Question 21.SE.6: A rock contains 0.257 mg of lead-206 for every milligram of ...
A rock contains 0.257 mg of lead-206 for every milligram of uranium-238. The half-life for the decay of uranium-238 to lead-206 is 4.5 × 10^9 yr. How old is the rock?
Analyze We are asked to calculate the age of a rock containing uranium-238 and lead-206, given the half-life of the uranium-238 and the relative amounts of the uranium-238 and lead-206.
Plan Lead-206 is the product of the radioactive decay of uranium-238. We will assume that the only source of lead-206 in the rock is from the decay of uranium-238, which has a known half-life. To apply first-order kinetics expressions (Equations 21.19 and 21.20) to calculate the time elapsed since the rock was formed, we first need to calculate how much initial uranium-238 there was for every 1 mg that remains today.
k=\frac{0.693}{t_{1 / 2}} (21.19)
\ln \frac{N_t}{N_0}=-k t (21.20)
Solve Let’s assume that the rock currently contains 1.000 mg of uranium-238 and therefore 0.257 mg of lead-206. The amount of uranium-238 in the rock when it was first formed therefore equals 1.000 mg plus the quantity that has decayed to lead-206. Because the mass of lead atoms is not the same as the mass of uranium atoms, we cannot just add 1.000 mg and 0.257 mg. We have to multiply the present mass of lead-206 (0.257 mg) by the ratio of the mass number of uranium to that of lead, into which it has decayed. Therefore, the original mass of {}_{92}^{238}U was
\begin{aligned}\text{Original}\,\,{}_{92}^{238}U &=1.000\, mg +\frac{238}{206}(0.257 \,mg ) \\&=1.297 \,mg\end{aligned}
Using Equation 21.19, we can calculate the decay constant for the process from its half-life:
k=\frac{0.693}{4.5 \times 10^9 yr}=1.5 \times 10^{-10}yr ^{-1}
Rearranging Equation 21.20 to solve for time, t, and substituting known quantities gives
t=-\frac{1}{k}\ln \frac{N_t}{N_0}=-\frac{1}{1.5 \times 10^{-10}yr ^{-1}}\ln \frac{1.000}{1.297}=1.7 \times 10^9 yr