How old is a specimen with 75% Potassium-40 (⁴⁰K) remaining?
⚡ Instant Calculated Answer
A specimen containing 75% remaining Potassium-40 (⁴⁰K) has an age of exactly 517.97 Ma (representing 0.42 half-lives elapsed since closure).
517.97 Ma
Primary Metric
0.42 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 = 75% = 0.75.
- Take natural logarithm of fraction: ln(0.75) = -0.28768.
- Apply decay constant lambda = ln(2) / T_1/2 = 5.554e-10 yr⁻¹.
- Compute age: t = --0.28768 / (5.554e-10) = 517.97 Ma.
📊 Nearby Parameter Reference Table
| Isotope Remaining | Half-Lives Elapsed | Calculated Age |
|---|---|---|
| 90.0% | 0.15 | 189.7 Ma |
| 82.5% | 0.28 | 346.36 Ma |
| 75.0% | 0.42 | 517.97 Ma |
| 67.5% | 0.57 | 707.67 Ma |
| 60.0% | 0.74 | 919.73 Ma |
Want to test custom numbers?
Launch the full interactive simulator pre-loaded with these exact calculation parameters.