# An animal dies in the forest. How long will it be until only 7/8 of the original amount of Carbon 14 remains?

## Recall that the half life of Carbon 14 is 5568 years. Round to the nearest year. NOTE: The half-life is incorrect. It is 5730 years. - Truong-Son

Mar 19, 2018

$1073$ years

#### Explanation:

All thanks to Truong-Son Nguyen, aka his method!

We use one of the radioactive decay equation, which states that

$\left[A\right] = {\left[A\right]}_{0} {e}^{- k t}$

• $\left[A\right]$ is the amount of substance right now

• ${\left[A\right]}_{0}$ is the original amount

• $- k$ is the decay constant, and it is negative because the system experiences decay

• $t$ is the time in years

We also have half-life, ${t}_{\frac{1}{2}}$, related by the equation,

${t}_{\frac{1}{2}} = \frac{\ln 2}{k}$

From the original equation, we have $\left[A\right] = \frac{7}{8} \left[{A}_{0}\right]$, and so:

$\frac{A}{{A}_{0}} = \frac{7}{8} = {e}^{- k t}$

$\therefore k t = - \ln \left(\frac{7}{8}\right) = 0.1335$

Therefore, the decay constant becomes,

$k = \frac{\ln 2}{{t}_{\frac{1}{2}}} = \frac{0.693}{5568} = 1.24 \cdot {10}^{-} 4 \setminus {\text{years}}^{-} 1$

And so, we get,

$t = \frac{0.1335}{k} = \frac{0.1335}{1.24 \cdot {10}^{-} 4 \setminus {\text{years}}^{-} 1} \approx 1073$ years