How many days are required for 200. grams of radon-222 to decay to 50.0 grams?
(1) 1.91 days (3) 7.64 days(2) 3.82 days (4) 11.5 days

The answer is (3) 7.64 days. The remaining mass of radon-222 is one fourth of the original mass, so two half lives have elapsed (because one half time one half equals one fourth). Each half life of radon-222 is about 3.82 days, so the total time elapsed is 3.82 * 2 = 7.64 days

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Dejanrasid Dejanrasid · Answer 2

The answer is: 3) 7.64 days.

The most stable isotope of radon (Rn) is radon-222 with a half-life of 3.82 days.

After first half life (3.82 days) = 0.5 · 200 g.

After first half life (3.82 days) = 100 g of radon-222.

After second half life (7.64 days) = 0.5 · 100 g.

After second half life (7.64 days) = 50 g of radon-222.

Half-life is the time required for a quantity (in this example number of radioactive nuclei of radon-222) to reduce to half its initial value and is independent of initial concentration.