How old is a specimen with 50% Uranium-235 (²³⁵U) remaining?
⚡ Instant Calculated Answer
A specimen containing 50% remaining Uranium-235 (²³⁵U) has an age of exactly 703.8 Ma (representing 1.0 half-lives elapsed since closure).
703.8 Ma
Primary Metric
1.0 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 = 50% = 0.5.
- Take natural logarithm of fraction: ln(0.5) = -0.69315.
- Apply decay constant lambda = ln(2) / T_1/2 = 9.849e-10 yr⁻¹.
- Compute age: t = --0.69315 / (9.849e-10) = 703.8 Ma.
📊 Nearby Parameter Reference Table
| Isotope Remaining | Half-Lives Elapsed | Calculated Age |
|---|---|---|
| 60.0% | 0.74 | 518.68 Ma |
| 55.0% | 0.86 | 607.03 Ma |
| 50.0% | 1.0 | 703.8 Ma |
| 45.0% | 1.15 | 810.78 Ma |
| 40.0% | 1.32 | 930.37 Ma |
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