How old is a specimen with 40% Carbon-14 (¹⁴C) remaining?
⚡ Instant Calculated Answer
A specimen containing 40% remaining Carbon-14 (¹⁴C) has an age of exactly 7,574 years (representing 1.32 half-lives elapsed since closure).
7,574 years
Primary Metric
1.32 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 = 40% = 0.4.
- Take natural logarithm of fraction: ln(0.4) = -0.91629.
- Apply decay constant lambda = ln(2) / T_1/2 = 1.210e-04 yr⁻¹.
- Compute age: t = --0.91629 / (1.210e-04) = 7,574 years.
📊 Nearby Parameter Reference Table
| Isotope Remaining | Half-Lives Elapsed | Calculated Age |
|---|---|---|
| 48.0% | 1.06 | 6,067 yrs |
| 44.0% | 1.18 | 6,786 yrs |
| 40.0% | 1.32 | 7,574 yrs |
| 36.0% | 1.47 | 8,445 yrs |
| 32.0% | 1.64 | 9,419 yrs |
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