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?
Homework Answers
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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