ACADs (08-006) Covered
Keywords
Description
Supporting Material
1.
(16 pts) Complete the following table regarding the characteristics of radiation.
A
C
D
Radiation
Type
Charge
Travel
Distance
α
+2
2-4 in.
β
+/-1
10-12 ft./
Mev
γ/x
0
∞
Pb, concrete water,
High Z material
n
0
∞
Water, concrete,
petroleum products,
Hydrogenous material
2.
3.
4.
Shielding
Material
Paper, dead skin,
anything
Al, Cu, wood, tin, Low Z
material
Hazard
(Internal/External/Both)
Internal
Both
External – lens of the
eye
Both
Predominantly internal
Both
Predominantly internal
The following notation results in what type of radiation being emitted?
A
Z
X ZA24X  _____
a.
b.
c.
d.
e.
alpha
negatron
positron
gamma/x-ray
neutron
The following notation results in what type of radiation being emitted?
A
Z
X  Z A1 X  _____
a.
b.
c.
d.
e.
alpha
negatron
positron
gamma/x-ray
neutron
The following notation results in what type of radiation being emitted?
A
Z
a.
b.
c.
d.
e.
X * ZAX  _____
alpha
negatron
positron
gamma/x-ray
neutron
B
5.
Equilibrium is said to exist when the daughter product is decaying faster than it is
being produced by the parent.
a. True
b. False
B
6.
Isomeric transition results in the formation of a new radioisotope.
a. True
b. False
B
7.
In nature, the three natural decay series exist in transient equilibrium.
a. True
b. False
B
8.
Most Compton electrons are inner shell electrons.
a. True
b. False
B
9.
Most photoelectrons are valence electrons.
a. True
b. False
C
10. As a rule of thumb, secular equilibrium is reach in __________ half-lives.
a.
b.
c.
d.
B
2
4
6
8
11. As a rule of thumb, transient equilibrium is reach in __________ half-lives.
a.
b.
c.
d.
2
4
6
8
12.
(9 pts) List and explain the methods by which charged particles can interact with matter and
transfer energy.
Ionization — The process of ionization involves the complete removal of orbital atomic
electrons as a result of the Coulomb force between the charged particle and the orbiting
electrons, which are themselves negatively charged. This process removes charge from a
neutral atom and so it becomes an ion. The combination of the removed electron and the
residual positive ion is called an ion pair, both of which can go on to produce additional
ionizations.
Excitation — The energy loss mechanism termed excitation is also a consequence of the
Coulomb force between the charged particle and atomic electrons. In this case, in contrast to
ionization, insufficient energy is transferred to the orbiting electron to break the electrical binding
force so the electron merely jumps up to a higher atomic energy level rather than leaving the
atom entirely. Note that in the case of excitation, the electron is still bound to the atom so that
the electrical neutrality of the atom is not disturbed. This process does NOT lead to the
formation of ion pairs and no free charge is released into the absorber. In most cases, the
excited electron will fall back to its original orbit and give up its excess energy in the form of
characteristic x-rays.
Bremsstrahlung — Bremsstrahlung is a word of German extraction that translates literally as
“braking radiation,” that is, radiation generated when the charged particle puts on the brakes.
The causative agent is again the Coulomb force, which produces a deflection in the path of the
charged particle. This change in direction is, in the physics sense, a negative acceleration (a
deceleration if you prefer) because the velocity vector changes with the directional change and
the speed of the particle is reduced due to the energy loss. The radiation emitted by the particle,
the bremsstrahlung, is electromagnetic in nature. It usually has an energy that puts it in the xray region of the overall electromagnetic spectrum.
13.
(10 pts) List and explain the energy transfer mechanisms for photonic radiation. Which of these
mechanisms is the most likely to occur and why?
Photoelectric Effect — The photoelectric effect is an all or nothing interaction. It either
happens or it doesn’t. The incoming photon transfers its full energy to an inner shell orbital
electron, almost always one in the K shell. The chance of an L shell electron participating is less
by a factor of 1000. The energized electron uses the excess energy to escape from the atom,
(i.e., it gives up the binding energy), and escapes the atom with the remaining energy given to it
by the photon. The photoelectric effect is probable only for photons with energies < 1 MeV.
Compton Scattering — Compton Scattering requires photon energies of 200 keV – 5 MeV. In
Compton scattering, the incoming photon transfers a portion of its energy to an orbital electron.
A lower energy photon then leaves, in a different direction, with the remaining energy.
Practically speaking, any orbital electron which has a binding energy less than about 10% of the
photon energy is available to interact by a Compton process. Thus virtually all the orbital
electrons except the two in the innermost K shell are available. (Generally, the K shell binding
energy is too high for them to “scatter”.)The interaction does not remove all the energy from the
incoming photon — a residual “Compton photon” is produced. The Compton electron is ejected
from the atom with the energy it receives minus the electron binding energy.
Pair Production — In the vicinity of the nucleus of an absorber atom, the incoming photon
suddenly disappears and in its place appears a pair of particles — an electron and a positron.
This pair have been created out of the pure energy carried by the photon. Einstein predicted this
possibility in his famous E = mc2 formula. If the entire rest mass of an electron is converted into
energy, 0.511 MeV is produced. The positron, being the anti-particle of the electron, also
“weighs” 0.511 MeV of mass-energy. Since fractions of particles do not exist, the minimum
amount of mass-energy that is needed to make the complete pair of particles is 0.511 MeV +
0.511 MeV = 1.022 MeV. This energy must be carried by the incoming photon, as a minimum, in
order for the interaction to proceed by the process of pair production. The term “threshold” is
applied to the process to indicate this requirement.
Compton Scattering is the most likely of the three interactions to occur due to its low threshold
energy, large energy band, and the fact that it does not interact with inner shell electrons.
14.
(2 pts) What minimum conditions must exist in order for a parent and daughter to be in secular
equilibrium?
The parent must have a half-life much longer than any other nuclide in the decay series.
A sufficiently long of time must have elapsed to allow for decay product in-growth.
15.
(6 pts) List and describe the methods by which thermal neutrons can interact with matter. As a
minimum, include in your discussion the nuclear formula that demonstrates the reaction, if
applicable.
Radiative Capture — The neutron is captured by the nucleus of an absorber atom, which then
leads to the emission of a photon (the capture gamma ray) to rid the atom of the excess energy.
A
Z
P  01n AZ1D *  AZ1D  
Charged Particle Emission — The neutron is captured by the nucleus of an absorber atom,
which then leads to the emission of a charged particle (proton, deuteron, alpha) to rid the atom
of the excess mass and energy.
A
Z
P  01n AZ1D*  AZ1D  particle ( p, d ,  )
Fission — involves the absorption of a thermal (low energy, slow speed) neutron by substances
such as U-235 or Pu-239. Such materials are said to be fissile. When the mass-energy of the
slow neutron is added to the fissile nuclide, the resultant nuclide is unstable and decays by
separating into two pieces, called fission fragments, which fly away from each other due to the
strong Coulomb repulsion. In the case of U-235, about 208 MeV of energy is released in the
process. Of the total, 83% goes to kinetic energy of the fission fragments, 6% is immediately
released as gamma rays and the remaining 11% is eventually released in the various
radioactive decays of the fission fragments.
16.
(4 pts) Define, and discuss the differences between, conversion and Auger electrons.
Conversion electrons are produced via the process of electron capture. In this type of
radioactive decay, the nucleus captures an orbital electron, almost always from the K shell.
Conversion electrons refer to orbital atomic electrons that are sometimes ejected from the atom
in place of the gamma ray that would normally be emitted. This process of internal conversion is
often thought of as a type of “internal photoelectric effect” in which the gamma ray is emitted by
the nucleus, but hits an orbital electron on its way out of the atom. The electron is ejected and is
then called a conversion electron to distinguish it from beta particles which might also be
emitted. The chief difference between a conversion electron and a beta particle is the fact that
internal conversion electrons are monoenergetic, whereas beta particles exhibit a spectrum of
energies.
Auger electrons are very similar to conversion electrons. Instead of a characteristic x-ray that
would normally be produced when an inner shell electron is captured by the nucleus, the atom
ejects one of its inner shell orbital electrons.
17.
(4 pts) Define and discuss the methods by which fast neutrons can interact with matter.
Elastic scattering — all the neutron energy remains in the form of kinetic energy of motion.
This is the familiar case of billiard balls bouncing off each other. The maximum energy transfer
in such collisions occurs when both objects that collide have the same mass. To most effectively
stop fast neutrons it is necessary to allow them to collide with target atoms having a nuclear
mass equal to the mass of a neutron. The best material is thus something rich in the element
hydrogen since the mass of a proton (hydrogen nucleus) closely approximates the neutron
mass. Shields of water, wax, concrete and various plastics are commonly used. In collisions
with the protons in these materials, the neutrons lose half of their energy, on the average, per
interaction. It turns out that elastic scattering is the mechanism responsible for about 80% of fast
neutron dose to human tissue.
Inelastic scattering — occurs if some of the collision energy is used to raise the target nucleus
into a higher nuclear energy level. Subsequently the nucleus will de-excite by emitting a
gamma. This is considered desirable, from a shielding standpoint, as the photon is easier to
attenuate than a high speed neutron.
18.
(2 pts) Define absorbed dose and list its units of measurement.
Absorbed dose measures the energy that is actually deposited in some given mass. It is
officially defined as the quotient of the deposited energy by the mass. It was originally defined
with the unit rad. One rad is the dose delivered to anything that receives 100 ergs of energy
deposited per gram of material. The new SI system unit is the joule per kilogram. In radiation
protection work, the “common unit” acknowledged by the SI system is the gray, abbreviated Gy.
19.
(2 pts) Define equivalent dose and list its units of measurement.
Modifying the absorbed dose by the quality factor produces a new quantity, the dose equivalent.
The formal definition is that “the dose equivalent,” is the product of the absorbed dose in tissue,
quality factor, and all other necessary modifying factors.” Note that the dose equivalent is
measured in tissue. Even though both a gray and a sievert are defined as 1 joule of deposited
energy per kilogram, note that the Sv definition includes the modifying factor Q so that the
numerical value of absorbed dose and dose equivalent will not be equal unless Q has the value
of 1. Note also that 1 Sv = 100 rem in the old system, and that Q is a dimensionless quantity
without units.
20.
(2 pts) Discuss how dose rate varies with distance from a plane source.
10 ft.
 @ < 1/d
2
 @ < 1/d
10 ft.
>10 ft.
When distance to plane source is small compared to longest dimension, exposure rate falls off a
little slower than 1/d (i.e. not as quickly as a line source.
As distance from plane source increases, exposure rate drops off at a rate approaching 1/d2.
21.
Perform the following unit analysis and conversion problems. SHOW ALL WORK!
a.
0.0025 mCi = ___2.5E3____________ nCi
2.5𝐸 − 3 𝑚𝐶𝑖
1 𝐶𝑖
1𝐸9 𝑛𝐶𝑖
2.5𝐸6 𝑛𝐶𝑖
(
)(
)(
)=
= 2.5𝐸3 𝑛𝐶𝑖
1𝐸3 𝑚𝐶𝑖
𝐶𝑖
1𝐸3
b.
32 pCi = ___1.184 Bq____________ Bq
32 𝑝𝐶𝑖
(
)(
c.
1 𝐶𝑖
3.7𝐸10 𝐵𝑞
1.184𝐸12 𝐵𝑞
)(
)=
= 1.184 𝐵𝑞
1𝐸12 𝑝𝐶𝑖
𝐶𝑖
1𝐸12
6.5E7 TBq = ____1.757E9________ Ci
6.5𝐸7 𝑇𝐵𝑞 1𝐸12 𝐵𝑞
𝐶𝑖
6.5𝐸19 𝐵𝑞
(
)(
)(
)=
= 1.757𝐸9 𝐶𝑖
𝑇𝐵𝑞
3.7𝐸10 𝐵𝑞
3.7𝐸10
d.
6.5E7 TBq = __3.942E21__________ dpm
6.5𝐸7 𝑇𝐵𝑞 1𝐸12 𝐵𝑞
𝐶𝑖
2.22𝐸12 𝑑𝑝𝑚
1.459𝐸32 𝑑𝑝𝑚
(
)(
)(
)(
)=
= 3.942𝐸21 𝑑𝑝𝑚
𝑇𝐵𝑞
3.7𝐸10 𝐵𝑞
𝐶𝑖
3.7𝐸10
e.
317,500 dpm = __1.43E-4_____________ mCi
3.175𝐸5 𝑑𝑝𝑚
𝐶𝑖
1𝐸3 𝑚𝐶𝑖
3.175𝐸8 𝑚𝐶𝑖
(
)(
)(
)=
= 1.43𝐸 − 4 𝑚𝐶𝑖
2.22𝐸12 𝑑𝑝𝑚
2.22𝐸12
22. (2 pts) A 600 mCi source emits two gammas—1099.2 keV (85%) 1291.6 keV (100%). What would
be the dose rate at 10 meters from the source? SHOW ALL WORK!
𝑰 = 𝟎. 𝟓𝑪𝑬𝑵
𝐼 = (0.5)(0.6)[(1.0992)(0.85) + (1.2916)(1)]
𝐼 = (0.5)(0.6)[(0.9343) + (1.2916)]
𝐼 = (0.5)(0.6)(2.226)
𝐼 = 0.6678 𝑟𝑒𝑚⁄ℎ𝑟 @ 1 𝑚
𝐼1 𝐷12 = 𝐼2 𝐷22
(0.6678)(1)2 = 𝐼2 (10)2
100𝐼2 = 6.678𝐸 − 1
𝐼2 = 6.678𝐸 − 3 𝑟𝑒𝑚⁄ℎ𝑟 @ 10 𝑚 = 𝟔. 𝟔𝟕𝟖 𝒎𝒓𝒆𝒎⁄𝒉𝒓 @ 𝟏𝟎 𝒎
23.
A 22-foot-long pipe containing radioactive liquid has a dose rate of 32 mrem/hr at 31 feet.
a.
(2 pts) What is the dose rate at the point at which the pipe becomes a point source?
SHOW ALL WORK!
𝐼1 𝐷12 = 𝐼2 𝐷22
(32)(31)2 = 𝐼2 (11)2
(32)(961) = 121𝐼2
3.075𝐸4 = 121𝐼2
𝐼2 = 𝟐𝟓𝟒. 𝟏 𝒎𝒓𝒆𝒎⁄𝒉𝒓 @ 𝟏𝟏 𝒇𝒕.
b.
(2 pts) What is the dose rate 6 inches from the pipe? SHOW ALL WORK!
𝐼1 𝐷1 = 𝐼2 𝐷2
(254.1)(1.32𝐸2) = 𝐼2 (6)
3.354𝐸4 = 𝐼2 (6)
𝐼2 = 𝟓. 𝟓𝟗𝑬𝟑 𝒎𝒓𝒆𝒎⁄𝒉𝒓 @ 𝟔 𝒊𝒏. = 𝟓. 𝟓𝟗 𝒓𝒆𝒎⁄𝒉𝒓 @ 𝟔 𝒊𝒏.
24.
Perform the following unit conversions. SHOW ALL WORK!
a.
1.15 µGy = __1.15E-10__ Mrad
1.15 𝜇𝐺𝑦
1 𝐺𝑦
1𝐸2 𝑟𝑎𝑑 1𝐸 − 6 𝑀𝑟𝑎𝑑
1.15𝐸 − 4
(
)(
)(
)(
)=
= 1.15𝐸 − 10𝑀𝑟𝑎𝑑
1𝐸6 𝜇𝐺𝑦
𝐺𝑦
𝑟𝑎𝑑
1𝐸6
b.
6.9 mCi = __2.553E-4__ TBq
6.9 𝑚𝐶𝑖
1 𝐶𝑖
3.7𝐸10 𝐵𝑞 1𝐸 − 12 𝑇𝐵𝑞
2.553𝐸 − 1𝑇𝐵𝑞
(
)(
)(
)(
)=
= 2.553𝐸 − 4 𝑇𝐵𝑞
1𝐸3 𝑚𝐶𝑖
𝐶𝑖
𝐵𝑞
1𝐸3
c.
7.5 rem = __𝟕. 𝟓𝑬𝟒 𝝁𝑺𝒗 __ µSv
7.5 𝑟𝑒𝑚
1 𝑆𝑣
1𝐸6 𝜇𝑆𝑣
7.5𝐸6𝑇𝐵𝑞
(
)(
)(
)=
= 7.5𝐸4 𝜇𝑆𝑣
1𝐸2 𝑟𝑒𝑚
𝑆𝑣
1𝐸2
25.
(2 pts) A nuclide has been on site for 4 months and has gone through 6 half-lives. What is its
half-life? SHOW ALL WORK!
𝑁𝑜. 𝐻𝑎𝑙𝑓 − 𝑙𝑖𝑣𝑒𝑠 =
𝐸𝑙𝑎𝑝𝑠𝑒𝑑 𝑇𝑖𝑚𝑒
Τ1/2
Τ1/2 =
𝐸𝑙𝑎𝑝𝑠𝑒𝑑 𝑇𝑖𝑚𝑒
𝑁𝑜. 𝐻𝑎𝑙𝑓 − 𝑙𝑖𝑣𝑒𝑠
Τ1/2 =
4 𝑚𝑜𝑠.
6
Τ1/2 =
4 𝑚𝑜𝑠.
6
Τ1/2 = 0.667 𝑚𝑜𝑠.
26.
(6 pts) A 6-foot-high pencil tank containing a 201Tl solution has a dose rate of 24 mrem/hr at
13 feet. How much 201Tl is contained in the tank? SHOW ALL WORK!
Dose rate at 3 ft.
𝐼1 𝐷12 = 𝐼2 𝐷22
(24)(13)2 = 𝐼2 (3)2
(24)(169) = 9𝐼2
𝐼2 =
4.056𝐸3
= 450.7 𝑚𝑟𝑒𝑚
9
Dose rate at 1 ft.
𝐼1 𝐷1 = 𝐼2 𝐷2
(450.7)(3) = 𝐼2 (1)
𝐼2 = 1.352𝐸3 𝑚𝑟𝑒𝑚⁄ℎ𝑟 = 1.352 𝑟𝑒𝑚⁄ℎ𝑟
Activity
𝐼 = 6𝐶𝐸𝑁
𝐶=
𝐼
6𝐸𝑁
𝐶=
1.352
6[(0.1674)(1) + (0.1353)(1)]
𝐶=
1.352
1.352
=
6(0.3027) 1.8042
𝐶 = 𝟎. 𝟕𝟒𝟗 𝑪𝒊