1 Radioactive Decay

Background

In September 1991 the famous Iceman (Ötzi), a mummy from the Neolithic period of the Stone Age found in the ice of the Ötztal Alps (hence the name Ötzi) in Southern Tyrolia near the Austrian-Italian border, caused a scientific sensation.

Problem

When did Ötzi approximately live and die if the ratio of carbon ${}_{6}C^{14}$ to carbon ${}_{6}C^{12}$ in this mummy is 52.5%?

Physical Information

In the atmosphere and in living organisms, the ratio of radioactive ${}_{6}C^{14}$ (made radioactive by cosmic rays) to ordinary ${}_{6}C^{12}$ is constant. When an organism dies, its absorption of ${}_{6}C^{14}$ by breathing and eating terminates. Hence one can estimate the age of a fossil by comparing the radioactive carbon ration in the fossil with that of the atmosphere. To do this one needs to know the half-life of ${}_{6}C^{14}$ , which is 5715 years.

Solution

Radioactive decay is governed by the ODE $y^{\prime}=ky$ . By separation and integration (where $t$ is time and $y_{0}$ is the initial ratio of ${}_{6}C^{14}$ to ${}_{6}C^{12}$ )

\frac{dy}{y}=k\,dt,\qquad\ln|y|=kt+c,\qquad y=y_{0}e^{kt}

Next we use the half-life $H=5715$ to determine $k$ . When $t=H$ , half of the original substance is still present, thus

y_{0}e^{kH}=0.5y_{0}.\qquad e^{kH}=0.5.\qquad k=\frac{\ln 0.5}{H}=-\frac{0.693% }{5715}=-0.0001213.

Finally, we use the ration 52.5% for determining the time $t$ when Ötzi died (actually was killed),

e^{kt}=e^{-0.0001213t}=0.525,\qquad t=\frac{\ln 0.525}{-0.0001213}=5312\qquad% \text{Answer: 5300 years ago}