Answer
$\approx {\bf17100}\, \text{years}$
Work Step by Step
We need to determine the age (in years) of a bone based on the $ ^{14}\text{C}/^{12}\text{C} $ ratio, which has decayed to $ 1.65 \times 10^{-13} $.
The decay of $ ^{14}\text{C} $ follows an exponential relationship. After a certain amount of time $ t $, the ratio $ R $ of $ ^{14}\text{C}/^{12}\text{C} $ in the sample can be described by:
$$
R = R_0 \left( \frac{1}{2} \right)^{t / t_{1/2}}
$$
where:
- $ R $ is the current $ ^{14}\text{C}/^{12}\text{C} $ ratio, given as $ 1.65 \times 10^{-13} $.
- $ R_0 $ is the initial ratio of $ ^{14}\text{C}/^{12}\text{C} $ at the time of death, $ 1.3 \times 10^{-12} $.
- $ t_{1/2} $ is the half-life of $ ^{14}\text{C} $, $ 5730 $ years.
Solve for $ t $:
$$
\frac{R}{R_0} = \left( \frac{1}{2} \right)^{t / t_{1/2}}
$$
Take the Natural Logarithm of Both Sides:
$$\ln\left[ \frac{R}{R_0} \right]=\ln\left[\left( \frac{1}{2} \right)^{t / t_{1/2}}\right]$$
$$\ln\left[ \frac{R}{R_0} \right]=\dfrac{t}{t_{_{1/2}}}\ln\left[ \frac{1}{2} \right]$$
$$t=t_{_{1/2}}\;\dfrac{\ln\left[ \frac{R}{R_0} \right]}{ \ln\left[ \frac{1}{2} \right]} $$
Substitute the known values:
$$t=(5730 )\dfrac{\ln\left[ \dfrac{(1.65 \times 10^{-13})}{( 1.3 \times 10^{-12})} \right]}{ \ln\left[ \dfrac{1}{2} \right]} $$
$$
t = \color{red}{\bf17064}\, \text{years}
$$
The age of the bone is approximately 17,100 years.
