Radioactive Dating — Question 3

Question 3

A stable reference isotope corrects for inherited daughter

A mineral contains parent P, daughter D and nonradiogenic stable reference S. At closure $D_0/S=.200$. Today $D/S=1.20$ and $P/S=.500$. S is conserved and all parent decays produce D.

1. Find the radiogenic daughter-to-parent ratio.

2. Find the age in units of parent half-life and compare it with the result obtained by mistakenly treating all D as radiogenic.

Question 3 - Solution

A stable reference isotope corrects for inherited daughter

(a) Method and calculation

$$D_*/S=(D-D_0)/S=1.20-.200=1.00.$$$$\boxed{D_*/P=(D_*/S)/(P/S)=1.00/.500=2.00}.$$

(b) Interpretation and further calculation

$$\boxed{t/T_{1/2}=\ln(1+2)/\ln2=1.585}.$$$$\boxed{t_{\mathrm{naive}}/T_{1/2}=\ln(1+1.20/.500)/\ln2=1.766}\quad\text{too old}.$$

Physical check

The stable isotope tracks inherited daughter without itself decaying. Its conservation and known initial daughter ratio are essential to the correction.