Q 3.2-25E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 23–27, assume that the rate of decay of a radioactive substance is proportional to the amount of the substance present. The half-life of a radioactive substance is the time it takes for one-half of the substance to disintegrate. Carbon dating is often used to determine the age of a fossil. For example, a humanoid skull was found in a cave in South Africa along with the remains of a campfire. Archaeologists believe the age of the skull to be the same age as the campfire. It is determined that only 2% of the original amount of carbon-14 remains in the burnt wood of the campfire. Estimate the age of the skull if the half-life of carbon-14 is about 5600 years.

Step-by-Step Solution

Answer

The estimated age of the skull is 31,606 years.

Step 1: Analyzing the given statement

Given that the rate of decay of a radioactive substance is directly proportional to the amount of the substance present.  Let the present amount of the radioactive substance be N.

Therefore, $\frac{\mathbf{d}\mathbf{N}}{\mathbf{d}\mathbf{t}}\mathbf{\propto }\mathit{N}$

Given that there is only 2% of the original amount of carbon-14 remains in the burnt wood of the campfire. We have to estimate the age of the skull if the half-life of carbon-14 is about 5600 years.

Step 2: Determining the formula with the help of the given proportionality relation, to solve the question

Given, 

$$\begin{gathered} \frac{dN}{dt}\propto N \\ \frac{dN}{dt}=-\lambda N \end{gathered}$$

where, $\lambda$ is the constant of proportionality.

 $$\begin{gathered} \frac{dN}{N}=-\lambda \\ \int \frac{dN}{N}=-\lambda \int dt \\ lnN=-\lambda t+ln{N}_0 \end{gathered}$$

  where,  N₀ is an arbitrary constant.

 $$\begin{gathered} lnN-ln{N}_0=-\lambda t \\ ln\left(\frac{N}{N_0}\right)=-\lambda t \\ \frac{N}{N_0}=e^{-\lambda t} \\ N=N_0{e}^{-\lambda t}······\left(1\right) \end{gathered}$$

One will use this formula to solve the question.

Step 3: Determining the value of λ

The half-life of carbon-14 is given as 5600 years. The formula for finding the half-life is, 

 $t_{\frac{1}{2}}=\frac{\ln 2}{\lambda }$

Here, $t_{\frac{1}{2}}=5600 years$

 

Thus,$5600=\frac{\ln 2}{\lambda }$

 $\lambda =\frac{\ln 2}{5600}······\left(2\right)$                 .

One will use this value of $\lambda$in step4 to find the estimated age of the skull.

Step 4: Finding the estimated age of the skull

Let the original amount of carbon-14 be N₀ and let the amount of remaining carbon-14 in the burnt wood of the campfire be N, which is given as 2% of the original amount,

 i.e., N = 0.02 N₀

 

Using the equation (1),

$$\begin{gathered} \left(0.02\right)N_0=N_0{e}^{-\lambda t} \\ 0.02=e^{-\lambda t} \\ e^{\lambda t}=\frac{100}{2} \\ e^{\lambda t}=50 \\ \lambda t=ln50 \\ t=\frac{ln50}{\lambda } \end{gathered}$$

 

Now, using the value of $\lambda$from equation (2),

  $$\begin{gathered} t=\frac{ln50}{ln2}·\left(5600\right) \\ t=31606years \end{gathered}$$                                                                                                                   

Hence, the estimated age of the skull is 31,606 years.