Chapter 1, Problem 1.25P

To determine

(a)

The value of the parameter b and plot the function by using half-life of carbon-14.

Answer to Problem 1.25P

The value of the parameter $b$ is $1.2603\times {10}^{-4}$ and the plot of function is shown in Figure 1.

Explanation of Solution

Given:

Half-life of carbon -14 is 5500 years

Concept Used:

Fraction of carbon -14 remaining time at $\text{'}t\text{'}$ is $\frac{C(t)}{C(0)}$

The exponential decay, $\frac{C(t)}{C(0)}=e^{-bt}$.

Calculation:

Obtain $b$ using half-life.

$\begin{array}{c}\frac{C(t)}{C(0)}=\frac{1}{2}\\ =0.5\end{array}$

Thus, $\frac{C(t)}{C(0)}=e^{{-}^{bt}}=0.5$

Substitute carbon half-life, 5500 years for $t.$

$\begin{array}{c}0.5=e^{-b(5500)}\\ \text{ln(0}\text{.5)=}-b\text{(5500)}\\ -0.693=-5500b\\ b=1.2603\times {10}^{-4}\end{array}$

Thus the value of the parameter $b$ is $1.2603\times {10}^{-4}$

Now the exponential decay function $\frac{C(t)}{C(0)}=e^{-1.2603\times {10}^{-4i}}$

Enter the following code in MATLAB:

$\begin{array}{l}>>\text{t=0:10:5500;}\\ \text{>>c=exp(-1}\text{.2603E-4}\text{.*t);}\\ >>\text{plot(t,c)}\\ \text{>>xlabel t}\\ \text{>>ylabel fractiondecay}\end{array}$

The following is the plot of the exponential decay function:

Figure 1.

Conclusion:

Thus, the value of the parameter $b$ is $1.2603\times {10}^{-4}$ and the plot of function is shown in Figure 1.

To determine

(b)

The time when the organism died.

Answer to Problem 1.25P

Organism died $836\text{years}\text{ago}$.

Explanation of Solution

Given:

Half-life of carbon -14 is 5500 years.

Concept Used:

Fraction of carbon -14 remaining time at $\text{'}t\text{'}$ is $\frac{C(t)}{C(0)}$.

The exponential decay, $\frac{C(t)}{C(0)}=e^{-bt}$.

Calculation:

We have, $$

$\begin{array}{c}\frac{C(t)}{C(0)}=0.9[because90\%]\\ \text{Also, }\frac{C(t)}{C(0)}=e^{-1.2603\times {10}^{-4}t}\end{array}$

Solve for t.

$\begin{array}{c}{e}^{-1.2603\times {10}^{-4}t}=0.9\\ \text{ln (0}\text{.9)=}-\text{1}\text{.2603}\times {\text{10}}^{-4}t\\ t=\frac{\ln \text{(0}\text{.9)}}{-1.2603\times {10}^{-4}}\\ \approx 836\text{years}\end{array}$.

Conclusion:

Thus,organism died $836\text{years}$ ago.

To determine

(c)

The effect of b on the age obtained in part (b).

Answer to Problem 1.25P

The organism died age is reduced to $760\text{years}$ ago.

Explanation of Solution

Given:

Half-life of carbon-14 is 5500 years and $b$ is off by $\pm 1\%$.

Concept Used:

Fraction of carbon - 14 remaining time at $\text{'}t\text{'}$ is $\frac{C(t)}{C(0)}$

The exponential decay, $\frac{C(t)}{C(0)}=e^{-bt}$.

Calculation:

Determine $b$ for an off by $\pm 1\%$

$b=1.1(1.2603\times {10}^{-4})$

Determine the time estimate

$\begin{array}{c}t=\frac{-\ln \text{(0}\text{.9)}}{b}\\ =\frac{-\text{ln(0}\text{.9)}}{(1.1)(1.2603\times {10}^{-4})}\\ =760\text{years}\end{array}$.

Conclusion:

Thus, the organism died age is reduced to $760\text{years}$ ago.