How old is a specimen with 50% Carbon-14 (¹⁴C) remaining?
⚡ Instant Calculated Answer
A specimen containing 50% remaining Carbon-14 (¹⁴C) has an age of exactly 5,730 years (representing 1.0 half-lives elapsed since closure).
5,730 years
Primary Metric
1.0 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 = 50% = 0.5.
- Take natural logarithm of fraction: ln(0.5) = -0.69315.
- Apply decay constant lambda = ln(2) / T_1/2 = 1.210e-04 yr⁻¹.
- Compute age: t = --0.69315 / (1.210e-04) = 5,730 years.
📊 Nearby Parameter Reference Table
| Isotope Remaining | Half-Lives Elapsed | Calculated Age |
|---|---|---|
| 60.0% | 0.74 | 4,222 yrs |
| 55.0% | 0.86 | 4,942 yrs |
| 50.0% | 1.0 | 5,730 yrs |
| 45.0% | 1.15 | 6,600 yrs |
| 40.0% | 1.32 | 7,574 yrs |
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