The half-life of Rn-222 is 3.8 days. Starting with 1.0 g of Rn-222, how much will be left after 10 half-lives? Recall that radioactive decays obey first-order kinetics.

Strategy All radioactive decays obey first-order kinetics. Therefore, its half-life is independent of the initial concentration.

Question

The half-life of Rn-222 is 3.8 days. Starting with 1.0 g of Rn-222, how much will be left after 10 half-lives? Recall that radioactive decays obey first-order kinetics.

Strategy All radioactive decays obey first-order kinetics. Therefore, its half-life is independent of the initial concentration.

Accepted Answer

After one half-life, the amount of Rn left is 0.5 × 1.0 g, or 0.5 g. After two half-lives, only 0.25 g of Rn remains. Generalizing the fraction of the isotope left after n half lives as [latex](1/2) ^{n}[/latex] , where n = 10, we write

quantity of Rn-222 left = [latex]1.0 g imes (\frac{1}{2} )^{10}[/latex]

[latex]=9.8 imes 10^{-4} g[/latex]

An alternative solution is to calculate the first-order rate constant from the half-life. Next, use Equation (13.3) to calculate the concentration of radon after 10 half-lives. Try it.

[latex]\ln \frac{[A]_t}{[A]_0}=-kt[/latex] (13.3)

Question 20.3: The half-life of Rn-222 is 3.8 days. Starting with 1.0 g of ...

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