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The activity of a sample of gas obtained from a basement containing radon-222 was found to...

The activity of a sample of gas obtained from a basement containing radon-222 was found to be 8 pCi/L. This isotope has a half-life of 3.8 days. If no additional radon-222 entered the basement, how long would it take for the activity to decline to 1 pCi/L?

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Answer #1

Given:

Half life = 3.8 days

use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k

= 0.693/(half life)

= 0.693/(3.8)

= 0.1824 days-1

we have:

[A]o = 8 pCi/L

[A] = 1 pCi/L

k = 0.1824 days-1

use integrated rate law for 1st order reaction

ln[A] = ln[A]o - k*t

ln(1) = ln(8) - 0.1824*t

0 = 2.079 - 0.1824*t

0.1824*t = 2.079

t = 11.4 days

Answer: 11.4 days

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