How old is a specimen with 25% Carbon-14 (¹⁴C) remaining?
⚡ Instant Calculated Answer
A specimen containing 25% remaining Carbon-14 (¹⁴C) has an age of exactly 11,460 years (representing 2.0 half-lives elapsed since closure).
11,460 years
Primary Metric
2.0 Half-Lives
Secondary Metric
📐 Mathematical Formula & Step-by-Step Substitution
- Identify the isotope half-life: T_1/2 = 5,730 years.
- Calculate remaining fraction N(t)/N_0 = 25% = 0.25.
- Take natural logarithm of fraction: ln(0.25) = -1.38629.
- Apply decay constant lambda = ln(2) / T_1/2 = 1.210e-04 yr⁻¹.
- Compute age: t = --1.38629 / (1.210e-04) = 11,460 years.
📊 Nearby Parameter Reference Table
| Isotope Remaining | Half-Lives Elapsed | Calculated Age |
|---|---|---|
| 30.0% | 1.74 | 9,952 yrs |
| 27.5% | 1.86 | 10,672 yrs |
| 25.0% | 2.0 | 11,460 yrs |
| 22.5% | 2.15 | 12,330 yrs |
| 20.0% | 2.32 | 13,304 yrs |
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