Chapter 7, Problem 55RE

To determine

To Calculate:

To calculate the age of a sample charcoal in which $90\%$ of the carbon-14 that was originally present has decayed.

Answer to Problem 55RE

The age of the sample charcoal is about $18,935$ years.

Explanation of Solution

Given information:

In a sample charcoal $90\%$ of the carbon-14 that was originally present has decayed.

Formula:

Half-life formula

  $A=A_0{\left(\frac{1}{2}\right)}^{\frac{t}{k}}$ Where $A$ -Final amount; $A_0$ - Initial amount; $t$ -number of years; $k$ -half life of the substance.

It is given that $90\%$ has decayed which means $10\%$ remains.

Now we use the Half-life formula

  $A=A_0{\left(\frac{1}{2}\right)}^{\frac{t}{k}}$ Where $A$ -Final amount; $A_0$ - Initial amount; $t$ -number of years; $k$ -half life of the substance.

So, let $A=0.1A_0$

Carbon-14 has a half-life of $5700$ years so let $k=5700$ .

Substitute these values into the half life formula and solve for $t$ ,

  $\begin{array}{l}A=A_0{\left(\frac{1}{2}\right)}^{\frac{t}{k}}\\ 0.1A_0=A_0{\left(\frac{1}{2}\right)}^{\frac{t}{5700}}\\ 0.1={\left(\frac{1}{2}\right)}^{\frac{t}{5700}}\\ {\log }_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(0.1\right)=\frac{t}{5700}\\ 5700\left({\log }_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(0.1\right)\right)=t\\ \Rightarrow t=18,935\end{array}$

(Or)

  $\begin{array}{l}{\log }_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(0.1\right)=\frac{t}{5700}\\ \frac{\ln 0.1}{\ln \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{t}{5700}\\ \frac{5700\ln 0.1}{\ln \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=t\\ t\approx 18,935\end{array}$

Therefore, the age of the sample charcoal is about $18,935$ years.