How old is a specimen with 75% Carbon-14 (¹⁴C) remaining?
⚡ Instant Calculated Answer
A specimen containing 75% remaining Carbon-14 (¹⁴C) has an age of exactly 2,378 years (representing 0.42 half-lives elapsed since closure).
2,378 years
Primary Metric
0.42 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 = 75% = 0.75.
- Take natural logarithm of fraction: ln(0.75) = -0.28768.
- Apply decay constant lambda = ln(2) / T_1/2 = 1.210e-04 yr⁻¹.
- Compute age: t = --0.28768 / (1.210e-04) = 2,378 years.
📊 Nearby Parameter Reference Table
| Isotope Remaining | Half-Lives Elapsed | Calculated Age |
|---|---|---|
| 90.0% | 0.15 | 870 yrs |
| 82.5% | 0.28 | 1,590 yrs |
| 75.0% | 0.42 | 2,378 yrs |
| 67.5% | 0.57 | 3,249 yrs |
| 60.0% | 0.74 | 4,222 yrs |
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