Chapter 5.8, Problem 12AYU

Estimating Age A fossilized leaf contains $70\%$ of its normal amount of carbon- $14.$ How old is the fossil?

To determine

To calculate: Theage of a fossil, if the fossilized leaf contains $70\%$ of its normal amount of the carbon- $14$.

Answer to Problem 12AYU

Solution:

The fossil is about $2948.5$ years old.

Explanation of Solution

Given Information:

A fossilized leaf contains $70\%$ of its normal amount of the carbon- $14$.

Formula used:

The amount $A$ of carbon- $14$ present at time $t$ is $A\left(t\right)=A_0{e}^{kt}$, where $A_0$ is the original amount of carbon- $14$ present, and $k$ is a negative number.

Calculation:

Half-life of the carbon- $14$ is $5730$ years.

By substituting value of $A\left(t\right)=\frac{1}{2}{A}_0$ and $t=5730$ in $A\left(t\right)=A_0{e}^{kt}$,

$\Rightarrow \frac{1}{2}{A}_0=A_0{e}^{k\left(5730\right)}$

Now, simplify this equation for $k$,

$\Rightarrow \frac{1}{2}=e^{5730k}$

By taking logarithm of both sides,

$\Rightarrow \ln \left(\frac{1}{2}\right)=5730k$

$\Rightarrow k=\frac{1}{5730}\cdot \ln \left(\frac{1}{2}\right)=-0.000120968$

Now, substitute this value of $k=-0.000120968$ in $A=A_0{e}^{kt}$,

$\Rightarrow A\left(t\right)=A_0{e}^{-0.000120968t}$

If the amount $A$ of carbon- $14$ now present is $70\%$ of the original amount,

$\Rightarrow 0.70A_0=A_0{e}^{-0.000120968t}$

By dividing both sides by $A_0$,

$\Rightarrow 0.70=e^{-0.000120968t}$

By taking logarithm on both sides,

$\Rightarrow \ln \left(0.70\right)=\ln \left(e^{-0.000120968t}\right)$

$\Rightarrow -0.35667494=-0.000120968t$

$\Rightarrow t=2948.5$

Thus, the fossil is about $2948.5$ years old.