An archaeologist has found a fossil in which the ratio of ${ } ^ { 14 } \mathrm { C }$ to ${ } ^ { 12 } \mathrm { C }$ is $\frac { 1 } { 9 }$ the ratio in the atmosphere. Approximately how old is the fossil? The half-life of ${ } ^ { 14 } \mathrm { C }$ is 5,730 years. Round to the nearest whole year.
A) 18,164 years
B) 18,194 years
C) 25,785 years
D) 51,570 years

Question

Quizplus · Accepted Answer

The decay of ${ }^{14}\mathrm{C}$ follows a logarithmic decay formula, $N = N_0 \times (\frac{1}{2})^{\frac{t}{T}}$, where $N$ is the final amount, $N_0$ is the initial amount, $t$ is the time, and $T$ is the half-life. Given the ratio of ${ }^{14}\mathrm{C}$ to ${ }^{12}\mathrm{C}$ is $\frac{1}{9}$ of its original value, we can set up the equation $(\frac{1}{2})^{\frac{t}{5730}} = \frac{1}{9}$. Solving for $t$ gives approximately 18,164 years.

An Archaeologist Has Found a Fossil in Which the Ratio 14C{ } ^ { 14 } \mathrm { C }14C

An Archaeologist Has Found a Fossil in Which the Ratio ${ } ^ { 14 } \mathrm { C }$