Paleontology & Geochronology

How old is a specimen with 90% Uranium-235 (²³⁵U) remaining?

⚡ Instant Calculated Answer

A specimen containing 90% remaining Uranium-235 (²³⁵U) has an age of exactly 106.98 Ma (representing 0.15 half-lives elapsed since closure).

106.98 Ma Primary Metric
0.15 Half-Lives Secondary Metric

📐 Mathematical Formula & Step-by-Step Substitution

\[ t = -\frac{T_{1/2}}{\ln(2)} \cdot \ln\left(\frac{N(t)}{N_0}\right) = -\frac{703800000}{\ln(2)} \cdot \ln(0.9) = 106.98 Ma \]
t = -(703800000 / ln(2)) * ln(0.9) = 106.98 Ma
  1. Identify the isotope half-life: T_1/2 = 703,800,000 years (703.8 Ma).
  2. Calculate remaining fraction N(t)/N_0 = 90% = 0.9.
  3. Take natural logarithm of fraction: ln(0.9) = -0.10536.
  4. Apply decay constant lambda = ln(2) / T_1/2 = 9.849e-10 yr⁻¹.
  5. Compute age: t = --0.10536 / (9.849e-10) = 106.98 Ma.

📊 Nearby Parameter Reference Table

Isotope Remaining Half-Lives Elapsed Calculated Age
99.0% 0.01 10.2 Ma
90.0% 0.15 106.98 Ma
81.0% 0.3 213.96 Ma
72.0% 0.47 333.55 Ma
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