Q 3.2-26E - 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.

To see how sensitive the technique of carbon dating of Problem 25 is

(a) Redo Problem 25 assuming the half-life of carbon-14 is 5550 yr.

(b) Redo Problem 25 assuming 3% of the original mass remains.

Step-by-Step Solution

Answer

(a) The estimated age of the skull is 31323 years.

(b) The estimated age of the skull is 28330 years.

Step1: Given data

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

Step 2: Analyzing the given statement

(a)

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 are 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 5550 years.

Step2: Determining the formula with the help of the given proportionality relation, to solve the question

Given, 

$\frac{dN}{dt}\propto N$

$\frac{dN}{dt}=-\lambda N$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, $ln{N}_0$is an arbitrary constant.

 

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

 One will use this formula to solve the question.

 

Step3: To determine the value of λ

The half-life of carbon-14 is given as 5550 years. The formula for finding the half-life is,

${\mathit{t}}_{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{=}\frac{\mathbf{l}\mathbf{n}\mathbf{2}}{\mathbf{\lambda }}$.

Here, $t_{\frac{\mathbf{1}}{\mathbf{2}}}=5550years$ 

Therefore, $5550=\frac{ln2}{\lambda }$

 

 $\lambda =\frac{ln2}{5550}$                                                                                       …… (2)

 

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

Step 4: To find 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=0.02N₀

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 $\mathit{\lambda }\mathbf{ }$from equation (2),

 $$\begin{gathered} t=\frac{ln50}{ln2}·\left(5550\right) \\ t=31323years \end{gathered}$$

Hence, the estimated age of the skull is  31323 ears.

Step 6: Analyzing the given statement

(b)

 

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.

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

Given that there are only 3% 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 7: Determining the formula with the help of the given proportionality relation, to solve the question

Given, 

$\frac{dN}{dt}\propto N$

$\frac{dN}{dt}=-\lambda N$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, In N₀ is an arbitrary constant.

 

    $$\begin{gathered} lnN-ln{N}_0=-\lambda t \\ ln\left(\frac{N}{N_0}\right)=-\lambda t \\ \frac{N}{N_0}=e^{-\lambda t} \\ N=N_0{e}^{-\lambda t} \end{gathered}$$                                                                                              …… (2)

 One will use this formula to solve the question.

 

Step 8: To determine the value of λ

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

${\mathit{t}}_{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{=}\frac{\mathbf{l}\mathbf{n}\mathbf{2}}{\mathbf{\lambda }}$

Here, $t_{\frac{\mathbf{1}}{\mathbf{2}}}=5600years$

Accordingly,$5600=\frac{ln2}{\lambda }$

 $\lambda =\frac{ln2}{5600}$                                                                                              …… (3)

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

Step 9: To find 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 3% of the original amount, i.e.,$\mathrm{N}=0.03 {\mathrm{N}}_0$

Using the equation (2),

 $$\begin{gathered} \left(0.03\right)N_0=N_0{e}^{-\lambda t} \\ 0.03=e^{-\lambda t} \\ e^{\lambda t}=\frac{100}{3} \\ \lambda t=ln\left(\frac{100}{3}\right) \\ t=\frac{1}{\lambda }ln\left(\frac{100}{3}\right) \end{gathered}$$

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

 $$\begin{gathered} t=\frac{ln\left(\frac{100}{3}\right)}{ln2}·\left(5600\right) \\ t=28330years \end{gathered}$$

So, the estimated age of the skull is 28330 years.