How old is a specimen with 12.5% Carbon-14 (¹⁴C) remaining?
⚡ Instant Calculated Answer
A specimen containing 12.5% remaining Carbon-14 (¹⁴C) has an age of exactly 17,190 years (representing 3.0 half-lives elapsed since closure).
17,190 years
Primary Metric
3.0 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 = 12.5% = 0.125.
- Take natural logarithm of fraction: ln(0.125) = -2.07944.
- Apply decay constant lambda = ln(2) / T_1/2 = 1.210e-04 yr⁻¹.
- Compute age: t = --2.07944 / (1.210e-04) = 17,190 years.
📊 Nearby Parameter Reference Table
| Isotope Remaining | Half-Lives Elapsed | Calculated Age |
|---|---|---|
| 15.0% | 2.74 | 15,682 yrs |
| 13.8% | 2.86 | 16,402 yrs |
| 12.5% | 3.0 | 17,190 yrs |
| 11.2% | 3.15 | 18,060 yrs |
| 10.0% | 3.32 | 19,034 yrs |
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