Radioactive Dating — Question 8

Question 8

Relative isotope uncertainty becomes an absolute age uncertainty

An ideal radioactive clock has known half-life $T=5730$ y and measured surviving fraction $f=.250\pm.010$, where the uncertainty is one standard deviation and is small enough for linear propagation.

1. Derive the age uncertainty from $t=-T\ln f/\ln2$.

2. Find age and uncertainty, then determine the precision in f required for an age uncertainty of 100 y at the same fraction.

Question 8 - Solution

Relative isotope uncertainty becomes an absolute age uncertainty

(a) Method and calculation

$$\frac{dt}{df}=-\frac{T}{f\ln2}\Rightarrow\boxed{\sigma_t=\frac{T}{\ln2}\frac{\sigma_f}{f}}.$$$$\boxed{t=11460\,\mathrm y,\quad\sigma_t=330.7\,\mathrm y}.$$

(b) Interpretation and further calculation

$$\sigma_f=f(\ln2)\sigma_t/T.$$$$\boxed{\sigma_f=.250(\ln2)(100)/5730=.003024}.$$

Physical check

This uncertainty includes only the measured fraction. Half-life, calibration and contamination uncertainties would require additional terms or a fuller model.