Chapter 9.1, Problem 70STP

To determine

How long ago did the person die.

Answer to Problem 70STP

The person died $44028.6$ years go.

Explanation of Solution

Given information: Given $t_{\frac{1}{2}}=5760$

Calculation:

The activity of carbon as function of time is $A=A_0{e}^{-\lambda t}$

The half life is found out as given below when $t=t_{\frac{1}{2}}$

  $\begin{array}{l}\frac{A_0}{2}=A_0{e}^{-\lambda t_{\frac{1}{2}}}\\ e^{-\lambda t_{\frac{1}{2}}}=\frac{1}{2}\\ -\lambda t_{\frac{1}{2}}=-\ln 2\\ \lambda =\frac{\ln 2}{t_{\frac{1}{2}}}\end{array}$

When the activity becomes $0.5\%$ then plug the value.

  $\begin{array}{l}0.005A_0=A_0{e}^{-\lambda t}\\ \ln \left(0.005\right)=-\lambda t\\ t=-\frac{\ln \left(0.005\right)}{\lambda }\\ t=-\frac{\ln \left(0.005\right)}{\frac{\ln 2}{t_{\frac{1}{2}}}}\text{ Since }\lambda =\frac{\ln 2}{t_{\frac{1}{2}}}\\ t=-\frac{\ln \left(0.005\right)}{\ln 2}\cdot t_{\frac{1}{2}}\\ t=-\frac{\ln \left(0.005\right)}{\ln 2}\left(5760\right)\text{ Since }{t}_{\frac{1}{2}}=5760\\ t=44028.6\end{array}$

Thus, the person died $44028.6$ years go.