Question

A rock from Australia contains 0.438g Pb-206 to every 1.00g U-238. Assuming that the rock did not contain any Pb-206 at the time of its formation, how old if the rock?

Answer 1

Since the formation rock U-238 gradually converts to Pb-206.

The half-life for the conversion of U-238 to Pb-206 is, t1/2 = 4.50 x 10⁹ years

Now t1/2 = 0.693 / k = 4.50 x 10⁹ years

=> k = 0.693 / 4.50 x 10⁹ years = 1.54 x 10^-10 year^-1

Given the mass of Pb-206 formed = 0.438

Hence moles of Pb-206 formed = mass / molecular mass = 0.438 / 206 = 0.0021262 mol Pb

The radioactive disintigration reaction is

U-238 ---- > Pb-206

1 mol ---------1 mol

1 moles of U-238 forms 1 mol of Pb-206.

Hence 0.0021262 mol Pb that would be formed by the moles of U-238 = 0.0021262 mol U-238

Hence mass of U-238 = (0.0021262 mol U-238) x (238 g/mol U-238) = 0.506 g U-238

Hence initial amount of U-238, N₀ = 1.506 g

Current amount of U-238, Nt = 1.00 g

Now applying the integrated equation for first order reaction

ln(N₀/Nt) = kt

=> t = (1/k) x ln(N₀/Nt) = (1 / 1.54 x 10^-10 year^-1) x ln(1.506 g / 1.00 g) = 2.66 x 10⁹ years (answer)