How old is a specimen with 20% Carbon-14 (¹⁴C) remaining?
⚡ Instant Calculated Answer
A specimen containing 20% remaining Carbon-14 (¹⁴C) has an age of exactly 13,304 years (representing 2.32 half-lives elapsed since closure).
13,304 years
Primary Metric
2.32 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 = 20% = 0.2.
- Take natural logarithm of fraction: ln(0.2) = -1.60944.
- Apply decay constant lambda = ln(2) / T_1/2 = 1.210e-04 yr⁻¹.
- Compute age: t = --1.60944 / (1.210e-04) = 13,304 years.
📊 Nearby Parameter Reference Table
| Isotope Remaining | Half-Lives Elapsed | Calculated Age |
|---|---|---|
| 24.0% | 2.06 | 11,797 yrs |
| 22.0% | 2.18 | 12,516 yrs |
| 20.0% | 2.32 | 13,304 yrs |
| 18.0% | 2.47 | 14,175 yrs |
| 16.0% | 2.64 | 15,149 yrs |
Want to test custom numbers?
Launch the full interactive simulator pre-loaded with these exact calculation parameters.