Question

1. (4) you find a bone in your back yard and it looks very old. you buy a c - 14 testing kit and calculate that there is 15% of the c - 14 remaining in the bone. if the half - life of c - 14 is 5730 years, how old is the bone?

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 radioactive substance at time $t$, $N_0$ is the initial amount of the radioactive substance, $T_{1/2}$ is the half - life of the radioactive substance. We know that $\frac{N}{N_0}=0.15$ (since 15% of C - 14 remains) and $T_{1/2}=5730$ years.

Step2: Substitute the values into the formula

Substituting into $N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}$, we get $0.15 = (\frac{1}{2})^{\frac{t}{5730}}$.

Step3: Take the natural logarithm of both sides

$\ln(0.15)=\ln((\frac{1}{2})^{\frac{t}{5730}})$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we have $\ln(0.15)=\frac{t}{5730}\ln(\frac{1}{2})$.

Step4: Solve for $t$

First, note that $\ln(\frac{1}{2})=-\ln(2)$. Then $t = 5730\times\frac{\ln(0.15)}{-\ln(2)}$.
We know that $\ln(0.15)\approx - 1.897$ and $\ln(2)\approx0.693$.
$t = 5730\times\frac{- 1.897}{-0.693}=5730\times2.737$.
$t\approx15683$ years.

Answer:

Approximately 15683 years.