How old is a specimen with 60% Potassium-40 (⁴⁰K) remaining?
⚡ Instant Calculated Answer
A specimen containing 60% remaining Potassium-40 (⁴⁰K) has an age of exactly 919.73 Ma (representing 0.74 half-lives elapsed since closure).
919.73 Ma
Primary Metric
0.74 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 = 60% = 0.6.
- Take natural logarithm of fraction: ln(0.6) = -0.51083.
- Apply decay constant lambda = ln(2) / T_1/2 = 5.554e-10 yr⁻¹.
- Compute age: t = --0.51083 / (5.554e-10) = 919.73 Ma.
📊 Nearby Parameter Reference Table
| Isotope Remaining | Half-Lives Elapsed | Calculated Age |
|---|---|---|
| 72.0% | 0.47 | 591.47 Ma |
| 66.0% | 0.6 | 748.13 Ma |
| 60.0% | 0.74 | 919.73 Ma |
| 54.0% | 0.89 | 1.109 Ga |
| 48.0% | 1.06 | 1.321 Ga |
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