# In 1951, the Dead Sea Scrolls were dated using carbon-14 dating. They were found to still contain 30% of their original carbon 14. If the half life of carbon 14 is 5,730 years, how old were the scrolls?

Jun 7, 2018

Roughly 10,000 years old according to the data in the question. But probably it is the decay that is 30%, in which case the age would be 2900 years old.

#### Explanation:

"Half life" means the amount of time that elapses before 1/2=50% of the amount of carbon 14 has disappeared. (Sorry for the unscientific expression.

The rate of decay is -5730 years
Let the start mass be ${A}_{0}$ at the first year.
The decay rate is geometric, i.e.
${A}_{t} = {A}_{0} {e}^{k t}$ or $\frac{{A}_{t}}{{A}_{0}} = {e}^{k t}$
$\implies k t = \ln \left(\frac{{A}_{t}}{{A}_{0}}\right)$
For $t = 5730$ $\frac{{A}_{t}}{{A}_{0}} = \frac{1}{2}$
$\implies 5730 k = \ln \left(\frac{1}{2}\right)$
$k = \ln \frac{\frac{1}{2}}{5730} = - 0.000120968 = - 1.20968 \cdot {10}^{-} 4$

From this we get:
$- 1.20968 \cdot {10}^{-} 4 \cdot t = \ln \left(0.3\right) = - 1.20397280433$
$t = \frac{- 1.20397280433}{- 1.20968 \cdot {10}^{-} 4} = 9953$

Note that the actual age of the Dead Sea Scrolls are more like 1800 to 2300 years old, so the amount of C14 given in the question is probably way too low. Should I guess, the decay has been 30%, i.e. that there is still 70% of the original C14 left. With that as the starting point we get:
$t = \ln \frac{0.7}{- 1.20968 \cdot {10}^{-} 4} = 2949$

This is a figure which is much more in agreement with the observed results, so I suspect you have misunderstood the given information.