# Carbon-14 has a half life of 5730 years. If a shell is found and has 52% of its original Carbon-14, how old is it?

Aug 16, 2016

Approximately $5406$ years

#### Explanation:

$- 5730 \cdot {\log}_{2} \left(\frac{52}{100}\right) = - 5730 \cdot \log \frac{\frac{52}{100}}{\log} \left(2\right) \approx 5406$ years

The proportion of ""^14C remaining, compared with the original is modelled by the function:

$p \left(t\right) = {2}^{-} \frac{t}{5730}$

where $t$ is measured in years.

Given that $p \left(t\right) = \frac{52}{100}$, we have:

$\frac{52}{100} = {2}^{- \frac{t}{5730}}$

Taking logs base $2$ we have:

${\log}_{2} \left(\frac{52}{100}\right) = - \frac{t}{5730}$

So multiplying both sides by $- 5730$ and transposing we get the formula:

$t = - 5730 \cdot {\log}_{2} \left(\frac{52}{100}\right)$

Footnote

The complexity of radiocarbon dating comes from two factors:

• The proportion of ""^14C to ""^12C is very small to start with.
• The proportion has been dramatically affected by factors like the industrial revolution, which replaced a significant portion (one third? one quarter?) of the $C {O}_{2}$ in the atmosphere with carbon dioxide generated from fossil fuels - which contain virtually no remaining ""^14C. As a result the method needs calibration to provide accurate dates.
Aug 16, 2016

$\textsf{5409 \textcolor{w h i t e}{x} \text{yr}}$

#### Explanation:

The expression for radioactive decay is:

$\textsf{{N}_{t} = {N}_{0} {e}^{- \lambda t}}$

$\textsf{{N}_{0}}$ is the initial number of undecayed atoms

$\textsf{{N}_{t}}$ is the number of undecayed atoms at time $\textsf{t}$

$\textsf{\lambda}$ is the decay constant

Taking natural logs of both sides:

$\textsf{\ln {N}_{t} = \ln {N}_{0} - \lambda t}$

$\therefore$$\textsf{\ln \left({N}_{t} / {N}_{0}\right) = - \lambda t}$

We can get the value of $\textsf{\lambda}$ using the expression:

$\textsf{\lambda = \frac{0.693}{t} _ \left(\frac{1}{2}\right) = \frac{0.693}{5730} = 1.209 \times {10}^{- 4} \textcolor{w h i t e}{x} {a}^{- 1}}$

Putting in the numbers:

$\textsf{\ln \left(\frac{52}{100}\right) = - 1.208 \times {10}^{- 4} \times t = - 0.6539}$

$\therefore$$\textsf{t = \frac{0.6539}{1.209 \times {10}^{- 4}} = 5409 \textcolor{w h i t e}{x} y r}$