Numerical Problems — Chapter 6: Nuclear Physics
Key Formula:
After n half-lives:
N = N₀ × (1/2)ⁿ = N₀ / 2ⁿ
Where:
- N₀ = initial amount
- N = amount remaining
- n = number of half-lives = total time / t₁/₂
Problem 1. The half-life of a radioactive substance is 5 years. If we start with 100 g, how much will remain after 20 years?
- Half-life, t₁/₂ = 5 years
- Initial amount, N₀ = 100 g
- Total time = 20 years
Number of half-lives, n = 20/5 = 4
Amount remaining, N = N₀ × (1/2)ⁿ = 100 × (1/2)⁴ = 100 × 1/16 = 6.25 g
Verification:
- After 5 years: 100 → 50 g
- After 10 years: 50 → 25 g
- After 15 years: 25 → 12.5 g
- After 20 years: 12.5 → 6.25 g ✓
Problem 2. The half-life of iodine-131 is 8 days. If a sample initially contains 160 g of I-131, how much will remain after 32 days?
- t₁/₂ = 8 days
- N₀ = 160 g
- Total time = 32 days
n = 32/8 = 4 half-lives
N = 160 × (1/2)⁴ = 160 × 1/16 = 10 g
Problem 3. A radioactive substance reduces from 200 g to 25 g in 60 years. Find its half-life.
- N₀ = 200 g; N = 25 g
- Total time = 60 years; t₁/₂ = ?
N = N₀ / 2ⁿ
25 = 200 / 2ⁿ
2ⁿ = 200/25 = 8 = 2³
Therefore n = 3 half-lives
t₁/₂ = Total time / n = 60/3 = 20 years
Problem 4. The half-life of Carbon-14 is 5730 years. A wooden artifact is found to contain only 25% of the original C-14. How old is the artifact?
- t₁/₂ = 5730 years
- N = 25% of N₀ = N₀/4
- Age = ?
N = N₀ / 2ⁿ
N₀/4 = N₀ / 2ⁿ
2ⁿ = 4 = 2²
n = 2 half-lives
Age = n × t₁/₂ = 2 × 5730 = 11,460 years
Problem 5. A radioactive element has a half-life of 10 minutes. After how many minutes will only 6.25% of the original sample remain?
- t₁/₂ = 10 minutes
- N = 6.25% of N₀ = N₀/16
- Time = ?
N = N₀ / 2ⁿ
N₀/16 = N₀ / 2ⁿ → 2ⁿ = 16 = 2⁴ → n = 4
Time = n × t₁/₂ = 4 × 10 = 40 minutes
Half-Life Decay Table
| Number of Half-lives | Fraction Remaining | % Remaining |
|---|---|---|
| 0 | 1 | 100% |
| 1 | 1/2 | 50% |
| 2 | 1/4 | 25% |
| 3 | 1/8 | 12.5% |
| 4 | 1/16 | 6.25% |
| 5 | 1/32 | 3.125% |
| 10 | 1/1024 | ~0.1% |
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