Question

Radiocarbon dating of a sample of wood from the tomb of an Egyptian pharaoh found that in 1 of the sample there were 9.843 × 10−15 of C-14 and 1.202 × 10−2 of C-12. The C-14 isotope is radioactive and undergoes − decay with a half-life of 5,730 years; the C-12 isotope is stable (does not decay). If the initial ratio of these isotopes is known to be −14 = 1.20 × 10−12, how old is the sample of wood in the tomb?

Answer 1

Let C be the total amount of C-14, after a time t from the origin of the piece of wood. If the initial amount of C-14 is C_0, and the half-life of C-14is τ, then, the formula that relates these variables, is:

$C=C_0\times2^{-t/\tau}$

Isolate t from the equation:

$\begin{gathered} \Rightarrow\frac{C}{C_0}=2^{-t/\tau} \\ \Rightarrow\log _2(\frac{C}{C_0})=-\frac{t}{\tau} \\ \Rightarrow-\tau\cdot\log _2(\frac{C}{C_0})=t \end{gathered}$

Therefore:

$t=-\tau\cdot\log _2(\frac{C}{C_0})$

Since the ratio r between the initial amount of C-14 and C-12 is given, we can find the initial amount of C-14:

$\begin{gathered} \frac{C_0}{C_{12}}=r \\ \Rightarrow C_0=r\times C_{12} \end{gathered}$

Substitute r=1.20*10^-12 and C_12=1.202*10^-2 mol to find the initial amount of C-14:

$\begin{gathered} C_0=1.2\times10^{-12}\times1.202\times10^{-2}\text{mol} \\ =1.4424\times10^{-14}\text{mol} \end{gathered}$

Substitute τ=5730y, C_0=1.4414*10^-14 mol, and C=9.843*10^-15 mol into the formula to find t, which is the age of the wood from the sample:

$\begin{gathered} t=-\tau\cdot\log _2(\frac{C}{C_0}) \\ =-5730y\times\log _2(\frac{9.843\times10^{-15}\text{mol}}{1.4424\times10^{-14}\text{mol}}) \\ =3158.956\ldots y \end{gathered}$

Therefore, to the nearest ten, the age of the sample of wood in the tomb, is:

$3160\text{ years}$