5. The isotope C-14 is radioactive and will decay (brealodown) at a steady rate of half the amount every 5,730 years (Half-Life). In living organisms, approximately \( 0.1 \% \) of thelr carbon atoms are the radioactive C-14 isotope, whereas \( 99.9 \% \) of their carbon atoms are either C-12 or C-13. A fossized plant is uncovered, and its Carbon atoms are composed of \( 0.0125 \% \) C-14 and 99 . 9875 C-12 and C-13. Approximately how old is the fossilized plant specimen?

To determine the age of the fossilized plant specimen based on the amount of C-14 remaining, we can use the concept of half-lives. 1. **Initial C-14 Percentage**: In living organisms, approximately \( 0.1\% \) of their carbon atoms are C-14. 2. **C-14 Percentage in Fossil**: The fossilized plant has \( 0.0125\% \) C-14. 3. **Calculate the Fraction of C-14 Remaining**: \[ \text{Fraction of C-14 remaining} = \frac{\text{C-14 in fossil}}{\text{C-14 in living organisms}} = \frac{0.0125\%}{0.1\%} = \frac{0.0125}{0.1} = 0.125 \] 4. **Determine the Number of Half-Lives**: The fraction of C-14 remaining can also be expressed in terms of half-lives. The formula for the remaining amount of a radioactive substance after \( n \) half-lives is: \[ \text{Remaining fraction} = \left(\frac{1}{2}\right)^n \] Setting this equal to the fraction we calculated: \[ 0.125 = \left(\frac{1}{2}\right)^n \] We can express \( 0.125 \) as a power of \( \frac{1}{2} \): \[ 0.125 = \frac{1}{8} = \left(\frac{1}{2}\right)^3 \] Therefore, \( n = 3 \). 5. **Calculate the Age of the Fossil**: Since each half-life of C-14 is \( 5,730 \) years, the age of the fossil can be calculated as: \[ \text{Age} = n \times \text{Half-life} = 3 \times 5,730 \text{ years} = 17,190 \text{ years} \] Thus, the approximate age of the fossilized plant specimen is **17,190 years**.