Chapter 7.4, Problem 11E

To determine

To calculate : the age of the parchment

Answer to Problem 11E

The age of the parchment is $2489.13years$

Explanation of Solution

Given information : the question is

The bombardment of the upper atmosphere by cosmic rays converts nitrogen to a radioactive isotope of carbon $C$ with a half life of about $5730$ years

A parchment fragment was discovered that about $74\%$ as much $C$ radioactivity as does plant material on the earth

Calculation:

  $\begin{array}{l}y\left(t\right)=y\left(0\right)\times e^{kt}\\ y\left(5730\right)=100\times e^{k\times 5730}\\ \frac{50}{100}=e^{k\times 5730}\\ \ln \left(\frac{1}{2}\right)=k\times 5730\\ k=\frac{-\ln \left(2\right)}{5730}\end{array}$

Let $y\left(0\right)$

  $=100$ because at $t$ =0 years , there is $100$

Let $y\left(5730\right)=$

  $50$ because at $t$ =5730 years there is $50$

  $\begin{array}{l}y\left(t\right)=y\left(0\right)\times e^{kt}\\ y\left(t\right)=100\times e^{\frac{-\ln \left(2\right)}{5730}\times t}\\ y\left(t\right)=100\times e^{-\ln \left(2\right)\frac{t}{5730}}\\ y\left(t\right)=100\times 2^{-\frac{t}{5730}}\end{array}$

Rule 1: $y\times \ln \left(x\right)=\ln \left(x^y\right)$

Rule2: $e^{\ln \left(x\right)}=x$

  $\begin{array}{l}y\left(t\right)=100\times 2^{-\frac{t}{5730}}\\ 74=100\times 2^{{}^{-\frac{t}{5730}}}\\ 0.74=2^{-\frac{t}{5730}}\\ \log \left(2\right)\left(0.74\right)=-\frac{t}{5730}\\ t=2489.13years\end{array}$