The half-life of 131 I is 8.04 days. (a) Convert the half-life to seconds. s (b)
ID: 1745714 • Letter: T
Question
The half-life of 131I is 8.04 days. (a) Convert the half-life to seconds.s
(b) Calculate the decay constant for this isotope.
s-1
(c) Convert 0.46 µCi to the SI unitthe becquerel.
Bq
(d) Find the number of 131I nuclei necessaryto produce a sample with an activity of 0.46 µCi.
131I nuclei
(e) Suppose the activity of a certain 131I sample is6.90 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.
half-lives
mCi (a) Convert the half-life to seconds.
s
(b) Calculate the decay constant for this isotope.
s-1
(c) Convert 0.46 µCi to the SI unitthe becquerel.
Bq
(d) Find the number of 131I nuclei necessaryto produce a sample with an activity of 0.46 µCi.
131I nuclei
(e) Suppose the activity of a certain 131I sample is6.90 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.
half-lives
mCi
Explanation / Answer
The half-life of 131I is t = 8.04 days. = 8.04 * 24 * 60 * 60 = 694656 seconds = 8.04 * 24 * 60 * 60 = 694656 seconds Decay constant, = 0.693 / t = 0.693 / 694656 = 9.976 * 10 -7 sec -1 Convert 0.46 µCi to the SI unitthe becquerel = 0.46 * 10 ^ -6 Ci = 0.46 * 10 ^ -6 * 37 * 10 ^ 9 Bq = 17.02 * 10 ^ 3 Bq (d). Activity A = N So, No.of 131I nuclei necessary to produce a sample with anactivity of 0.46 µCi is N = A/ = 0.46 Ci /9.976* 10^ -7 i.e. N = 0.46 * 10 ^ -6 * 37 * 10 ^ 9 / ( 9.976 * 10 ^-7 ) = 1.70609 * 10 ^ 10 Convert 0.46 µCi to the SI unitthe becquerel = 0.46 * 10 ^ -6 Ci = 0.46 * 10 ^ -6 * 37 * 10 ^ 9 Bq = 17.02 * 10 ^ 3 Bq (d). Activity A = N So, No.of 131I nuclei necessary to produce a sample with anactivity of 0.46 µCi is N = A/ = 0.46 Ci /9.976* 10^ -7 i.e. N = 0.46 * 10 ^ -6 * 37 * 10 ^ 9 / ( 9.976 * 10 ^-7 ) = 1.70609 * 10 ^ 10Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.