1. Suppose a radioactive isotope has a half-life of tu/2 22 days. How long will
ID: 636228 • Letter: 1
Question
1. Suppose a radioactive isotope has a half-life of tu/2 22 days. How long will it take for a sample of the isotope to decay to 25% of its original activity? a. 0.0315 days b. 0.0437 days ?. 44 days d. 22 days e. 0.0630 days The next two questions (2 and 3) refer to a radioactive isotope for which t/2 92.6 min. 2. How long will it take for 5.44 g of sample to decay to 1.21 g? a. 0.016 min b. 200 min c. 812 min d. 3.3 min e. 104 min 3. If 55.6 mCi sample of the isotope decays for 60 min, how many mCi are left? d. 3.2 mCi e. 15.3 mCi a. 35.5 mCi b. 72 mCi c. 0.025 mCiExplanation / Answer
1)
Given:
Half life = 22 days
use relation between rate constant and half life of 1st order reaction
k = (ln 2) / k
= 0.693/(half life)
= 0.693/(22)
= 3.15*10^-2 days-1
we have:
[A]o = 100 (Let initial concentration be 100)
[A] = 25 (Remaining is 25 %)
k = 3.15*10^-2 days-1
use integrated rate law for 1st order reaction
ln[A] = ln[A]o - k*t
ln(25) = ln(1*10^2) - 3.15*10^-2*t
3.219 = 4.605 - 3.15*10^-2*t
3.15*10^-2*t = 1.386
t = 44.0 days
Answer: c
2)
Given:
Half life = 92.6 min
use relation between rate constant and half life of 1st order reaction
k = (ln 2) / k
= 0.693/(half life)
= 0.693/(92.6)
= 7.484*10^-3 min-1
we have:
[A]o = 5.44 g
[A] = 1.21 g
k = 7.484*10^-3 min-1
use integrated rate law for 1st order reaction
ln[A] = ln[A]o - k*t
ln(1.21) = ln(5.44) - 7.484*10^-3*t
0.1906 = 1.694 - 7.484*10^-3*t
7.484*10^-3*t = 1.503
t = 200 min
Answer: b
3)
Given:
Half life = 92.6 min
use relation between rate constant and half life of 1st order reaction
k = (ln 2) / k
= 0.693/(half life)
= 0.693/(92.6)
= 7.484*10^-3 min-1
we have:
[A]o = 55.6 mCi
t = 60.0 min
k = 7.484*10^-3 min-1
use integrated rate law for 1st order reaction
ln[A] = ln[A]o - k*t
ln[A] = ln(55.6) - 7.484*10^-3*60
ln[A] = 4.018 - 7.484*10^-3*60
ln[A] = 3.569
[A] = e^(3.569)
[A] = 35.5 mCi
Answer: a
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