How old is a specimen with 30% Carbon-14 (¹⁴C) remaining?
⚡ Instant Calculated Answer
A specimen containing 30% remaining Carbon-14 (¹⁴C) has an age of exactly 9,952 years (representing 1.74 half-lives elapsed since closure).
9,952 years
Primary Metric
1.74 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 = 30% = 0.3.
- Take natural logarithm of fraction: ln(0.3) = -1.20397.
- Apply decay constant lambda = ln(2) / T_1/2 = 1.210e-04 yr⁻¹.
- Compute age: t = --1.20397 / (1.210e-04) = 9,952 years.
📊 Nearby Parameter Reference Table
| Isotope Remaining | Half-Lives Elapsed | Calculated Age |
|---|---|---|
| 36.0% | 1.47 | 8,445 yrs |
| 33.0% | 1.6 | 9,164 yrs |
| 30.0% | 1.74 | 9,952 yrs |
| 27.0% | 1.89 | 10,823 yrs |
| 24.0% | 2.06 | 11,797 yrs |
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