How old is a specimen with 90% Potassium-40 (⁴⁰K) remaining?
⚡ Instant Calculated Answer
A specimen containing 90% remaining Potassium-40 (⁴⁰K) has an age of exactly 189.7 Ma (representing 0.15 half-lives elapsed since closure).
189.7 Ma
Primary Metric
0.15 Half-Lives
Secondary Metric
📐 Mathematical Formula & Step-by-Step Substitution
- Identify the isotope half-life: T_1/2 = 1,248,000,000 years (1.25 Ga).
- Calculate remaining fraction N(t)/N_0 = 90% = 0.9.
- Take natural logarithm of fraction: ln(0.9) = -0.10536.
- Apply decay constant lambda = ln(2) / T_1/2 = 5.554e-10 yr⁻¹.
- Compute age: t = --0.10536 / (5.554e-10) = 189.7 Ma.
📊 Nearby Parameter Reference Table
| Isotope Remaining | Half-Lives Elapsed | Calculated Age |
|---|---|---|
| 99.0% | 0.01 | 18.1 Ma |
| 90.0% | 0.15 | 189.7 Ma |
| 81.0% | 0.3 | 379.4 Ma |
| 72.0% | 0.47 | 591.47 Ma |
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