Radioactive Dating — Question 5

Question 5

Modern contamination biases a radiocarbon result toward younger ages

A measured sample has carbon-14 fraction $f_m=.300$ relative to modern carbon. A known fraction c=.100 of its present carbon is modern contamination with fraction 1. The rest is original sample carbon with unknown fraction $f_s$. Use half-life 5730 y.

1. Solve the mixing equation for the original sample fraction.

2. Find the corrected age and the age that would be reported if contamination were ignored.

Question 5 - Solution

Modern contamination biases a radiocarbon result toward younger ages

(a) Method and calculation

$$f_m=(1-c)f_s+c\Rightarrow\boxed{f_s=(f_m-c)/(1-c)}.$$$$\boxed{f_s=(.300-.100)/.900=2/9}.$$

(b) Interpretation and further calculation

$$\boxed{t_s=5730\frac{\ln(9/2)}{\ln2}=12433\,\mathrm y}.$$$$\boxed{t_m=5730\frac{\ln(1/.300)}{\ln2}=9953\,\mathrm y}\quad\text{younger than }t_s.$$

Physical check

Fractions mix linearly by carbon amount; ages do not. The stated contamination fraction refers to the carbon currently measured.