Q.5.55 - An Introduction to General, Organic, and Biological Chemistry

Question

Carbon dating of small bits of charcoal used in cave paintings has determined that some of the paintings are from 10000 to 30000 yr old. Carbon- 14 has a half-life of 5730 yr. In a 1-$\mu g$ sample of carbon from a live tree, the activity of carbon-14 is 6.4 $\mu$Ci. If researchers determine that 1 $\mu g$of charcoal from a prehistoric cave painting in France has an activity of 0.80$\mu Ci$, what is the age of the painting? 

Step-by-Step Solution

Answer

The age of the painting is $17188.6$years.

Part (a) Step 1: Given Information

Carbon-$14$has a half-life of$5730$yr.

 In a 1$\mu g$-sample of carbon from a live tree, the activity of carbon-$14$is $6.4\mu Ci$

 1 $\mu g$of charcoal from a prehistoric cave painting in France has an activity of $0.80\mu Ci$.

Finding the age of the painting.

Step 2: Explanation

The half-life of Carbon-14 =${\mathrm{t}}_{\frac{1}{2}}=5730\text{ years }$

The activity of carbon-14 =${\alpha }_o=6.4\mu \text{ Cidecay/sec }$and the activity of painting =${\alpha }_t=0.8\mu \text{ Cidecay/sec }$

Using the formula,

${\mathrm{t}}_{\frac{1}{2}}=\frac{0.693}{\lambda }\Rightarrow \lambda =\frac{0.693}{t_{{\displaystyle \frac{1}{2}}}}=\frac{0.693}{5730}=1.21\times {10}^{-4}\text{decay}$

$\mathrm{t}=-\frac{1}{\lambda }\ln \left(\frac{{\alpha }_{\mathrm{t}}}{{\alpha }_o}\right)$$\Rightarrow \mathrm{t}=\frac{-1}{1.21\times {10}^{-4}}\times 2.303\times \log \left(\frac{0.8}{6.4}\right)=17188.6\text{yrs}$

Hence the age of the painting is t =$171886.6$ years