Mastering Radioactive Decay: From Basics to Parallel & Successive Decay
Your complete guide to decay laws, half-life vs mean life, and solving complex JEE problems with radioactive chains.
Why Radioactive Decay is Crucial for JEE
Radioactive decay appears in every JEE paper with 3-4 questions carrying significant marks. Mastering this topic gives you an edge in Modern Physics section.
🎯 JEE Focus Areas
- Basic decay laws and formulas (2-3 marks)
- Half-life vs mean life calculations (3-4 marks)
- Successive decay problems (4-5 marks)
- Parallel decay and equilibrium (3-4 marks)
🚀 Quick Navigation
1. Fundamental Laws of Radioactive Decay
The Decay Law
Radioactive decay follows first-order kinetics, meaning the decay rate is proportional to the number of undecayed nuclei.
Mathematical Formulation
Differential Form:
Integrated Form:
Activity:
Where:
- $N$ = Number of undecayed nuclei at time $t$
- $N_0$ = Initial number of nuclei
- $\lambda$ = Decay constant
- $A$ = Activity (decays per second)
- $A_0$ = Initial activity
Example: Basic Decay Calculation
Problem: A radioactive sample has an initial activity of 8000 Bq. After 2 hours, its activity reduces to 1000 Bq. Find the decay constant.
Solution:
Step 1: Use the activity formula: $A = A_0 e^{-\lambda t}$
Step 2: Substitute values: $1000 = 8000 e^{-\lambda \times 7200}$ (time in seconds)
Step 3: Simplify: $\frac{1}{8} = e^{-7200\lambda}$
Step 4: Take natural log: $\ln(\frac{1}{8}) = -7200\lambda$
Step 5: Solve: $\lambda = \frac{\ln(8)}{7200} = \frac{2.079}{7200} = 2.89 \times 10^{-4} \text{ s}^{-1}$
2. Half-Life vs Mean Life: Key Differences
| Parameter | Half-Life ($T_{1/2}$) | Mean Life ($\tau$) |
|---|---|---|
| Definition | Time for half the nuclei to decay | Average lifetime of a nucleus |
| Formula | $T_{1/2} = \frac{\ln 2}{\lambda}$ | $\tau = \frac{1}{\lambda}$ |
| Relationship | $T_{1/2} = \tau \ln 2 \approx 0.693\tau$ | |
| After 1 mean life | $N = \frac{N_0}{e} \approx 0.368N_0$ (36.8% remain) | |
Important Relationships
$\lambda = \frac{\ln 2}{T_{1/2}}$
Decay constant from half-life
$\tau = \frac{T_{1/2}}{\ln 2} \approx 1.443 T_{1/2}$
Mean life from half-life
Example: Half-Life Calculation
Problem: A radioactive element has a half-life of 10 days. What fraction of the original sample remains after 30 days?
Solution:
Method 1: Using half-life concept
Number of half-lives = $\frac{30}{10} = 3$
Fraction remaining = $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$
Method 2: Using decay formula
$\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{10} = 0.0693 \text{ day}^{-1}$
$\frac{N}{N_0} = e^{-\lambda t} = e^{-0.0693 \times 30} = e^{-2.079} = \frac{1}{8}$
3. Successive Radioactive Decay
Radioactive Series Decay
When a parent nuclide decays into a daughter nuclide that is also radioactive, we have successive decay.
Typical Successive Decay Chain
Bateman Equations
Daughter Nucleus Amount:
Special Case - Transient Equilibrium ($\lambda_1 < \lambda_2$):
Special Case - Secular Equilibrium ($\lambda_1 << \lambda_2$):
Example: Successive Decay Problem
Problem: A → B → C (stable). If $T_{1/2}(A) = 10$ min and $T_{1/2}(B) = 2$ min, find the time when daughter B reaches maximum activity.
Solution:
Step 1: Calculate decay constants
$\lambda_A = \frac{\ln 2}{10} = 0.0693 \text{ min}^{-1}$
$\lambda_B = \frac{\ln 2}{2} = 0.3466 \text{ min}^{-1}$
Step 2: Time for maximum daughter activity
$t_{max} = \frac{\ln(\lambda_B/\lambda_A)}{\lambda_B - \lambda_A}$
$t_{max} = \frac{\ln(0.3466/0.0693)}{0.3466 - 0.0693} = \frac{\ln(5)}{0.2773} = \frac{1.609}{0.2773} = 5.8$ min
Step 3: Verify this is reasonable
Since B decays faster than A, maximum occurs relatively quickly.
4. Parallel Radioactive Decay
Multiple Decay Modes
Some nuclides can decay through multiple independent channels simultaneously.
Parallel Decay Scheme
Key Formulas for Parallel Decay
Total Decay Constant:
Branching Ratio:
Partial Half-Life:
Where $f_i$ is the branching fraction for mode i.
Example: Parallel Decay Calculation
Problem: A nuclide decays 60% by α-emission and 40% by β-emission. The half-life for α-decay alone would be 100 years. Find the actual half-life.
Solution:
Step 1: Find decay constant for α-decay
$\lambda_α = \frac{\ln 2}{100} = 0.00693 \text{ year}^{-1}$
Step 2: Relate branching ratio to decay constants
Branching ratio $= \frac{\lambda_α}{\lambda_{total}} = 0.6$
Therefore: $\lambda_{total} = \frac{\lambda_α}{0.6} = \frac{0.00693}{0.6} = 0.01155 \text{ year}^{-1}$
Step 3: Calculate actual half-life
$T_{1/2}^{total} = \frac{\ln 2}{\lambda_{total}} = \frac{0.693}{0.01155} = 60$ years
The actual half-life is 60 years due to multiple decay channels.
5. JEE Practice Problems
Test Your Understanding
Problem 1: The half-life of a radioactive substance is 20 days. What percentage decays in 10 days?
Problem 2: A radioactive sample has mean life of 100 seconds. Calculate the time taken for 75% decay.
Problem 3: In the decay A → B → C (stable), if λ_A = 0.1 min⁻¹ and λ_B = 0.4 min⁻¹, find when B's activity is maximum.
Problem 4: A nuclide can decay by both α and β emission with partial half-lives of 50 years and 25 years respectively. Find total half-life.
📋 Formula Quick Sheet
Basic Decay
- $N = N_0 e^{-\lambda t}$
- $A = \lambda N = A_0 e^{-\lambda t}$
- $\lambda T_{1/2} = \ln 2$
- $\tau = 1/\lambda$
Advanced Concepts
- Successive decay: Use Bateman equations
- Parallel decay: $\lambda_{total} = \sum \lambda_i$
- Branching ratio: $f_i = \lambda_i/\lambda_{total}$
- Equilibrium: $\lambda_1 N_1 = \lambda_2 N_2$
🎯 JEE Exam Strategy
Spend 2-3 minutes on basic decay problems, 4-5 minutes on successive decay.
Quickly identify if it's basic decay, successive, parallel, or equilibrium problem.
Ensure time units match between half-life and the given time.
Use $\ln 2 \approx 0.693$, $1/e \approx 0.368$ for quick calculations.
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