Question 43.3: An Old Rock Another isotope used in radioactive dating is ^2......
An Old Rock
Another isotope used in radioactive dating is ^{238}U, which decays into ^{206}Pb and has a half-life of 4.46 × 10^9 yr. Suppose a rock contains a ratio of lead-206 to uranium-238 nuclei of N_{Pb} /N_U = 0.65. How old is the rock? (Assume the rock was originally all uranium-238.)
INTERPRET and ANTICIPATE
Because a substantial amount of the uranium has decayed into lead, we expect that the age of the rock will be close to the uranium half-life of several billion years.
SOLVE
At t = 0, the sample contained N_0 nuclei of uranium. Today’s number N_U is given by Equation 43.8.
The number of lead nuclei N_{Pb} in the rock today is the number of original uranium nuclei minus the number of uranium nuclei today.
N_{ Pb }=N_0-N_{ U } \quad \quad (1)Substitute Equation 43.8 into Equation (1).
\begin{aligned}& N_{ Pb }=N_0-N_0\left(\frac{1}{2}\right)^{t / T_{1 / 2}} \\& N_{ Pb }=N_0\left[1-\left(\frac{1}{2}\right)^{t / T_{1 / 2}}\right] \quad \quad (2)\end{aligned}Use Equations 43.8 and (2) to write a ratio for N_{ Pb } / N_{ U }.
\begin{aligned}& \frac{N_{ Pb }}{N_{ U }}=\frac{N_0\left[1-\left(\frac{1}{2}\right)^{t / T_{1 / 2}}\right]}{N_0\left(\frac{1}{2}\right)^{t / T_{1 / 2}}} \\& \frac{N_{ Pb }}{N_{ U }}=\left(\frac{1}{2}\right)^{-t / T_{1 / 2}}-1\end{aligned}Substitute values and solve for t.
\begin{aligned}& 0.65=\left(\frac{1}{2}\right)^{-t / T_{1 / 2}}-1 \\& \left(\frac{1}{2}\right)^{-t / T_{1 / 2}}=1.65 \\& t=\frac{T_{1 / 2} \ln 1.65}{\ln 2}=\frac{\left(4.46 \times 10^9 yr \right) \ln 1.65}{\ln 2} \\& t=3.2 \times 10^9 yr\end{aligned}CHECK and THINK
As expected, our answer is several billion years. When working such a problem, you shouldn’t expect to find rocks that are older than the age of the Earth (about 4.5 billion years).