The half-life of 222 Rn is 3.83 days. (a) Convert the half-life to seconds. 1 s
ID: 1758691 • Letter: T
Question
The half-life of 222Rn is 3.83 days. (a) Convert the half-life to seconds.1 s
(b) Calculate the decay constant for this isotope.
2 s-1
(c) Convert 0.53 µCi to the SI unitthe becquerel.
3 Bq
(d) Find the number of 222Rn nuclei necessaryto produce a sample with an activity of 0.53 µCi.
4222Rn nuclei
(e) Suppose the activity of a certain 222Rn sample is6.30 mCi at a given time. Find the numberof half-lives the sample goes through in 40.2 d and the activity atthe end of that period.
5 half-lives
6 mCi (a) Convert the half-life to seconds.
1 s
(b) Calculate the decay constant for this isotope.
2 s-1
(c) Convert 0.53 µCi to the SI unitthe becquerel.
3 Bq
(d) Find the number of 222Rn nuclei necessaryto produce a sample with an activity of 0.53 µCi.
4222Rn nuclei
(e) Suppose the activity of a certain 222Rn sample is6.30 mCi at a given time. Find the numberof half-lives the sample goes through in 40.2 d and the activity atthe end of that period.
5 half-lives
6 mCi
Explanation / Answer
a)Half life: 3.83 days 3.83 days * 24 hours * 60 min * 60 s = 330912 s b) decay constant = 0.693 / T1/2 = 0.693 / 330912 s = 2.0942 x10-6 s-1 c) Given : 0.53 Ci = 0.53 x 10-6 Ci = 0.53 * x 10-6 * 3.7 x 1010 Bq = 19,610 Bq d) Weknow that
Numberof nuclei = Activity / decayconstant = 19610 / 2.0942 x10-6 = 9.363 x109 222Rn nuclei
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