lodine-125 (251) is used to treat, among other things, brain tumors and prostate
ID: 103366 • Letter: L
Question
lodine-125 (251) is used to treat, among other things, brain tumors and prostate cancer. It decays by gamma decay with a half-life of 54.90 days. Patients who fly soon after receiving 125I implants are given medical statements from the hospital verifying such treatment because their radiation could set off radiation detectors at airports. If the initial decay rate (or activity) was 532 Ci, what will the rate be after 3690 days? ACi How many months after the treatment will the decay rte be reduced to 13.5% of the original value? Assume months of 30 days monthsExplanation / Answer
1)
we have:
Half life = 54.9 days
use relation between rate constant and half life of 1st order reaction
k = (ln 2) / k
= 0.693/(half life)
= 0.693/(54.9)
= 1.262*10^-2 days-1
we have:
[I]o = 532.0 uCi
t = 369.0 days
k = 1.262*10^-2 days-1
use integrated rate law for 1st order reaction
ln[I] = ln[I]o - k*t
ln[I] = ln(532) - 1.262*10^-2*369
ln[I] = 6.2766 - 1.262*10^-2*369
ln[I] = 1.6188
[I] = 5.047 uCi
Answer: 5.05 uCi
2)
we have:
[I]o = 100 (Let 100 be initial decay rate)
[I] = 13.5 (final decay rate is 13.5 % of original)
k = 1.262*10^-2 days-1
use integrated rate law for 1st order reaction
ln[I] = ln[I]o - k*t
ln(13.5) = ln(100) - 1.262*10^-2*t
2.6027 = 4.6052 - 1.262*10^-2*t
1.262*10^-2*t = 2.0025
t = 154 days
t = 154/30 months
t = 5.13 months
Answer: 5.13 months
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.