Question 19.6: using uranium/Lead Dating to Estimate the Age of a Rock A me...
using uranium/Lead Dating to Estimate the Age of a Rock
A meteor contains 0.556 g of Pb-206 to every 1.00 g of U-238. Assuming that the meteor did not contain any Pb-206 at the time of its formation, determine the age of the meteor. Uranium-238 decays to lead-206 with a half-life of 4.5 billion years.
| SORT You are given the current masses of Pb-206 and U-238 in a rock and asked to find its age. You are also given the half-life of U-238. | GIVEN m_{U-238} = 1.00 g; m_{Pb-206} = 0.556 g; t_{1/2} = 4.5 \times 10^{9} yr FIND t |
| STRATEGIZE You use the integrated rate law (Equation 19.3) to solve this problem. However, you must first determine the value of the rate constant (k) from the half-life expression (Equation 19.1).
t_{1/2} = \frac{0.693}{k} [19.1] ln\frac{ N_{t}}{N_{0}}= -kt [19.3] Before substituting into the integrated rate law, you also need the ratio of the current amount of U-238 to the original amount (N_{t}/N_{0}). The current mass of uranium is sim ply 1.00 g. The initial mass includes the cur rent mass (1.00 g) plus the mass that has decayed into lead-206, which can be deter mined from the current mass of Pb-206. Use the value of the rate constant and the initial and current amounts of U-238 along with integrated rate law to find t. |
CONCEPTUAL PLAN
t_{1/2} → k . t_{1/2} = \frac{0.693}{k} g Pb-206 → mol Pb-206 → mol U-238 → g U-238 . \frac{1molPb}{206gPb} \frac{1molU}{1molPb} \frac{238gU}{1molU} k, N_{ t},N_{0} → t . ln\frac{ N_{t}}{N_{0}}= -kt |
| SOLVE Follow your plan. Begin by finding the rate constant from the half-life. Determine the mass in grams of U-238 that is required to form the given mass of Pb-206. Substitute the rate constant and the initial and current masses of U-238 into the inte grated rate law and solve for t. (The initial mass of U-238 is the sum of the current mass and the mass that is required to form the given mass of Pb-206. |
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t_{1/2} = \frac{0.693}{k}
k = \frac{0.693}{t_{1/2}}= \frac{0.693}{4.5 \times 10^{9} yr} = 1.\underline{5}4 \times 10^{-10}/yr
0.556 \cancel{g Pb-206} \times\frac{\cancel{1 mol Pb-206}}{206 \cancel{g Pb-206}}\times\frac{\cancel{1 mol U-238}}{\cancel{1 mol Pb-206}}\times\frac{ 238 g U-238}{\cancel{1 mol U-238}}
= 0.64\underline{2}4 g U-238
ln\frac{ N_{t}}{N_{0}}= -kt
t = –\frac{ln\frac{N_{t}}{N_{0}}}{k} = -\frac{ln\frac{ 1.00 \cancel{g}}{1.00 \cancel{g} + 0.64\underline{2}4 \cancel{g}}}{1.\underline{5}4 \times 10^{-10}/yr}
= 3.2 × 10^{9} yr
CHECK The units of the answer are correct. The magnitude of the answer is about 3.2 billion years, which is less than one half-life. This value is reasonable given that less than half of the uranium has decayed into lead.