How old is a specimen with 10% Carbon-14 (¹⁴C) remaining?
⚡ Instant Calculated Answer
A specimen containing 10% remaining Carbon-14 (¹⁴C) has an age of exactly 19,034 years (representing 3.32 half-lives elapsed since closure).
19,034 years
Primary Metric
3.32 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 = 10% = 0.1.
- Take natural logarithm of fraction: ln(0.1) = -2.30259.
- Apply decay constant lambda = ln(2) / T_1/2 = 1.210e-04 yr⁻¹.
- Compute age: t = --2.30259 / (1.210e-04) = 19,034 years.
📊 Nearby Parameter Reference Table
| Isotope Remaining | Half-Lives Elapsed | Calculated Age |
|---|---|---|
| 12.0% | 3.06 | 17,527 yrs |
| 11.0% | 3.18 | 18,246 yrs |
| 10.0% | 3.32 | 19,034 yrs |
| 9.0% | 3.47 | 19,905 yrs |
| 8.0% | 3.64 | 20,879 yrs |
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