Question

1. if carbon - 14 has a half - life of 5,730 years, how old is a wooden log that has 12.5% of the c - 14 remaining? * 4 points

a. 2,865 years
b. 5,730 years
c. 11,460 years
d. 17,190 years

Explanation:

Step1: Determine number of half - lives

We know that after each half - life, the amount of the radioactive substance is halved. Starting with 100% of the substance, after 1 half - life, 50% remains; after 2 half - lives, 25% remains; after 3 half - lives, 12.5% remains. So the number of half - lives $n = 3$.

Step2: Calculate the age of the wooden log

The age $t$ of the sample is given by the formula $t=n\times T_{1/2}$, where $T_{1/2}$ is the half - life. Given $T_{1/2}=5730$ years and $n = 3$. Then $t=3\times5730$ years.
$t = 17190$ years

Answer:

D. 17,190 years