A rock contains 0.275 mg of lead-206 for each milligram of uranium-238. The half-life for the decay of uranium-238 to lead-206 is 4.5 × 10⁹ yr. The rock was formed __________ yr ago.

Question

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Let t be the time since the rock was formed. Then we have: (lead-206 amount)/(uranium-238 amount) = 0.275 mg/mg We can use the fact that the ratio of lead-206 to uranium-238 is related to the age of the rock through the following equation: (lead-206 amount)/(uranium-238 amount) = e^(λt) where λ is the decay constant, which is related to the half-life as follows: λ = ln(2)/t(1/2) Substituting the given half-life of uranium-238, we get: λ = ln(2)/(4.5 × 10⁹ yr) ≈ 1.54 × 10⁹ /yr Substituting this and the given lead-206 to uranium-238 ratio into the equation above, we get: 0.275 mg/mg = e^(1.54 × 10⁹ /yr * t) Taking the natural logarithm of both sides and solving for t, we get: t = ln(0.275)/ (1.54 × 10⁹ /yr) ≈ 1.79 × 10⁹ yr Therefore, the rock was formed approximately 1.79 × 10⁹ yr ago.

A Rock Contains 0