Problem#1
If a 6.13-g sample of an isotope having a mass number of 124 decays at a rate of 0.350 Ci, what is its half-life?
Answer:
The mass of a single nucleus is 124mp = 2.07 x 10-25 kg
│dN/dt│ = 0.350Ci = 1.30 x 1010 Bq
N = (6.13 x 103 kg)/(2.07 x 10-25 kg) = 2.96 x 1022
Then we use
│dN/dt│ = λN
1.30 x 1010 Bq = (2.96 x 1022)λ
λ = 4.39 x 10-13/s
so that
T1/2 = ln 2/λ = ln 2/(4.39 x 10-13/s) = 1.58 x 1012 s
T1/2 = 5.01 x 104 y
Problem#2
Radioactive isotopes used in cancer therapy have a “shelf-life,” like pharmaceuticals used in chemotherapy. Just after it has been manufactured in a nuclear reactor, the activity of a sample of 60Co is 5000 Ci. When its activity falls below 3500 Ci, it is considered too weak a source to use in treatment. You work in the radiology department of a large hospital. One of these 60Co sources in your inventory was manufactured on October 6, 2004. It is now April 6, 2007. Is the source still usable? The half-life of 60Co is 5.271 years.
Answer:
The amount of elapsed time since the source was created is roughly 2.5 years.
Can be written as
N = N0(2)-t/T1/2
The current activity is
N = (5000Ci)(2)-(2.5y)/(5.271 y) = 3000 Ci
Problem#3
The common isotope of uranium, 238U, has a half-life of 4.47 x 109 years decaying to 234Th by alpha emission. (a) What is the decay constant? (b) What mass of uranium is required for an activity of 1.00 curie? (c) How many alpha particles are emitted per second by 10.0 g of uranium?
Answer:
From the known half-life, we can find the decay constant, the rate of decay, and the activity.
Given: T1/2 = 4.47 x 109 y = 1.41 x 1017 s and 1Ci = 3.70 x 1010 decays/s
(a) the decay constant is
λ = ln 2/T1/2 = (ln 2)/(1.41 x 1017 s) = 4.92 x 10-18/s
(b) │dN/dt│ = λN
3.70 x 1010 decays/s = (4.92 x 10-18/s)N
N = 7.52 x 1027 nuclei.
The mass m of uranium is the number of nuclei times the mass of each one.
m = N(238mp) = (7.52 x 1027)(238)(1.67 x 10-27 kg) = 2.99 x 103 kg.
(c) N = (10.0 x 10-3 kg)/(238 mp) = 2.52 x 1022 nuclei
│dN/dt│ = λN = (4.92 x 10-18/s)(2.52 x 1022) = 1.24 x 105 Bq
Problem#4
Radiation Treatment of Prostate Cancer. In many cases, prostate cancer is treated by implanting 60 to 100 small seeds of radioactive material into the tumor. The energy released from the decays kills the tumor. One isotope that is used (there are others) is palladium (103Pd), with a half-life of 17 days.
If a typical grain contains 0.250 g of 103Pd, (a) what is its initial activity rate in Bq, and (b) what is the rate 68 days later?
Answer:
From the half-life and mass of an isotope, we can find its initial activity rate. Then using the
half-life, we can find its activity rate at a later time.
(a) λ = ln2/T1/2 = (ln 2)/[(17 days)(24 h/day)(3600 s/h)] = 4.7 x 10-7/s
N = (0.250 x 103 kg)/(103mp) = 1.45 x 1021
Then,
│dN/dt│ = λN = (4.7 x 10-7/s)(1.45 x 1021) = 6.8 x 1014 Bq
(b) 68 days is 4T1/2 so the activity is (6.8 x 1014 Bq)/(24) = 4.2 x 1013 Bq
Problem#5
A 12.0-g sample of carbon from living matter decays at the rate of due to the radioactive 14C in it.
What will be the decay rate of this sample in (a) 1000 years and (b) 50,000 years?
Answer:
the activity A = │dN/dt│ obeys the same decay
Given: A0 = 180.0 decays/min and T1/2 = 5730 y = 1.81 x 1011 s.
(a) for t = 1000 y,
λ = ln 2/T1/2 = (ln 2)/(5730y) = 1.21 x 10-4/y
we use
A = A0e-λt = (180.0 decays/min)e-(1.21 x 10-4/y)(1000y) = 159.5 decays/min
(a) for t = 50,000 y,
A = A0e-λt = (180.0 decays/min)e-(1.21 x 10-4/y)(50,000y) = 0.43 decays/min
Problem#6
Radioactive Tracers. Radioactive isotopes are often introduced into the body through the bloodstream. Their spread through the body can then be monitored by detecting the appearance of radiation in different organs. 131I, a β– emitter with a half-life of 8.0 d, is one such tracer. Suppose a scientist introduces a sample with an activity of 375 Bq and watches it spread to the organs. (a) Assuming that the sample all went to the thyroid gland, what will be the decay rate in that gland 24 d (about 3½ weeks) later? (b) If the decay rate in the thyroid 24 d later is actually measured to be 17.0 Bq, what percentage of the tracer went to that gland? (c) What isotope remains after the I-131 decays?
Answer:
The decay rate decreases by a factor of 2 in a time of one half-life.
(a) 24d is 3T1/2 so the activity is (375 Bq)/(23) = 46.9 Bq.
(b) The activity is proportional to the number of radioactive nuclei, so the percent is
17.0 Bq/46.9 Bq = 36.2%
(c) isotope remains after the I-131 decays is
131I53 → -1e0 + 131Xe54
The nucleus 131Xe54 is produced.
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