Chapter 10.8, Problem 2E

(a)

To determine

Find the ratio of C-14 in the sample now to C-14 in the sample when it was part of a living animal.

Answer to Problem 2E

  $\frac{N}{N_0}=\frac{0.08}{{10}^{12}}=8\times {10}^{-14}$

Explanation of Solution

Given:

A scientist finds that a bone sample contains C-14 and C-12 in the ratio 0.08 to ${10}^{12}$ . Atmospheric C-14 fraction was the same then as it is now.

Calculation:

Since when an organism dies, the amount of C-12 it contains remains the same, but the C-14 decays. But it is given that atmospheric C-14 fraction was the same then as it is now.

So, the ratio of C-14 in the sample now to C-14 in the sample when it was part of a living animal is :

  $\frac{N}{N_0}=\frac{0.08}{{10}^{12}}=8\times {10}^{-14}$

(b)

To determine

Find the number of years from the animal was alive.

Answer to Problem 2E

249,300.84 years.

Explanation of Solution

Formula Used:

Half-life decay formula:

  $N=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{5730}}$ where N is the amount of C-14 in the sample discovered , $N_0$ is the amount of C-14 in the sample when it was alive.

Calculation:

From part (a) , the ratio of C-14 in the sample now to C-14 in the sample when it was part of a living animal is :

  $\frac{N}{N_0}=\frac{0.08}{{10}^{12}}=8\times {10}^{-14}$

Let the time passed since the animal was alive = t $\begin{array}{l}N=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{5730}}\\ \Rightarrow \frac{N}{N_0}=2^{-\frac{t}{5730}}\\ \Rightarrow {\log }_2(8\times {10}^{-14})=-\frac{t}{5730}\mathrm{.................}[take{\text{ log}}_2\text{ each side ; }\log a^b=b\log a]\\ \Rightarrow t=-5730({\log }_2{2}^3+{\log }_2{10}^{-14})\mathrm{......}[\log ab=\log a+\log b]\\ \Rightarrow t=-5730(3-14{\log }_{2}10)\mathrm{...............}[\log a^b=b\log a;{\log }_{a}a=1]\\ \Rightarrow t=-5730(3-14\cdot 3.322)\\ \Rightarrow t=-5730(3-46.508)\\ \Rightarrow t=249300.84\end{array}$

So, the time passed since the animal was alive is 249,300.84 years.