Paleontology & Geochronology

How old is a specimen with 30% Rubidium-87 (⁸⁷Rb) remaining?

⚡ Instant Calculated Answer

A specimen containing 30% remaining Rubidium-87 (⁸⁷Rb) has an age of exactly 84.764 Ga (representing 1.74 half-lives elapsed since closure).

84.764 Ga Primary Metric
1.74 Half-Lives Secondary Metric

📐 Mathematical Formula & Step-by-Step Substitution

\[ t = -\frac{T_{1/2}}{\ln(2)} \cdot \ln\left(\frac{N(t)}{N_0}\right) = -\frac{48800000000}{\ln(2)} \cdot \ln(0.3) = 84.764 Ga \]
t = -(48800000000 / ln(2)) * ln(0.3) = 84.764 Ga
  1. Identify the isotope half-life: T_1/2 = 48,800,000,000 years (48.8 Ga).
  2. Calculate remaining fraction N(t)/N_0 = 30% = 0.3.
  3. Take natural logarithm of fraction: ln(0.3) = -1.20397.
  4. Apply decay constant lambda = ln(2) / T_1/2 = 1.420e-11 yr⁻¹.
  5. Compute age: t = --1.20397 / (1.420e-11) = 84.764 Ga.

📊 Nearby Parameter Reference Table

Isotope Remaining Half-Lives Elapsed Calculated Age
36.0% 1.47 71.928 Ga
33.0% 1.6 78.054 Ga
30.0% 1.74 84.764 Ga
27.0% 1.89 92.182 Ga
24.0% 2.06 100.474 Ga
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