison of Radiation Types
| Property | α | β⁻/β⁺ | γ |
|---|---|---|---|
| Nature | ⁴He nucleus | Electron/Positron | Photon |
| Charge | +2e | ∓e | 0 |
| Mass | 4 u | ~0.0005 u | 0 |
| Speed | ~0.05c | ~0.9c | c |
| Penetration | Lowest | Medium | Highest |
| Stopped by | Paper | Al (few mm) | Pb (many cm) |
| Ionization | Highest | Medium | Lowest |
| Deflection in field | Small | Large | None |
| A change | -4 | 0 | 0 |
| Z change | -2 | ±1 | 0 |
Radioactive Decay Law
The Fundamental Law
:::box Decay Law
$$N(t) = N_0 e^{-λt}$$where:
- N(t) = number of nuclei at time t
- N₀ = initial number of nuclei
- λ = decay constant (characteristic of isotope)
- t = time :::
Meaning: Radioactive decay is:
- Random (can’t predict individual decay)
- Spontaneous (doesn’t need external trigger)
- Exponential (constant fraction decays per unit time)
Activity (Rate of Decay)
:::box
$$A(t) = -\frac{dN}{dt} = λN(t) = λN_0 e^{-λt}$$ $$A(t) = A_0 e^{-λt}$$Units:
- SI: Becquerel (Bq) = 1 decay/second
- Old: Curie (Ci) = 3.7 × 10¹⁰ Bq :::
Half-Life (T₁/₂)
Time for half the nuclei to decay:
$$N(T_{1/2}) = \frac{N_0}{2}$$:::box Half-Life Formula
$$T_{1/2} = \frac{\ln 2}{λ} = \frac{0.693}{λ}$$After n half-lives:
$$N = N_0 \left(\frac{1}{2}\right)^n$$where $n = \frac{t}{T_{1/2}}$ :::
Interactive Demo: Visualize Nuclear Decay
Watch radioactive nuclei decay exponentially over time, demonstrating half-life concepts.
Mean Life (τ)
Average lifetime of a nucleus:
:::box
$$τ = \frac{1}{λ} = \frac{T_{1/2}}{\ln 2} = 1.44 \times T_{1/2}$$:::
Relation:
$$T_{1/2} : τ : \frac{1}{λ} = 0.693 : 1 : 1$$Interactive Demo: Decay Calculator
const RadioactiveDecaySimulator = () => {
const [isotope, setIsotope] = useState('C14');
const [time, setTime] = useState(5730);
const [N0, setN0] = useState(1000000);
const halfLives = {
'C14': 5730, // years
'U238': 4.47e9, // years
'Ra226': 1600, // years
'I131': 8.02, // days
'Co60': 5.27 // years
};
const T_half = halfLives[isotope];
const lambda = 0.693 / T_half;
const N_t = N0 * Math.exp(-lambda * time);
const A_t = lambda * N_t;
const A_0 = lambda * N0;
const n_halfLives = time / T_half;
const percentRemaining = (N_t / N0) * 100;
return (
<div>
<h3>Radioactive Decay Simulator</h3>
<label>Isotope:</label>
<select value={isotope} onChange={(e) => setIsotope(e.target.value)}>
<option value="C14">Carbon-14 (T½=5730 y)</option>
<option value="U238">Uranium-238 (T½=4.47 Gy)</option>
<option value="Ra226">Radium-226 (T½=1600 y)</option>
<option value="I131">Iodine-131 (T½=8 d)</option>
<option value="Co60">Cobalt-60 (T½=5.27 y)</option>
</select>
<label>Initial nuclei: {N0.toExponential(2)}</label>
<input
type="range"
min="100000"
max="10000000"
step="100000"
value={N0}
onChange={(e) => setN0(Number(e.target.value))}
/>
<label>Time elapsed: {time} {isotope === 'I131' ? 'days' : 'years'}</label>
<input
type="range"
min="0"
max={T_half * 3}
value={time}
onChange={(e) => setTime(Number(e.target.value))}
/>
<div className="results">
<h4>Decay Parameters:</h4>
<p>Half-life: {T_half} {isotope === 'I131' ? 'days' : 'years'}</p>
<p>Decay constant: {lambda.toExponential(3)} per year</p>
<p>Mean life: {(1/lambda).toFixed(0)} {isotope === 'I131' ? 'days' : 'years'}</p>
<h4>At time t = {time}:</h4>
<p>Number of half-lives: {n_halfLives.toFixed(2)}</p>
<p>Remaining nuclei: {N_t.toExponential(2)}</p>
<p>Percent remaining: {percentRemaining.toFixed(1)}%</p>
<p>Initial activity: {A_0.toExponential(2)} Bq</p>
<p>Current activity: {A_t.toExponential(2)} Bq</p>
<div className="visual">
<div style={{
width: '100%',
height: '30px',
backgroundColor: '#ccc',
position: 'relative'
}}>
<div style={{
width: `${percentRemaining}%`,
height: '100%',
backgroundColor: 'blue',
transition: 'width 0.3s'
}} />
</div>
<p>Visual: Fraction remaining</p>
</div>
{percentRemaining < 1 && (
<p style={{color: 'red'}}>
⚠ Less than 1% remaining - essentially all decayed!
</p>
)}
</div>
</div>
);
};
Decay Series (Radioactive Chains)
Some isotopes decay through multiple steps:
Uranium-238 Series
²³⁸U (4.5 Gy)
↓ α
²³⁴Th (24 d)
↓ β⁻
²³⁴Pa (1.2 min)
↓ β⁻
²³⁴U (246,000 y)
↓ α
... (many steps)
↓
²⁰⁶Pb (stable)
Total: 14 steps, ending at stable lead-206!
Other Series
- Thorium-232 series → ²⁰⁸Pb
- Uranium-235 series → ²⁰⁷Pb
- Neptunium-237 series → ²⁰⁹Bi (extinct in nature)
Memory Tricks 🧠
“ALPHA BETA GAMMA” Comparison
Alpha: All Atomic mass lost (4 units) Large and heavy (He nucleus) Paper stops it High ionization Always from heavy nuclei
Beta: Basically electron Energy spectrum continuous Transmutation (Z changes by 1) Aluminum stops it
Gamma: Go through everything And no charge Mass-less photon Maximum penetration After other decays usually
Decay Formula Memory
$$N = N_0 e^{-λt}$$“Number equals N-zero times e to the minus lambda-t”
Half-life:
$$T_{1/2} = \frac{0.693}{λ}$$“Point-six-nine-three over lambda”
Half-Life Pattern
After 1 T½: ½ remains
After 2 T½: ¼ remains
After 3 T½: ⅛ remains
After n T½: (½)ⁿ remains
"Every half-life, half remains!"
Common Mistakes ⚠️
❌ Mistake 1: Half-life formula
Wrong: T₁/₂ = 1/λ Right: T₁/₂ = 0.693/λ (or ln2/λ)
❌ Mistake 2: Activity vs number
Wrong: A = N Right: A = λN (activity is rate, not number!)
❌ Mistake 3: Decay direction
Wrong: α decay increases Z Right: α decay decreases Z by 2, A by 4
❌ Mistake 4: Beta decay mass
Wrong: β⁻ decay decreases mass number Right: β⁻ keeps A same, only Z changes
❌ Mistake 5: Exponential vs linear
Wrong: N = N₀ - λt (linear decay) Right: N = N₀e^(-λt) (exponential decay)
Important Half-Lives
Common Isotopes
| Isotope | Type | Half-Life | Use |
|---|---|---|---|
| ²³⁸U | α | 4.47 × 10⁹ y | Dating rocks |
| ²³⁵U | α | 7.04 × 10⁸ y | Nuclear fuel |
| ⁴⁰K | β⁻, EC | 1.25 × 10⁹ y | Dating |
| ¹⁴C | β⁻ | 5,730 y | Carbon dating |
| ²²⁶Ra | α | 1,600 y | Historical source |
| ⁶⁰Co | β⁻, γ | 5.27 y | Medical |
| ⁹⁰Sr | β⁻ | 28.8 y | Fission product |
| ¹³¹I | β⁻, γ | 8.02 d | Medical |
| ²²²Rn | α | 3.82 d | Radon gas |
| ²¹⁰Po | α | 138 d | Curie’s poison! |
Carbon-14 Dating
The Principle
- Cosmic rays create ¹⁴C in atmosphere
- Living organisms maintain constant ¹⁴C/¹²C ratio
- After death, ¹⁴C decays (T₁/₂ = 5730 y)
- Measure remaining ¹⁴C → determine age!
:::box
$$t = \frac{T_{1/2}}{\ln 2} \ln\left(\frac{N_0}{N}\right)$$or
$$t = \frac{5730}{\ln 2} \ln\left(\frac{A_0}{A}\right)$$where A₀ is activity in living material, A is current activity. :::
Useful range: Up to ~50,000 years
Example: Dead wood has ¼ the ¹⁴C activity of living wood.
$$t = \frac{5730}{0.693} \ln(4) = 8267 \times 1.386 = 11,460 \text{ years old}$$Equilibrium in Decay Series
Secular Equilibrium
Parent half-life » daughter half-life:
$$T_{1/2}^{parent} >> T_{1/2}^{daughter}$$Eventually:
$$λ_p N_p = λ_d N_d$$Daughter decays as fast as it’s produced!
Example: ²³⁸U → ²³⁴Th system
Transient Equilibrium
Parent half-life > daughter half-life (but comparable):
$$\frac{A_d}{A_p} = \frac{T_{1/2}^d}{T_{1/2}^d - T_{1/2}^p}$$Practice Problems
Level 1: JEE Main Basics
Q1. Complete the decay: $^{238}_{92}U → ? + ^4_2He$
Solution:
α decay: A decreases by 4, Z decreases by 2
A: 238 - 4 = 234
Z: 92 - 2 = 90
²³⁴₉₀Th (Thorium-234)
Q2. A radioactive sample has half-life 10 days. What fraction remains after 30 days?
Solution:
n = t/T₁/₂ = 30/10 = 3 half-lives
Fraction = (1/2)³ = 1/8
Q3. The decay constant of a sample is 0.01 per year. Find the half-life.
Solution:
T₁/₂ = 0.693/λ = 0.693/0.01 = 69.3 years
Level 2: JEE Main/Advanced
Q4. ²²⁶Ra (T₁/₂ = 1600 y) decays to ²²²Rn (T₁/₂ = 3.8 d). Initially pure Ra sample. After long time, what is ratio of Ra to Rn decay rates?
Solution:
At secular equilibrium:
λ_Ra × N_Ra = λ_Rn × N_Rn
Decay rates: A_Ra = A_Rn
Ratio = 1:1
Even though numbers are different (N_Rn << N_Ra),
decay rates are equal at equilibrium!
Q5. A sample’s activity decreases to 6.25% of original in 2 hours. Find: (a) Half-life (b) Decay constant
Solution:
A/A₀ = 6.25/100 = 1/16 = (1/2)⁴
4 half-lives in 2 hours
T₁/₂ = 2/4 = 0.5 hours = 30 minutes
λ = 0.693/T₁/₂ = 0.693/0.5 = 1.386 h⁻¹
Q6. A ¹⁴C sample has activity 15 disintegrations/minute/gram. Living carbon has 16 dpm/g. Find age of sample. (T₁/₂ = 5730 y)
Solution:
A/A₀ = 15/16
t = (T₁/₂/ln2) × ln(A₀/A)
t = (5730/0.693) × ln(16/15)
t = 8267 × ln(1.0667)
t = 8267 × 0.0645
t ≈ 533 years
Level 3: JEE Advanced
Q7. Two radioactive samples A and B have equal activities initially. After 4 days, activity of A is 1/16 and B is 1/4 of initial. Find: (a) Ratio of half-lives (b) Ratio of activities after 8 days
Solution:
(a) For A: A_A/A₀ = 1/16 = (1/2)⁴
4 days = 4 half-lives of A
T_A = 1 day
For B: A_B/A₀ = 1/4 = (1/2)²
4 days = 2 half-lives of B
T_B = 2 days
T_A : T_B = 1:2
(b) After 8 days:
For A: n = 8/1 = 8 half-lives
A_A = A₀(1/2)⁸ = A₀/256
For B: n = 8/2 = 4 half-lives
A_B = A₀(1/2)⁴ = A₀/16
A_A/A_B = (1/256)/(1/16) = 16/256 = 1/16
Q8. ²³⁸U decays to ²⁰⁶Pb with T₁/₂ = 4.5 × 10⁹ years. A rock has U:Pb ratio of 1:1. Find age of rock.
Solution:
Initially: N₀ of U, 0 of Pb
After time t: N of U, (N₀-N) of Pb
Given: N/(N₀-N) = 1/1
∴ N = N₀ - N
∴ 2N = N₀
∴ N = N₀/2
One half-life has passed!
Age = T₁/₂ = 4.5 × 10⁹ years
Actually, should verify:
N/N₀ = e^(-λt) = 1/2
λt = ln2
t = (ln2)/λ = T₁/₂ ✓
Q9. A radioactive sample undergoes two types of decay with constants λ₁ and λ₂. Show that effective half-life is:
$$T_{eff} = \frac{\ln 2}{λ_1 + λ_2}$$Solution:
Effective decay constant: λ_eff = λ₁ + λ₂
(Each decay mode contributes independently)
Total decay rate:
dN/dt = -(λ₁ + λ₂)N = -λ_eff N
Solution: N = N₀ e^(-λ_eff t)
Half-life: N₀/2 = N₀ e^(-λ_eff T_eff)
1/2 = e^(-λ_eff T_eff)
ln(1/2) = -λ_eff T_eff
-ln2 = -λ_eff T_eff
T_eff = ln2/λ_eff = ln2/(λ₁ + λ₂) ✓
Physical meaning: Multiple decay paths
→ faster overall decay → shorter half-life
Q10. Show that mean life τ = 1/λ equals the average lifetime of radioactive nuclei.
Solution:
Mean life = ∫₀^∞ t × P(t) dt
where P(t) = probability of decay at time t
P(t)dt = (fraction surviving to t) × (prob of decay in dt)
= (N/N₀) × (λ dt)
= e^(-λt) × λ dt
τ = ∫₀^∞ t × λ e^(-λt) dt
Integration by parts:
Let u = t, dv = λe^(-λt)dt
du = dt, v = -e^(-λt)
τ = [-te^(-λt)]₀^∞ + ∫₀^∞ e^(-λt) dt
= 0 + [-e^(-λt)/λ]₀^∞
= 0 - (-1/λ)
= 1/λ ✓
Also: τ = T₁/₂/ln2 = T₁/₂/0.693 ≈ 1.44 T₁/₂
Applications of Radioactivity
1. Medical
- Diagnosis: PET scans (¹⁸F), X-rays
- Treatment: Cancer therapy (⁶⁰Co, ¹³¹I)
- Sterilization: γ-rays kill bacteria
2. Dating
- Carbon-14: Organic materials (50,000 y)
- Uranium-Lead: Rocks (millions of years)
- Potassium-Argon: Volcanic rocks
3. Industrial
- Thickness gauges: β-rays
- Smoke detectors: ²⁴¹Am (α-emitter)
- Tracers: Follow fluid flow
4. Research
- Rutherford scattering: α-particles
- Tracer studies: Biological pathways
- Neutron activation: Element analysis
Nuclear Stability
Stable vs Unstable
Stable nuclei:
- Balanced N/Z ratio
- “Valley of stability” on N-Z plot
- ~270 stable isotopes
Unstable nuclei (radioactive):
- Too many neutrons → β⁻ decay
- Too many protons → β⁺ decay or EC
- Too heavy → α decay
- Excited state → γ decay
N-Z Plot (Stability Curve)
N
│ Heavy
│ (α decay)
│ ╱
│ ╱ Neutron-rich
│ ╱ (β⁻ decay)
│ ╱
│╱______ Stable
│ ╱ (N≈Z for light,
│ ╱ N>Z for heavy)
│ ╱ Proton-rich
│ ╱ (β⁺ or EC)
│ ╱
└─────────────→ Z
Light
Cross-Links to Related Topics
- Nuclear Structure - Composition of nucleus
- Binding Energy - Why some nuclei unstable
- Nuclear Reactions - Induced transmutations
- Rutherford Model - α-scattering experiments
Quick Revision Checklist ✓
- α decay: A-4, Z-2, ⁴He emitted
- β⁻ decay: A same, Z+1, e⁻ emitted
- β⁺ decay: A same, Z-1, e⁺ emitted
- γ decay: A and Z same, photon emitted
- Decay law: N = N₀e^(-λt)
- Activity: A = λN
- Half-life: T₁/₂ = 0.693/λ
- After n half-lives: N = N₀(1/2)^n
- Mean life: τ = 1/λ = 1.44T₁/₂
- Penetration: α < β < γ
- Ionization: α > β > γ
- ¹⁴C dating: T₁/₂ = 5730 years
Final Tips for JEE
- Master decay equations: Know what changes in each type
- Exponential decay: Not linear! N = N₀e^(-λt)
- Half-life formula: T₁/₂ = 0.693/λ (not 1/λ!)
- Activity = rate: A = λN (dN/dt)
- Conservation laws: Check A and Z before/after
- Units: Bq (SI) vs Ci (old)
- Carbon dating: Know T₁/₂ = 5730 years
- Penetration order: α < β < γ (remember!)
- Multiple half-lives: Use (1/2)^n
- Physical meaning: Random, spontaneous, exponential
Last updated: May 8, 2025 Previous: Binding Energy Next: Nuclear Reactions