Question

A sample of paper from an ancient scroll was found to contain 39.5% 14C content as compared to a present-day sample. The t1/2 for 14C is 5720 years. What is the age of the scroll?

Answer 1

Answer

7665 years

Procedure

Let N₀ be the amount of carbon-14 present in a living organism. According to the radioactive decay law, the number of carbon-14 atoms, N, left in a dead tissue sample after a certain time, t, is given by the exponential equation:

N = N₀e^(-λt)

where λ is the decay constant which is related to half-life (T1/2) by the equation:

$\lambda=\frac{ln(2)}{t_{\frac{1}{2}}}$

Here, ln(2) is the natural logarithm of 2.

The percent of carbon-14 remaining after time t is given by N/N₀.

Using the first equation, we can determine λt.

The half-life of carbon-14 is 5,720 years, thus, we can calculate λ using the second equation, and then find t.

$\lambda=\frac{ln(2)}{5720}=1.211\times10^{-4}$

Solving the second equation for t, and using the λ we have just calculated we will have

t= 7665 years