Radioactive Dating — Question 10

Question 10

A late loss of parent can make a daughter clock appear too old

A mineral initially has no daughter, and each parent decay makes one stable daughter. After exactly one parent half-life, an instantaneous event removes 20 percent of the remaining parent but no daughter. Measurement follows immediately.

1. Find measured parent and daughter numbers in units of the original $P_0$.

2. Calculate the apparent closed-system age from D/P and its error in half-life units.

Question 10 - Solution

A late loss of parent can make a daughter clock appear too old

(a) Method and calculation

$$\text{Before loss: }P=P_0/2,\quad D=P_0/2.$$$$\boxed{P_{\mathrm{meas}}=.8(P_0/2)=.4P_0,\quad D_{\mathrm{meas}}=.5P_0}.$$

(b) Interpretation and further calculation

$$\boxed{D/P=1.25,\quad t_{\mathrm{app}}/T=\log_2(1+1.25)=1.16993}.$$$$\boxed{(t_{\mathrm{app}}-t_{\mathrm{true}})/T=.16993}\quad\text{an overestimate}.$$

Physical check

The daughter-to-parent ratio rose without additional radioactive decay. This is why a measured ratio must be interpreted with a geological closure model.