Iodine-125 (125^|) is used to treat, among other things, brain tumors and prosta
ID: 966870 • Letter: I
Question
Iodine-125 (125^|) 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 125^| 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 522 muCi, what will the rate be after 389.0 days? How many months after the treatment will the decay rate be reduced to 11.5% of the original value? Assume months of 30 days.Explanation / Answer
1)
Radio active decay is a first order reaction.
For first order recation,
half life t1/2 = 0.693 /k where k is rate constant
k = 0.693/ t1/2 --- Eq (1)
k = 1/t ln { [A]o/[A]t} -----Eq (2)
From Eqs (1) and (2),
0.693/ t1/2 = (1/t) ln {[A]o/ [A]t} ------Eq (3)
Given that
half life of I-125 t1/2= 54.9 days
time t = 389 days
Initial decay rate of I-125 [A]o = 522 µCi
Final decay rate of I-125 [A]t = ?
Substitute all the values in Eq (3),
0.693/ t1/2 = (1/t) ln {[A]o/ [A]t}
0.693/54.9 days = (1/389 days) ln {522/ [A]t}
ln {522/ [A]t} = (0.693/54.9) x 389
On solving,
[A]t = 3.85 µCi
Final decay rate of I-125, [A]t = 3.85 µCi
Therefore,
decay rate after 389 days = 3.85 µCi
2) 0.693/ t1/2 = (1/t) ln {[A]o/ [A]t} ------Eq (3)
Given that
half life of I-125 t1/2= 54.9 days
time t = ?
Initial decay rate of I-125 [A]o = 100 %
Final decay rate of I-125 [A]t = 11.5 %
Substitute all the values in Eq (3),
0.693/ t1/2 = (1/t) ln {[A]o/ [A]t}
0.693/ 54.9 days = (1/t) ln {100/ 11.5}
t = (54.9/0.693) xln {100/ 11.5}
= 171.3 days
= 5.7 months
t = 5.7 months
Therefore,
after 5.7 months , decay rate will be reduced to 11.5 % of its original value.
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