How old is a specimen with 60% Uranium-235 (²³⁵U) remaining?
⚡ Instant Calculated Answer
A specimen containing 60% remaining Uranium-235 (²³⁵U) has an age of exactly 518.68 Ma (representing 0.74 half-lives elapsed since closure).
518.68 Ma
Primary Metric
0.74 Half-Lives
Secondary Metric
📐 Mathematical Formula & Step-by-Step Substitution
- Identify the isotope half-life: T_1/2 = 703,800,000 years (703.8 Ma).
- 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 = 9.849e-10 yr⁻¹.
- Compute age: t = --0.51083 / (9.849e-10) = 518.68 Ma.
📊 Nearby Parameter Reference Table
| Isotope Remaining | Half-Lives Elapsed | Calculated Age |
|---|---|---|
| 72.0% | 0.47 | 333.55 Ma |
| 66.0% | 0.6 | 421.9 Ma |
| 60.0% | 0.74 | 518.68 Ma |
| 54.0% | 0.89 | 625.66 Ma |
| 48.0% | 1.06 | 745.25 Ma |
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