Chem 481
D. Miller
1.
Name
Answer Key
24
Na emits a 1369-keV and a 2754-keV gamma ray in each of its
decays. Calculate the expected dose rate (mrad/hr) exactly two
feet from a 425-mCi 24Na point source.
(5)
2630 mrad/hr
How much lead shielding (thickness in cm) is required to reduce
the gamma dose from this source to 2.0 mrad/hr at two feet.
Assume that the half thickness of lead is 1.35 cm.
(7)
φ = 2.0 mrad/hr
φ0 = 2628 mrad/hr
μ = (ln 2)/1.35 cm = 0.513 cm-1
14 cm
2.
Short answer
Define each of the following term.
(3 ea)
coincidence losses -
rem -
These are signals in a counting system that are not
recorded because they occur so close in time to
another signal that some component of the sytem
does not have sufficient time to reset.
This is a unit of dose equivalent. 1 rem = 1 rad x QF where QF
(quality factor) accounts for differences in energy transfer by
various forms of radiation.
„erenkov radiation -
This is electromagnetic radiation emitted when a
charged particle moves through a medium faster
than the speed of light in that medium.
bremsstrahlung -
This is the emission of radiation (photons) when an
electron is accelerated in the electric field of a
nucleus.
Chem 481
D. Miller
2.
Name
Answer Key
(continued)
What are the major sources of natural background radiation for
the average person?
(5)
internal radioactivity (inhaled 222Rn and decay products, 40K, 14C)
building materials and soil
cosmic rays
If a worker at Chernobyl received 15,000 mrem of whole-body
exposure as a result of the accident, will this result in serious
health problems for this individual? Support your answer with
references to U.S. legal limits of exposure, natural exposure
levels and health effects at different exposure levels.
(6)
15,000 mrem = 15 rem
Although this is 3 times the annual legal occupational exposure permitted in the U.S. and
almost 100 times a typical annual natural exposure level, this dose is well below that
needed to produce a noticeable physical effect (~50 rem will produce a decrease in
white cell count). This worker is in no immediate danger, although his/her long-term
cancer risk might be slightly increased.
3.
A 15-mCi sample of 99Mo (t1/2 = 65.94 h) was adsorbed onto an ion
exchange column. 99Mo decays to 6.01-hour 99mTc. After 48.0 hours
the 99mTc is completely "milked out" (eluted) from the 99Mo ionexchange resin "cow". Based on this information, how many
millicuries of 99mTc were present in the eluate?
(10)
This is an example of transient equilibrium. At the time of elution
λD = (ln 2)/6.01 h = 0.115 h-1
λP = (ln 2)/65.94 h = 0.0105 h-1
10. mCi
Chem 481
D. Miller
Name
Answer Key
4.
For each of the following multiple-choice questions CIRCLE THE
LETTER preceding the correct answer. Only one answer will be
accepted for each.
(3 ea)
The range of a beta particle in air is typically how many times
larger than that for an alpha particle of the same energy.
A.
5
B.
50
C.
500
D.
5000
As the atomic number of the gamma-ray absorber increases,
A.
photoelectric absorption becomes more likely.
B.
Compton scattering becomes more likely.
C.
pair production becomes more likely.
D.
all of the above are true.
E.
none of the above are true.
As the gamma-ray energy increases,
A.
photoelectric absorption becomes more likely.
B.
Compton scattering becomes more likely.
C.
pair production becomes more likely.
D.
all of the above are true.
E.
none of the above are true.
Chem 481
D. Miller
5.
Name
Answer Key
A 137Cs gamma standard (1.06 :Ci on 11-15-80) was counted this
morning for 100 sec and a total of 25995 counts were observed.
The counter background was found to be 3730 counts in 200 sec.
Calculate the net count rate and the 1F uncertainty in this count
rate.
(5)
net count rate = 260 - 18.6 ± (1.62 + 0.312)½ = 241 ± 2 cps
241 ± 2 cps
Calculate the counting efficiency of this detector system.
(137Cs has a half life of 30.17 y and emits a gamma ray in 85.1%
of its decays.)
(5)
R = 241 cps and I = 0.851
t = 28 yr + 138 days = 28.38 yr
0.0139 (1.39%)
Chem 481
D. Miller
6.
Name
Answer Key
Briefly explain why the resolving time of a Geiger-Müller counter
is much larger than that for a gas-filled proportional counter.
(5)
The signal generated by a gas-filled proportional counter results from amplification (via a
Townsend avalanche) of the initial ion pairs produced by radiation. Because the gas
amplification is confined to a very small distance from a fine wire anode, and since only
the electrons are collected for the output signal, the output pulse has a fast rise time and
the detector has a short dead time and short resolving time. Gas amplification in a G-M
counter is not localized, but proceeds along the entire length of the anode. In order for
the detector to again produce detectable output pulses, the extensive positive ion sheath
must migrate to the cathode, a process that can take 100-300 microseconds. Thus, G-M
counters have a much longer resolving time than proportional counters.
Why is a halogen gas often incorporated into the gas filling of a
Geiger-Müller counter?
(4)
Br2 and Cl2 are added as quenchers to suppress the repeated discharge of the counter.
They limit the extent of secondary electron emission as cations are neutralized at the
cathode by becoming ionized and ultimately dissociated. Halogen atoms can then
recombine so the quencher lifetime is extended.
7.
Calculate the specific activity in MBq/mmol of a carrier-free
I-labeled sample. The half life of 125I is 59.4 days.
125
(4)
8.13 x 107 MBq/mmol
Explain why gamma radiation is so much more penetrating than
alpha or beta radiation.
(3)
Because gamma rays are chargeless and have zero rest mass, they have a specific
ionization that is much less than that of alpha and beta radiation and hence are able to
penetrate an absorber much further.
Briefly explain how a thermoluminescence dosimeter works.
(3)
A TLD consists of a small crystal of CaF2 or LiF. When radiation interacts with the
crystal it excites electrons in the crystal lattice to higher energy states, where they stay
trapped due to impurities in the crystal. Heating the crystal causes the electrons to drop
back to their ground state, releasing photons. This photon emission is measured and is
proportional to the amount of radiation exposure received by the crystal.