External Radiation Exposure
Control
ACADs (08-006) Covered
Keywords
Description
Supporting Material
HPT001.017
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NUCLEAR TRAINING
TRAINING MATERIALS COVERSHEET
RADIOLOGICAL PROTECTION TECHNICIAN INITIAL TRAINING
PROGRAM
FUNDAMENTALS TRAINING
HPT001
COURSE
COURSE NO.
EXTERNAL RADIATION EXPOSURE CONTROL
HPT001.017
LESSON TITLE
LESSON PLAN NO.
INPO ACCREDITED
YES
X
NO
MULTIPLE SITES AFFECTED
YES
X
NO
PREPARED BY
Ralph G. Wallace/Brian K. Fike
PROCESS REVIEW
David L. Stewart
LEAD INSTRUCTOR/PROGRAM MGR. REVIEW
R. L. Coleman
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PLANT CONCURRENCE
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TVAN CONCURRENCE (If applicable)
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BFN
SQN
WBN
CORP
Receipt Inspection and Distribution:
Training Materials Coordinator
Standardized Training Material
Copies to:
TVA 40385 [NP 6-2003] Page 1 of 2
/
Date
HPT001.017
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NUCLEAR TRAINING
REVISION/USAGE LOG
Rev. #
Description of Changes
0
Initial Issue.
1
Program was inactive.
Reviewed and revised to reactivate.
2
Program was inactive.
Reviewed and revised to reactivate.
TVA 40385 [NP 6-2003] Page 2 of 2
Date
Pages Affected
Reviewed By
ALL
3/27/90
ALL
ALL
Ralph G. Wallace/
Brian K. Fike
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I.
PROGRAM: Radiological Control (RADCON) Individualized Instruction
II.
COURSE:
III.
TITLE:
IV.
LENGTH OF LESSON:
V.
TRAINING OBJECTIVES
A.
Fundamentals Training
External Radiation Exposure Control
16 hours
Terminal Objective
Upon completion of this module, participants will demonstrate knowledge and
understanding of external radiation exposure control and the methodology used to
calculate the exposures. A score of > 80% must be achieved on a written
examination.
B.
Enabling Objectives
Standards and conditions apply to all enabling objectives. They include the training
participant’s ability to utilize, under the examination ground rules (i.e. without the use
of training materials or outside assistance), the information presented in this lesson
plan
1.
Identify the three methods used to control radiation exposures.
2.
Define dose and dose rate and tell the difference between the two.
3.
Apply the equation defining the relationship between time, dose, and dose rate
to perform calculations of ‘stay time.’
4.
Define point and line sources and perform calculations using the Inverse
Square Law and the line source equation.
5.
Apply the equation DR = 6CE to calculate dose rates.
6.
Define the specific gamma ray constant and be able to use it in calculating
radiation exposures.
7.
Define ‘bremsstrahlung’ and tell its impact in shielding considerations.
8.
Identify the types and characteristics of materials that can best shield neutrons.
9.
List three (3) factors influencing the attenuation of photons as they pass
through a material.
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10.
Identify the differences among the following:
a.
‘linear attenuation coefficient,’
b.
‘mass attenuation coefficient,’
c.
‘energy absorption coefficient.’
11.
Apply the basic shielding equations to calculate exposure levels and shield
thicknesses.
12.
Define radiation ‘buildup’ and describe its impact on shielding considerations.
13.
Define
a.
‘half-value layer (or thickness)’
b.
‘tenth-value layer (or thickness)’
and use these concepts to calculate dose rates and shielding needs.
VI.
List the rule of thumb tenth-value layer values for lead, steel, concrete, and
water.
15.
Define ‘skyshine’ and describe its impact on shielding considerations.
TRAINING AIDS
A.
B.
C.
VII.
14.
Whiteboard with markers.
Networked computer and overhead projector.
Laser Pointer (Optional)
TRAINING MATERIALS:
A.
Appendices
1.
Handouts
a.
b.
c.
d.
e.
f.
HO-01 – Enabling Objectives
HO-02 – Gamma Energy & Occurrence and Specific Gamma Ray
Constant for Selected Radionuclides
HO-03 – Attenuation Coefficients
HO-04 – Mass Attenuation For Selected Elements (Graph)
HO-05 – Tenth and Half-Value Thicknesses
HO-06 – Equations
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B.
ATTACHMENTS
1.
Power Point Transparencies, Slide show located at P\Training\Technical
Programs and Services\Radcon\Initial Program\Lesson Plan Library\Power
Point Files
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
m.
n.
o.
p.
q.
r.
s.
t.
u.
v.
w.
x.
y.
z.
aa.
bb.
cc.
dd.
ee.
ff.
gg.
hh.
ii.
jj.
kk.
ll.
mm.
TP-01 - External Radiation Exposure Control
TP-02 - Enabling Objectives – 1
TP-03 - Enabling Objectives – 2
TP-04 - Enabling Objectives – 3
TP-05 - Enabling Objectives – 4
TP-06 – Radiation Exposure Control Methods
TP-07 – Dose & Dose Rate
TP-08 – Dose Example
TP-09 – Stay Time
TP-10 – Stay Time Solution
TP-11 – Types of Radiation Sources
TP-12 – Inverse Square Law
TP-13 – Inverse Square Law Calculations
TP-14 – Inverse Square Law Equation
TP-15 – Problem # 1
TP-16 – Problem # 1, Cont’d
TP-17 – Problem # 2
TP-18 – Approximation of Exposure for Gamma Emitters
TP-19 – Limitations of Equation
TP-20 – Problem # 3
TP-21 – Problem # 4
TP-22 – Problem # 4 Solution
TP-23 – Problem # 5
TP-24 – Problem # 5 Solution
TP-25 – Specific Gamma Ray Constant
TP-26 – Specific Gamma Ray Constant Example
TP-27 – Problem # 6
TP-28 – Problem # 7
TP-29 – Line or Parallel Source
TP-30 – Problem # 8
TP-31 – Problem # 9
TP-32 – Problem # 10
TP-33 – Problem # 10 Solution, Part 1
TP-34 – Problem # 10 Solution, Part 2
TP-35 – Shielding
TP-36 – Bremsstrahlung
TP-37 – Bremsstrahlung (Diagram)
TP-38 – Bremsstrahlung & Shielding
TP-39 – Attenuation
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TP-40 – Factors Affecting Attenuation of Photons
TP-41 – Attenuation Model
TP-42 – Linear Attenuation Coefficient
TP-43 – Shielding Equation
TP-44 – Problem # 11
TP-45 – Problem # 11 Solution
TP-46 – Problem # 12
TP-47 – Problem # 12 Solution
TP-48 – Total Linear Attenuation
TP-49 – Mass Attenuation Coefficient
TP-50 – Mass Attenuation Coefficient Graph
TP-51 – Problem # 13
TP-52 – Problem # 13 Solution
TP-53 – Buildup Factor
TP-54 – Buildup Factor Figure
TP-55 – Energy Absorption Coefficient
TP-56 − Energy Absorption Coefficient Equation
TP-57 – Problem # 14
TP-58 – Problem # 14 Solution
TP-59 – Quick Shielding Estimates
TP-60 – TVL & HVL for 1 MeV Photons
TP-61 – Rule of Thumb – TVL Nuclear Plant Environment
TP-62 – Number of Tenth-Value Thicknesses
TP-63 – Problem # 15
TP-64 – Problem # 16
TP-65 – Problem # 16 Solution
TP-66 – Problem # 17
TP-67 – Problem # 17 Solution
TP-68 – TVL/HVL Equations
TP-69 – Problem # 18
TP-70 – Problem # 19
TP-71 – Calculate Number of TVLs
TP-72 – Problem # 20
TP-73 – Problem # 21
TP-74 – Problem # 21 Solution
TP-75 – Shield Placement
TP-76 – Skyshine
TP-77 – Summary – 1
TP-78 – Summary – 2
TP-70 – Summary – 3
TP-80 – REMEMBER!
2.
(HO-07) Practice Problems (13 pages)
3.
(HO-08) Practice Problems – Solutions (13 pages)
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REFERENCES:
A.
ACAD 93-008, “Guidelines for Training and Qualification of Radiological Protection
Technicians,” National Academy For Nuclear Training, August 1993.
B.
Basic Radiation Protection Technology, 2nd Edition, Daniel A. Gollnick, Pacific
Radiation Corporation, Altadena, CA, 1988.
C.
Radiological Health Handbook, U.S. Department of Health, Education, and Welfare,
Public Health Service, Rockville, MD, 1970
D.
The Health Physics and Radiological Health Handbook, Bernard Shleien, Editor,
Scinta, Inc., Silver Spring, MD, 1992.
E.
Principles of Radiation Protection, Morgan and Turner, John Wylie & Sons, Inc.,
New York, 1967.
F.
Handbook of Mathematical Tables and Formulas, Richard Stevens Burlington,
Handbook Publishers, Inc., Sandusky, OH, 1958.
G.
NRC/NEU Exam Study Guide, RP-4, Radiation Protection, at,
http://www.nukeworker.com/study/hp/neu/index.shtml.
H.
The Southern Company, Farley Nuclear Plant Procedure RAD-30102C, Exposure
Rate Determination, August 2001.
I.
The Southern Company, Farley Nuclear Plant Procedure RAD-30304, Shielding
Fundamentals, August 2001.
J.
http://www.inform.umd.edu/Campusinfo/Departments/EnvirSafety/rs/material/tmsg/r
s6.html.
K.
http://www.ehrs.upenn.edu/training/unc-med/intro2.html
L.
www.paralog.com/wiki/?RadiationSafety
M.
http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html.
N.
http://en.wikipedia.org/wiki/Bremsstrahlung
O.
http://www.bennymak.com/Attenuation.htm.
P.
http://www.chattanoogastate,edu/classes/nm200/handouts/half%20value%20layer%2
0problems.doc.
HPT001.017
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Q.
Other web sites of potential interest:
1.
2.
3.
4.
http://www.eas.asu.edu/~holbert/eee460/gammashielding.pdf.
http://www.triumf.ca/safety/rpt/rpt.html.
http://epa.gov/radiation/understand/protection_basics.htm.
http://osha.gov/SLTC/radiationionizing/introtoionizing/slidepresentation/
slide5.html.
HPT001.017
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IX
INTRODUCTION:
Handout # 01
The operation of nuclear power facilities inherently involves some risk
of exposure to radiation and radioactive materials. Radiation
protection is concerned with the protection of individuals, their
progeny and mankind as a whole from these risks while still allowing
necessary activities from which radiation exposure might result. Other
lessons address the concept of maintaining exposures As Low As is
Reasonably Achievable (ALARA), monitoring for exposures to
radiation, and the physical measures which may be taken to reduce
radiation exposures. In this lesson, we will consider the procedures
used to calculate radiation exposure levels at various distances from
radiation sources and the impacts of shielding on the exposure levels.
TP-01
TP-02
TP-03
TP-04
TP-05
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X.
LESSON BODY
A.
There are potential hazards associated in working with
radioactive materials. Applying basic radiation control
measures can control the external dose which one receives.
The principal objective of radiation protection is to ensure that
the dose received by any individual is as low as reasonably
achievable (ALARA), while not exceeding the maximum
permissible limit. Any one, or a combination, of the following
methods may achieve this objective:
Reference L
1.
TP-06
2.
3.
B.
INSTRUCTOR NOTES
Limit the time of exposure. For illustrative purposes, a
person entering a relatively high radiation field of 1000
millirem/hr, but for only 30 seconds, would receive a
relatively low dose of 8 millirem.
By increasing the distance between the source of
exposure and an individual, the dose received can be
significantly reduced. When an individual doubles
his/her distance from a source, the dose will usually be
reduced by approximately three-fourths.
Absorbing material, or shields, can be incorporated to
reduce exposure levels. The specific shielding material
and thickness is dependent on the amount and type of
radiation involved. Lead, concrete, and steel are
examples of the types of materials used for shielding in
nuclear power plants.
Time
1.
Ask the students to
identify the 3
exposure control
methods.
Objective 1
Reference G
The concept of using time to limit radiation dose is
easy to understand if we think of what happens when
we attempt to get a suntan. Early in the season, we limit
the amount of time we spend in the sun to avoid a
severe sunburn. The same holds true for radiation
exposure. We have learned that radiation measurement
is usually specified as a rate, i.e., mrem per hour,
disintegrations per minute, etc. These; measurement
units explain the concept of time for radiation
protection. If we were to enter an area where the
radiation dose rate was 200 mrem per hour, we would
receive a dose of 200 mrem in one hour. If we limited
our time in this same area to only a 1/2 hour, our dose
would be 1/2 of 200 mrem or 100 mrem. We can see
that the longer a person remains in a radiation area, the
larger his dose.
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X.
LESSON BODY
2.
3.
INSTRUCTOR NOTES
Dose is the total amount of radiation absorbed. Dose
rate is the rate at which the radiation is absorbed. This
is usually specified as mrem per hour or depending on
the instrument, mrad per hour. If we look at a survey
instrument and it reads 50 mrem per hour, we know
that we are being exposed to a dose of 50 mrem in one
hour, 100 mrem in two hours, and 150 mrem in 3
hours. Limiting our time in the area can limit our
radiation dose.
Dose and Dose Rate
a.
Dose is the total amount of radiation absorbed.
Dose rate is the rate at which the radiation is
absorbed. This is usually specified as mrem per
hour or depending on the instrument, mrad per
hour. If we look at a survey instrument and it
reads 50 mrem per hour, we know that we are
being exposed to a dose of 50 mrem in one
hour, 100 mrem in two hours, and 150 mrem in
3 hours. Limiting our time in the area can limit
our radiation dose.
b.
If our survey instrument is sensitive to beta and
gamma radiation, its reading in mrem per hour
is equivalent to mrem per hour as the quality
factor for both of these radiations is 1.
c.
These concepts, dose and dose rate, help control
the dose an individual can receive while
working in a radiation area. There is a formula
that related dose and dose rate:
Dose = Time x Dose Rate
The radiation dose a person receives is equal to
the time he spends in the area multiplied by the
dose rate within the area.
Ask the students to
distinguish between
dose and dose rate.
Objective 2
TP-07
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X.
LESSON BODY
d.
INSTRUCTOR NOTES
Example:
The Instrumentation Department
needs to calibrate an instrument in an area
where the dose rate is 50 mrem per hour. They
believe it will take 2 hours to calibrate the
instrument. What will be the total dose the
worker receives?
Dose = Time x Dose Rate
Dose = 2 hours x 50 mrem/hour
Dose = 100 mrem
e.
The next step is to determine whether a dose of
100 mrem is a problem. If 100 mrem is below
plant administrative limits, then there is no
problem. If it is above the administrative limits,
then more than one person must be used to perform
the work. One person goes into the area, performs
the first part of the job, and leaves the are before
the administrative limits on absorbed dose are
exceeded. Another person performs the next
portion of the job, and this sequence continues
until the job is completed. this does not limit the
collective exposure required to complete the job; it
only guarantees that no individual exceeds the
administrative limits.
f.
Stay Time
The formula can be used to calculate the length of
time a person stays in an area (Stay Time) so he
does not exceed the administrative or federal limits
for the whole-body dose. If both sides of the
equation are divided by dose rate, the following
formula is used:
Dose
Time 
Dose Rate
(Equation # 1)
TP-08
Objective 3
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X.
LESSON BODY
g.
INSTRUCTOR NOTES
Now the length of time a person can remain in a
specific area without exceeding his allowable dose
limit (Stay Time) can be calculated.
Example:
The maintenance force needs to
TP-09
replace a filter in an area where the dose rate is 100
mrem per hour. The station administrative limits
specify that the individuals involved cannot exceed
a dose of 300 mrem per week. At least 8 hours will
be required to complete the filter changeout. How
long can personnel remain in the area and how
many people will be required to complete the job?
Dose
Time 
Dose Rate
Time 
300mrem
100mrem per hour
Time = 3 hours
h.
Each individual assigned to the job could remain in
the area for 3 hours without exceeding the
administrative limit. If 8 hours are required to
complete the job, we need:
1) 8 hours/3hours per individual = 2 and 2/3rds
people or three people to complete the job.
2) This assumes that only one individual will be
in the area at any one time. If it takes two
individuals working together, the total number
of individuals required to complete the job
doubles.
i.
This example demonstrates that planning is
essential to radiation protection. You should know
exactly what it is you will do while in the radiation
area, so that valuable time is not lost unnecessarily.
TP-10
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X.
LESSON BODY
j.
C.
INSTRUCTOR NOTES
If necessary, workers can practice on a mockup in
an unrestricted area so that they can work as
efficiently as possible when performing the actual
work.
Distance
Reference L
It is common sense to spend as little time as possible in areas
where you are exposed to radiation, it is also common sense to
stay as far away from a radiation source as possible. We can
demonstrate this by looking at a light bulb. If we are very close to
the bulb, the light appears very bright, as we move away, the
brightness of the light appears to be reduced. The same holds true
for radiation. The farther we are away from the source of
radiation, the less our exposure will be. The reduction in dose
depends on the type of radiation emitted and on the physical size
of the source itself.
There are four types of radiation we are concerned with in a
nuclear power station: alpha, beta, gamma, and neutron radiation.
Depending on the type of radiation, distance effects can vary
dramatically. An alpha particle, because it is a large heavy
particle, will interact very quickly, i.e., it has a high specific
ionization, and its total path length will be short. This is true in air
as well as in any other material medium.
For example, alpha radiation will only travel a few centimeters in
air. Beta radiation, because it is 1/1840th the mass of a proton,
will travel several meters in air. Gamma radiation can travel up to
a thousand meters or more in air. In a nuclear power station we
are primarily concerned with gamma radiation and we will limit
our discussion to gamma radiation concerns.
Dose rate reduction of gamma radiation depends on the relative
physical size of the object emitting the radiation. If the source is
physically small, the reduction in dose rate can be very large as
the distance increases. If, on the other hand, the source is very
large, the dose rate will not decrease as rapidly. A large source
such as a tank of radioactive water is called a plane source
because the radiation appears to come from a large plane not just
a single point.
TP-11
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X.
LESSON BODY
INSTRUCTOR NOTES
1.
Point Isotropic Source – A source of radiation which acts
as though all of its radiation were emitted from a point and
equally in all directions.
2.
Inverse Square Law
Reference M
Any point source which spreads its influence equally in all
directions without a limit to its range will obey the inverse
Objective 4
square law. This comes from strictly geometrical
considerations. The intensity of the influence at any given
radius r is the source strength divided by the area of the
sphere. Being strictly geometric in its origin, the inverse
square law applies to diverse phenomena. Point sources of
gravitational force, electric field, light, sound or radiation
obey the inverse square law. It is a subject of continuing
debate with a source such as a skunk on top of a flag pole;
will its smell drop off according to the inverse square law? TP-12
a.
This can be represented by the equation:
S
I = —————————
4 π r2
Where:
I = The intensity of the radiation at distance r
S = The source strength in photons/cm2-sec
π = 3.1416
r = The distance (cm) from the source to the
point of the intensity measurement.
b.
Now, let’s substitute some real numbers into the
formula:
First, assume a source strength of
106 photons/cm2-sec
Now, let’s look at the intensity at various distances
from the source:
S = photons/cm2-sec
Represents the
number of photons
on a 1 cm2 area on
the surface of a
sphere in one
second. 4 π r2
represents the
volume of the
sphere. Therefore,
the formula
calculates the
number of photons
on the fractional
area of the sphere at
distance r.
TP-13
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X.
LESSON BODY
INSTRUCTOR NOTES
Distance from Source
Photons/cm2-sec
10 cm (x)
20 cm (2x)
30 cm (3x)
40 cm (4x)
50 cm (?x)
c.
796
199 (1/4 * 796)
88 (1/9 * 796)
50 (1/16 * 796)
? (1/? * 796)
Ask the students to
supply the numbers
that go in place of
the ?.
5x, 32 (1/25 * 796)
We can see from the pattern that as the distance
increases by a factor of two, the intensity decreases
by the square of the distance. This is called the
“Inverse Square Law.” In general, the Inverse
Square Law may be represented by the equation:
TP-14
I1 * d12 = I2 * d22
I1
—
I2
or (Equation # 2)
d22
=
d12
—
or,
I1 * d12
I2 = ——————
d22
or
I2 = (I1 * d12)/ (d22)
(Equation # 3)
Where:
I1 = Dose equivalent rate at distance d1 from the
source.
I2 = Dose equivalent rate at distance d2 from the
source.
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X.
LESSON BODY
INSTRUCTOR NOTES
I1 and I2 can be expressed as any time rate that
measures radiation intensity at a point, such as
dps, dpm, cpm, R/hr, mR/hr, etc., and d1 and d2
can be expressed as any distance (cm, in, ft, m,
etc.) so long as both distances are expressed in the
same units. You should always use the SelfChecking Error Prevention Tool to ensure that all
units in an equation are compatible. Be especially
careful with R and mR and with m and cm.
d.
Problem # 1
We have a Ra-226 source that produces a dose of
10,000 µR/hr at 1ft. We want to use this source to
calibrate a pressurized ionization chamber. What
will be the source strength at distances 10, 20, 25,
30, and 40 feet from the source?
I1 * d12
Using the equation: I2 = ——————
d22
Where:
I1 = 10,000 µR/hr
d1 = 1 ft
For d2 = 10 ft, we have:
(10,000µR/hr) * (1 ft2)
I2 = ——————————
(10 ft)2
or I2 = (10,000 µR/hr)/100 or,
I2 = 100 µR/hr at 10 feet
Instructor Note:
Stress Error
Prevention Tools
Self-Checking
S top
T hink
A ct
R eview.
TP-15
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X.
LESSON BODY
e.
INSTRUCTOR NOTES
Using this same procedure to calculate the
exposure rate at the other distances, we have:
d2
—
10 feet
20 feet
25 feet
30 feet
40 feet
f.
TP-16
I2
—
100 µR/hr
25 µR/hr
16 µR/hr
11 µR/hr
6.25 µR/hr
Problem # 2
TP-17
A source reads 125 rem/hr at 1 foot. How far
from the source must you get to lower the dose
equivalent to 1 rem/hr?
Given:
I1
I2
d1
d2
=
=
=
=
125 rem/hr
1 rem/hr
1 foot
?
I1 * d12 = I2 * d22
(125 rem/hr) * (1 ft) 2 = (1 rem/hr) * (d2 ft) 2
d22 = 125 ft2
d2 = 11.2 ft
NOTE: Again, remember to be unit consistent
when performing these calculations. Feet or
inches, mrem, rem, or rad must all be the same
throughout the calculations. Also, measurements
are always taken from the source to the point of
measurement. REMEMBER YOUR ERROR
PREVENTION TOOLS!
Instructor note:
Again stress the
use of the Error
Prevention Tools:
Self-Checking
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X.
LESSON BODY
3.
INSTRUCTOR NOTES
An approximation of the radiation exposure level (R/hr)
at one foot from a point source can be made by using the
6 CEn equation, or
Handout-02
DR = 6 CEn
Objective 5
(Equation # 4)
TP-18
Where:
DR = Dose Rate, R/hr at 1 foot from a point source
C = Activity of the source in Curies
En = Total effective gamma ray energy (Mev) per
disintegration.
Note: This formula is to be used as an approximation
only (within ± 20%) and can be used only for x-rays and
gamma energies between 0.07 and 2 MeV. The gamma
ray energies are available in many reference sources,
including Reference C. Selected values are presented in
Handout # 02.
TP-19
a.
TP-20
Problem # 3
Determine the exposure rate from a point source
containing 10 curies of Cesium-137.
DR = 6 CEn
C = 10 Ci
En = 0.66 MeV (From Handout # 02)
DR = 6 * 10 * 0.66 = 39.6 R/hr at 1 foot
b.
Problem # 4
TP-21
Determine the exposure rate 12 feet from a point
source containing 50 curies of cobalt-60.
C = 50 Ci
En = 1.17 + 1.33 = 2.50 (Handout # 02)
DR = 6 * 50 * 2.5 = 750 R/hr at 1 foot
TP-22
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X.
LESSON BODY
INSTRUCTOR NOTES
I2 = (I1d12)/d22 = (750 R/hr * 1 ft2)/(12 ft)2
I2 = 750 R/hr-ft2/144 ft2 = 5.2 R/hr at 12 ft.
c.
Problem # 5
TP-23
Determine the exposure rate from a point source
containing 2.5 curies of iron-59.
C = 2.5 Ci
Fe-59 has 4 gamma emissions as shown below:
0.143 MeV emitted 1.0% of the time
0.192 MeV emitted 3.1% of the time
1.099 MeV emitted 56% of the time
1.292 MeV emitted 43% of the time, therefore:
En = (0.143 * 0.01) + (0.192 * 0.03) +
(1.099 * 0.56) + (1.292 * 0.43), or
TP-24
En = 1.18 MeV
4.
DR = 6 * 2.5 * 1.18 = 17.7 R/hr at 1 foot
Determining the dose rate for a specific gamma-emitting
radionuclide of known radioactivity content.
The gamma exposure rate in R/hr at 1 cm from a 1 mCi
source is specific for each gamma-emitting radionuclide.
This value, known as the specific gamma ray constant or
the specific gamma ray dose constant ( Γ ), has been
tabulated for many radionuclides and is presented in
Reference C, as well as in other publications. The data
for selected radionuclides are presented in Handout # 02.
a.
The specific gamma ray constant ( Γ ) for a
nuclide emitting gamma radiation, is defined as
the product of exposure rate at a given distance
from a point source of that nuclide and the square
of the distance divided by the activity of the
source, or
Γ = R – cm2 / h – mCi, or
Objective 6
TP-25
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a more useful form of the formula is:
Γ/10 = R/hr at 1m for each curie of activity.
b.
These values can be used to determine the dose
rate when the activity is known. Let’s look at
some examples.
1) What is the exposure rate at 1 meter from a
1 Ci Ra-226 source?
TP-26
From Handout # 02 we find that the specific
gamma ray constant for Ra-226 is
8.25 R – cm2 / h – mCi
Dividing this value by 10 gives
0.825 R/hr at 1m for each curie of activity,
and since we have 1 Ci of Ra-226, the dose
rate for this source is:
0.825 R/hr at I meter.
2) Problem # 6
Determine the exposure rate from 2 Ci of
Ra-226 at a distance of 5 meters from the
source.
From the example above, we have:
DR = 0.825 R/hr at 1 m/Ci, or
DR = 1.65 R/hr at 1 m for 2 Ci
TP-27
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Now, using the Inverse Square Law, we can
determine the dose rate at 5 meters:
I1 * d12
Using the equation: I2 = ——————
d22
(1.65 R/hr) * (1 m)2
I2 = ——————————
(5 m)2
I2 = (1.65 R - m2/hr)/25 m2, or
I2 = 0.066 R/hr at 5 meters for 2 Ci
3) Problem # 7
Determine the exposure rate 6 meters from
3 Ci of Co-60.
From Handout # 02 for Co-60,
Γ/10 = 1.32 R/hr at 1m for each curie of
activity.
For 3 Ci, Γ/10 = 3.96 R/hr at 1m
Using the Inverse Square Law,
I2 = (I1d12)/d22 =
I2 = (3.96 R/hr) * (1 m)2/(6 m)2
I2 = (3.96 R-m2/hr)/(36m2), and
I2 = 0.11 R/hr at 6 meters for 3 Ci
TP-28
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5.
INSTRUCTOR NOTES
Determining the exposure rate from a line or parallel
source.
a.
b.
Not all radiation sources are point sources. We
must consider other types of sources such as
pipes and storage tanks to make better
approximations for adequate radiation protection.
Some sources are considered parallel or line
sources, in that the longest dimension of the
source is used for calculation. A direct linear
relationship is used out to a distance of one-half
the longest dimension of the source:
The formula for calculating this decrease is:
I1d1 = I2d2
(Equation # 5)
Where:
I1 = dose rate at perpendicular distance d1
I2 = dose rate at perpendicular distance d2
c.
The formula may also be written as:
I2 = I1d1/d2
d.
(Equation # 6)
One can see that the dose rate decreases as
distance increases in a linear fashion. The line
source formula is applicable for use when both d1
and d2 are less than or equal to one half the
length of the line source (L/2). From the point
where the distance exceeds L/2, the Inverse
Square Law applies.
Objective 4
Ask the students
for an example of
a line or parallel
source.
Answer: Pipe
TP-29
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e.
INSTRUCTOR NOTES
Problem # 8
TP-30
A survey meter reading at a distance of 2 feet
from a 20-foot section of pipe is 100 mrem/hr.
What is the dose equivalent rate at a distance of 4
feet from the pipe?
Using the formula, I2 = I1d1/d2
I2 = (100 mrem/hr)(2 ft)/(4 ft), or
I2 = 50 mrem/hr
f.
Problem # 9
TP-31
A pipe passing through a room where some
maintenance work is to be done reads 2 R/hr on
contact. How far away should the workers stay to
avoid a dose rate of 200 mR/hr (0.2 R/hr)?
1)
Assume the contact reading was made at
1 inch from the surface of the pipe.
2)
Using the formula
I1d1 = I2d2
or
d2 = I1d1/ I2
then
d2 = (2 R/hr)(1 in)/0.2 R/hr
d2 = 10 inches
or
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g.
INSTRUCTOR NOTES
Problem # 10
A storage tank measuring 6 ft. long and 2 ft. in
diameter measures 15rem/hr at 1 ft. What will be
the dose rate at 20 ft?
Reference H
TP-32
NOTE: This problem must be completed in two
steps.
1)
Determine the dose rate at distance L/2 using
the formula.
I2 = I1d1/d2
I2 = (15 rem/hr)(1 ft)/(3 ft), or
I2 = 5 rem/hr at 3 feet.
TP-33
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2)
Determine the dose rate at 20 feet using the
Inverse Square Law formula:
TP-34
I2 = (I1 * d12)/ (d22)
I2 = (5 rem/hr)(3 ft)2 (20 ft)2
I2 = 45 rem-ft2/400 ft2, or
I2 = 0.1125 rem/hr or 112.5 mrem/hr
D.
Shielding
Reference J
1.
When reducing the time or increasing the distance may not
be possible, one can choose shielding material to reduce
the external radiation hazard. The proper material to use
depends on the type of radiation and its energy.
TP-35
2.
Alpha particles are easily shielded. A thin piece of paper
or several cm of air is usually sufficient to stop them.
Thus, alpha particles present no external radiation hazard.
Beta particles are more penetrating than alpha particles.
Beta shields are usually made of aluminum, brass, plastic,
or other materials of low atomic number to reduce the
production of bremsstrahlung radiation. Appendix IV
gives the range of beta radiation for selected radionuclides
in air and plastic.
a.
b.
Bremsstrahlung, German for braking radiation, is
electromagnetic radiation produced by the
acceleration of a charged particle, such as an
electron, when deflected by another charged
particle, such as an atomic nucleus. The term is
also used to refer to the process of producing the
radiation. Bremsstrahlung has a continuous
spectrum.
TP 36
TP-37
Bremsstrahlung is a type of "secondary radiation,"
in that it is produced as a reaction in shielding
material by the primary radiation (beta particles).
In some cases, the bremsstrahlung produced by
some sources of radiation interacting with some
types of radiation shielding can be more harmful
than the original beta particles would have been.
Reference K
Objective 7
TP-38
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3.
Neutrons are best shielded by a material consisting of light
elements, such as polyethylene, paraffin, and water.
Neutrons lose more energy when interacting with light
elements, such as hydrogen, than with heavier elements,
such as lead. The highly energetic neutrons in the plant are
almost exclusively in the reactor. Because the reactor is
shielded with water, neutrons lose their energy in
collisions with light atoms in the water and, therefore, are
generally contained in or around the reactor.
4.
Gamma radiation is the most difficult to shield against
and, therefore, presents the biggest problem in the plant.
The penetrating power of the gamma is due, in part, to the
fact that it has no charge or mass. Therefore, it does not
interact as frequently as do the other types of radiation.
The three methods of gamma interaction all involve
interactions near the nucleus or interactions with the
electrons around the nucleus. For this reason, more
gamma interactions occur in a dense material that has
many electrons. One such material is lead. Lead is very
dense and a lead atom has 82 electrons. Thus, a gamma
would interact more times in passing through 8 inches of
lead than in passing through the same thickness of a light
material, such as water. As with other types of radiation,
gamma radiation loses energy as it interacts. When it has
lost all of its energy, it disappears and is no longer a
problem.
5.
When a beam of X-ray or gamma photons passes through
an object, interactions occur that result in a decrease of the
number of transmitted (non-interacting) photons. This
process is called attenuation, which is defined as the
lessoning of the amount, force, magnitude or value of:
weakening: a reduction in the severity, vitality or
intensity of…
INSTRUCTOR NOTES
Objective 8
Reference O
TP-39
TP-40
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There is no specific value for the range of a photon in
matter; rather, there is a probability that a photon will
interact within a specified thickness of absorbing material.
This probability is called the attenuation coefficient, u.
The value of u is primarily a function of the energy of the
photons and the nature of the absorber. Denser absorbers
are better for photons; lead is the material of choice. The
ratio I/Io is the fraction of photons remaining after a beam
passes through an absorber of thickness x.
There are a number of factors that can affect the
attenuation of the photons such as:
6.
a.
The energy of the radiation. For example, a 10
Mev gamma will travel farther than a 2 MeV
gamma.
b.
The material of which the shield is made. A
material with many orbital electrons, like lead is
also a heavy, dense material. Water is also used
for gamma shielding, but there are not as many
orbital electrons available, so it is not as efficient
as a shield of lead, that is the gamma radiation will
travel further through the material before it
interacts with an atom.
c.
The thickness of the absorber will also impact the
distance the gamma radiation will travel before
interacting with the material.
If a narrow beam of single-energy photons is directed
through a substance, energy will be removed from the
beam by Compton scatter, photoelectric effect and pair
production. Since we assume a narrow beam, a Compton
scattering process will be the equivalent of a photoelectric
absorption; that is, the scattered photon is also removed
from the beam. Let us measure the relative intensity (I/IO)
of the beam as it passes through various thicknesses of an
absorber.
Objective 9
TP-41
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X and gamma ray absorption is an exponential process,
e.g., there is a constant fractional decrease in intensity per
unit thickness of the substance. The value of this constant
is denoted by Mu, (), and is called the total linear
attenuation coefficient. Consider a narrow beam of
single-energy photons incident upon a thin slab of uniform
material with thickness dx as seen in Figure 1. If a
number ‘I’ of photons are incident on the material and
have a probably of interaction  per centimeter, then the
number of photons which have not interacted, (the
intensity of the beam that penetrates the shield) is given by
the equation:
I = I0e-µx,
where
TP-42
Handout # 03
Objective 10
TP-43
(Equation # 7)
Objective 11
I0 =
I =
e =
µ =
x =
The intensity of the radiation entering the shield.
The intensity of the gamma radiation emerging
from the shield.
Base of natural logarithms
The total linear attenuation coefficient.
The thickness of the shielding material.
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a.
INSTRUCTOR NOTES
Problem # 11
TP-44
The exposure rate from a 1 MeV gamma source is
500 mR/hr. You package the source in a container
with 2 inches of lead around the source. What is
the exposure rate outside the package?
Using equation # 7, I = I0e-µx, where
I0 = 500 mR/hr
x = 2 inches (or 2.54 cm/in * 2 in = 5.08 cm) of
lead
Determine µ from Handout #03
(1 MeV and 5.08 cm)
µ = 0.804 cm-1
Then, I = (500 mR/hr) (e-(0.804)(5.08))
I = (500 mR/hr) (e-4.08), or
I = (500 mR/hr) (0.0168)
I = 8.42 mR/hr
TP-45
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b.
INSTRUCTOR NOTES
Problem # 12
TP-46
What thickness of water is needed to reduce a
1 MeV gamma dose rate from 100 mR/hr to
10 mR/hr?
Again using equation # 7, but rearranging it to
solve for x, we have,
I = I0e-µx , or I/ I0 = e-µx
TP-47
Taking the natural logarithm of each side yields,
ln (I/ I0) = ln e-µx, or
ln (I/ I0) = -µx, and
x = [ln (I/ I0)]/-µ
(Equation # 8)
Now,
I0 = 100 mR/hr
I = 10 mR/hr
µ = 0.0707 cm-1
x = ln(10/100)/-0.0707
x = ln(0.1)/-0.0707, or
x = (-2.303)/-0.0707
x = 32.57 cm
TP-48
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Figure 2 Total Linear Attenuation Coefficients For Lead
c.
The total linear attenuation coefficient (µ) is
defined as the sum of the probabilities of
interaction per unit path length by each of the three
scattering and absorption processes, photoelectric
effect, Compton effect, and pair production. The
dimensions of µ are inverse length (for example,
cm-1). The reciprocal of µ is defined as the mean
free path which is the average distance the photon
travels in an absorber before an interaction takes
place.
Reference I
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d.
Because linear attenuation coefficients are
proportional to the absorber density, which usually
does not have a unique value but depends
somewhat on the physical state of the material, it is
customary to use "mass attenuation coefficients"
which removes density dependence:
Objective 10
e.
The mass attenuation coefficient (µm) is the linear
attenuation coefficient divided by the density of
the material (ρ).
TP-49
µm = µ/ρ
Handout # 04
For a given photon energy, µm does not change
with the physical state of a given absorber. For
example, it is the same for water whether present
in liquid or vapor form. If the absorber thickness is
in cm, then µm will have units of:
Note: This graph
can be used to
obtain coefficients
and will be used in
Problem # 13.
cm-1/(cm2/g)
Values for mass attenuation coefficient and density
for selected materials are given in Handout # 03.
Substituting into equation # 7 gives:
I = I0e-(µm)(ρ) x,
where
(Equation # 9)
Rearranging the equation to solve for x yields
x = [ln (I/ I0)]/-(µm)(ρ)
(Equation # 10)
µm = Mass attenuation coefficient (cm2/g)
ρ = Density, g/cm3,
and
µm * ρ = µ
TP-50
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INSTRUCTOR NOTES
Problem # 13
TP-51
A source is to be shipped in a wooden box. The
gamma reading at the surface of the box is 1 R/hr.
What thickness of lead lining is required to reduce
the exposure rate at the surface of the box to 2
mR/hr if the energy level is 0.66 MeV? Use the
mass attenuation coefficient from Handout # 04.
TP-52
Using equation 10,
x = [ln (I/ I0)]/-(µm)(ρ)
x = [ln (2 mR/hr ÷ 1,000 mR/hr)]/-(0.105cm2/g)
(11.35 g/cm3)
x = ln(0.002)/-1.192 cm-1
x = -6.2115/-1.192 cm-1
x = 5.21 cm
7.
If a beam of gamma rays is projected at a lead wall, we
find that the intensity of the beam decreases exponentially
over the width of the wall. This is due to the gamma
interactions occurring in the wall. However, some gamma
interactions, such as pair production and Compton
scattering, result in other gamma rays of lower energy
being given off. For this reason, the actual decrease in the
intensity of a beam of gamma rays passing through a wall
is less than the theoretical decrease. There is more
buildup of scattered radiation through an iron shield than a
lead shield. For relatively thick shields of iron and steel,
the scattered radiation is a greater contributor to the total
dose rate than is the uncollided flux which penetrates the
shield. This increase in the intensity of the exiting beam is
called the “buildup,” and is taken into account in the
following formula by the Buildup Factor, “B.”
TP-53
Objective 12
TP-54
Reference I
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I = BI0e-µx,
where
B =
I0 =
Buildup Factor,
The intensity of the radiation entering the
shield.
The intensity of the gamma radiation
emerging from the shield.
Base of natural logarithms
The linear attenuation coefficient.
The thickness of the shielding material.
I =
e =
µ =
x =
(Equation # 11)
In this case, one can calculate the true intensity I if the
value of the buildup factor is known. The buildup factor
corrects for the underestimation of scattered radiation
reaching the detector. Tables of buildup factors for a
number of materials can be found in the
References C and D.
8.
a.
Buildup factors are determined by a series of
involved computer calculations. Tabulations of
buildup factors for some situations can be found in
various references, including References C and D.
b.
The application of these factors can be quite
complicated and are beyond the scope of this class.
A third equation that is often used to determine the
intensity of radiation passing through a shield is:
I = I0e-µex,
I0 =
I =
e =
µe =
x =
where
(Equation # 12)
The intensity of the radiation entering the
shield.
The intensity of the gamma radiation emerging
from the shield.
Base of natural logarithms
The energy absorption coefficient.
The thickness of the shielding material.
This equation provides an approximation of the values that
would be obtained by the use of Equation # 11.
Reference E
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a.
The total energy absorption coefficient, for a
parallel beam or non-point source of specified
radiation, µe, represents the fraction removed by
attenuation in passing through a thin layer of a
substance, or the fraction of incident radiant
energy absorbed per unit thickness of an absorber.
The sum of the absorption coefficient and the
scattering coefficient is the attenuation coefficient.
It is a function of the energy of the radiation and is
expressed as cm-1. The coefficient may also be
given as the mass absorption coefficient (µe/ρ).
TP-55
b.
Values for µe are not widely available; therefore
they are not presented in this lesson plan.
However, Problem # 14 does involve the use of the
energy absorption coefficient.
TP-56
c.
Problem # 14
TP-57
Given a box containing a non-point parallel source
of Radium-226 with an exposure rate of 0.75 R/hr
and a 0.8 MeV gamma. Determine the amount of
lead required to reduce the box surface reading to
2 mR/hr.
Using Equation # 12, and reordering it to solve for
x,
x = [ln (I/ I0)]/-µe
Where µe = 0.5727 cm-1
x = [ln (2 mR/hr ÷ 750 mR/hr)]/-0.5727 cm-1
x = ln(0.00267)/- 0.5727 cm-1
x = (-5.926)(-0.5727 cm-1)
x = 10.35 cm
TP-58
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9.
INSTRUCTOR NOTES
For quick shielding estimates, we can use multiples of
half-value or tenth-value thicknesses. One half-value
thickness (or layer, HVL) is the thickness of material
required to reduce the photon intensity to 1/2 the initial
value, assuming no buildup. One tenth-value thickness (or
layer, TVL) is the thickness of material required to reduce
the photon intensity to 1/10 the initial value, assuming no
buildup.
TP-59
a.
Reference K
Both the tenth and half-value thicknesses are
dependent upon the energy of the photon and the
material it passes through. The following table
lists tenth and half-value thicknesses of 1 Mev
gamma rays for various materials.
Objective 13
Reference I
TP-60
Handout # 04
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Table 1
TENTH AND HALF-VALUE THICKNESSES FOR 1 Mev  RAYS
1/10 VALUE
1/2 VALUE
LEAD
1.15 inches
0.33 inches
CONCRETE
6.5 inches
1.9 inches
WATER
13.5 inches
4.0 inches
b.
c.
The values listed in the above table are only valid
for 1 Mev gamma rays. Since most gamma and
X-ray radiation in a nuclear power plant is
polyenergetic, several rules of thumb have been
developed for reasonable shielding estimates using
tenth-value thicknesses. These thumb rules are as
follows:
Objective 14
1)
2)
3)
4)
TP-61
The TVL for lead is about 2 inches
The TVL for steel or iron is about 4 inches
The TVL for concrete is about 12 inches
The TVL for water or polyethylene is about
24 inches
Using rules of thumb, we can calculate the shield
thickness of various materials to get 1/4, 1/2, 1, 2,
3 or even more tenth thicknesses.
Number of
Tenth-Thickness
¼
1/2
1
2
3
Water
6
12
24
48
72
Shield (inches)
Concrete
Steel
3
6
12
24
48
1
2
4
8
12
Lead
1/2
1
2
4
6
TP-62
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d.
INSTRUCTOR NOTES
Problem # 15
TP-63
The dose rate from a valve is 1,200 R/hr. If 4
inches of lead is used to shield the valve, what will
be the shielded dose rate?
4 inches of lead represents 2 tenth-value layers.
For a dose rate of 1,200 R/hr,
Addition of:
1 TVL
2 TVL
Reduces the dose rate to:
120 R/hr
12 R/hr
Therefore, the shielded dose rate will be 12 R/hr.
e.
Problem # 16
A source with a contact dose rate of 200 mR/hr is
laying under 24 inches of water. What is the dose
rate at the surface of the water?
24 inches of water represents one tenth-value
layer, therefore, the dose rate at the surface of the
water would be:
(200 mR/hr) * (0.1) = 20 mR/hr.
TP-64
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f.
INSTRUCTOR NOTES
Problem # 17
TP-66
The dose rate from a component is 10 R/hr. If 3
half-value layers of shielding is placed around the
component, what would be the shielded dose rate?
For 10 R/hr,
TP-67
Addition of:
1 HVL
Reduces the dose rate to:
5 R/hr
2 HVL
2.5 R/hr
3 HVL
1.25 R/hr,
Therefore, the shielded dose rate is 1.25 R/hr.
g.
In addition, half-value thicknesses and tenth-value
thicknesses are tabulated for selected radionuclides
in the following table: (Reference D)
Half- and Tenth- Value Layers for Selected Gamma-Ray Sources
Radionuclide Half-Life
Gamma
Energy,
MeV
Cesium-137
Cobalt-60
Radium-226
0.662
1.17 & 1.33
0.047 – 2.4
27 years
5.24 years
1622 years
Half-Value Layersa
Concrete Steel Lead
(cm)
(cm) (cm)
4.8
6.2
6.9
a. Approximate values obtained with large attenuation.
1.6
2.1
2.2
0.65
1.20
1.66
Tenth-Value Layersa
Concrete Steel Lead
(cm)
(cm) (cm)
15.7
20.6
23.4
5.3
6.9
7.4
2.1
4.0
5.5
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h.
INSTRUCTOR NOTES
Two basic formulas for calculating the number of
tenth-value thicknesses and half-value thicknesses
are:
Tp-68
D = Do (1/10)N
(Equation # 13)
Where D = Final dose
Do = Initial dose
N = Number of tenth thickness
and,
D = Do (1/2)M
(Equation # 14)
Where D = Final dose
Do = Initial dose
M = Number of half-thicknesses
i.
Problem # 18
A source reading 900 R/hr is shielded by 7 TVLs
of iron. What is the shielded dose rate?
D = Do (1/10)N
D = 900 R/hr (1/10)7 = 900 R/hr (1)7/(10)7, or
D = 900 R/hr (1 * 10-7) = 9.0 * 10-5 R/hr, or
D = 0.09 mR/hr
TP-69
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j.
INSTRUCTOR NOTES
Problem # 19
TP-70
A source with a dose rate of 400 R/hr is shielded
by 5 half-value layers of lead. What is the shielded
dose rate?
D = Do (1/2)M
D = 400 R/hr (1/2)5, or 400 R/hr (0.03125)
D = 12.5 R/hr.
k.
The number of tenth-value thicknesses can be
calculated by rearranging Equation # 13 to solve
for N, such that:
D = Do (1/10)N, and D/Do = (1/10)N
In order to solve for N, we need to take the
logarithm of each side of the equation, so,
log(D/Do) = log(1/10)N
From the rules for the use of logarithms,
log(M)N = N * log(M), therefore,
log(D/Do) = N * log(1/10) or N * log(0.1)
Since log(0.1) = -1,
log(D/Do) = -N, or
N = -log(D/Do)
(Equation # 15)
TP-71
HPT001.017
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X.
LESSON BODY
l.
INSTRUCTOR NOTES
Problem # 20
TP-72
How many tenth value thicknesses are required to
decrease a dose rate from
300 rem/hr to 2 mrem/hr?
Using Equation # 15,
N = -log(D/Do), or
N = -log(2 mrem/hr ÷ 300,000mrem/hr)
N = - log(0.000006667) = 5.18 tenth value
thicknesses.
10.
Besides the type and amount of shielding, the placement
of the shield is also a concern.
Reference G
a.
TP-73
Problem # 21
Assume that the radiation level from a pump is 30
mR/hr one foot from the pump. If a shield of lead 2
inches thick is placed so that the outside edge of
the lead is one foot from the pump, calculate the
readings at a distance of 10 feet from the pump.
Step 1: Calculate the reading through the shield:
2 inches of lead is 1 TVL, so the dose rate is
3 mR/hr through the shield.
Step 2: Calculate the readings 10 feet from source:
Using Equation # 3, I2 = (I1 * d12)/ (d22)
I2 = (3 mR/hr)(1 ft)2/(10 ft)2
I2 = (3 mrem-ft2/hr)/(100 ft2)
I2 = 0.03 mR/hr
TP-74
HPT001.017
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X.
LESSON BODY
b.
c.
E.
INSTRUCTOR NOTES
As the example shows, the location does not affect
the thickness of the shield because a given
thickness always provides the same fraction of
reduction. However, the farther away from a
source we get, the more the radiation diverges.
Although the shield thickness is the same, the
height and width of the shield have to be larger. As
shown in Figure RP-4-2, it is usually best to place
the shield as close to the source as possible to keep
the shield as small as possible and thus minimize
its cost.
TP # 75
Another consideration is the phenomenon referred
to as “skyshine.” Gamma rays interact with the
atoms in the air and produce Compton scatter
gamma rays. This gives the gamma rays the
appearance of ‘turning corners.’ The phenomenon
is illustrated in TP # 76. The name was coined to
reflect the fact that the gamma rays appear to shine
down from the sky if adequate shielding is not
provided above the source. To illustrate the
potential problem, a 50 Ci cobalt-60 source behind
a 12-foot high concrete wall with no top, will
produce a skyshine level of about 17 mR/hr three
feet outside the wall.
Reference B
Practice Problems
Reference G
TP # 76
Objective 15
Handout # 07,
Attachment 2
Work the practice problems contained in Attachment 2. The
solutions are presented in Attachment 3.
Handout # 08,
Attachment 3
HPT001.017
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Page 45 of 51
X.
LESSON BODY
XI
SUMMARY
INSTRUCTOR NOTES
TP-77
Radiation protection is concerned with the protection of individuals, their
progeny and mankind as a whole from the risks received from radiation
and radioactive materials, while still allowing necessary activities from
which radiation exposure might result. Three basic components of a
radiation protection program are decreasing time of exposure, increasing
distance from radiation and radioactive materials, and providing shielding
from radiation. These activities are critical in controlling doses and dose
rates associated with exposures to radiation and radioactive materials.
The relationship between dose, dose rate, and time allows us to determine
the optimum time to receive the minimum exposure.
We have seen that alpha radiation can be shielded by something as
simple as air or paper while beta requires thicker shielding materials, like
wood or aluminum. Neutrons are shielded by materials with a low
Atomic Number, such as water or polyethylene, but gamma and X-rays
require shields with high Atomic Numbers, like lead, concrete and steel.
There are a number of mathematical principles and equations used to aid
in the control of radiation exposures. The Inverse Square Law equation
defines the relationship between dose rates at varying distances from a
point source and the line or parallel source equation is used to calculate
doses from longer sources. The simple equation, DR = 6CE will
approximate dose rates at 1 foot from a point source if the activity and
gamma ray energy are known. Attenuation coefficients can be used to
calculate dose rates from radiation that has passed through shielding
material. The concepts of half-value and tenth-value layers provide a
mechanism to quickly estimate doses from radiation passing through
shielding material.
TP-78
Additional considerations in estimating radiation doses are the effects of
Bremsstrahlung and photon buildup, in which the number of photons
exiting a shield can be greater that calculated, and may even exceed the
incident intensity. Skyshine can also result in increased exposures to
radiation as it passes over a shield and interacts with the atoms in the air
to increase the number of photons reaching an individual.
TP-79
Remember that Self-Checking is especially important when performing
calculations. We must constantly be aware of the units of the parameters
we are using in the equations and diligent to use the correct numbers for
all input values. In addition, we must ensure that we are using the proper
equation that applies to the problem we are trying to solve. Use of the
Questioning Attitude will aid us in selecting the proper methodology.
TP-80, Emphasize
Error Prevention
Tools!
HPT001.017
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Page 46 of 51
Handout # 01
Enabling Objectives
1.
Identify the three methods used to control radiation exposures.
2.
Define dose and dose rate and tell the difference between the two.
3.
Apply the equation defining the relationship between time, dose, and dose rate to perform
calculations of ‘stay time.’
4.
Define point and line sources and perform calculations using the Inverse Square Law and
the line source equation.
5.
Apply the equation DR = 6CE to calculate dose rates.
6.
Define the specific gamma ray constant and be able to use it in calculating radiation
exposures.
7.
Define ‘bremsstrahlung’ and tell its impact in shielding considerations.
8.
Identify the types and characteristics of materials that can best shield neutrons.
9.
List three (3) factors influencing the attenuation of photons as they pass through a material.
10.
Identify the differences among the following:
a.
‘linear attenuation coefficient,’
b.
‘mass attenuation coefficient,’
c.
‘energy absorption coefficient.’
11.
Apply the basic shielding equations to calculate exposure levels and shield thicknesses.
12.
Define radiation ‘buildup’ and describe its impact on shielding considerations.
13.
Define
a.
‘half-value layer (or thickness)’
b.
‘tenth-value layer (or thickness)’
and use these concepts to calculate dose rates and shielding needs.
14.
List the rule of thumb tenth-value layer values for lead, steel, concrete, and water.
15.
Define ‘skyshine’ and describe its impact on shielding considerations.
HPT001.017
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Page 47 of 51
Handout # 02
Gamma Energy & Occurrence and Specific Gamma Ray Constant for Selected
Radionuclides
1
Nuclide
Half-Life
Mn-54
C0-60
312.7 d
5.3 y
Fe-59
44.6 d
Zn-65
244.4 d
I-131
8.0 d
Cs-137
Ra-226
30.1
1600 y
Gamma Energy
MeV
0.835
1.17
1.33
0.143
0.192
1.099
1.292
0.551
1.115
0.284
0.364
0.637
0.662
0.186
+ daughters
Gamma
Occurrence, %
100
100
100
1.0
3.1
56.5
43.2
2.8
50.8
6.1
81.2
7.3
90
3.3
Γ is given as R-cm2/hr-mCi. Divide Γ by 10 to obtain R/hr at 1 meter/Ci.
Γ1
4.7
13.2
6.4
2.7
2.2
3.3
8.25
HPT001.017
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Handout # 03
Attenuation Coefficients
A. Mass Attenuation Coefficients, cm2/g
Energy, MeV
Material
Density
0.1
0.2
0.5
0.8
1.0
2.0
5.0
ρ, g/cm3
Al
2.70
0.171
0.122
0.0844
0.0684
0.0613
0.0432
0.0284
Fe
7.86
0.370
0.146
0.0840
0.0669
0.0599
0.0425
0.0314
Cu
8.96
0.461
0.157
0.0836
0.0660
0.0589
0.0420
0.0318
Pb
11.35
5.40
0.991
0.161
0.0885
0.0708
0.0455
0.0424
H2O
1.000
0.171
0.137
0.0968
0.0786
0.0707
0.0494
0.0303
Concrete
2.35
0.179
0.127
0.0877
0.0709
0.0637
0.0448
0.0290
1.29E-3
0.154
0.123
0.0870
0.0707
0.0636
0.0445
0.0275
Air
B. Linear Attenuation Coefficients, cm-1
Energy, MeV
Material
Density
0.1
0.2
0.5
0.8
1.0
2.0
5.0
ρ, g/cm
3
Al
2.70
0.462
0.329
0.228
0.185
0.166
0.117
0.0767
Fe
7.86
2.908
1.148
0.660
0.526
0.471
0.334
0.247
Cu
8.96
4.131
1.407
0.749
0.591
0.528
0.376
0.285
Pb
11.35
61.29
11.25
1.827
1.004
0.804
0.516
0.481
H2O
1.000
0.171
0.137
0.0968
0.0786
0.0707
0.0494
0.0303
Concrete
2.35
0.421
0.298
0.206
0.167
0.150
0.105
0.0682
1.29E-3
1.99E-4
1.59E-4
1.12E-4
9.12E-5
8.20E-5
5.74E-5
3.55E-5
Air
HPT001.017
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Handout # 04
Mass Attenuation Coefficients
HPT001.017
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Handout # 05
TENTH AND HALF-VALUE THICKNESSES
A.
B.
TENTH AND HALF-VALUE THICKNESSES FOR 1 Mev  RAYS
1/10 VALUE
1/2 VALUE
LEAD
1.15 inches
0.33 inches
CONCRETE
6.5 inches
1.9 inches
WATER
13.5 inches
(Approx. 1 foot)
4.0 inches
NUMBER OF TENTH-VALUE THICKNESSES FOR SELECTED MATERIALS:
RULE OF THUMB NUMBERS BASED ON MIXED ENERGIES AS FOUND
IN AN OPERATING NUCLEAR POWER PLANT
Number of
Tenth-Thickness
1/4
1/2
1
2
3
Shield (inches)
Water Steel
Lead
6
12
24
48
72
1
2
4
8
12
1/2
1
2
4
6
C. Half- and Tenth- Value Layers for Selected Gamma-Ray Sources
Radionuclide Half-Life
Cesium-137
Cobalt-60
Radium-226
Gamma
Energy,
MeV
Half-Value Layersa
Concrete Steel Lead
(cm)
(cm) (cm)
27 years
0.662
5.24 years 1.17 & 1.33
1622 years 0.047 – 2.4
4.8
6.2
6.9
1.6
2.1
2.2
a. Approximate values obtained with large attenuation.
0.65
1.20
1.66
Tenth-Value Layersa
Concrete Steel Lead
(cm)
(cm) (cm)
15.7
20.6
23.4
5.3
6.9
7.4
2.1
4.0
5.5
HPT001.017
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Handout # 06
Equations
Equation #
Equation
Used For:
1
Time = Dose/Dose Rate
Stay Time.
2
I1 * d12 = I2 * d22
Inverse Square Law.
3
I2 = (I1 * d12)/ (d22)
Inverse Square Law rearranged to
solve for Intensity 2.
4
DR = 6 CEn
Dose rate, R/hr at 1 foot, from
a known gamma energy.
5.
I1d1 = I2d2
Linear distance relationship for
beam or line sources.
6
I2 = I1d1/d2
Linear distance equation rearranged
to solve for Intensity 2.
7
I = I0e-µx,
Point source shielding equation.
8
x = [ln (I/ I0)]/-µ
Point source shielding equation
rearranged to solve for shield thickness
9
I = I0e-(µm)(ρ) x,
Point source shielding equation using
mass attenuation coefficient.
10
x = [ln (I/ I0)]/-(µm)(ρ)
Equation # 9 rearranged to solve for
shield thickness.
11
I = BI0e-µx,
Shielding equation incorporating the
Buildup Factor.
12
I = I0e-µex,
Approximation of Buildup equation
using the energy absorption coefficient
13
D = Do (1/10)N
Dose using tenth-value thickness.
14
D = Do (1/2)M
Dose using half-value thickness.
15
N = -log(D/Do)
Calculate number of tenth-value layer.
HPT001.017
Revision 2
Attachment 2
Page 1 of 13
Handout # 07
Attachment 2
Practice Problems
1.
You have volunteered to go into a 50 rem/hr field to save a life. The emergency limit
for this activity is 75 rem. What is your calculated Stay Time or the maximum length
of time you can remain in the 50 rem/hr location?
2.
If a point isotropic source reads 10 rem/hr 2 feet away, what will be the reading at 5
feet?
3.
A source reads 100 rem/hr at 1 foot. How far from it must you get to lower the dose
equivalent rate to 1 rem/hr?
HPT001.017
Revision 2
Attachment 2
Page 2 of 13
4.
A survey meter reads 100 mrem/hr 2 feet from a 20 ft. long pipe. Determine the dose
equivalent rate at 4 feet.
5.
Induced Na-24 activity in a cooling water line passes through a small diameter pipe in
an access room 20 ft. wide. The door to the room is in the center of the 20 ft. wall, at a
distance of 10 ft. from the pipe. If the dose equivalent rate in the doorway is 3.24
mrem/hr, what is the dose equivalent rate midway between the pipe and the door at a
distance of 5 ft. from the pipe?
6.
1 Ci of Ra-226
Γ = 8.25 R-cm2/hr-mCi, and Γ/10 = 0.825 R/hr at 1 m for each Ci.
What is the exposure rate at 4 m?
HPT001.017
Revision 2
Attachment 2
Page 3 of 13
7.
What is the exposure rate 6 m from a 1 Ci Co-60 source?
8.
100 mrem is the maximum dose equivalent allowed for performing a certain job. The
area dose equivalent rate is 2 rem/hr. Calculate the Stay Time.
9.
What thickness of lead is needed to reduce a 1 MeV reading from 500 mr/hr to
60 mr/hr?
HPT001.017
Revision 2
Attachment 2
Page 4 of 13
10.
What thickness of water is needed to reduce a 1 MeV gamma dose rate to one tenth its
original value, ie, what is the TVL of water for 1 MeV gamma? (Handout # 5)
11.
Using the results from Problem # 10, complete the following problem:
A person is standing at the edge of a swimming pool containing spent fuel and observes
a dose rate of 5 mR/hr (assume 1 MeV gamma). What would the reading be if the
water level dropped 1 ft? What would the rate be if the water dropped 5 ft?
12.
A Ra-226 source is to be shipped in a wooden box lined with lead. The gamma reading
at the surface of the box is 1 R/hr. What thickness of lead lining is required to reduce
the reading at the surface to 2 mR/hr if the gamma energy is 0.8 MeV?
HPT001.017
Revision 2
Attachment 2
Page 5 of 13
13.
It is desired to reduce a 1 MeV beam of γ rays from 640 mR/hr to 40 mR/hr. How
many cm of lead are required? Use the µm and density values from Handout # 3.
14.
How many TLVs of water are necessary to bring an exposure rate for 1 MeV γ from
6000 mR/hr to 6 mR/hr?
15.
You have a shielding situation with a non-point parallel source. How much lead must
be used to line a box containing Ra-226 to reduce the box surface reading from 1 R/hr
to 2 mR/hr? Assume a gamma energy of 0.8 MeV and µe = 0.5727.
HPT001.017
Revision 2
Attachment 2
Page 6 of 13
16.
Lead is used to shield a 2.0 MeV γ source which reads 60 R/hr at 3 ft with no shield.
a. What is the exposure rate at this distance when a 4 cm lead shield is in place?
Assume B = 2.51. The density of lead is given in Handout # 3.
b. Is there any advantage to placing the shield closer to the source than to the exposed
individual? If so, what?
HPT001.017
Revision 2
Attachment 2
Page 7 of 13
17.
A lead lined box is being used to ship a Co-60 source. What is the thickness of lead
(cm) required to reduce the box contact dose rate from 70 R/hr to 7 mR/hr?
a. Calculate using the point source shielding equation (assume µ = 0.681).
b. Calculate using TVL.
18.
It is desired to reduce a 1 MeV γ beam from 480 mR/hr to 30 mR/hr using lead as a
shield. How many HVLs are required? How many cm of lead does this represent?
(HVL for 1 MeV γ in lead is given in Handout # 5.)
HPT001.017
Revision 2
Attachment 2
Page 8 of 13
19.
Using the data given in Problem # 18, determine the number of TVLs required. How
many cm of lead is this?
20.
After working on a small radioactive fitting for two hours, you discover that your
electronic dosimeter display is blank. A survey indicates that the dose rate at 1 ft from
the fitting is 900 mR/hr. You estimate that your average working distance was 2 ft
from the fitting. Calculate the dose received.
21.
A 5 Ci point source emits a 1.5 MeV γ per disintegration. We want to put up a lead
shield 5 ft from the source so that a maintenance worker there will not be exposed to
more than 100 mrem if he works behind the shield for 2 hours. Assume that
µ = 0.587 cm-1. What would be the minimum thickness of the lead shield?
HPT001.017
Revision 2
Attachment 2
Page 9 of 13
22.
An operator is 15 ft from a small radioactive crud deposit where the dose rate is
measured at 600 mR/hr. He must operate a valve 5 ft beyond the source. How long
can he linger at the valve before he will receive a total dose of 100 mrem? Assume the
crud deposit is in the form of an isotropic point source.
23.
A Ra-226 source will be contained at the center of a wooden box. The dose rate at the
surface of the box is 4 R/hr. What thickness of lead is required to reduce the dose rate
to 1 mR/hr? Assume B = 2.75. The value for µ is given in Handout # 3.
24.
What is the exposure rate 3 m from a drain which gives a contact reading of 4 R/hr on a
teletector?
HPT001.017
Revision 2
Attachment 2
Page 10 of 13
25.
A survey meter measures a dose rate of 2 R/hr at a 10-ft perpendicular distance from a
40-ft long pipe carrying radioactive material. What is the reading 1 ft from the pipe?
26.
An unshielded 5 Ci source of Ra-226 is in your work area. Calculate Stay Time at a
distance of 10 ft. (Your RWP limit is 500 mrem).
27.
A radioactive source produces an exposure rate of 15 mR/hr outside a 1 HVL shield. If
an additional two half-value layers are added, what will be the resultant exposure rate?
HPT001.017
Revision 2
Attachment 2
Page 11 of 13
28.
An unshielded Tc-99m vial is generating an exposure rate of 8850 mR/hr. What will
the exposure rate be when the vial is placed in a shield with a thickness of 10 HVLs?
29.
The half-value layer of lead is 0.27 mm for Tc-99m. A dose rate of 5300 mR/hr was
measured before the source was shielded. What will be the exposure rate after the vial
is placed in a shield made with 0.90 mm of lead?
30.
A thicker shield is to be used for the above source. The shield is made of 3.8 mm of
lead. What will be the exposure rate after shielding?
HPT001.017
Revision 2
Attachment 2
Page 12 of 13
31.
A shielded source of Tl-201 registers as 25 mR/hr on an ionization survey meter held 4
inches from the surface. If the HVL for lead is 0.2 mm, what will be the dose reading
at 4 inches after the shielded source is placed in a lead pig constructed of 1.35 mm thick
lead?
32.
A Mo-99 generator is located next to a wall shared by a public hallway. The reading in
the hallway is 6.4 mR/hr. How much shielding must be added to decrease the exposure
rate too 2 mR.hr? The HVL for Mo-99 is 0.7 cm of lead.
33.
A survey meter reading taken outside the shield surrounding a radioactive storage area
gives a reading of 12.8 mR/hr. Only Tc-99m is being stored. The HVL for lead is 0.27
mm. How many mm of lead must be added to bring the reading down to a background
reading of 0.02 mR/hr?
HPT001.017
Revision 2
Attachment 2
Page 13 of 13
34.
What is the dose rate at 1 foot from 30 mCi of I-131? I-131 has 5 gammas.
0.08 MeV
0.284 MeV
0.364 MeV
0.637 MeV
0.723 MeV
35.
2.6 % of the time
6.1 % of the time
81.2% of the time
7.3 % of the time
1.8 % of the time.
A pipe carrying radioactive material runs vertically through a room that has a height of
12 feet. The dose rate measured 1 foot from the pipe is 18 R/hr. What is the exposure
rate at the doorway 18 feet from the pipe? (Hint: What is L/2?)
HPT001.017
Revision 2
Attachment 3
Page 1 of 13
Handout # 08
Attachment 3
Practice Problems - Solutions
1.
You have volunteered to go into a 50 rem/hr field to save a life. The emergency limit
for this activity is 75 rem. What is your calculated Stay Time or the maximum length
of time you can remain in the 50 rem/hr location?
Time = D/DR, where D is the dose limit and DR is the dose rate.
Time = 75 rem ÷ 50 rem/hr = 1.5 hr
2.
If a point isotropic source reads 10 rem/hr 2 feet away, what will be the reading at 5
feet?
I1d12 = I2d22
(10 rem/hr)(2 ft)2 = (I2)(5 ft)2
I2 = 40 rem-ft2/hr ÷ 25 ft2
I2 = 1.6 rem/hr
3.
A source reads 100 rem/hr at 1 foot. How far from it must you get to lower the dose
equivalent rate to 1 rem/hr?
I1d12 = I2d22
(100 rem/hr)(1 ft)2 = (1 rem/hr)(d22)
d22 = 100 ft2
d22 = 10 ft
HPT001.017
Revision 2
Attachment 3
Page 2 of 13
4.
A survey meter reads 100 mrem/hr 2 feet from a 20 ft. long pipe. Determine the dose
equivalent rate at 4 feet.
I1d1 = I2d2
(100 mrem/hr)(2 ft) = I2(4 ft)
I2 = 200 mrem-ft/hr ÷ 4 ft
I2 = 50 mrem
5.
Induced Na-24 activity in a cooling water line passes through a small diameter pipe in
an access room 20 ft. wide. The door to the room is in the center of the 20 ft. wall, at a
distance of 10 ft. from the pipe. If the dose equivalent rate in the doorway is 3.24
mrem/hr, what is the dose equivalent rate midway between the pipe and the door at a
distance of 5 ft. from the pipe?
I1d1 = I2d2
(3.25 mrem/hr)(10 ft) = I2(5 ft)
I2 = (32.5 mrem – ft/hr)/5 ft
I2 = 6.48 mrem/hr
6.
1 Ci of Ra-226
Γ = 8.25 R-cm2/hr-mCi, and Γ/10 = 0.825 R/hr at 1 m for each Ci.
What is the exposure rate at 4 m?
I1d12 = I2d22
(0.825 R/hr)(1 m)2 = (I2)(4 m)2
I2 = (0.825 R-m2/hr)/16 m2
I2 = 0.052 R/hr or 52 mR/hr at 4 meters
HPT001.017
Revision 2
Attachment 3
Page 3 of 13
7.
What is the exposure rate 6 m from a 1 Ci Co-60 source?
Γ/10 = 1.32 R/hr at 1 m
I1d12 = I2d22
(1.32 R/hr)(1 m)2 = (I2)(6 m)2
I2 = (1.32 R-m2/hr)/(36 m2)
I2 = 0.0367 R/hr or 36.7 mR/hr at 6 m
8.
100 mrem is the maximum dose equivalent allowed for performing a certain job. The
area dose equivalent rate is 2 rem/hr. Calculate the Stay Time.
Time = D/DR
Time = 0.1 rem ÷ 2 rem/hr = 0.05 hr or 3 min
9.
What thickness of lead is needed to reduce a 1 MeV reading from 500 mr/hr to
60 mr/hr?
From Handout # 3, µ = 0.804 cm-1
I = I0e-µx
60 mR/hr = 500 mr/hr * e-0.804 cm-1 * x
0.12 = e-0.804 cm-1 * x
ln(0.12) = -0.804 cm-1 * x
x = (-2.12)/(-0.804 cm-1)
x = 2.64 cm
HPT001.017
Revision 2
Attachment 3
Page 4 of 13
10.
What thickness of water is needed to reduce a 1 MeV gamma dose rate to one tenth its
original value, ie, what is the TVL of water for 1 MeV gamma? (Handout # 5)
From Handout # 5, TVL for water for a 1 MeV gamma = 13.5 inches, or approximately 1 ft
11.
Using the results from Problem # 10, complete the following problem:
A person is standing at the edge of a swimming pool containing spent fuel and observes
a dose rate of 5 mR/hr (assume 1 MeV gamma). (a) What would the reading be if the
water level dropped 1 ft? (b) What would the rate be if the water dropped 5 ft?
a. D = D0(1/10)M, Since we are reducing shielding, we must solve the equation for D0
A 1 ft drop in water level represents approximately 1 TVL, therefore,
5 mR/hr = D0(0.1)1
D0 = (5 mR/hr)/(0.1) = 50 mR/hr
b. A 5 ft drop in water level represents approximately 5 TVL, therefore,
5 mR/hr = D0(0.1)5
D0 = (5 mR/hr)/0.00001 = 5 E-5 mR/hr, or 500 R/hr
12.
A Ra-226 source is to be shipped in a wooden box lined with lead. The gamma reading
at the surface of the box is 1 R/hr. What thickness of lead lining is required to reduce
the reading at the surface to 2 mR/hr if the gamma energy is 0.8 MeV?
From Handout # 3, µ for 0.8 MeV γ in lead is 1.004 cm-1
I = I0e-µx
2 mR/hr = (1,000 mR/hr)(e-1.004 cm-1 * x)
ln(2 mR/hr/1,000 mR/hr) = (-1.004 cm-1) * x
x = 6.19 cm
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Revision 2
Attachment 3
Page 5 of 13
13.
It is desired to reduce a 1 MeV beam of γ rays from 640 mR/hr to 40 mR/hr. How
many cm of lead are required? Use the µm and density values from Handout # 3.
From Handout # 3, µm = 0.0708 cm2/g, and ρ = 11.35 g/cm3
I = I0e-µm * ρ * x
40 mR/hr = (640 mR/hr)e-0.0708 cm2/g * 11.35 g/cm3 * x
ln(40 mR/hr ÷ 640 mR/hr) = (-0.804 cm-01) * x
x = (ln0.0625)/(-0.804 cm01)
x = (-2.773)(-0.804 cm01) = 3.45 cm
14.
How many TLVs of water are necessary to bring an exposure rate for 1 MeV γ from
6000 mR/hr to 6 mR/hr?
D = D0(1/10)N
6 mR/hr = (6,000 mR/hr)(0.1)N
0.001 = (0.1)N
log(0.001) = N * log(0.1)
-3 = N * (-1), or N = 3 TVLs
15.
You have a shielding situation with a non-point parallel source. How much lead must
be used to line a box containing Ra-226 to reduce the box surface reading from 1 R/hr
to 2 mR/hr? Assume a gamma energy of 0.8 MeV and µe = 0.5727.
I = I0e-µe *
x
2 mR/hr = (1,000 mR/hr)e-0.5727 cm-1* x
ln(2 mR/hr ÷ I,000 mR/hr) = -0.5727 cm-1* x
x = (-6.215)/(-0.5727 cm-1) = 10.85 cm
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Revision 2
Attachment 3
Page 6 of 13
16.
Lead is used to shield a 2.0 MeV γ source which reads 60 R/hr at 3 ft with no shield.
(a) What is the exposure rate at this distance when a 4 cm lead shield is in place?
Assume B = 2.51. The density of lead is given in Handout # 3.
From Handout # 3, µ = 0.0455 cm2/g and ρ = 11.35 g/cm3
I = BI0e-µm * ρ * x
I = (2.51)(60 R/hr)e-(0.0455 cm2/g * 11.35 g/cm3 * 4 cm)
I = 150.6 R/hr * e-2.066
I = (150.6 R/hr)(0.127)
I = 19.1 R/hr
c. Is there any advantage to placing the shield closer to the source than to the exposed
individual? If so, what?
A smaller shield would be required if it were placed closer to the source, saving cost
and weight of the shield.
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Revision 2
Attachment 3
Page 7 of 13
17.
A lead lined box is being used to ship a Co-60 source. What is the thickness of lead
(cm) required to reduce the box contact dose rate from 70 R/hr to 7 mR/hr?
a. Calculate using the point source shielding equation (assume µ = 0.681 cm-1).
I = I0e-µx
7 mR/hr = (70,000 mR/hr)(e-0.681 cm-1 * x)
0.0001 = e-0.681 cm-1 * x
ln(0.0001) = (-0.681 cm-1 * x)
x = (-9.21)/(-0.681 cm-1), or x = 13.5 cm
b. Calculate using TVL.
D = D0(1/10)N
7 mR/hr = 70,000 mr/hr (0.1)N
log[(7 mR/hr)/(70,000 mR/hr)] = N * log(0.1)
(-4) = (-1) N, or N = 4 TVLs
From Handout # 5C, the TVL for Co-60 in lead is 4 cm, therefore 4 cm * 4 TVLs = 16 cm
18.
It is desired to reduce a 1 MeV γ beam from 480 mR/hr to 30 mR/hr using lead as a
shield. How many HVLs are required? How many cm of lead does this represent?
(HVL for 1 MeV γ in lead is given in Handout # 5.)
D = D0(1/2)M
40 mR/hr = (640 mR/hr)(1/2)M
0.0625 = )(1/2)M
log(0.0625) = M * log(0.5), or -0.301M = -1.204
M = 4 HLVs
From Handout # 5, the HVL value for Pb is 0.33 inches
4 HVLs * 0.33 in = 1.32 in
1 in = 2.54 cm, therefore, 1.32 in * 2.54 cm/in = 3.35 cm
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Revision 2
Attachment 3
Page 8 of 13
19.
Using the data given in Problem # 18, determine the number of TVLs required. How
many cm of lead is this?
D = D0(1/10)N
0.0625 = (0.1)N
N * log(0.1) = log(0.0625
-1N = -1.204, or N = 1.204 TVLs
From Handout 5, the TVL for lead is 1.15 in, therefore,
1.204 TVLs 8 1.15 in = 1.38 in of Pb,, or 1.38 in * 2.54 cm/in = 3.51 cm
20.
After working on a small radioactive fitting for two hours, you discover that your
electronic dosimeter display is blank. A survey indicates that the dose rate at 1 ft from
the fitting is 900 mR/hr. You estimate that your average working distance was 2 ft
from the fitting. Calculate the dose received.
I1d12 = I2d22
(900 mR/hr)(1 ft)2 = I2(2 ft)2
4 I2 ft2 = 900 mR-ft2/hr, or I2 =225 mR/hr
225 mR/hr * 2 hr = 450 mR
21.
A 5 Ci point source emits a 1.5 MeV γ per disintegration. We want to put up a lead
shield 5 ft from the source so that a maintenance worker there will not be exposed to
more than 100 mrem if he works behind the shield for 2 hours. Assume that
µ = 0.587 cm-1. What would be the minimum thickness of the lead shield?
a.
Calculate the exposure rate from 5 Ci of a 1.5 MeV γ from the equation DR = 6CE
DR = 6(5 Ci)(1.5 MeV) = 45 R/hr at 1 foot
b.
Calculate the unshielded exposure rate at 5 ft from the equation I1d12 = I2d22
(45,000 mR/hr)(1 ft)2 = I2(5 ft)2, or 25 I2 ft2 = 45,000 mR-ft2/hr
I2 = (45,000/25) mR/hr = 1800 mR/hr
c.
Calculate shielded exposure rate at 5 ft using equation I = I0e-µx, where µ=0.587 cm-1
Limit of 100 mR for 2 hr = 50 mR/hr, therefore, 50 mR/hr = (1800 mR/hr)e-0.587 cm-1 * x
e-0.587 cm-1 * x = 0.0278, or -0.587 cm-1 * x = ln0.0278
-0.587 cm-1 * x = -3.58, or x = 6.1 cm
HPT001.017
Revision 2
Attachment 3
Page 9 of 13
22.
An operator is 15 ft from a small radioactive crud deposit where the dose rate is
measured at 600 mR/hr. He must operate a valve 5 ft beyond the source. How long
can he linger at the valve before he will receive a total dose of 100 mrem? Assume the
crud deposit is in the form of an isotropic point source.
I1d12 = I2d22
(600 mR/hr)(15 ft)2 = I2(5 ft)2
I2(25 ft2) = 135,000 mR-ft2/hr, or I2 = 5400 mR/hr at 5 ft
Time = Limit/DR = 100 mR/5400 mR/hr = 0.02 hr
0.02 hr * 60 min/hr = 1.2 min
23.
A Ra-226 source will be contained at the center of a wooden box. The dose rate at the
surface of the box is 4 R/hr. What thickness of lead is required to reduce the dose rate
to 1 mR/hr? Assume B = 2.75. The value for µ is given in Handout # 3.
I = BI0e-µx, where µ = 1.004 cm-1
(1 mR/hr) = 2.75(4,000 mR/hr)e-1.004 cm-1 * x
e-1.004 cm-1 * x = 0.000091
-1.004 cm-1 * x = ln(0.000091) = -9.31
x = 9.27 cm
24.
What is the exposure rate 3 m from a drain which gives a contact reading of 4 R/hr on a
teletector?
Assume contact reading at 1 cm
I1d12 = I2d22
(4,000 mR/hr)(1 cm)2 = I2(300 cm)2
I2 = (4,000 mR-cm2/hr)/(90,000 cm2) = 0.044 mR/hr
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Revision 2
Attachment 3
Page 10 of 13
25.
A survey meter measures a dose rate of 2 R/hr at a 10-ft perpendicular distance from a
40-ft long pipe carrying radioactive material. What is the reading 1 ft from the pipe?
L/2 = 20 ft, therefore the distance, 10 ft is less than L/2
I1d1 = I2d2
(2 R/hr)(10 ft) = I2(1 ft)
I2 = 20 R-ft/hr/1 ft = 20 R/hr at 1 ft.
26.
An unshielded 5 Ci source of Ra-226 is in your work area. Calculate Stay Time at a
distance of 10 ft. (Your RWP limit is 500 mrem).
From Handout # 2, Γ/10 = 0.825 R/hr at 1 m for each Ci of activity, therefore
(0.825 R-m/hr-Ci)(5 Ci) = 4.125 R/hr at 1 meter
1 m = 3.28 ft, therefore DR = 4.125 R/hr at 3.28 ft
I1d12 = I2d22, or (4.125 R/hr)(3.28 ft)2 =I2(10 ft)2
I2(100 ft2) = 44.38 R-ft+/hr, or I2 = 0.44 R/hr
Assume 1 R = 1 rem
Time = 500 mrem/440 mrem/hr = 1.14 hr
27.
A radioactive source produces an exposure rate of 15 mR/hr outside a 1 HVL shield. If
an additional two half-value layers are added, what will be the resultant exposure rate?
D = D0(1/2)M
D = (15 mR/hr)(0.5)2
D = 3.75 mR.hr
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Revision 2
Attachment 3
Page 11 of 13
28.
An unshielded Tc-99m vial is generating an exposure rate of 8850 mR/hr. What will
the exposure rate be when the vial is placed in a shield with a thickness of 10 HVLs?
D = D0(1/2)10
D = (8850 mR/hr)(0.5)10, = 8850 mR/hr)(0.00098)
D = 8.6 mR/hr
29.
The half-value layer of lead is 0.27 mm for Tc-99m. A dose rate of 5300 mR/hr was
measured before the source was shielded. What will be the exposure rate after the vial
is placed in a shield made with 0.90 mm of lead?
Shielding = (0.90 mm)/(0.27 mm/HVL) = 3.33 HVLs
C , = 5300 mR/hr(0.5)3.33
D = 5300 mR/hr * 0.0099 = 527 mR/hr
30.
A thicker shield is to be used for the above source. The shield is made of 3.8 mm of
lead. What will be the exposure rate after shielding?
Shielding = (3.8 mm)/(0.27 mm/HVL) = 14 HVLs
D = D0(1/2)M , = 5300 mR/hr(0.5)14
D = 5300 mR/hr * 0.000061 = 0.32 mR/hr
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Revision 2
Attachment 3
Page 12 of 13
31.
A shielded source of Tl-201 registers as 25 mR/hr on an ionization survey meter held 4
inches from the surface. If the HVL for lead is 0.2 mm, what will be the dose reading
at 4 inches after the shielded source is placed in a lead pig constructed of 1.35 mm thick
lead?
Shielding = (1.35 mm)/(0.2 mm/HVL) = 6.75 HVLs
D = D0(1/2)M , = 25 mR/hr(0.5)6.75
D = 25 mR/hr * 0.0093 = 0.23 mR/hr
32.
A Mo-99 generator is located next to a wall shared by a public hallway. The reading in
the hallway is 6.4 mR/hr. How much shielding must be added to decrease the exposure
rate too 2 mR.hr? The HVL for Mo-99 is 0.7 cm of lead.
D = D0(1/2)M
(2 mR/hr) = (6.4 mR/hr)(1/2)M
(0.5)M = 0.3125
M(log 0.5) = log 0.3125
-0.301 M = -0.505, or M = 1.68 HVL
1.68 HVLs * 0.7 cm/HVL = 1.17 cm
33.
A survey meter reading taken outside the shield surrounding a radioactive storage area
gives a reading of 12.8 mR/hr. Only Tc-99m is being stored. The HVL for lead is 0.27
mm. How many mm of lead must be added to bring the reading down to a background
reading of 0.02 mR/hr?
D = D0(1/2)M
(0.02 mR/hr) = (12.8 mR/hr)(0.5)M
(0.5)M = 0.00156, and M * log(0.5) = log(0 00156)
-0.301 M = -2.806, or M = 9.33 HVLs
9.33 HVLs * 0.27 mm/HVL = 2.52 mm
HPT001.017
Revision 2
Attachment 3
Page 13 of 13
34.
What is the dose rate at 1 foot from 30 mCi of I-131? I-131 has 5 gammas.
0.08 MeV
0.284 MeV
0.364 MeV
0.637 MeV
0.723 MeV
Total γ energy
x
x
x
x
x
2.6 % of the time =
6.1 % of the time =
81.2% of the time =
7.3 % of the time =
1.8 % of the time. =
=
0.002
0.017
0.296
0.047
0.013
0.375 MeV
DR = 6CE
DR = 6(0.03 Ci)(0.375 MeV) = 0.0675 R/hr at 1 foot, or
DR = 67.5 mR/hr at 1 foot
35.
A pipe carrying radioactive material runs vertically through a room that has a height of
12 feet. The dose rate measured 1 foot from the pipe is 18 R/hr. What is the exposure
rate at the doorway 18 feet from the pipe? (Hint: What is L/2?)
a. Calculate the dose rate at L/2 using the equation I1d1 = I2d2
(18 R/hr)(1 ft) = I2(6 ft)
I2 = (18 R-ft/hr)/(6 ft) = 3 R/hr at L/2 = 6 ft
b. Calculate the dose rate at 18 ft using the equation I1d12 = I2d22
(3 R/hr)(6 ft)2 = I2(18ft)2
324 ft2 * I2 = 108 R-ft2/hr
I2 = 0.33 R/hr at 18 ft.