Question 25.8: Kennewick Man In 1996, ancient human bones were discovered n...
Kennewick Man
In 1996, ancient human bones were discovered near Kennewick, Washington, and their owner was dubbed Kennewick man. Analysis of the bones indicated a { }^{14} C /{ }^{12} C \text { ratio of } 4.34 \times 10^{-13}. What’s Kennewick man’s age?
ORGANIZE AND PLAN The current \text { it }{ }^{14} C /{ }^{12} C \text { ratio is } 1.20 \times 10^{-12} \text {. Since } { }^{12} C \text { is stable while }{ }^{14} C \text { decays, the }{ }^{14} C /{ }^{12} C ratio falls at the same rate as the ratio N / N_{0} \text { for }{ }^{14} C \text {. Thus, }
\frac{{ }^{14} C /{ }^{12} C (\text { old })}{{ }^{14} C /{ }^{12} C (\text { current })}=\frac{N}{N_{0}}=e^{-\lambda t}.
Knowing this ratio, we can find the time t.
\text { Known: } t_{1 / 2}=5730 \text { years for }{ }^{14} C ;{ }^{14} C /{ }^{12} C \text { ratio }(\text { old })= 4.34 \times 10^{-13} ;{ }^{14} C /{ }^{12} C \text { ratio }(\text { current })=1.20 \times 10^{-12} .
\frac{{ }^{14} C /{ }^{12} C \text { (old) }}{{ }^{14} C /{ }^{12} C (\text { current })}=\frac{4.34 \times 10^{-13}}{1.20 \times 10^{-12}}=0.362.
\text { Thus } 0.362=e^{-\lambda t} \text {. Taking the natural logarithm of both sides: } \ln \left(e^{-\lambda t}\right)=-\lambda t=\ln (0.362)or
t=-\frac{\ln (0.362)}{\lambda}=-\frac{\ln (0.362)}{\ln (2)} t_{1 / 2}=-\frac{\ln (0.362)}{\ln (2)}(5730 y ).
= 8400 y.
REFLECT Corrections for varying atmospheric { }^{14} C concentration revised this age to 9200 years.