Question

jim is an archaeologist who wants to determine the age of a fossil. the half - life of carbon - 14 is 5730 years. jim determines there must have been 6.6 grams of carbon - 14 in the fossil when the animal was alive. now there is only 1.2 grams of carbon - 14 remaining. what is the age of the fossil?

14,092.54

16,345.18

13,119.45

15,540.97

Explanation:

Step1: Recall the radioactive - decay formula

The formula for radioactive decay is $N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}$, where $N$ is the amount of the substance at time $t$, $N_0$ is the initial amount of the substance, $T_{1/2}$ is the half - life of the substance. We know that $N_0 = 6.6$ grams, $N = 1.2$ grams, and $T_{1/2}=5730$ years.

Step2: Substitute the values into the formula

Substituting the values into the formula $1.2 = 6.6(\frac{1}{2})^{\frac{t}{5730}}$. First, divide both sides by $6.6$: $\frac{1.2}{6.6}=(\frac{1}{2})^{\frac{t}{5730}}$. Simplify $\frac{1.2}{6.6}=\frac{2}{11}$. So, $\frac{2}{11}=(\frac{1}{2})^{\frac{t}{5730}}$.

Step3: Take the natural logarithm of both sides

$\ln(\frac{2}{11})=\ln((\frac{1}{2})^{\frac{t}{5730}})$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln(\frac{2}{11})=\frac{t}{5730}\ln(\frac{1}{2})$.

Step4: Solve for $t$

We know that $\ln(\frac{2}{11})=\ln(2)-\ln(11)\approx0.693 - 2.398=- 1.705$ and $\ln(\frac{1}{2})=-\ln(2)\approx - 0.693$. Then $t = 5730\times\frac{\ln(\frac{2}{11})}{\ln(\frac{1}{2})}=5730\times\frac{-1.705}{-0.693}$.
$t = 5730\times2.46032\approx14092.54$ years.

Answer:

14,092.54