Paleontology & Geochronology

How old is a specimen with 10% Potassium-40 (⁴⁰K) remaining?

⚡ Instant Calculated Answer

A specimen containing 10% remaining Potassium-40 (⁴⁰K) has an age of exactly 4.146 Ga (representing 3.32 half-lives elapsed since closure).

4.146 Ga Primary Metric
3.32 Half-Lives Secondary Metric

📐 Mathematical Formula & Step-by-Step Substitution

\[ t = -\frac{T_{1/2}}{\ln(2)} \cdot \ln\left(\frac{N(t)}{N_0}\right) = -\frac{1248000000}{\ln(2)} \cdot \ln(0.1) = 4.146 Ga \]
t = -(1248000000 / ln(2)) * ln(0.1) = 4.146 Ga
  1. Identify the isotope half-life: T_1/2 = 1,248,000,000 years (1.25 Ga).
  2. Calculate remaining fraction N(t)/N_0 = 10% = 0.1.
  3. Take natural logarithm of fraction: ln(0.1) = -2.30259.
  4. Apply decay constant lambda = ln(2) / T_1/2 = 5.554e-10 yr⁻¹.
  5. Compute age: t = --2.30259 / (5.554e-10) = 4.146 Ga.

📊 Nearby Parameter Reference Table

Isotope Remaining Half-Lives Elapsed Calculated Age
12.0% 3.06 3.817 Ga
11.0% 3.18 3.974 Ga
10.0% 3.32 4.146 Ga
9.0% 3.47 4.335 Ga
8.0% 3.64 4.548 Ga
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