How old is a specimen with 60% Carbon-14 (¹⁴C) remaining?
⚡ Instant Calculated Answer
A specimen containing 60% remaining Carbon-14 (¹⁴C) has an age of exactly 4,222 years (representing 0.74 half-lives elapsed since closure).
4,222 years
Primary Metric
0.74 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 = 60% = 0.6.
- Take natural logarithm of fraction: ln(0.6) = -0.51083.
- Apply decay constant lambda = ln(2) / T_1/2 = 1.210e-04 yr⁻¹.
- Compute age: t = --0.51083 / (1.210e-04) = 4,222 years.
📊 Nearby Parameter Reference Table
| Isotope Remaining | Half-Lives Elapsed | Calculated Age |
|---|---|---|
| 72.0% | 0.47 | 2,715 yrs |
| 66.0% | 0.6 | 3,434 yrs |
| 60.0% | 0.74 | 4,222 yrs |
| 54.0% | 0.89 | 5,093 yrs |
| 48.0% | 1.06 | 6,067 yrs |
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