Question

The amount of the radioactive isotope carbon-$14$present in small quantities can be measured with a Geiger counter. Carbon-$14$is replenished in live organisms, and after an organism dies the carbon-$14$in it decays at a rate proportional to the amount of carbon-$14$present in the body. Suppose $C\left(t\right)$is the amount of carbon-$14$in a dead organism $t$years after it dies.

(a) Set up a differential equation describing $\frac{dC}{dt}$, and solve it to get a formula for $C\left(t\right)$. Your answer will involve two constants.

(b) The half-life of carbon-$14$is $5730$years. (See part (b) of the previous problem for the definition of half-life.) Use this half-life to find the value of the proportionality constant for the model you found in part (a).

(c) Suppose you find a bone fossil that has $10\%$of its carbon-$14$ left. How old would you estimate the fossil to be?

Short answer

Part $\left(a\right)$: A formula for $C\left(t\right)$is, role="math" localid="1649407936304" $C\left(t\right)=A{e}^{-kt}$.

Part $\left(b\right)$: The value of the proportionality constant for the model is, $0.00012$.

Part $\left(c\right)$: The fossil is approximately$19188$years old.

Step-by-step solution

Part a Step 1. Given information

Carbon-$14$decays at a rate proportional to the amount of carbon-$14$present in the body.

$C\left(t\right)$is the amount of carbon-$14$ in a dead organism $t$years after it dies.

Part a Step 2. Note that the amount of carbon-14 is continuously decreasing with time.

This means that the rate of change of Carbon$-14$is negative. Hence, the mathematical equation representing this model will follow the model of negative growth rate.

From the given information, if $C\left(t\right)$is the amount of carbon$-14$remaining after $t$years, then the decay rate $\frac{dC}{dt}$of carbon$-14$is proportional to $C\left(t\right)$.Hence, the model representing the decay rate of carbon$-14$is,

$$\begin{gathered} \frac{dC}{dt}\alpha C \\ =-kC \end{gathered}$$

Part a Step 3. In the above relation k is the decay constant.

Now integrate the obtained equation on both sides,

$$\begin{gathered} \int \frac{dC}{C}=-k\int dt \\ \ln \left|C\right|=-kt+C_1 \\ C=e^{-kt+C_1} \\ =A{e}^{-kt}(e^{C_1}=A) \end{gathered}$$

Not that the solution contains two constants namely the decay constant$k$and the constant$A$.

If the initial amount of carbon$-14$present in the body is taken as $C_0$,then the constant $A$becomes $C_0$and the obtained model becomes,

$C=C_0{e}^{-kt}$.

Part b Step 1. Suppose that the quantity of carbon-14 present initially is C0,

Then the half life of $5730$implies that the amount of carbon$-14$remaining in the body after $5730$years will be $\frac{C_0}{2}$. So, substitute $t=5730,C\left(t\right)=\frac{C_0}{2}$in the obtained model .

$$\begin{gathered} \frac{C_0}{2}=C_0{e}^{-5730k} \\ {e}^{-5730k}=2 \\ k=\frac{1}{5730}\ln 2 \\ \approx 0.00012 \end{gathered}$$

Part c Step 1. Substitute the value of k obtained in part b to get the solution of the decay model as,

$C\left(t\right)=C_0{e}^{-0.00012t}$

From the given information,

The fossil found has $10\%$of its carbon$-14$left. So, $C\left(t\right)=0.1C_0$.

Equate both the obtained equation.

$$\begin{gathered} 0.1C_0=C_0{e}^{-0.00012t} \\ {e}^{0.00012t}=10 \\ 0.00012t=\ln 10 \\ t=\frac{1}{0.00012}\ln 10 \\ t=19188.2 \\ t\approx 19188 \end{gathered}$$