Problem#1
The unstable isotope 40K is used for dating rock samples. Its half-life is 1.28 x 109 y. (a) How many decays occur per second in a sample containing 1.63 x 10–6 g of 40K. (b) What is the activity of the sample in curies?
Answer:
Given: T1/2 = 1.28 x 109 y
(a) λ = (ln 2)/T1/2 = (ln 2)/( 1.28 x 109 y x 3.156 x 107 s/y) = 1.715 x 10-17/s
The mass of one 40K atom is approximately 40u, so the number of 40K nuclei in the sample is
N = (1.63 x 10–9 kg)/(40u) = (1.63 x 10–9 kg)/(40 x 1.66054 x 10-27 kg) = 2.454 x 1026.
Then,
│dN/dt│ = λN = (1.715 x 10-17/s)(2.454 x 1026) = 0.421 decays/s
(b) the activity of the sample in curies is
│dN/dt│ = (0.421 decays/s)(3.70 x 1010 decays/s) = 1.14 x 10-11 Ci
Problem#2
As a health physicist, you are being consulted about a spill in a radiochemistry lab. The isotope spilled was 500 µCi of 131Ba which has a half-life of 12 days. (a) What mass of 131Ba was spilled? (b) Your recommendation is to clear the lab until the radiation level has fallen 1.00µCi. How long will the lab have to be closed?
Answer:
The mass of one 131Ba nucleus is about 131 u.
(a) –dN/dt = 500 µCi = 500 x 10-6 x 3.70 x 1010 decays/s = 1.85 x 107 decays/s
λ = ln 2/T1/2 = (ln 2)/(12 x 86,400 s/d) = 6.69 x 10-7 s
dN/dt = -λN
1.85 x 107 decays/s = (6.69 x 10-7 s)N
N = 2.77 x 1013 nuclei
The mass of this many Ba nuclei 131 is
m = (2.77 x 1013 nuclei)(131 x 1.66 x 10-27 kg/nucleus) = 6.0 x 10-12 kg = 6.0 ng
(b) A = A0e-λt
1µCi = (500µCi)e-λt
ln (1/500) = –λt = – (6.69 x 10-7 s)t
t = 9.29 x 106 s = 108 days
Problem#3
Measurements on a certain isotope tell you that the decay rate decreases from 8318 decays/min to 3091 decays/min in 4.00 days. What is the half-life of this isotope?
Answer:
Given: A0 = 8318 decays/min, A = 3091 decays/min and t = 4 days.
A = A0e-λt
ln (A/A0) = –λt, while λ = ln2/T1/2, then
ln(A/A0) = –(ln2/T1/2)t
ln(3091/8318) = –(ln2/T1/2)(4.00 days)
T1/2 = 2.77 days/0.9899 = 2.8 days
Problem#4
The isotope 226Ra undergoes α decay with a half-life of 1620 years. What is the activity of 1.00 g of 226Ra. Express your answer in Bq and in Ci.
Answer:
1 mol of 226Ra has a mass of 226 gram. 1 Ci = 3.70 x 1010 Bq.
λ = (ln 2)/T1/2 = (ln 2)/(1620 y x 3.15 x 107 s/y) = 1.36 x 10-11/s
N = 1 g[6.022 x 1023 atoms/226g] = 2.665 x 1025 atoms
Then
│dN/dt│ = λN = (1.36 x 10-11/s)(2.665 x 1025) = 3.62 x 1010 Bq
│dN/dt│= (3.62 x 1010 Bq)(1 Ci/3.70 x 1010 Bq) = 0.98 Ci
Problem#5
The radioactive nuclide 199Pt has a half-life of 30.8 minutes. A sample is prepared that has an initial activity of 7.56 x 1011 Bq. (a) How many 199Pt nuclei are initially present in the sample? (b) How many are present after 30.8 minutes? What is the activity at this time? (c) Repeat part (b) for a time 92.4 minutes after the sample is first prepared.
Answer:
Given: │dN/dt│= 7.56 x 1011 Bq
(a) λ = (ln 2)/T1/2 = (ln 2)/(30.8 min x 60 s/min) = 3.75 x 10-4/s
│dN/dt│ = λN0
7.56 x 1011 Bq = (3.75 x 10-4/s)N0
N0 = 2.02 x 1015 nuclei.
(b) The number of nuclei left after one half-life is N0/2 = 1.01 x 1015 nuclei, and the activity is half:
│dN/dt│ = 3.78 x 1011 Bq
(c) After three half-lives (92.4 minutes) there is an eighth of the original amount, so N = 2.53 x 1014 nuclei, and an eighth of the activity:
│dN/dt│ = 9.45 x 1010 Bq
Problem#6
Radiocarbon Dating. A sample from timbers at an archeological site containing 500 g of carbon provides 3070 decays/min. What is the age of the sample?
Answer:
Given: λ = 1.21 x 10-4/y
The activity of the sample is
(3070 decays/min)/[(60 sec/min)(0.500 kg)] = 102 Bq/kg,
while the activity of atmospheric carbon is 255 Bq/kg (see Example 43.9). The age of the sample is then
t = –ln(102/225)/(1.21 x 10-4/y) = 7573 y.
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