How old is a specimen with 40% Uranium-235 (²³⁵U) remaining?
⚡ Instant Calculated Answer
A specimen containing 40% remaining Uranium-235 (²³⁵U) has an age of exactly 930.37 Ma (representing 1.32 half-lives elapsed since closure).
930.37 Ma
Primary Metric
1.32 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 = 40% = 0.4.
- Take natural logarithm of fraction: ln(0.4) = -0.91629.
- Apply decay constant lambda = ln(2) / T_1/2 = 9.849e-10 yr⁻¹.
- Compute age: t = --0.91629 / (9.849e-10) = 930.37 Ma.
📊 Nearby Parameter Reference Table
| Isotope Remaining | Half-Lives Elapsed | Calculated Age |
|---|---|---|
| 48.0% | 1.06 | 745.25 Ma |
| 44.0% | 1.18 | 833.6 Ma |
| 40.0% | 1.32 | 930.37 Ma |
| 36.0% | 1.47 | 1.037 Ga |
| 32.0% | 1.64 | 1.157 Ga |
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