a.) The half-life for the radioactive decay of Ce141 is 32.5 days. If a sample h
ID: 530774 • Letter: A
Question
a.) The half-life for the radioactive decay of Ce141 is 32.5 days. If a sample has an activity of 3.3 Ci after 162.5 d have elapsed, what was the initial activity, in microcuries, of the sample?
b.)As part of her nuclear stress test, Simone starts to walk on the treadmill. When she reaches the maximum level, Paul injects a radioactive dye with 201Tl with an activity of 74 MBq . The radiation emitted from areas of her heart is detected by a scanner and produces images of her heart muscle. After Simone rests for 3 h, Paul injects more dye with 201Tl and she is placed under the scanner again. A second set of images of her heart muscle at rest is taken. When Simone's doctor reviews her scans, he assures her that she had normal blood flow to her heart muscle, both at rest and under stress.
1b)If the half-life of 201Tl is 3.0 days, what is its activity, in megabecquerels after 3.0 days?
2b)What is its activity in megabecquerels after 12.0 days?
c.)At a distance of 34 ft , an ionizing radiation source delivers 5.5 rem of radiation. How close could you get to the source and still have no biological effects?
Explanation / Answer
a)
Radio active decay is a first order reaction.
For first order reaction,
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 Ce-141, t1/2 = 32.5 days
time t = 162.5 days
Initial activity of Ce-141, [A]o = ?
Final amount of Ce-141 [A]t = 3.3 Ci
Substitute all the values in Eq (3),
0.693/ t1/2 = (1/t) ln {[A]o/ [A]t}
[(0.693)/(32.5 days)] = (1/162.5 days) ln{[A]o/3.3 Ci }
Then,
[Ao] = 105.5 Ci
Therefore,
initial activity of Ce-141 = 105.5 Ci
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