How old is a specimen with 90% Carbon-14 (¹⁴C) remaining?
⚡ Instant Calculated Answer
A specimen containing 90% remaining Carbon-14 (¹⁴C) has an age of exactly 871 years (representing 0.15 half-lives elapsed since closure).
871 years
Primary Metric
0.15 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 = 90% = 0.9.
- Take natural logarithm of fraction: ln(0.9) = -0.10536.
- Apply decay constant lambda = ln(2) / T_1/2 = 1.210e-04 yr⁻¹.
- Compute age: t = --0.10536 / (1.210e-04) = 871 years.
📊 Nearby Parameter Reference Table
| Isotope Remaining | Half-Lives Elapsed | Calculated Age |
|---|---|---|
| 99.0% | 0.01 | 83 yrs |
| 90.0% | 0.15 | 870 yrs |
| 81.0% | 0.3 | 1,741 yrs |
| 72.0% | 0.47 | 2,715 yrs |
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