ｾ＠
...... ｯＺｾＬ＼＠
, ...... ｟ＢＧｾ＠
... ｾ＠ ....,......,. ·; .... :N',.j..... ···T·' ....."·,.. ,- ;.•
f｜ｾＮＧ＠
,:.
i
｜ｾＮｻＬ［ｌ＼Ｚ＠
｟ＮＧｾＢ＠
\
"
• :,..;;"
ＮｾＬ＠
... '
........ ｾ＠ ...... " .. -......... ｾＮ＠
". ' •. :,"_"
'
.:.'Y,··.'·;,,· Ｌｾ＠ ,;' ,. ｾＬＮＺ＠ :.:. i.::')"., ,:.': ＮＢｾ＠
ＧｩＢｾ＠ ,r...ｾ＠ ... ＧﾷｾＢ＠
," ;0'.•.• 'r'''' ｾＮ＠
. ' _.
ＬＧｾＮ＠
'."
i· . '. " "I .... - ｾ＠ ... ｲｾＧＮﾷ＠
ＧＢｾ＠
.. , .... ..,..• r.r:> Ｎｾ＠
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...... ＮＬｴＢｾ＠
•.
ｾＬ＠ ... ＢＮｾＬ＠
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,
•. - ..... ,. " -.•. ｾＮ＠
\ 79"27,4 \tAf If' ＼Ｏｾ＠
ｾＢＮＬＧ＠
A
,., ＧｾＢＡＺ＠
:. "ＺＭｾ＠
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•.
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The Health PhYSiCS
Introduction to Health Physics
Problems Made Easy
Herman Cember, Ph.D.
Professor Emeritus, Northwestern University and
Visiting Professor, Purdue University
Thomas E. Johnson, Ph.D.
Assistant Professor, Department of Preventive Medicine and Biometrics
Uniformed Services University of the Health Sciences
PS&E Publications
Silver Spring, Maryland
Copyright © 1999 by Herman Cember and Thomas E. Johnson
All rights reserved. No part of this book may be reproduced in any form without
written permission.
Published by:
PS&E Publications
(an imprint of Bartleby Press)
P.O. Box 1516
Silver Spring, MD 20915
Library of Congress Catalog Card Number: 98-68614
ISBN 0-9625963-6-1 .
Review questions in this book are from Introduction to Health Physics, Third
Edition by Hennan Cember, published by McGraw Hill and are reprinted with
permission.
The authors gratefully acknowledge the assistance of Professor W.P. Roach at the
Unifonned Services University of the Health Sciences.
Contents
Introduction
v
Solutions for Chapter 2
Review of Physical Principles
1
Solutions for Chapter 3
Atomic and Nuclear Strucure
39
Solutions for Chapter 4
Radioactivity
59
Solutions for Chapter 5
Interaction of Radiation with Matter
107
Solutions for Chapter 6
--'Radiation Dosimetry
149
Solutions for Chapter 9
Health Physics Instrumentation
209
Solutions for Chapter 10
External Radiation Protection
243
Solutions for Chapter 11
Internal Radiation Protection
279
. Solutions for Chapter 12
Criticality
313
Solutions for Chapter 13
Evaluation of Protective Measures
341
Solutions for Chapter 14
Nonionizing Radiation
395
To our wives, Sylvia and Melissa
!
-'
vi
INDRODUICTION
agencies, such as the American National Standards Institute (ANSI Standards),
American Society for Testing Materials (ASTM), American Conference of Gov';
emmental Industrial Hygienists (ACGIH) These recommendations may be incorporated by reference directly into regulations, or may be applied in the interpretation of the regulations.
If the principles on which radiation safety is based are well understood, then
the details of the continuing revisions may be easily applied to practical situations.
The main purpose of these problems is to help to clarify these basic principles.
Introduction
he problems in the textbook Introduction to Health Physics, 3rd edition, are
based mainly on Dr. Cember's experience as a practicing health physicist
and teacher over a span of more than 40 years. They were formulated
mainly to illustrate practical problems and the principles underlying the problem and
its solution.
Although only one solution is shown for each problem, many of the problems
have an answer that can be reached by more than one path.
As is true with all aspects of environmental and occupational health and
safety, standards and calculational methods are continuously evolving and undergoing changes. Although the authors may attempt to do so, by practical necessity it
is not possible for a textbook to keep current with the very latest recommendations,
standards and computational methodology.
The purpose of a textbook, in contrast to a handbook or to a yearbook, is to
present the basic ideas and principles on which standards an methodology are based.
Thus, for example, in the field of ionizing radiation, the nomenclature, calculational methodology, the biological pharmaco-kinetic models (such as the successive lung models recommended by the NCRP and by the ICRP) and retention functions are undergoing continuous re-examination as more observational and scientific data become available. Even the basic radiation safety philosophy which forms
the very basis for radiation safety standards has undergone a radical change from
the concept of a "tolerance dose" to the zero threshold concept.
In the case of non-ionizing radiation too, we have changing methodologies.
We recognize three radiation zones for the purpose of safety evaluation of RF and
microwave radiation: the near, intermediate, and far fields. There are no sharp dividing lines that show where the near field ends and the far field begins. Rather, the
transition from one to the other is very gradual. There is no definitive formulation,
although there are several different formulas, for estimating the distance from an
antenna to the end of the near field or to the beginning of the far field. Similarly,
since there is no sharp boundary at the edge of a laser beam, and also because of the
nonuniform (generally Gaussian) energy distribution across the beam, there are
several different definitions of the beam diameter.
In the actual practice of health physics, the health physicist must be knowledgeable regarding the latest recommendations of the several non-governmental
T
v
Solutions for Chapter 2
REVIEW OF PHYSICAL PRINCIPLES
Two blocks, of mass 0.1 kg and 0.2 kg, approach each other along a frictionless surface, at velocities 0.4 and 1 mis, respectively. If the 'blocks collide, and
remain together, calculate their joint velocity after the collision.
2.1
0.1 kg
1 mls
0.4 mls
0.2 kg
First, solve for the momentum of each block.
p=mv
0.1 kg x 0.4 mls = 0.04 kg·mls
0.2 kg x 1 mls = 0.2 kg·mls
Since the blocks are moving in opposite directions (remember that velocity has a
direction associated with it), subtract· the momentums:
0.04 kg·mls - 0.2 kg·mls = -0.16 kg·mls
The negative sign indicates that after the blocks collide and join, they will have
momentum to the left. Since the momentum of the 'combined' block is known,
the velocity of the 'combined' block can be found.
The mass of the combined block is
0.1 kg + 0.2 kg =0.3 kg
The momentum of the combined block is -0.16 kg·mls
p
m
v=-=
-0.16 kg . m / s
0.3 kg
-053 m
s
So the 'combined' block would be moving at 0.53 mls to the left.
1
2•1
2
2.2
THE HEALTH PHYSICS SOLUTIONS MANUAL
2.2 A bullet whose mass is 50 g travels at a velocity of 500 m/s. It strikes a
rigidly fixed wooden block, and penetrates a distance of 20 cm before coming to a
stop.
(a) What is the deceleration of the bullet?
1 meter
r=20cm=20cmx 00
1 cm
v = 500 m/sec
0.2m
.
1 kg
In = 50 g = 50 g x 1000 g = 0.05 kg
Equation 2.5
1
21m
W =2" InV = 2 x 0.05 kg x ( 500-;-)
2.
3
= 6.25 x 10
kg· m
2
S2
Equation 2.6A;
2
6.25 X 103 kg· m
2
f= =
sec
r
0.2 m
w
= 3.125 x 10 kg· ｾ＠ = 3.125 x 10 N
4
4
sec
Using equation 2.1,
3.125 X 10 4 kg· m
a=f =
sec 2
In
0.05 kg
= 6.25 x 105
m 2 is the deceleration of the bullet.
sec
4
(b) What was the decelerating force? 3.125 x 10 N as calculated in part (a).
(c) What was the initial momentum of the bullet?
Equation 2.77 is modified to use the initial velocity of the bullet, instead of the
speed of light, so that
kg·m
p= mv =0.05 kg x 500 m/s = 25 -=-sec
SOLUTIONS FOR CHAPTER 2
3
(d) What was the impulse of the collision?
Equation 2.10A gives the impulse:
f D.t = mD.v
Since mD.v is simply the change in momentum, and the final velocity is zero, the
impulse is:
. kg·m
25
per second, which can be written as 25 N'sec
sec
Compute the mass of the earth, assuming it to be a sphere of 25,000 miles
circumference, if at its surface it attracts a mass of 1 g with a force of 980 dynes.
2.3
lxl0-5 N
f = 980 dynes x Id yne
1 kg
9.8 x 10-3 N = 9.8 x 10-3 kg· m
sec 2
-3
m,= 1 g x 1000 g = 1 x 10 . kg
Find the radius of the earth:
Circumference is 25,000 miles:
5280 ft 0.3048 m
7
25000 miles x 1 mile x
1 ft
= 4.02 x 10 meters
Circumference = C = 211: r
C
4.02 X 10 7 meters
r- - - - - - - - - = 6.4 x 106 meters is the radius of the earth.
- 2n 2n
Solving equation 2.28 for m2 , the mass of the earth:
i
L.
2.3
THE HEALTH PHYSICS SOLUTIONS MANUAL
4
F
ｇｭｬｮｾ＠
2
r
.
N·m 2
11
G = 6.67 X 102
kg
Fr2
m --- =
2 - Gnl
l
2.4
mr
9.8 x 10-3 N X (6.4 X 106
= 6.02 X 1024 kg
N
2
x(lx10- 3 kg)
6.67 X 10- ll Ｎｾ＠
kg
2.4 An automobile weighing 2000 kg, and going a speed of 60 kmlhr, collides
with a truck weighing 5 metric tons that was moving at right angles to the
direction of the auto at a speed of 4 kmlhr. If the two vehicles become joined in
the collision, what is the magnitude and direction of their velocity after the
collosion?
2000 kg
Car
c=J
60 kmlhr
Angle =?
..
ｾＮａ＠
4 kmlhr
5000 kg
Truck
Calculate the initial momentum for each vehicle:
p=mv
p car = 2000 kg x 60 kmlhr = 120,000 kg·knilhr
Ptruek = 5000 kg x 4 kmlhr = 20,000 kg·kg/hr
5
SOLUTIONS FOR CHAPTER 2
Recalling that momentum is a vector, we can add the momentum for each vehicle
as we would the sides of a triangle:
Combined momentum of car
and truck represented by
hypotenuse
car momentum of
120,000 kg-kmlhr
truck momentum of 20,000 kg-kmlhr
Adding together the momenta, we obtain:
.j(l20,000)2 + (20,000)2 = 121655
kg·km
hr
Because the car and truck 'stuck together', the combined mass of the truck and
car must be used:
2000 kg + 5000 kg = 7,000 kg
Knowing the momentum, we can find the velocity of the 'combined' truck and
car after the collision using the momentum equation in a rearranged form:
121655 kg· km
v =..E. = ___ＭＮＺｨｾｲ｟＠
m
7000 kg
= 17.38 km is the velocity of the combined truck and
hr
car after the collision_
The direction of the car and truck is determined through geometry, using the
same momentum vectors_ The angle corresponding to the tangent of the opposite
over the adjacent sides would yield the angle 8:
20,000 kg -km
tan e = opposite side =
hr
= 0.1667
adjacent side 120,000 kg· km
hr
Taking the inverse tangent, we obtain: tan -I (0.1667) = 9.46
0
THE REALTIl PHYSICS SOLUTIONS MANUAL
6
:.5
2.5 A small electrically charged sphere of mass 0.1 g hangs by a thread 100 cm
long between two parallel vertical plates spaced 6 cm apart. If 100 volts are
9
across the plates, and if the charge on the sphere is 10- coulombs, what angle
does the thread make in the vertical direction?
m = 1 . 10-4 kg
L= 1 m
d=0.06m
V= 100 V
9
q = 10- C
F e = qt:
<P= ?
y direction
0.06m
x direction
At static equilibrium, the sum of the horizontal forces, F H =0, and the sum of the
vertical forces, Fv =O. The only forces acting on the sphere are:
1. The force of gravity in the negative y direction (Fg = -mg)
2. The electrical force exerted by the plates (Fe = q£)
3. The tension force on the string (T), which can be resolved into two components,
one in the y direction {T( case)} and a second in the x direction { T( sine) }.
"LFH = Fe -T(sine) = 0
"LFv = -mg + T(cose) = 0
Dividing "LFH by I.FV' we have
qE
sinO
-=--=tanO
mg cosO
According to equation 2.39
V
E=-
d
Substituting into the equilibrium equation gives
SoumoNs FOR CHAPTER 2
7
Lx V =tan8
mg
d
tan 8 =
1 X 10-9 C
1 x 10-4 kg x 9.8 ｾ＠
x
100 V
= 1.7 X 10-3
0.06 m
s
e = 0.00170 radians = 0.092°
Note: for very small angles, the tangent of
e = e radian
2.6 A capacitor has a capacitance of 10 J-lF. How much charge must be removed
to cause a decrease of 20 volts across the capacitor?
2.6
10 ｾｆ＠
= 10 x 10--6 F
Use equation 2.66 to solve for the charge:
Q = C x V = lOx 10--6 F x 20 volts = 2 x 10 -4 C
2.7 A small charged particle whose mass is 0.01 g remains stationary in space
when it is .placed in an upward directed electric field of 10 VIcm. What is the
charge on the particle?
First, convert to SI units:
e= 10 V/cm = 1000 Vim
m = 0.01 g = 0.00001 kg
q = ? coulombs
The force acting in the downward direction is equal to the force acting in the
upward direction. The electric force is 'pushing' the particle up, while the force
of gravity is 'pulling' it down.
f = electric force
mg = force from gravity
2.7
8
THE HEALTII PHYSICS SOLUTIONS MANUAL
Since the particle is not moving, the forces are equal. Equation 2.36 gives the
force from the electric field:
f= eq
Equating the electric force and the gravitational force:
eq=mg
Solving for q;
m
0.00001 kg x 9.8 - - 2
q =m x g =
sec
E
1000 V
m
2.8
9.8 x 10-8 C
2.8 A 1 micron diameter droplet of oil, whose specific gravity is 0.9, is introduced
into an electric field between two large parallel plates, separated by 5 mm, across
which is placed a potential difference V volts. If the oil droplet carries a net charge
of 100 electrons, how many volts must be placed across the plates if the droplet is
to remain suspended between the plates?
To accomplish the desired outcome, the downward force of gravity, mg, acting on
the drop must equal the upward electrical force, qe, acting on the drop.
Find the volume of the droplet:
1 x 10-6 m
. 1 micron = 1 micron x .
1 rrucron
Volume of sphere =
4
3
4
3 nr = 3 n (
6
1 x 10- meter diameter
6)3
1 x 102
19
3
= 5.24 x 10- m
Find the mass of the droplet using the specific gravity:
mass = volume x density
6
5.24 x 10-19 m 3 x 1 x 10
[
1m
ｾｭＳ＠
Jx [0.9
ｾ＠ x kg J= 4.7 10- kg
cm
1000 g
X
16
9
SOLlmoNS FOR CHAPTER 2
The gravitational force can be found using equation 2.1;
Fg =ma=mg
g
= 9.8 m/s2
The electrical force, Fe' is given by equation 2.36,
and the electric field intensity, E, is related to the plate separation distance, d, and
the voltage across the plates, V, by equation 2.39,
V
E=-
d
Substituting the expression for E, and equating the upward electrical force to the
downward gravitational force, we have
V
q-=mg
d
m
d
V = mg =
5 x 10-3 m x 4.7 x 10- 16 kg X 9.8-2
sec
100 electrons x1.6 x 10-
q
19
C
electron
v = 1.44 volts
2.9 A diode vacuum tube consists of a cathode and an anode spaced 5 rom apart.
If 300 volts are placed across the electrodes,
(a) What is the velocity of the electron midway between the electrodes, and at the
instant of striking the plate, if the electrons are emitted by the cathode with zero
velocity?
KE='!'mv 2
2
V=
ｾＲ＠
KE
xm
r-'''' ＧＢＭＺＮｾ［＠
I ｜｟ＺＬ［ＮＭＨＩｾ＠
l
L
L.._. ___ . .ＬＮ｟ｾＺ｜Ｂ＠
ｬＢＮＺＭ［ｾ＠
ｾ＠ ［ＺＧｾＢＬ＠
［ｾＮＢ＠
I_:>i:,.:.;,;:..ｾ｜＠ . .:' : ..r. :,':; ＬＮＭＡｾ［Ｂ＠
"Of'
)Y)."
I
ＮＺｾＧ＠ ....__ . . __. _.1
2.9
THE HEALTH PHYSICS SOLUTIONS MANUAL
10
The kinetic energy of an electron after having been accelerated to 300 V is 300 eV,
or 300 eV x 1.6 x 10-19J/eV = 4.8 x 10- 17 1. Substituting the values for the kinetic
energy and for the electron mass and solving for v gives the velocity at the instant
of striking the plate:
17
v(300 eV) =
2 x 4.8 X 10- J = 1.03 X 107 m
ＹＱｸｬＰｾ＠
kg
s
12
Since v is proportional to (KEi , and since potential, and hence the electron's
kinetic energy midway between the plates is V2 the total voltage drop, the
electron's velocity at that point is
v(150 eV) =
T1
x 1.03 107 m = 0.73 x 10 7 m
ｾＲ＠
s
.s
X
(b) If the plate current is 20 rnA, what is the average force exerted on the anode?
m = mass of each electron = 9.1 x 10-31 kg
Now calculate the number of electrons striking the plate in a one second time
period;
1C
20 rnA x __I_A
__ x _s_e_c x _6_.2_x_l0_1_8_e_le_c_t_ro_n_s = 1.24 x 1017 electrons per sec
1000rnA
lA
1C
From part (a);
The electron's velocity is 1.03 x 107 rnIs and changes to zero after it impacts the
plate.
Rearranging equation 2.10a;
f&=ml1v
ml1v
f = -- _
I1t -
[9.1 x 10-31
7
kg x' (1.24 X 1017 electrons)] x (1.03 x 10 m
electron
sec)
sec
1
f= 1.16 x ｉｾｎ＠
2.10
2.10 Calculate the ratios
proton.
%and 7mo for a 1 MeV electron and for a 1 MeV
11
SOLUTIONS FOR CHAPTER 2
Energy of the electron
m
31
=
9.11 x 10- kg
a
8
c= 3 x 10 mlsec
E = 1 MeV = 1 MeV x
1.6 X 10-13 J
MeV
= 1.6 x 10-13 J
Using equation 2.20
E=mc
a
2
Using the respective values:
1.6 x 10- J = (9.11 X l(T" kg) x (3 X 10' mls)' x
13
ｾｉ｟＠
-1
1
v
2
c2
%= 0.941
Find the electron mass ratios using equation 2.4;
m
mo
1
=
ｖＱＭｾ＠
ｾ＠
1
= -;::::===
.J1-0.941
2
=2.96
For a 1 MeV proton
27
rna = 1.67 x 10- kg
8
c = 3 X 10 mlsec
1.6 X 10-13 J
E = 1 MeV = 1 MeV x
MeV
= 1.6 x 10-13 J
Using the relativistic expression for the kinetic energy, equation 2.20
I
__
i ..
12
THE HEALTH PHYSICS SOLUTIONS MANUAL
Substituting in the values:
g
l v,-I
1-2
c
%=0.046
Find the proton mass ratios using equation 2.4;
m
mo
=
g1-;1
1
=-;====
1.001
J1-0.9412
·c
2.11
Assuming an uncertainty in the momentum of an electron to be equal to one
half its momentum, calculate the uncertainty in position of a 1 MeV electron.
2.11
mo = 9.11 x 10- kg
8
c = 3 x 10 mlsec
31
1.6 X 10- 13 J
E = 1 MeV = 1 MeV x
Usirig equation 2.20
MeV
= 1.6 x 10-13 J
2
E=moc
Substituting align the values into the relativistic energy equation, we have
r,
•
13
SOLUTIONS FOR CHAPTER 2
1.6 X 10-13 J = 9.11 X 10-31 kg X (3 X 108m1s)2 X
%=0.941
8
8
v = 0.941 x 3 x10 m1sec = 2.823 x 10 mls
Find the electron mass using equation 2.4;
31
m = --==m=o= = 9.11 x 10- = 2.7 x 10-30 kg
2
.J1-0.9412
1-c2
H
The momentum of an electron is:
8
v = 2.823 X 10 mls
8
30
22
p= mv = 2.7 x 10- kg x 2.823 x 10 mls = 7.6 X 10- kg . mlsec
Assuming the uncertainty associated with the momentum is one half the momentum:
I'::!p =112 x 7.6 x 10-22 kg· mlsec = 3.8 x 10-22 kg· mlsec
According to Heisenberg's Uncertainty Principle, equation 2.81,
h
III x I1p '? 2 Tt
h
Therefore, the minimum uncertainty = III = 2 TtD.p
2
6.6 X 10-34 kg· m
sec
13
&t=
kg·m =2.77x10- m
22
2 x Tt x 3.8 X 10- --=-sec
.1. ....
14
2.12
THE HEALTH PHYSICS SOLUTIONS MANUAL
2.12 If light quanta have mass, they should be attracted by the earth's gravity. To
test this hypothesis a parallel beam of light is directed horizontally at a receiver 10
miles away. How far would the photons have fallen during their flight to the
receiver if their quanta have mass?
Calculate the time it takes for light to travel 10 miles:
.
4
5280 ft x 0.3048 m _
10 mIles x 1 mile
1 ft
- 1.61 x 10 m
8
4
Since light travels at 3 x 10 mlsec and 10 miles.is 1.61 x 10 m
sec
.
4
5
1.61 x 10 m x 3 x 108 m'= 5.36 x 10- sec is the time it takes to travel 10 miles.
Velocity in the vertical direction at time zero is zero, so the distance fallen is given
by:
m2 x (5 .3 6 x 10 _ 5
)2
sec
= 1.4 x 10-8 meters
Y = -1 gt2 = -1 x 98
. -2
2
sec
2.13
2.13 The maximum wavelength of UV light observing the photoelectric effect in
tungsten is 2730 angstroms. What will be the kinetic energy of photoelectrons
produced by UV radiation of 1500 angstroms?
Use equation 2.76,
h = 6.614 X 10-34 J . sec
8
c = 3 x 10 mlsec
A = 1500 angstroms = 1.5 x 10-7 meters
c
E=h -
A
E = 6.614 X 10. J·s x
34
3xl0 8 m
sec x
7
1.5xlO· m
V
= 8.27 eV
e
19
1.6xlO·
J
Thus 8.27 e V is the total energy carried by the incident photon. Since the energy
can1ed by the 2730 angstrom UV photon is the energy, <1>, that binds the electron
to the atom,
r----
SOLUTIONS FOR CHAPTER 2
15
gm
c
3x10 eV
34
<p == h- == 6.614 X 10- 1. sec x
sec x 16 10- 19 1 == 4.54 eV
A
2.73 x 10-7 m
. x
The kinetic energy of the photoelectron (Epe) is equal to the difference between the
photon energy
[h ｾ｝＠
and the energy required to remove the electron from the
atom (the binding energy, <1».
c
E==h--<p
pe
A
So the energy available to the photoelectron is the energy of the 1500 angstrom
light less the energy of ionization:
c
.
E ==h,\ -<p==8.27eV-4.54eV=3.73eV
pe
f\.,
Calculate the uncertainty in position of an electron that was accelerated
across a potential difference of 100,000 ± 100 volts.
2.14
Since the total energy is equal to the kinetic energy of the electron plus the rest
mass energy;
To calculate the uncertainty in momentum (/1p), relativistic effects must be
considered. The relationship among total energy, rest mass energy and momentum under relativistic conditions can be represented geometrically as a right triangle:
pc
E==Ek +E0
= mc
2
E= ｅｾＫｬ｣Ｒ＠
Replacing E obtain:
.Ik:...
2.14
16
THE HEALTIl PHYSICS SOLUTIONS MANUAL
2
2
2
(Ek + E) = Eo + P c
2
Expanding the expression:
Collecting terms and solving for p;
From equation 2.32, the kinetic energy of the electron is equal to the charge times
the applied voltage ( Ek = Vq), so replacing Ek;
Differentiate momentum (P) with respect to voltage,
1 -1 (V 2q2 +2(
dp =-x
Vq) Eo
c
2
)_! (l2V + 2qE )dV
2
0
Where:
8
c = 3 X 10 m1sec
5
V= 1 x 10 volts
dV = 100 x 2 = 200 V (accounts for the ±100 volts)
19
q = 1.6 x 10- C (charge on electron)
31
mo = 9.11; 10- kg (Appendix A)
c = 3 x 10 m1sec (Appendix A)
17
SOLUTIONS FOR CHAPTER 2
m)2
-14 kg. m
E =m/=9.11x10- kgx ( 3x10 =8.2xlO
2
o
0
sec
sec
3I
2
8
Placing values into the equation to find the change in momentum:
1 1 (V 2 q2 +2(
dp=-xVq) Eo )_.!.2 (l2V+ 2qE)dV
c 2
0
Please note that the following equation is in three lines:
dp =
(
1
1
x-x
3x 108 m
2
sec
ｫｧＮｾＲＩＭ＠
(IX105vf x(1.6x10-19CY +2x((lxl05V)x1.6xl0-19C)x8.2xl0-14
.
sec
I
x
25 kg·m
d = 1.94 x 10- -=-P
sec
According to equation 2.81 where:
h
III x flp "? 21t
2
6.6 X 10-34 kg· m
seck
= 5.4 X 10-10 m
III "? 2 hfl =
1t p 2 x 1t x 1.94 X 10-25 g. m
sec
What voltage is required to accelerate a proton from zero velocity to a
velocity corresponding to a de Broglie wavelength of 0.01 angstroms?
2.1S(a)
First, find the momentum associated with this wavelength with equation 2.79:
·L_
2.15
18
THE REALlli PHYSICS SOLUTIONS MANUAL
o
A= 0.01 A= 0.01 A x
1 x 10-lO m
12
= 1 X 10- meters
1A
h 6.63x10-34 J·sec
kg·m
mv = - =
= 6.63 x 10-22 - - " - 12
A
1 x 10- m
sec
The equation from example 2.19 must be used to allow for relativistic effects. To
determine whether relativistic effects are important, substitute the reset mass of
27
the proton, mo = 1.673 x 10. kg into the equation for momentum in example
2.19 and solve for {3:
Replacing values:
8
m
1.672x10-27kg {3
22 kg· m
6.63 X 10=
I
x x (3 x 10 - )
sec
" 1 - B2
sec
A.
v
tJ = -
c
3
= 1.32 x 10-
v = 1.32 X 10-3 x (3 X 108 m1sec) =3.95 x 105 m1sec
Since {3 < 0.1, relativistic effects are negligible, and
1. 2
KE= -mv
2
5 2
m
)
Ｑｾ＠
KE = -1 x 1.673 x 10-27 kg x (
3.95
X 10 X
19 =816 eV
2
s
1.6 x 10- J
Since q(proton) = q(electron), accelerating voltage = 816 V
(b) What would be the kinetic energy of an electron with this wavelength?
The momentum associated with this wavelength would be the same as for the
proton, the mass and velocity would be different, however:
SOLUTIONS FOR CHAPTER 2
-22
mv =6.63 x 10
19
kg· m
sec
The equation from example 2.19 must be used t6 allow for relativistic effects;
8
c = 3 X 10 mlsec
31
mo = 9.11 x 10- kg
v
ｾ］＠
I
-
c
p=mv
Replacing values:
22
6.63 x 10-
kg· m
sec
=
31
9.11 x 10- kg
J1-
ｾＲ＠
A
8
m
x t-' X ( 3 x 10 - )
sec
v
ｾ］＠
- =0.92
c
Since f3 > 0.1, relativistic effects must be considered
Substitute f3 into. equation 2.20 and solve for the kinetic energy of the electron,
using
2( J1-1ｾＲ＠ -lJ = 0.511 MeV x (-J1-1
0.85
E = moc
(c) What is the energy of an X-ray photon whose wavelength is 0.01 angstrom?
34
8
hc
6.614X10- J.secx(3x10 m)
E= =
sec = 1.98 x 10-13 J
12
A
1 x 10- m
1
l _ __
ｬＭﾷｾ＠
ﾷｾＺ［ＧＭ］ｬ＠
I . . . :;.:
u'::'·,c;.>';;: ＧＮＺｾｽ［＠
f
｜Ｇｾ＠
｜ｾ＠ ＯｾＺ＠ ［｜ｾ＠
ｾ＠
.:":::';'"
¥.
ｾﾷ＠
.::_:u,) )..::-')_0
ｾＡＺＮｲ＠
••
,',oj
＼Ｎｾ＠
_ _ . , , .., ...... , . " . _ . . . . . w ..
I
ｾ＠
"t
20
THE HEALTI-I PHYSICS SOLUTIONS MANUAL
J3
eV
6
1.98x10- lx1.6x10- 19 1 = 1.24 x 10 eV=1.24MeV
2.16
2.16 A current of 25 rnA flows through 25 gage wire, 0.0179 in. (17.9 mils) in
22
3
diameter. If there are 5 x 10 free electrons per cm in copper, calculate the
average speed with which electrons flow in the wire.
Find the number of electrons per cm:
d = 17.9 mils = 17.9 mils x
2.54 x 10-3 cm
'1
= 0.0455 cm
llTIl
2
3
Area of circle = n/ = n x{0.5 x 0.0455 cm }2 = 1.62 X 10- cm
5 x 10 22 electrons
3
- - - - 3 - - X 1.62 x 10ｾｮ＠
2
cm = 8.12 x 10
19
electrons
cm
Find the number of electrons traveling per second:
Ｑｾ＠
1A
sec
1 electron
17 electrons
=
1.5625
x
10
x-- x
25 rnA x
sec
1000 rnA 1 A
1.6 X 10-19 C
Combine the two above results:
1m
5
1 cm
1.5625 x 1017 electrons
19
sec
x 100 cm = 1.92 x 10- m1sec
8.12 x 10 electrons x
2.17 An electron starts at rest on the negative plate of a parallel plate capacitor,
and is accelerated by a potential of 1000 volts across a gap of 1 cm.
a. With what velocity does the electron strike the positive plate?
SOLUTIONS FOR CHAPTER 2
, :
21
0.01 m
V== 1000 V
d==O.Olm
31
m == 9.109 x 10- kg
19
q == 1.602 X 10- C
In this problem, the electron moves from a position of high potential energy, to a
place of zero potential energy, converting the potential energy into kinetic energy.
The kinetic energy can be expressed as:
KE == yz m l
The potential energy of a charged particle at ail electrical potential is given by
equation 2.38:
w== Vq
N ow we "have 2 expressions, one for the kinetic energy and another for potential
energy. Setting them equal to each other:
\
ｾ＠
＠ｾ
t
f
YZml=Vq
,
ii
r
>,
!
Substituting in values to fmd velocity:
19
V=
2 x 1000 V x 1.602 X 10- C = 1.88 X 10 7 m
9.109 x 10-31 kg
sec
Since the velocity is less than 10% of the speed of light, relativistic corrections
are not required.
b. How long does it take the electron to travel the 1 cm distance?
L
22
THE HEALTII PHYSICS SOLUTIONS MANUAL
We cannot use the end velocity to find the time since the electron is continuously
, accelerating. However, we can find the acceleration and, knowing the distance,
compute the time.
First, find the force acting on the electron. Using equation 2.36:
/=eq
and combining this with equation 2.38:
Now replace £ in equation 2.36 with V/d:
v
f= -q
d
We can compute the acceleration of the particle across the 1 cm gap with equation 2.1:
/=ma
Since we have an expression for force if = (V/d) q), install that into the classic
f = ma equation and solve for acceleration;
Vq _ 1000 V x 1.6 X 10-19 c
16 m
31
a = md - 9.109 x 10- kg x 0.01 m = 1.76 x 10
7
Since velocity = acceleration x time, and knowing the final velocity is
7
1.88 x 10 mlsfrom part a:
t ］ｾ＠
a
1.88 X 10 7 m
s
16
1.76 x 10 ｾ＠
1.07 X 10-9 sec
S
2.18
2.18 A cylindrical capacitor is made of two coaxial conductors - the outer one
has a diameter of 20.2 mm and the diameter of the inner one is 0.2 mm. The
inner conductor is 1000 volts positive with respect to the outer conductor. Repeat
SOLlITIONS FOR CHAPTER 2
23
parts (a) and (b) of problem 17, and compare the results to those of problem 17.
(a) With what velocity does the electron strike the positive plate?
V= 1000 V
q = 1.6 X 10-19 C
31
m = 9.11 x 10- kg
The kinetic energy of the electron depends only on the potential difference across
which it was accelerated. From equation 2.33:
1 2
Vq = -mv
2
= ｾＲｖｱ＠
=
m
V
19
2 x (1000 V) x 1.6 X 1O- C _
1
9.11 x 10-31 kg
- 1.9 x 10 m1sec
Since this is less than 10% of the speed of light, relativistic effects are ignored.
(b) How long does it take the electron to travel the 1 cm distance?
The travel time is calculated from
velocity = acceleration x time
Since we already know the velocity (from part a), we must calculate the acceleration. From equations 2.1 and 2.36, we have
!=ma= Eq
Eq
a=m
However, the electric field intensity, E varies radially in the case of coaxial
electrodes. According to equation 2.40, E at a radial distance r, is
THE HEALTIl PHYSICS SOLUTIONS MANUAL
24
Where r b is the larger radius and ra is the smaller radius. Substituting for E in the
equation for a, we have
sq
q
V
m
m
a = - = - x ---;--:-
rx
In(:')
dr q
Thus, v = at = - = - x
dt In
V
() x t
r.
r x In .J?...
ra
V
q
rdr =In
-
q
'b
Jrdr = -
ra
J
rb tdt
In ra
(
V
Inl
r
( ) Jt dt
rb 0
ra
n-
Integrating between these limits, we have
1
2
t=
q
V
-x--In
In(rb/ra)
Substituting the numerical values
19
q = 1.6 X 10- C
31
In = 9.11 X 10- kg
V= 1000 V
r = 1 em = 0.01 m
25
SOLUTIONS FOR CHAPTER 2
'a=
0.2 mm
-4
2
=0.lmm=1xl0 m
20.2mm
2
= 10.1 mm = 1.01 x 10- m
b
2
and solving for t, we have
,=
1
2
2
((1.01 x 10- mf -(1 x 10--4mf)
t=
1.6 X 10-19 C
1000 V
9.11 X 10- kg
In(1.01 X 10-2 m/l X 10--4 m)
-----x
-;-------------:31
9
t = 1.65 X 10- sec
Two electrons are initially at rest, separated by 0.1 nm. After both electrons
are released, they repel each other. What is the kinetic energy of each electron
when they are 1.0 nm apart?
2.19
The change in the potential energy of a system of 2 electrons separated by an
initial distance r 1 which increases to r 2 as they repel each other is given by
equation 2.31:
(-1 -1)
/),. W = kq]q2 -;; - ｾ＠
This decrease in potential energy of the system appears as kinetic energy of the
two particles as they move apart. If we substitute the following values into the
above equation, we find that
N·m 2
k = 9 X 10
C2
(equation 2.25)
9
19
q = 1.6 X 10- C
r l = 0.1 nm = 0.1 x 10-9 m
9
r 2 = 1.0 nm = 1 x 10- m
Note that the equation is split into two lines
2
N· m
19
19
/),. W = 9 x 10
C 2 x 1.6 x 10- C x 1.6 x 10- C
9
2.19
26
THE HEALlH PHYSICS SOLUTIONS MANUAL
xCx;;·'m ｏＮｬｸｾﾷＧ＠
DW = 2.07 X 10-18 J
Since this is the total kinetic energy in the system, for each electron
18
KE = Y2 x 2.07 X 10-18 J = 1.04 X 10- J
2.20
2.20 A cyclotron produces a 100 microamp beam of 15 MeV deuterons. If the
cyclotron were 100% efficient in converting electrical energy into kinetic energy
of the deuterons, what is the minimum required power input, in kilowatts?
Ｑｾ＠ sec
15 MeV 1.6 x 10-13 J 1 W
deuteron
10O JlA x
x
x .
x
x= 1500 W
1 x 10 6 J.lA 1.6 x 10- 19 C deuteron
1 MeV
I.!.
s
2.21
2.21 A 1 ｾｆ＠
capacitor is fully charged to 100 V by connecting it across the
terminals of a 100 volt battery. Then, a 2 ｾｆ＠ capacitor·is charged to 100 V in the
same ma.J1ner. The two charged capacitors are then removed from the batteries,
and connected as shown below. What is the charge on each capacitor?
++++++++++
+++++++++
/
-'
Use equation 2.66:
c] = 1 ｾｆ＠
6
= 1 x 10- F
Vj = 100 volts
I
SoumONS FOR CHAPTER 2
27
Q=CV
Q = 1 x 10-6 F xl 00 V = 1 x 10-4 C
C2 =2 JlF =2 x 10- F
VI = 100 V
6
Q = CV = 2 x 10-6 F xl 00 V = 2 x 10-4 C
2.22(a) What voltage must be applied across two oppositely charged parallel
plates, 2 cm apart, in order to have an electron, starting from rest, strike the
8
opposite plate in 10- second?
(b) With what speed will the electron strike the plate?
. Finding the velocity (speed) first, using ｣Ｑ｡ｳｩｾｬ＠
under constant acceleration:
physics since the electron is
tu = 2 cm = 0.02 m
t = 10-8 sec
vo =0
1
tu = v x t = - (v + v )t
2
I
0
Solve for vi
2Ax
t
plate.
2 x (0.02 m)
6
10-8
= 4 x 10 m1sec is the speed the electron strikes the
sec
vI = - - =
To find the required voltage, use equation 2.33:
1
Vq = -mv
2
2
31
m = 9.11xl0- kg
19
q = 1.6 X 10- C
6
V = 4 x 10 m1sec
2.22
28
2.23
THE HEALTH PHYSICS SOLUTIONS MANUAL
2.23 When hydrogen "bums" it combines with oxygen according to
and emits about 2.3 x 105 Joules of heat energy in the production one mole of
water. By what fraction is the mass of the reactants reduced in this reaction?
5
E = 2.3 X 10 J
8
C = 3 x 10 mfs
To find the mass reduction, calculate the mass lost:
E=111C
111
2
.E 2.3x105J
.
= -2
=
= 2.56 x 10-12 kg = 2.56 x 10-9 g IS. the total mass reductIOn
3x 108 m
sec
for one mole. The mass of one mole of water is:
C
2.016 g H2 + 16 gO:::: 18.016 grams in one mole of water
To find the fraction of mass reduction, divide the mass reduction by the total
mass:
9
2.56 X 10- g
10
18.016 g = 1.42 x 10- is the fraction mass reduced
2.24
2.24 (a) A 1000 MW(e) nuclear power plant operates at a thermal efficiency of'
33% and at 75% capacity for 1 year. How many kilograms of nuclear fuel are
consumed?
To solve this problem, the value "1000 MW(e)" must be understood. MW(e)
means that this is how much electrical energy is produced from some thennal
energy input into the plant. No power plant is 100% efficient, and the plant in
this problem converts thermal energy to electrical energy with a 33% efficiency.
Ｎｾ＠
"
r-..
SOLUTIONS FOR CHAPTER 2
29
Therefore, the plant must make more thermal heat than it makes electrical
energy. The energy input required to make 1000 MW in this plant is:
100%
1000 MW(e) x 33% = 3030.3 MW(t)
It is important to note that MW(t) stands for megawatts of thermal energy
produced. Since the plant only operates at 75% capacity, the average MW(t)
power would be:
3030.3 MW(t) x 75% = 2272.73 MW(t) during a year.
Converting into watts and subsequently, joules,
I
r,
t
I
I
1 lIs = 1 W
1(1/ )
7'
6
Isec
3.1536
x 10 sec
10
W =.
7 17 x 10 16 11yr pro227 2 .73 MW()
t x
x
x
W
yr
IMW
duced by the power plant
The mass equivalent of this energy is
E
m- -2 -
7.17 x 10 16 J
- c (3 x108 m)
-
0.8 kg nuclear fuel/yr.
2
sec
2.24(b) If a coal fired plant operates at the same efficiency and capacity factor,
how many kilograms of coal must be burned during the year if the heat content
of the coal is 27 MJlkg (11,000 Btu/pound)?
Using the energy from part (a), the number of kilograms can be calculated.
7.17 X 1016 1 are produced by the power plant burning coal (since the same
amount of electricity was produced at the same efficiency and capacity).
Multiplying to obtain the kg of coal burned:
7.17 X 10 161 x 27 1X kg
10 6 1
L.
2
9
.7 x 10 kg
30
THE HEALTII PHYSICS SOLUTIONS MANUAL
Since there are 1000 kg per metric tonne,
2.7 x 106 tonnes of coal are burned in a year.
2.25
2.25 The solar constant is defined as the rate at which solar radiant energy falls
on the earth's atmosphere on a surface normal to the incident radiation. The
mean value for the solar constant is 1353 W/m2, and the mean distance of the
8
earth from the sun is 1.5 x 10 km.
(a) At what rate is energy being emitted from the sun?
The surface area of a sphere at the distance the earth is from the sun is the area
throughout which the sun's energy is distributed. By taking the sun's power
divided by the area over which it is distributed, the solar constant is found.
= 1.5 x 10 km = 1.5 x 10 m
8
r
A h
ｾ＠
11
= 4nr2 =4 x n x (1.5 x 1011 m)2 = 2.83 x 1023 m2
.
Sun power = surface area x power per unit area
23 2
W 2
26
Sun power = 2.83 x 10 m x 1353 1m = 3.83 x 10 W
(b) At what rate in tonnes per second, is the sun's mass being converted to
energy?
Ｑｾ＠
27
26 W
sec
MeV
1 amu
1.66 x 10- kg 1 tonne
x--x
x
x
x--3 .83 x 10
W
1.6xl0-13 J 931 MeV
1amu
1000 kg
6
=4.3 x 10
2.26
tonnes
sec
2.26 What is the energy of a photon whose momentum is equal to that of a 10
MeV electron?
Since the electron has such high energy, relativistic equations must be used. To
find the momentum associated with the electron, its mass and velocity must be
determined.
The electron's total energy is
31
SOLUTIONS FOR CHAPTER 2
(10 MeV + 0.51 MeV) x 1.6 x 10-
13
J
MeV
-
m
29
m = 1.868 x 10- kg
The electron's velocity at this energy is calculated by solving equation 2.4 for v:
m=g2
1-2
c
Substituting
29
m = 1.868 x 10- kg
31
9.109 x 10- kg
8
c =3 X 10 mlsec
mo=
we find that
v
=(c (1- ::)) ｾ＠ =((3 10' :)'(1- ｾＺ＠
2
X
ｾＺ＠ ｾＡｎ＠
8
v = 2.996 X 10 mls
Using these values, we find that the electron's momentum
8
p = mv = 1.868 x 10-29 kg x 2.996 X 10 mls
kg·m
p = 5.597 X 10s
21
The energy of the photon that has this momentum is found by combining equations 2.76 and 2.78 to give
E =p c
y
L
y
32
THE HEALTIl PHYSICS SOLUTIONS MANUAL
5597 X 10-21 kg· m x 3 X 108 m
=
E
s
s
1.6 X 10- 13 _1_
y
MeV
E = 10.5 MeV
y
2.27
2.27 What is the wavelength of
(a) an electron whose kinetic energy is 1000 eV?
Since the electron;s KE« 15 keV, relativistic effects are negligible. Therefore
2
1
KE=-:-mv
2
1
2 x 0.001 MeV x 1.6 x 10-13 _1_ 2
_ _-=M.:. .=. : :. -V,e .:. - =1.87 x 107 mls
v= _ _ _ _ _ _ _
9.11 X 10-28 kg
The de Broglie wavelength is found with equation 2.78:
34
h = 6.614 X 10- 1 . sec
o
h .
A=-=
mv
34
6.614 X 10- J . S
.
rn
9.11 x 10-28 kg x 1.87 X 107 -
-11
=3.87x 10
S
(b) a 10-8 kg oil droplet falling at a rate of 0.01 mlsec?
The de Broglie wavelength, equation 2.79, is:
A = ｾ＠
34
= 6.614 X 10- 1 . sec
mv
1 x 10-8kg x 0.01 rn
sec.
6.614 x 10-24 rn
rnx
1A
= 0.387 A
10
1x10- rn
r
SOLUTIONS FOR CHAPTER 2
33
(c) a 1 MeV neutron?
Determine whether the 1 MeV neutron is relativistic. According to equation 2.20
KE
The
of the neutron is 4.7 Me V when v = 0.1 c. Therefore, we have a nonrelativistic situation in the case of a 1 MeV neutron. The neutron's KE = Y2 ml
and it's momentum therefore is
1
p=mv=m(
KE)2
2x
m
.
27
The neutron's mass = 1.67 x 10- kg, and
13
its KE = 1 MeV x 1.6 x 10. JlMeV = 1.6 x lo·13 J
Substituting these values into the momentum equation above yields
P
p
= 1.67 ｾ＠ 10-27 k (2 x 1.6 x 10-
1
13
1]:2
g 1.67 X 10-27 kg
kg·m
sec
= 2.31 x 10-20 -.::::..-
The de Broglie wavelength is
34
A =!!.- = 6.614 X 10- J . sec = 2.86 x 10-14 meters
p
2.31 x 10-20 kg· m
sec
2.28 (a) The specific heat of water in the English system of units is 1 Btu per
pound degree F.; in the cgs system, it is 1 calorie per gram degree C. Calculate
the number of joules per Btu, if there are 4.186 joules per calorie.
1Btu x lIb x 9°F x(X)_J- x 1 calorie = 1 calorie
Ib·oF 454 g 5°C
Btu 4.186 J
g'oC
2.28
THE HEALTIl PHYSICS SOLUTIONS MANUAL
34
x = 1.05 X 10 J in one Btu.
3
(b) What is the specific heat of water, in J/kg.oC
1 calorie 4.186 J 1000 g
J
gram.. C x calorie x kg = 4186 kg·· C
2.29
2.29 The maximum amplitude of the electric vector in a plane wave in free space
is 275 Vim.
(a) What is the amplitude of the magnetic field vector?
In free space, the magnetic field vector is related to the electric field vector by
equation 2.56
V
E =275o
m
7
llo = 4 x 1t X 10-
c2
A2 ·m2
Substituting these values into equation 2.56:
(b) What is the rms value of the electric vector?
I
35
SOLUTIONS FOR CHAPTER 2
275 V
Maximum amplitude
V
RMS value =
(;:;2
= _:==-m!;.. = 194.5 -
J2
vL
m'
2
(c) \Vhat is'the power density, in mW/cm , in this electromagnetic field?
Using the values from part (a) in equation 2.68:
V
E = 275 -
m
o
275 V
m
Z = Eo =
A
Ho
0.73
= 377 ohms
m
Solving with equation 2.71:
I
ｾＮ＠
2
2 = 10 mW/cm
= 100 -W
I
I
t
ｾＺ＠
t
J
m
2
What is the free space power density, in milliwatts per cm , of a 2450
MHz electromagnetic wave whose maximum electric intensity is 100 millivolts/m?
2.30(a)
p=
(7t)' = ＨｏＮＱｾＯｊＲｲ＠
Z
=
377
1.33 10-5 ":
x
m
(b) What is the maximum magnetic field intensity in this wave?
2.30
36
THE HEALTH PHYSICS SOLUTIONS MANUAL
Equation 2.56:
mV
E = 100 = 100 x 10-3V
-
m
o
m
2
C
N .m2
-12
co=8.85 x 10
C2
7
1l0 =4xnxl0-
E
Ho=
ｏｾｴＮ＠
r;:
2
A ·m
2
2
3
100 X 10- V X
m
ｾ＠
.,J;: =
8.85 X 107
4 x n x 10-.
12
C 2
N·m
C'
2
A ·m
2
A
Ho = 2.65 x 10 -4
m
2.31
A radio station transmits at a power of 50,000 W. Assuming the electromagnetic energy to be isotropic ally radiated (in the case of a real radio transmitter,
emission is not isotropic),
(a) What is the mean power density at a distance of 50 km?
2.31
50 km = 50,000 meters
10
2
A sphere = 4nr2 = 4n(50000 m)2= 3.14 x 10 m
50000 W
-6
2
-7
2
3.14 X 1010 m 2 = 1.59 x 10 W/m = 1.59 x 10 mW/cm
(b) What is the maximum electric field strength at this distance?
The mean power density and the maximum electric field strength are related by
equation 2.71
SOUJTIONS FOR CHAPTER 2
v
37
_V
m
E =0.035-=3.5xl0 2 m
o
(c) What is the maximum magnetic field strength at that distance?
Solving for Ho:
2 x (1.59 X 10-6 W)
ｾ＠
2P
H --=
0-
Eo
0.035 V
m
2
2.32 The mean value for the solar constant is 1.94 calories per cm • Calculate the
electric and magnetic field strengths corresponding to the solar constant.
2
P = 1.94 calories x 4.186 J x 1 min x 10000 cm x 1 W = 1353.5 W
2
min·cm
1m2
Ｑｾ＠
m2
calorie 60 sec
sec
The maximum electric field strength is found by substituting equation 2.74
p=
(i)' (Eo/J2)'
Z
-
E=
lz
= 1353.5 W/m'
377
E
= 714 Vim
The average power level, in terms of RMS values for E and H is, according to
equation 2.73
P=ExH
2.32
THE HEALTH PHYSICS SOLUTIONS MANUAL
38
W
A
135352
H =
m =1.89714 V
m
m
3
2.33
2.33 What volume of water, m , must fall over a dam 10m high to generate
enough electricity to keep a 100 W electric bulb lit for 1 year, if the overall
efficiency of the hydroelectric plant is 20%?
E = Power x time
J/
.
l/sec 1
365 d 24 hr 3600 sec
E = 100 W x W x yrx lyr x d x
hr
9
3.154 X 10 J is the
energy required over one year to light the light bulb. This energy required is equal
to the amount of potential energy the water has due to its height behind the dam;
9
PE = 3.154 x 10 J
g = 9.8 mls2 (gravitational acceleration)
h = 10 meters
Potential energy = PE = mgh
Solving for the mass of water required:
9
•
•
m = -PE = 3.154 x 10 J = 3.22 x 107 kg IS
the mass of water reqUIred
to
gh
m
9.8-x10m
sec 2
produce 100 watts for 1 year.
However, since the plant is only 20% efficient,
8
3.22 X 10 7 kg
0.2
= 1.61 x 10 kg of water will be actually used.
3
The water density is 1000 kg/m :
5
3
1.61 X 108 kg x 1000 kg = 1.61 X 10 m is the volume of water required.
ｾＭ
Solutions for Chapter 3
ATOMIC AND NUCLEAR STRUCURE
What is the closest approach that a 5.3 MeV alpha particle can make to a
gold nucleus?
3.1
3.1
Gold has 79 protons, giving it a net charge on the nucleus of :
q1
=79 protons x
proton
1.264 X 10-17 C
An alpha particle has 2 protons, giving it a net charge on the nucleus of:
q2 = 2 prot ons x
1.6 x 10-19 C
proton
-19 C
- 3.2 X 10
1x106 eV 1.6x10-19 J
13
W = 5.3 MeV = 5.3 MeV x 1 MeV x
leV
= 8.48 x 10- J
Equation 2.24 is the equation for electrical force between two charged particles:
I
/ =ko qlq2
r
2
•
And equation 2.2 gives the equation for work:
W=fr
The work done by bringing these two widely separated charged particles to a
separation r is given by combining equation 2.2 and 2.24;
39
THE HEALTII PHYSICS SOLUTIONS MANUAL
40
Solving for r, the closest approach;
3
3.2
3.2 Calculate the number of atoms per cm of lead, given the density of lead is
11.3 g/cm3 and its atomic weight is 207.21.
23
22 atoms
11.3 g
1 mole
6.02 x 10 atoms
x
1
3.29
x
10
3 x 207 21
cm
. g
mo e
cm 3
3.3
3.3 A ｾＭ
10
meson has a charge of - 4.8 xl0- sC and a mass 207 times that of a
resting electron. If a proton should capture a 1-1- to form a "mesic" atom, calculate
(a) the radius of the first Bohr orbit
fmd the ground state orbit radius using equation 3.6;
n=1
34
h = 6.625 X 10- J·sec
28
31
m
= 207 x 9.11 x 10- kg = 1.89 xl0- kg
meson
Z=1
-10
e = 4.8 x 10
-10
sC = 4.8 X 10
1C
.
-19
sC x 3 X 109 sC = 1.6 xl0 C
r
12 X (6.625 X 10-34 J . sec
r
41
SOLUTIONS FOR CHAPTER 3
11
r = 2.49 xl0- J3 m = 2.49 x 10- cm
(b) the ionization potential equals the energy, eV, in orbit n i , equation 3.10.
2n2ko2mZ2e41
E=
h2
n2
n=l
34
h = 6.625 x 10- }xsec
28
m
=207 x 9.11 x 10-31 kg = 1.89 X 10- kg
meson
Z=1
10 ｎｾＧｲ＠
2xrr' X(9 X
9
X1.89xlO-28kgX(l')x(1.6XIO-19C)'
E=
(6.625 x l0-34 J .SeCr
E =4.5 x 10-16} = 4.5 x 10-16J
1
x12
eV
3
= 2.8 x 10 eV
x 1.6 X 10-19 J
3.4 Calculate the ionization potential of a singly ionized 4He atom.
The ionization potential represents the binding energy of an electron; it represents the work that must be done in order to remove the electron from the atom.
For a one electron atom, such as a singly ionized helium atom, the energy level
of an electron in orbit np and hence the binding energy , is given by equation
3.10.
I•
I
1
,f
t·
I
1
ｾ＠
I!
-L.
E=-2n
2k 2 Z2 4
om e 1
h
2
n i2
where,
n = 1 = ground state
h'= 6.625 x 10-34 J ·sec
m electron = 9.11 x 10-31 kg
3.4
42
THE HEALTII PHYSICS SOLUTIONS MANUAL
Z=2
N·m 2
kO = 9 x 10
2
C
e = 1.62 x 10- 19 C
9
Substituting these values into equation 3.10 we have:
2X1t' X(9 X 109
ｎｾＧｲ＠
E=
x9.11xlO-31kgx(2')x(1.6xlO-19C)'
r
(6.625 X 10-34 J . sec
1
x12
E = 8.68 X 10- 18 J
In electron volts, we have
-18
E = 8.68 x 10
3.5 .
eV
J x 1.6 X 10-19 J 54.4 eV
3.5 Calculate the current due to the hydrogen electron in the ground state of
hydrogen.
First, find the radius of the ground state orbit using equation 3.6;
where,
n=l
34
h = 6.625 x 10- J 'sec
31
me1ectron = 9.11 X 10- kg
Z=l
SOLlITIONS FOR CHAPTER 3
43
e = 1.6 x 10-19 C
r =
2
2
4n: me Zko
12 X (6.625 X 10-34 J . sec
r
11
r= 5.17 x 10- m
Use equation 3.3 to find the velocity of the electron:
nh
v =- - =
2n:rm
(1) x 6.625 x 10-34 J . sec
= 2.24 X 106 mlsec
2 x n: x (5.17 x 10-11 m) x 9.11 x 10-31 kg
The distance that the orbital electron in hydrogen atom travels is the circumfer11
10
ence of its orbit circumference = 2n:r = 2 x n: x (5.17 x 10- m) = 3.25 x 10Next, compute the time to make a revolution around the hydrogen atom.
10
d = 3.25 X 10- m
6
v = 2.24 X 10 mlsec
d
3.25 x 10-10 m
t =- =
v
2.24 x 106 m
16..
.
= 1.45 x 10- sec IS the time for a complete revolutIOn.
sec
Current is measured as the flow of electric charge per unit time:
1 electron
1.6 x 10-19 C I A
1000 rnA
- - - - . ,16- - x
x - - x - - - - = 1.1 rnA
1.45 x 10- sec
1 electron
1ｾ＠
1A
sec
Calculate the ratio of the velocity of a hydrogen electron in the ground state
to the velocity of light.
First, find the radius of the ground state orbit using equation 3.6;
n=1
34
h = 6.625 X 10- J·sec
31
meIectron = 9.11 X 10- kg
Z=1
3.6
3.6
44
THE HEALTH PHYSICS SOLUTIONS MANUAL
e = 1.62 x 10- 19 C
n 2h 2
r=--4n 2 me 2 Zk o
12 X (6.625 X 10-34 J . sec
r
(9 10 _N_._m_J
2
r= 4 x ,,' x (9.11 x 10-31 kg) x (1.62 x 10-" C)' x 1 x
X
9
C
r = 5.17 x 10-
11
2
m
Use equation 3.3 to find the velocity of the electron:
nh
(1) x 6.625 x 10-34 J . sec
2nrIn
2 x n x (5.17 x 10-11 m) x 9.11 x 10-31 kg
v =- - =
= 2.24 X 106 m/sec
8
The speed of light is 3 x 10 m/sec
2.24 x 10 6
-3-x-10-'-g- = 0.0075
3 ｾ＠ 7
3.7 Calculate the Rydberg constant for deuterium.
Equation 3.12 is where the elements of the equation for Rydberg's constant for a
single-electron atom is found:
8
c = 3 x 10 m/sec
n=l
34
h = 6.625 X 10- J·sec
meIectron ｾ＠ 9.11 X 10-31 kg
Z=l
N·m 2
ko =9 X 10
C2
9
e = 1.6 x 10-19 C
Substituting these values:
R
2n(ko) 2 mZ 2 e 4
SOLUTIONS FOR CHAP1ER 3
2X,,2 X(9 X
R==
10 ｎｾＲｲ＠
9
45
x9.l1xlO-31 kgx(1)2 x(1.6XIO-19
C)'
r
(3 X 108 m) x (6.625 X 10-34 J . sec
sec
3.8 What is the uncertainty in the momentum of a proton inside a nucleus of
27
AI? What is the kinetic energy of this proton?
The maximum uncertainty in position of a proton inside the nucleus is the
diameter of the nucleus. Find the radius of the 27 Al nucleus to determine its
position with equation 3.1
A == 27
15
r== 1.2 x 10- A1f3 == 1.2 X 10-
15 X (27)1/3 == 3.6 X 10-15 meters
Since the furthest one can be away when determining the position is across the
nucleus, the diameter is the maximum uncertainty:
3.6 x 10-15 meters X 2 == 7.2 x 10- 15 meters
From equation 2.81 find the momentum uncertainty:
34
h = 6.625 X 10- J·sec
x =7.2 X·10- 15 meters
>
h
LlP
- 21t ｾ＠
A
I1p ｾ＠
= 6.625 X 10-34 J . sec
kg· m
=1.46 x 10-20
2 x 1t x (7.2 x lO-15 m )
sec
1.46 X 10-15 g. cm
sec
Since the mass of the proton is
mproton = 1.676 x 10-27 kg, its velocity is
L
3.8
46
THE HEALTH PHYSICS SOLUTIONS MANUAL
p
1.46 x 10-20 kg·m
= _ _ _ _ＭＮＺｳ［･｣ｾ
m
1.676 x 10-27 kg
V = -
8.76 x 106 m1sec
Now find the kinetic energy of the proton (equation 2.3)
1
E = -mv
2
2
= -1 x 1.676 X 10-27 kg x ( 8.76 X 106 m ) 2 = 6.42 x 10-14 J
2
ｾ｣＠
-14
MeV
6.42 x 10 J x 1.6 X 10-13 J = 0.4 MeV
3.9
3.9 A sodium ion is neutralized by capturing ale V electron. What is the
wavelength of the emitted radiation if the ionization potential of Na is 5.41 volts?
Since it is an electron experiencing the 5.41 V, the electron must lose not only the
energy of ionization but also the 1 e V of kinetic energy. Thus, the energy lost in
the ionization process is 6.41 eY. Equation 2.76 can be rearranged to find the
wavelength:
h = 6.625 X 10-34 J·sec
8
c = 3 x 10 m/sec
1.6 x 10-19 J
E = 6.41eV x
1 eV
= 1.03 x 10-18 J
6.625 X 10-34 J . sec(3 x 108 m)
., __ _hc __
sec
7
/I"
------- - - - = 1.93 x 10- m = 193 nm
18
E
1.03 x 10- J
3.10
3.10(a) How much energy would be released if one gram deuterium were fused
to form helium according to the equation 2H +2H---74He + Q?
Find the number of 2H atoms present in 1 g 2H:
SOLUTIONS FOR CHAPTER 3
47
23
23
mole 6.02 x 10 atoms
1 g (H) x -2- x
1
1
= 3.01 x 10 atoms, however it takes two
moe
g
2
23
atoms of 2H to make one atom 4He, so only 1.505 x 10 atoms will be made from
this reaction.
Sum the mass of the deuterium atoms and subtract that from the mass of the 4He
atom fonned, to determine the mass deficit:
2H = 2.0140 amu
4He = 4.0026 amu
2 x (2.0140 amu) - 4.0026 amu = 0.0254 amu per atom of 4He fonned
Since 1 amu= 931 MeV
0.0254 amu 931 MeV 1.6 x 10-13 J
II
1.505 x 10 atoms x
x
x
MeV
= 5.7 x 10 J
atom
amu
23
(b) How much energy is necessary to drive the two deuterium nuclei together?
The work done in overcoming the repulsive force when bringing the two nuclei
together until they touch, at a center to center distance of the two nuclei radii, is
given by equal 2.31:
where
ql =q2= 1.6 x 10-19 C
The distance rl is calculated with equation 3.1
15 I13
r l = 2r= 2 x 1.2 X 10· A = 2 x 1.2 X 10-15 X i 13 = 3 x 10-15m
W=
I
I
I
L
ｫｏｱｬＲＨｾＭＩ＠
rl
r2
14
14
W = 7.68 x 10- J = 7.68x 10- J x
1 MeV
13
= 0.48 MeV
6
1. x10- J
48
3.11
THE HEALTH PHYSICS SOLUTIONS MANUAL
3.11 The density of beryllium, atomic number 4, is 1.84 g/cm3 , and the density
3
of lead, atomic number 82, is 11.3 g/cm • Calculate the density of a 9Be and a 208Pb
nucleus.
A(Be)
=9
15
r = 1.2 x 10- A
1/3
15
= 1.2 X 10-15 x 91/3 = 2.5 x IO- m
Volume of the nucleus (assume sphere):
Now divide the atomic mass of beryllium. by the volume of the nucleus:
9 amu
6.545 X 10-44 m
3 X
3
1m
14 g
1.66 x 10-27 kg 1000 g
X
X
-6
3 = 2 3 x 10
•
1 amu
1 kg
1 X 10 cm
cm 3
For lead:
A(Pb) = 207
15
15
r = 1.2 x 10-15 Al/3 = 1.2 X 10- X 20i/3 = 7.1 x 10- m
Volume of the nucleus (assume sphere):
4 3 4
15 3
-42 3
V=-nr =-xnx(7.1x10- m)=1.5x10 m
3
3
Now divide the atomic mass of beryllium by the volume of the nucleus:
3
27
207 amu x 1.66 x 10- kg x 1000 g x
1m
= 2.3 x 1014 _g_
6
3
1.5x10-42 m 3
1 amu
1 kg
1x10 cm
cm 3
Note that the nuclear density is generally independent of the atomic number.
3.12 Determine the electronic shell configuration for aluminum, atomic number 13.
3.12
from table 3.1
Shell number
N umber of electrons in shell
K
2
L
8
M
3
SOLUTIONS FOR CHAPTER 3
49
3.13 What is the difference in mass between the hydrogen atom and the sum of
the masses of a proton and an electron? Express the answer in energy equivalent
(eV) of the mass difference.
3.13
"Masses
IH = 1.007825035423 amu
Ip = 1.00727647012 amu
e- = 5.48579903 X 10-4 amu
(1.00727647012 amu + 5.48579903 X 10-4 amu) - 1.007825035423 amu
8
= 1.46 x 10- amu
9
1.07356 X 10- amu = 1 eV
-8
eV
1.46 x 10 amu x 1.07356 x 10-9 amu = 13.6 eV
3.14 If the heat of vaporization of water is 540 calories per gram at atmospheric
pressure, what is the binding energy of a water molecule?
3.14
H 20 = 18 g per mole
540 cal 4.186 J
eV
18 g (H 2 0)
1 mol"
19
g
x cal x 1.6 x 10- J x
mol
x 6.02 x 10 23 molecules - 0.42 eV
3.15 The ionization potential of He is 24.5 eV.
(a) What is the maximum velocity with which an electron is moving before it can
ionize an unexcited He atom?
me= 9.11 x 10-31 kg
E = 24.5 eV = 24.5 eV x
1.6 X 10-19 J
eV
=3.92 x 10-18 J
Solve for velocity using equation 3.7:
. )2E
V
= --;;; =,
2 x 3.92 X 10-18 J
9.11 X 10-31 kg
=2.93 10 mlsec
6
X
3.15
····f=-:
50
THE HEALTH PHYSICS SOLUTIONS MANUAL
(b) What is the minimum wavelength of a photon in order that it ionize the He
atom?
Equation 2.76:
h = 6.625 X 10-34 J·sec
8
e = 3 x 10 m/sec
E = 24.5 eV = 24.5 eV x
1.6 X 10-19 J
eV
=3.92 x 10-18 J
m
he 6.625 x 10-34 J . sec x 3 x 108 A=- =
sec =50.7x 10-9 m=50.7nm
18
E
3.92 x 10- J
3.16
3.16 In a certain 25 watt mercury-vapor ultraviolet lamp, 0.1 % of the electrical
energy input appears as U.v. radiation of wavelength 2537 angstroms. What is
the photon emission rate, per second, from this lamp?
Find the energy of each photon with equation 2.76:
h = 6.625 x 10-34 J 'sec
8
e = 3 x 10 m/sec
o
A = 2537 Ax
1 x 10-10 m
0
_
2.537 X 10-9 m
1A
6.625 X 10-34 1· sec x 3 x 108 _m_
_ _ _ _ _ _ _---::-_ _---'-se_c = 7.834 X 10-19 l/photon
2.537 x 10-9 m
Find the actual amount of energy given off in the UV range:
1
11
25 W x 0.1 x sec = 0.025100 1 W
sec
SOLUTIONS FOR CHAPIER 3
51
Multiply the energy of the photons by the UV energy to find the photon emission
rate:
16 photons·
0.025 J
photon
19
sec
sec x 7.834 x 10- J = 3.19 x 10
3.17 The atomic mass of tritium is 3.017005 amu. How much energy in MeV is
required to dissociate the tritium into its component parts?
3.17
Sum the individual components which make up tritium and determine the mass
deficit:
n = 1.008665 amu
p = 1.007276 amu
e = 5.485 x 10-4 amu
Tritium has 2 neutrons, 1 proton and 1 electron:
(2) x 1.008665 amu + (1) x 1.007276 amu + (1) x 5.485 X 10-4 amu
= 3.0251545 amu
Subtract from this the mass of tritium to find the mass deficit:
3.0251545 amu - 3.017005 amu = 8.1495 x 10-3 amu
931 MeV = 1 amu
3
8.1495 x 10- amu x
931 MeV
= 7.587 MeV is required to dissociate the tritium.
amu
Compute the wave length frequency, and energy (electron volts) for the
second and third lines in the Lyman series.
3.18
Equation 3.2 to solve for the wavelength:
R = 1.097 X 107 m- I
3.18
52
THE HEALTII PHYSICS SOLUTIONS MANUAL
n1 = 1
n 2 = 3 (since the fIrst line would be formed between the 1st and 2nd orbits)
n: - ｮｾ＠ 1J = 1.097 x 10 (1Ｑｾ＠ - Ｓｾ＠ 1J = 9.76 x 10 m-
1 = R(1
A
7
6
I
A, the wavelength of the 2nd line, = 1.023 x 10-7 m
Equation 2.57 describes the frequency and wavelength relationship:
m
.
·3x10 8 15
f = ｾ＠ =
sec = 2.93 x 10 sec-I is the frequency of the 2nd line.
A 1.023 x 10-7 m
nd
The energy of the 2 line is found using equation 2.76A;
h = 6.625 X 10-34 J·sec
34
1
E = hf = 6.625 X 10- J·sec x 2.93 x 1015 = 1.94 X 10-18 J
sec
1 eV
18
1.94 X 10- J x 1.6 X 10-19 J
12.11 eV is the energy of the 2nd line.
The third line is calculated using equation 3.2 to solve for the wavelength:
R = 1.097 X 107 m-1
c= 3x 108 m
s
n1 = 1
n2 =4
"
SOLUTIONS FOR CHAPTER 3
53
A = 9.7 X 10-8 m is the wavelength of the 3rd line.
Equation 2.57 describes the frequency and wavelength relationship:
3 X 108 m
f -- ｾ＠ A -- 9.7 X 10-sec
8m
3.085 x lOis sec-I is the frequency of the 3fd line.
The energy of the 3rd line is found using equation 2.76A;
34
h = 6.625 x 10- J 'sec
34
E = hf= 6.625 X 10- J·sec x 3.085 x 10
eV
18
2 .044 x 10- J x - - - 191.6 X 10- J
15
1
-
sec
= 2.044 X 10-18 J
12.78 eV is the energy of the 3fd line.
3.19 Using the Bohr atomic model,. calculate the velocity of the ground state
electrons in hydrogen and in helium.
First, find the ground state orbit radius for hydrogen using equation 3.6;
n=l
34
h = 6.625 x 10- J 'sec
meIectron = 9.11 X 10-31 kg
Z=l
N·m 2
ko = 9 X 10
C2
9
e
= 1.62 x 10- C
19
n2h 2
r=--4n 2 me 2 Zk o
3.19
54
THE HEALTH PHYSICS SOLUTIONS MANUAL
r = 5.17 x 10- JI m
Use equation 3.3 to find the velocity of the electron:
(1) x 6.625 x 10-34 J . sec
6
= 2.2 X 10 m1sec is the
v = -- =
31
11
2nrm 2 x n x (5.17 x 10- m) x 9.11 x 10- kg
nh
velocity of the hydrogen electron.
For helium, find the ground state orbit radius using equation 3.6;
n=l
h = 6.625 X 10- J·sec
31
me1eClron = 9.11 X 10- kg
Z=2
34
N ·m
ko=9x10 -C29
2
19
e = 1.62 x 10- C
r
12 X (6.625 X 10-34 J . sec
ｲ］Ｍｾ
ｾＬ［ＧＩ＠
4 x ,,' x (9.11 X 10-31 kg) x (1.62 X 10-1' C)' x 1 x (9 X 10' N
11
r = 2.58 x 10- m
Use equation 3.3 to find the velocity of the electron:
(1) x 6.625 x 10-34 J . sec
6
v =- - =
= 4.49 X 10 m1sec is the
2nrm 2 x n x (2.58 x 10-11 m) x 9.11 x 10-31 kg
nh
velocity of the helium electron.
3.20
3.20 The heat of combustion when H2 combines with 02 to form water is 60
kcallmole water. How much energy (electron volts) is liberated per molecule of
water produced?
60 kcal x
1 mole·
x 1000 cal x 4.186 J x
leV
= 2.6
eV
19
23
kcal
cal
1.6 x 10- J
molecule
mole
6.02 x 10 molecules
SOLUTIONS FOR CHAPTER 3
55
The atomic weightcs of 1610' 170h and 18? are .15 ·994915, 16.999131, and
17.999160 respectively. alcu ate t e atomIC weig h t of oxygen.
3.21
3.21
The abundance of each isotope can be found in the CRC handbook.
Type Oxygen
Weight
Abundance
Abundance x weight
16
15.994915
0.99759
15.956367
17
16.999131
0.00037
0.006289678
18
17.999160
0.00204
0.036718286
Sum
15.999377522
3.22 Calculate the molecular weight of chlorine, C12 , using the exact atomic
weights of the CI isotopes given in the reference sources (CRC).
Type Chlorine
Weight
Abundance
Abundance xWeight
35
34.968852
0.7577
26.49589916
37
36.965903
0.2423
8.956838297
Sum
35.45273746
3.23 If 9 grams of NaCI were dissolved in 1 liter of water, what would be the
concentration, in atoms per mL, of each of the constituent elements in the
solution?
Na = 22.99 g/mole
CI =35.45 g/mole
NaCI = 22.99 g/mole + 35.45 g/mole = 58.43 g/mole
Calculate the number of atoms ofNa added to the solution in the 9 grams:
6.02 x 10 23 atoms Na
1L
mol NaCI
1 mol Na
9 g NaCI
--=--- x
x
x
x--IL
58.43 g NaCI 1 mol NaCI
1 mol Na
1000 mL 9.27 x 10
19 atoms Na
mL
Calculate the number of atoms of CI added to the solution in the 9 grams:
6.02 x 10 23 atoms CI
1L
9 g NaCl
mol NaCI
1 mol CI
---x
x
x
x
=
1L
58.43 g NaCI 1 mol NaCI
1 mol CI
1000 mL
3.22
3.23
56
THE HEALTI-I PHYS1CS SOLUTIONS MANUAL
9.27 x 10
19
atoms CI
mL
H = 1 g/mole
= 16 g/mole
H 20 = 2(1 g/mole) +16 g/mole = 18 g/mole
°
ｾ＠
Calculate the number of atoms of H in 1 milliliter of water (assume 1 liter = 1000
g water):
1000 g HiO mol H 2 0
2 mol H
6.02 x 10 23 atoms H
1L
----x
x
x
x
=
18 g H 2 0 1 mol H 2 0
1 mol H
1000 mL
1L
atoms H
6.69 x 1022 - - mL
Calculate the number of atoms of
water):
°in 1 milliliter of water (1 liter = 1000 g
°
1000 g H 0 mol H 2 0
1 mol
6.02 x 10 23 atoms O I L
- - - -2 x
x
x
x_
1L
18 g H 2 0 1 mol H 2 0
1 mol
1000 mL 22
3.34 x 10
3.24
atoms
mL
°
°
3.24 The visual threshold of the normal human eye is about 7.3 x 10- 15 W /cm2
for light whose 'A = 556 nm. What is the corresponding photon flux in photons/
2
cm /sec?
'A = 556nm = 556 x 10-9 m
Find the energy of the photons in joules using equation 2.76;
m
c=3x 108 sec
34
h = 6.625 X 10- J·sec
3 X 108 m
E == h ｾ＠ = 6.625 X 10- J·sec x 556 x ＱＰｾＧＺｮ＠
34
photon.
19
= 3.57 x 10- J is the energy per
r
ｾ＠
SOL1JI10NS FOR CHAPTER 3
57
Multiplying the energy per unit area by the number of photons per joule (inverse
of 3.57 x 10-19 J /photon):
1_1_
sec
1 photon
photon
7.3 x 10-15 W
4 -=---_---:--x
X
=
2
x
10
cm 2
1W 3.57 x 10-19 1
cm 2 sec
3.25 What is the binding energy of the last neutron in 17O?
Find the mass deficit between 160 plus 1 neutron, and 170 , which is the energy of
removing one neutron, to determine the binding energy:
From CRC:
17
0 = 16.999131 amu
16 0 = 15.994915 amu
In = 1.008665 amu
16
0 + In =15.994915 amu + 1.008665 amu = 17.00358 amu
The difference in masses is:
17.00358 amu - 16.999131 amu = 0.004449 amu
931 MeV = 1 amu
0.004449 amu x
931 MeV
= 4.14 MeV
1 amu
3.25
·t
I"
iｾＧＬ＠
ri
\
Solutions for Chapter 4
RADIOACTIVITY
Carbon-14 is a pure beta emitter that decays to 14N. If the exact atomic
masses of the parent and daughter are 14.007687 and 14.007520 atomic mass
units, respectively, calculate the kinetic energy of the most energetic beta particle?
-4.1
4.1
14.007687 - 14.007520 = 1.67 x 10-4 amu
931 MeV = 1 amu
-4
931 MeV
= 0.156 MeV
1.67 x 10 amu x 1
amu
4.2 If 1.0 MBq (27 JlCi) 1311 is needed for a diagnostic test, and if 3 days elapse
between shipment of the radioiodine and its use in the test, how many Bq must
be shipped? To how many JlCi does this correspond?
A = 1.0 MBq (Final quantity desired)
t = 3 days
T = 8.04 days (Half life of 1-131)
Equation 4.21
') = 0.693 = 0.693
- = 0.086 d-
Il.
T
I
8.04 d
Using equation 4.18;
Ao = ｾ＠
l
A
e
=
1 MBq
-0086 x 3
e·
.,
.
= 1.3 MBq IS reqUIred to be shIpped.
59
4.2
60
THE REALTII PHYSICS SOLUTIONS MANUAL
To convert to curies:
1 Ci
10 6 Bq 10 6 J-lCi
1.3 MBq x 3.7 x 10 10 Bq x MBq x 1 Ci = 35.135 f..lCi
4.3
4.3 The gamma radiation from 1 mL of a solution containing 370 Bq (0.01 f..lCi)
198Au and 185 Bq (0.005 /lCi) 1311 is counted daily over a 16 day period. Assume
an equal detection efficiency of the scintillation counter of 10% for all the
quantum energies involved. What will be the relative counting rates of the 131 1
and 198Au at time t = 0, t = 3 days, t=8 days, and t=16 days. Plot the daily total
counting rates on semi-log paper, and write the equation of the curve of total
count rate vs. time.
198Au emits gammas in 99.5% of the transformations (from 1CRP 38) and has a
half life of 2.7 days.
1 trans·
370 Bq x
sec x 99.5 x 10 count = 36.8 counts per sec at time zero
Bq
100 100 trans
The 1311 decay scheme is on page 83 of "Introduction to Health Physics."
1 trans
185 Bq x
sec x -107 x 10 count = 19.8 counts per sec at time zero
Bq
100 100 trans
So the total number of counts detected per second at time zero would be:
36.8 + 19.8 = 56.6 counts per sec total
So the ratio of counts for each at time zero is:
36.8
198Au = - - = 0.65 = 65% of counts were due to 198Au
565
131
19.8
1 = 565 = 0.35= 35% of counts were due to 131 1
At t =3 days;
•
SOLUTIONS FOR CHAPTER 4
I
98Au
T1f2 = 2.7 days
Equation 4.21
A = 0.693 = 0.693
T
=0.257 d-1
2.7 days
Equation 4.18:
t= 3 days
A = 0.257 days-1
Ao = 36.8 cps
A = Ao e-At = 368
. x e-0.257X3 =170
. cps
131I
Tlf2
= 8.05 days
Equation 4.21
A= 0.693 = 0.693
T
= 8.6 x 10-2 d- 1
8.05 days
Replacing the values in equation 4.18:
t= 3 days
A = 8.6 X 10-2 d- 1
Ao = 19.8 cps
-At - 19 8
15
A -A
. 3 cps
0 e
. x e- 0.086 x 3 -
Total counts per second on 3fd day:
I
17.1 cps + 15.3 cps = 32.4 cps
61
:r
,t
62
17.1
Au = 32.4
198
I31
ｾ＠
THE HEALTH PHYSICS SOLUTIONS MANUAL
= 0.52 = 52% of counts were due to
198
Au
15.2
131
1 =- - = 0.479 = 48% of counts were due to 1
32.4
At t = 8 days;
ｾＮ＠
Jｾ＠
l
,
I
,
t,:,'
ｾ＠
198
Au
Replacing the values in equation 4.18:
ｾ＠
ｾ＠
t
t = 8 days
A = 0.257 d- I
Ao = 35.96 cps
ｾ＠
t·
p
F
ｾ＠
f
A = Ao e-AI = 36.8 x e- O.257 x8 = 4.7 cps
131
1
Equation 4.18:
t = 8 days
A = 8.6 X 10-2 d- 1
Ao = 19.8 cps
A =A 0 e-AI = 19 .8 x e- 0.086 x 8 =9 .9 cps
Total counts per second on 8 day: '4.7 cps + 9.9 cps = 14.6 cps
th
4.7
.
198
Au = 14.6 = 0.32 = 32% of counts were due to Au
198
131
9.9
131
1 = 14.6 = 0.68 = 68% of counts were due to 1
At t =16 days;
t
l'
ｾＮ＠
t·
SOLUTIONS FOR CHAPTER 4
63
198Au
Replacing the values in equation 4.18:
t=16days
A =0.257 d- I
Ao = 36.8 cps
...:)..t - 0 60
. cps
0 e
- 36 .8 x e- 0.257 x 16 A -A
131
1
Equation 4.18:
t = 16 days
A = 8.6 X 10-2 days-I
Ao = 19.8 cps
. x e-0.086 x16 = 5 cps
A =A 0 e-At = 198
Total counts per second on 16 day: 0.6 cps + 5 cps = 5.6 cps
th
198
0.6
Au = -6
5.
= 0.11 = 11 % of counts were due .to 198Au
1311 = 56 = 0.89 = 89% of counts were due to 1311
5.
The decay of the activity of the mixture is given in the table below and plotted
graphically:
t
i .
I
t
Day
ACt), cps
% 198Au
%131 1
0
3
8
16
56.6
32.4
14.6
5.6
65
52
32
11
35
48
68
89
64
THE HEALTH PHYSlCS SOLl.JIlONS MANUAL
60
50
1\
\
40
.,
fr 30
ｉｾ＠
'" ｾ＠
20
ｾ＠
--
r-- r--
10
12
I-----
10
o
o
2
4
8
6
16
14
18
Days
The equation for the curve of the total activity as a function of time is the sum of
the activities of each of the two components, 198Au and 131"1, The initial count
rates, A o' and the transformation constants for each isotope are:
198
131
Au
1
Ao
36.8 cps
19.8 cps
l
slope, A 0.257 d0.086 d- I
A(t) = A 01 e -I'l' + A 02 e -1..21
A(t) = 36.8 e -O:2571 + 19.8e-O·0861 .
4.4
4.4 The following counting rates were obtained on a sample that was identified
as a pure beta emitter.
Day
o
1
2
3
5
10
20
cpm
5500
5240
5000
4750
4320
3400
2050
SOLUTIONS FOR CHAPTER 4
65
(a) Plot the data on semi-log paper.
10000
E
c.
<.>
ＱｾＭ＠
o
5
10
15
20
25
Day
(b) Determine the half life from the graph.
Looking at the graph, it can be seen that the number of counts has decreased by
half, to 2750 counts, at approximately 14.3 days.
(c) What is the value of the transformation constant, per day?
Use equation 4.21,
0.693
. A= T
0.693
-2
-1.
(d) Write the equation for the decay curve.
Equation 4.18 is replaced by the values:
ｾＭ
.
= 14.3 days - 4.85 x 10 d IS the transformatIOn constant.
66
THE REALTIl PHYSICS SOLUTIONS MANUAL
Ao = 5500
A = 4.85 X 10-2 0 1
A =A
e- = 5500 x e-D·0485xt
At
o
(e) What is the radionuclide?
Looking in the RHH in the table of beta emitters by increasing energy, and for an
it can be seen that the isotope is 32p.
.
isotope that emits only a ｢･ｾ｡Ｌ＠
4.5
4.5 If we start with 5 mg 21Dpb, what would be the activity of this sample 10
years later?
TI12
= 22 years
Equation 4.21,
. 0.693
0.693
2
I
A =T = 22 yr = 3.15 x 10- yr- is the transformation constant.
Equation 4.18 is replaced by the values:
t = 10 yr
Ao=5 mg
A =3.15 X 10-2 yr- I
A =Aoe....A t =(5 mg) x e-D·0315xIO = 3.65 mg is the amount of 21 Dpb left after 10
years.
Calculating the specific activity of 21Dpb using equation 4.31;
ARa= 226
TRa = 1620 yr
A p b-21O = 210
Tpb-210 = 22 yr
=
226 x 1620 yr _
Ci .
. . ' ..
21
210 x 22 yr - 79.25
1S the spec1f1c activ1ty of Dpb.
Calculating the activity after 10 years:
g
SOLUTIONS FOR CHAPTER 4
67
g
x 79.25 Ci _
._
.
3.65 mg x 1000 mg
g
- 0.2893 Cl - 298.3 mCl
3.7 X 10 10 Bq
0.2893 Ci x
Ci
= 1.07 x 1010 Bq = 10.7 GBq
10
4.6 The decay constant for 235 U is 9.72 X 10- per year. Compute the number of
transformations per second in a 500 mg sample of 235U.
Iv::;;: 9.72 X 10-10 per year
Find the half life with equation 4.21
8
0.693 _
0.693
T=-'A--9.72x10-10 yr- 1 =7.13x10 yr
Using the specific activity equation (4".31):
A Ra = 226
TRa = 1620 yrs
A U- 235 = 235
8
TU- 235 = 7.13 X 10 yr
SA -
226 x 1620 yrs
A Ra X TRa
6
= 2.19 X 10- Ci/g U-235
Ai x r; - 235 x (7.13 x 108 yrs)
Calculate the number of curies present in 500 mg;
6
1 g x 2.19 x 10- Ci _
-6
.
500 mg x 1000 mg
g
- 1.093 x 10 Cl
Converting curies to transformations per second:
3.7 X 10 10 transformations
1.093 x 10-6 Ci x
second
1 Ci
sec
= 4 x 104 transformations per
4.6
68
THE HEALTH PHYSICS SOLUTIONS MANUAL
Alternately;
Activity = AN, .where N = number of radioactive atoms.
6.02 X 10 23 atoms
N=
mole x 05 g = 1.28 X 10 21 atoms
g
235-mole
10
A=
9.72 X 10- y.1
ＳＶＵｾ＠
Y
x 8.64 x 10
= 3.082 X 10-17 S·1
4
ｾ＠
d
21
4
Activity = 3.08 x 10. s·! x 1.28 X 10 atoms = 4 x 10 dps
17
4.7
Two hundred MBq (5.4 mCi) 2!Dpo are necessary for a certain ionization
source. How many grams 21Dpo does this represent?
4.7
Tp o-210 = ＱＳｾ＠
days
A pO-210 =210
A Ra = 226
TRa = 1620 yrs
Use the specific activity equation (4.31):
SA -
-
A Ra X TRa
Ai x I;
226 x 1620 yrs
1
= 4.61 X 103 Ci/g Po-21D
- 210 x 138 d x yr
365 d
Converting activity to mCi:
5.4 mCi = 5.4 x 10- Ci
3
Now convert Ci into grams:
5.4 X 10-3 Ci x 4.61
ｾ＠ ｾＰＳ＠
Ci = 1.17 x 10-6 g = 1.17 j.lg
SOLUTIONS FOR CHAPTER 4
69
4.8 How long would it take for 99.9% of 137Cs to decay, if its half-life is 30
4.8
years?
T== 30 years
Equation 4.21
0.693
T
A ==-- ==
0.693
1
== 0.0231 yr30yr
What fraction is remaining if 99.9% has disappeared?
1.000 - 0.999 == 0.001 is all that would remain.
So that the ratio between what remains and what was started with is:
final
Ao == initial
A
0.001
Using equation 4.21 and solving for t:
A
A == e-A1 == 0.001
o
6.91
t == -
A
== 299 years
4.9 How long will it take for each of the following radioisotopes to decrease to
0.0001 % of its initial activity?
TIrJ. == 66 hr
(a) 99Mo
Equation 4.17 is used to find the number of half lives it takes for any isotope to
decay to 0.0001 % of its initial value:
n == number of half lives
I
6
/== 0.000001==10.
·0
\ . -
4.9
70
I
THE HEALTII PHYSICS SOLUTIONS MANUAL
T
[
1
Replacing values
6
n = log(2) = 20 half lives are needed to decrease to 0.0001 % of initial activity.
So for 99Mo: T1I2 = 66 hr; 66 hr x 20 = 1320 hr =55 days
(b) 99mTc: TJ/2 =6 hr; 6 hr x 20 = 120 hr = 5 days
TJI2 = 8 days; 8 days x 20 = 160 days
(c) 13JI :
(d) J25I : T1I2 = 60 days; 60 days x 20 = 1200 days
4.10
4.10 For use in carcinogenesis studies, benzo(a)pyren(! is tagged with 3H to a
specific activity of 4 X lOll Bq/millimole. If there is only 1 tritium atom on a
tagged molecule, what percentage cif the benzo( a)pyrene is tagged with 3H ?
Remember that tritium is a hydrogen atom with 2 neutrons, and is written 3H.
First, fmd the specific activity of carrier free. tritium, using equation 4.31:
ARa = 226
.TRa = 1620 years
A H _3 = 3
TH _3 = 12.3 years
226 x 1620 yr
SA H_3 = AHTH =
= 9922 Cilg Tritium
3x 12.3 yr
ARaTRa
9922 Ci
3g
1 mo1 3H
3.7 x 10 JO Bq
--- x
x
x
----..:;.
g 3H
mol 3H 6.02 x 10 23 atoms 3H
Ci
-9
Bq
= 1.83 x 10 atom 3H
"',.
. J
SOLUTIONS FQR CHAPTER 4
71
Next, calculate the proportion of benzo(a)pyrene molecules tagged:
4 x 1011 Bq
romol Benzo.
------=-- x
::: 6.64 x 10-10
1000 mmol
1 mol
x
23
.
1 mol
6.02 x 10 molecules
Bq
Benzo. molecules
Divide the number of Bq/Benzo atom by Bq/3H atom to fmd the fraction of
benzo(a)pyrene atoms tagged:
10
6.64 X 10- Bq}
{ Benzo. atom
9
1.83 x 10- Bq}
{ atom (3H)
=0.363 is the fraction of benzo(a)pyrene molecules tagged
Thus, 36.3% of the benzo(a)pyrene molecules are tagged with tritium.
4.11 How many alpha particles are emitted per minute by 1 cm3 222Ra at a
temperature of 27°C and a pressure of 100,000 Pa?
100,000 Pa = 0.987 atmospheres
27 + 273 = 300 K
Assume that the radon is an ideal gas.
Recalling the ideal gas law where PV=nRT,
P = pressure in atmospheres
V = ｖｯｬｾｭ･＠
in liters
n = number of moles
R= gas constant = 0.082 L·atmlmol· K
T = absolute temperature, kelvin
n
P
V
RT
- =-
!
ｾＭＮ＠
=
0.987 atm.
moles
of radon gas
L
= 0.04
0.082
·atm x 300 K
L
mole·K
4.11
72
THE HEALTI-I PHYSICS SOLUTIONS MANUAL
Now calculate the specific activity of 222Rn (equation 4.31):
ARa = 226
TRa = 1620 years
A Rn-222 = 222
TRn-222 = 3.82 days = 0.01 year
226 x 1620 yr
= 222 xO.01 yr = 1.65 x lOS Ci/g 222Rrt
ARaTRa
SA Rn-222
= A RnlRn
'T
Calculate the number of transformations per minute using the specific activity;
and the number of moles per liter of 222Rn calculated above. (Note the equation
is split into two lines);
1L
1.65 x 105 Ci 222 g 222RI1l 0.04 mol 222Ra
-----x
x
x
x
g 222Rft
mol
L 222Ra
1000 cm 3
x
1 cm 3 222Ra
sample
= 3.25 X 10
4.12
15
2.22 x 1012 trans
min
1a
x-x Ｍｾ＠
1 Ci
trans
a per minute emitted from the sample
4.12 Calculate the number of beta particles emitted per minute by 1 kg KCI, if
4°K emits 1 beta particle per transformation.
Find the specific activity of 4°K (equation 4.31):
A Ra = 226
TRa = 1620 yr
A K-40 =40
9
TK-40 = 1.29 x 10 yr
ARaTRa
SA · AO
K-.
= A
'T
K1K
-
226 x 1620 yr
40x1.29x10 9 yr = 7.1 x 1ｾ＠
Natural abundance of 4°K is 0.0117 % of all K.
K = 39.1 grams per mole
CI = 35.457 grams per mole
Cilg 4°K
SOLUTIONS FOR CHAP1ER 4
73
KCI = 39.1 + 35.457 = 74.557 grams per mole
Solve for the number of cunes of 4°K per sample, using the specific activity:
1000 g KCI 1 mol KCI
1 mol K
----x
x
x
sample
74.6 g KCl 1 mol KCl
(
0.0117) I 40K
100 rna
40 g40K 7.1 x 10-6Ci
x
x---1 mol K
1 mol 40 K
g40K
7
== 4.49 x 10- Ci 4°K per sample
Convert to betas per minute:
4.49 x 10-7 Ci
sample
12 trans
1 f3
f3
2.22 x 10 - . x
rrun x - - = 9.96 X 105 - . - in one kg sample
Ci
trans
rrun
4.13 Iodine 125, a widely used isotope in the practice of nuclear medicine, has a
half life of 60 days.
(a) How long will it take for 4 MBq (- 1f.lCi) to decrease to 0.1 % of its initial
activity?
Tlf2
= 60 days
Equation 4.21
'1 ==
A
0.693 = 0.693 = 0 01155 d- 1
T
60 d
.
Equation 4.21:
t =60 days
A =0.01155 d- 1
A
Ao = 0.001 = e-A
A
InA
t=-_o
-A
t
InO.OOI
-0.01155 d- I
=598 d
4.13
74
THE HEALTH PHYSICS SOLUTIONS MANUAL
(b) What is the mean life of 1251?
A =0.01155 d- I {From part (a)}
Equation 4.24
1
't
4.14
1
= A = 0.01155 d- ' = 86.6 days
4.14 If uranium ore contains 10% U 30 8, how many metric tons are necessary to
produce 1 g radium if the extraction process is 90% efficient?
First, find the ratio (in grams) of uranium to the radium in the ore using the
specific activities of each. Assume that the radium is in secular equilibrium with
the ,uranium, and radium has the specific activity of 1 curie per gram.
Use equation 4.31
ARa =226
TRa = 1620 years
Au = 238
9
Tu = 4.51 X 10 years
226 x 1620 yr
7
= 238x4.51xl09 yr = 3.4 X 10- Ci/g
Obtain the number of grams of radium per gram of uranium, knowing that in ore,
. .m eqUl'libnum
.
J:
R a IS
WI·th 238U . Therelore,
we h ave:
226
7
3.4 X 10- g radium / g uranium
= 118.8 moles U 30/ton
SOLUTIONS FOR CHAPTER 4
Ｍ］ｾ＠
118.8 mol U 3 0 8
tonne ore
x
3 mol 238 U
1 mol U 3
°
x
8
75
238 g 238 U
3.4 x 10-7 g Ra
1 mol 238 U
X --------'-1 g 238 U
= 0.029 g radium/tonne ore
Inverting this solution to find how many tons ore per one gram:
tonne ore
0.029 g Radium = 35 tonnes per gram of radium
However, the process is only 90% efficient (according to the question), so:
35 tonns ore 100
1 g radium x 90 = 39 tonnes of ore are needed to yield 1 gram of radium.
4.15 How much 234 U is there in one ton of the uranium ore containing 10% U 0 ?
3
8
. The activity of 234U is equal to the activity of 238 U since 234 U is in secular equilib-
.
. h 238U .
num
WIt
SA ( 238 U) =
1620 Y x 226 . = 3.4 X 10-7 Ci
45 x 10 9 y x 238
g
In one ton (2000 lbs.) uranium ore, we have 200 lbs. U 0 ' or
3
200 lbsx ＴＵＳＮＶｾ＠
lb
= 9.07 x 10
4
S
g U 3 0 S and the amount of 238 U in this amount of
U 30 S is
238
0.993 x 9.07 x 10 4 g U 3 0 8 x
3.4 x 238
g U = 7.64 X 104 g 238U
3.4 x 238+ 8 x 16 g U 3 8
The activity in this quantity of 238U is
°
7.64 X 10 4 g 238U x 3.4 X 10-7
g
The specific activity of 234 U is
;! U
= 2.6 x 10-2 Ci
4.15
76
THE HEALTH PHYSICS SOLUTIONS MANUAL
SA (234
U) =
1620 Y x 226 = 6.3 x 10-3 Ci
g
4.58 x 105 Y x 234
Since the activities of the two uranium isotopes are equal, the weight of the 234U
IS
Alternatively;
234U = 5.5 X 10-5 is the fractional abundance of 234U in natural uranium
Calculate the number of grams per mole of U 30 S:
Uranium
238.0289 x 3 = 714.0867 g/mole
Oxygen
15.9994 x 8 = 127.9952 g/mole
U 30 S == 714.0867 g/mole + 127.9952 g/mole = 842.0819 glmole
Note that the equation is split into two lines below;
x
3 mol U"
1 mol U 30 S
x
5.5 x 10-5moIC34 U)
1 mol U
= 0.0178 moles 234U per ton ore
0.0178 mol 234U 234 g 234 U
------ x
=4.18 g 234U per ton of ore
ton" ore
1 moe34 U
4.16
4.16 Compare the activity of the 234 U to that of the 235U and the 23S U in the ore of
problems 4.14 and 4.15.
SOLUTIONS FOR CHAPTER 4
77
The natural abundance of 235 U = 0.72%, and its half-life =7.13 X lOs yr. Therefore, in 1 ton (2000 lbs) of 10% U30 g ore, we have
(2000xO.l)lbsU30s ｸＴＵＳｾ＠
ton
lb
(3x238)"gU
(3 x 238 + 8 x 16) g U 3
°
8
x7.2xlO-3 g 235U =554 g 235U
gU
ton
The specific activity of 235U (eq. 4.31) is
ARa =226
TRa = 1620 yr
AU_235 = 23 5
S
TU_235 = 7.13 x 10 yr
A Ra X TRa
226 x 1620 yr
6
SA = Ai x I; = 235 x (7.13 x 108 yr) = 2.19 x 10- Cilg U-235
2
235U activity in one ton ore is 2.6 x 10. Cilton (from problem 4.15)
234U activity = 23SU activity (secular equilibrium)
Therefore the total activity in 1 ton of this ore is
1.21 x 10.3 + 2(2.6 x 10.2) = 5.32 X 10.2 Ci/ton
2
234U = 2.6x 10- x 100= 48.87%
5.32 x 10-2
235
2
= 1.21 X 10- x 100 = 2.27%
U 5.32 x 10-2
23SU actIvIty
.. = 234U actIvIty
.. = 48 .8701
'l0
Total activity = 48.87% + 2.27% + 48.87% = 100%
Altemati vely,
234
U
ARa = 226
TRa = 1620 yrs
AU_234 = 234
TU_234 = 2.47 x 105 yr
Equation 4.31:
i;
78
SA =
235
THE HEALTH PHYSICS SOLUTIONS MANUAL
226 x 1620 yr'
3
Ajl; = 234 x (2.47 x 105 yr) = 6.33 x 10- Ci/g U-234
ARaTRa
U
ARa = 226
TRJi = 1620 yr
AU_235 = 235
g
TU- 235 =7.13 x 10 yr
Replacing numbers in equation 4.31:
A Ra X TRa
SA = Ai xl;
226 x 1620 yr
= 235 x (7.13 x 108 yr) = 2.19 x 10--6 Ci/g U-235
ＲｾＸｕ＠
226
. TRa = 1620 yr
AU- 238 = 238
TU_238 = 4.51 x 109 yr
ARJi =
Equation 4.31:
SA --
A Ra X TRa
226 x 1620 yr
7
Aj x r; - 238 x 4.51 x 109 yr = 3.41 X 10- Ci/g U-238
Percent abundance
U = 0.0055
235U = 0.720
238U = 99.2745
234
From problem 4.14 there are 118.8 moles U 30/ton ore
118.8 mol U 3 0 g
3 mol U
ton ore
x 1 mol U
3
° = 356.4 moles Uranium per ton ore.
8 .
SOLUTIONS FOR CHAPTER 4
79
356.4 mol U 0.0055 mol 234 U
234 g
6.33 x 10-3 Ci
----- x
x
x
---ton are
100 mol U
1 mol 234 U
g
== 0.029 Ci 234 U per ton ore
Similarly for U-235
356.4 mol U
ton ore
----- x
235 g
2.19 x 10-6 Ci
0.720 mol 235 U
x
x ----100 molU
1 mol 235U
g
== 0.00132 Ci 235 U per ton ore
And U-238
238 g
3.41 x 10-7 Ci
356.4 mol U 99.2745 mol 238 U
----- x
x
x
---ton ore
100 mol U
1 mol 238 U
g
=0.0287 Ci
238
U per ton ore
4.17 Calculate the activity in Bq and /-lCi in each of the uranium isotopes in 1 g
of natural U, and then, using these results together with the values for the isotopic abundance's, calculate the activity of 1 g of natural U.
238
U
The sample activity is equal to the SA times the mass of the sample. The specific
activity is (equation 4.31):
ARa= 226
TR1J. = 1620 yr
AU_238 = 238
9
TU_238 = 4.5 x 10 yr
SA - A
ARaTRa
T.
U-238
U238
=
226 x 1620 yr = 3.4 x 10-7 Ci
238 x 4.5 x 109 yr
g
The isotopic abundance of 238U is 99.276% of all uranium, so the activity in one
·
. due to 238 U IS:
gram 0 f naturaI uramum
4.17
80
THE REALlli PHYSICS SOLUTIONS MANUAL
99.276 g 238U . 3.4 X 10-7 Ci 10 6 flCi
4
1gx
100 g U
x
g 238U
x
Ci
= 0.34 IlCi 0.26 x 10 Bq) in one
gram of natural uranium due to 238U.
For 235U we have
ARa= 226
TRa = 1620 yr
AU_235 = 235
TU_235 = 7.13 x 108 yr
The isotopic abundance of 235 U is 0.7196% of all uranium, so the activity in one
·
.
due to 235U IS:
gram 0 f naturaI uramum
7.2x10-3 g 235U 2.19x10-6Ci 10 6 flCi
19X
19U
x
g
xCi =0.016 IlCi (5.9 x 102Bq) in
one gram of natural uranium due to 235U.
Since 234U is in secular equilibrium with 238U, the two activities are equal, 0.34
IlCi (1.26 x 104 Bq)
Adding the activities together, 238U + 235U+234U
0.341lCi + 0.0161lCi + 0.34 IlCi = 0.7 IlCi = 2.6 x 104 Bq in 1 g natural uranium.
4.18
4.18 What will be the temperature rise after 24 hours in a well insulated 100 mL
aqueous solution containing 1 gram Na2 35S04, if the specific activity of the sulfur ｾｊ＠
12
is 3.7 X 10 Bq/gram (100 Ci/gram)?
First, fmd the specific activity of 35S and compare it to the specific activity listed
in the problem to find the fraction of sulfur atoms that are 35S;
Equation 4.31
I
r
I
SOLUTIONS FOR CHAPTER 4
81
ARa== 226
TRa == 1620 yr
AS-35 = 35
TS-35 = 87.44 days = 0.24 yr
SA = ARaTRa
= 226'x 1620 yr = 4.36 x 104 Ci
A S - 35 TS - 35
35 x 0.24 yr
g
Now calculate the fraction of the mass of the tagged (*S) sulfur in the compound
35
.
that is . S:
Ci
100- Tagged 35S
g
1 g35 S
---=----- = - - 4.36x 10 4 Ci S
439 g.S
g
Since only a very small fraction of all the sulfur atoms is tagged as 35S , using 32
grams for the atomic weight of the tagged sulfer is a reasonable assumption.
The atomic weights are as follows:
Na=23
S =32
0= 16
Na2 *S04 = (2 x 23) + 32 + (4 x 16) = 142 g/mole
35S activity concentration
IgNa 2 *S04 x
ＳＲｧＪｾ＠
ｸｬＰｾＳＮＷｏｂｱ］ＸＴＹ＠
100 mL
142 g Na 2 S04
If C5S) =0.049 Me V per beta
Specific heat of water = 4.19 J/mL·oC
Temperature rise:
-:; -.
g *S
Ci
Bq
mL
82
THE HEALTII PHYSICS SOLUTIONS MANUAL
ｾ＠
13
8.34 X 10 9 Bq x 1 sec x 0.049 MeV x 1.6 x 10- J x 8.64 X 104 sec
ｾ＠
Ｑｍｾ＠
d
mL Ｑｾ＠
J
cal
4.l9-x1--cal
mL·oC
= 1.3 °C (or K) rise in temp.
4.19
4.19 The mean concentration of potassium in crustal rocks is 27 g/kg. If 4°K
constitutes 0.012% of potassium, what is the 4°K activity in one metric ton of
rock?
Calculate the specific activity of 4°K using equation 4.31:
ARa = 226
TRa = 1620 yr
A K-40 =40
9
TK-40 = 1.3 X 10 yr
1620 yr x 226
40x1.3x10 9 yr =7.04x ｬｾｃｩＯｧ＠
1 tonne rock
10 3 kg rock
27 g K
0.00012 g4°K 7.04 x 10-6Ci 1 x 10 6 ｾｃｩ＠
--=------ x
x
x
x Ｍｾ
tonne rock
1 kg rock
1gK
g4°K
1 Ci
= 23 IlCi /tonne rock
4.20
4.20 A solution of 203Hg is received with the following assay: 1MBq/mL on 1
March 1981 at 8:00 am. It is desired to make a solution whose activity will be
0.1 MBq/mL on 1 April 1981. Calculate the dilution factor to give the desired
"
T 112 203Hg = 46 days.
activIty.
ｾ＠
t = 31 days
Equation 4.21:
0.693
1
A = 46 days = 0.015 d-
,:'g.
SOLUTIONS FOR CHAPTER 4
83
Find the activity that would be present on 1 April 1981 using equation 4.18;
A = Ao e-At = 1 MBq/ mL x e-O.015x31 = 0.628 MBql mL is the activity left in the
solution after 31 days.
To find the quantity needed to obtain 0.1 MBq/mL:
v1 = 1 mL
V2 =?mL
CI = 0.628
MBq/mL
.
C2 = 0.1 MBq/mL
CV
MBq
0.628-- x 1 mL
V = _ 1_1 =
mL
C2
0.1 MBq
mL
2
= 6.28 mL
6.28 to 1 is the dilution factor.
4.21 In a mixture of two radioisotopes, 99% of the activity is due to 24Na and 1%
is due to 32p' At what subsequent time will the two activities be equal?
24Na Tin = 15 hours = O. 6 25 days
Using equation 4.21
A
Na=
32
0.693
-1
0.625 d = 1.11 d
P TII2 = 14.3 days
A = 0.693 = 0.0485 d- 1
p
14.3 d
Use equation 4.18:
A = A oe-;!.!
4.21
84
THE HEALTII PHYSICS SOLUTIONS MANUAL
Since the activity of the 24Na is 99 times the 32p activity, the two activities will be
equal
99 x e-l.Il x t = 1 x e-O· 0485 x t
In 99
t = - - = 4.3 days = 104 hr
1.06
.
4.22
4.22 Low level waste from a biomedical laboratory consists of a mixture of 100
Jl.Ci (3.7 MBq) 131 I and 10 Jl.Ci (0.37 MBq) 1251. Plot the decay curve for the total
activity over a period of 365 days and write the equation for the decay curve.
1000
100
..
•
•
ｾ＠
••
••
••
•
••
•••
ｾ＠
-+
ｾ＠
•
••
•• •
--
0.1
o
50
100
150
200
Days
250
300
•••
350
400
I
- SOLUTIONS FOR CHAPTER 4
85
The decay constant for each of the isotopes can be found using equation 4.21:
For 1311 ,
T= 8.05 d
0.693
A=T=
0.693
8.05d
0.087 d- I
For 1251,
T= 60.14 days
0.693
0.693
A=T = 60.14 d = 0.0115 d
_I
The total activity is the combination of the decay of the 131 1 and 1251 decay equations using equation 4.18;
A =A oe-AI=A I_l3l e-AI + A 1-125 e...)..1 = 100 x e--0.087 I + 10 x e--O.01I5xl
X
4.23 ThB is transformed to The at a rate of 6.54% per hour, and The is trans-
formed at a rate of 1.15% per min. How long will it take for the two isotopes to
reach their equilibrium state?
ATbB = 0.0654 hr- I
\1)C
= 0.0115 min-I =
0.0115 60 min
mm x hr
= 0.69 hr- I
Use equation 4.57 to find the time at which the The activity reaches its maximum activity:
In(AThCJ
1 ( 0.69 )
J:;:; = n 0.0654 = 3.78 hr
trn=
AThC - AThB
0.69 - 0.0654
So 3.78 hours after the ThB is isolated the activity will be at a maximum, however, looking at the drawing on page III of the third edition, it can be seen that
,i,'
f
ｾ｟＠
_ __
k&:- -- .
4.23
86
THE HEALW PHYSICS SOLUTIONS MANUAL
they are not in equilibrium until the activity of the ThC moves off of it's "peak",
which will occur roughly after 6 to 7 half lives (60.5 min times 7, approximately
twice the time it took to reach peak activity), and the two isotopes will then be in
"equilibrium".
4.24
4.24 How many grams of 9Gy are there when 9Gy is equilibrated with 10 mg 90Sr?
90
Sr
TI/2
= 28 yr
Calculate the specific activity of 90Sr using equation 4.31:
ARa = 226
TRa = 1620 yr
ASr =90
TSr = 28 yr
ARaTRa
SA = Asr'Fsr
226 x 1620 yr
= 90 x 28 yr = 145.3 CiJg = 145.3 mCiJmg
At secular equilibrium, the 90y activity = 90S r activity,
10 mg x 145.3 mCiJmg = 1453 mCi = 1.453 Ci
Calculate the specific activity of 90y using equation 4.31:
ARa = 226
TRa = 1620 yr
Ay =90
226 x 1620 yr
= 90 x 7.3 X 10-3 yr = 5.57 x lOS CiJg
,
Calculating the 9Gy weight after equilibrium:
'.,'
1.453 Ci. =2.6 X 10-3 mg =2.61lg
5.57 x 10 5 Cl
g 90y
4.25
4.25 Radiogenic lead constitutes 98.5% of the element as found in lead ore. The
isotopic constitution of lead in nature is: 204Pb , 1.5%; 206Pb , 23.6%; 207Pb , 22.6%;
208Pb , 52.3%. How much uranium and thorium decayed completely to produce
985 mg of radiogenic lead?
ｪ
.
....1'
.
"
Ｇｾｴ＠
.ＭＮｾ＠ ,
SOLUTIONS FOR CHAPTER 4
87
Assume the total sample had 1000 mg of lead, with only 985 mg radiogenic, so
the number of mg of each type of lead would be:
2°6pb, 23.6% x 1000 mg = 236 mg
207Pb, 22.6% x 1000 mg = 226 mg
208Pb, 52.3% x 1000 mg = 523 mg
206pb is the end product of the 238U decay chain,
207pb , 22.6% x 1000 mg = 226 mg
207Pb is the end product of the 235U decay chain,
235
1 mo·
1207Pb 1 mol 235 U 235 g 235 U
207
U: 226 mg Pb x
207 g x 1 mol207Pb x 1 moe38 U - 256.6 mg
2°Spb, 52.3% x 1000 mg = 523 mg
208Pb is the end product of the 232Th decay chain,
1 mol 208 Pb 1 moe32 Th 232 g 232 Th
232m. 523
208
x
x
583 '
In:
mg Pb x 208 g
1 mol 208 Pb 1 mol232Th =
mg
4.26 How long after 1 kg of 241Pu is isolated will the 241 Am activity be at its
maximum? What will the activity be at that time?
241
Pu: TIn = 13.2 yr
Using equation 4.21: APu
241
Am: TIn = 458 yr
=
0.693
T
0.693
I
-13-.2-yr- = 0.0525 yr-
4.26
88
THE REALTII PHYSICS SOLUTIONS MANUAL
0.693 0.693
I
Using equation 4.21: Apu = -T- = 458 yr = 0.0015 yrThe time of maximum activity is found using equation 4.57:
I (0.0015)
..
n 0.0525
70
------ =
yr to max. actIvIty
0.0015 - 0.0525
Calculate the number of atoms of 241Pu in 1 kg:
1000 g 1 mol 6.02 x 10 23 atoms
24
= 2.5 x 10 atoms Pu
NAo = 1 kg Pu x 1 kg x 241 g x
1 mol
To find the activity after 70 years, use equation 4.53:
Am
I
I
.A N
24
= 0.0015 yr- x 0.0525 yr- x 2.5 x 10 atoms (e -0.0525 x 70 _ e -0.0015 x 70)
Am
0.0015 yr- I -0.0525 yr- 1
21
AAmN Am = 3.37 X 10 transfonnations/yr
3.37 X 10 21 transformations
1 yr
1d
1 hr
1 Bq
x--x
x-.
x
yr
365 d 24 hr 3600 sec trans = 1.1 x 1014 Bq
sec
4.27
4.27 How long after 14'rninute 146Ce is isolated will the activity of the 24 minute
Pr daug hter be equal to that of the parent.?
146Ce :
Tin = 14 min
146
Using equation 4.21:
Pr·.
146
T In = 24 ffil·n
"I
!'I.Ce
0.693
. -I
= 14 min = 0.0495 mm
SOLUTIONS FOR CHAPTER 4
Using equation 4.21:
A = 0.693
Pr
24 min
89
= 0.0289 min-I
The activity of the I46Pr daughter (B), after isolation of its parent is give by
equation 4.53
A A N
B A
AO (e-AAt _e-ABt)
AB (t ) = AB N B = A-A
B
A
Since AANAo =A AO ' the activity of the daughter may be rewritten as
AB AAO (-AA1
AB (t)- A _ A
B
e
- e
-ABt)
A
Setting the activity of I46Ce (A) equal to that of I46Pr (B);
e -0.0495t
=
0.0289
(e -0.04951 _ e -0.02891 )
0.0289 - 0.0495
t = 26 min.
4.28 Thirty seven MBq (1 mCi) 99"Tc are milked from a 99Mo "cow". What will
be the activity of the 99"Tc daughter, 99Tc , 1 year after the milking?
Since the half-life of 99m.yC is 6 hours (2.16 x 104seconds), all of it will have
decayed to 99Tc atoms one year after milking. The half-life of 99Tc is 2.13 x 105
12
years (6.72 x 10 seconds). Since activity is given by
A=AN
we can calculate the number of 99myc atoms in 37 MBq (37 x 106 disintegrations
per second)
4.28
90
THE HEALTII PHYSICS SOLUTIONS MANUAL
0.693
xN
2.16 X 10 4 s
N = 1.16 X 10 12 atoms Ｙｾｃ＠
The activity of the 99Tc, therefore is
37 X 10 6 S-l =
A=
4.29
AN = 6.720.693
116 10 12
I
x
atoms = 0.12 s- =0.12 Bq
X 1012 s x.
4.29 Calculate the specific activity of 85Kr (T 1/2
3
J.!Cilcm at 25°C and 760 mm Hg.
= 10.7 years) in Bq/m3 and
L·atm
I K
moe·
P = 760 mm Hg = 1 atm
R = 0.082
The ideal gas equation is used to find the number of moles per liter of 85 K gas:
P
n
-=- =
V
RT
-2
1 atm
L t
= 4.1 x 10 moleslL
0.082 . a m x 298 K
mole·K
Calculating the specific activity with equation 4.31:
ARa = 226
TRa = 1620 yr
A Kr =85
TIV = 10.7 yr
ARaTRa
226 x 1620 yr
SA = AKrTKr = 85 x 10.6 yr =406.3 Cilg
406.3Ci
g
85g 4.1 x 10-2 mol
L
1x10 6 !-lCi
x-- x
x
x
= 1.4 X 106 J.!Cilcm3
mol
L . 1000 cm 3
1 Ci
'SOLUTIONS FOR CHAPTER 4
91
1.4x]06 JlCi x 3.7x10 4Bq x lx10 6 cm 3
--- = 5.19 X 1016 Bq/m3
3
3
cm
1 J.lCi
m
4.30 Calculate the specific power of 3SS and of 14C in
4.30
(a) watts per MBq,
The maximum energy of the 3SS beta particle is 0.167 MeV, but the average
energy of the beta is 0.0488 MeV,
35S= 0.049 MeV/transformation
1W
-1-J/--:-- x
/sec
1 trans
sec
1x106 Bq 0.0488MeV 1.6x10- 13 J
-9 ｾ＠
Bq x MBq x
trans
x
MeV
= 7.8 x 10 MBq
The maximum energy of the 14C beta particle is 0 .156 MeV, but the average
energy of the beta is 0.0494 Me V,
14C= 0.0494 MeV/transformation
1 trans
6
1.6x10-13 J
lW
sec x 1x10 Bq x 0.0494MeV x
---1 J/ x Bq
MBq
trans
MeV
-9
7.9 x 10
Isec
(b) Watts per kg.
The specific activity of 35S is calculated with equation 4.31 :
A Ra = 226
TRa = 1620 yr
A s =35
T = 88 day x
s
'A-
Sr1 -
1 yr
= 0.24 yr
365 d
ARaTRa
226 x 1620yr
4 .
9
AsYs = 35 x 0.24 yr = 4.34 x 10 Cilg = 1.61 x 10 MBq/g
Applying the information from part (a),
W
MB
.
q
'-'
92
I:
THE HEALTH PHYSICS SOLUTIONS MANUAL
9
9
7.8xI0- W 1.61xl0 MBq 1000g
4 'V
35
----X
X
= 1.3 x 10 - of S
ｾ＠
ｾ＠
MBq
g
The specific activity of 14C is calculated with equation 4.31:
ARa = 226
TRa = 1620 years
A s = 14
Ts = 5730 years.
ARaTRa
SA= A T
c c
226 x 1620 yr
= 14 X 5730 yr = 4.56 Ci/g
1 MBq
4.56 Ci 3.7 x 10 1°Bq
g
x
Ci
x 1 x 106Bq = 1.69 x 105 MBq/g
Applying the infonnation from part (a),
7.9 X 10-9W 1.69 x 105MBq 1000 g
14
W
---- X
X
= 1.3 - of C
ｾ＠
ｾ＠
MBq
g
4.31
Calculate the specific power of 90Sr in
(a) watts per MBq,
4.31
The maximum energy of the 90Sr beta particle is 0.546 MeV, but the average
energy of the beta is 0.1958 MeV. However, 90Sr typically is in equilibrium with
its short lived (64 hr) daughter, 9Gy,which emits a beta whose average energy is
0.9348 MeV
9Gy= 0.9348 MeV/transfonnation
Summing the two average energies gives the average beta energy per 90Sr transfonnation.
0.1958 + 0.9348 = 1.13 MeV/transfonnation
1 trans
6
lW
sec
1 x 10 Bq 1.13 MeV 1.6 x 10-13 J
---,--- x--'='-=- x
x
x---I JI
Bq
MBq
trans
MeV
-7
1.8 x 10
W
MBq
Isec
1
j
SOLUTIONS FOR CHAP1ER 4
93
(b) Watts per kg.
The specific activity of 90Sr is calculated with equation 4.31:
ARa = 226
TRa = 1620 yr
A Sr =90
TSr = 28 yr
226x 1620 yr
= 90 x 28 yr = 145.3 Ci/g
1 MBq
145.3 Ci 3.7 x 10 10 Bq
6
g
x
Ci
x 1 x 106Bq = 5.4 x 10 MBq/g
Applying the information from part (a),
W
90
1.8 X 10-7 W 5.4 x 106MBq 1000 g
-----x
X
= 972- of Sr
MBq
g
kg
kg
4.32 How many joules of energy are released in 3 hours by an initial volume of 1
liter 41 Ar at oce and 760 mm Hg?
V=lL
T= OCC = 273 K
L'atm
R = 0.082 moe·
1 K
P = 760 mm Hg = 1 atm
The ideal gas equation is used to find the number of moles of gas present:
n = _P_ =
1atm
= 0.0447 mol
RT 0.082 L·atm x 273 K
L
V
mole·K
Converting to grams of 41 Ar per liter;
4.32
94
THE HEALTH PHYSICS SOLUTIONS MANUAL
0.0447 mol x 41 g = ＱＮＸＳＲＷｾ＠
L
1 mol
L
The maximum energy of the 41 Ar beta particle is 1.198 MeV, but the average
energy of the beta is 0.4593 MeV. Also, 41 Ar emits a y with each transformation,
of 1.293 MeV, which contributes to the energy output of 41 Ar. Thus, the two
energies are summed;
1.293 + 0.4593 = 1.75 MeV/transformation (total)
The specific activity of 41Ar is calculated with equation 4.31:
ARa = 226
TRa = 1620 yr
A AI =41
T
AI
Id
lyr
-4
= 1.83 hr x -x
=
2 08 x 10 yr
24 hr 365 d
.
Combining the specific activity with the number of grams per liter of 41 Ar;
10
t
.
Ys
7'
3.7 x 10 6 0-13 J 1 83
1W
4 .3 x lOCI
X
S x 1. 7 MeV x 1. x 1
x· g x __ = 7.9 x 105 _s
g
Ci
t
MeV
L
1 ｾ＠
L
s
ｈｯｾ･ｶｲＬ＠
the 41Ar is transformed to stable potassium with a half-life of 110
minutes. The rate of energy emission from the 41 AI will decrease according to the
radioactive transformation.
J =J e-t..1
o
where A is the 41 Ar transformation constant
A =' 0.693 =
ｾＱＲ＠
0.693 =0.0063 min- 1
110 min
The total energy release during 3 hours, therefore, is calculated by integrating the .
energy release rate over 180 minutes (3 hours)
SOLUTIONS FOR CHAPTER 4
95
1 = 10 fe-At = 1 0 (1- e-'J...t)
o
"-
Substituting the approximate values into this equation yields
7.9 x 10 5 ｾ＠
1=
x 60 sec
sec
min x (1 - e-0.0063 x ＱＸｾ＠
0.0063 min- I
= 5.1 X 109 J
4.33 (a) Calculate the specific power, in watts/kg, of 41Ar.
4.33
In problem 4.32, we found that 41Ar emits 1.75 MeV per transformation, and that
4
its specific activity is 4.3 x 10 Ci/g. The specific power of 41Ar is calculated with
these values:
3 7 1010 trans
4.3xl07Ci 1000g
. x
sec
1.75MeV 1.6xl0-13 J 1 W
----x
x
x
x
x-g
kg
1 Ci
trans
MeV
Ｑｾ＠
sec
(b)
What is the specific power of 41 Ar 4 hours after the 41Ar is isolated in a
bottle?
The specific power of 41 Ar does not change, since specific power is measured on
a per kg basis. However, the quantity of 41 Ar will decrease to about 22% of its
initial value after 4 hours, and the total power output will decrease accordingly.
4.34 What volume of radon 222 (at O°C and 760 torr) is in equilibrium with 0.1
gram radium 226?
ｾ＠
Since the specific activity of 226Ra is 1 Ci/g, and since the 222Rn is in secular
equilibrium with the 0.1 Ci 226Ra, we have 0.1 Ci 222Rn. The specific activity of
222Rn is calculated with equation 4.31 :
.--"
4.34
96
THE REALlli PHYSICS SOLUTIONS MANUAL
ARa = 226
TRa = 1620 yr
As = 222
T = 3.82 d x
s
ARaTRa
SA= A T
Rn
Rn
1 yr
= 0.0104 yr
365 d
226 x 1620 yr
= 222 x 0.0104 yr = 1.58 x 1()5 Ci/g
-."
1
"
rium with 226Ra
I
According to the ideal gas law, the volume of any gas at 0 °C and 760 nun Hg is
22.4 L (2.24 x 104rnL). The volume ofRn in equilibrium with 0.1 Ci Ra is
I
t,.:.•.•.·
1 g222Rn
1 mole
0.1 Ci Rn x 1.58 x 105Ci x 222 g222Rn = 2.8 x 10-9 mole 222Rn present in equilib-
ｾ＠
ｾ＠ ,
ｾ＠
I:r
ｾＬ＠
Vol (Rn) = 2.8 x 10- mole x 2.24 x 104rnL/mole = 6.4 x 10-5rnL
9
4.35
,
4.35 One hundred milligrams radium as RaBr2 (specific gravity = 5.79) is in a
platinum capsule whose inside dimensions are 2 mm diameter x 4 cm long. What
will be the gas pressure, at body temperature, inside the capsule 100 years after
manufacture if it originally contained air at atmospheric pressure at room temperature (25°C)?
Careful examination of table 4.3 reveals that when 226Ra decays, its progeny
emits alpha particles and since each alpha particle is a helium nucleus, it will
eventually capture 2 electrons and form helium gas, which is what will produce
the pressure in the capsule.
226Ra emits one alpha (T 1I2 = 1620 year)
222Rn emits one alpha (T1I2 =3.825days)
2l8po emits one alpha (T
= 3.05 min)
1/2
214pO emits one alpha (T I12 = 1.64 X 10-4 sec)
21OpO emits one alpha (T 1/2 = 138.4 days)
Pure radium when isolated does not quickly attain transient equilibrium with all
of its progeny. The half life of 210pb (19.4 years) limits the rate that determines
the establishment of equilibrium between radium and all its progeny. The fust
SOLUTIONS FOR CHAPTER 4
97
four alpha emitters in the chain of radium progeny attain secular equilibrium
within a very short period of time relative to the 100 years required for 21Dpb to
attain (for practical purposes) equilibrium. Thus, assume that 226Ra emits 4 alpha
particles per decay (since in secular equilibrium almost immediately), and there
is a buildup with a fifth alpha particle (from 21DpO) which is also emitted.
The fraction of alphas from 21Dpo can be determined by comparing the area under
the "built up"curve to the total area of the dashed rectangle.
Activity
1
0.9
0.8
0.7
......>s:
:;:
0.6
0.5
CJ
oct
0.4
0.3
0.2
0.1
0
0
20
40
60
80
100
120
years
Fraction of a from 21Dpo contributing to the helium over 100 years:
If AE = equilibrium activity, then
140
160
98
THE REALTII PHYSICS SOLUTIONS MANUAL
rAE (1- e-
A1
)dt
.
area un dercurve = -...:......
0
_____ =
f ractIon area rectangle
A ExT
AE [Io dt - Io e-
AT
dt]
Integrating from T = 0 to T = 100:
T
-[i(l- ･ｾａｔＩ｝＠
fraction
100 - [
T
1 (1- e -{).0359(100) )]
0.0359
100
0.73
Thus the effective number of a particles emitted over 100 years is 4 + 0.73 =
.
4.73 a particles per decay of 226Ra .
Now calculate the amount of 226Ra ｴｲ｡ｮｳｦｯｾ･､＠
during the 100 year period. Since
equation 4.18 gives the amount left:
A = A e-At
o
To find the fraction that decays, modify equation 4.18;
A decayed =A 0 (l_e- Af )
Rearrange this equation;
-A f
Adecayed
Ao
-(I-e)
,
f·
Put in the values:
0.693
-1.
A -1620 yr = 0.000428 yr (equatIOn 4.21)
T= 100 years
Adecayed
Ao
.
= ( 1 - e-Af) = ( 1 - e-D.OOO428 x 100) = 0.042
0.042 is the fraction of the 226 Ra activity which decays over 100 years. So the
quantity of radium decaying over 100 years is:
SOLUTIONS FOR CHAPTER 4
99
0.042 x 100 mg = 4.2 mg
Converting this to decays, and knowing that there are 4.73 a particles emitted
per decay (from above) on average, over 100 years;
4.73a
1g
1 mol 226 Ra 6.02 x 10 23 atoms decayed
x
x
4 .2 mg x 1000 mg x
226
226 g
1 mol Ra
atom decay
19
== 5.3 X 10 atoms of helium formed over 100 years.
Converting into moles:
19
1 mol
-s
5.3 x 10 atoms He x 602 10 23
= 8.8 x 10 - moles He formed
.
x
atoms
Volume of the capsule is:
L=4cm
r= 1 mm= 0.1 cm
2
2
3
lL
-4
V = Lnr = 4 x n x 0.1 = 0.126 cm x 1000 cm 3 = 1.26 x 10 L
The volume that the RaBr2 occupies is now calculated to determine what volume
the gas may occupy;
Molecular Weight of RaBr2 = 226 + 2 x .79.916 = 385.832 grams per mole
PRaBr2
I
I
t
= 5.79 (given in problem)
385.832 g RaBr 2 0 1 226R
------x
. g
a
5
226 gRa
- - - - " - - - - - - - - - - :3- - = 2.95 x 10- L is the RaBr2 volume
g RaBr
cm
5.79
3 2 x 1000-cm
L
Thus the volume available for the gas is:
1.26 x 10-4 L - 2.95 X 10-5 L = 9.65 X 10-5 L
The ideal gas equation is used to find the pressure of the helium gas formed at
room temperature in the needle;
100
THE HEALTI-I PHYSICS SOLUTIONS MANUAL
T= 298 K
V =9.65 X 10-5 L
5
n = 8.8 x 10- moles He
R =0.082
L·atm
I K
moe·
L·atm
x 298 K
nRT
8.8 x 10-5 mol x 0.082
P - - - - _ _ _ _ _ _ _----::m=o.::...;l:..;:.e_·=K:......-_ _ = 22.3 atm
9.65x 10-5 L
- V Thus 22.3 atmospheres is the pressure due to the helium gas formed by the decay
of the radium alone. The initial gas pressure when the needle was. manufactured
must also be considered.
22.3 + 1 = 23.3 atm.
4.36
3
4.36 A volume of 10 cm tritium gas 3H2 at NTP dissipates 3.11 joules per hour.
(a) What is the mean activity of the tritium?
3
V = 10 cm = 0.01 L
T= O°C = 273 K
L·atm
R =0.082 moe·
1 K
P =760 torr = 1 atm
The ideal gas equation is used to find the number of moles of gas present:
PV
1 atrnx 0.01 L
n- = 4.47 X 10-4 moles tritium gas present
- RT - 0.082 L·atm x 273 K
mole·K
-4
6.02 x 1023 molecules
2 atoms
20
4.47 x 10 mol x
x
= 5.38 X 10 atoms. tritium
1 mol
1 molecule
The decay constant for tritium is found using equation 4.21:
TIn = 12.3 years
SOLUTIONS FOR CHAPTER 4
101
0.693
1 yr
I 0.0563
= 0.0563 yr- = y rx .32 x 107 sec = 1.76 X 10-9 sec-I
AH = 123
• yr
Activity = AN = 1.76 X 10-9 sec-I x 5.38 x 1020 atoms tritium = 9.47 x lOll Bq
(b) What is the mean beta ray energy, if one beta particle is emitted per transition?
1 hr
1 MeV
1 sec
MeV
3.11 J
--x
x
x
= 5.7 X 10-3 - 13
hr
3600 sec 1.6 x 10- J 9.47 X 1011 trans
trans
4.37 Barium 140 decays to ＱＴｾ｡＠
with a half-life of 12.8 days, and the ＱＴｾ｡＠
decays to stable 140Ce with a half-life of 40.5 hours. A radiochemist, after
precipitating 140Ba, wishes to wait until he has a maximum amount of ＱＴｾ｡＠
before
separating the ＱＴｾ｡＠
from the ＱＴｾ｡Ｎ＠
(a) How long must he wait?
B
A
Using equation 4.21:
= 0.693
'I
I\,
I
=
0.054 d12.8 d
- -
Ba
A = 0.693 = 0.41 d- 1
La
1.69 d
Equation 4.57:
tm=
In(AABaLa] = ｉｮＨｾＩ＠ 0.054
ALa -ABa
= 5.69 d = 136.5 hr
0.41-0.054
(b) If he started with 1000 MBq (27mCi) before separating the ＱＴｾ｡＠
ＱＴｾ｡Ｌ＠
how many micrograms ＱＴｾ｡＠
will he collect?
from the
Equation 4.53 relates the activity of the daughter (A L ) to the initial activity of the
parent (AB)at time t after isolation of the parent.
4.37
-. ｾＱ＠
102
THE HEALTH PHYSICS SOLUTIONS MANUAL
1
A
La
= 27 mCi
0.41 d(e-O.OS4XS.69 _e-{).41XS.6) = 19.9 mCi
0.41 d -1 _ 0.054 d- 1
The specific activity of ＱＴｾ｡＠
(T I12 = 1.69 d) is
d
1620 y x 365- x 226
SA =
y
= 5.65 x lOs Ci
1.69 d x 140
g
Therefore the weight of ｉＴｾ｡＠
19.9 x 10- ｾｩ＠
3
is
=3.5 x 10-8 g
5.65 x 10 5 Cl
g
Alternatively:
From part (a)
1
A =0.054 d- =
Ba
0.054
d
-7
-1
1d
1 hr
x--x
=
6.25 x 10 sec
24 hr 3600 sec
t m = 136.6 hr = 4.92 x 10 sec
5
Placing the terms in equation 4.53:
ANLa = 7.36 X 108 Bq
SOLUTIONS FOR CHAPTER 4
7.18 X 10
8
B
103
1 Ci
2
140.q x 3.7 x 1010Bq = 2.0 x 10- Ci is the maximum activity of the La.
Next, convert to grams:
From above,
ALa =4.75 X 10-6 sec-1
(AN) =7.36 X 108 Bq
La
Solve for NLa
7.36x 10 8 Bq
NLa=
Ala
7.36 x 10 8 atoms
= ____--=s=-=e.. .:;,.c_ = 1.5 x 10 14 atoms 14<ta
4.75 x 1O--{i sec
140 g140La
1 mol 140La
x -"---=----1.5 x ] 0 14 atoms 140LaX
23
602
. x 10 " atoms 140L"1
a
mo1140La
3.5 X 10-8 g 14<ta
4.38 Strontium 90 is to be used as a heat source for generating electrical energy
in a satellite, (a) How much 90Sr activity is required to generate 50 watts of
electrical power, if the conversion efficiency from heat to electricity is to 30%?
Since the process of energy conversion is only 30% efficient, the required power
(gross) needed is:
100
50 W x 30 = 166.67 Ware actually needed.
The maximum energy of the 90Sr beta particle is 0.546 MeV, so the average energy
of the beta would be roughly 113 of 0.546 MeV, as described in chapter 4,
EpCOSr)"= 0.546 MeV
ＨｾＩ＠
= 0.182 MeV/transformation
However, since 90Sr typically is in equilibrium with its short lived (64 hr) daughter, 90y , it's beta is also taken into account, as it is has 2.27 MeV and contributes
significantly to the power,
EpCoy) 2.27 MeV
ＨｾＩ＠
= 0.76 MeV/transformation
Summing the two energies,
4.38
104
THE REALTII PHYSICS SOLUTIONS MANUAL
0.182 + 0,76 = 0.94 MeV/transformation
1 trans
13
-13 W
0.94 MeV x 1.6 x 10- J _
1W
sec
MeV
- 1.5 x 10 Bq
1 J / x Bq x trans
/sec
1 Bq
1 Ci
4
166.67 W x 1.5 X 10-13 W x 3.7 X 10 10 Bq = 3.0 X 10 Ci 90S r
(b) Weight of the isotopic heat source is an important factor in the design of the
power source. If weight is to be kept to a minimum, and if the source is to generate
50 watts after 1 year of operation, would ｴｾ･ｲ＠
be an advantage to using 2IOpO?
The energy of the 21OpO alpha particle is 5.3 'MeV, but as noted on p.77 of the text,
including the recoil energy yields 5.4 MeV/transformation; the 2lOpO half-life is
138 days.
"
.;
1 trans
1W
＠ｾ
1 J / x Bq
Isec
5.4 MeV 1.6 x 10-13 J
x trans x
MeV
13
8.64 X 10-
W
Bq
1 Bq
1 Ci
3
166.67 W x 8.64 X 10-13 W x 3.7 X 10 10 Bq = 5.2 X 10 Ci is needed at the end of
ａｯ］ｾ＠
the year to allow for decay. At the beginning of the year the required 21OpO
activity according to equation 4.18 is
A
e
5.2 X 10 3 Ci
4
.
= -0.693
=3.3 x 10 CI needed at the start of the year, so that at
- x365d
ＬＡｾ＠
ＬＮｾＨ＠
.'
e 138d
the end of the year, there is enough activity present to provide the 166.67 (gross)
watts of power to the unit.
Calculate the number of grams of 21OpO to which this quantity corresponds:
Tp o-210 = 138 days
A po-210 =210
ARa = 226
TRa = 1620 yrs
.
,
SOLUTIONS FOR CHAPTER 4
105
Substituting these values into the equation for specific activity equation (4.31),
we have:
226 x 1620 yr
=4.61 x 103 CilgPo-210
210x(138dx 1 yr)
365 d
1g
4
3.3 x 10 Ci x 4.61 x 10 3 Ci = 7.2 g
210
Po needed at the start of the year.
Now calculate how many grams of 90Sr would be needed, based on the
4
3.0 x 10 Ci 90Sr calculated in part (a), and assuming that the decay would"be
negligible over a one year period, since 90Sr has a 28 year half life:
The specific activity of 90Sr is calculated with equation 4.31 :
ARa = 226
TRa = 1620 yr
A s =90
Ts = 28 yr
ARaTRa
226 x 1620 yr
SA = A T. = 90 x 28 yr = 145.3 Cilgram
Sr
Sr
3.0 X 104 Ci 90Sr x
1g
= 206.5 g 90Sr required.
145.3 Ci
Compare the 7.2 g 21OpO with 206.5 g 90Sr required, and there is an advantage to
•
21On.
usmg roo
4
4.39 Carbon 14 is produced naturally by the 14N(n,pi C interaction of cosmic
14
radiation with the nitrogen in the atmosphere at a rate of about 1.4 x 10 Bq/
year. If the half life of 14C is 5700 years, what is the steady state global inventory
of 14C?
Q = global inventory
rate of formation of 14C = 1.4 x 10 15 Bq/yr
rate of decay of 14C = AQBq/yr
Equation 4.21
4.39
--,."
--j
106
THE HEALTII PHYSICS SOLUTIONS MANUAL
0.693
0.693
A-- -T- -- 5700 yr = 1.2 X 10-4 yr- 1 14C
Under steady state conditions
Rate of formation = Rate of decay
15
I
1.4 X 10 Bq/yr = Ayr- x Q Bq
1.4 X 1015 Bq
yr
=1.15 x 1019 Bq
Q -----"--1.2 X 10-4 yr
4.40
4.40 The global steady state inventory of naturally produced tritium from the
interaction of cosmic rays with the atmosphere is estimated by the United
Nations Scientific Committee on the Effects of Atomic Radiation to be 1.26 x
18
6
10 Bq (34 x 10 Ci). If the half life of tritium if 12.3 years, what is the annual
production of natural tritium?
Q = global inventory of 3H = 1.26 x 10 18 Bq/yr
A = rate constant of 3H
T1I2
= 12.3 yr
Equation 4.21
0.693
0.693
=5.6x 10-2 yr- 13H
AB -- -T- - 12.3 yr
Under steady state conditions,
Production per year = decay per year
2
16 Bq .
5.6 X 10IS the
Production per year = A Q =
x 1.26 X 10 18 Bq = 7.1 x 10 yr
yr
rate of production of 3H in the atmosphere.
Solutions for Chapter 5
IN1ERACTION OF RADIATION WITH MATIER
3
5.1 The density of Hg is 13.6 grams/cm and its atomic weight is 200.6. Calcu3
-5.1
late the number of Hg atoms/cm .
22
3
13.6 g
mol
6.02 x 10 23 atoms
3
X 2006
x
l
I
=
4.08
x
10
atoms/cm
cm
. gmo
3
5.2
5.2 The density of quartz (Si02) crystals is 2.65 gmlcm • What is the atomic
density (atoms/cm3 ) of silicon and oxygen in quartz?
Atomic Weights: Si = 28.09,
-
° = 16
28.09 + (2 x 16) = 60.09 g/mol Si02
First, Silicon:
2.65 g 1 mol Si0 2
1 mol Si
6.02 x 1023 atoms
---'x
x
x
65 1022
/
3
cm 3
60.09 g
1 mol Si0 2
mol
= 2. x
atoms cm
Oxygen:
°
2 mol
6.02 x 1023 atoms
2.65 g 1 mol Si0 2
-x
x
x
5 3 1022
/
3
mol
= . x
atoms cm
cm 3
60.09 g
1 mol Si0 2
5.3 Compare the electronic densities of a piece of aluminum 5 rom thick and a
piece of iron of the same density thickness.
t =5 mm = 0.5 cm
,
ｾＭ
107
5.3
__ ",..r.
108
THE HEALTH PHYSICS SOLUTIONS MANUAL
3
PAl = 2.7 g/cm
Density thickness of 0.5 cm thick AI:
2.7 g
2
0.5 cm x - = 1.35 g/cm
3
cm
Computing the electronic density (knowing that z = 13 and molecular ｷｾｩｧｨｴ＠
27):
=
23
23
/ 2
1.35 g 1 mol 6.02 x 10 atoms 13 electrons
X
x 1
Al = 3.913 x 10 electrons cm
2 X
cm
27 g
1 mol
atom
The density thickness of 5 rum of aluminum was computed above as 1.35 g/cm2.
Computing the electron density of a piece of iron with the same density thickness,
1.35 g/cm2 (the iron would, of course, not be as thick).
Fe = 55.85 g/mole
Fe = 26 electrons per atom
Fe electronic density:
1.35 g
cm
---=
2 X
5.4
23
1 mol
6.02 x 10 atoms 26 electrons
23
/ 2
X
X
= 3.78 x 10 electrons cm
55.85 g
1 mol
1 atom Fe
t..:.
f,
5.4 In surveying a laboratory, a health physicist wipes a contaminated surface,
and runs an absorption curve using a thin end window counter and aluminum
absorbers. The range of the beta rays (no gamma rays were found) was found to
2
be 800 mg/cm aluminum. What could the contaminant be?
2
Compute the energy using equation 5.3, where R is in units of mg/cm
Ｎｾ＠
R = 800 mglcm2
,
SOLUTIONS FOR CHAPTER 5
109
In E = 6.63 - 3.2376 (10.2146 -In R)'I2
In E = 6.63 - 3.2376 (10.2146 -In (800))V2
E = 1.73 MeV
The Radiological Health Handbook (1970) has a table of energies of beta emitters
arranged by energy on page 91 which can ·be used to determine the isotope from
the beta energy. A maximum of 1.73 MeV beta particle corresponds to P-32.
5.5. A ｨ･Ａｬｴｐｙｾ､ｩ｣ｳＮｦｮ＠
an ｵｮｫｾｷ＠
cohntaminant thbat ｰｲｯｾ･ｳ＠
to be a pudre beta
enutter. .1O e p 1 entl y t e contanunant, e runs an a sorptIOn curve to etermine the maximum energy of the beta rays. He uses an end window G.M.
2
he [mds
counter whose mica window (density = 2.7 gmlcm ) is 0.1 mm thick, ｡ｮｾ＠
that 1.74 mm AI stops all the beta particles. ｔｨｾ＠
distance between the sample and
the G.M. counter was 2 cm. What was the energy of the beta particle? What is the
contaminant?
Calculate the density of each of the components in the beta particle's path:
the air, aluminum, and the mica window:
Air
thickness of air = 2 cm
3
PruT. = 1.293 x 10- g/cm3
1.293 X 10-3 g
-----=--3--'- X
cm
2
2 cm = 0.002586 g/cm
Aluminum
thickness of aluminum = 1.74 mm
PAl = 2.7 g/cm3
lcm
2
g
1.74 mm x 2.7 cm 3 x 10 mm = 0.4698 g/cm
Mica
thickness of mica = 0.1 mm
Pniica = 2.7 g/cm3
5.5
ＭｾＧＮ＠
110
THE HEALTH' PHYSICS SOLUTIONS MANUAL
0.1 mm x
1 cm
g
2
x 2.7 - 3 = 0.027 g/cm
10mm
cm
ＭｾＺ＠
t
!
The density thicknesses are now added together to fmd the total amount of material that stops the beta:
R = 0.002586 g/cm2 + 0.4698 g/cm2 + 0.027 g/cm2 = 0.4994 g/cm2
2
Convert this to mg/cm
R-
0.4994 g 1000 mg
2
x
= 499.4 mg/cm
1 cm 2
1g
2
Putting R = 499.4 mg/cm into the range to energy equation (5.3) to find the
energy:
In E =6.63 - 3.2376 (10.2146 -In (R»1I2 .
k-
In E = 6.63 - 3.2376 (10.2146 -In (499.4»112
E= 1.17 MeV
The table for ascending beta emitters in the RHH shows that Bi-210 emits a 1.17
MeV beta with no gamma
5.6
5.6 A 5 MeV photon produces a positron electron pair in a Pb shield. If both
particles are of equal energy, how far will they travel in the shield?
From the text in Chapter 5, it can be seen that the mass energy equivalence of a ,
positron is 0.511 MeV and 0.511 MeV for the electron. Subtracting the energy of :".
formation from the initial photon energy:
5 MeV - (2) 0.511 = 3.98 MeV is the energy left to contribute to the kinetic
energy of the positron and electron. If each particle has equal energy, the remaining energy is divided between the two particles:
.
3.98 MeV
- - 2 - - = 1.99 MeV is the kinetic energy associated with each particle.
SOLUTIONS FOR CHAPTER 5
111
The range of an electron can be found using equation 5.2, since an electron with
kinetic energy is indistinguishable from a beta;
R = 412E 1.265 - O.0954(lnE)
R = 412 (1.99) 1.265-0.0954(ln(1.99» = 940 ｭｾ＠
em
The density of lead is 11.35 g/cm3
940 mg
1g
1 em 3
cm 2 x 1000 mg x 11.35 g - 0.083 cm is the range in lead of the electrons.
Note that the annihilation radiation from the positron is not addressed in this
problem.
5.7 A Compton electron that was scattered straight forward (<1> = 0 0 ) was
completely stopped by an aluminum absorber 460 mg/cm2 thick.
a) What was the kinetic energy of the Compton electron?
2
Using the range energy equation (5.3), where R = 460 mg/cm
In E = 6:63 - 3.2376 (10.2146 -In R)In
1
E =In- {6.63 - 3.2376 (10.2146 -In (460))112} = 1.09 MeV
b. What was the energy of the incident photon?
Since the Compton electron is scattered directly forward (<1> = 00 ), the photon had
to be scattered directly backward, = 1800 (since this reaction can be thought of
as a classical physics collision problem), so that momentum is conserved. Also,
conservation of energy applies in this reaction, so that the energy of the Compton
electron (1.09 MeV from part a) , and the energy of the scattered photon E',
added together must equal the energy of the incident photon (E):
e
E = 1.09 + E'
Using this information, as well as equation 5.31A,
I
ｾＭ
5.7
112
THE HEALTI:I PHYSICS SOLUTIONS MANUAL
·fe,.
Solve for E' in the equation E = 1.09 + E'
E' =E -1.09
Now substitute in for E'
E -1.09
E
Since it is an electron which is scattered, the mass energy equivalence of an
electron (0.51 MeV)can be used in the equation, and = 180 0 • Substituting
values:
e
E
E-l 09 M
1+(EI051 eV)x(1-cos 180)
.
Solving for E,we get the quadratic equation
(3.6 x E)- (3.92 x E) - 1= 0
This is of the form ax + bx + c, ｷｨｾｲ･＠
2
a = 3.6
b= - 3.92
c=-1
Solve for E
2
E=
-b±-Jb2 -4ac = 3.9 ± J-3.9 -4x(3.6)x(-I)
- - - ' ' - - - - - ' - - - - - - - = 0.542 ± 0.76
2a
2 x (3.6)
E = 1.30 MeV is the solution which satisfies this equation.
5.8
5.8 Monochromatic O.IMeV gamma rays are scattered through an angle of 1200
by a carbon block.
SOLUTIONS FOR CHAPTER 5
113
(a) What is the energy of the scattered photon?
E=O.lMeV
moc2 = 0.511 MeV
= 120°
e
Equation 5.31A:
E' _ _ _---:E2_ _ _ _ = _ _ _ _ _0_.1_ _ _ _ _ = 0.077 MeV
1 + (E / moc )(1- cose)
1 + (0.1 /0.511) x (1- cos(120))
(b) What is the kinetic energy of the Compton electron?
E-E' = Ee0.1 MeV - 0.077 MeV = 0.023 MeV
5.9 A 1.46 MeV ｧ｡ｭｾ＠
from naturally occurring 40K is ｳｾ｡ｴ･ｲ､＠
two times: first
through an angle of 30 and then through an angle of 150 .
(a) What is the energy of the photon after the second scattering?
Calculating the energy of the first scattered photon:
E= 1.46 MeV
= 30°
2
moc = 0.51 MeV (the rest mass energy of an electron)
e
Using this information, 5.31A,
_ _ _ _ _1_.4_6_ _ _ _ = 1.056 MeV
1 + (1.46 / 0.511) x (1- cos(30))
This is the energy of the first scattered photon. Now it is scattered again:
Substitute in the variables for the second scatter (Equation 5.31A),
E" = _ _ _ _ _1_.0_5_6_ _ _ _ _ = 0.22 MeV
1 + (1.056/0.511) x (1- cos(150))
5. 9
114
THE HEALTH· PHYSICS SOLUTIONS MANUAL
(b) What is the energy of the photon if the angular sequence is reversed?
Using the same fonnulas, only reversing that the order the angles are input:
E'= _ _ _ _ _I_.4_6 _ _ _ __
1 + (1.46/0.511) x (1- cos(150»
=0.23 MeV
This is the energy of the first scattered photon. Now it is scattered again:
0.23
1 + (0.23/0.511) x (1- cos(30»
E"
= 0.22 MeV is the energy of the second
scattered photon.
5.10
5.10 What is the energy of the Compton edge for the 0.661MeV gamma from Cs-
137?
E=0.661 MeV
e = 180° for maximum energy Compton electron (definition of Compton edge)
2
m oc = 0.51 MeV (the rest mass energy of an electron)
Using this infonnation, (Equation 5.31A)
E'
E
= _ _ _ _ _0_.6_6_1_ _---,-__ =0.184 MeVis
l+(E Imoc )(1-cos8) 1+(0.661/0.511) x (1-cos(l80»
2
the energy of the scattered photon.
U sing the relationship that
Ee_ = E - E'= 0.661- 0.184= 0.48 MeV for the Compton edge.
5.11
The energy of a scattered photon is 0.2 MeV after it was scattered through
an angle of 135°. What was the photon'S energy before the scattering collision?
5.11
E;= 0.2 MeV
moc2 = 0.51 MeV
= 135°
e
SOLUTIONS FOR CHAPTER 5
115
Equation 5.31A:
E"
0.2
E
1 + (E 1 moc 2 )(1- cose)
E
1+(E 10511) x (1-cos(135»
E=0.61 MeV
5.12 What is the energy of the Compton edge for the following gammas?
(a) 0.136 MeV from 57CO ,
The Compton edge occurs when the maximum amount of energy is transferred
from the incoming photon to the scattered electron. The scattered photon is
0
scattered at an angle of 180 to the scattered electron. There are many ways to
solve for the energy of the Compton edge electron; for this solution one technique
will be used, and answer (b) will be solved using another technique.
Use equation 5.30:
!lA = 0.0242 ( 1 - cos 8) A = 0.0242 ( 1 - cos(180» = 4.8 x 10-2 A is the change
in wavelength of the incident photon compared to the scattered photon.
Calculating the initial wavelength of the incident photon:
A = 12400 =
EeV
12400
6
= 9.12 X 10-2 A
0.136 MeV x 1 x 10 eV
MeV
Adding the change to ｴｨｾ＠
2
ｴｾＢ＠
4.84 X 10-
.-
___ .
initial wavelength:
A + 9.12 X 10-2 A = 0.1396 A
5.12
---:l
116
THE HEALTH PHYSICS SOU.J110NS MANUAL
Converting the scattered photon wavelength back to MeV:
12400
E'=--
A
12400
0.1396 = 88840 eV = 0.089 MeV is the energy of the scattered
photon. Now subtract it from the initial energy of the photon (incident photon):
Ecompton eIectron =E(y)- E"(y)= 0.136 MeV - 0.089 MeV = 0.047 MeV is the energy
of the most energetic (Compton edge) electron.
58
(b) 0.811 MeV from C0,
Using equation 5.31A to fmd the energy of the scattered photon (E'):
2
moc = rest mass energy equivalence of electron = 0.51 MeV
= 1800
E=0.811 MeV
e
E' =
= ｟Ｍ［ｾＰＮＸＱ＠
E
1+
ＨｾＩＱＭ
cose)
_ _ __
1+
moc
Ｈ｟ＰｾＱＩ＠
0.194 MeV is the
x (1- cos(180))
energy of the scattered photon.
Subtract it from the initial energy of the photon (incident photon):
E()
Y - E"C)-E
Y - compton electron
0.811 - 0.194 =0.617 MeV is the energy of the most energetic (Compton edge)
electron.
(c) 1.33 MeV from 60CO .
Using equation 5.31A to find the energy of the scattered-photon (E'):
2
moc = rest mass energy equivalence of electron = 0.51 MeV
= 1800
E = 1.33 MeV
e
SOLUTIONS FOR CHAPTER 5
E
E' =
1+
ＨｾＩＱＭ moc
1.33
= ＭＺｾ
cose)
117
0.214 MeV is the
1 + (1.33) x (1- cos(l80))
051
energy of the scattered photon.
Subtract it from the initial energy of the photon (incident photon):
Ecompton eIectron = E(y) - E'(y) = 1.33 - 0.214 = 1.12 MeV is the energy of the most
energetic (Compton edge) electron.
5.13
The following gamma-ray absorption data were taken with lead absorb-
ers:
Absorber
o 2 4 6
thickness, mm
20 25
8 10 15
Counts
per minute
1000 880 770 680 600 530 390 285 210
(a) Determine the linear, mass, and atomic attenuation coefficients.
The attenuation coefficient may be detennined graphically. Plot the data on semilog paper, with the count rate on the logarithmic scale. If the data fall on a single
straight line, as in this case, then the attenuation coefficient is the slope of the
line. The absorber half thickness (HVL) is determined from the curve and the
slope is calculated. From the absorption curve, we find the half thickness to be
11 mm (1.1 cm).
ｾｴ＠
= 0.693 = 0.693 = 0.63 cm·!
HVL
1.1 cm
5.13
118
THE HEALTH PHYSICS SOLUTIONS MANUAL
ＱＰｾＭ＠
I
-!
I
-I
E
Q.
U
h.
I
-I
I
!
I
thickness in mm
Calculating the mass attenuation coefficient:
P Pb = 11.35 gJcm3
3
II
Il/inear
__
r
p
m
__
0.63 x cm
cm 11.3 g
0.056 cm2Jg
Calculating the atomic attenuation coefficient:
Since the atomic weight ofPb is 207.21, there are 207.21 g Pb per mole.
=
fl
a
.. N
fll
atoms
em 2
0.63 em- l
19.2 X 10-24 cm2Jatom
6.02 x 10 23 atoms
mole
.
g -x em
207.21-
mole
3
11.3 g
(b) What was the energy of the gamma ray?
Iｾ］＠
SOLUTIONS FOR CHAPTER 5
119
Check in Appendix E, under lead attenuation, to detenrune the energy. From part
(a), the mass attenuation coefficient is 0.055 cm2Jg. Looking carefully in Appendix
E, a value of 0.0569 cm 2Jg is close to the value calculated in part (a), and this
corresponds to a value of 1.25 MeV. So the gamma ray energy is approximately
1.25 MeV.
5.14
The following absorption data were taken with aluminum absorbers:
Absorber
Thickness. ern 0
Counts
per minute
5.14
0.02
1000 576
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.2
0.4
0.8
1.5
2
2.8
348' 230
168
134 120
107
96
95
90
82
68
60 50
(a) Plot the data. What types of radiations do the curve suggest?
1000
[
100
<.l
10
0
0
C')
0
0
<0
C\J
0
0>
C\J
l()
C')
l()
0
0
0
<0
<0
I"-
0
0;
0
'<t
C!
I"-
,....
a
ｾ＠
C')
-.:t:
<0
""!
0>
C!!
C\J
IX1
l()
ｾ＠
<0
a
N
'<t
N
C')
N
N
I"'<t
N
a
<0
N
C')
I"-
N
Thickness in cm
.r.. -' ..... Ｂ＼ｾＺ［＠
e:::..
____ ______________________________ "ｾＢ［Ｇ＠
ｾ＠
.......
,
! ..
ﾷｾＺ［Ｍ＠
t::.,.c....":','.' !•.:,:,r........: Ｂｾ＠ ! :> '•.ｾＧＭ＠ .. Ａｾ＠
.' -' . . .
L'_ _ _ _ｾ＠
. ｾﾷｬ＠
... ;: •.••(.-:--"
/-' ) •.-:.
11:
-'1"','i
120
THE REALTII PHYSICS SOLUTIONS MANUAL
Because the curve falls rapidly for the fIrst 0.08 cm, it appears that there is ｾ＠ ｾ＠
emitter present. Since counts above background are still detected after all the betas
have been stopped, a y must also be emitted.
The data points, after the betas have been stopped, fallon a straight line on semilog paper; these points probably represent a gamma ray.
The line intersects the y axis at 100 cpm, the gamma count rate from the sample
due to the gamma emitter is 100 cpm.
(b) If a beta particle is present, what is its energy?
A line representing the beta portion of the absorption curve intersects the gamma
portion at 0.16 cm. This thickness represents the range, in aluminum, of the beta
radiation.
Expressing the 0.16 cm range in aluminum in units of density thickness:
Density of aluminum, p = 2.7 g/cm3
0.16 cm x 2.7 ｾ＠ x 1000 mg
cm
1g
432 mg/cm2
and using the range equation (5.3) to fmd the energy of the ｾＺ＠
In E = 6.63 - 3.2376 x (10.2146 -In R)1/2
In E = 6.63 - 3.2376 x (10.2146 -In (432»112
E = IMeV is the approximate energy of the ｾ＠
(c) If a gamma ray is present, what is its energy?
The slope of the "hard" straight line component of the absorption curve, which
represents the gamma component, is
ｾ］Ｍ＠
0.693
HVL
ＲＮＸ｣ｭｸＷｾ＠
0.693
I
ｾＮ＠
' c m2
=0.92g
cm
From Appendix E, we fInd the gamma energy to be 0.4 MeV for a mass attenuation coefficient of 0.0922 cm2/g
J
SOLUTIONS FOR CHAPTER 5
]21
(d) Looking up the 0.6 MeV ｾ＠ in a table of energy vs. isotope (in the Radiological Health Handbook, 1970), Au-198 is found to be the most likely isotope
because it emits a OAI MeV gamma and a 1.37 MeV beta. Computer programs
also exist which can be used to determine the isotope based solely on energies.
(e) \Vrite the equation that fits the absorption data.
The absorption curve can be resolved into 2 components. First, the "hard" portion,
which is a straight ·line representing gamma radiation, is extrapolated to the y
axis. Then, the various y values of the extrapolated straight line are subtracted
from the total absorption curve, to give the y values for the "soft" component,
which represents possible beta radiation. In this ·manner, the intercept of the
gamma component is 100 cpm and its slope is 0.25 cm- I . The intercept of the
beta component is 1000 - 100, or 900 cpm, and its slope is 24 cm- I . The equation
of the absorption curve, therefore is
A = 900e-24t + 100e-o.2St
The frequency of betas and gammas from 198Au is about equal. The higher beta
count rate is due to the higher counter efficiency for betas than for gammas.
A collimated gamma ray beam consists of equal numbers of 0 .1 MeV and
1. .0 MeV photons. If the beam enters a 15 cm thick concrete shield, what is the
relative portion of 1 MeV photons to 0.1 MeV photons in the emergent beam?
5.15
=0.169 cm /g
2
ｾｏＮｬｍ･ｖ＠
(Appendix E)
ｾｬＮｏｍ･ｖ＠
= 0.0635 cm /g (Appendix E)
Pconcrete= 2.35 g/cm3 (Table 5.2)
2
Convert the thickness to density thickness:
2.35 g
2
= 15 cm x cm3 = 35.25 g/cm
.
t
Use equation 5.19:
III = e-)lt
ｾＭＮ＠
o
5.15
122
THE HEALTH PHYSICS SOLUTIONS MANUAL
I
I MeV
=
I
O.IMeV
5.16
-J.lI Mev
(
e-II
e
t )
rO.1 MeV
t
=
(e--{)·0635.35.25)
e--{)·169.35.25
= 41.3
5.16 Three collimated gamma-ray beams of equal flux, whose quantum energies
are 2,5, and 10 MeV, respectively, pass through a 5 cm thickness of lead. What
is the ratio of the emergent fluxes?
J.l
J.l
2MeV
= 0.0455 cm2/g
SMeV
= 0.0424 cm /g
J.lJOMeV
PPb
ｾＺ＠
2
= 0.0484 cm2/g
= 11.35 g/cm
3
Use equation 5.24A:
For 2 Me V photons
0 0756
e-J.lt =e-0.0455 em21g x 5 em x 11.35 glcm3 =e-2.58 =.
For 5 MeV photons:
-/.It
e
=e
-0.0424 em21g x 5 em x 11.35 g/em3
-2.41
0 09
=.
-2.75
0 064
=.
=e
For 10 MeV photons:
-/.It
e
-0,0484 em2/g x 5 em x 11.35 g/em3
=e
=e
ｾＮ＠
Calculating the ratios:
0.0756 0.0756 0.0756
0.0756 0.09 : 0.0641 = 1 : 0.8 : 1.2
5.17
A collimated beam of 0.2 Me V gamma radiation delivers an incident
energy flux of 2 J/m2/sec to a Pb shield 1 g/cm2 thick.
(a) What is the incident photon flux, photons/cm2/sec?
5.17
i
J
SOLUTIONS FOR CHAPTER 5
123
2
1 photon x
MeV
x 2 J x 1m
= 6.25 x 10 9 pho!ons
cm . s
0.2 MeV 1.6 x 10-13 J m 2 • s 10 4 cm 2
(b) What is the rate of energy absorption in the shield, ergs/g/sec and J/kg/sec?
cm 2
II
= 0.821-r'en
g is the energy absorption coefficient for Pb for 0.2 MeV gamma
(Appendix F)
1 =2
o
J
2
m .s
=2xl0
-4
J
cm 2 .s
Incident energy absorbed = 10 - 1 = 10 - 10 e'pt = 10 (1 - e-pt)
cm2
xI_g-J
-D.821_
J (
2
M=2x10-4-1-e
g
cm
2
cm ·s
J
J
= 1.12 X 10-4- = 0.112-Ll1
g. s
kg.s
AT
7
J
1 X 10 erg
ergs
1.12 x 10- - - X
= 1.12 X 10 3 - kg's
J
g.s
1
5.18 Calculate the thickness of Al and Cu required to attenuate narrow, colli-
mated, monochromatic beams of 0.1 MeV and 0.8 MeV gamma rays to
(a) one half the incident intensity (HVL),
(b) one-tenth the incident intensity (TVL). Express your answer in cm and
2
grams/cm.
(c) What is the relationship between a half-value layer and a tenth-value layer?
Gamma ray attenuation is given by equation 5.19
The linear attenuation coefficients, ｾ＠ cm-\ as listed in Table 5.2, and the mass
2
attenuation coefficients, !J.m cm /g, as listed in Appendix E are:
5.18
124
THE HEALTII PHYSICS SOLUTIONS MANUAL
!J, cm j
/lm cffi2/g
0.1 MeV
0.8 MeV
0.1 MeV
0.8 MeV
AI
0.435
0.185
0.161
0.068
Cu
3.8
0.58
0.427
0.065
By definition, a half value layer, HVL, is that absorber thickness that transmits
one half the incident radiation, and a tenth value layer, TVL, transmits one tenth
the incident radiation.
ＧＮｾ＠
I _ 1 -!U
TVL: I-10 e
o
ｉｮＨｾＩ＠
In (1..)
t(HVL)=--Jl
t(TVL) =
10
-Jl
- .
t
HVL= 0.693
Jl
TVL
HVL
TVL= 2.3
Jl
2%
X
= 69
.
Therefore, TVL = 3.3 HVL
If we substitute the appropriate values for the attenuation coefficients into the
equations above, we have for 0.1 MeV, the linear (HVL j ) and the mass (HVLm)
ｾ｡ｬｦ＠
value layers and tenth value layers, in AI:
HVL =
£
0.693 = 1.59 cm
0.435 cm'!
The TVL = 3.3 HVL
Therefore,
TVL j = 3.3 x 1.59 cm = 5.25 cm
HVL m =
0.693
g
2 = 4.3-2
0.161 cm
cm
g
TVLm = 3.3 x 4.3 g/cm2 = 14.2 g/cm2
Other HVL's and TVL's are calculated in a similar manner, and the results are
tabulated below:
SOLUTIONS FOR CHAPTER 5
125
0.1 MeV
HVL
0.8 MeV
Linear, cm
Mass, crri/g
Linear, cm
Mass, crri/g
AI
1.59
4.3
3.75
10.2
Cu
0.18
1.62
1.12
10.7
0.1 MeV
TVL
0.8 MeV
Linear, cm
Mass, crri/g
Linear, cm
Mass, crri/g
AI
5.25
14.2
12.4
33.7
Cu
0.59
5.35
3.7
35.3
5.19 The mass attenuation coefficient for muscle for 1 MeVy radiation is 0.070
2
cm /g. What is the mean free path of a 1 Me V photon in muscle?
5.19
The mean free path is simply the reciprocal of the attenuation coefficient.
cm 2
II =0.070r'm
g
g
1 -3
Pmcm
1
1
·MFP = - =
2
II 0.07 cm X Ｑｾ＠
g
14.29 cm
cm 3
5.20 A laminated shield consists of two layers each of alternating thickness of
2
aluminum and lead, each layer having a density thickness of 1.35 g/cm • The
shield is irradiated with a narrow collimated beam of 0.2 MeV photons.
(a) What is the overall thickness of the laminated shield in cm?
The shield consists of 1 sheet AI, a sheet of Pb, a sheet of AI, and a sheet of Pb.
This makes a total of 2 layers of each type of material in the shield. Since there
are two layers of each material, multiply each density thickness by two;
5.20
126
THE HEALTH PHYSICS SOLUTIONS MANUAL
2
2
1.35 g/cm x 2 = 2.7 g/cm is the total density thickness contributed by each
material layer.
U sing the density of each material, the thickness of each layer of material can be
determined:
Aluminum has a density of 2.7 g/cm3
g
2.7--2
cm = 1 cm is the thickness of the aluminum in the shield.
g
2.7--3
cm
Lead has a density of 11.3 g/cm
3
g
2.7-2
__c=m=- =0.24 cm is the thickness of the lead in the shield.
g
11.3-3
cm
Adding together the two layers:
1+ 0.24 = 1.24 cm thick is the total thickness
(b) Calculate the shield attenuation factor when the (1) aluminum layer is first,
(2) lead layer is first.
From table 5.2:
JlAI = 0.324 cm-1
1
J,lpb = 10.15,c'mFrom part (a), the thicknesses are:
tAI = 1 cm Al
tPb = 0.24 cm Pb
Combining this into equation 5.19;
1110= e-Ilt = eＭｻｾｬ＠
x 0.324) + (0.24 x ＱｏＮｾＵﾻＩ］＠
0.063
PUtting the lead or aluminum first does not make a difference in the result.
SOLUTIONS FOR CHAPTER 5
127
Calculate the probability that a 2 MeV photon in a narrow collimated beam
will be removed from the beam by each of the following shields,
(a) Lead, 1 cm thick
5.21
1
I = Probability of transmission
o
1 - ｾ＠
10
= 1 - eＭｾｴ＠ = Probability of removal
t= 1 cm
= 0.516 cm- (table 5.2)
ｾｐ｢＠
I
11 = 1- e _ = 1 _ e-0.516x 1 = 0.403
1-'1
o
(b) Iron, 1 cm thick
t= 1 cm
= 0.335 cm- I (table 5.2)
ｾｆ･＠
1 1
1=
-e
-Ill
= l_e-0.535xl = 0.285
o
(c) Lead, 1 g/cm2 thick
PPb=11.35 g/cm3
cm 3
2
= 8.8 x 10- cm
t = --2 X
cm
11.35 g
1g
= 0.516 cm- I (table 5.2)
ｾｐ｢＠
1 _
1-1e
_1-'1
= 1 - e-0.516 x 0.088 = 0.044
o
ｬ
(d) Iron, 1 g/cm2 thick
.. -
-
5.21
128
THE HEALTII PHYSICS SOLUTIONS MANUAL
1g
cm
t=--2 x
ｾｐ｢＠
cm 3
=0.127cm
7.86 g
= 0.335 cm- l (table 5.2)
I
1=
1- e-J.II = 1 - e-0.535 0.127 =0.0416
x
o
5.22
5.22 Calculate the neutron threshold energy for the reaction llC(n,y) 12C, if the
prompt capture gamma ray is 21.5 MeV.
Calculating the mass lost by the neutron absorption (From CRC), and adding the
energy equivalent of the prompt gamma ray to the 12C;
IlC = 11.01143 amu
In = 1.008665 amu
12C = 12.00000 amu
931 MeV = 1 amu
amu
-2
Y = 21.5 MeV x 931 MeV = 2.309435 x 10 amu
2
12.00000 amu + 1.008665 amu + En = 11.01143 amu + 2.309345 x 10- amu
En = 2.998 X 10-3 amu
931 MeV
3
- - - x 2.998 x 10- amu = 2.8 MeV
amu
.
5.23
5.23 X-rays are generated as bremsstrahlung by causing high-speed electrons to
be stopped by a high atomic numbered target. If the electrons are accelerated by
a constant high voltage of 250 k V, and if the electron beam current is lOrnA,
calculate the X-ray energy flux at a distance of 1 m from the tungsten target.
Neglect absorption by the glass tube, and assume that the bremsstrahlung are
emitted isotropically.
SOLUTIONS FOR CHAPTER 5
129
For mono--energetic electrons, equation 5.11b applies to detennine the fraction of
energy in the electron beam which is converted into bremsstrahlung:
E = 250 ke V = 0.25 Me V, the kinetic energy of the electron hitting the target
Z =74 (the atomic number for Tungsten)
3
3
f = 1 X 10- Z E = 1 X 10- x 74 x 0.25 = 0.0185 is the fraction of the electron's
energy converted to photons.
Calculating the energy input:
P = power in watts
I = current in amps = 10 rnA = 10 X 10-3 A
E = Voltage = 250 kV = 250 X 103 V
3
P = I E = 10 X 10- A x (250 X 103 V) = 2500 W
Therefore, the rate at which the energy is converted to bremsstrahlung is:
2500 W x 0.0185 = 46.25 W is the rate energy is converted to bremsstrahlung.
Since the bremsstrahlung is spread over the surface of a sphere whose radius is
100 cm:
46.25 W x 10 3 mW
W0.37
· mW/cm2
X-ray energy flux = ----(---2-=
4 x n x 100 cm)
5.24 If the most energetic photon results from the instantaneous stopping of an
electron in a single collision, what voltage must be applied across an X-ray tube in
order to generate X-rays whose shortest wavelength approaches 0.124 angstroms?
A =0.124 angstroms =0.124 x 10-8 cm
h = 6.63 x 10-27 erg/sec
c = 3 X 10 em/sec
10
Equation 2.76;
27
MeV
_
6.63 x 10- erg· sec x 3 x 10 10 -em x 6.25 X 10 5 _
sec
erg
he
E
- -- -A
= O.l.MeV
0.124 x 10-8 cmx 1.6 x 10-12 _erg
eV
Since 1 e V is the energy gained by an electron accelerated by a potential of 1 volt,
a potential of 100,000 V the required wavelength.
5.24
,.
-1"···'•.":
130
5.25
THE HEALTH PHYSICS SOLUTIONS MANUAL
5.25 A beta particle whose ldnetic energy is 0.159 MeV passes through a
2
4 mg/cm window into a helium-filled Geiger tube. How many ion pairs will the
beta particle produce inside the tube?
First, determine the range of the J3 (0.159 MeV) using equation 5.2;
R = 412 E 1.265-0.09541nE = 412 X (0. 159)1.265-0,0954X !n (0.159)= 29.145 mg/cm2 is the range
of the electron.
Since the thin window partly attenuates the J3, subtract the thin window density
2
(4 mg/cm ) thickness from the total range:
2
2
29.145 mg/cm - 4 mg/cm = 25.145 mg/cm2
Now find the fraction of energy left in the beta after going through the window;
25.145 ｭｾ＠
em
mg = 0.863
29.145-2
em
Multiplying by the initial energy:
0.863 x (0.159MeV) = 0.137 MeV is the energy left after going through the
window.
'From table 5.1, 4He consumes 41.5 eV per ion pair:
0.137 MeV x
5.26
1 x 106eV
1 MeV
x
ion pair He
41.5 eV
3305 ion pairs
6
5.26 If the neutron emission rate from 252Cf is 2.31 x 10 neutrons per second per
J.lg, and the transformation rate constant for alpha emission is 0.25 per year, what
is the neutron emission rate per MBq and per J.lCi?
A = 0.25 yr- 1
0.693
TCf= 0.25 = 2.772 yr
SOLUTIONS FOR CHAPTER 5
131
Calculate the specific activity of 252Cf, use equation 4.31 in order to determine
in 1 ｾｧ＠ :
the number of ｾｃｩ＠
ARa= 226
TRa = 1620 yr
ACf = 252
TCf = 2.772 yr
SA = ARaTRa = 226 x 1620 yr
ACf TCf
252 x 2.772 yr
524.12 Cilg = 524 ｾｃｩｉｧ＠
To find the neutron emission rate per ｾｃｩＺ＠
2.31 X 10 6 neutron
sec
x
1 Jlg
524 JlCi
3
4.4x10
n
%
sec
JlCi
4.4 x 10 3 neutrons
.
sec
27 JlCl
5
n
----=--=--=-- x
= 1.2 x 10 ｾ＠
JlCi
MBq
sec .
MBq
5.27 Calculate the speed of a "slow" neutron whose kinetic energy is 0.1 e V. To
what temperature does this energy correspond?
Substituting neutrons kinetic energy and mass (Appendix A) into the equation
for kinetic energy (equation 2.3).
1
Kinetic Energy = - mv 2
2
J
1
0.1 eV x 1.6 x 10-19 - = - x 1.675 x 10-27 v 2
eV 2
3
v =4.4 X 10 mls =4.4 X 105 cmls
Alternatively, knowing that the velocity of a 0.025 e V neutron = 2200 mls .
(p.l52), and that
5.27
132
THE HEALTH PHYSICS SOLUTIONS MANUAL
Calculating the velocity with equation 5.42:
E = 0.025 eV
ｅｾ＠ = 0.1 eV
6 2 2
vJ = (2200 mJS)2 = 4.84 X 10 m /s
J
_ｾ＠ ｾ＠ £,v; _ 0.1 x (4.84
x 10 = 4.4 10' mls = 4.4 10' emls
0.025
v
6
)
2
-
X
X
-
Use Equation 5.42
E=kT
eV
k= 8.625 x 10-5 1( (p.152)
E=O.leV
T= E =
0.1 eV
k 8.6 X 10.5 eV
1159 K = 8860C
K
5.2·8
5.28 When 9Be is irradiated with deuterons, neutrons are produced according to
the reaction !Be (d,n) ｊｾｂ＠
.The cross section for this reaction for 15 MeV
deuterons is 0.12 bams. What is the neutron flux at a distance of 25 cm from a 1
g beryllium target that is irradiated with 100 ｾａ＠
beam of deuterons, 1.13 cm
diameter, assuming an isotropic distribution of neutrons?
Find the number of Be atoms available to react with:
N = 1 g Be x
1 mol
9 gBe
x
6.02 x 1023 atoms Be
1 mol·
22
= 6.69 X 10 atoms 9Be in the target
Next, determine the number of deuterons available to react with the target:
ＱＰｾｸＶ＠
ｾｸＱ＠
1A
sec
J4
----'·-----19-C..:....:.../=------=::-==-=- = 6.25 x 10 deutlsec
1.6 x 10 7 deuteron
SOLUTIONS FOR CHAPTER 5
133
Sinee the area of the deuteron beam is
1[21[
2
2
A= -d = -0.13 em) = 1 em
4
4
The deuteron flux at the target is 6.25 x 1014 deutlsee/cm2
Use equation 5.59, and ｩｧｮｯｲｾｴｨ･＠
production of the neutrons:
decay portion CCJ3 is stable) to find the rate of
Reactions per second = N<j>cr = neutrons produced per second
22
N = 6.69 X 10 atoms 9Be in the target
cr = 0.12 b
14
2
<I> = 6.25 x 10 deutlsec/cm
2
1 X 10-24 cm
14
6.69xl022 atomsx6.25xl0
deut xO.12bx
atom
2
cm sec
1b
= 5.0175 X 1012 neutrons/sec
Sinee the neutrons are isotropieally distributed; the neutron flux at a distance of
25 em will be distributed over a surface area of
S =411:,-2 = 4 x 11: X 25 2= 7854 cm2 and the neutron flux is
5.0175 X 10 12 neutrons
_ _ _ _ _-:-'2s;...;;;e...;:;,.c_ = 6.4 x 108 neutrons/cm2 per sec
7854 em
5.29 What is the thickness of Cd that will absorb 50% of an incident beam of
thennal neutrons? The capture cross section for the element Cd is 2550 barns for
thennal neutrons; the specific gravity of Cd is 8.65 and its atomic weight is 112.4.
cr = 2550 b = 2550 X 10-24 cm2
8.65 g
1 mol
6.02 x 10 23 atoms
22'
N = cm3 x 112.4 g x
1 mol
= 4.63 x 10 atoms Cd
Use equation 5.43;
IIIo = e-crNt
5.29
134
THE HEALTII PHYSICS SOLUTIONS MANUAL
1110 = 0.5 = e -2550 x 10-24 cm 2 x 4.63 x 10 22 atoms Cd x t
t = 0.0059 cm thick
5.30
5.30 A small 124Sb gamma ray source, whose activity is 3.7 x 10 10 Bq (1 Ci), is
completely surrounded by a 25 g sphere of beryllium. Calculate the number of
neutrons per second from the 9Be (y, n) 8Be reaction if the cross section is 1
millibarn, and if the diameter of the spherical cavity enclosing the gamma ray
source is 1 cm. The density of Be = 1.8 g/cm3.
The neutron production rate, R, is
R = N<I>cr,
Where
N = number of target atoms
<I> = photon flux
cr = reaction cross section
However, since the gamma source is inside the Be sphere, the gamma flux will
continuously decrease as the gammas penetrate into the sphere (Neglecting the
very small fraction of the photons that react with a Be nucleus, since cr = 1 mb).
We must therefore calculate the neutron production rate, dr, in each" successive
volume dV (= 4n / dr), and integrate these rates over the thickness of the Be
sphere. The neutron production rate dR in an infinitesimal volume dV, is given by
2
'"
photons
cm
d'R = d'N atoms x 'I'
2 X a-cm·s
atom
23
6.02 X 10 atoms Be 18 g
3
dN =
9 B
x . - - 3 X dV cm
g e
cm
23
dN = 1.2 x 10 atoms 9Be x ( 4 x n x / ) dr
dr
The sphere's outer radius is:
4
3
1t(r
3
_
053)cm3 x ＱＮＸｾ＠
r = 1.51 cm
cm
= 25 g
SOLUTIONS FOR CHAPTER 5
135
S photons/s
<1>=
4 rtr 2
S = gamma ray source activity, photons per second
2
(4 2d) 3 Sphotons/scm
dR = 1.2 x 10 23 atoms
x cr - 3 x nr r cm x
cm
4nr 2 cm 2
atom
Since the threshold for the (y, n) reaction is 1.67 MeV, only photons whose energy
is equal to or greater than 1.67 MeV can initiate the reaction. Examination of the
124Sb decay scheme (ICRP 38) shows that only 57% of the photons have energies
greater than or equal to 1.67 MeV. The effective transformation rate in the 1 Ci
124Sb source is
0.57 x 3.7 x 10 10 photons x 1 Ci = 2.11 x 1010 photons
s·Ci
s
The total neutron production rate is
R=
T(l.2
X
0.5
2
s
10 23 atorr: x 2.11 x 10 10 photons x 1 x 10-27 cm
cm
s
atom
R = 2.53 X 10 6
)dr
Tdr
2.6 x 10 neutrons
0.5
s
=
6
Cadmium is used as a thennal neutron shield in an average flux of 1012
2
neutrons per cm /sec. How long will it take to use up 10% of the Il3Cd atoms at
20 D C?
5.31
reactions
sec
th
- - - = 'I'
neutrons
crn 2
N
x cr
x target atoms
2
target atom
cm . sec
dN
.
- = -<pan
dt
dN = -(<pan )dt
Integrating between the original and final numbers of 113Cd atoms.
N=Ne-tjlcrt
o
5.31
136
THE HEALTII PHYSICS SOLUTIONS MANUAL
2
neutrons/cm /sec
a = 20600 b = 2.1 X 10-20 cm2
ｾ＠
'I' = 10
12
Equation 5.57
NINo = 0.9
N= N o e-4crt
NINo = e-4crt
0.9 = e-IOEI2X(2.1E-20)X t
6
t = 5.02 X 10 sec
6
hr
d
5.02 x 10 sec x 3600 sec x 24 hr = 58.1 days
5.32
5.32 The cross section for the 32 S(n, p) 32p reaction is 300 millibarns for neutrons
greater than 2.5 MeV. How many microcuries of 32p activity can we expect if 100
2
mg of 32S is irradiated in a fast flux of 10 neutron/cm2 sec for 1 week?
2
1x10-24 cm.
cm2
1
b
-_ 3 x 10-25 - a= "300 mb x
3
x ___Ｍ］｡ｴｯＢＧｭｾ＠
1 x 10mb
1b
atom
1g
1 mol 6.02 x 10 23 atoms
1 mol
= 1.88 x 1021 atoms 32S
n = 100 mg x 103mg x 32 g x
ｾ＠
2
2
'I' = 10" neutrons/cm
sec
32p half life is 14.3 days:
0.693
A = 14.3 = 0.0485 d
t =7 days
Using equation 5.59:
-1
SOLUTIONS FOR CHAPTER 5
137
2
25
2
10 neutrons 3 x 10- cm
21
-00485 x 7
AN =
x
x 1.88 xl0 atoms x ( 1 - e·
)
2
atom
cm . sec
'AN = 0.016 dlsec = 0.016 ｾ＠
x
sec
rs: = 4.4 x 10- [lCi
Ci
I[l
3.7 x 104
7
sec
5.33 If the absorption coefficient of the high energy COinponent of cosmic
3
radiation is 2.5 x 10- per meter water, calculate the reduction in intensity of
these cosmic rays at the bottom of the ocean, at the depth of 10,000 m.
5.33
Equation 5.19
Jl = 2.5 X 10-3 m- I (given)
t = 10,000 m
1/10 = e- Ilt = e-(2.5E-3) x (10,000) = 1.39 x 10-11
5.34 If deuterium is irradiated with 2.62 MeV gamma rays from 208T1 (ThC"), the
nucleus disintegrates into its component parts of 1 proton and 1 neutron. If the
neutron and proton each has 0.225 MeV of kinetic energy, and if the proton has a
mass of 1.007593 atomic mass units, calculate the mass of the neutron.
5.34
First, write the reaction:
2.62 MeV + ｾ＠ H ｾ＠
ｾ＠ P + ｾ＠ n + 2 x 0.225 Me V
Based on Appendix A, there are 931 Me V per amu
J
[0.225 MeV
2.62 MeV
I
+2.013928 amu = 1.007593 +on + 2 931 MeVI
[ 931 MeVI
II amu
II amu
J
I
on = 1.0089 amu
5.35 A beam of fast neutrons includes two energy groups. One group, of 1 Me V
neutrons, includes 99% of the total neutron flux. The remaining 1% of the
neutrons have an energy of 10 MeV.
5.35
...Ｍｾ＠
i
138
THE HEALTH PHYSICS SOLUTIONS MANUAL
The removal cross sections are in barns as follows:
1 MeV
4.2b
8b
5.5b
H
°
Ph
10 MeV
0.95b
1.5b
5.lb
(a) What will be the relative proportions of the two groups after passing through
25 cm of water?
In this problem we wish to find the shielding effectiveness (l1I;J of Pb and H 20.
Calculate the number of atoms per cm3 for hydrogen:
3
Water has a density of I g/cm , and the molar density of water is 18 g/mole.
1 g H 2 0 1 mol H 2 0
2 mol H
6.02 x 10 23 atoms H
N =
x
x
x------H
cm 3
18 g H 2 0
1 mol H 2 0
mol H
NH =6.69 X 1022 atoms H / cm3
A similar calculation is done for oxygen:
°
23
N = 1 g H 2 0 X 1 mol H 2 0 x 1 mol
x 6.02 x 10 atoms
3
18 g H 2 0
1 mol H 2 0
mol
cm
°
°
°
NH = 3.34 X 10 atoms 0/ cm
22
3
Information given in the problem:
G H, I MeV
= 4.2 b = 4.2 X 10-24 cm2
.
-24
2
= 0.95 b = 0.95 x 10 cm
-24
2
=8 b = 8 x 10 cm24 2
Go, 10 MeV = 1.5 b = 1.5 X 10- cm
GH,IOMeV
.
GO, I MeV
t = 25 cm
Equation 5.43 can now be solved for 1.0 and also 10 MeV ｮ･ｵｴｲｯｾ＠
attenuation factor.
I
-{JNt
I =e
_
o
Substituting in values:
to obtain the
SOLUTIONS FOR CHAPTER 5
139
Solve for the exponent first:
1 MeV neutrons
24
2
atoms H
3
x 25 em = 7.0245
em
22
aNtH = 4.2 x 10- em x 6.69 x 10
24
2
22
aNto = 8 x 10- em x 3.34 x 10
atoms H
3
x 25 em = 6.68
em
aNtH + aNto = 7.0245 + 6.68 = 13.7045
Replacing the exponent in the equation:
I
I = e-aNt = e-13.7045 = 1.12 x 10-6 for 1 MeV neutrons
o
Performing a similar calculation for 10 MeV neutrons:
24
2
22
aNtH = 0.95 x 10- cm x 6.69 x 10
-24
aNto = 1.5 x 10
2
cm x 3.
34
022
x1
atoms H
. 3
x 25 cm = 1.59
cm
atoms H
x 25 cm = 1.25
cm 3
aNtH + aNto = 1.59 + 1.25 = 2.84
Replacing the exponent in the equation:
I
5 8 X 10-2 clor 10 MeV neutrons
I = e-aNt = e-aNt = e-2.84 =.
-
o
Multiply each by its attenuation factor and initial abundance:
1.12 x 10-6 for 1 MeV neutrons x 0.99 = 1.11 x 10-6
5.8 X 10-2 for 10 MeV neutrons x 0.01 = 5.8 x 10-4
The proportion of the 1 MeV to lOMeV neutrons in the emergent beam is
1.11 X 10-6
- - - = 0.0019 : 1 is the ratio.
5.8 x 10-4
What would be the relative proportion of the two groups after passing.
through a slab of lead of the same density thickness?
Find the density thickness of 25 cm of water:
5.35(b)
140
THE HEALTII PHYSICS SOLUTIONS MANUAL
ｾ＠ cm x 25 cm = 25 g/cm2 is the density thickness of the water shield.
3
Now calculate the linear thickness of a 25 g/cm2lead shield using equation 5.1.
t = density thickness = 25 g/cm
2
11.3 g/cm 3
density
= 2.21 cm
23
22
3
11.3 g 1 mol Pb 6.02 x 10 atoms Pb
3
N = cm x 207.21 g x
1 mol Pb
= 3.28 x 10 atoms/cm Pb
The attenuation factor is given by equation 5.43:
I
-aNI
I =e
_
o
Using the given removal cross sections for Ph for 1 and 10 MeV neutrons, we fmd
the value for the exponent to be:
N
22
24
2
(1 M V) 3.2 x; 10 atoms x 5.5 b x 1 X 10- cm x 2.21 cm = 0.4
b
cr t
e =
cm 3
Ncr t (10 MeV)
22
24
3.28 X 10 atoms. 5 1 b 1 X 10- cm 2
- - - - 3- - x.
X
x 2.21 cm = 0.37
cm
b
The Pb attenuation factors, therefore are:
I
[(1 MeV) = e...(JNt = ･ｾﾷＴ＠
o
I
[(10 MeV) = e...(JNt = ･ｾＮＳＷ＠
o
= 0.67
= 0.69
Now multiply by the proportion ·given in the problem:
1 MeV
0.67 x 0.99 = 0.6633
10 MeV
0.69 x 0.01 =0.0069
Dividing the two values to obtain the ratio: 0.6633/0.0069 = 96: 1
SOLUTIONS FOR CHAPTER 5
141
5.36 Boral is an aluminum boron carbide alloy used as a shield against thermal
neutrons. If the boron content is 35% by weight, and if the density of boral is 2.7
g/cm3, calculate the half-thickness of boralfor thermal neutrons at room temperature. The capture cross sections are; boron = 755 barns, aluminum = 230
millibams, carbon = 3.2 millibams.
The AS ME Boiler and Pressure Vessel Code, Section II, Materials, gives the
carbon content of boral as 2%, and an aluminum content of 63%.
Use equation 5.43:
I
_
I o· = e
-aNt
The half value layer means that the intensity would be reduced by one half, so
I
that the ratio of I
o
= 0.5" giving the formula:
0.5 = e-ONt
Solve for t;
0.693
t=--
aN
a is given in the question, so find the number of atoms for each component of
boral, and then solve for the thickness in the above equation:
23
2.7 g 1 mol B 6.02 x 10 atoms B
22 atoms B
NB = cm 3 x 10.82 g x
1 mol B
x 0.35 = 5.26 x 10
cm 3
23
2.7 g 1mol Al 6.02 x 10 atoms Al
22 atoms AI
3
NA]= cm x 26.98 g x
1 mol Al
x 0.63 = 3.8 x 10
cm 3
2.7 g 1 mol Al 6.02 x 10 23 atoms Al
atoms C
x
x 0.02 = 2.7 x 1021 - -3N = - -3x
c cm
12.011 g
1 mol AI
cm
Now calculate aN, the macroscopic cross section, L, for each material:
aNB =
5.26 x 10 22 atoms B 755 b
1 x 10-24 cm 2
x
x
= 39.7 cm-I
cm 3
atomB
b
3.8 x 1022 atoms Al 230mb 1 x 10-24 cm 2
x
x
aNAl =
cm 3
atomB
1000 mb
= 0.0087 cm- I
5.36
142
THE HEALTII PHYSICS SOLUTIONS MANUAL
2.7 X 10 21 atoms Al 3.2 mb 1 x 10-24 cm 2
aN =
x
x
= 8.64 X 10-6 cm- I
3
c
cm
atom B
1000 mb
Adding together the contributions:
L = aN = 39.7 cm-
t
\
5.37
I
0.693
0.693
--- =0.0175cm
112 aN - 39.7 cm-1
Examination of the cross sections and abundance of each of the elements in boral
shows that AI and C contribute very little to the total attenuation, and could have
been ignored. The complete calculations are included here for reference.
5.37 The scattering cross sections for N and
°
for thermal neutrons are 10 and
4.2 barns, respectively. (a) Calculate the macroscopic cross section for air at
STP. Air consists of 79 volume percent nitrogen and 21 volume percent oxygen.
2
1 X 10-24 cm
.
cm 2
a = 10 b x
atom = lOx 10-24 - - N
1b
atomN
3 g
N 2 2 atoms 6.02 x 10 23 molecules 1 mol
NN=1.293 x 10- - -3 x 0.79- x
x
mol
x 28 g =
air . molecule
cm
atoms
=
4.39 X 1019 - N
cm
N
3
2
cm2
1 X 10-24 cm
24
to =4.2 X 10- - - a o = 4.2 b x
1 bam
atom
°
3 g
02
2 atoms 6.02 x 10 23 molecules 1 mol
N o =1.293 x 10- cm 3 x 0.21 air x molecule x
mol
x 32 g
N
0=
1.02 x 10
19
atoms
cm3
From example 5.11:
J
il
SOLUTIONS FOR CHAPTER 5
143
cm 2
atoms
cm 2
atoms
1x10-23
x 4.39x1019
3 +4.2x1O-24
xl.02x10 19 - atom N
cm
atom 0
cm 3
L.rur = 4.82 X 10-4 cm- l
b) What is the scattering mean free path of thermal neutrons in air?
1
1
MFP = L = 4.82 X 10-4 cm-l
=2075 cm =21 meters
5.38 How many scattering collisions in graphite are required to reduce the energy
of 2.5 MeV neutrons to
(a) 0.1 % of the initial energy?
If the mean logarithmic energy decrement is Sper collision (equation 5.46), the
number of collisions required to reduce a neutron from an initial energy E i ' to a
final energy Ef-is
In E; -In.E f
n=
ｾ＠
= .!.In Ei
c;
Ef
and the mean logarithmic energy decrement is given by equation 5.47
aln(a)
s= 1 +-...:....-.:...
I-a
a = [(M -m)/(M +m)f
E. = 2.5 MeV
E f = 0.001 x 2.5 MeV = 0.0025 MeV
M (Carbon) = 12 amu
m (neutron) = 1.008665 amu
I
Using equation 5.47 information and information in the text following equation
5.47, solving for a:
M - m 2 ( 12 -1.008665)2
(
a = M + m) = 12 + 1.008665 = 0.714
s= 1 + aln(a) = 1 + (0.714) x In(0.714) = 0.158
I-a
1-0.714
5.38
144
THE HEALTI-l PHYSICS SOLUTIONS MANUAL
With these values we find the logarithmic energy decrement, S, to be 0.158 and
the number of collisions to be:
! ｬｾ＠
n = <;
E; = _1_ x In 25 MeV
..
E
0.158
0.0025 MeV = 44 collISIOns
f
(b) 0.25 eV?
0.025 eV = 0.025 x 10-6 MeV
,
! In E; = _1_ x In
n = <;
5.39
Ef
25 MeV
..
0.025 x 10-6 MeV - 117 collISIOns
0.158
5.39 A cobalt foil, 1 cm diameter x 0.1 mm thick, is irradiated in a mean thermal
2
flux of 1 x lOll neutrons/cm per second for a period of 7 days. If the activation
cross section is 36 barns, and if the density of cobalt is 8.9 grams/cm3 , what is
the activity, in Bq and in microcuries, at the end of the irradiation period? Note
that natural cobalt is 100% 59CO .
v = hnr2 =0.01 cm x n x (0.5 cm)2 =7.854 x 10-3 cm3
59
23
8.9 grams 79 10-3 3 1 mole Co 6.02 x 10 atoms
n=
3
x. X
cm x
x
cm
59 g
mole
20
n = 7.17 x 10 atoms
2
ｾ＠ = 1 X lOll neutrons/cm sec
cr = 36 b = 36 X 10-24 cm2/atom
0.693
-1
A= 5.27 yr = 0.1315 yr
1 yr
t = 7 d x 365 d = 0.0192 yr
Substituting these values into the neutron activation equation 5.59:
AN =
1 x lO ll neutrons 36 x 10-24 cm2
x
x 7.17 x 1020 atoms x ( 1 _ e-0.1315 x 0.0192 )
. cm 2sec
atom
SOLUTIONS FOR CHAP1ER 5
145
6
AN = 6.5 X 10 trans/sec = 6.5 x 10 Bq = 6.5 MBq
6
6
Ci
1 x 10 ｾｃｩ＠
10
x
1 Ci
3.7 x 10 Bq
6
6.5 x 10 B q x
= 175 Ci
J.l
5.40 Type 304 stainless steel consists of 71 weight percent Fe, 19% Cr, and 10%
Ni. The isotopic abundance, percentage, and the respective 2200 rnls capture cross
sections (barns) are given below:
Fe
Fe
Fe
Cr
Cr
Cr
Ni
Ni
Ni
A
Abund
crc
A
Abund
crc
A
Abund
crc
54
5.84
2.9
50
4.31
17
58
67.76
4.4
56
91.68
2.7
52
83.76
0.8
60
26.16
2.6
57
2.17
2.5
53
9.55
18
61
1.25
2
58
0.31
1.1
54
0.38
0.38
62
3.66
15
64
1.16
1.5
a) Calculate the macroscopic capture cross section.
To calculate the macroscopic cross section, the macroscopic cross section is
calculated for each element, and then each macroscopic cross section for each
element is then multiplied by the abundance of each element (see example 5.11).
L =P x
1 mole
x (J x (percent abundance of element)
gram atomic wt
%
23
1 X 10-24 em
g
6.02 x 10 atoms b
atom
L = cm3 x
g/
x x
b
x fractional abundance
7mol
Multiply column 2 by (l/column 3) by column 4 by column 5 by Avogadro's
number to obtain column 6.
/'
/.'
I
1_(_'__ vc ... , ,,:0<,' ,',(
t
(
.
\
ｾｊＱ＠
•
,
! \
ｾ［Ｍ
t·
"
\
( ｾＭ ｾ＠ '!:..
I
h'
'
...
N
I
! .
I
I
,-
')
..'
Q,..j ! \ . .
I '"
I
)
I..
5.40
146
THE HEALTH PHYSICS SOLUTIONS MANUAL
Column ]
Fe
Fe
Fe
Fe
Cr
Cr
Cr
Cr
Column 2
Column 3
Column 4
Column 5
Column 6
DensIty SS
P, gjCITf
A
Abood
gjIIDle
7.8196
7.8196
7.8196
7.8196
54
56
57
58
Cap. Cross
(j, barns
2.9
2.7
2.5
Macro
(ea)
'0.014764
7.8196
7.8196
7.8196
7.8196
50
52
53
54
0.0584
0.9168
0.0217
0.0031
0.20808
0.00448
0.000277
0.227601
Fe total
sum x 0.71
0.161597
17
0.8
18
0.38
sum
0.068982
0.06066
0.15268
0.000788
0.283111
Cr total
sum x 0.19
0.053791
4.4
0.24198
2.6
0.053363
1.1
sum
0.0431
0.8376
0.0955
0.0238
Column 7
Ni
Ni
7.8196
7.8]96
58
60
0.6776
0.2616
Ni
7.8196
61
0.0125
2
0.001929
Ni
7.8196
62
0.0366
15
0.041683
Ni
7.8196
64
0.0116
1.5
sum
0.00128
0.340235
sum x 0.1
Ni total
0.034024
Sum Total
0.249411
Yielding a macroscopic cross section of 0.249 cm-
1
(b) If a 1 cm diameter collimated beam of 2200 mls neutrons is incident on a 2
mm thiCk sheet of type 304 stainless steel, how many neutrons per sec will be
2
captured if the flux is 5 X lOll n/cm per second?
112
10= 5 x 10 n/cm sec
t= 0.2 cm
L =0.249 cm- I
Reactions J..
- - - = ",an
s
Find the amount of flux that is transmitted first, using equation 5.43:
I =I e--aNt
o
Remembering that L = Ncr (from example 5.11)
II I o = e-r.t = 5 x lOll e-O.249 x (0.2) = 0 .95
SOLUTIONS FOR CHAPTER 5
147
1- 0.95 = 0.05 are captured.
1 cm diameter beam = 0.7854 cm2 area, SO;
0.05 x 5 x 1011
n
2
n
10
x 0.7854 cm = 1.96 x 10 captured
cm ·sec
sec
2
5.41 AIM solution of boric acid, H3B03, is irradiated for 7 days in a thermal
11
flux 10 n/cm2/sec at a temperature of 40°C. What is the concentration of Li,
moleslliter, after the irradiation?
4010 b
1 x 10-24 cm 2
(JB-IO = 4010 b =
lOB x
b
atom
at 293K
2
-24
4010 x 10
cm .
.
- - IS the cross sectIOn
atom
Calculate the cross section at 40°C using a modification of equation 5.53, because
<\> is assumed to be Maxwellian distribution, and not a monoenergetic beam of
neutrons. The average cross section is given below:
a
TO = 293 K (Chapter 5, reference temperature for thermal neutrons)
T=273+40=313K
cr ］ｾｔＢ＠
= 4010 X 10- em' x ｾＲＹＳ＠
24
1.128
T
1.128
K = 3.44 X 10-'1 em'
313 K
atom
Only 19.6% of all the boron atoms are ｉｾＬ＠
and the other boron atoms cross
sections for absorption are not significant. Calculating the number of ｉｾ＠
atoms
in the solution:
n=
1 mol H 3B0 3
L
x
6.02 x 1023 molecules B
1 mol H 3B0 3
19.6 atomslOB
x----100 atoms B
.23 atoms lOB
n=1.18x10
L
Each In reaction with boron results in the production of one lithium atom:
-
'-'-
-
5.41
148
10
5
THE HEALlli PHYSICS SOLUTIONS MANUAL
B+ 0In---* 73 Li+ 24 He
Equation 5.58 can be modified so that the rate of production can be represented.
Since Li does not decay, it reduces from
number reactions
time
'"
- - - - - - = 'fan
2
em
cr =3.44 x 10- - atom
21
atomsIOB
23
n = 1.18 x 10 --L-II
n
1 x 10 - 2 - em ·sec
ｾ］＠
t
= 7 d =7 d x
24 hrs 3600 sec
1d x
1 hr
number reactions = ｾ｣ｲｮＨ＠
1 x 10
II
6.05 x 105 sec
time) =number of Li atoms
lOB
2
n
-21 em
23 atoms
5
2
x 3.44 x 10 - - x 1.18 x 10
x 6.05 x 10 sec =
em . sec
atom
L
number of Li atoms = 2.46 x 10
19
19
atoms Li
L
2.46 X 10 atoms Li
mol
4 1 10-5 mol Li .
x 6 02 023
=.
x
L IS the final concentraL
. x 1 atoms
tion.
Solutions for C·hapter 6
RADIATION DOSIMETRY
'6.1 A 50 JlClkg ＨｾＲＰｭｒＩ＠
pocket dosimeter with air equivalent wall has a
sensitive volume whose dimensions are 0.5 in. diameter and 2.5 in. long; the
volume is filled with air at atmospheric pressure. The capacitance of the dosimeter
is 10 pPd. If 200 V are required to charge the chamber, what is the voltage across
the chamber when it reads 50 JlClkg ＨｾＲＰ＠
mR)?
ｾｑ＠
=50 JlC
m
kg
Calculate the volume of the chamber first:
2.54 cm = 1 in.
L = 2.5 in = 6.35 cm
r=
0.5 in
cm
x 2.54-.- = 0.635 cm
2
III
3
2
V = 1[; r2 L = 1[; X 0.635 x 6.35 = 8.04 cm
The density of standard air is:
3
3
Pair = 1.293 X 10- g/cm3 = 1.29 x 10-{) kg/cm (Table 5.2) .
Therefore, the amount of charge produced in the chamber, ｾｑＬ＠
ｾｑ＠
= 50 JlC x 1 X 10-6 C x 1.293 X 10- kg air x 8.04 ｣ｭｾｲ＠
6
kg air
1 j.lC
1 ｣ｭｾ＠
is :
= 5.2 x 10-10 C
chamber
C == Capacitance = 10 pP = 10 x 10- p
12
The initial voltage = Vi = 200 V
Putting values into equation 6.8;
149
'6.1
150
THE REALm PHYSICS SOLUTIONS MANUAL
. 5.2 X 10-10 C
(200 V- Vf) = lOx 10-12 F
Vf = 148 V
6.2
6.2 An air ionization chamber whose volume is 1 liter is used as an environmen-
tai monitor at a temperature of 27°C and a pressure of 700 torr. What is the
. exposure rate, in J-lClkg per hour and in mRihr if the saturation current is 10-13
amperes?
T = 27°C + 273°C = 300 K
Current = 1 x 10-13 A = 1 X 10-13 C/sec
Volume = 1 liter
Convert to Clkg. This is simply charge per unit mass, so divide the charge produced in the air by the mass of the air. Since the air is given as a volume, it is
converted into mass, using the density of air, and multiplying by the appropriate
correction factors to account for the different pressure and temperature:
Density of standard air: 1.293 x 10-6 kg/cm3 (Table 6.2)
Calculating the mass of air in 1 liter. first, in kg:
3
1000 cm 3 1.293 x 10-6 kg
1 LXI L
=1.293 x 10- kg air in 1 liter volume
x
cm3
Convert. to the new temperature and pressure:
3
273 K 700 torr
3
1.293 x 10- kg x 300 K x 760 torr = 1.1 x 10- kg at the new temp. and press.
Now divide the charge, in coulombs, by the mass of air in which the charge was
produced:
ＱｸＰＭＳｾ＠
sec
1.1x10- kg air
- - - =3 ' - - X
3600 sec
= 0.33 x 10-6 C/kg per hr = 0.33 J-lC/kg per hr
hr
SOLUTIONS FOR CHAPTER
151
6
C
1000 mR
0.33 X 10-6 kg X 2.58 X 10-4C / kg = 1.3 mRlhr
6.3 A beam of 1 MeV gamma rays and another of 0.1 MeV gamma rays each
produce the same ionization density in air. What is the ratio of 1 : 0.1 MeV
photon flux?
6.3
Equal ionization density implies equal energy absorption per cc of air.
3
Energy absorbed per cm = ｾ＠
MeV
photons
2
X E h
x ｾ＠
cm . sec
p oton
-1
a cm
where fla = energy absorption coefficient
J.
'1'0.1
photons 0 1 MeV
-1 _ J. photons
10 MeV
-1
2
X.
X !J.O.1 cm
- '1'1
2
x.
X !J.I.O cm
cm sec
photon
cm see
photon
From Appendix F;
I
!J.O.1 = 0.0233 cm1
!J.I.O = 0.0280 cmPerforming algebra:
ＭｊＮｬ｟｣ＺＧｰｭＢｨｏＬＲｾ］ｳ＠
ｾ＠
photons
= ＭＮＡｰＺｾ｟ｮﾷｬ＠
'I'
0.1
cm 2sec
0.1 MeV x 0.0233 cm- 1
0.1 MeV x !J.Ol cm- 1
ｰｾｯＺｮ＠
=
1.0
x !J.! 0 cm photon·
1.0
_ = 0.083
photon
x 0.0280 em 1
6.4 Assuming a specific heat of the body of 1 calorie/g, calculate the temperature
rise due to a total body dose of 5 Gy.
5GY X l!kgx lkg xlcaloriex lOe =1.19xlO-30e
Gy 1000 g 4.186 J
calorie
gram
6.4
152
6.5
THE HEALTIl PHYSICS SOLUTIONS MANUAL
6.5 Compute the exposure rate, in mGyIhr at a distance of 50 cm from a small
vial containing 10 mL of an aqueous solution of
(a) 2 GBq (54.1 mCi) 51 Cr,
From basic principles, equation 6:15 gives the exposure rate
J x A -tps
S
MeV
m ·1
-x E -x1
. 6 x 10- - - x 3600 -xJl
f -phot
a
13
X=
t
phot
MeV
MBq
h
4 x 1t X d 2 m 2 x p kg x 1 }kg
m3
Gy
f = 0.09 photlt
E= 0.323 MeV/phot
9
9
A = 2 x 10 Bq = 2 x 10 tps = 2000 MBq
d = 0.5 m
p = 1.293 kg/m3
2
cm
-3
g
cm
Jl = 0.0288 - - x 1.293 x 10 - 3 X 100- = 3.7 X 10-3 m- I (Appendix F)
cm
m
a
g
x 3.7 x 10-3 m- I
0.09 phot x 0.323 MeV x 1.6 x 10-13 _1_ x 2 X 10 9 ! x ＳＶＰｾ＠
X= ___t_ _ _ｾｰ｟ｨｯｴ＠
_ _ _ _ _ _M_e_V_ _ _ _s _ _ _h_r________
4 x 11: x (0.5 m)2 x 1.293 kg xI }{g
m3
Gy
'.
X = 3 X 10-5 Gylhr =0.03 mGylhr
Alternatively, the exposure rate can be calculated with the specific gamma ray
emission constant, Table 6.3, which gives the exposure rate at 1 meter from a unit
quantity of activity. The exposure rate from any quantity of activity at any
distance is then calculated from
.
C/ m 2
/kg'
xAMBq
X = _M_B_q.!...·_h_r_ __
(d, m)2
SOLUTIONS FOR CHAPTER
%
1.11 x 10kg·
10
X(5ICr) =
6
153
2
m
x 2 x 103 MBq
C/
= 8.88x 10-7 ｾ＠
MBq ·hr
(0.5 m)2
hr
To convert to mGy/hr
X(51Cr) = 8.88 x 10-7 %g x 34 Gy x 10' mGy =0.03 mGy
hr
%g
Gy
hr
(b) 2 GBq (54.1 mCi) 24Na, based on the transformation schemes ｳｨｯｾｮ＠
C/ .m 2
For 24 Na, r = 12.8 x 10-9 /kg
MBq·hr
Alternatively, the exposure rate may also be calculated from basic ｰｲｩｮ｣ｬ･ｾＬ＠
J
tps
s
If.E.'tl. x 1.6 x 10- 13 - - x A GBq X 10 9 - - x 3600X = __'_'_·_ _ _ _ _M_eV
_ _ _ｾＭｇｂＮＺｱｨ＠
4nd 2 m 2 xl
){g
Gy
ｾ＠
= 100% = 1
fz = 100% = 1
EI = 2.75 MeV/phot
E2 = 1.37 MeV/phot
9
9
A = 2 x 10 Bq = 2 x 10 tps = 2000 MBq
d= 0.5 m
p = 1.293 kg/m3
below:
154
THE HEALTII PHYSICS SOLUTIONS MANUAL
2
2
cm
33 g
m
Il = 0.0212-- x 10 . - x I 4 2 = 2.12 X 10-3 m2/kg
al
g
kg
10 cm
_
cm
1la2 - 0.027 g
2
X
10 33 ｾ＠
y
t
kg
X
1
m2
3
2
10 4 cm2 = 2.7 x 10- m /kg
m2
kg
MeV
y
M e V m2
Y
kg
y
t
"f.E.1I . = I-x 2.75--x 0.00212-· +1-x1.37--xO.0027L..
I
I t"" az
"f·E.11
L.. I It""al. = 953 X 10
X=
953 X 10-3
-3
MeV· m 2
t
MeV· m
2
t
J
tps
s
x 1.6 X 10-13 - - x 2 GBq X 10 9 - - x 3600MeV
GBq
hr
4
x = 3.5
6.6
3
X 10- Gy/hr
X" x (05 m)' xl
Kg
Gy
= 3.5 mGy/hr
6.6 What is the dose rate to the flesh during exposure to 25.4 IlC/hr (100mRlhr)
of 0.5 MeV gamma radiation?
Exposure in C/kg is a measure of dose to air. Tissue absorbs more energy, by a
. )l energy (tisSUJ()
factor of
II
(air) , than does air from the same exposure.
t"" energy
The tissue dose rate is calculated by converting the exposure to air dose, and then
multiplying the air dose by the tissue factor (J-l for 0.5 MeV from Table 5.3).
2
0.0327 cm /
b = 25.4 Ikg x 34 )lGy x
ｾ＠ g x 1 mGy
2
.
hr
)lC I
0.0297 cm /
1000 )lGy
)lCI
..
D=0.95 mGy/hr
Ikg
/g
SOLUTIONS FOR CHAPTER
6
155
6.7 A collimated beam of 0.3 MeV gamma radiation, whose energy flux is
5 J/m2/s, is shielded by 2 cm Pb.
2
(a) What is the incident particle flux, photons/cm /sec?
2
photon x
MeV
x ; 1 x
1m
= 1.04 x 1010 ｰｨｾｴｯｮｳ＠
13
O.3MeV 1.6x10- J m ·sec 1x104cm2
cm ·sec
(b) What is the exposure rate in mRlhr and C/kg/hr, in the incident and emergent
beams?
Equation 6.9 is used to find the incident exposure rate:
E=0.3 MeV
<\>
= 1.04 x 1010 photons/cm2/sec
2
!-La = 0.0288 cm /g (Table 5.3 for absorption)
<I>
photons E MeV 1.6 x 10-13 _1_ ｾ＠ cm
cm 2 • sec photon
MeV a g
2
34}{g
5{g
2
1.04 X 1010 ｰｨｾｴｯｮｳ＠
x 0.3 MeV x 1.6 x 10-13 _1_ x 0.0288 cm x 1000 g
X=
cm . sec
photon
MeV
g
kg
34}{g
5{g
X = 4.23 X 10-4 ｾ＠
ci
sec
= 4.23 X 10-4 ｾ＠
CI
sec
x 3600 sec = ＱＮＵＲｾ＠
hr
C/
is the incident
hr
exposure rate.
Converting to Rfhr with equation 6.7;
5{g 3881 R 1000 mR
6 mR
1.52-- x eel x
= 5.91 x 10 - - is the incident exposure rate
hr
1 jkg
1R
hr
1
To find the emergent beam exposure rates, use equation 5.19;
t=2cm
!-L = 4.02 cm-I (table 5.2 for attenuation)
6.7
156
THE HEALTH PHYSICS SOLUTIONS MANUAL
.!..- = e-J.l1 = e-4·02cm-1 2cm = 3.22 X 10-4 is the reduction in intensity.
x
10
So the emergent exposure rates will be:
1.52 J{g x 3.22 X 10-4 = 4.9 X 10-4 J{g
hr
hr
5.91 X 10 6 mR x 3.22 X 10-4 = 1.9 X 10 3 mR
hr
hr
(c) What is the tissue dose rate, mGy/hr, in the incident and emergent beams?
To convert to tissue dose rates, use equation 6.11, incident dose first;
ｾｭ＠
ｾ｡＠
= 0.0317 (Table 5.3)
= 0.0288 (Table 5.3)
Gy 1000 mGy
4 mGy
.
=
5.7 x 10 - - is the incident beam exposure rate.
D = 57 -.- x
hr
1 Gy
hr
Calculating the emergent beam exposure rate with equation 6.11:
i> = 34 x 0.0317 x 4.9 x 10-4J{g x 1000 mGy = 18 mGy
0.0288
hr
1 Gy
hr
SOLUTIONS FOR CHAPTER
6
157
6.8 The exposure rate in a beam of 100 ke V gamma rays is 25.8 ｾｃｬｫｧ＠
(100 mR)
per hour. What is
2
(a) photon flux, photons/cm /sec?
Equation 6.9 can be used to solve for the photon flux. However, the linear absorption coefficient fl a, which is divided by the air density, Pa in equation 6.9, is the
mass absorption coefficient for air. Equation 6.9 may therefore be rewritten as
<I>
photons x ＳＶＰｾ＠
cm 2 • sec
x E MeV x 1.6 x 10-13 _J_ x Jla cm x ＱＰｾ＠
hr
phot
MeV
g
2
kg'
34/kg
%g
2
Table 5.3 lists the mass energy absorption as fl (air, 0.1 MeV) = 0.0231 cm /g.
Substituting this, the photon energy, and the exposure rate into the equation yields
25.8 X 10-6
%g
=
hr
<I>
phot ｸＳＶＰｾｏＮｬ＠
cm 2 • sec
2
MeV x1.6xl0- 13 - J - x0.231 cm
phot
MeV
g
hr
ｸｬＰｾ＠
kg
34}{g
%g
<I> = 6.6 xl 0 5
photons
cm 2 • sec
2
2
(b) Power density, W/m and mW/cm ?
6.6 X 10' photons x 0.1 MeV x 1.6x 10- J x ｾ＠
photon
MeV
1J
cm 2 • sec
13
l
4
sec
x 1 x 10 em' = 1.06 x 10-4 W
m2
m2
6.8
158
6.9
THE HEALTH PHYSICS SOLUTIONS MANUAL
6.9 In an expenment, a 250 g rat is injected with 10 JlCi 203 Hg in the form of
Hg(NO)2· The rat was counted daily in a total body counter, and the following
equation was fitted to the whole body counting data
Y =0.55e-o.345t + 0.45e-o.o346t,
where Y is the fraction of the injected dose retained t days after injection. If the
long lived component of the curve represents clearance from the kidneys, while the
short lived component represents clearance from the rest of the body, calculate the
radiation absorbed dose to the whole body and the kidneys. Assume each kidney
weighs 0.7 g, and that the Hg is uniformly distributed in the kidneys and in the
body. Base the calculation on the transformation scheme given in Fig. 6.13.
Dose to the body from distributed 203 Hg :
The dose to the body includes two components: The dose due to the 203Hg in the
body, and the dose to the body fromthe 203Hg concentrated in the kidneys.
First find the dose rate for Body ｾ＠ Body. The decay scheme and the Input Data
in Fig. 6.13 show that in 203Hg, a 0.213 MeV beta and a 0.279 MeV gamma are
emitted from the nucleus in each transformation. The Output Data show that only
81.7% of the 0.279 MeV gammas are seen; The other 18.3% are internally
converted. Thus, in addition to the betas, whose mean energy is 0.058 MeV, we
also have 3 groups of monoenergetic conversion electrons. We also have characteristic x-rays from the internal conversion electrons. Energy is absorbed from
each of these radiations. In the case of the betas, conversion electrons, and the
very low energy (average =0.011 MeV) of the characteristic L X-rays, it is
assumed that all the energy is absorbed, i.e. the absorbed fraction ｾ＠ = 1. Assuming
that the 250 g rat can be approximated by a thick ellipsoid, interpolation and
extrapolation of the data in Table 6.5 to 0.25 kg shows that only about 9% of the
gamma and characteristic x-ray energy is absorbed. The energy of each of these
radiations, their frequency, and the absorbed fraction are listed ill the table below.
The amount of energy absorbed from each of these radiaitons is the product of
these 3, and the amount of energy absorbed per 203Hg transformation is the sum of
the individual contributions.
SOLUTIONS FOR CHAPTER
Radiation
MeV x
0.058
0.2791
Y
Conv. e- K 0.1936
Conv. e- L 0.2648
Conv. e-M 0.2761
.X-rays, K a . 0.0722
X-rays, KI) 0.0833
0.01l2
X-rays, L
ｾ＠
f
1.00
0.817
0.132
0.039
0.0117
0.0984
0.0286
0.0533
x
6
¢
159
Absorbed Energy
1
0.09
1
1
1
0.09
0.09
1
Total
0.05800
0.02052
0.02556
0.01033
0.00323
0.00064
0.00021
0.00060
0.12 MeV/t
The given retention equation shows that 0.55 of the activity is deposited in the
body, and that it is cleared from the body at a rate of 0.345 per day. The activity
initially deposited in the body is:
qbody =
5
0.55 x 10 JlCi = 5.5 JlCi = 2.035 x 10 Bq
The dose rate due to this activity is calculated with equation 6.76.
J
s
tps
MeV
q Bq x 1 - x E - - x 1.6 X 10-13 - - x 8.64 X 10 4 = ______ｂｾｱ＠ ___e___t ________ｾＭｍ･ｖ､｡ｹ＠
b
1J)
Ｆｾ＠
mkgx- Gy
kg
.
D
Ｆｾ＠
｢ｂｏ､ｙｾＦ＠
body.
1
s
tps
. MeV
2.035 X 10 5 Bq x 1 - x 0.12-- x 1.6 x 10-13 - - x 8.64 X 10 4 Bq
t
MeV
day
］Ｍｾ＠
11)
0.25 kg x - , Gy
kg
3
= 1.35 x 10- Gy/day is the initial dose rate to the body from 203Hg in the
160
THE HEALTII PHYSICS SOLUTIONS MANUAL
AE = 0.345 d- 1 is the effective clearance rate for the body (as given)
1.35 x 10-3 Gy
day
3.91 X 10-3 Gy =0.391 rad
1
0.345--
day
Now find the 203Hg dose rate from Body+-Kidneys. Consider only the y and K xrays (since no betas or L X-rays are considered to escape).
To estimate the dose to the rat's body from the 203Hg in the kidneys, let us approxi3
mate the rat as a 250 g sphere whose density is 1 g/cm , and therefore has a
radius of 3.9 cm. Furthermore, let us assume the activity in the kidneys to be a
"point" source in the center of the "sphere." We will consider only the gamma
and K x-rays, since the betas and the L X-rays will be absorbed within the
kidneys. Under these assumed conditions, the fraction of the emitted energy, ｾＬ＠
that is absprbed from the source is given by
ｾ＠ = Absorbed Energy = 1- e- IV
Emitted Energy
Where J.l is the energy absorption coefficient (Table 5.3). For the 0.279 MeV
gammas
ＭｻＩＮＰＳＱＲｾｸＹｧ
ｾＨＰＮＲＷＹ＠
MeV) = 1- e
2
2
g
cm
= 0.12
The initial dose rate to the body due to the 0.279 MeV gamma from the kidneys is
(Equation 6.76). (Note: The given retention equation shows that 0.45 of the
activity is deposited in the kidneys.)
b( )
Y Body+-Iddney
=
tps
MeV
J
s
q Bq x 1 - x E - - x 1.6 X 10-13 - - x 8.64 X 10 4 Bq
e
t
MeV
J )
mkgx 1- Gy
kg
day
SOLUTIONS FOR CHAPTER
6
161
1>(Y) Body+-kidney =
.
tps
y
MeV
J
s
0.45 x 3.7 x 10 5 Bq x 1 - x 0.817-x 0.279-- x 0.12 x 1.6 x 10-13 - - x 86400Bq
t
Y
MeV
d
0.25 kg x 1ｾＩｇｙ＠
kg
DBodYf-kidney = 2.5 x 10-4Gy/day is the initial dose rate to the body from
kidneys:·
For the K x-rays, mean energy = 0.075 MeV and frequency = 0.127, the energy
2
absorption coefficient (Table 5.3) is 0.027 cm /g, and the absorbed fraction is:
cm 2
g
g
cm
-0.027-x3.9-
4>(0.075 MeV) = 1- e
2
= 0.1
The initial X-ray dose rate to the body from the kidneys is estimated as
D(X- ray) Body+-kidney =
tps
y
MeV
J
s
0.45 x 3.7 x 105 Bq x 1 - x 0.127 - x 0.075-- x 0.1 x 1.6 x 10-13 - - x 86400Bq
t
y
MeV
d
ｏＮＲＵｫｧｘｬＺｾ＠
b(x-raY)BOdyf--kidney =
8.77 x 10-6 Gy/d
The 203Hg is cleared from the kidney, according to the retention equation, at a rate
of 0.0345 per day. The total dose to the body from the 203 Hg in the kidney is
D
.
Body+-Kldneys
D
(2.5 x 10-4 + 8.8 x 10-6 ) Gy
= _A0E =
0.0346d -I
d
= 7.5 x 10-3 Gy
The total dose to the body from both the bodily deposited nuclide and kidney
deposited nuclide is:
162
THE HEALTH PHYSICS SOLUTIONS MANUAL
3
3
3.91 X 10- Gy + 7.50 X 10- Gy = 1.14 X 10-2 Gy = 1.14 rads
Dose to the kidneys
A value of 1 for <I> is used since it is characteristic radiation (low energy). Assume
all the y's escape from the small volume of the kidneys and consider only the betas
and conversion electrons. From Figure 6.13;
MeV x
f x
0.058
1.00
0.1936
0.132
0.2648
0.039
0.2761
0.0117
f3
K ee-
Lee-
Mee--
Absorbed Ener
0.058
0.0256
0.0103
0.0032
0.097 MeV/t
1
1
1
1
Total
Ee = 0.097 MeV/t
The retention equation shows that 0.45 of the activity is deposited in the kidneys
and that it is cleared from the kidney at a rate of 0.0345 per day. The activity
initially deposited in the kidneys is:
Qkjdneys
= 0.45 x 10 JlCi = 4.5 JlCi:= 1.665 X 105 Bq
mkidneys =
3
1.4 g = 1.4 x 10- kg
The dose rate due to an internally deposited radionuclide is 'given by equation
6.76:
Q Bq x I_tP_s x E
iJkidneYf-kidney =
Bq
e
_M_eV_ x 1.6 x 10-13 _J_ x 8.64 x 10 4 ｾ＠
t
MeV
d
1
mkgx- Gy
kg
J)
SOLUTIONS FOR CHAPTER
6
163
tps
MeV
1
S
5
1.665 X 10 Bq x 1 - x 0.097-- x 1.6 x 10-13 - - x 8.64 X 10 4 _
Bq
t
MeV
d
DkidneY+-kidney = ---------=------.--1-=1-;-:----------1.4 x 10-3 kg x Gy
kg
Dkidney(;-kidney
=0.159 Gy/d is the initial dose rate to the kidneys from nuclide in
the kidneys.
Using equation 6.58 to find the dose commitment for the kidneys;
AE =0.0346 d- is the term for the kidneys (as given)
1
Dkidney+-kidney
= ｾｯ＠
fIv E
0.159 Gy
=
d = 4.61 Gy = 461 rad is the committed dose to the
1
0.0346d
kidneys.
6.10 Iodine is deposited in the thyroid at a rate of 0.139 per hour. If the radioac-
tive half life of 1231 is 13 hours, what is the deposition half life?
Equation 4.21 is used to solve for the radioactive decay constant:
TR = 13 hours
T .= 0.693 =
B
AB
0.693 = 5 hr
0.139 hr- 1
Putting values into equation 6.54:
T = TR X TB = 13 hr x 5 hr =3.6 hr
TR+TB 13hr+5hr
E
6.10
164
6.11
THE HEALTH PHYSICS SOLUTIONS MANUAL
A patient with cancer of the thyroid has been found to have a thyroid iodine
uptake of 50%. How much 131 1 must be injected to deliver a dose to the thyroid,
which weighs 30 g, of 15 grays (1500 rad) in 3 days?
Since no iodine retention time is given for this cancer patient, we will assume the
1CRP 28 value and calculate the effective half life using 6.54.
6.11
TR = 8.05 d
TB = 138 d (1CRP 28)
T = TR X TB = 8.05 d x 138 d = 7.6 d, effective half life 1311 in the body.
E
'TR+TB 8.05d+138d
Converting to effective elimination constant, using equation 6.52:
A = 0.693 = 0.693 = 0.091 d- 1
E
T£
7.6 d
Find the initial dose rate (equation 6.57):
D = 15 Gy
AE = 0.091 d- 1
t = 3 days
D'A.
15 Gy x 0.091d-1
D=
E
= (1- e -(O.091)X3) = 5.71 Gy/d is the initial dose rate to the
o (1- e -A£t )
.
thyroid
Example problem 6.13 in the text demonstrates how to calculate the average
energy 1311 imparts per transformation, 0.230 MeV/t.
Using equation 6.47, and knowing that the initial dose rate is 5.71 Gy/d:
m =30 g = 0.03 kg thyroid mass (appendix 3)
D"= q Bq x ltps/ Bq x E MeV / t x 1.6 X 10-13 J / MeV x 8.64 x 104 sec / d
J
kg
mkgx 1-/Gy
SOLUTIONS FOR CHAPTER
6
165
1tps
MeV
J
q Bq x x 0.230 - - x 1.6 X 10-13 - - x 8.64 X 104sec / day
5.71 Gy =
Bq
t
MeV
day
0.03 kg x 1ｾ＠
/ Gy
kg
q Bq = 54 x 106 Bq = 54 MBq
Since the thyroid uptake is 50% of the activity twice the activity must be administered to deliver the proper radiation dose:
54 x 2 = 108 MBq must be injected to deliver 15 Gy in 3 days.
6.12 The mean concentration of potassium in seawater is 380 mg/kg. What is the
dose rate, in milligrays per year and in millirads per year, in the ocean depths due
to the dissolved 4OK ?
4°K comprises approximately 0.0119% of all potassium (CRC). Please note that
the RHH (1970) incorrectly lists 0.118% of all potassium as 4°K.
The specific activity of4°K will also be needed:
Use equation 4.31
A Ra = 226
TRa = 1620 years
A =40
K
9
TK = 1.26 x 10 years
3.7 X 10 to trans
sec
380 mg K
1gK
0.0119 g 4°K 7 x 10-6Ci
-----:'------- X
X
X
X -------....:=1 kg seawater 1000 mg K
100 g K
1 g4°K
1 Ci
= 11.7 (t/s)/kg
6.12
166
THE HEALTII PHYSICS SOLUTIONS MANUAL
l1.7-tsec
3600 sec x24
hrx 365 day =.
3 69 x 108 (t/)/k
--....::..::..:::......-x
-year g seawater
kg seawater
hr
day
year
4°K emits the following:
1.1 % of the time.
1.46 MeVy
(.l- is emitted in 89% of the decays.
0.509 MeVaverage J-'
>
The average energy therefore is
E = (0.11 x 1.46) + (0.89 x 0509) = 0.614 MeV
t
In an infmite medium, the energy emitted =energy absorbed
3.69 X 108 t/ yr 0.614 MeV 1.6 x 10-13 J
----------x
x-------kg
y
MeV
l:g%y/
6.13
5
3.6 X 10- Gy/yr = 3.6 mrads/yr
6.13 Calculate the annual radiation dose to a reference person from the 4°K and
from the 14C deposited in his body. The specific activity of carbon is 0.255 Bq (6.9
pCi) per gram. Assume in both instances, that the radioisotopes are uniformly
distributed throughout the body.
Performing the calculations for potassium first:
In problem 6.12, we calculate the specific activity of4°K to be 7.04 x 10-6 Cilg
CWK)
From appendix 3, table 2, the amount of potassium found in a reference person is
.
40
140 grams. The natural abundance of K is 0.0119% (Table 4.5). Please note that
the RHH (1970) incorrectly lists 0.118% of all potassium as 4°K. Calculating the
number of transformations of Ｔｾ＠
per second in the body:
SOLUTIONS FOR CHAPTER
6
10 t
3
4
140 g K 0.0119 g °K 7.04 x 10---{jCi. .7 x 10 ｾ＠
---=-- x
X
40
X
body
100 g K
g K
Ci .
167
= 4.3 X 103 _t_ of 40
sec
K
in body
40
K emits the following
A
= 0.509 MeV in 89% of all transformations
jJ average
Y= 1.46 MeV (11 % of all transformations)
For uniformly distributed 1.5 MeV gamma, Table 6.8 shows the absorbed fraction
to be 0.302.Therefore the total amount of energy absorbed per transformation is:
Ee = Ee(Y) + ｅ･ＨｾＩ＠
= (1.46 MeV x 0.302 x 0.11) + (0.509 MeV x 0.89)
Ee = 0.502 MeV/trans
Calculating the dose due to potassium using equation 6.76;
m = 70 kg (appendix 3, table 1)
3
t
40
of K
q = 4.3 x 10 .
sec
Ee= 0.502 Me V/trans
q Bq x 1 tps x Ee MeV x 1.6 x 10-13 _J_ x 8.64 x 10 4 sec x 365 days
D=
Bq
trans
MeV
day
yr
1I)
mkgx- Gy
kg
4.3 X 103 trans x 0.502 MeV x 1.6 x 10-13 _J_ x 8.64 X 10 4 sec x 365 day
b= ________s_ec________tr_a_n_s__________M__eV___________､｟｡ｾｹ＠ ____ｾｹ･｟｡ｲ＠
70 kg x _1
J)GY
kg
mGy
D. = 1.55 x 10-4 -Gy = 0.155 - f rom the 40K
yr
yr
168
THE HEALTII PHYSICS SOLUTIONS MANUAL
Performing the calculations for 14C which is a pure beta emitter where the average
beta energy is 0.05 MeV.
From appendix C, the amount of carbon found in a reference person is 16,000
grams. Calculating the number of transformations of 14C per second in the body,
using the specific activity given in the problem:
t
1t
of 14C in body
q = 16000 g C x 0.255 Bq x sec = 4080 body
gC
1Bq
sec
Calculating the dose due to 14C using equation 6.76;
m = 70"kg (appendix 3, table 1)
q = 4080 Bq of 14C
E e= 0.05 MeV/trans
ｄ］Ｍｾ＠
.
tps
MeV
13
J
4 sec
d
qBqx1-xE --x1.6x10- --x8.64x10 -x365Bq
e trans
"MeV
day
yr
mkgx_1_J /Gy
kg/'
M_eV_ x 1.6 X 10-13 _J_ x 8.64 X 104 _se_c x 365_d_ay_
4080-t- x 0.05_
iJ _____s_e_c______tr_an__s _________M-=-e_V-:--________d_ay::.-.-__ｾｹ｟･｡ｲ＠
J /Gy
70 kg x 1__
kg/,
"
Gy
b = 1.53 X 10-5 -
mGy
= 0.0153 - - from 14C
yr
yr
Summing the doses from both yields:
mGy
mGy
mGy
0.155-- from the 4CX + 0.015-- from the 14C =0.17 - " yr
yr
yr
SOLUTIONS FOR CHAPTER
6
169
6.14 A thin walled carbon wall ionization chamber, whose volume is 2 cm3 , is
filled with standard air at O°C and 760 torr and is placed inside a tank of water to
make a depth dose measurement. A 24 MeV betatron beam produces a current of
0.02 J.1A in the chamber. What was the absorbed dose rate?
According to the Bragg-Gray relationship (equation 6.14)
energy absorbed
unit mass
---==------- = P w J
m
Mass stopping power ratios for electrons generated by x-rays may be obtained
from ICRU 14, Table A.3.
Pm = mass stopping power ratio of water for 24 MV X-rays = 1.08
To calculate the number of ion pairs per gram of air:
density of air = 1.293 x 10-3 g/cm3 (Table 5.2)
1 ion pair = 1.6 x 10-19 coulombs
0.02 /lC x
1C
x 1 ion pair
6
sec 1 x 10 /lC 1.6 x 10-19 C
ion pairs
= 4.8 x 10 13 ion pairs
J = --=-- = -------'-----::----3
gram air . sec
gram air
2
3
1.293 x 10- gair
cm air x
3
cm air
For betas (and electrons), w = 34 electron volts are expended per ion pair in air
(Table 5.1).
1.6 x 10-12 erg = 1 eV
1 rad = 100 ergs/gram
Substituting these values into the Bragg-Gray equation and calculating the dose
rate, we have
ergx 1 rad
eV J -ipx 16
. x 10-12 -D· =p xw-x
m
ip
g. s
eV 100 ergs
g
D = 1.08 x 34 eV x 4.8 x 10 13 ip x 1.6 X 10-12 erg x 1 rad
ip
g·s
D = 28.4 rads/s = 284 mGy/s
eV
100 ergs
g
6.14
170
6.15
THE HEALTII PHYSICS SOLUTIONS MANuAL
6.15 An aluminum ionization chamber containing 10 cm3 air at 20°C and 760 torr
operates under Bragg Gray conditions. After a I-hour exposure to 60CO gamma
9
rays, 3.6 x 10- coulomb of charge is collected. If the relative mass stopping
power of Al for the electrons generated by the 60CO gammas is 0.875, what was
the dose to the aluminum?
The Bragg Gray rule applies in this problem:
energy absorbed
=p w J
unit mass
m
Pm = 0.875
density of air = 1.293 x 10-3 g/cm3 (Table 5.2)
1 ion pair = 1.6 x 10- 19 C
W ｾ＠ 34 eV per ion pair in air (Table 5.1)
Converting the volume of air in the chamber to STP, assuming it is an ideal gas:
Standard Temp = 273 K
3 273 K
3
10 cm x 293 K =9.32 cm is the volume in the chamber at STP
3.6 X 10-9 ｾ＠
J = ion pairs
gram air
w=
x 1 ion pair
hr 1.6x10-19 C
1.6 x 10-12 erg
34 eV
ion pair
x
eV
1.87 x 1012 ion pairs
. gram air
erg
= 5.44 X 10-11 - - ion pair
Equation 6.14 gives the Bragg-Gray relationship:
eV
lp
1rad
ip
g .s
100 ergs
g
D=p x w - x J - x - - m
D = 0.875 x 5.44 x 10-11 erg x 1.87 x 10 12 ip x 1 rad
ip
g
100 ergs
g
D = 0.889 rads = 8.9 mGy
.
SOLUTIONS FOR CHAPTER
6
171
3
6.16 An ion chamber made of 50 grams copper has a 10 cm cavity filled with air
at STP. The temperature of the copper rose 0.002 °C after exposure to 60CO
gamma rays. If the mass stopping power of Cu is 0.753 relative to air, and if the
specific heat of Cu is 0.092 calories per gram per degree C, calculate
(a) The absorbed dose to the copper,
4.184 J = 1 calorie
0.092 calories 4.184 J
x l ' x 50 g x 0.002°C = 0.0385 J deposited in the chamber
gram· 0 C
ca one
.
.
.
0.0385 J 1000 g 1Gy
---x
x - - =0.77 Gy deposIted III chamber
50g
kg
Ｑｾ＠
kg
(b) The amount of charge (in coulombs) formed by ionization in the cavity during
exposure.
Pm = 0.753
34 eV
1.6 x 10-19 J
J
---x
= 5.44 X 10-18 .
.
IOn paIr
ion pair
eV
Equation 6.14 gives the Bragg Gray relationship:
D= M
absorbed
11m
Pm wi
·
544 x 10-18 joules x ] ipx1-Gy
D =p x
m'
ip
kg Jig
Dose determined in part (a) was 0.77 Gy, and the quantity of coulombs is needed,
which is a part of the "1" term, so solve for "1"
0.77 Gy = Pm w]
6.16
172
THE HEALTII PHYSICS SOLUTIONS MANUAL
.
0.77 Gy = (0.753) x 5.44 x 10-
18
I
JOU e
•
ｾ＠
xJ
5
1.293 x 10- kg
•
IOn paIr
ｾｹ＠
x ---,-
joule/
/kg
Total charge = 2.43 x 10 ip x 1.6 X 10-19 Clip = 3.9 x 10-7 C
12
6.17
6.17 An aqueous suspension of virus is irradiated by X-rays whose half value
layer is 2 rom Cu. If the exposure was 355 C/kg (1.3 x 106 R), and if the depth of
the suspension is 5 rom, what was the absorbed dose, and what was the mean
ionization density?
I
I
I
x
. Do
.!
dx
I
Dxl
I
d----..
Use equation 5.18B to determine the linear attenuation of the copper;
In /1/0 =- J..lt
Rearranging;
J..l = In(05) = 0.347 rom- I = 3.47 cm- I is the linear attenuation of Cu .
.-2 rom
Look up in table 5.2 the energy that this attenuation coefficient corresponds to,
and it is found that the approximate average energy of the x-rays is
0.112 MeV.
SOLUTIONS FOR CHAPTER
6·
To convert roentgens to rads using equation 6.12,
ｉｬｾｷ＠
=0.0258 cm /g (Table 5.3 for 0.112 MeV)
2
ｉｬｾ｡＠
= 0.0237 cm2/g (Table 5.3 for 0.112 MeV)
Ilj{
_ 87.7 x
Pw roentgen _ 87.7 x 0.0258 R
- 100 0.0237
rads - 100 Ilal
.
/Pa
0.95 rad = 1 R
6
6
Do = 1.3 x lOR = 1.3 x 10 R x
0.95 rad
6
1 R = 1.24 x 10 rads
The dose at depth x is
Let
I = integral dose, gram-rads
.
2
A = irradiated area. Since dose is an intensive property, we may use A = 1 cm
p
Ｍｾ＠
= density of suspension, 1 ｾ＠ cm
Jl (0.112 MeV, H 2 0) =0.0258 cm- (Table 5.3)
1
The absorbed energy, dI gram-rads, in the thickness dx is:
Integrating to find the integral dose through the 5 mm (0.5 cm) of suspension;
0.5
f dI = P x (A) x Do f ･Ｍｾ､ｸＮ＠
o
1=
Px (A) x Do x (1- ･ＭｾＩ＠
Il
173
174
THE HEALTII PHYSICS SOLUTIONS MANUAL
Ｑｾ＠
1 = cm
2
xl cm x 1.24 x 106 rads
0.0258 cm-
1
x (1- e -O.0258X(O.5») = 6.2 x 105 g·rads
Dividing by the mass of the unit area irradiated:
The number of ion pairs formed per eV in water can be found in "Medical Physics
Handbook 15, Fundamentals of Radiation Dosimetry", Second edition, table 2.5,
published by Adam Hilger, 1985 as 29.6 eV per ion pair;
1.24 x 104Gy x
6.18
J/k
1 jk
.
.
k
eV
IOn
rur
18 .
•
gx
19
X
P
X
g = 2.6 x 10 IOn parrs/g
1 Gy 1.6 x 10- J 29.6 eV 1000 g
6.18 A child drinks 1 liter of milk per day containing 131 r at a mean concentration
of 33.3 Bq (900 pCi) per liter over a period of 30 days. Assuming that the child
has no other intake of ＱＳｲｾ＠
calculate the dose to the thyroid at the end of the 30
days ingestion period, and the dose commitment.
Activity
in
thyroid
o
stop intake
Increasing ｔｩｭ･ｾ＠
SOLUTIONS FOR CHAPTER
6
175
First calculate the effective halflife of the 1-131 in the body, use equation 6.54:
TR = 8.05 d
TB = 138 d (ICRP 28)
T E-
TR X TB
TR + TB
8.05 d x-138 d
= 8.05 d + 138 d = 7.6 d effective half life of 1311 in body.
Converting to effective elimination constant, using equation 6.52:
A = 0.693 = 0.693 = 0.091 d- I
E
TE
7.6 d
The average energy of each 131 1 beta particle is found (Figure 6.11), and the yield
from each decay is also tabulated:
Energy, Me VIt
0.0701
Yield,!
0.016
0.0955
0.1428
0.1917
0.2856
0.069
0.005
0.904
0.006
The mean ｾ＠ energy/transfonnation is:
ｅ･ＨｾＩ＠
=
1:.E X f!3i = (0.0701 x 0.016) + (0.0955 x 0.069) + (0.1428 x 0.005) +
(0.1917 x 0.904) + (0.2856 x 0.006)
ｅ･ＨｾＩ＠
= 0.184 MeV/t
Calculating the contribution due to the y, the specific absorbed fraction is found in
appendix 4 for an adult. Assume all the 1311 is deposited in the thyroid. So for this
case, the contribution from the y is not significant-and can be ignored, especially
since the child's thyroid is small (-2-5g, 1CRP 53).
I
L
176
THE HEALTH PHYSICS SOLUTIONS MANUAL
MeV
f
Spec. Abs
MeV/t
0.723
0.016
0.00166
1. 92E-05
0.637
0.069
0.00166
7.3E-05
0.503
0.003
0.00166
2.5E-06
0.326
0.002
0.00155
1.01E-06
0.177
0.002
0.00155
5A9E-07
0.365
0.853
0.00155
0.000483
0.284
0.051
0.00155
2.25E-05
0.08
0.051
0.0429
0.000175
0.164
0.006
0.00155
1.53E-06
Sum
0.000778
The intake is q = 33.3 Bq/day, however, only one-third of the iodine is directly
deposited in the thyroid (ICRP 30).
K = 33.3 Bq x 1 deposited = 11.1 Bq
d
3 ingested
d
Some of the iodine is eliminated daily, so find the concentration in the thyroid at
anytime:
dq d
..
d·
.
- lsappearance
.dt
= epOSltIon
dq
-=K-A
dt
. eff q
Separating the variables, we have
After integration, and solving for q as a function of t;
SOLUTIONS FOR CHAPTER
As t ｾ＠
6
177
00, q approaches
Bq
K
day
qoo = - =
= 122 Bq
A 0.091 day
11.1
m = 20 g (for adult, from appendix C), for a child, assume 10% of adult mass (10
CPR 20),2 g.
Putting values into equation 6.76:
D.
q Bq x ltps/
x E MeV / t x 1.6 x 10-13 J / MeV x 8.64 x 10 4 sec __
/d
__Bq __________________________________
］ｾｯＮ＠
ｾ＠
J
00
mkg x 1--/Gy
kg
122 Bq x ltps/ Bq x 0.184 MeV It x 1.6 X 10-13 J / MeV x8.64 x 10 4 sec / day
2gx
J
x 1-- / Gy
kg
1000 g
kg
D = 1.55 X 10-4 Gy = 0.155 mGy/d to child's thyroid is the dose rate after an
day
co
infinite ingestion period at 33 Bq/d.
The total dose at the end of t days of continuous intake is given by
t
D = f D(t)dt
o
Since D(t) is proportional to q(t), the expression for V(t) is analogous to that for
q(t).
Therefore the dose rate after t days of intake is:
178
THE HEALTH PHYSICS SOLUTIONS MANUAL
The above equation is integrated with respect to time to find the dose for the 30
day period of intake. The dose rate at an infinite time (the equilibrium dose) is
known from the earlier calculation to be 0.155 mGy/day.
I
D = J1>(t)dt
o
Dose =
f1> . dt = 1>00 f(1 - e }it
-AI
o
0
Integrating yields:
And the accumulated dose at 30 days is:
iJ =0.155 mGy/day
A : 0.091 d- I
t = 30 d
D = 0.155 mGy x [30 d +
1
x (e -O.091x(30) -1)] - 3 G
d
0.091d-1
m y
3 mGy is the accumulated dose at the end of the 30 day period.
The dose commitment is the sum of the dose accumulated during intake, and then
during elimination (washout).
SOLUTIONS FOR CHAPTER
6
179
Dose
rate
D
00
o
30 days
Time after start ingestion
Find the dose rate at t = 30 days
b =b
30
(1- e-A1 ) = 0.155 mGy x (l_e--{)·091X(30») = 0.145 mGy
d
00
.
0.145
D = D30 =
A
iJ
00
.
day
mGy
d
1.59 mGy is the dose after ingestion stops.
0.091.!
d
--1------.___---------.-.::::....-:,=--------------..--.--.---.------____
/
Dose
rate
o
30 days
Time after start ingestion
The total dose, from the time intake started to the end of the first 30 days, plus the
dose after the intake stopped will be;
3 mGy + 1.6 mGy = 4.6 mGy total dose to the childs thyroid.
180
6.19
THE HEALTII PHYSICS SOLUTIONS MANUAL
6.19 A patient who weighs 50 kg is given an organic compound tagged with 4
MBq (108 !lCi) 14C. On the basis of bioassay measurements, the following whole
body retention data were inferred:
Day
MBq
o
1
1
2.94
2
2.32
3
1.9
4
1.6
5
1.4
6
1.2
8
0.9
10
0.8
12
0.6
14
0.5
(a) plot the retention data. and write the equation for the retention curve as a
function of time.
10
ｾ＠
.,.
1
0.1 - t - - - - - - - . - - - - - - - , , - - - - - - - - , - - - - - - - . - - - - , . - - - : - r - - - - - , - - - - - ,
o
4
2
6
10
12
16
8
14
Days
Days are labeled on the x-axis. MBq on the y-axis. The curve appears to have
two components to it, a long lived and short lived. To resolve these two components, and determine the equation, draw a line along the long lived portion to the y
axis;
SOLUTIONS FOR CHAPTER
6
181
10
0.1 - + - - - - - - - , - - - . . - - - - - - - r - - - - , - - - - - - - , - - - - - - . - - - - - , - - - - - - - ,
2
4
o
6
8
10
12
14
16
Days
Since the line intercepts the y axis at 2.2 ｍｾｱＬ＠
that is the amount deposited in the
long lived compartment. The time for the activity to decrease to one half of the
initial activity, to 1.1 MBq, is found from the graph to be 6.5 days. The slope of
the line, therefore, is
and the equation for the long lived component is
R (t, LL) = 2.2e-O.ilt
The short lived component of the curve is found by subtracting the long lived
activity from the total activity. Thus, for t = 0, we have
182
THE HEALTH PHYSICS SOLUTIONS MANUAL
4 MBq - 2.2 MBq = 1.8 MBq
The long lived activity for days 1 - 5 are calculated from the equation for the long
lived component, and then are subtracted from the total activity. These differences,
which are tabulated and plotted below, fall on a straight line. The half-time for this
short lived component is 1.1 days, and the slope is 0.69311.1 days = 0.63 per day.
The equation for the short lived component is therefore
R (t, SL) = 1.8e-o.63t • The equation for the retention curve, which is the sum of the
two components is
R(t) = 1.8e-O.63 t + 2. 2e-o.II I
Day
Total
Long lived
Short lived
0
4
2.2
1.8
1
2.94
1.97
0.97
2
2.32
1.77
0.55
3
1.9
1.58
0.32
4
1.6
lA2
0.18
5
1.4
1.27
0.13
10
tT
m
:0
1
0.1.j------,------r---.,...------,------r---.,...-----,------,
16
14
4
12
6
8
10
o
2
Days
SOLUTIONS FOR CHAPTER
6
183
(b) Assuming the 14C to be uniformly distributed throughout the body, calculate
the absorbed dose to the patient at day 7 and day 14 after. administration of the'
drug.
E avg = 0.04947 MeV
m = 50 kg
Using equation 6.47 to find the initial dose rate from q Bq:
1 trans
.
qBqx
sec x E MeV
- x 1. 6 x 10- 13 - 1- x 864
. x 104 _sec
Bq
trans
MeV
d
D=
11)
mkgx- Gy
kg
1 trans
q Bq x
ｄ］Ｍｾ
.
sec x 0.04947 MeV x 1.6 x 10-13 _1_ x 8.64 x 104 sec
Bq
trans
MeV
d
__
50 kg x _11 )GY
kg
D= 1.38
X
10-11 x ｱｾ＠
d/Bq
The sum of the doses from each component gives the total dose. The dose for each
component is given by equation 6.57, where Do is the initial dose rate for that
component and Ais·the clearance rate for that component.
184
THE HEALTIl PHYSICS SOLUTIONS MANUAL
Note equation is split into two lines.
(1.38 X 10-11 X 1.8 X 106) Gy
_ _--'-d_ x {1- e -0.63 x (7)} +
D= _ _ _ _ _ _ _
1
0.63 d(1.38 X 10-11 x 2.2 X 10 6 ) ｾｹ＠
.
_ _ _ _ _ _ _ _ _. ;;.;.d_ {1- e -0.11 x (7)}
.
0.11 d- 1
D =0.19 mGy is the absorbed dose to the patient after 7 days.
Calculating the dose after 14 days in a similar fashion (Note equation is split into
two lines)
G .
(1.38 X 10- x 1.8 x 106)--.r
11
--=d_ x {1- e-O•63X (14)} +
D= ________
1
0.63 d.
(1.38 X10-11 x 2.2 x 106) Gy
.
d {1_e-O·11x(l4)}
0.11 d-1
D =0.26 mGy is the absorbed dose to the patient after 14 days.
(c) What is the dose commitment from this procedure?
SOUJTIONS FOR CHAPTER
6
185
D = 0.32 mGy is the dose commitment.
6.20 A 2 Me V electron beam is used to irradiate a sample of plastic whose
2
thickness is 0.5 g/cm . If a 250 /-lamp beam passes through a port I cm in diameter to strike the plastic, calculate the absorbed dose rate.
First, calculate the range of a 2 MeV electron using equation 5.2;
E=2MeV
R =412EL265-0.0954xlnE =412 X (2)L265-0.0954Xln(2) =945.8 ｭｾ＠
is the range of the 2
. cm
Me V electron. An approximation is then madethat there is a linear relationship
between energy and range.
range of material
Energy deposited = range 0 f'1m"t'1aleIectron x Initial Energy
Energy deposited =
mg x 2 MeV = 1.06 MeV per electron is the energy
945.8cm 2
deposited in the material-using this linear approximation.
Converting the Me V associated with each electron to joules:
250 /-lA beam (given)
·6.20
186
THE HEALTH PHYSICS SOLUTIONS MANUAL
1 A = 1 C/sec
19
1.6 X 10- C/electron (Charge on electron)
13
1.06 MeV 1.6 x 10- J
1 electron
250
electron x
MeV
x 1.6 x 10-19 C x.
ｾａ＠
1 C / sec
x 1 x 106 ｾａ＠
= 265 J/sec is
the energy deposited.
Since dose is simply energy deposited per unit mass of material, calculate the mass
of the plastic disc:
d= 1 cm
r = 0.5 cm
2
Area of plastic = n r2 = n x (0.5/ =0.785 cm
Calculating the mass:
2
0.5 g
3
0.785 cm x - - 2 = 0.392 g = 0.392 x 10- kg
cm
.
J
265sec
ＭｾＺ］ＢＧＳｘｊ
0.392 x 10- kg
1 Gy
= 6.75 x 105 Gy/sec
1kg
Alternatively, stopping power may also be used to calculate the answer:
MeV
Stopping power in Lucite = 1.81 g/cm2
MeV
Energy lost in plastic = 1.81 g/cm2 x. 0.5
25 X 10-4 A x 1
o/s x
A
C!2 = 0.905 MeV/e-
1 ex 0.905 MeV x 1.6 x 10-1 _J_
1.6 X 10-19 C
eMeV
0.392 x 10 _3 kg x 1J-) Gy
kg
= 5.75 X 105 Gy/sec
SOLUTIONS FOR CHAPTER
6
187
The different answers are probably due to the fact that range energy equations
used in the previous solution are curve fitted experimental equations.
6.21 Calculate the average power density, in watts per kg, of an aqueous solution
of 60CO, at a concentration of 10 MBq per liter, in
(a) An infinitely large medium,
Cobalt 60 emits a 1.17 MeV and a 1.33 MeV gamma ray with each decay, and a
0.314 MeV (max) energy beta. The average energy of the beta is calculated using
the approximation from Chapter 4, by taking one third of the maximum energy:
1
0.314 MeV x "3 = 0.1047 MeV (average)
So that the energy that 60CO looses per decay is:
1.17 + 1.33 + 0.1047 = 2.6 MeV
10 6 trans
10 MBq
sec
2.6 MeV 1.6 x 10-13 1 1 W
-----=- X
X
X
X -L
1 MBq
trans
MeV
1_1_
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _-=se:::...=....c = 4.2 x 10-6 W
Ｑｾ＠
ｾ＠
L (water)
(b) a 6 liter spherical tank.
The absorbed energy due to the beta particle is calculated first in the same manner
as in part (a);
10 MBq x
------'
L
10 6 trans
sec x 0.104 MeV x 1.6 x 10-13 1 x _
1W
_
1 MBq
trans
1 kg
L (water)
MeV
1_1_
sec
1.66 X 10-7 W
kg
6.21
188
THE HEALTII PHYSICS SOLUTIONS MANUAL
Now calculate the absorbed energy due to the photons, however, some of the
photons escape without imparting any energy to the solution. The absorbed
fraction of energy is found using table 6.5 and interpolating for a 6 liter (kg)
sphere with an average photon energy of 1.25 Me V. A value of 4>=0.223 is interpolated. So, the average photon energy imparted to the sphere per decay will be:
(1.17MeV + 1.33 MeV) x 0.223 = 0.5575 MeV.
Finding the power density, using the same format as in part (a);
10 6 trans
10 MBq
sec
0.5575 MeV 1.6 x 10-13 J 1 W
------=- x
x
x
x -L
1 MBq
trans
MeV
I_J_
_ _ _ _ _ _ _ _ _ _-:--_ _ _ _ _ _ _ _.: :. :se:: . =-c = 8.92 X 10-7 W
I kg
kg
L (water)
Summing the power density from the beta and the gamma-rays;
-7 W
. -7 W
-6 W
8.92 x 10 - + 1.66 x 10 - = 1.06 x 10 - is the power density for a 6 liter
kg.
kg
kg
spherical tank.
6.22
6.22 A 20 liter sealed polyethylene cylinder contains 3700 MBq (100 mCi) 137Cs
waste uniformly dispersed in concrete. Neglecting absorption by the cover,
estimate the dose rate at the top of the container, and at 1 meter over the top.
g
p = 2.35-3 (Table 5.2)
cm
1000 cm 3
4
3
Vcym
l' d = 20 L x
= 2 x 10 cm
er
1L
II
t"'concrete
= 0.2 cm- l (Table 5.2) .
SOLUTIONS FOR CHAPTER
6
189
Energy of 137CS gamma ray =0.662 MeV 85% of the time.
The dose rate in an infinite medium can be found first (similar to equation 6.76);
t
1-
9
3.7 x 10 Bq x _S_ x 0.662 MeV x 0.85 y x 1.6 X 10-6 erg x ＳＶＰｾ＠
4
3
2x10 cm
1Bq
y
t
MeV
g
2.35-- x 100 ergs rad
3
cm
g
hr =2.55 rad
I
r
hr
Since the concern is the surface dose rate, half of an infmite medium is a surface,
so divide the infinite medium in half to obtain the surface dose rate;
rad
rad
mGy
2.55- x 0.5 = 1.27- = 12.7 - - is the dose rate at the surface.
hr
hr
hr
The dose rate at 1 meter is calculated assuming that the cylinder's diameter is the
same as its height:
d=h
1t
2
V=-xd xd
4
4
3
1t
2
2x 10 cm ="4xd xd
d = 29.42 cm
So the diameter of the 20 liter cylinder is 29.42 cm. Now find the area of the lid:
1t
2
1t
2
A = -x d = -x (29.42 cm) = 679.8 cm
4
4
2
190
THE HEALTII PHYSICS SOLUTIONS MANUAL
Absorption coefficient for concrete (Appendix F) and density of concrete (Table
5.2)
2
cm 2
cm
g
= 0.06815 cm- I .
ｾ＠ =0.029- =0.029- x 2.35g
g
cm 3
x
=29.42 cm height of cylinder
Find the apparent areal concentration of the activity in the 20 liter container by
integrating over the volume of the container with respect to height: (equation.
10.14)
x
Ca = JCV e-lUdx
o
0.005 mC i
.
3
C = Cv (1- e-fIX) =
cm
x (1- e -o.06815X(29.14)) = 0.064 mCl
Jl
0.06815 cm- 1
cm 2
a
To find the apparent activity on the lid, multiply the surface area of the lid (determined above) by the concentration:
mCi
2
0.064 - - 2 x 679.8 cm = 43.16 mCi
cm
Plane Source Calculation using equation 10.10;
R·m 2
(Table 6.3)
Cl·hr
r w-137 = 0.33.
h= 1 m
r=0.147m
Ca = 0.64 Cilm2
2
2
2
R . m2 064 Ci I (0.147 +1 )
H1m --1t X r X Ca X In(r2 +h
) = 1t X 033
2
•
X.
-2 X n
2
h
Him· =0.014 Rlhr ｾ＠ 0.13 mGy/hr
ｃｩﾷｾ＠
m
1
SOLUTIONS FOR CHAPTER
6
191
Alternatively, assuming the apparent activity is approximated by a point source,
calculate the dose rate at 1 meter using equation 10.1;
R·m 2
r Cs-J37 = 0.33.
(Table 6.3)
CI·hr
s= 1 m
n = 43.16 mCi = 0.043 Ci
J
R 2
0.043 Ci x (0.33__.. m_
CI·hr
X = nr =
(l m/
S2
］ＰＮＱＴｾ＠
hr
Using equation 6.12 to convert to rads;
J.l7rm = 0.0326 crJl/g (table 5.3)
J.ljp'a = 0.0296 cm /g (table 5.3)
2
X ］ＰＮＱＴｾ＠
hr
J.l;:
rads = 87.7 x
100
Pm X roentgens = 87.7 x 0.0326 x ＰＮＱＴｾ＠
100 0.0296
J.ljp'a
0.0137 rads 10 mGy
hr
x 1 rad = ＰＮＱＴｾ＠
= "0.0137 rads
hr
hr
mGy
is the dose rate at 1 meter from the surface.
Alternative Solution for Surface Dose
Please note that the apparent areal concentration could have also been used to
calculate the dose rate on the surface of the lid, rather than assuming an infinite
medium. Half of the radiation from the apparent surface activity is assumed to be
going up, and half down. The surface dose rate is given by the following equation:
192
THE HEALTH PHYSICS SOLUTIONS MANUAL
7
t
· 3.7 x 10 M V
2
m C1
S
1
e
--6 erg
cm
C -x
x
f
x
E
x
1.6
x
10
x
Jl
x 3600
2
a cm
1 mCi
t
1
MeV
enmuscle
g
. 1
D = - x ----------------.:.------:-----------=---"-
ＱＰ･ｾｳｲ｡､＠
2
7 t
M V
2
C · 3.7 x 10 L
x
0.064 mIx
s
x
0.85
0.662_ex
1.6
x
10--6
erg
x
0.032
cm
x 3600
cm2
1Ci
t
1
MeV
g
1
Ｍｸｾ｟Ｌ
100 ･ｾｳ＠
2
rads
ｾｲ｡､＠
.
= 1.22 - h is the estimated dose rate on the surface of the container (calculated
.
r
using the apparent areal concentration)
It can be seen that the 1.22 radslhr obtained here compares favorably with the
1.27 radslhr calculated using an infinite medium (within 5% of each other).
6.23
18
6.23 A nuclear bomb is exploded at an altitude of 200 m. Assuming 10 fissions,
in the explosion. 6 fission gammas of 1 MeV each and 2 prompt neutrons of 2
MeV each. estimate the dose from the gammas and from the neutrons at 1500 m
from ground zero. Neglect the shielding effect of the air and scattering from the
ground.
The distance from the explosion to a point 1500 meters away is:
.J15002 + 200 2 = 1513.3 m
Calculating the gamma dose first using equation 6.10;
18
ｾｭ＠
4> =
2
10 fissions
x 61 x
1m
= 2.08 x 107 _1_
4
2
2
4 x 7t x (1513.3 mr fission 1.0 x 10 cm
cm
= 0.0308 cm
g
2
SOLUTIONS FOR CHAPTER
ｾ｟ｙｅｍ･ｖＱＮＶｸＰＭＳｬｭ＠
2
cm
2
y
MeV
1/
/kg
Gy
Dy=
cm
. g
6
ｸＱＰＳｾ＠
193
kg
Dy =
2.08 X 10 7 ｾ＠
2
x 1.0 MeV x 1.6 x 10-13 _1_ x 0.0308 cm x 10 3 xl
cm
y
MeV
g
0.001 kg x /(g
cm3
Gy
= 1 X 10-4 Gy Y
Next, calculate the neutron dose. The scattering cross sections for 2 Me V neutrons
are approximately the same as for 5 MeV neutrons (cross sections are relatively
constant between 1 and 10 MeV). Using the information from example 6.16 and N
andjfrom table 6.12, Synthetic Tissue Composition:
Element
0
C
H
N
Na
a
cm2/atom
1.55 X 10-24
1.65 x 10-24
1.50 x 10-24
1.00 x 10-24
2.3 x 10-24
2.8 x 10-24
(J
N,atoms/kg
2.69E+25
6.41E+24
5.98E+25
1.49E+24
3.93E+22
1.70E+22
j
0.111
0.142
0.5
0.124
0.08
0.053
sum
Ncif
4.628
1.502
44.85
0.1848
0.007231
0.002523
51.17
cm2
+. = 51.17"L.. N.(J.
, .),
k
g
i
ｾ＠
=
18
10 fissions
. 4xnx(1513. 3m
Equation 6.103;
r
2
x 3 neutrons x
1m
= 1.04 x 107 neutrons
4
2
fission
1.0x10 cm
cm 2
194
THE HEALTH PHYSICS SOLUTIONS MANUAL
2
1.04 X 107 neutrons x 2 MeV x 1.6 x 10-13 _1_ x 51.17 cm
2
MeV
kg = 1.7 x 10-4Gy
cm
D=
l){g.GY
neutron dose
6.24
6.24 An unmarked, unshielded vial containing 370 MBq (10 mCi) 2-Na is left in a
hood. A radiochemist not knowing of the presence of the 2-Na, spends 8 hours at
his bench, which is 2 meters from the 24Na. Based on the 24Na transformation
scheme shown in problem 6.5, calculate
(a) the dose rate at 2 meters from the 370 MBq source.
(Equation 10.1)
A = 10mCi
d = 2 meters
R·m 2
r = 1.84 Ci. hr (table 6.3)
R·m 2
1 Ci
3
. r A 1.84. x 10 mCi x 3 ·
Cl· hr
10 mCI = 4.6 x 10- Rlhr = 0.046 mGylhr is
X = -2 =
d
(2mt
the initial dose rate.
(b) The dose commitment from the 8-hour exposure.
2-Na decays with a half life of 15 hours. The dose rate therefore is continuously
changing. The total dose is therefore;
mGy
.
8hr
b
0.046--(
_oo693 X8hr )
.
f b . e -At dt = -"_0 (1- e -At) =
hr 1- e 15 hr
D= JD(t)·dt= ° °
A
(0.693)
,
15 hr
D=0.31 mGy
SOLUTIONS FOR CHAPTER
.
d .h
6
195
. h 197.H g or 203 H g IS
· used diagnostically in
6.25 ChI ormero dnn tagge elt er WIt
studies of renal function. Calculate the dose to the kidneys, for the case of normal
uptake, from injection of 3.7 MBq (100IlCi) of each of the radioisotopes. Assume
very rapid kidney deposition, followed by unusually rapid elimination with a
biological half time of 6.5 hours.
Calculating 197Hg first:
Table 6.11 gives information on how much Hg will be deposited in the kidneys,
and with normal uptake, 35% (0.35) of the injected activity will be deposited in
the kidneys.
AS<O) = (0.35) x 3.7 MBq = 1.295 MBq = 1.3 x 10 Bq
6
The biological halflife is given in the question, so calculating the effective halflife
for 197 Hg using equation 6.54:
TR = 65 hr
TB = 6.5 hours
= 65 hr x 6.5 hr = 5.9 hr
65 hr+6.5 hr
Finding the effective elimination constant by equation 6.52:
A = 0.693 = 0.693 = 0.117 hr- I
E
TE
5.9 hr
The cumulated activity is found with equation 6.91;
6
1.3 x 10 Bq
_
A (0) =
10
0.117 hr'! x 1 hr = 4 x 10 Bq·sec
A(kid) = {
E
3600 sec
Find the value now for S Ｈｫｩ､ｮ･ｹｳｾＩ＠
from NMJMIRD Pamphlet No. 11,
"S," Absorbed Dose per Unit Cumulated Activity for Selected Radionuc1ides and
Organs, Oct. 1975, page 246 and 247 to calculate the dose to the kidneys from the
activity deposited in the kidneys;
6.25
196
THE HEALTH PHYSICS SOLUTIONS MANUAL
S Ｈｫｩ､ｮ･ｹｳｾＩ＠
=5.5 x 10 J-lC'1- hr
ｾ＠
rad
Converting units:
=
S Ｈｫｩ､ｮ･ｹｳｾＩ＠
5.5 x 1ｏｾ＠ rad
1 Gy
----x
J-lCi· hr
100 rad
X
1 J-lCi
1 hr
4 3 0Gy
4
X
= .1 xl 14
3.7 x 10 Bq 3600 sec
Bq . sec -
Now substitute the values in equation 6.97 to find the dose to the kidneys from the
activity deposited into the kidneys;
D Ｈｫｩ､ｮ･ｹｳｾＩ＠
=AkidneyS X S( kidneys ｾ＠
Ｈｫｩ､ｮ･ｹｳｾＩ＠
D
kidneys)
14
=4 x 10 Bq·sec x (4.13 x 10- BqGy. sec)
10
= 1.65 x 10- Gy is the dose to the kidneys from activity
D Ｈｫｩ､ｮ･ｹｳｾＩ＠
deposited in the kidneys.
3
Table 6.11 shows that although activity is deposited in the kidneys, it is also
deposited in the liver, spleen and whole body in significant quantities. They also
contribute dose to the kidneys from the activity deposited in these other places.
The spleen will be ignored in this problem since the deposition in it is so small
(0.02). Calculating the dose from the liver to the kidneys {S Ｈｫｩ､ｮ･ｹｳｾｬｶｲＩｽ＠
next:
From table 6.11, 15% (0.15) of the activity deposited into the liver:
AlO) = (0.15) x 3.7 MBq = 0.555 MBq = 5.55 X 105 Bq
From Table 6.11, we find the biological half life of Hg in the liver to be 324 hours
(13.5 days). Therefore,
T
E
= 65 h x 324 h = 54 h
65h+324h
and
SOLUTIONS FOR CHAJYIER
6
197
A = 0.693 = 0.693 = 0.0128 h- 1
TE
54 h
E
The cumulated activity (equation 6.91) is
AI'rver (0)
A(liver)
=
AE
0.0128 h -1 X
1h
= 1.56 X lOll Bq·sec
3600 sec
To calculate the dose to the kidneys from the activity deposited in the kidneys, first
find the value for ｓＨｫｩ､ｮ･ｹｳｾｬｶｲＩ＠
from NMlMIRD Pamphlet No. 11, "S,"
Absorbed Dose per Unit Cumulated Activity for Selected Radionuc1ides and
Organs, Oct. 1975, page 246 and 247.
-16
Gy
= 2.7 x 10-6 ｾｃｬ＠ rad
.
=
2.03 X 10
. hr
Bq . sec
S Ｈｫｩ､ｮ･ｹｳｾｬｶｲＩ＠
Now substitute the values in equation 6.97 to find the dose to the kidneys from the
activity deposited into the liver;
D Ｈｫｩ､ｮ･ｹｳｾｬｶｲＩ＠
D Ｈｫｩ､ｮ･ｹｳｾｨｶｲＩ＠
=Auver x S (kidneys f-liver)
"
) =3.17 x 10-5 Gy
= 1.56 x 1011 Bq·sec x (2 .03 x 10-16 BqGy
·sec
.is the dose to the kidneys from the 197Hg in the liver.
Calculating the dose from the total body 197Hg to the kidneys {S
. Ｈｫｩ､ｮ･ｹｳｾ｢ｯＩｽＺ＠
According to Table 6.11, if 35% + 15% + 2% are deposited in the kidneys, liver,
and spleen, then 48% of 3.7 MBq or 1.8 x 106 Bq are distributed throughout the
body. The biological halflife in the body is given as 240 hours (10 days). The
effective halflife in the body is
198
THE HEALTH PHYSICS SOLUTIONS MANUAL
T = 65 h x 240 h
E
51 hand
65 h + 240 h
A = 0.693 = 0.693 = 0.0136 h- I
E
TE
51 h
The cumulated .activity in the body is
6
AbOdY(O) 1.8 X 10 Bq
II
o Y = AE
= 0.0136 hr- 1 = 4.765 x 10 Bq·sec
A(b d)
Find the value now for S(kidneysf-body) from NMlMIRD Pamphlet No. 11, "S,"
Absorbed Dose per Unit Cumulated Activity for Selected Radionuc1ides and
Organs, Oct. 1975, page 246 and 247 to calculate the dose to the kidneys from the
activity deposited in the body;
S (kidney sf-body) = 3.3 x 10
-6
rad
-16
Gy
.
=
2.48 x 10
Bq . sec
/lCI . hr
Now substitute the values in equation 6.97 to ｾｭ､＠
activity deposited into the body;
the dose to the kidneys from the
D (kidney Sf-body) = Abody x S (kidneysf-body)
.
11
16
Gy
D (kidney Sf-body) = 4.765 x 10 Bq·sec x 2.48 x 10.
.
Bq . sec
D (kidney Sf-body) = 1.182 x 10-4 Gy
Adding the doses together:
1.65 X 10-3 Gy (kidneYf-kidney) + 3.17 x 10-5 Gy (kidneyf-liver)+
+ 1.18 x 10-4 Gy (kidneyf-body) = 1.80 x 10-3 Gy to the kidneys
SOLUTIONS FOR CHAPTER
6
199
1 X 105 mrad
Gy
= 180 mrads to kidneys from the 3.7 MBq 197 Hg
1.8 x 10- Gy X
3
tagged chlormerodrin.
Since the deposition in other organs is not listed or small, they do not contribute
significantly to the dose to the kidneys. Even if the dose from the liver had been
omitted the net dose would not have changed significantly.
N ow calculate the dose from the same quantity of 203 Hg activity to the kidneys
using the same format and equations:
Table 6.11 shows that 35% (0.35) of the injected activity will be deposited in the
kidneys.
AlO) = (0.35) x 3.7 MBq = 1.295 MBq = 1.3 x ＱＰｾ＠
Bq
Using the abnormally short biological half life given in the question, the effective
half life for 203 Hg, using equation 6.54 is:
TR = 46.9 days = 1125.6 hr (fig 6.13)
TB = 6.5 hours
1125.6 hr x 6.5 hr
1125.6 hr + 6.5 hr
=-------
6.46 hr
Finding the effective elimination constant with equation 6.54:
A = 0.693 = 0.693
-I
E
TE
6.46 hr = 0.107 hr
Placing values into equation 6.91 to find the cumulated activity;
6
A(kid) = As(O) = __1_._3_x_lO__B--"q'---_=4.37 x 10IOBq'sec
AE
0.1 07 ｾ＠ x 1 hr
hr 3600 sec
from table 6.9 to calculate the dose
Find the value now for S Ｈｫｩ､ｮ･ｹｳｾＩ＠
to the kidneys from the activity deposited in the kidneys;
200
THE HEALTIl PHYSICS SOLUTIONS MANUAL
S Ｈｫｩ､ｮ･ｹｳｾＩ＠
= 8.1 x 10
-4
rad
" -14 Gy
.
= 6.08 x 10
JlCI . hr
Bq . sec
Now substitute these values into equation 6.97 to find the dose to the kidneys from
the activity deposited into the kidneys;
D Ｈｫｩ､ｮ･ｹｳｾＩ＠
= Akidneys X S Ｈｫｩ､ｮ･ｹｳｾＩ＠
D Ｈｫｩ､ｮ･ｹｳｾＩ＠
Gy )
=4.37 x 10 Bq·sec x (608
. x 10Bq . sec
D Ｈｫｩ､ｮ･ｹｳｾＩ＠
= 2.66 x 10- Gy = 266 mrads
14
"10
3
The contributions of the 203 Hg activity in the liver and body to the total dose to the
kidneys are calculated in the same manner as for 197Hg. For each of these compartments, as well as for the kidneys (whose calculations are detailed above), the
numerical values for the parameters were calculated in the same manner as for
197Hg , and are tabulated below
Day
Total
Long lived
Short lived
0
4
2.2
1.8
1
2.94
1.97
0.97
2
2.32
1.77
0.55
3
1.9
1.58
0.32
4
1.6
1.42
0.18
5
1.4
1.27
0.13
The total dose to the kidneys from the 3.7 MBq 203Hg chlormerodrin is the sum of
the contributions from each compartment, 3.89 x 103 Gy, or 389 mrads.
Hg dose to kidneys = 180 mrads
201 Hg dose to kidneys = 389 mrads
197
SOLUTIONS FOR CHAPTER
6
201
6.26 A nuclear medicine procedure used to evaluate pulmanary perfusion uses
intravenously injected Ｙｾｃ＠
tagged microspheres that are rapidly taken up by the
lungs, where they are temporarily trapped in a small fraction of the capillaries.
Absence of radioactivity in a part of the lung means decreased perfusion of that
activity
part, and suggests a possible pulmonary embolism. Some 60% of the Ｙｾｃ＠
is transferred out of the lung with a biological T 112 of 1.8 hours, and the other 40%
has a biological TI12 of 36 hours.
(a) What is the mean intrapulmonary residence time of the Ｙｾｃ＿＠
First calculate the effective haIflife of each compartment using equation 6.54:
Transferred out of the lung (60%)
TB = 1.8 hr
TR = 6 hr
""-6 hr x 1.8 hr
..{) hr + 1.8 hr
1.385 hr
Since this is the effective half life of 60%,
1.385 hr x 0.60 = 0.831 hr is ｴｾ･＠
ment.
"weighted" effective half life for this compart-
The "other" 40%;
TB = 36hr
TR = 6 hr
T, _
E -
TRTB
6 hr x 36 hr
TR +TB = 6hr+36hr = 5.143 hr
Since this is the effective half life of 40%,
5.143 x 0.40 = 2.0572 hr is the "weighted" effective half life for this compartment.
Combining the two "weighted" halftimes:
0.831 hr + 2.0572 hr = 2.89 hr
6.26
202
THE HEALTH PHYSICS SOLUTIONS MANUAL
However, the mean time is found by equation 4.24:
-
1 - --.,...1
__ ｾ＠
4 16 h r IS
. the mean mtrapu
.
I monary
= 2.89 hr =.
0.693
0.693
- Aeff - 0.693L
t - -
/TE
residence time.
(b) What is the dose to the lung from the intrapulmonary activity per MBq injected?
.
1 x 106Bq x 1 x 10 6/-lCi
C1
3.7 x 10lOBq - 27 fl
1 MBq x 1 MBq
A = activity x mean residence time = 27 flCi x 4.159 hr = 112.3 flCi·hr
From NMlMIRD Pamphlet No. 11, October, 1975, page150, for 99m.rC, or using
Table 6.10;
S ＨｬｵｮｧｾＩ＠
5
= 5.2 x 10-
/-l
rad
C· h
l'
r-
Using equation 6.97 with the above values:
ｄＨｬｵｮｧｾＩ＠
= A x ｓＨｬｵｮｧｾＩ＠
ｄＨｬｵｮｧｾＩ＠
= (112.3 flCi·hr) x 5.2 x 10- /-lC·l' hr - 5.84 x 10- rad
=5.84 mrad = 58.4 flGy
ｄＨｬｵｮｧｾＩ＠
5
6.27
rad
3
6.27 Three mCi 99m.rC labeled sulfur colloid is injected to visualize the liver. 60%
of the injectate is deposited in the liver, 30% in the spleen, and 10% in the red
bone marrow. Calculate the absorbed dose to:
_
(a) the liver
The radiological half life, TR , for 99m.rC =6 hours. This is very much less than the
biological half time in these organs (ICRP 53), therefore, TE = TR , and
A = 0.693 = 0.693 = 0.12 h -I
E
T,E
6h
SOLUTIONS FOR CHAPTER
6
203
a. Dose to liver
Since the 3,000 /-lCi (3 mCi) are injected, and 60% is deposited in the liver,
0.6 x 3000 JlCi = 1800 JlCi are deposited in the liver.
The radiation dose, in traditional units, is given by equation 6.100
D( target) =
S Ｈｬｩｶ･ｲｾＩ＠
D( liver ｾ＠
Ao ( source) JlCi
.1
AE h
5
= 4.6 x 10·
liver) =
X S( target ｾ＠
source)
rad
.
JlCl· hr
rad
9
C· h for 91'c is found in Table 6.10
Jl 1· r
0.6 x 3000 JlCi
rad
1
x 4.6 x 10-5
= 0.690 rad
0.12 h·
JlCi· hr
The spleen's contribution to the liver dose is
D(liver ｾ＠
spleen) =
Ao (spleen) JlCi
-I
AE h
0.3 x 3000 !lCi
ｄＨｬｩｶ･ｲｾｳｰｮＩ］＠
0.12h
-I
x S(liver ｾ＠
rad
spleen)--.!lCl·hr
x9.8x10-7
rad
.
-7.35x10-3 rad
!lCI·hr .
The red bone marrow's (RBM) contribution to the liver dose is.
D(liver ｾ＠
RBM)
D(liver ｾ＠
RBM) =
Ao(RBM) !lCi
(.
x S lIver ｾ＠
AE h .
) rad
RBM - - !lCi . hr
0.1 x 3000 JlCi
rad
.
_ 2.3 X 10-3 rad
JlCl· hr
--'----'-1--
0.12 h
-1
x 9.2 X 10-7
D (Liver) = 0.690 + 0.0074 + 0.0023 = 0.7 rad
The dose to the spleen and to the RBM is calculated in a similar manner, using the
parameter values and the results tabulated below:
204
THE HEALTH PHYSICS SOLUTIONS MANUAL
Liver as target Ｈｬｩｶ･ｲｾｳｯｵ｣Ｉ＠
Source
Liver
spleen
RBM
Activity, /-lCi
1800
900
300
AE' h- I
0.12
0.12
·0.12
rad
S f.lCi· h
4.6 x 10-5
9.8 x 10-7
9.2 x 10-7
Dose(liver)
Dose, rad
0.6900
0.0074
0.0023
0.7
(b)
Spleen as target Ｈｳｰｬ･ｮｾｯｵｲ｣Ｉ＠
Source
Liver
spleen
RBM
Activity, /-lCi
AE' h- I
1800
900
300
0.12
0.12
0.12
rad
S f.lCi· h
9.2 x 10-7
3.3 x 10-4
9.2 x 10-7
Dose(spleen)
Dose, rad
0.0138
2.4750
0.0023
2.5
(c)
RBM as target (RBM ｾｳｯｵｲ｣･Ｉ＠
Source
Activity, /-lCi
AE' h- I
Liver
spleen
RBM.
1800
900
300
0.12
0.12
0.12
rad
S f.lCi· h
1.6 x 10-6
1.7 x 10-6
3.1 x 10-5
Dose(spleen)
6.28
Dose, rad
0.0240
0.0128
0.0775
0.1
6.28 A patient is treated for Graves disease with 111 MBq (3 mCi) 1-131. Uptake
studies with a tracer dose of 1-125 showed a thyroid uptake of 60% and a biological TI12 = 2 days. Assuming a very rapid thyroid uptake, calculate the dose to the
thyroid from this treatment.
.
First calculate the effective half life using equation 6.54:
TB =2d
TR = 8.05 d
SOLUTIONS FOR CHAPTER
6
205
= _T.-. :. :.R----=::TB_ = 8.05 d x 2 d = 1.6 d = 38.45 hr
T
TR+TB
E
8.05d+2d
Find the effective clearance constant for the thyroid with equation 4.21;
= 0.693
A
eif
T
0.693
-2
-1
= 38.45 hr = 1.8 x 10 hr
Equation 6.91 is used to find A ;
3
3 mCi x 1 x 10 J.lCiJ x 0.6
A = activity x fraction depos. = (
1 mCi
A
1.8 x 10-2 hr- 1
2
S Ｈｴｨｹｲｯｩ､ｾＩ＠
= 2.2 x 10-
5
1 X 10 J-lCi·hr
rad
C' h (From MIRD pamphlet 11)
J.l l ' r
Putting values into equation 6.97;
= A x S Ｈｴｨｹｲｯｩ､ｾＩ＠
D Ｈｴｨｹｲｯｩ､ｾＩ＠
D Ｈｴｨｹｲｯｩ､ｾＩ＠
.
(
2
rad J
5
=1 x 10 J-lCi·hr x 2.2 x 10- J.lCi. hr = 2200 rads = 22 Gy
is the dose to the thyroid.
An alternative method is to calculate the initial dose rate, Do' using the effective
energy deposited in the thyroid as calculated in Chapter 6,0.227 MeV per transformation for 1311, and the mass of the thyroid, 20 grams, and a 60% deposition:
6
13
60 x III MBq x 1 x 10 t x 0.227 Mev x 1.6 x 10. J x 3600 sec
D = 100
sec
t
MeV
hr
= 0.44 Gy
1 Ji
hr
20 x 1 kg x/kg
g 1000 g
Gy
The total dose is then found using the effective elimination constant calculated
earlier:
206
THE HEALTII PHYSICS SOLUTIONS MANUAL
This is close to the value calculated more formally of 22 Gy.
6.29
6.29 A well insulated water sample is irradiated with gamma rays at a rate of 10
Gy (1000 rads) per hour. What is the rate of temperature rise in the water, °Clhr.
1 Gy. =
hr
IXkg by definition.
hr
The specific heat of water is I°C/l03cal/kg
10 Gy x
6.30
6.30 A laboratory worker who weighs 70 kg was accidentally exposed for several
hours in an atmosphere containing tritiated water vapor. Urine analysis for tritium
were made for 7 weeks, starting 1 day after exposure, and the following data were
obtained on a 24 h urine samples:
Day
Bq
2· 3
5
7
10
1
524 485 450 402 342 293
14
227
21
147
28
94
35
60
42
40
49
26
According to reference man, 47% of the daily water output is via the urine.
Assuming that the tritium is uniformly distributed throughout the body's water,
(a) plot the data and write the equation for the clearance curve,
SOLUTIONS FOR CHAPTER
6
207
1000
ＱＰｾＭ
ＱＰｲＭｾ
o
5
10
15
20
25
30
35
40
45
Day
The clearance curve, when plotted on semi-log paper, fonns a straight line.
Therefore, the equation of the curve is
By extrapolating the line to t = 0, Ao is found to be 555 Bq/day. The clearance half
time is found from the graph to be 10.8 days. The slope of the curve, A, is
Iv = 0.693 = 0.693 = 0.064 d- 1
1'..12
10_8 d
The equation for the clearance curve is
A(t) = 555e-o.064t Bq I day
(b) calculate the worker's dose commitment from this accidental exposure.
To calculate the dose commitment from this accidental exposure from basic
.principles with equation 6.47, where E eH) = 5.7 x 10-3 MeV/transfonnation:
208
THE HEALTII PHYSICS SOLUTIONS MANUAL
Since 47 % of the daily water output is via the urine, the total extrapolated initial
daily water output is
.
555 Bq
Ao =
= 1181 Bq/day.
0.47
The total amount of 3H that was absorbed is equal to the total excreted, which is
given by the area under the clearance curve.
.
A
1181 Bq
d
4
A = f A(t)dt = AoS e-Atdt = _0 =
.1 = 1.85 X 10 Bq
o
0
A 0.064 d
00
• •
00
The initial dose rate due to this 3H activity is
q Bq x
Ｚｾ＠
/ x E MeV x 1.6 x lO"13J x 86400 sec
%q
t
MeV
day
Ｍｾ］＠
mkgx lKg/Gy
.
. 04 B
1. 85 x 1
qx
Ｚ･ｾ＠
70 kg x 1Kg/GY
Gy
day
Now fmd the dose commitment using equation 6.58:
Do =2.1 X 10- Gy/day
8
AE = 0.064 d- l (see part a)
.'
day
/
5.7 x 10.3 Mev 1.6 x 10-13 J 86400 sec
%q x
.
t
x
MeV
x ---d-ay--
ｄｯ］Ｍｾ
.
Do
= 2.1 x 10-8
Gy
2.1 X 10-8 Gy
Do
d
-7·
D = 1 E = 0.064 d- l = 3.4 x 10 Gy = 0.034 mrad
Solutions for Chapter 9
HEALTH PHYSICS INSTRUMENTATION
9.1 If a certain counting standard has a mean activity of 400 cpm,
- 9.1
(a) What is the probability of observing exactly 400 counts in one minute?
Find the standard deviation of 400 counts in one minute. Since we can approximate this using the Poisson distribution, equation 9.36 is used:
n=400
cr =
!:r = J 400 = 20
Use equation 9.31 to obtain the probability of exactly 400 counts in one minute.
n=400
cr = 20
pen) =
1
｣ｲｾＲｮ＠
ｾ＠
e
｟ＨｮｾＩＲ＠
/20- 2
= p(400) = ＲＰｸｾｮ＠
1
,-;;- e
-(400-400)2/2(20)2
= 0.02
(b) What is the probability of measuring 390 - 410 counts in 1 minute?
Calculate the number of standard deviations away 410 is from the mean, using
equation 9.49;
n=400
cr = 20 (from part a)
n-n 410 - 400 = 0.5 is the number of standard deviations from the mean.
20
cr
Look this number up in a table of values, the area under the normal distribution
curve from the mean to 0.5 standard deviations away is 0.1915.
t=--
Calculating t for 390,
209
210
THE REALlli PHYSICS SOLUTIONS MANUAL
390-400
---=-0.5
20
Since the normal curve is symmetric, the area is the same, 0.1915. The probability of a count between 390 and 410 is given by the area under the normal curve
between - O.Sa and +0.5a =0.38
t
9.2
9.2 A sample counted 560 counts in 10 min., while the background counted 390
counts in 15 minutes.
(a) What is the standard deviation of the gross and background counting rates?
ng = 560
t g = 10
Equation 9.37 gives us the solution:
Fn
n
.r± a = -t ±tr
r ± a = 560 ± J560
g
g
10
10
56 ± 2.37 cpm for the sample gross count rate
Background is calculated the ｳ｡ｭｾ＠
nb= 390
t b= 15
way, using equation 9.37:
Fn
n
r+a=-±b b
t
t
rb
390 J390
±
= 26 ± 1.32 cpm for the background count rate
15
15
± ab = -
(b) What is the standard deviation of the net counting rate?
Using equation 9.43 and the results from part (a);
SOLUTIONS FOR CHAPTER 9
211
(c) What are the 90% and 99% confidence limits for the net counting rate?
Using the infonnation from part (a), the net counting rate is:
rg - rb
= 56 - 26 = 30 cpm
The confidence limits are found by multiplying the standard deviation of the net
counting rate, as found in part (b), by the number of standard deviations associated with the 90% confidence level. The 90% confidence level is associated with
1.645 standard deviations (as found in chapter 9), giving:
rn ± 1.645cr
30 ± 1.645 x (2.7)
30 ± 4.44
34.44 to 25.56 cpm is the 90% confidence interval
For the 99% confidence limits, 2.575 standard deviations are required;
rn ± 2.575cr
30 ± 2.575 x (2.7)
30±7
37 to 23 cpm is the 99% confidence level
9.3 A 10 min sample count yielded 1,000 counts. A 10 min background mea-
surement gave 250 counts. Assuming negligible radioactive decay of the sample
during the counting period, what is the sample net count rate and its 95% confi. dence interval?
ng = 1000 counts
t g . = 10 min
nb = 250 counts
9.3
212
r =
g
r
b
=
THE HEALTII PHYSICS SOLtmONS MANUAL
1000 counts
.
= 100 cpm
10mm
250 counts
. =25 cpm
10mm
rn = rg - rb = 100 cpm - 25 cpm = 75 cpm
Using equation 9.43 to find the standard deviation of the net counts;
Chapter 9 gives the number of standard deviations that the 95% confidence
interval must cover as 1.960', so, the solution would be;
rn ±1.96O'
substituting in values,
75 ± (1.96 x 3.5) cpm
75 ± 6.9 cpm
9.4
9.4 A 10 minute sample count was 756 and a 40 min background was 600
counts.
(a) What ｩｾ＠ the net count rate and its standard deviation?
ng =756 counts
t g = 10 min
nb= 600 counts
tb= 40 min
rg =
756 counts
10 min
=
b'
600 counts
= 15 cpm
40 min
r
75.6 cpm
SOLUTIONS FOR CHAPTER 9
213
rn = rg -li, = 75.6 cpm -15 cpm = 60.6 cpm
Using equation 9.43 to find the standard deviation of the net count rate;
an =
15
10+ 40 = 2.8 cpm
75.6
60.6 ± 2.8 cpm
(b) What is the precision of the measurement, expressed as %?
The precision of this measurement expressed as percent is defined as the standard deviation divided by the net count rate times 100,
an 100
2.8
precision = = - - x 100 = 4.65%
rn
60.6
9.5 A background counting rate of 30 cpm was determined by a 60 min count. A
sample that was counted for 5 min gave a gross count of 170.
(a) At the 90% confidence level, is there activity in the sample?
t b = 60
ng= 170
t g= 5
r g = 170/5 = 34 cpm
rb = 30 cpm
r1 =M1 = 34
r2 =M2 =30
Using equation 9.49
9. 5
214
THE HEALTH PHYSICS SOLUTIONS MANUAL
t = 1.47 standard deviations difference between the two means. A one tail test
must be used for this problem since the question is asking only whether the
sample is greater than background. For a 1 tailed test, the critical value for t
(90%) = 1.28. Since we found t = 1.47, we conclude that activity is present at the
90% confidence level.
(b)
Is there activity in the sample at the 95% confidence level?
Since t (the number of standard deviations) was calculated in part (a) as_1.47, and
the critical value for t (95%) = 1.645 for a 1 tailed test, we conclude that there is
not activity present at the 95% confidence level.
9.6 As a test of the operation of a certain counter, two measurements were made
on the same long-lived sample. The first gave 10,210 counts in 10 minutes, and
the second gave 4995 in 5 minutes. Is the counter operating satisfactorily?
= 10210
t1 = 10
121
122 = 4995
t2 = 5
r1 =M1 =
10210
10 = 1021cpm
4995
r2 =M2 = -5- = 999cpm
U sing equation 9.49
IM 1 -M 2 1
t=
'i
r
tl
t2
2
-+-
11021 -9991
999
= -1021
-+- = 1.27
10
5
A two tail test is used for this problem since we are only looking for differences
between the two ｭ･｡ｳｵｲｮｴｾ＠
The critical value for t (90%) for a 2 tailed test
is 1.645. Since the measurements show that t = 1.27, we conclude that there is
no -difference between the two means, and that yes, the counter is operating
properly.
SOLUTIONS FOR CHAPTER 9
215
9.7 A 1 minute count shows a gross activity of 35 counts. If the background is
1560 in 60 min, how long must the sample be counted in order to be within
±10% of the true activity at the 95% confidence level?
tg = 1
r=35
cpm
g
nb= 1560
tb = 60
n
1560
r=2.=--=26 cpm
b
tb
60
Find the net c<?unting rate:
rg - rb= 35 - 26 = 9 cpm
The estimated true counting rate is 9 cpm. If it is desired to be within 10% of 9
cpm, the estimate must be within 0.1 x (9 cpm) = 0.9 cpm
To find the net standard deviation associated with 0.9 cpm at 95% confidence
interval:
1.96 standard deviations are associated with a 95% confidence level (Chapter 9).
0.9 = 1.96cr
cr = 0.459
Now use equation 9.43 to solve for the time the sample should be counted (t):
g
(J n
=
0.459 =
35
26
tg
60
-+-
tg = -160
Since the solution is negative, it is not
possible to obtain the stated accuracy and precision in the problem, unless the
background counting time is increased.·
216
9.8
THE HEALTII PHYSICS SOLUTIONS MANUAL
9.8 A sample that had been counted for 15 min showed a counting rate of 32
cpm. The background, counted for 10 min, was 15 cpm.
(a) What is the net counting rate, at the 95% confidence limit?
r=
32 cpm
g
tg= 15 min
rb= 15 cpm
tb= 10 min
rnet = rg - rb= 32 - 15 = 17 cpm is the net count rate
Use equation 9.43 to fmd the standard deviation associated with the net count
rate:
cr n =
32
15
15 + 10 = 1.91cpm
Chapter 9 gives the 95% confidence level as 1.96 standard deviations, so the
result is:
rn ± 1.96 cr
17 ± 1.96 x (1.91)
17 ± 3.7 cpm
(b) What is the coefficient of variation (relative error at ± 1 standard deviation)
of the net counting rate?
Use equation 9.39b and data from part (a);
n= 17
cr = 3.7
cr
1.91
% relative probable error =100 x -;;- =100 x 17 = 11 %
9.9
9.9 . A sample has an estimated gross counting rate of 35 cpm (based on a 2
minute count). The background, determined by a 1 hour count, is 10 counts per
SOLUTIONS FOR CHAPTER 9
217
minute. How long should the sample be counted if we want to be 95% certain
that the net counting rate is within ± 5% of the true net counting rate?
rg = 35 cpm
tg = 2 min
rb = 10 cpm
tb = 60 min
Estimate the net count rate first:
rne t = rg - rb = 35-10 = 25 cpm
The estimated true net counting rate is 25 cpm. If it is desired to be within 5% of
25 cpm, the estimate must be within 0.05(25cpm) = 1.25 cpm
To find the net standard deviation associated with 1.25 cpm at 95% confidence
interval: 1.645 standard deviations are associated with a 95% confidence level
for a one tail test. A one tail test is used since we wish to see whether the sample
is greater than the background.
1.25 = 1.645cr
cr = 0.76
Now use equation 9.43 to solve for the time the sample should be counted (t):
g
cr n =
35 10
0.76= - + t8
60
tg = 85 minutes.
9.10 (a) The gross 1 minute count on a sample was 100, and the background,
counted during 1 minute was 50 counts. What was the net counting rate at the
90% confidence level?
rg = 100 cpm
tg =lmin
rb = 50
l
9.10
218
THE HEALTH PHYSICS SOLUTIONS MANUAL
rnet = rg -rb= 100-50=50cpm
Use equation 9.43 to fmd the standard deviation associated with the net count rate:
rg rb
JMOO
50
cr = -+- = - + - = 1225
n
tg tb
1
1
.
The number of standard deviations associated with 90% is 1.645 (from chapter
9)
rn ± 1.645an = 50 ± (1.645) x 12.25 = 50 ± 20 cpm
9.1 O(b) If the sample and background were each counted for 10 minutes, and
. gave counting rates of 100 and 50 cpm respectively, what was the net counting
rate at the 95% confidence interval?
r=
100cpm
g
tg = 10 min
rb = 50
tb =10min
rnet = rg - rb= 100 - 50 = 50 cpm
Use equation 9.43 to fmd the standard deviation associated with the net count
rate:
cr. ｾ＠
r, + rb ｾｴｯ＠
tg
tb
10
+ 50 ｾＳＮＸＷ＠
10
The number of standard deviations associated with 95% is 1.96 (from chapter 9)
rn ± 1.96an = 50 ± (1.96) x 3.87 = 50 ± 8 cpm
9.11
9.11 A shielded low background counter has an average counting rate of 2 cpm.
What is the probability that a 1 min counting period will record?
(a) 2 counts,
(b) 4 counts,
Use equation 9.34
(c) 0 counts.
SOLUTIONS FOR CHAPTER 9
219
n=2
n=2
pen) -
Ｈｾｲ＠
･Ｍｾ＠ =p(2)
n!
e-2
2! = 0.271
22
(b) 4 counts,
n=2
n=4
24 e-2
4! = 0.09
(c) 0 counts,
-
n=2
n=O
pen) -
Ｈｾｲ＠
･Ｍｾ＠
n!
= p(O)
2° e-2
O! = 0.135
A sample of river water was taken near the waste discharge pipe of an
isotope laboratory, and another sample was taken upstream of the discharge
point. Each sample was counted for 10 min, and gave 225 and 210 cpm, respectively. At the 99% confidence level, is the downstream water more radioactive
than the water upstream?
9.12
tg
9.12
= 10
tb= 10
rg = 225 cpm
rb = 210 cpm
r1 =M1 =225
r 2 = M 2 = 210
ｲﾷＧｷｾＺﾻＬ［ＭＩｬ＠
ｾ＠
ｾ＠
ｲｾ＠
ｬＮＺｊＧｾ｜Ａ＿ｲｴ＠
ｴｬＢｾ＼ｖ＠
ｬ［ｩＮｹＺﾷＩｾＧ＠
__ｾＢＧｬ＠
_'.t
'(y ＬｾＧ［ｦ＠
｜ＬＧＺＱＮｴＩｪﾷｾ＠
ＮＡＺｾｌ＠
. )_f)a","'s.
_. ｾ＠ ..［Ｚｾｴｦ＠
• ...,. ...........,.-...... ,'__ .__ ......._ ..ｾＮＬ＠ .....
# ......
ｴＺｾ＠
..... "' ... ,..-....
ｾ＠
i
f
J
220
THE HEALTH PHYSICS SOLUTIONS MANUAL
U sing equation 9.49 without absolute values, because there is a reason for the
ordering of the values,
t=
Mj-M 2
r)
r
t)
t)
- + -2
=
225 - 210 =2.27
225 210
-+10
10
t = 2.27 standard deviations difference between the two means. A one tail test is
used since we are only checking to determine whether the activity downstream is
greater than upstream t (99%, 1 tail) = 2.327. Since the measurement t < 2.327,
the downstream water is not more radioactive than the upstream.
9.13
9.13 A certain counting standard has a true mean counting rate of 50 cpm.
(a) What is the probability of observing exactly 50 counts in 1 minute?
Find the standard deviation of 50 counts in one minute. Since we can describe
the distribution of count rates by the Poisson distribution, equation 9.36 is used:
a = Jii = Eo = 7.07
Since the mean, 50>30, we can use either the Poisson distribution, equation 9.34,
or the normal distribution, equation 9.31 to obtain the probability of exactly 50
counts in one minute. Assuming the normal distribution we have:
n=50
a = 7.07
(b) What is the probability of measuring 43-57 counts in 1 min?
43 - 57 =mean ± 1a
Since 68% of the area under the normal curve is included between ± la, then,
is 0.68.
the probability of measuring 43-57 ｾｯｵｮｴｳ＠
(c) What is the probability of finding more than 57 counts in 1 min?
SOLUTIONS FOR CHAPTER 9
221
Since 50% of the area of a normal probability curve lies above 50 counts (the
mean), and 57 counts is one standard deviation from the mean (34% of the area),
the remaining area would be 50% - 34%=16%.
The probability of obtaining greater than 57 counts is 0.16.
9.14 A counting system has a background of 360 counts during a 20 minute
counting period. What is the lower limit of detection with this system for counting times of
(a) 2 min.
nb = 360 counts
tb = 20 min
rb =
360 counts
20 min
18 cpm
ts = 2 min
Equation 9.53 is used;
LLD =
ＳＮＲＹＢｲ｢ｴｳＨｉＫｾＩ＠
+3
ｾ＠
tb
ts
=
ＲｭｩｾＩ＠
3.29X,(18cou.nts) x2minx(1+
+3
mm
20 mm
2 min
LLD = 12 cpm
(b) t = 20 minutes
s
+3
3.29 x '(18 cou.nts) x 20 min x (1 + 20 ｾｮＩ＠
ｾ＠
mm
20mm
=4.6cpm
20 min
(c) ts = 200 min.
LLD=
LLD=
3.29 x (18 cou.nts) x 200 min x (1 + 200 ｾｩｮＩ＠
ｾ＠
mm
20mm
200 min
LLD = 3.3 cpm
--
i=-....
+3
9.14
222
9.15
THE HEALTII PHYSICS SOLUTIONS MANUAL
9.15 To detennine possible low level contamination, smears are counted for 10
minutes at an overall counting efficiency of 10%. The smear is considered
positive if its activity exceeds the MDA. If the blank gives 400 counts in 10
minutes, what is the MDA for this counting system.
Equation 9.54 is used;
k = 10% =0.1 counting efficiency
tb = 10 min
=400
nb
MDA =
4.66F; + 3 _ 4.66 x J400 + 3
kt
-
0.1 x (10)
- 96.2 dpm
96.2 d 1 min
1 Bq
--x
x---'-=MDA = min
60 sec Id/ = 1.6 Bq
Isec
9.16
9.16 A blank in an alpha counter records 28 counts in 2 hours. Calculate the
lower limit of detection for a 1 hour and a 2 hour sample count.
Calculating for 1 hour;
nb = 28 counts
tb
= 2 hr = 120 min.
28 counts
rb = 120 min
0.23 cpm
ts = 1 hr = 60 min
Equation 9.53 is used;
3.29 x
il4
co:ts) xl hr x (I ＫｾＩ＠
1 hr x 60 min
hr
LLD =0.30 cpm above background
For t s = 2 hours, we can use either equation 9.53, as above
+3
SoumoNS FOR CHAPTER 9
x co;;tS) x2 hr xＨｬＫｾＩ＠
3.29 il4
LLD=
223
+3
2 hr x 60 min
hr
LLD = 0.23 cpm above background
or, since ts = tb, we can use equation 9.50
LLD=
LLD =
4.66Fn + 3
4.66cr b + 3
ts
=
4.66-fi8 +3
120 min
0.23 cpm above background
9.17 A counting standard was counted for 5 minutes before an experiment, and
gave mean count rate of 5965 cpm. After the experiment, the standard was
counted again, and gave 6070 cpm during a 2 minute check. Was the counting
system operating as expected?
Use the two tail test (since we are only looking for a difference between the two
counts) to determine if there is a difference between the two counts, equation 9.49:
M1 = 5965 cpm
M2 = 6070 cpm
r1 = 5965 cpm
r2 = 6070 cpm
t=5min
1
t2 = 2 min
t.=
IMl - M21 = 15965-60701 = 1.615
,t;+r:rl
r2
ｾＭＵＫＲ
5965
6070
"t" is less than 1.96, so there is no difference between the two count rates at the
95% confidence level.
Yes, the counter is operating properly.
9.17
224
9.18
THE HEALlH PHYSICS SOLUTIONS MANUAL
9.18 An unsealed air wall ionization chamber whose volume is 275 cm3 is
calibrated at an atmospheric pressure of 760 torr and a temperature of 20°C. A
measurement is made at an altitude of 7000 ft (2120 m), where the pressure is
589 torr and the temperature is 25°C. The meter reading was 10 mRJhr (2. 58 IlCI
kg/hr). What is the corrected exposure rate?
The air is less dense at measurement, due to lower pressure and higher temperature, than at calibration. When corrected for pressure and temperature, we have
°
mR
760
h
589
X (corrected) = 1 0 - x -
9.19
x
mR
= 13.1273 + 20
h
273+25
9.19 A 0.0025 IlCi (92.5 Bq) 14C source is placed into an air filled current
ionization chamber whose detection efficiency is 40%. The ionization current
produces a 10 mV drop when it flows through a 10120hm load resistor. What is
the mean energy of the 14C beta particles?
Basic electricity equation;
V=10mV=0.0IV
12
R = 10 0hm
I = V = 0.01 V _
R
10 12 0hm - 1
x
-14
10 A
Table 5.1 gives the average energy lost by beta particles in air as 33.7 eV per
electron produced by ionization.
100 eO produced 33.7 eV 53 10 6 eV
1 x 10-14A· x lo/sec x electron
x
x
=.x19
1A
1.6 X 10- C
40 e- detected
electron
sec
t
92.5 Bq = 92.5 sec
6
sec
3
5.3 x 10 eV
x
5 = 57 X 10 eV = 57 keY is the mean energy.
sec
92. t
SOLlJI10NS FOR CHAPTER 9
225
9.20 A survey meter whose time constant is 6 sec is used to measure the scat-
tered radiation from an x-ray machine. After a 0.2 second exposure time, the
meter read 10 mR (258 J.lClkg)lhr. What was the actual exposure rate?
9 20
•
Meter reading, X = 10 mR
t = 0.2 sec
RC = 6 sec
If the measurement time, t, is small relative to the time constant, RC, then the
actual exposure rate, X0' is related to the meter reading, X by
x = X 1 - e-v7?c)
0(
X =
o
.
X
10
(1- e -tl RC)
=
mR
hr
(1- e -O.2sec/6sec )
= 305 mR
hr
9.21 What is the gamma threshold energy for a Cerenkov counter whose index
of refraction is 1.6?
Using equation 9.7 to find the energy of the electron that produces the Cerenkov
radiation:
n = 1.6
R
1
E=0.51·
1-n2
-1
h
1_1
=0.51 x
1-2
= 0.143 MeV
16
So a 0.143 MeV electron would produce the desired effect. What energy y ray
would produce a 0.143 MeV Compton electron when back scattered 180 0 (a
180 0 back scattered electron would have the most energy that could be transferred by a y ray to an electron) ?
EcornPton(e) = E(y)initial - E(ytauered = 0.143 MeV
Solve for the scattered photon energy next, E(y).IDltJ. 'al- 0.143 MeV = E(y) scattered
9.21
226
THE HEALTII PHYSICS SOLUTIONS MANUAL
Replace E' in equation 5.31A with theabove tenn for the scattered photon;
E'=
ＱＫＨｾｊｉＭ｣ｯｳ･Ｉ＠ moc
E
.
E
E -0.143 MeV ］Ｍ［ｾ
1+
ＨｾｊＱＭ moc
cose)
8 = 1800
2
moc = 0.511 MeV (energy equivalence of an electron)
E-O.143 MeV
E
·1+( 0.511EMeVＩｘＨｉＭｃｏｾＱＸＰ｝＠
E = 0.275 MeV is the threshold gamma ray energy (Note you will need to use the
quadratic equation to solve for E)
9.22
9.22 An ionization chamber has a window thickness of 2 mg/cm2. If a 0.01 J.lCi
(370 Bq) 21<»0 source is located 1 cm in front of the window, so that the counting
geometry is 25%, calculate the saturation ionization current.
Equation 9.10 is used to solve for the current:
N particles x E eV x 1.6 x 10-19 ｾ＠
sec
particle
ion
1= -------!;.--=-=------- = current in amps
eV
wion
The energy of the 21<»0 a will be degraded by traveling through the air and the
window.
First, calculate the range of the 5.3 MeV a in air, using equation 5.14:
E=5.3 MeV
R cm = 1.24E MeV - 2.62 = 1.24 x (5.3 MeV) -2.62 = 3.95 cm range in air.
Convert the range of the a into aerial density units (mg/cm2):
SOLUTIONS FOR CHAPTER 9
227
.
mg
R = 1.293 - - 3 X 3.95 cm = 5.1 mg/cm2
em
The range is reduced by passing through the 1 cm of air, so convert the 1 cm air
to aerial density units by multiplying by the density of air:
Pair = 1.293 x 10-3 g/cm3 = 1.293 mg/cm3
1.293 X 10-3 g 1000 mg
1 em x
3
x
g
cm
1.293 mg/cm2
2
Since the window is given as 2 mg/cm thick, it is already in aerial density units,
it is now possible to sum the two materials that the a must pass through, the air
and the window:
1.293 'mg/cm2(air)+ 2 mg/cm2(window) = 3.293 rng/crn2
Using a ratio of the range of the a in air (5.1 mg/cm2 as calculated above),
calculate the energy of the a after passing through 3.293 mg/cm2 of materials:
mg
mg
5.1-3.2932
2
cm
cm x 5 .3 MeV = 1.88 Me V is the energy of the a after passing
5.1 ｭｾ＠
cm
through the air and window
N = 370 a per second from 21OpO (given) x 0.25 (only 25% efficient) = 92.5 a/
sec
w = 35.5 eVlion (based on information in Chapter 9)
Substituting to find the current:
92.5 particles x 1.88 MeV x 1.6 xl 0- 19 ＭＺｾ＠
I = Ｍｾ･ｖＱｍＷ
sec
particle
IOn = 7.8 x 10-13 C/sec
35.5-x 6
ion 10 eV
228
9.23
THE HEALlli PHYSICS SOLUTIONS MANUAL
9.23 An air wall, air-filled ionization chamber, whose volume is 100 cm3, gives
12
a saturation current of 10- A when placed in an X-ray field. If the temperature
was 27°C, and the atmospheric pressure was 740 mm Hg, what was the radiation exposure rate?
9
3 X 10 sC = 1 C
3
1 statcoulomb/cm = 1 roentgen
. 12
Ｑｾ＠
1
X 10- A
sec 3 x 109 sC X -3600
sec = 10.8 sC I hr
-----:- X - - X
__
3
100 cm 3
100cm
lA
C
Ｑｾ＠
Note that roentgens are measured at a standard temperature of 273 K and 760
mm, so the exposure rate must be temperature corrected:
\
27°C = 300 K
3
Convert statcoulombs per cm into roentgens by correcting for temperature and
pressure. Both the temperature and pressure, in this case, act to decrease the
number of molecules with which the radiation can interact. Therefore both
temperature and pressure correction factors must be greater than one.
10.8sC/hr
100 cm
9.24
3
x
lR
760mm 300K
C x 0
x
= 0.122 RIhr or 122 mRlhr
l_s_ 74 mm 273 K
cm 3
9.24 (a) What value resistor, to be placed in series with the ion chamber, in
problem 23, is required to generate a voltage drop of 10 mY?
For DC circuits:
V = 10mV = 10 X 10-3 V
1=1 X 10-12 A
10
10 X 10-3 v
12
R = I = 1 x 10- A = 1 x 10 ohms is the value of resistor required.
V
(b) If the capacity of the chamber is 250 ｾｆＬ＠
detector circuit?
what is the time constant of the
229
SOLUTIONS FOR CHAPTER 9
As explained in Chapter 9, the product RC is the time constant, 'T. The resistance
was calculated in part (a):
R = 1 x 1010 0 hrus
12
C = 250 IlIlF = 250 x 10- F
'T
= RC = 1 X 1010 ohms x 250 x 10-12 F =2.5 sec
(c) How much time is required before the meter will read 99% of the saturation
current?
RC = 2.5 sec (from part b)
I is the saturation current
V=IR
Using equation 9.9:
v = IR (l_e tIRc)
where V is the voltage at some time t, and IR is the saturation current times
resistance of the circuit, or Vo'
To fmd when 99% of the saturation voltage is reached, ｾ＠
substituting in the values,
0.99 = (1 _ etf2 .5 sec)
t = 11.5 sec
Vo
= 0.99,
230
9.25
THE HEALTII PHYSICS SOLUTIONS MANUAL
9.25 A pocket dosimeter has a capacitance of 5 f.lf.lF and a sensitive volume of
3
1.5 cm • What is the charging voltage if it is to be used in the range 0 - 200 mR
(0 - 51.5 f.lC/kg) and the voltage across the dosimeter should be one-half the
charging voltage when the dosimeter reads 200 mR?
c= 5 f.lf.lF = 5 x 10-12 F
density of air = 1.293 x 10-6 kg/cm3 (Table 5.2)
Q=
51.5/-lC
kg air
1.293 x 10-6 kg air
X
3
cm air
15
X
•
3
cm x
1C
6
1 x 10 /-lC
=9.988 X 10-11 C
Use equation 2.66
9.988 X 10-11 C
/),. V = - =
=20 Volts for the difference in charge, from fuil to half
C
5x10- 12 F
charge.
Q
Therefore the charging voltage = 2 x /),. V = 40 volts
9.26
9.26 A Geiger tube has a capacitance of 25 f.lf.lF. The time 'required to collect all
the positive ions is 221 x 10-6 sec. In order to produce sharp output pulses,- it is
desired to limit the time constant of the detector to 50 f.lsec.
(a) What is the value of the series resistor?
't
= 50 f.lsec = 50 x Ｑｾ＠
sec
C = 25 f.lf.lF = 25 x 1Q-12F
As explained in Chapter 9, the product RC is the time constant, 'to
't = RC
1:
50 X 10-6 sec
R= C = 25 X 10-12 F = 2 X 106 Q (ohms) is the value of the series resistor
(b) If 108 ion pairs are formed per Geiger pulse, what is the upper limit of the
output voltage pulse?
SOLUTIONS FOR CHAPTER 9
231
First, find Q:
108 ions
Q = pulse x
1.6x10-19 C
= 1.6 X 10- C
11
ion
C = 25 IlllF = 25 x lO-12F (given in question)
Use equation 9.1:
Q
v=-C
1.6 x 10-lI e
= 25 x 10-12 F = 0.64 V is the maximum output voltage.
9.27 A Geiger counter has a resolving time of 250 J...lsec. What fraction of counts
is lost due to the counter's dead time if the observed counting rate is 30,000
cpm?
Using equation 9.5, where:
9.27
Ro = 30,000 cpm
't
= 250 J...lsec = 250Jl sec x
1 sec
1 min
x--10 Jl sec 60 sec
6
Ro
30000
Rt = 1- Ro't = 1- (30000) x 4.17 x 10-6
4.17 x lQ-6 min
= 34,286 cpm
Finding the fraction of counts lost:
1 - 30000 ｾ＠ 0.125
34286
12.5% of the counts are lost.
9.28 The fact that the gas multiplication in a proportional counter is very much
less than that in a Geiger counter means that a pulse amplifier for use with a
proportional counter must have a lower input sensitivity than one used with a
9.28
232
THE HEALTH PHYSICS SOLUTIONS MANUAL
Geiger counter. Calculate the input sensitivity for an amplifier to be used with a
2 in. dia. hemispherical windowless gas-flow proportional counter whose capacitance is 20 J..lJ..lF and which is operated to give a gas amplification of 5 x 103 •
Assume that the output pulse is "clipped" to one-half its maximum height.
Use equation 9.1:
12
C = 20 J..lJ..lF = 20 x 10- F
Q=
5xl0 3 ionpairs
event
1.6xl0- c
16
x..
=
8 x 10- C
IOn paIr
19
Q
8 X 10-16 c
-5
v=-=
=4xlO
V
C 20x 10-12 F
Since the pulse is "clipped" by 50% ,
4 x 10-5 V x 50% = 2 x 10-5 V
9.29
9.29 What is the sensitivity of a thermal neutron detector whose volume is 50
3
cm , and is filled with 96% enriched BF3 to a total pressure of 70 cm Hg? (Assume the temperature is 293 K).
Use the Ideal gas law to find the number of moles of BF3 gas in the detector:
1 atm
P =. 70 cm Hg x 76 cm 'H g - 0.921 atm
lL
v= 50cm x 1000cm
3
3
0.05 L
L'atm
R=0.082 - - mole·K
T= 293 K
Substituting in the values:
PV
-
RT
=n
=
0.921 atm x 0.05 L
L
=1.92 X 10-3 moles gas
0.082· atm x 293 K
mole·K
SOLUTIONS FOR CHAPTER 9
233
Calculate the number of atoms of lOB in the detector, since the thermal neutron
cross section for absorption for the other atoms in the gas is orders of magnitude
smaller:
0.96 moleslOB 6.02 x 10 23 atoms lO B
N = 1.92 x 10- moles gas x
x -----:-:--1 mole gas
1 molelOB
N = 1.1 x 1021 atoms lOB gas
3
The cross section for thermal neutron absorption for lOB can be found in Chapter
9 as:
(j
= 4010 b = 4010 X 1024 cm2/atom
Sensitivity is defined by equation 9.18,
CR
Sensitivity =
T
and is calculated with equation 9.17
Ncr O = CR
1.128
<I>
Substituting:
2
Ncr
1.1 X 10 21 atoms x 4010 x 10-24 cm
Sensitivity = 1.128 =
atom
1.128
Sensitivity = 3.95
ｾ
ｰｳ＠
neutron
cm 2 ·sec
9.30 If the BF3 tube of problem 29 is used as a current ionization chamber, what
saturation current would result from a thermal flux of 109 neutrons per cm2/sec ?
Calculate the amount of energy released by the thermal neutrons per neutron
reaction:
9.30
234
THE HEALTIf PHYSICS SOLUTIONS MANUAL
M COB) + m (n) =M (7Li) + M (4He) + Q
The nuclear masses of each isotope is used in this mass energy balance equation.
10.016114 (B-10) + 1.008982 (neutron) = 7.018185 (Li-7) + 4.003873 (He-4)
Q = 11.025096 - 11.022058 = 3.038 x 10-3 AMU/reaction
931 MeV = 1 AMU
3
3.038 x 10- AMU x
931 MeV
AMU = 2.83 MeV/reaction
7Li emits a 0.48 MeV capture gamma 6% of the time, which is assumed to escape
the chamber, so the average kinetic energy released per neutron capture is::
2.83 MeV x 0.94 = 2.66
(2.83 - 0.48) x 0.06 =0.141
0.141 + 2.66 = 2.8 MeV average kinetic energy imparted by each thermal
neutron reaction. This energy will be expended in producing ions in the BF3 gas
at a rate of 1 ion pair per 35.6 eV.
From problem 9.29:
21
N = 1.1 X 10 atoms
ih
9
2
'I' = 10 neutrons/cm /sec
aD =4010 barns =4010 x 10-24 cm2
Equation 9.17 is used to fmd. the neutron reaction rate, RR, in the detector:
21
Ncr 0 '" 1.1 X 10 atoms x 4010 x 10-24 cm 2 0 9 neutrons
RR= - - ' f ' =
xl
2
1.128
1.128
RR = 3.91 x 10 reactions/sec
9
Equation 9.10
cm / sec
SOLUTIONS FOR CHAPTER 9
RR reactions x E eV
x 1.6 x 10-19 ｾ＠
I =
sec'
reaction
ion x 1
w(eV / ion)
A!
235
C
s
3.91 X 10 9 reations x 2.8 x 10 6 eV x 1.6 x 10-19 ｾ＠
x 1ｾ＠
sec
.
reaction
ion
C/ s
ｉ］Ｍｾ＠
35.6 ｾｖ＠
IOn
9.31 How long would it take for the sensitivity of the BF3 detector of problem 30
to decrease by 10%?
9.31
From problem 9.29:
21
10
1.1 x 10 atoms B
9
3.91 x 10 neutron interactions/sec
For the sensitivity to decrease by 10%, the number of lOB atoms must decrease
by 10%, that is
21
20
10
1.1 x 10 atoms x 0.1 = 1.1 x 10 atoms B must be consumed by neutron
captures.
9
At the reaction rate of 3.91 x 10 reactions/second, the time to produce 1.1 x 1020
reactions is
20
T
1.1 x 10 reactios
-------::---------= 28. x 1010 sec
9
3.91 x 10 reactions / sec
10
1hr
1d
1yr
T= 2.8 x 10 sec x 3600 sec x 24 hr x 365 d = 892 yr
9.32 What is the sensitivity for 1 MeV and for 10 MeV neutrons (amps per
2
neutron per cm /second) of an ion chamber that is filled with CH4 gas to a .
pressure of 760 mm Hg, if its volume is 500 cm3 ?
Calculate the number of H atoms in the ion chamber:
9.32
236
THE HEALTII PHYSICS SOLUTIONS MANUAL
P =760 mm Hg = 1. atm
V = 500 cm3 = 0.5 L
T = 20°C = 293 K
R = 0.082
PV
n=-
RT
L'atm
I K
moe·
ｯＮｾ＠
LxI atm
= 0.0208 moles CH in the ion chamber
4
0.082 . atm x 293 K
.
mole·K
Since CH4 has 4 atoms of hydrogen per molecule,
·.
1.25 x 10n x 4 = 5.02 x 1On hd
y rogen atoms
are ill the chamber.
Find the neutron capture cross section for each of the energies. Carbon is not
considered important in this problem since it is not as effective at transferring
energy due to its comparatively smaller cross section. We cannot use equation
5.53 to determine the cross section at energies higher than 1000 eV (see pages
157,380), since capture energies tend to become flat at very high energies. The
cross section can be found in the "Barn Book", 1955 or from Brookhaven
National Laboratories database.
An average of 50% of a neutron's energy is lost in each collision with H, and
assuming that each neutron only collides once with a hydrogen atom. We may
assume that only one collision occurs is because the mean free path of a neutron
is significantly larger than the chamber:
MFP = _ _ _ _ _,--1_ _ _-:-:-_ _
cm 2
5.02 x 1022 atoms
45xl0-24
.------atom H 500 cm3 chamber
2213 cm
Using equation 9.17 to determine the number of neutrons per second reacting
(Note that the beams are monoenergetic, so a correction for the most probable
energy is not required):
SOLUTIONS FOR CtJAPTER 9
RR
7
't'
ｾ］ＰＮＲＶ＠
RR
237
2
cm
= 5.02 x 10 atom x 4.5 x 10- - atom
22
24
reacti%.ns
sec
neutron
cm 2 • sec
w = 28.2 eV/ion pair in CH4 (Attix)
-
6
E = 1 MeV = 1 x 10 e V
Using equation 9.9 and accounting for only 50% of the neutron's energy being
transferred, the current is,
RR reactions E
eV
1.6 x 10-19 ｾ＠
x1 A
IOn
C/ sec
sec
reactions
1= 0.5 x
eV
wion
reacti%nS
x 1A
0.226 sec
x 05 x 10 6 eV x 1.6 x 10-19 ｾ＠
neutrons
reaction
IOn C I sec
2
1= Ｍｾ＠
__ｾ］Ｍ cm • sec
28.2 ｾｖ＠
IOn
16
I (amps) = 6.4 x 10- amp per neutron per cm2
For 10 MeV neutrons, a similar calculation is performed:
Using equation 9.17 to determine the number of neutrons per second reacting
(Note that the beams are monoenergetic, so a correction for the most probable
energy is not required):
238
THE HEALTII PHYSICS SOLUTIONS MANUAL
reacsti%nS
2
cm
7= 5.02 x 10 atom x 1 x 10- - - = 0.0502
'I'
atom
CR
22'
24
neutron
cm 2 • s
w =28.2 eV/ion pair CH4
Using equation 9.9 multiplied by 50% (as explained earlier):
RR reactions E eV 1.6 x 10- 19 ｾ＠
x 1A
I = 0.5 x
s
reaction
ion Cis
w(eV I ion)
ｲ･｡｣ｴｩｾｮｳ＠
0.0502
s
ｉ］Ｍｾ
t
neu rons
cm 2 • s
x 5 X 10 6
eV x 1.6 x 10-19 ｾ＠
x 1A
reaction
IOn Cis
28.2 ｾｖ＠
IOn
15
2
I =1.4 X 10- amp per neutron per cm per second for 10 MeV neutrons
9.33 A 1000 MBq (27 mCi) 60CO source is lost. At what distance can the lost
source be detected with a survey meter whose sensitivity is 0.013 J.lC/kg per hour
(0.05 mRlhr) above background?
x= 0.05 mRlhr = 0.05 x 10- Rlhr
3
Ci
2
A =27 mCi = 27 x ｾＰＭＳ＠
I (amps) =1.4 x 10- amps per neutron per cm per second for 10 MeV neutrons
R·m 2
r = 1.32
(table 6.3)
Ci·hr
Equation 10.1:
.
f'A
x=,d
2
0.05 x 10-3 R I hr
= 26.7 meters
SOLUTIONS FOR CHAPTER 9
239
9.34 The thennal neutron flux from a moderated 252 Cf neutron calibration source
is determined by irradiating a gold foillem diameter x 0.013 cm thick for a
period of 7 days at a distance of 100 em from the source. The foil was counted
immediately after the end of the irradiation period, and found to have an activity
of 100Bq (2.7 nCi).). What was the thennal flux at the point where the foil was
irradiated? The activation cross section for gold is 98.5 barns.
Volume of the gold foil:
1cm
2
t= 0.013 em
r=--=0.5em
v = nr t = n x 0.5 x 0.013 = 0.0102 cm3
2
2
P = 19.32 g/cm3
Calculating the number of atoms of gold in the foil:
23
3 19.32 g 1 mole 6.02 x 10 atoms
mole
= 6.03 x 1020 atoms in foil
0.0102 cm x em3 x 197 g x
Use equation 5.59:
AN = 100 sec· l
24
2
() = 98.5 barns = 98.5 x 10- cm /atom
20
n = 6.02 x 10 atoms in the foil
t = 7 days
In(2)
0.693
A -- -t1l2- = 2.698 days = 0.257 d- l
2
cm
100 sec-I = <I> x 98.5 X 10-24 - - x 6.03 x 102°atoms in foil x ( 1 - e-{)·257X7)
atom
3
<I> = 2 X 10
2
neutrons/cm /sec at the 100 em irradiation point
240
9.35
THE HEALTH PHYSICS SOLUTIONS MANUAL
9.35 A thermal neutron counter 1 cm diameter X 10 cm long is filled with BF3
gas at atmospheric pressure and 20°C. What is the counting rate when the
counter is in a thermal nOeutron flux (Emp = 0.025 e V) of 1000 neutrons per square
cm per second?
°
The fluorine atomic cross section for thermal neutrons (Emp = 0.025 eV) is so
small (0.01 barns) compared to the cross section for boron (753 barns) that only
boron will be considered for this problem (cross sections from NRHH). Calculate
the number of moles of gas in the counter, using the ideal gas law:
P = 1 atm
VCY1 = (length x (diameter/ x : = 10 cm x 12 cm2 x 0.785 = 7.85 cm3
3
1L
VCY1 =7.85 cm x 1000 cm 3
3
=7.85 x 10- L
L·atm
R=0.082 - - mole·K
T= 20°C = 293 K
1 atm x 7.85 x 10-3L
-4
n = RT =
L. t
= 3.26 x 10 moles gas in the counter
0.082
am x 293 K
mole·K
PV
Converting moles gas into atoms of boron in the counter:
N = 3.26xlQ-4 molex
<I>
6.02 x 10 23 molecules
mo
1
= 1000 neutrons/cm21 sec
(J' =
753 b for thermal neutrons for boron
x
1 atom B
molecule
= 1.96 X 1020
SOLUTIONS FOR CHAPTER 9
Since thermal neutrons are used, equation 9.17 is used to find the number of
counts per second;
ｎｾ｣ｲ＠
0
CR = 1.128
1 96 10 20
B 1000 neutrons 753 b 1 x 10-24 cm 2
•
x
atoms x
2
x
X -----
= ________ＭＢＧ｣］ｭＯｳ･｟ＮＺ［｡ｴｯｂｬ｢ｾ＠
1.128
CR = 132 cps
241
Solutions for Chapter 10
EXTERNAL RADIATION PROTECTION
7
10.1 A Po-Be neutron source emits 10 neutrons per second, of average energy 4
MeV. The source is to be stored in a paraffm shield of sufficient thickness to
2
reduce the fast flux at the surface to 10 neutrons per cm /sec. Consider paraffin to
be essentially CH 2 (for the purpose of this problem) and to have a density of 0.89
3
g/cm.
(a) What is the minimum thickness of the paraffin shield?
The unshielded fast flux at a distance equal to the shield thickness is
n
<I>
2
cm ·sec
s isec
=
471:[t cmf
With a shielding thickness t and a buildup factor B, <p at the shield's surface is
The attenuation coefficient, Jl, for fast neutrons from a Po-Be source is 0.126 cmI (RHH), and B for thick (20 cm) hydrogenous shields is 5. Substituting these
values into the shielding equation gives
10
n
2
cm • sec
=
5x 10 7 nl
7 sec e-O.l 26t
471:[t cmf
This equation is most easily solved by assuming a value for t, and then calCulating the resulting <p . This process is repeated until the value for t is found that
gives the desired <p. By this method of iteration,
243
10.1
244
THE HEALTH PHYSICS SOLUTIONS MANUAL
2
t = 44 cm results in <P = 8 nlcm /sec, which is close to the desired maximum of
Ｈ｢ｾ＠
If the slowing down length is 6 cm, the thermal diffusion length is 3 cm, and
the diffusion coefficient is 0.381 cm, what will be the thermal neutron leakage at
the surface of the field?
The thermal neutron leakage is given by equation 5.51:
ｾ＠
Se- R/ L
= 41tRD
Where
7
S = 10 nlsec
R=44cm
D = 0.381 cm
L=3cm
Substituting these values into the equation, we have
ｾｉｨ＠
10 7 ｾ＠
=
sec
e
4 x 1t X 44 cm x 0.381 cm
44 em
3em
n
ｾ＠ Ih = 0.08 -cm---=--2s-e-c
(c) What is the gamma ray dose rate, due to the hydrogen capture gammas, at
the surface of the shield?
The capture of a thermal neutron by a hydrogen atom results in the emission of a
2.26 MeV gamma ray (example 10.11). The paraffm, therefore, acts as a
distributed source of gamma radiation. The surface dose rate due to the hydrogen
capture gammas in the paraffin shield is calculated with equation 10.37: This
equation treats the hydrogen capture gammas as if the paraffin shield contained a
uniformly distributed gamma emitter.
SOLUTIONS FOR CHAPTER 10
245
The concentration C of the gamma emitter is equal to the difference between the
source emission rate, S, and the neutron escape rate from the surface, divided by
the volume of the shield
10 n _ 4rc(44 cm)2(8 + 0.08) ｾ＠
Y/
"B "
lsec
C=
s
cm ·s =27S
= 27S-q3
4
cm 3
-rc(44)3
cm
3
C = 27.5 X 10-6 "MBq"/cm3
7
r is calculated using equation 6.5 and 6.17
2
Gy
(Cjk )m
g
·x 10 4 cm x 34-r = 3.69 x 10- x 2.26 MeV
2
MBq·h
ｾ＠
Cjkg
2
9
2
r = 2.8 x 10-3 Gy· cm = 2.8 mGy· cm
MBq·h
2
MBq·h
The energy absorption coefficient for paraffin for 2.26 Me V gamma's is calculated by interpolating between 2 and 3 MeV data for water (Table 5.3): ｾ･ｮＨＲＮＶ＠
2
MeV, H20) = 0.0252 cm /g. Since H 2C == H 20 for gamma energy absorption, we
calculate the linear energy absorption coefficient for paraffin with !lm for water.
!ll = !lm x P
2
!ll = 0.252 cm /g x 0.89 g/cm3 = 0.022 cm- 1
246
THE HEALTII PHYSICS SOLUTIONS MANUAL
Inserting these values into the dose rate equation, we have
. 1 27.5xl0-{;MBq ＲＮＸｭｇｹｾ｣＠
D= - x
x
3
cm
2
x
hr·MBq
4x1t
(
1- e-O·022x (44)
)
0.022cm-1
. _ 0.014 mGy = 1.4 mrads
Dhr
hr
10.2
10.2 An x-ray therapy machine operates at 250 kVp and 20 rnA. At a target to
skin distance of 100 em, the exposure rate is 20 Rlmin. The workload is 10,000
rnA min/week. The x-ray tube is constrained to point vertically downward. At a
distance of 4 m from the target is an uncontrolled waiting room. Calculate the
thickness of lead to be added to the wall if the total thickness of the wall (which
is made of hollow tile and plaster, density 2.35 glcm3) is 2 in.
Since the primary beam is directed downward, the wall shields only scattered
and leakage radiation. We calculate the shielding requirement separately for each
of these. If they differ by < 1 TVL, the thicker one. is adequate. IF they differ by
< 1 TVL, then we add 1 HVL to the thicker barrier.
The scattered radiation attenuation requirement is calculated with equation 10.23:
K = ｾＨ､＠
ux
aWT
sea
)2(d sec )2 400
F
where
P =0.002 R (uncontrolled area)
W = 10,000 rnA min/week @ 1 meter
a = ratio of scattered intensity, at 1 m and 90° scatter, to primary intensity =
0.0019 (from table 10.3)
T = Occupancy Factor = 1 (table 10.1)
dsea = 1 m
d =4m
ｆ･ｾ＠ area (assumed in this case) of beam in scattering plane =400 cm2
SOLUTIONS FOR CHAPTER 10
247
Substituting these values into the equation, we find the attenuation factor to be
K ux = 1.68 X 10- RlmA·min == 1 meter.
3
Use figure 10.14 on the 250 kVp curve to locate the required thickness of lead,
which is 4.5 mm.
The leakage radiation from a therapy housing:S; 500 kV = 1 Rlhr at 1 meter. The
required attenuation factor to give an average exposure of P Rlwk at a distance
of d meters from a tube whose beam current is I rnA, is given by equation! 0.29:
2
B
LX
Pxd x 601
=-----
WT
Substituting the respective values into the equatIon, we have
,.' The number of HVL's required to attain this value is:
j,i
.• ?!'
3.84 X 10-3 = ｾ＠
2n
n=8
One HVL (Pb, 250 kV) = 0.88 mm (Table 10.2), therefore,
the required thickness = 8 HVL x 0.88 InIivHVL = 7.04 mm Pb
The difference between the barrier thickness for leakage and for scattered
radiation is
dt = 7.04 mm - 4.5 mm = 2.54 mm
Since 1 TVL = 2.9 mm Pb (Table 10.2) > 2.54 mm, we add 1 HVL to the greater
thickness inaccordance with NCRP49. The total required thickness is
t
=7.04 mm + 0.88 mm = 7.92 mm Pb = 9 HVL's
The wall thickness is 2" (5.08 cm) plaster. Since the plaster has the same density
as concrete, we may use the HVL of concrete to fmd the wall equivalent
THE HEALTII PHYSICS SOLUTIONS MANUAL
248
T(wall) =
5.08 cm = 1.81 HVL
2.8cm/HVL
The additional lead requirement is
T(Pb) =9 HVL - 1.8 HVL = 7.2 HVL
T (Pb) = 7.2 HVL x 0.88 mmlHVL = 6.3 mm Pb
From Table 10.5, we find that 6.35 mm, or one quarter inch lead sheet which is
commercially available.
10.3
10.3 A 7.4 x 10
13
Bq (2000 Ci) 60CO teletherapy unit is to be installed in an
existing concrete room in the basement of a hospital so that the source is 4 m
from the north and west walls - which are 30 in. thick. Beyond the north wall is a
fully occupied control room. Beyond the west wall is a parking lot. The useful
beam is to be directed toward the north wall for a maximum of 5 hr per week.
The useful beam is to be directed at the west wall 1 hr per week. Considering
only the radiation from the primary" beam, how much additional shielding, if any,
is required for each of the walls?
Control Room P = 100 mrems/wk
30"(0.76 m)
Public Parking Lot
P = 2 mrems/wk
5 hr/wk
30"(0.76 m)
1 hr/wk
ｾＭｴＢ＠
4m
4.76 m
ＭｾＧ
I
Consider the 60CO to be a point source. Use equation 10.1 to determine the
exposure rate unshielded at the boundary:
d =4 m + 0.76 m = 4.76 m
2
R·m
lco-6o = 1.32 Ci. hr (table 6.3)
SoumoNS FOR CHAPTER 10
249
A = 2000 Ci
2
.
Ar
X = -2 =
d
116.4 R
hr
problem)
2000Ci x (1.32 R· m
Ci·hr
(4.76 m)
2
J
= 116.5 R / hr
5 hr
x -wk = 582 Rlwk (Directed to control room wallS hrs/wk as stated in
According to p.434, 5000 mreml50 weeks a year, 0.1 Rlwk is allowed. Also
according to NCRP 49, page 6, the maximum permissible occupational exposure
(controlled room) is 0.1 Rlwk. Assume 100% occupancy for the control room.
The broad beam HVL for 60CO photons of concrete in Table 10.2 is used to
calculate the broad beam attenuation coefficient;
•
I-l = 0.693 = 0.693 = 0.1 12 cm-1
HVL
6.2 em
Since the attenuation coefficient is for broad beam geometry, no buildup factor is
required.
Calculate the shielding required under conditions of broad beam geometry, the
thickness required for the north wall (control room) with equation 5.19;
I-.l = 0.112 cm3
10 = 582 Rlwk = 582 x 10 mRlwk (as calculated above)
1= 100mRIwk
1
In
In(:J [ＵＸＲｘＱＰＳｾ＠
mR]
100wk
t=--- =
-0.112 cm- I
= 77.4 em concrete needed for the north wall.
Since we need 77.4 em, and 76 em is installed, 1.4 em of concrete is needed for
the control room.
250
THE HEALTH PHYSICS SOLUTIONS MANUAL
#HVL=
thickness
HVL
1.4 cm concrete
- 0.225 HVL required (Table 10.2)
6 .2 cm concrete HVL
0.225 HVL of lead is 3.1 mm thick for 60CO photons (Table 10.2). The next
available commercial lead thickness (Table 10.5) is 1/8" sheet, which must be
added to the control room wall shielding.
For the parking lot; The maximum permissible dose for uncontrolled areas
(parking lot), is 100 mremlyear (2mremlwk), according to 10CFR20.1301.a.1.
NCRP Report 102 indicates a limit of 2 mremlwk as well for the general public.
An occupancy factor of 114 from table 10.1 is also applied, so that the maximum
exposure in the parking lot must not exceed:
mrem 4
mrem
limit thus applies to the parking lot
x -(Occupancy factor) = 8
wk
1
wk
wall.
2
Calculate the shielding required under conditions of broad beam geometry, the
thickness required for the parking lot with equation 5.19;
1 =1oe-J.1I
!l = 0.112 cril- 1 (from fIrst part)
R·
3 mR
R 1hr
1 = 116.5-x-=116.5-=116.5x10 (one hour towards this wall, as
o
hr wk
wk
wk
stated in problem)
1= 8 mremlwk
(J
In[
ＸＺｾｭ＠
rr:
f fi
J
In 1
116.4 x 103
t = __/...:..0_ = _ _ _ _ _ _.:.:..::.:_ = 85.6 cm concrete needed for the parking
-0.112 cm-1
lot.
Since 76 cm is present, 85.6 cm -76 cm = 9.6 cm concrete needed.
thickness
#HVL= HVL
9.6 cm concrete
6.2 cm concrete HVL
1.55 HVL needed
SOLUTIONS FOR CHAPTER 10
251
1.55 x 12 nun (Pb HVL) = 18.5 mm Pb needed. The nearest commercially
available lead sheet is 1" thick.
ｾｏＮＴ＠
10.4
9
A radiochemist wants to carry a small vial containing 2 x 10 Bq (50 mCi)
Co solution from one hood to another. If the estimated carrying time is 3 min,
what would be the minimum length of the tongs used to carry the vial in order
that his dose not exceed 60 llGy (6 mrad) during the operation?
By definition, 6 mrads gamma radiation = 6 mrems. For radiation safety purposes, 6 mR = 6 mrems. Therefore, 6 mrads gamma = 6 mR. According to
equation 10.1
2
r R·m x A Ci
i-J =
Ci· hr
(dmf
For a 3 minute exposure time, the radiation exposure rate is
6 mrems = i-J mrems/hr x 3/60 hr
i-J = 120 mrems/hr = 0.12 remlhr
Substituting
i-J = 0.12 remlhr
R·m 2
(Table 6.3)
r CO- 60 = 1.32.
CI·hr
A = 0.05 Ci (50 mCi)
into the equation above, and solving for d, we find
2
d =
1.32 R· m x 0.05 Ci x 1 rem
Ci· hr
R =0.74 m =74 cm
0.12 rem
hr
,\
252
10.5
THE HEAL1H PHYSICS SOLUTIONS MANUAL
10.5 A viewing window for use with an isotope that emits I-MeV gamma rays is
to be made from a saturated aqueous solution of KI in a rectangular battery jar.
What will be the attenuation factor, assuming conditions of good geometry, if the
solution thickness is 10 cm, and if the glass walls are equivalent in their attenuation property to 1 mm lead? A saturated solution of KI may be made by adding
30 g KI to 21 mL water to give 30 mL solution at 25 °e. Total attenuation cross
sections for 1 MeV gamma rays for the elements in the solution are:
The attenuation factor is calculated from equation 5.19:
The linear attenuation of the KI solution is calculated with the aid of equation
5.23
lh
where N. is the concentration of the i atom or molecule per cc, and 0'. is its cross
section. For the components of the solution, we have:
I
I
K (A = 39.100) = 4 barns
H (A = 1) = 0.2 barns
I (A= 126.91)= 12 barns
o (A = 16) = 1.7 barns
KI (MW = 166.01) = 16 barns
H 20 (MW = 18) = 2.1 barns
The concentrations, N., are
I
30 g
= 6.02 X 10-3 mol KI
cm 3
KI: 166.01 g/mol x 30 cm 3
21
.
2 molH/
----g---:-=3.89xl03
H20: 18 g/mol x 30 cm
cm
°
23
Since there are 6.02 x 10 molecule/mole and 10-24 cm2Jb
III (solution) = 6.02xlQ23x lO-24(6.02xl0-3xI6+3.89xl0-2x2.1) = 0.107 cm- l
III (Pb, 1 MeV) = 0.771 cm- 1 (Table 5.2)
ｾＮ＠
253
SOLUTIONS FOR CHAPTER 10
ｾ＠
= e --0.107 em· 1 xlO em X e -0.771 em· 1 xO.! em = 0.32
10
10.6 Lead foil consists of an alloy containing 87% Pb and 12% Sn, 1% Cu. Its
specific gravity is 10.4. If the mass attenuation coefficients for these three
elements are 3.50,1.17, and 0.325 cm2/g respectively, for X-rays whose wavelength is 0.098 angstroms
(a) calculate the mass and linear absorption coefficients for lead foil,
Mass;
Il mass ｾｊ［＠
Il i
fraction present, f
0.87
0.12
0.01
EleIrent
Pb
Sn
Cu
Il (linear) = ｐｾｊ［＠
Il i = ｐｾｊ［＠
Attenuation coef.,jJ
3.5
1.17
0.325
Sum
fx M
3.045
0.1404
0.00325
3.19 cm2/g
Il (mass)
2
.1
g
cm
Il (linear) = 10.4 - - 3 X 3.19 - - = 33.2 cm
g
cm
(b) What thickness of lead foil would be required to attenuate the intensity of
57CO gamma rays E(gamma) = 0.123 MeV, by a factor of 25?
Equation 5.19;
Il = 33.16 cmIlIa = 1/25 = 0.04
I
I{:J In(O.04) =
t.= - - =
-33.12 cm- I
-I-l
0.097 cm
10.6
254
10.7
THE HEALTH PHYSICS SOLUTIONS MANUAL
10.7 A hypodennic syringe that will be used in an experiment in which 90Sr
solution will be injected has a glass barrel whose wall is 1.5 mm thick. If the
3
density of the glass is 2.5 g/cm • how many mm thick ｭｾｓｴｗ･｡ｫ＠
a Lucite
sleeve that will fit aroull<nlie syringe if no beta particles are to come through the
Lucite? The density of Lucite is 1.2 g/cm3 •
Strontium 90. which emits a 0.55 MeV beta. decays to 9'Y, which emits a 2.27
MeV beta (RHH). The shield must therefore be designated to stop the 2.27 MeV
beta. Using equation 5.2 to find the range of the beta:
R = 412(E MeV )1.265-0.09541n(E)
R = 412 X (2.27)1.265-0.o9541n(2.27)= 1090 mg/cm2 =1.09 g/cm2is the density thick...
ness required.
.-"
Already there is glass shielding pre·sent. whose density thickness is:
2.5g
2
O.15 cmx-= 0.375 g/cm
3
cm
Calculating the amount of additional shielding required:
2
1.09 g/cm2 - 0.357 g/cm = 0.715 g/cm2 needed.
3
Lucite is 1.2 g/cm • so calculating the additional thickness required:
3
0.715 g x cm x 10 mm
.
2
1
2
1
=
6
mm
Luclte
cm
. g
cm
13
10.8 A room in which a 7.4 x 10 Bq (2000 Ci) 137Cs source will be exposed has
the following layout. Calculate the thickness t of concrete so that the exposure
rate at the outside surface of the wall does not exceed 0.64 I-lC/kg (2.5 mR) per
hour.
SOLUTIONS FOR CHAPTER 10
255
Consider the 137Cs to be a.point source. Use equation 10.1 to determine the
exposure rate unshielded at the boundary:
d=10ft=3.048m
R·m 2
r Cs-137 = 0.33.
(table 6.3)
Cl' hr
A = 2000Ci
.
Ar
X = -2 =
d
2000Ci x (0.33 R· m2)
Ci·hr
.
- 71 RIhr is the exposure rate unshielded.
(3.048 my
The desired exposure rate is:
2.5 mR
1R
hr x 1000 mR = 0.0025 Rlhr
Determine the required attenuation;
ＰＮＲＵｾ＠
Required fraction attenuation =__-.. . ::::h:;:..r = 35 x 10-5
ＷＱｾ＠
hr
The broad beam HVL of concrete for 137Cs is listed as 4.8 cm (Table 10.2). The
number of HVL's needed to reduce the transmitted radiation by this factor is
-log (3.5 x 10-5 )
n=
log 2
=14.8
Therefore, the required thickness = 14.8 HVL's x 4.8 cmIHVL = 71 cm = 28 in.
256
THE HEALTIl PHYSICS SOLUTIONS MANUAL
Alternatively, if the broad beam attenuation data for the nuclide of concern are
not available, the following procedure may be utilized.
Calculate the shielding required under conditions of "good" or collimated
geometry first, to obtain an estimate of the minimum thickness required with
equation 5.19;
and then recalculated using a buildup factor, B, equation 10.17
IIIo = Be-Ill
l
/1= 0.184 cm- (table 5.2)
10 =71 Rlhr
1= 0.0025 Rlhr
(1;I)
In(0.0025 R / hr)
In- - - =
71 R / hr
=55.7 cm concrete needed under conditions of
t=
l
-0.184 cm"good" geometry.
Now estimate the required thickness when the buildup factor is included by
adding one Half Value Layer (HVL) to the value for thickness under conditions.
of "good" (collimated) geometry. The HVL for concrete is calculated
HVL = ·0.693 =
/.l
0.693 = 3.8 cm
0.184 cm- l
55.7 + 3.8 = 59.5 cm as a first estimate of the thickness required.
The number of relaxation lengths, pt, for this thickness will be:
pt = 59.5 x (0.184) = 11 relaxation lengths.
In the absence of specific data for B, B may be estimated as
SOLUTIONS FOR CHAPTER 10
257
Recalculating the exposure rate using equation 10.17,
Ilt = 11
t = 59.5 cm,
. B = 1 + 11 = 12
10= 71 Rfhr
( R)
R
1 = BI e- JlI = 12 x 71- x e-11 = 0.0142o
hr
hr
This is approximately six times higher than the required 0.0025 Rfhr. Add 3
more HVL's,
59.5 + 3(3.8) = 71 cm as a second estimate of the thickness required.
The number of relaxation lengths, Ilt, for this thickness is:
Ilt = 71 x (0.184) = 13 relaxation lengths.
B = 1 + Ilt = 14
Recalculating the exposure rate using equation 10.17,
Ilt =13
t = 71 cm,
B= 14
10= 71 Rfhr
( R)
R
1= BI e- JlI = 14 x 71- x e- 13 = 0.0023o
hr
hr
Thus, the shield is 71 cm or 28 inches concrete, which is the same value that was
calculated using the broad beam HVL's in Table 10.2 or NCRP 49.
10.9 What minimum density thickness must a pair of gloves have to protect the
hands from 32p radiation?
Maximum p energy for 32p is 1. 708 MeV
Input this into equation 5.2 to find the maximum range of the p;
10.9
258
. THE HEALTIJ PHYSICS SOLUTIONS MANUAL
R = 412( E MeVr265-0.0954xln(EMeV)= 412 x (1.708 MeVr26S-0.0954Xln(1.708MeV)= 789 ｭｾ＠
cm
10.10
10.10 When a radium source containing 50 mg Ra encapsulated in 0.5 mm thick
Pt is placed into a Pb storage container, the measured exposure rate at a distance
of 1 m from the source is 5.41 J-LC/kg (21 mR) per hour. If this same container is
used for storing 137Cs, how many MBq may be kept in it for a period of 4 h
without exceeding an exposure of 10.3 J-LC/kg (40 mR) at a distance of 50 cm
from the source?
.
Ar
X=d2
d= 1 m
2
R·m
rRa_226=0.825.
(table 6.3)
Cl· hr
1g
x 1 Ci =50x10-3 Ci
A -50mgRax
1000 mg 1 g Ra
. Ar
X = -2 =
d
R m2
50 x 10-3 Ci x (0.825__.. _ _ )
Cl·hr
(1 m)2
= 0.041 Rlhr =41 mRJhr is the exposure
rate unshielded.
The reduction in intensity due to the lead shield is
21mR
1 =
hr =0.5
10 41 mR
hr
Thus, one half value layer (HVL) for Ra is provided by the shielding. Since
broad beam geometry is involved, fmd the HVL for radium broad beam geometry in Table 10.2 as 1.66 cm of Pb. Thus, 1.66 cm of Pb is the thickness of the
lead in the shield.
SOLUTIONS FOR CHAPTER 10
259
We want the shielded 137Cs exposure rate to be 40 mRl4 hr = 10 mRfhr.at 0.5 m.
The unshielded gamma ray intensity of the stored I37Cs is calculated with equation 5.19, using the broad beam HVL from Table 10.2 (or NCRP 49) to determine the broad beam attenuation coefficient of the lead storage container.
= 0.693 =
HVL
!l
0.693 = 1.066 cm- 1
0.65 cm
Substituting into the attenuation equation, we have
rnR
-1
10_=Ioe-l.066cm
x1.66cm
hr
10 = 58.7 mRlhr
Use equation 10.1 to determine the activity;
d= 0.5 m
r CS- 137
= 330
rnR . m2
C· h (table 6.3)
1·
r
A=?
X = 58.7 mRlhr
. Ar
x=d
2
Solving for A (activity);
A - Xd
- r
2
58.7 X (0.5)2 = 0.043 Ci
- 330 mR·m 2
Ci·hr
_
3.7 x 104 MBq
A =0.043 Ci x
Ci
= 1.6 x 103 MBq
What is the maximum working time in a mixed radiation field consisting
of 6 mC/kg (20 rnR) per hour gamma, 40 ｾｇｹ＠
(4 mrad) per hour fast neutrons,
and ＵＰｾｇｹ＠
(5 mrad) per hour thermal neutrons, if a maximum dose equivalent of
3 mSv (300 mrem) has been specified for the job?
10.11
10.11
260
THE REALTII PHYSICS SOLUTIONS MANUAL
Since this problem involves legal units (rad, rem), for regulatory purposes, let 1
R= 1 rem.
The quality factors for converting rads to rems can be found in Table 7.8 and in
10 CFR20, Table 20.1004(b).1, (1994).
Qgamma =1
Qfast n = 10
Qthermal n
=2
Multiplying the quality factor times each type of radiation to obtain rems:
20 mradslhr x Qgamma = 20 x 1 = 20 mremslhr
4 mradslhr x Qfast n = 4 x 10 =; 40 mremslhr
5 mrads/hr x Qthermal n = 5 x 2 = 10 mremslhr
Total = 70 mrems/hr
Since only 300 mrems are allowed,
300 mrems x
10.12
1 hr
= 4.29 hours stay time allowed, or 4 hours 17 minutes
70mrems:
.
10.12 Maintenance work must be done on a piece of equipment that is 2 meters
from an internally contaminated (with 137CS) valve. The exposure rate at 30 cm
from the valve is 500 Rlhr (0.13 C/kg-hr). If 4 hours is the estimated repair time,
what thickness of lead shielding is required to limit the dose equivalent of the
maintenance persons to 1 mSv (100 millirems)?
Determine the exposure rate at 2 ·meters without shielding (equation 9.20):
II = 500 Rlhr
d l =30cm =0.30m
d2 =2m
2
500 X (0.3 ) _
RIh
22
-11.25
r
1 R of exposure = 1 rem of dose equivalent gamma rays for radiation safety
purposes.
SOLUTIONS FOR CHAP1ER 10
261
Since the repairs should only take 4 hours, and the maximum dose allowed is
100 mrem, the required dose rate after shielding is:
100 mrems
1 rem
---- x
= 0.025 remlhr
4 hours
1000 mrems
Use Table 10.2 to find the HVL for Cs-137 as 0.65 cm ｵｮ､ｾｲ＠
conditions of broad
beam geometry. Calculate the broad beam attenuation coefficient;
0.693
0.693
-1
!.l =HVL
- - = 065
= 1.07 cm
. cm
Equation 5.19 gives the shielding thickness:
10 = 11.25 remlhr
1 = 0.025 remlhr
!.l = 1. O7 cm
-1
I = I e-J.It
o
0.025 = 11.25 x e-1.07 x (t)
t= 5.7 cm
An alternative solution which could be utilized if broad beam attenuation curves
are not available but generalized curves of build up factors, such as figures 10.9
and 10.10 in the textbook are available.
Equation 5.19 gives an estimate for the shielding thickness under conditions of
"good" geometry:
10 = 11.25 remlhr
1 = 0.025 remlhr
1
Jl = 1.34 cm- (Table 5.2, for 0.661 MeV gamma and Pb)
1
HVL =0.69311.34 cm- = 0.52 cm
I = I oe-Ilt
0.025 = 11.25 x e-1.34X(t)
262
THE HEALTH PHYSICS SOLUTIONS MANUAL .
t =4.56 cm
Since the shield must be thicker than this to allow for buildup, add one half value
layer (HVL) to give 4.56 + 0.52 = 5.08 cm.
Now, find in Fig. 10.9 that for 0.66 MeV gamma in Pb, for
Jlt = 1.34 cm- l x 5.08 cm = 6.8, B = 2.
Using this increased thickness, with B = 2, in equation 10.17
I =IoBe- IJI
and solving for I, we find
1= l1.25x2xe-6.8 = 0.025,
which is the desired value. Thus, a Pb shield 5.08 cm (2 in.) thick is satisfactory.
This calculated thickness is in good agreement with the 4.56 cm thickness
previously calculated.
10.13
Calculate the exposure rate from a 100,000 MBq (2.7 Ci) 60Co "point"
source, at a distance of 1.25 meters, if the source is shielded with 10 cm Pb?
10.13
Equation 10.1 is used to first obtam ｴｾ･＠
exposure rate without shielding:
r e 6o = 1.32 (R-m2)1(Ci-hr) From Table 6.3
o-
A = 2.7 Ci
d= 1.25 m
x= r A2 = 1.32 x 2.7 = 2.28 Rlhr is the exposure rate unshielded.
d
(1.25)2
Next, calculate the shielded exposure rate, and account for the buildup factor by
using the broad beam HVL (Table 10.2) to calculate the broad beam attenuation
coefficient.
ｾ｢ｲｯ｡､＠
beam
0.693
0.693
l
= HVL = 1.2 cm Pb = 0.578 cm-
SOLUTIONS FOR CHAPTER 10
.263
10.14 A stainless steel bolt came loose from a reactor vessel. It is planned to pick
up the bolt with a remotely operated set of tongs and transport it for inspection
and study. The bolt had been in a mean thermal neutron flux of 2 X 10 12 nJ
2
cm ·sec for a period of 900 days, and will be picked up 21 days after reactor
shut-down. Calculate the gamma ray dose rate at a distance of 1 meter from the
bolt if the bolt weighs 200 grams and has the following composition by weight:
Fe
Ni
Mn
C
80%
19%
0.5%
0.5%
Check the NRHH to find which isotopes are produced and which are significant
and to obtain their cross sections for activation. In this alloy, the activated 56Mn
and 54Ni have very short half lives, and thus will have decayed away in 21 days.
4
Carbon is not activated CC emits only betas) and 55Fe emits no gammas. Iron 59,
9
TI12 = 44.6 days, which is produced by 58Fe (n,yi Fe is the only gamma emitter; it
emits a 1.1 MeV gamma in 56% and 1.29 MeV gamma in 44% of all its decays.
The isotopic abundance of 58Fe is 0.31 % and its activation cross section is l.28
barns (NRHH). Since the 59Fe half life is 44.6 days, saturation activity, equation
5.59, is attained:
A(sat) = 'AN = ｾ｣ｲｮ＠
n, the number of target 58Fe atoms, is calculated
80 g Fe 0.3158 Fe 1 mole 58 Fe 6.02 x 10 23 atoms 58 Fe
200
ngx
x
x
x-------_
100 g
100 g Fe
58 g 58 Fe
1 mole 58 Fe
atoms
Substituting these values into the activity equation:
A(sat) =
2 X 10 12 n
2
24
2
xl.28x10- cm x5.l5x10 2l atoms= l.32x 10 lOdps
cm ·sec
10
10
A(sat) = l.32 x 10 dps = l.32 X 10 Bq = l.32 X 104 MBq
After 21 days, the activity is
10.14
264
THE HEALTI:I PHYSICS SOLUTIONS MANUAL
_ 0.693 x2J d
A = 1.32 X 10 4 MBq e 44.6 d
3
A = 9.525 X 10 MBq
The dose rate at one meter from this activity is calculated with equation 10.1.
.
f'A
H=-2
d
f' = 3.65 x 10-9 "I.E. C/kg· m x ＳＴｾ＠
L....
MBq
2
I
I
C/kg
f' = 3.65 x 10-9 (0.561 x 1.1 + 0.44 x 1.29) x 34 = 1.4 x 10-7
Sv·m 2
.
MBq
2
l.4xl0-7 Sv·m x9.525 x l0 3 MBq
MBq·h
H=------"---------
(1
mr
H = 1.4 X 10-3 Sv = 1.4 mSv
10.15
10.15 A circular area 1 meter in diameter is accidentally contaminated with 10
MBq (270 f-lCi) l3lI. What is the maximum dose equivalent rate at a distance of 1
meter above the contaminated area?
The dose rate at a height of 1 meter is the sum of the gamma and beta dose rates.
The gamma dose rate is calculated with equation 10.10:
The area contaminated is:
A ; 'ItR2 = 'It x (0.5m)2 = 0.785 m 2
R = 0.5 m (radius of the circular area)
h = 1 m (height above circular area)
SOLUTIONS FOR CHAPTER 10
265
10 MBq = 12.73 ｍｾｱ＠
Ca = 0.785 m 2
m
X·m 2
r 1-131 = 1.53 X 10- MBq . hr (Table 6.3)
9
Using equation 10.10;
iI = 34 x 1t x 153 X 10-9 X· m
2
MBq· hr
•
x (12.73 MBq) x In((0.5 m)2 + (1 m)2J
m2
(1)2
-6
H = 0.46 x 10 Sv/hr = 0.46I1Sv/hr is the photon dose.
The beta dose is calculated next. Iodine emits 5 groups of betas. The mean value
of the maximum energies of these, Emax = 0.595 MeV, and the average energy of
all these betas is E = 0.195 MeV/transformation. The dose rate to the skin at a
distance d above the contaminated area is given by equation 6.31a.
3
E x e -11ad X e -1-1 t0007
D• b
= C x 1.3 x lOx
·
x!J. mrads/h
a t .
where
l1a = 16 (Em - 0.036r1.4 = 16 (0.585 - 0.036r I.4= 37.04 cm2/g (Equation 6.20)
Jl t = 18.6 (Em - 0.036r 1.37 = 18.6 (0.585 - 0.036r1.37 = 42.3 cm2/g (Equation 6.21)
Substituting C =
a
270/-lei
.
2 ' E = 0.195 MeV/transformatIOn, and
1t (100 cm)
2
3
d = 100 cm x 1.293 x 10- g/cm3 = 0.1293 g/cm2, and the respective values for Ila
and I1 t , we have
Db .= 2.3 mrad/hr = 23 !J.Sv/hr
The total dose rate is 23 + 0.46 = 24 !J.Sv/hr
266
10.16
THE HEALTH PHYSICS SOLUTIONS MANUAL
10.16 Design a sphericallead storage container that will attenuate the radiation
dose rate from 5 x 1010 Bq (1.35 Ci) 24Na to 100 !1Gy/hr (10 mradfhr) at a distance of 1 meter from the source. (The source is physically small enough to be
considered a "point.")
Sodium 24 emits 2 photons per transformation:
(1) 2.75 MeV gamma
5.2)
!l(Pb) =0.476 cm-! Ｈｔｾ｢ｬ･＠
HVL = 0.693/0.476 cm-! = 1.5 cm
(2) 1.37 MeV gamma
!1(Pb) = 0.621 cm-! (Table 5.2)
J
HVL = 0.693/0.621 cm- = 1.12 cm
r(
y
-
m2
G
24 Na) = 12.8x 10-9 C/kg
I
.
x 34MBq·h
C/kg
r(
24 Na
) = 4.4 x 10-
7
Gy·m 2
MBq·h
·
(Table 6.3)
Dose rate at 1 meter,
2
.
4
6 !lGy
7 Gy· m
x 5 x 10 MBq X 10 - D(l m) = 4.4 x 10MBq·h
Gy
D(l m) = 2.2 x 10 4 JlGy
h
The required attenuation:
The number of HVL's to attain this attenuation
SOLUTIONS FOR CHAPTER 10
267
n =7.8 = 8 HVL's
Each of the two gammas contribute to the shielded dose rate, but the lead shield
will attenuate the 1.37 MeV gamma much more than the 2.75 MeV gamma. Let
us first consider the attenuation only of the 2.75 MeV gamma by 8 HVL's plus 1
HVL to account for the 1.37 MeV gamma and for ｢ｵｩｬ､ｾｰＮ＠
The estimated shield
thickness thus is
t = 9 HVL x 1.5 cmlHVL = 13.5 cm..
The buildup factor, B, for the 2.75 MeV ｧ｡ｾ＠
is
B (Pb,. /If = 0.476 cm- x 13.5 cm =6.4) = 3 (Fig. 10.9)
l
The unshielded dose rate at 1 meter due to the 2.75 MeV gamma is (equation
6.17 and 6.5)
2
G
Do (2.75 MeV) = (3.65 x 10- x 2.75) C/kg ·m x 5 x 10 MBq x 34-y9
4
MBq·h
C/kg
Do (2.75 MeV) = 1.7 x 10-2 Gy
h
and the shielded dose rate is
D= 8.5 X 10-5 Gy = 85 ｾｇｹ＠
h
h
For the 1.37 MeV gamma, the unshielded dose rate at 1 meter is
.
C/kg·m 2
Gy
9
Do (1.37 MeV) = (3.65 x 10- x 1.37)
x 5x 10 4 MBq x 34-MBq·h
C/kg
Do (1.37 MeV) = 0.85 x 10-2 Gy
h
268
THE HEALTH PHYSICS SOLUTIONS MANUAL
and the shielded dose rate, with a value of
B (Pb, Jll =0.621 cm- x 13.5 cm = 8.4) = 3.8 is
1
•
•
2
Gy
D = DoBe-1'1 = 0.85 x 10- -
h
x 3.8 x e-8.4
iJ = 7.3 X 10-{) Gy = 7 IlG y
h
h
The total dose rate at 1 meter with 13.5 cm of lead shielding is
D = 85 + 7 = 92 /-lGy/h, which is compatible with the 100 /-lGy/h requirement.
10.17
10.17 What thickness of standard concrete is needed to reduce the intensity of a
collimated beam of 10 MeV X-rays from 104 mW/cm2 to an intensity corre3
sponding to 2.5 x 10- centisieverts per hour?
2
The exposure rate is measured in mW/cm because the roentgen and C/kg are
defmed only for x-rays or gammas whose energy is less than 3 MeV.
,
mW
4
Calculate the flux required to obtain 10 - - 2 ;
cm
10MeV
-x
y
ｾｯ＠
13
1.6 x 10- J x 1 W x 10 3 mW = 10 4 __
mW
x
2
cm ·sec
MeV
1 JI
W
cm 2
/sec
12
Incident flux, <I> = 6.25 x 10
o
photons
cm .sec
-=---2- -
Now calculate the flux required to produce 2.5 x 10-3 cSv/hr in tissue (equation
6.10). The absorption coefficient in tissue is utilized since the units are Sv. The
energy absorption coefficient for tissue is 0.0154 cm2/g (Table 5.3).
SOLtJTIONS FOR CHAPTER 10
13
<I>
25x 10-3 cSv =
Y
X 10 MeV x 1.6 x 10- J x 0.0154 cm
cm ·sec
y
MeV
g
269
2
2
[O.OOlfg IJ x i Gy
hr
7Gy
100cSv
Solving the above equation for photon flux;
The required attenuation factor is
I =
10
6
= 1.6 x 10-7
1xl0
12
6.25 X 10
The number of HVL's to attain this attenuation is
_1_ = 1.6 x 10-7
2n
n = 22.6 HVL's
The broad beam HVL of concrete for 10 MeV photons is listed in Table 10.2 as
11.9 cm. The required thickness therefore is
t = 22.6 HVL's x 11.9 cmIHVL = 269 cm
7
10.18 A Ra-Be neutron source emits about 1.2 x 10 fast neutrons (average
energy = 4 MeV) per gram Ra. What fraction of the dose equivalent from an
unshielded source is due to the neutrons?
Find the exposure from the Ra gamma rays first:
A = 1 g Ra = 1 Ci Ra (to yield 1.2 x 107 fast neutrons)
R·m 2
r Ra-226 = 0.825 Ci . hr (table 6.3)
10.18
-i
I
I
270
THE HEALTH PHYSICS SOLUTIONS MANUAL
The unshielded gamma ray exposure rate at the distance of 1 meter from 1 gram
of Ra is 0.825 Rlhr.
For radiation safety purposes, 1 R = 1 rem = 0.01 Sv, so no conversion factors
are applied.
The gamma dose rate is 0.825 remlhr = 8.25 mSv/hr.
Now calculate the neutron flux at one meter unshielded. Assume the neutrons are
isotropically distributed over a spherical area whose radius is 100 cm (1. m).
7
Using 1 Ci Ra, 1.2 x 10 fast neutrons are produced per second. Dividing this by
the spherical area at one meter will yield the neutron flux at one meter,
7
5
n
1.2 X 10 n/sec _
- - - - - - - : 2 : - - 95. - - 2 - 411:(100 cm)
cm . sec
From Table 9.5 it is found that for a 40 hour exposure to 4 MeV neutrons
9.2
= 1 mSv
n
cm
2
•
sec
wk
Using this information and the calculated neutron flux above, ratio the two
values to fmd the neutron dose rate from the Ra-Be neutrons
955
n
9.2
n
cm 2 • sec = ---=.=.=---=:.::;...:;..
cm 2 • sec
DII
1 mSv
wk
DII = 10.38 mSv
wk
Now calculate the fraction of dose equivalent due to the neutrons:
f
Dn
10.38 mSv
wk
=-:--------,--:..:..::.:..,.----,-
= Dn + Dy
f =0.03 = 3%
(8.25 mSv x 40 h) + (10.38 msv)
hr
wk
wk
.'
SOLUTIONS FOR CHAPTER 10
271
10.19 Transport ｲ･ｧｵｬ｡ｴｩｯｮｾ＠
for shipping a radioactive package specifya.maximum surface dose rate eqUIvalent of 2 mSv/hr (200 rnrernlhr) and a maXImum of
0.1 mSv/hr ( 10 mrernlhr) at 1 meter from the surface. If aqueous 137CS waste is
to be mixed with cement for disposal, what is the. maximum specific activity of
the concrete if it is to be cast in 20 liter cylindrical polyethylene containers 30
cm diameter for shipment to the waste burial site?
1 R = 1 rem for radiation safety purposes. The top surface of the cylindrical
container approximates a plane source. The dose rate would be found using
equation 10.11. In this case, solve equation 10.11 for Ca;
Ci
R·m 2
r2 +h2
H=1txr
xC - 2 xln
2
Ci.hr
am
h
.
r = 15 cm = 0.15 m
h= 1 m
2
. m (Table 6.3)
r Cs-I37 = 0.33 R
Ci . hr
if = 10 rnrems/hr = 0.01 remlhr
Substitute in the values;
rem
033 R . m2 x Ca - Ci x In(0.15
m)2 +(1 m)2
. 1-- = 1t x .
------'---'-00
Z
hr
Ci·hr
m
(lm)2
Ci
Ci
mCi
Ca -2 = 0.434which is the apparent areal concentration of
2 =0.0438-m
m
cm z
activity of the top of the cylindrical container.
Use equation 10.14 to find the volumetric concentration that results in this areal
concentration at the container top:
cm Z
g
J.1 = 0.0290- X 2.35= 0.068 cm-I(Appendix E, Table 5.2)
g
cm 3
mCi
C = 0.0438-a
cm 2
10.19
272
THE HEAL11l PHYSICS SOLUTIONS MANUAL
2 X 10 4 cm 3
t = n(15 cm)2 = 28.3 cm (height of container)
Substituting these values into the equation above, we fmd
mCi
5 )lCi .
.
137
.
.
3 - - 3 IS the concentratIOn of Cs whIch wIll produce a
cm
cm
dose rate of 10 mrems/h at a distance of 1 meter above the surface of the container.
Cv = 35 x 10
-3
--3 =
3 mCi
Cs concentration of 3.5 x 10- - - 3 is
cm
used. The surface dose rate must not exceed 200 mrems/h. To calculate the
surface dose rate, we will assume that the concrete cylinder IS infinitely thick,
and use equation 6.44a.
Now check the surface dose rate when a
137
b = l.1x 10 3 xCｶｾ＠ "f·E.
I
1
where
ECS_137 = 0.661 MeV/gamma
f es-137 = 0.85 gamma/transformation
Cv = concentration, /-lCilg
and density of concrete = 2.35 g/cm3
35)lCi/
b = 1.1 X 10' x
.
.
ｾｭＧ＠
235 g
x 0.85 x 0.661 = 920 mrems
cm
3
ｾ＠
Since this is too high, we must reduce Cv to
200 x 35 !-lCi = 0.76 !-lCi
920
cm 3
cm 3
The maximum activity that can be mixed into the 20 L concrete slug for disposal
IS
A = 0.76 !-lC! x 2 x 10 cm 3 = 15.2 x 10 3 !-lCi = 562 MBq
cm
4
SOLUTIONS FOR CHAPTER 10
273
10.20 A technician's job in a radiopharmaceutical laboratory involves simulta-
neous handling of 5000 MBq (135 mCi) 121:,4000 MBq (108 mCi) 198Au and
2000 MBq (54 mCi) 2'Na for 1 hour per day, 5 days per week for an indefinitely
long time. Her average dose equivalent during the other 7 hours will be 0.01
mSv (1 millirem). Her body will be 75 cm from the sources while she works with
them, and manipulators will be provided so that her hands will not be exposed
inside the shield.
(a) What is the source strength for each of the sources?
Traditional units will be used to solve this problem because the regulations of the
USNRC are written in traditional units. The source strength for each radioisotope
is calculated with the specific gamma ray constant given in Table 6.3. Since we
are dealing with gamma radiation, 1 mR = 1 mrem.
Isotope
r mrem·m 2
mCi·hr
A, rrCi
Source mrem·m 2
Strength
hr
if=
SS
(0.75 mr
1251
0.07
135
9.45
16.8 mremslhr
198Au
0.23
108
24.8
44.1 mremslhr
24Na
1.84
54
99.4
176.7 mremslhr
(b) What thickness of lead shielding is required if her weekly dose equivalent is
to he within ALARA guidelines, that is, at 1110 of the maximum permissible
dose?
The USNRC's annual dose limit is 5000 mrems, which is reduced by ALARA
policy to 500 mrems. For a nominal 250 day work year, the operational hourly
limit is 2 mrems/day. If the technician receives 1 mrem during 7 hours, she is
limited to 1 mrem while working with these 3 sources.
Let us calculate the lead shielding to reduce the 24Na dose rate from 177 to 1
mremlhr. Because the 2'Na gamma energy is so much higher than from the two
other sources, this shield thickness will, for practical purposes, absorb all the
other gammas. The number of half value layers, n, needed to attenuate the 2'Na
gammas is
I
1
1
-=-=-
n=7.5HVL
10.2
274
THE HEALTII PHYSICS SOLUTIONS MANUAL
To be conservative, the calculated thickness of a HVL will be based on the 2.75
MeV gamma. From Table 5.2, we find, by interpolation /.l (pb, 2.75 MeV) =
0.486 cm'!
.
HVL = 0.693 =_0_.6_93_ = 1.43 cm
11
0.486 cm'!
To account for buildup, add 1.5 HVL, for a total trial shield thickness of
t = (1.5 + 7.5) HVL x 1.42 cm/HVL = 12.9 cm
Now calculate the attenuation of the 2.75 MeV gamma. The unshielded dose rate
(equation 6.l8):
at 0.75 m is estimated with the specific gamma ray ｣ｾｮｳｴ｡＠
mrem·m 2
7--[' = OS'Lf.£. = 05x 2.75 = 1 . 3mCi.hr
I
I
(equation 10.1)
2
1.37 mrem·m x54 mCi
mCi· hr
iI = [' x A =
2
d
(0.75 mr
1315 mrems
hr
The buildup factor, B, for a lead shield 6.3 relaxation lengths thick (Jll = 0.486 x
12.9 = 6.3) is 3.5 (Fig. 10.9). The shielded dose rate is
iI = 1315 mrems x 35e-6·3 = 0.85 mrems
hr
hr
If we make a similar calculation for the 1.37 MeV gamma, we find the dose rate
to be 0.05 mremlhr. For the other two radionuclides, a 12.9 cm thick lead shield
reduces the radiation for practical purposes to zero. The required shield thickness
.
is 12.9 cm of lead (5").
275
SOU.JTIONS FOR CHAPTER 10
Design a spherical shield for a 1 x lOll Bq (2.7 Ci) 90Sr "point" source so
that the dose equivalent rate at the surface will not exceed 2 mSv (200 millirem)
per hour. What is the dose equivalent rate at a distance of 1 meter from the
shielded source?
10.21
Assume that 90y is in equilibrium with the 90Sr. 90y emits a 2.27 MeV beta.
Using equation 5.2 to find the range of the beta:
R = 412 x (EMeV)L265-0.0954In(E)
R = 412 X (2.27)1.265-0.0954XIn(2.27)= 1090 mg/cm2 = 1.09 g/cm2is the density thickness required.
Example 10.9 lists the density of polyethylene as: p = ＰＮＹＵｾ＠
cm
Calculating the thickness of polyethylene required to shield the 90Y betas;
3
1.09g x cm = 1.1cm
cm 2 0.95 g
A significant amount of bremsstrahlung is produced by the polyethylene shielding and a layer of lead is used to shield the bremsstrahlung. First, calculate the
amount of bremsstrahlung produced. The effective "Z" of polyethylene, assuming for this case, the formula for polyethylene is CH2, is calculated using equation 10.34;
23
3 0.95 g 1 mol 6.02 x 10 molecules 2 atoms 8 1022 atoms
N = 1 cm x
x
x
x
= X
H
cm 3
14 g
mol
molecule
cm 3
23
1 atoms 4 10 22 atoms
3 0.95 g 1 mol 6.02 x 10 molecules
x
x
x
= x
3
cm
14 g.
mol
molecule
cm 3
N c = 1 cm x
C=6
H= 1
10.21
276
THE HEALTII PHYSICS SOLUTIONS MANUAL
Bremsstrahlung production is now estimated using equation 5.11:
EY _90 = 2.27 MeV
!Y-90
= 35 X 10-4 ZE = 35 x 10-4 x (4.75) x 2.27 = 3.8 x 10-3
The average energy of the betas from the 90Sr - 90y source is 0.18 + 0.93 = 1.11
Me VItransformation.
1 trans
sec x 1.11 MeV = 1.11 x 1011 MeV
Bq
d
sec
E = 1 X 1011 Bq x
f3
The amount of this beta energy that is converted to x-rays is
MeV
J
Ex =! x Ef3 = 3.8 X 10-3 x 1.11 X 1011 - - x 1.6 X 10-13 _ _
MeV
s
-5 J
Ex =6.75x10 -
s
If the source is surrounded by 1.1 cm of plastic, then the x-ray energy flux at the
plastic surface is
ｾｘｅ＠
6.75 X 10-5
.!
=
; = 4.44 x 10-{)
4n(1.1 cm)
J
2
cm ·s
The dose rate from this flux is calculated from
cm 2
J
s
Jl en - - X 3.6 x 10 3 if = __c_m_·_s_ _---..:g:::::....-...._ _ _ _ _
h
ｾ＠
XE
2
X
0.001 Jig xl Gy x
1 Sv
Gy
10 3 mSv
Sv
Using Ilen (muscle, 2.27 MeV) = 0.0249 cm2/g (Table 5.3), we find that
SOLUTIONS FOR CHAPTER 10
277
.
mSv
H=398.l-h
We wish to reduce the dose rate with the lead shieid by a factor of
ｾ＠
= _2_ = 5.02 x 10-3
Ho 398.1
Combining both inverse square dispersion and attenuation by the shield, but
excluding buildup, we can calculate a minimum shield thickness t
I
Using Il (Pb, 2.27 MeV) = 0.506 cm- (Table 5.2) we fmd that t = 4.2 cm. To
account for buildup, increase t to 5 cm, and then using a buildup factor B (Pb,
2.27 MeV, Ilt = 2.53) = l.82 (RHH, Fig. 10.9), calculate the resulting attenuation
ｾ＠
Ho
= 1.82( 1.1 ) 2 e -2.53 = 4.7 X 10-3
5+ 1.1
This is slightly better than the design criterion of 5.02 x 10-3. Therefore, the
shielding requirements are
1.1 cm polyethylene and 5 cm lead.
Solutions for Chapter 11
INTERNAL RADIATION PROTECTION
35
A health physicist finds that a radiochemist was inhaling Ba S04 particles
that were leaking out of a faulty glove box. The radiochemist had been inhaling
the dust, whose mean radioactivity concentration was 3.3 MBq/m3 (9 x 10-5 mCi/
cm\ for a period of 2 hr. Using the ICRP 3 compartment lung model, calculate
the absorbed dose to the lung during the 13 week period and during the 1 year
period immediately following inhalation.
11.1
TR = 87 days
Since no particle size is given, the ICRP default value of 1 ｾｭ＠
AMAD particles
is used. Assuming conditions of light work, Appendix C gives a volume of 9.6
m3 of air inhaled in an 8 hour day. Assume that the person inhales approximately
10 m3 of air. Calculate the inhaled activity:
2 hrs x ＳＮｾ＠
MBq
1 x 106 Bq 10 m 3
6
x MBq x 8 hr = 8.25 x 10 Bq is the total inhaled activity
The elimination constant for each of the compartments is calculated next.
Table 8.5, Figure 8.4 or ICRP 26 give the following depositions of inhaled 1 ｾｭ＠
AMAD particles:
Deposition Fraction in lung
Total activity
Activity in lung
N·P
30%
0
8.25 x 106 Bq
0
TB
8%
0.08
8.25 x 106 Bq
6.6 X 105 Bq
P
25%
0.25
8.25 x 106 Bq
2.1 X 106 Bq
Total deposited in lung
2.7 X 106 Bq
279
11.1
280
THE HEALTII PHYSICS SOLUTIONS MANUAL
Clearance rates from the various lung compartments depend on the solubility
class. BaS04 is in class w, moderate solubility (ICRP, Health Physics. 12:173,
1966, 10 CPR 20 Appendix B). Class w particles are assigned biological removal
half times, TB , as shown in the table below. For all the particles in the TB region,
and for 40% of those in the P region, TB ｾ＠ 1 day. Since these clearance half times
«< 87 days, the effective clearance half time, TE is the same as TB • The effective
clearance rate constant for these particles is:
A = 0.693
T
E
E
For 60% of the particles deposited in the P region, whose TB = 50 days, the
effective TE is calculated from equation 6.54:
= 87 days x 50 days = 32 days
87 days + 50 days
A _ 0.693
E -
0.693
7;/2 = 32 day -0.022 d-
l
A
dol
Region
% cleared
As(O), Bq
TB , d
TB
50
0.5 x 6.6 x 105
0.01
TE'd
0.01
TB
50
0.5 x 6.6 x 105
0.2
0.2
3.47
P
40
0.4 x 2.61 x 106
1
1
0.693
P
60
0.6 x 2.1 x 106
50
32
0.022
E'
69.3
The dose from the activity in the lung is given by the product of the accumulated
activity, A Bq·d and the dose conversion factor, DCF GylBq·d.
Gy
D = A Bq·d x DCF, Bq.d
The-cumulated activity for a single compartment is (equation 6.91)
- = As (0) (l-e -A ' )
A
A
SOLUTIONS FOR CHAPTER 11
281
and for n compartments
During the first 13 weeks, all the compartments but the last will be emptied, that
M
is e- ::: for the first 3 compartments(Note that the equation is in 2 lines):
°
_
A(13 weeks) =
(O.5)x6.6xl0 5 Bq
69.3 d- 1
6
+
(0.5)x6.6xl0 5 Bq
6
3047 d- 1
+
(004)x2.1xl0 Bq (0.6)X2.IXI0 Bq(
-0022 91)
+
+
l-e . x
1
0.022 d0.693 d -1
A(13 weeks) = 5.1 x 10 7 Bq·d
After 1 year, e-0.022 x 365 == 0, and the last term becomes 0.6 x 2.1 x 106 Bq/0.022 d1, which leads to a 1 year cumulated activity
A(1 yr) = 5.9 x 10 7 Bq·d
Sulfur 35 emits a single beta whose mean energy is 0.049 MeV, and no gammas.
Using a lung weight of 1 kg (Appendix C), the DCF is calculated
mGy
=
Bq·d
I_t_x 8.64 x 10 4 ｾｸ＠
s·Bq
d
4.9 X10-2 MeV x 1.6 x 10-13 _1_
t
MeV
J/kg
1 Gy
1 k g x 1- - X -:---,'--Gy 10 3 mGy
7 mGy
DCF = 6.8 x 10- - Bq·d
Therefore
D(13wk) =5.1xl0 7 Bq·dx 6.8 X10-7 mGy = 35mGy
Bq·d
D(1 yr) = 5.9 x 10 7 Bq . d x 6.8 X 10-7 mGy = 40 mGy
Bq·d
282
11.2
THE HEALTH PHYSICS SOLUTIONS MANUAL
2
A tank, of volume 100 L, contained 85 Kr gas at a pressure of 10.0 kglcm •
(9.71 atmospheres). The specific activity of the krypton is 20 CiJg. The tank is in
an unventilated storage room, at a temperature of 27°C, whose dimensions are 3
x 3 x 2 m. As a result of a very small leak, the gas leaked out until the pressure
2
in the tank was 9.9 kg/cm • A man unknowingly then spent 1 h in the storage
room. Assume the half saturation time for krypton solution in the body fluids to
7
be 3 min. Henry's law constant for Kr in water at body temperature is 2.13 x 10 •
Calculate (a) the immersion dose, (b) the internal dose due to the inhaled krypton. The partition of Kr in water to Kr in fat is 1: 10.
11.2
First, calculate the activity of the escaped gas:
The original number of moles of labeled Kr in the tank is calculated with the
ideal gas law.
PV=nRT
n = PV = 9.71 atm x 100 liters
RT 0.082 liter . atm x 300 K
mole·K
39.4 moles
The amount of Kr gas that escaped, Q, is
kg
(10.0-9.9)-2
Q=
cm x 39.4 mol = 0.394 mole
ＱＰＮｾ＠
cm 2
The specific activity of the tagged Kr is 20 CiJg, while the specific activity of
. 85 Kr is calculated with equation 4.31.
. ARaTRa
SA- A
T
Kr-85 Kr-85
= 226x 1620 yr = 418.2 Ci
85xlO.3yr
The fraction of Kr that is 85 Kr is
f =
ＲＰｾ＠
CV = 0.0478
418.2 /g
g
---------------------
- - - -
SOLUTIONS FOR CHAPTER 11
283 '
The molecular weight of the tagged Kr is
(0.478 x 85) + 0.9522 x 83.8 = 83.86 g/mole
and the molar concentration of 85 Kr activity is 83.86 glmole x 20 Citg = 1677 Cit
mole.
The atmospheric concentration of 85 Kr activity in the room is
10 Bq
..
0.394mole x 3.7 x 10 B (
C.)
1
c= actIvIty =
Ci = 1.36 x 10123 36.7m3
m3
volume
3 mx 3 mx 2 mx
Krypton 85 emits one beta per transformation, Emax = 0.672 MeV, E = 0.246
MeV, and a gamma photon in only 0.41 % of tbe transformations. Thus, only the
beta will deliver a significant dose to the skin, which is calculated with equation
6.38:
The absorption coefficient, ｾ＠
is given by equation 6.21:
Ilb = 18.6(Emax - 0.036y
L37 = 34.6 cm /g
.
2
Bq
mGy
.
Db = 2.45x 10-7 X 1.35 X 10 12 - 3 X 0.246 MeVx e-34.6c (0.007) = 64x 10 3 - m
hr
•
3 rads
Db = 6 4x 10 hr is the external dose rate to skin.
.
rads
Db = Db X t = 6.4 x 10 3 - - X 1 hr = 6.4 x 10 3 rad
hr
SYSTEMIC INTERNAL DOSE
Calculate the dose to from 85 Kr in the body fluids (water in the body) and ｾ＠ the
body fat. Start by computing the molar concentration of the 85 Kr in the air:
284
THE HEAL'IH PHYSICS SOLUTIONS MANUAL
Ci
36.7
-3
1 3
85
_ _ ....::.m==-- 1 moIe 85Kr
m
= 1 03 10-9 mole Kr
.x
x
6
3
•
x
3
418.2 CI
85 g
-1 x 10 cm
cm
g
Now compute the concentration of natural krypton in the air. The concentration
of naturally occurring krypton in the atmosphere is approximately 1 ppm (CRC).
Using the ideal gas law, we fmd the molar volume of air at 25°C to be:
V RT
-=-=
n
P
0.082 liter· atm x 298 K
.
mole· K
= 24.4 lIters,
1 atm
mole
and the molar concentration of naturally occurring Kr is
1 mole Kr x mole air x 1 liter air = 4.1 x 10-11 mole Kr
10 6 mole air 24.4 liters air 1000 cm 3 air
cm 3 air .
The concentration in the room of the 85Kr is much greater than naturally occurring
Kr. Thus, naturally occurring Kr is ignored when calculating permeation of 85Kr
into body fat. Using values calculated earlier for 85Kr:
0.394 mole 85Kr escaped 24.4 liters air
m 3 air
-4
85
l
-----------"'--x
x
5.34 x 10 moles Kr
3
mole air
1000 liter air
18 m
mole air
Calculate the amount of 85Kr which permeates the water in the body. The saturation concentration of a gas in water is given by Henry's Law (equation 8.16)
p(gas) =K
ng
+ ns
Since the total pressure of a gas mixture equals the sum of the partial pressures
of its constituents, the partial pressure of the Kr in the air is
p(Kr)
760mm
5.3 X 10-4 mole Kr
=-------
1 mole air
J
I
SOLUTIONS FOR CHAPTER 11
285
p(Kr) = 0.41 mrn Hg
7
Henry's Law constant, K = 2.13 X 10 mm Hg for Kr at 38°C (Table 8.16). The
molar concentration of water is
1000 g H 2 0
n = ___I=it.=..=e-=-r_=55.56 moles H 2 0
s
18 g H 2 0
liter H 2 0
mole
Substituting these values into Henry's Law, we have
ng
0.41 mm=2.l3x 10 7 rrunx---=---n +55.56 moles
g
liter .
moles Kr
ng = 1.06 x 10 l·t
1 er b 0 dyflUl·d
-6
A 70 kg reference man's water content is 43 L, and the total 85 Kr content in the
man's body fluids is
moles Kr
43 liters bodyfluid
Ci
.
2 C·
x
x
1677
7
6
10b0 dY
1
moIe Kr - . X
1 er b 0 d yfl Ul·d
1.06 x 10-6 l·t
To find the quantity in the fat, recall from the problem that the partition ratio is
1: 10, yielding:
X moles Kr gas
kg fat
Ｍｾ］＠
1.06 X 10-6 moles Kr gas x 1 liter body fluid
liter bodyfluid
1 kg
-5
X = 1.06 x 10
10
1
moles gas
kg fat
From Appendix C, the quantity of fat in the body is 13.5 kg, so the total quantity
of activity in the fat is:
""-'
286
THE HEALlli PHYSICS SOLUTIONS MANUAL
-5
1.06 x 10
moles Kr x 13.5 kg x 1677
Ci
... .
J: t
b
d
I
Kr
=
0.24
CI
actIvIty III the fat.
k g la
0 y
mo es
Adding the activity from the body fluids (water) and fat:
7.6 X 10-2 Ci + 0.24 Ci = 0.316 Ci
3.7 x 1010 Bq
0.316 Ci x
Ci
10
1.17 x 10 Bq is the total activity deposited in the'
body.
Assume that the activity is uniformly distributed throughout the body. Calculate
the dose rate using equation 6.47:
I.!.
｢ｳＨｾＩ＠
(q)Bq x _s E MeV 1.6 x 10-13 _J_ x 3600 sec
=
Bq trans
J
MeV
hr
(m)kg x 1 - /Gy
.kg/,
Substituting
10
q = 1.17 X 10 Bq
jf= 0.246 MeV
m =70 kg (Appendix C)
into the equation and solving for the dose rate, we get
｢ｳＨｾＩ］ＳＮＹｸｬＰＭＴ＠
G,! =3.9xl0-2 ｲ｡ｾｳ＠
mm
Since the half saturation time is 3 minutes, the man's saturation dose rate, D s'
will not be reached until about 15 to 18 minutes after the start of inhalation. His
dose during inhalation, D l' therefore is given by equation 11.17
mm
where k, the saturation rate constant equals 0.693/3 minutes = 0.231 min-I.
rads x [
1
DJ = 3.9 x 10-2 -.60 .
mm . -I X (1- e-0231 x 60 ) ] = 2.17 rads
mm
0.231 mm
SOLUTIONS FOR CHAPTER 11
287
The dose from exhalation is described by equation 6.58;
rads
0.039D· s
.
D = _
mm = 0.17 rads
E
A - 0.231 min- 1
So that the total systemic internal dose is:
2.17 rads + 0.17 rads = 2.34 rads
DOSE TO LUNGS
Calculate the dose to the lungs from the 85Kr residing in the lungs. Appendix C
gives the mass of each lung as 1 kg, and the mean volume of air in the lungs as
2.7 liters {since the functional residual capacity is 2.2 liters (Appendix C) and
the person is performing light activity, 20 L/min and 20 breaths/min = 1 liter/
breath but a breath is air in and air out, so the mean breath is 0.5 liter.}
3
lxl0- m 3
12 Bq
9
q = 2.7 liters x
.
x 1.36x 10 - 3 = 3.6 x 10 Bq
1 lIter
m
E =0.246 MeV
m = 1 kg (Appendix C)
Calculate the dose using equation 6.47:
I!
(q)Bq x _s E MeV 1.6 x 10- 13 _1_ x 3600 s
MeV
D. (J3 ) = - - - - -Bq
- = : .t . . . - - - - , - - - -hr
-(m)kg x 1ｾ＠ lOy
kgl'
I!
9
b(J3)
3.6 x 10 Bq x _s x 0.246 MeV x 1.6 x 10-13 _1_ x 3600 s
Bq
t
MeV
hr
1 kgx Ｑｾ＠
lOy
kgl'
.( )
Oy
rads .
85
D J3 = 0.52 hr = 52 ｾ＠
IS the dose rate to the lungs from the Kr breathed
into the lungs. So that one hour in the 85 Kr atmosphere gives 52 rads to the lung.
288
THE HEALTII PHYSICS SOLUTIONS MANUAL
After inhalation of 85 Kr has ceased, the lung will continue to be irradiated as the
10
radiokrypton in the body, 1.17 x 10 Bq, is released at a rate ot 0.231 per
minute. The mean residence time in the lung of this activity is
1
= 4.33 min
t = ..!. =
k 0.231 min- I
The dose to the lung during the washout is
D(washout) = iJ x t
1.17 x 1010Bq x 60 tpm x 0.246 MeV x 1.6 x 10-13 _J_.
Bq
t
MeV 433 .
D(washout) =
x. mm
1 kgxI- Gy
. kg
'J;
D(washout) = 0.12 Gy = 12 rad
Total dose to the lung = 52 rads + 12 rad
D(lung) =64 rads
It should be noted that the external dose is very much greater than the internal
dose. This is true for all the rare gases except radon. Thus, the DAC for these
gases is based on submersion rather than inhalation.
11.3
11.3 If the man in problem 2 turned on a small ventilation fan of capacity 100 fe /
min as he entered the room, calculate his immersion and inhalation doses.
m3
ft3
m3
IOO-x
=2.83min 35.3 fe
mm
The volume of the room is 3 x 3 x 2 = 18 m3
Calculate the turnover rate;
k=
ＲＮＸＳｾ＠
3
ｾｮ＠
18 m
=0.16 room changes per minute
Finding the concentration at the end of the 60 minute (l hour) period with
equation 11.18:
SOLl.ITIONS FOR CHAPTER 11
289
Ci
C = 36.6 - 3 (from problem 11.2)
m
o
k = 0.16 min- 1
t=60min
C i0 ·16
Ci
3
. x 60 = 2.5 X 10- C -- Coe-kt-36
- .6_e3
3
m
m
Now calculate the average concentration in the room (note that on a semilog
graph, the decrease in concentration would describe a straight line. The average
concentration.):
-3 Ci
Ci
Ci
..
(log(Co) -IOg(C))
I
_
C = antIlog
2
= v Co x C = 36.6-3 x 2.5 x 10 - 3 = 0.3-3
m
m
m
Now find the ratio of this to the original concentration:
0.3 C!
3
Ｍ］ｾＱＺＮ＠
= 8.2 x 10- is the fraction of the unventilated dose that will be re36.63
m
ceived with fan on.
The external dose to the skin would then be (from problem 11.2):
3
3
6.4 X 10 rads x 8.2 X 10- =52.5 rads
The systemic body dose would be:
2.4 rads x 8.2 x 10-3 = 2 x 10-2 rads
The inhalation dose to the lungs would be:
52 rads x 8.2 x 10-3 = 0.4 rads
11.4 An accidental discharge of 89 Sr into a ｲ･ｳｶｯｩｾ＠
3
3
resulted in a contamination
level of 37 Bq ( 10- IJ,Ci) per cm of water.
(a) Using the basic radiological health criterion of the ICRP, would this water be
acceptable for drinking purposes for the general public if the turnover half time
of the water in the reservoir is 30 days?
(b) If the water were ingested continuously, what maximum body burden would
be reached?
(c) How long after ingestion started would this maximum occur?
(d) What would be the absorbed dose during the first 13 weeks of ingestion?
11.4
290
THE HEALTIJ PHYSICS SOLUTIONS MANUAL
(e) What would be the absorbed dose during the first year?
(f) What would be the absorbed dose during 50 years following the start of
ingestion?
Reservoir
Areservoir)
Man
Amall)
Excretion and Transformation
(b) (c) Find an expression for the quantity of activity (q) in man at any given
time from the activity in the reservoir.
Co = Reservoir concentration at time zero.
q = quantity in man
q = 0 at t = 0
Ar = Removal rate from the reservoir
Am =Removal rate from man
fw = 0.21 = Fraction of intake deposited in bone by ingestion. See ICRP 2 , p.
176. Note that FGR 11 (pA8) and ICRP 61 (p.ll) only address the fraction
absorbed into the blood from intake with this term and NOT the fraction absorbed in the bone.
mL x C0 x f w x e -A r t
daily intake = 2200day
daily elimination =Am X q
dq = daily intake _ daily elimination = (2200 mL x Co x e -Art X f w) - (A m X q)
ｾ＠
let K represent
. mL
K=2200- xeo xfw
day
Substituting for the value K;
､ｾ＠
SOUITIONS FOR CHAPTER 11
291
The above equation is the fonn
dy
-+Py= Q
dx
Where
=q
x = t
P = 1m
Q = Ke-A.rt
y
The solution of the differential equation, using the integrating factor method
described in the textbook on pages 108 and 109 is
The above expression will give the quantity of the activity in man (body burden)
at any given time after starting to drink the water.
The time after start of continuous ingestion to the maximum body burden is
given by equation 4.57:
Since the biological half life of 89Sr is is about 50 years (ICRP 2) and its radiological half life is 50.3 days (ICRP 2), its effective half life is also about 50.3
days, and the effective elimination constant is (equation 6.52):
Q693
Am =-T-
Q693
= 50.3 days
1
ｾ＠
= 0.0138 d- is the elimination rate of Sr from man.
E
The turnover half time of the reservoir (Tres ) is 30 days, so its effective turnover
half time, Tr (equation 6.54) is:
292
THE HEALTII PHYSICS SOLUTIONS MANUAL
ｾ･ｳ＠
X TB
30 X 50.3
Tr = T + T = 30 + 50.3 = 18.8 days
res
B
and its elimination rate constant
A = 0.693 = 0.693
ｾ＠
r
0.0368 d- I
18.8 days
Find t max ;
t
max
A )
In( ----E!...
1n (0.0138 daYS-I)
.
Ar
=
0.0368 days-I
= 42.6 days
(Am -A r ) (0.0138 days-l -0.0368 days-I)
(c) The maximum body burden will occur after 42.6 days of continuous intake.
Now fmd the maximum body burden, solving first for K;
C = 37 Bq
o
mL
fw = 0.21 (From ICRP 2, p.176) (Fraction reaching organ of reference by ingestion)
mL
mL
Bq
4 Bq
K = 2200- x C xf = 2200- X 37 x 0.21 = 1.7 x 10 initial uptake
day
mL
day
0
w
day
(b) Calculate the activity in the human body of Sr-89 at the time of maximum
body burden (42.6 days).
Bq
K = 17094 -d is the uptake
ay
Ar =0.0368 d- 1
Am =0.0138 d- I
t = 42.6 days
SOLUTIONS FOR CHAPTER 11
_
q-
293
17094 Bq
day
(-(000368days- 1)x4206days
-(000138 daYS-I) 4206daysJ
x e
-e
0.0138 days-J - 0.0368 days-l
X
The maximum body burden would be 2.6 x 105 Bq reached after 42.6 days.
Calculation of the absorbed dose
(d) The dose over a time period t to the skeleton from skeletal deposit of 89Sr
(99% of the body's Sr content is in the skeleton) is calculated from
t
H=
f if dt
o
The instantaneous dose rate if due to a skeletal burden of q Bq is:
q Bq x 1 tps x E MeV x 1.6 x 10- 13 _1_ x 8.64 x 104 sec
if =
Bq
t
MeV
day
1 ; Gyx-1 Gy
wkgx1kg
Sv
-
0
89
Substituting E ( Sr) = 0.583 Me V (NRHH) and w(skeleton) = 7 kg, and combing constants we find that
.
Sv
H=qBqxA-d·Bq
if Sv = 1.151 X 10-9 Sv X q Bq
d·Bq
d
Therefore
t
f
H = A q (t) x dt
o
Substituting the expression for q (t) into the integral yields
H=
AS
K
OJ... m -J... r
(e(-Ar)t -e-(Am)t)dt
0
-.,
294
THE HEALTI-l PHYSICS SOLUTIONS MANUAL
Integrating, we obtain
Substituting
4
Bq
K = 1.7 x 10 -d (uptake)
ay
Ar =0.0368 d- 1
Am =0.0138 d-1
-9
Sv
A = 1.15 x 10 B q. day
we have
H =-8.5 x 10-4 [ 27.17(1 - e-O.0368t) - 72.46(1 _ e-O.01381)]
Solving for t =91 days (13 weeks); we find
2
H (91 days) = 2.19 x 10- Sv = 21.9 mSv (2,190 mrems)
For t = 365 days (1 year) and for t = 50 years, e-O.03681 and e-O.0138t «< 1, therefore,
H (1 yr, 50 yr) =-8.56 x 10-4 [27.17 -72.46]
(e)(f) H (1 yr, 50 yr) = 3.88 x 10- Sv = 38.8 mSv (3,880 mrems)
2
The fact that the 1 year and 50 year doses are the same means that all the dose
was absorbed during the fIrst year.
(a) The basic radiological health criterion limits the dose to members of the
public to less than or equal to 1/50 of the occupational limit. For an individual
organ or tissue such as the skeleton, this corresponds to 1150 x 500 mSv = 10
mSv. In this case, the calculated dose is 39 mSv. Therefore, the water is not
acceptable for public consumption at this level of contamination.
. I
• I
I
11.5 Nickel carbonyl Ni(CO)4 has a maximum permissible atmospheric concen-
11.5
tration of 1 part per billion (ppb) based on its chemical toxicity. A chemist is
The specific activity of the nickel
going to use this compound tagged ｷｩｴｨﾷＶｾＮ＠
8
is 2.5 x 10 Bq/g (6.75 mCilg). The industrial hygienist is planning to limit the
J
295
SOLUTIONS FOR CHAPTER 11
atmospheric concentration of Ni(CO)4 in the lab to 0.5 ppb. Will this restriction
7
meet the requirement for the radioactivity DAC of 3 x 10- j.lCiJrnL?
Determine if enough Ni is 63Ni to change the molecular weight used in converting to moles.
The specific activity of 63Ni is given by equation 4.31, where;
= 226 x 1620 yr = 63.2 Ci
63x 92 yr
g
6.75 X 10-3 Ci
63.
SA •Ni
g
-4
f( NI) = S'A 63NI' Ｍ］ｾ＠
= 1.07 x 10
.t1
63.2 Ci
g
Very few of all Ni atoms in the compound are 63Ni. Therefore, the atomic weight
of *Ni will be about the same as Ni, 58.71.
Calculate the number of mCi per mole of compound;
mCi
58.71 g Ni x 6.75 mCi = 396.3
g Ni
mole NiCO 4
mole NiCO 4
·0.5 parts per billion, is the same as 0.5 moles of Ni(CO)4 per 1 billion moles air.
Converting to air concentration:
4
2.45 x 10 mL per mole air at STP, 760 mm Hg and 25°C
3
05 mol NiCO 4 x 396.3
mCi
x
mol air
x 10 /lCi = 8.1 x 10-9 /lCi .
4
9
10 mol air
mol NiCO 4· 2.45 x 10 mL
mCi
mL IS
the airborne concentration corresponding to 0.5 ppb. Yes, the restriction is met
with these requirements.
Chlorine 36 tagged chloroform, CHCI 3, whose specific activity is 100 j.lCiJ
mole, is to be used under such conditions that 100 mg/hr may be lost by evaporation. The experiment is to be done in a laboratory of dimensions 15'x10'x8'.
11.6
11.6
296
THE HEALTIf PHYSICS SOLUTIONS MANUAL
The lab is ventilated at a rate of 100 fe/min.
(a) Do any special measures have to be taken in order to control the atmospheric
concentration of the 36C1 to 10% of its DAC
6
(DAC = 1 x 10- JlCilmL)?
Calculate the volume of the room:
15 x lOx 8 = 1200 fe is the volume of the room.
100 ft3
- - - is the flow rate in the room. Converting to metric units;
mIll
100 ft 3
m3
m3
=2.83k
x
= min
35.314 ft 3
min"
First, find the specific activity of 36C1 and compare it to the specific activity listed
in the problem to find the fraction of chlorine atoms that are 36CI:
Equation 4.31 is used with the following data:
ARaTRa
SAe6CI) = - - - A CI - 36 TCI - 36
__
226 x 1620 yr __ 3.3 x 10-2 Ci
36 x 3.08 x 105 yr
g
6
2
3.3 X 10- Ci 1 x 10 JlCi
36 g
3 mole 36CI
356 I" 0 6 /lCi
Ci
x mole 36 CI x 1 mole CHCl =
x
mole if all
g
x
3
the CI atoms were 36Cl.
Computing the fraction of 36CI tagged:
100 /lCi
36
mole. = 2.81 X 10-5 moles Cl
356 x 106 /lCI
mole CI
mole
The molecular weight is not significantly affected by the 36Cl.
SOLUTIONS FOR CHAPTER 11
297
The molecular weight of CHCl 3 is (Appendix B);
C= 12
H=l
CI = (3) x 35.45 = 106.35
Total for CHCl3 = 12 + 1 + 106.35 = 119.35
Finding the steady state concentration in the room, with a loss rate of 100 mg/hr:
The rate of generation of the radioactive CHCl 3 vapor, G, is
G
100 mg x
g
x mole x 100 ｾｃｩ＠
x 1 hr = o.oold ｾｃｩ＠
hr
1000 mg 119.35 g
mole
60 min
min
The steady state concentration, where the rate of generation is equal to the rate
of removal is (equation 11.26):
1.4 x 10-3 ｾｃｩ＠
G
.
ｾ＠
C = - = _ _ _ _ _ _---=ffil:=.:3:::.::n::....-___ = 4.94 X 10-10 Q
ＱＰｾｘＨ＠
min
1m )
3.28 ft
x10 6 mL
m3
C.
I
is the steady state
mL
concentration in the room. This is much less than 0.1 x DAC (1 x 10- ｾＩＮ＠
6
Therefore no special controls are needed.
(b) To what concentration of chloroform, in parts per million, does the radiological DAC correspond for this compound? Compare this concentration to the
chemical PEL for chloroform.
1o
C·
Since 100 !-lCi/mole produces 4.94 x 10- !-l : ,calculate the number of moles
cm
per mL are in the room;
494 ＱＰＭｏｾ＠
. x
C·
3
cm 3 x 1cm = 4.94 x 10-12 moles CHCl 3 •
•
•
ｾｃｩ＠
mL
mL
IS the concentratIon III the
100-mole
room.
3
Standard air at 760 mm and 25°C has 24.5 x 10 mL/mole air. Combining this
information;
298
THE HEALTII PHYSICS SOLUTIONS MANUAL
10-7 moles CHCl 3
. x
4.94 x 10-12 moles CHCI,. x 245 x 103 mL . = 12
mL
mole aIr
. mole air
Just like when converting to percent, where you multiply the fraction by 100,
here where the result must be expressed in parts per million, the answer is
multiplied by 1 million:
moles CHCl 3
6
I'
x 1 x 10 =0.12 ppm, compare with the OSHA P.E.L of
mo e air
50 ppm (29 CPR 1910.1000)
7
1.2 x 10-
11.7
11.7 For the purpose of estimating hazards from toxic vapors or gases of high
molecular weight, it is sometimes incorrectly assumed that settling of the vapor
is determined by the specific gravity of the pure vapor, which is defined as
Molecular weight of the pure vapor
"Molecular weight" of air
instead of the correct specific gravity given by
Molecular weight of air and vapor mixture
"Molecular weight" of air
(a) If the vapor pressure of benzene (benzol), C 6H6, is 160 rom Hg at 20°C,
calculate the correct specific gravity of a saturated air mixture of benzene vapors
and compare it to the specific gravity of the pure vapor.
MW ofC 6H6 = 78
Air is approximately 20% oxygen and 80% Nitrogen
"MW" of air = (0.2 x 32) + (0.8 x 28) = 28.8
78
MWC6H6
Specific Gravity of the pure vapor = MW.
= 28.8 = 2.71
au
Specific Gravity of mixture =
VP H
C6 6
atmospheric pressure
MW
H
C6 6
- VP
p
+
aim
C6 H 6
atmospheric pressure
MWair
MW .
au
SOLUTIONS FOR CHAPTER 11
299
160 x 78 + 760-160 x 28.8
760
= 1.36
Specific Gravity of mixture = 760
28.8
(b) If the chemical PEL for benzene is 10 ppm by volume, calculate the specific
gravity of an air-benzene mixture of this concentration.
Specific Gravity of mixture
(PpmC6H6 x 106)Mwc6H6 +[I-(ppmc6H6 x 10
6
)]MW
air
MWair
6
....
.
(10xl0 )x78+[I-(10xl0 6 )]x28.8
SpecIfIc GravIty of mIxture - -------"---------''--28.8
= 1.00002
(c) What is the maximum specific activity of 14C-tagged benzene in order that
not be exceeded
one half the radiological DAC for 14C (DAC = 1 x 10-6 ｾｃｩｊ｣ｭＳＩ＠
by a benzene concentration of 10 ppm?
cm 3
Note that standard air contains 24.5 x 10
I . at 25°C and 760 mm Hg.
mo e aIr
3
Half of the DAC is calculated as;
IlCi
cm3
IlCi
3
1 x 10. - - 3 x 0.5 x 24.5 x 10
. = 0.01225----'---cm
mole arr
mole air
6
Convert concentration in ppm to moles;
-6 mole C 6 H 6
I'
10 ppm = lOx 10
mo e aIr
Combining the two;
0.01225 IlCi
_ _ _-..-:.m=o=l=e--=:;a=irC--- = 1.225 x 10 3 _ _Il_C_i__
mole C 6 H 6
10 x 10-6 mole C 6 H 6
mole air
IlCi
x
1 mole
= 1.2
IlCi
1.225 x 10 mole C 6H 6 1000 millimole
millimole C 6 H 6
3
300
11.8 .
THE HEALTII PHYSICS SOLUTIONS MANUAL
.
Iodine 131 is to be continuously released to the environment through a
chimney whose effective height is 100 m, and whose discharge rate is 100 m3/
min. The average wind speed is 2 m.p.s. and the lapse rate is stable.
(a) At what maximum rate may the radioiodine be discharged if the maximum
downwind ground level concentration is not to exceed 10% of the ICRP's DAC
of 700 Bq/m3 (2 x 10-8 mCilcm\
(b) How far from the chimney will this maximum occur?
11.8
Calculating part (b) first is required to answer part (a). Looking at table 11.10,
stable lapse rate and a 2 mlsec wind implies category D. The maximum ground
concentration occurs on plume centerline, using equation 11.7:
H= 100m
a = H = 100 m = 70.7 m
z
12
12
(b) Careful examination of figure 11.8 shows that a = 70.7 m at a downwind
3
.
Z
.
distance of 3.1 x 10 m.
(a) Calculating the maximum rate of discharge, Q, is done using equation 11.5
(assuming the 1-131 is a gas, and total reflection by the ground does occur);
x = 0.1 x (700
!;) = 70 !; is the maximum desired ground concentration.
X{X,y)
y = 0 since the maximum is on the centerline;
2
a y = 2.1 x 10 m (from fig. 11.7, at 3.1 x 103m; "D" stability category)
a Z =70.7 m (calculated above)
H = 100 m is the effective stack height
!l = 2 mlsec
Q
Ｍｾ］･＠
Ｚｾ＠
nX(2.1x102m)x(70.7m)x2 m
sec
+[
(2.1:2( ;:,:
1
SOLlTTIONS FOR CHAPTER 11
301
7
Q = 1.8 x 10 Bq is the maximum discharge rate of 1311 to meet 10% of the DAC.
sec
11.9 Inhalation exposure is often described as the product of atmospheric con3
centration and time, as in units of Bqxs/m • Using the ICRP assumptions that
23% of inhaled iodine is deposited in the thyroid, and that the thyroid weighs 20
3
131
g, calculate the dose corresponding to an acute exposure of 1 Bqxs/m of (a) 1,
(b) 1331. (c) Assuming that the other 77 % of the inhaled iodine is absorbed into the
blood and is bound to the protein, calculate the total body doses due to the'
protein-bound iodine.
Reference person (appendix C) breaths 20 liters per minute during light activity.
3
Bq.s 20 liters 1 min
1m
x-1- -3 x - - x
m
min 60sec 10 3 liters
3.33 X 10-4 Bq
The numerical values for the parameters needed to solve this problem, effective
half life, TE and effective absorbed energy per transformation, EE' may be found
in various sources, including several ICRP publications. Although the exact
numbers differ somewhat among the sources, all are estimates based on mathematical models, and are approximately the same. ICRP 2 concisely lists these
values for easy reference, and are tabulated below for use in the solution of the
problem. The ･ｦｾｴｩｶ＠
elimination rate constant, AE, is not listed in ICRP 2, but
is calculated with equation 6.52; AE = 0.693ITE •
131
133
1
1
TE,D
AE' d· 1
Ep MeV/t
Tp d
AE' d- I
Ep MeV/t
Thyroid
7.6
0.09
0.23
0.87
0.8
0.54
Body
7.6
0.09
0.44
0.87
0.8
0.84
The dose is calculated from equation 6.58
D= ｾＬ｡ｮ､＠
1\
E
Do' the initial dose rate, is calculated with equation 6.47:
11.9
302
THE HEALTII PHYSICS SOLUTIONS MANUAL .
qBqx1 tps XEE ｍ･ｾ＠
Bq
t
ｄｯ］ｾＭ＠
x1.6x10-13_J-x8.64x104 sec
MeV
day
ｭｫｧｘＱｾＩｇｙ＠
kg
For the case of 1311 in the thyroid:
tp s
M eV
(3.33 X10-4 x 0.23)Bq x 1__ x 0.23__ _ x 1.6 x 10-13 _J- x 8.64 X10 4 ｾ＠
Bq
t
MeV
d
.
ｄｯ］Ｍｾ
0.02 kg x 1 ｾＩｇｙ＠
kg
.
Do = 1.22 X 10- Gy
Jl
day
The dose to the thyroid is
.
1.22 X 10-11 Gy
131
Do
day
•
D (thyrOId, I) = - =
I
AE
0.09 day-
The dose to the body from
D=
ｸＭｾ
131
10
1.35 x 10- Gy
1 is (Note the equation is split into two lines)
1
x
0.09 d- I
tps
MeV
J .
s
(3.33>< 10-4 xO.77)Bq x l-xO.44--x 1.6 x 10-13 - - x 8.64 x 10 4 Bq
t
MeV
d
ＷＰｫｧｘＱｾＩｇｹ＠
kg
D (body,1311) = 2.48 x 10-13 Gy
For the case of 1331 (Note that the equation is split into two lines)
303
SOLUTIONS FOR CHAPTER 11
D( thyroid,
133
1
1 --x
0.8 d- I
tps
MeV.
J
. 4 s
(3.33 X 10-4 x 0.23)Bq x 1 - x 054-- x 1.6 x 10-13 - - x 8.64 X 10 Bq
t,
MeV
d
J)
x
0.02 kg x 1 - Gy
kg
lI
D (thyroid, 1331) = 3.57 x 10- Gy
1
D(body 133 1 =
x
0.8 d- I
,
M eV
(3.33 X 10-4 x 0.77)Bq x l_tp_s x 0.84__ _ x 1.6 x 10- 13 _J- x 8.64 X 10 4 ｾ＠
Bq
t
MeV
d
ｸＭｾ
70kgx ｬｾＩｇｙ＠
kg
14
D (body, 1331) = 5.32 x 10- Gy
In summary, the doses are
1331
13II
Thyroid
1.4 x 10- 10 Gy . 3.6 x 10- 11 Gy
Body
2.5 x 10- 13 Gy
5.32 X 10- 14 Gy
11.10 Disposal of animal carcasses in a biomedical research institution is by
incineration. If the incinerator requires 34 kg air per minute, how much 131 1
activity may be incinerated per 40 hour week, assuming all the iodine in the
10
animal carcasses is volatilized, if the 10 CFR 20 limit of 2 x 10- IlCilmL in the
effluent air is not to be exceeded?
The 10CFR20 atmospheric limits for inhalation are based on the volume of air
that a person inhales during a year, at a nominal temperature of 25°C. The
3
density of air at STP = 1.293 kg/m • At 25°C, this corresponds to
p(air, 2YC) = 1.293 kg x
. m3
273
273 + 25
= 1.185 kg3
m
11.10
304
THE REALlli PHYSICS SOLUTIONS MANUAL
3
34 kg X 1 m
X lxl06 mL =2.87xl07 mLair
min
1.185 kg
m3
min
10
7
3 8 /lCi
2 X 10- /lCi 2.87 x 10 mL air 60 min 40 hr
----'---x
x
x--=1 . _ _
rnL
min
hr
week
week
BUT, 10 CFR 20 effluent limits are based on continuous (168 hr/wk) exposure.
Therefore, for a week the maximum activity that may be sent up the chimney is
/lCi
hr
J.lCi
13.8 40h x 168 --k = 58 --k
r
wee
wee
11.11
A graphite moderated reactor is cooled by passing 680,000 kg air per hour
through the core. The mean temperature in the core is 300°C, and the thermal
2
neutron flux is 5 X lOB neutrons/cm /sec. If the air spends an average of 10 sec in
the reactor core, what is the rate of production of 41Ar? If the chimney through
which the air is discharged is 100 meters high and has an orifice diameter of 2
meters; and the temperature of the effluent air is 170°C, while the ambient
temperature is 30°C on a sunny day and if the mean wind velocity is 2 m/sec, at
what distance from the chimney will the ground level concentration of 41 Ar be a
maximum? What will be the value of this maximum concentration (in Bq/m3)?
How does this figure compare to the DAC for 41 Ar?
11.11
The production rate of 41 Ar activity is given by equation 5.59:
2
"l 7\1
em
atoms (1 - e -AI)
A = IWV
= 'I''" neutrons
X ()-- X n
2
.
cm . sec
atom
sec
The number, n, of 40Ar atoms that pass through the core is calculated: Argon
comprises 0.934 volume percent of air, and 99.6% of all Ar is 4°Ar. Therefore
consider all the Ar to be 4°Ar. We will use the mean molecular weight of all the
gases in air, 28.8, as the "molecular weight" of air.
6.8xl05 ｫｧｸＱＰＳｾ＠
h
n=
ＲＸＮＭｾ
g
mol air
kg x 9.34 X 10-3 mol Ar x 6.02 X 1023 atoms Ar x 1 h
mol air
mole Ar 3600 s
n = 3.7 x 1025 atomsI sec
SOLUTIONS FOR CHAPTER 11
305
The radioactive decay constant, A, for 41 Ar is
A =0.693 =
0.693
T
s
110 min x 60__.
mm
1.05 x 10-4 sec- I
The cross section for thermal neutron capture at 20°C (293K), 0"0' is 0.64 barns
(CRC). 0" at the core temperature of 300°C (573K) is calculated with equation
5.53, modified to account for the fact that the mean 0" of a thermal neutron
energy distribution is 1.1280"0.
_ cr 293 K ｾＲＷＳ＠
0"573K - 1.128
K = 0.64 X 10- em ｾＲＹＳ＠
K_
-25
2
T
1.128
573 K - 4.1 x 10 cm
24
2
Substituting these values into the activation equation, we have
2
A = 5 X 1013
n
x 4.1 x 10-25 cm x 3.7 x 1025 atoms (1- e-1.05XI0-4s0IxIOS)
2
cm • s
atom
s
ll
A = 8 x lO Bq/s
Maximum.ground level concentration, Xmax , occurs at the downwind distance
where (equation 11.7):
H, the effective chimney height, equation 11.6, is
v, the effluent velocity is given by
3
Qm
1
s2 =
V=
A
m 4 (2 mr
v = 75.5 mls
11:
6.8 x 105 kg
x
h
(1.293 x
273 ) kg ｸＳＶＰｾ＠
273+ 170 m 3
h
306
THE HEALTII PHYSICS SOLUTIONS MANUAL
Substituting into the equation for H gives
1.4
H= 100m+2m x
75.5 m
sec
2
m
K)
1 140
(
x + 443 K = 523 m
sec
H
523m
crz = ,,2
r;:; =
r;:; = 370 m
,,2
The given atmospheric conditions give stability category A. Figure 11.8, Curve
A, shows that crz = 370 m at a downwind distance of 800 m. The ground level
concentration is given by equation 11.5.
X=
Q
e
ＭｾＨＺ［Ｉ＠
ｾ＠
ncrycrzJl
crz = 370 m
cry = 170 m (Fig. 11.8)
=
X(80Q,0)
_.!. x (523 2)
8 x 1011 Bq
sec
n x (170 m) x (370 m) x Ｒｾ＠
xe
2
370 2
sec
5 Bq
X(
800,0
) =7.4 x 10 - 3
m
This does not consider decay during travel time. Considering only the 800 m
distance downwind, at a wind speed of 2 mis, the travel time is 400 seconds, or
6.7 minutes. The maximum ground level concentration, therefore is (equation
4.18);
Bq
Bq
- 0.693 x 6.7 min
=7.4xl05-xe
1I0min
=7.1xl05max
m3
m3
X
SOLUTIONS FOR CHAPTER 11
307
Compare this with the value in 10 CFR 20, Appendix B, table 2, column I, for
8
effluent concentrations, 1 x 10- /-lCiJmL, or the occupational DAC in 10 CFR
20, Appendix B, Table 1, column 3, of3 x 10-O/-lCiJmL. The effluent concentration exceeds the applicable NRC standards for these atmospheric conditions.
About 1013 Bq (270 Ci) of 14C waste is generated per year from biomedical
sources in the United States. If this waste will continue to be generated at the
same annual rate,
(a) What will be the resultant steady state quantity of 14C waste?
11.12
Under steady state conditions:
Amount generated per year = Amount decayed per year
0.693
0.693
7;/2
5720
-4
A= - - = - - = 1.21 x 10 yr
1013 Bq
16
A
SS
="iG = 1.21 10y09 y01 = 8026x 10 Bq
X
(b) How long will it take until 99% of the steady state inventory is reached?
The buildup towards secular equilibrium (steady state) is given by a variant of
equation 4.38
oAt)
A=A(1-e
ss
where Ass is the steady state activity, A is the activity at time t after the start of
buildup.
A
A = 0.99 = 1 - eo"l
ss
eo"l = 1 - 0.99 = 0.01
Solving for t, we have:
11.12
308
THE HEALTH PHYSICS SOLUTIONS MANUAL
In(O.OI)
4
t= -1.21 x 10-4yr-1 = 3.8 x 10 yr
11.13
Analysis of albacore in the Pacific Ocean for 137CS from nuclear bomb
fallout showed the mean concentration to be 2.74 Bq/kg (7.4 pCi/kg) wet weight
during the period 1965 to 1971. Calculate the committed dose equivalent due to
the consumption of 1 kg albacore .per week for 1 year.
11.13
..
2.74Bq x lkg x 1 wk =0391 Bq
Dady mtake- kg
wk 7 days
.
day
Cs-137 elimination follows a two compartment model (ICRP 30), one with a 2
day half life, the other having a 110 day half life. Ten percent of the ingested
activity is deposited in the 2 day half life compartment and 90% is deposited in
the long lived compartment.
The committed dose equivalent is
Sv
H=A Bq·dxS-Bq·d
First, we will calculate A
The 137CS body burden, q(t) at any time after the start of continuous intake is
given by a variant of equation 8.40
where Q01 and Q02 are the steady state activities in the respective compartments, and
0.693
A1 = 2 days = 0.347 day
-1
0.693
""2 = 110 days = 6.3 x 10-3day-l
At steady ·state:
309
SOUJTIONS FOR CHAPTER 11
Amount deposited per day = amount eliminated per day
An the ingested Cs is absorbed and deposited in the body. Therefore, the daily
deposited activity is
. = 2.74 Bq xl kg x 1 wk = 0.391 Bq
qm
kg
qOl
=0.97 Bq
q('f2 =
wk
7d
d
9 X qOl = 8.73 Bq
The total cumulated activity is
A = A(ingestion) + A(elimination)
A(ingestion) is given by a variant of equation 11.17:
A(ingestion) =
:1
qo[t; - (1- e-
JJ
)]
and A( elinmination) is given by a variant of equation 6.58
A(elinmination) =!l...
A
For compartment 1, the short lived compartment, q at the end of the 365 day
ingestion period will equal qo. Substituting the values for compartment 1, we have
A = 0.97 Bq x [365 day _
I
1
x (1- e -0.347 day-l x 365
0.347day-l
､｡ｹＩｾ＠
ｾ＠
+ [ 0.97 Bq ]
0.347day-l
310
THE HEALlH PHYSICS SOLUTIONS MANUAL
Al = 354 Bq . day
For compartment 2, q(365), the activity after 365 days ingestion is
q(365) = 8.73 Bq x ( 1- e ·0.0063 x (365») = 7 .85 Bq
Substituting the values for compartment 2, we have
A = 8.73 Bq[365 day _
2
1
(1- e -6.3xIO-3 day-I x 365 ､｡ｹＩｾ＠
3
6.3 X 10- day-I
ｾ＠
+
7.85 Bq
6.3 X 10-3 day-I
A2 = 3186 Bq·day
The total cumulated activity is
A =354 + 3186 = 3540 Bq·d
11
The S(bodyf-body) factor for 137Cs is 9.1 X 10. SvlBq·d (MIRD Pamphlet No.
11 or page 222 of the text). The dose equivalent due to eating the 137Cs contaminated seafood is calculated using equation 6.97;
-
3
H = A x S = 3.54 x 10 Bqxday x 9.1 x 10-
11
Sv
Bq·day
H = 3.2 X 10- Sv
7
11.14
11
11.14 Krypton gas, tagged with 85Kr to a specific activity of 1.3 x 10 Bq/mole,
(3.5 Cilmole), will be transferred from a tank into another vessel at a rate of 0.1
cm3/min (at 25°C and 760 torr) through plastic tubing. There is a remote possibility that the tubing connection will break, and the gas will escape into the
laboratory. If the laboratory dimensions are 3 m x 4 m x 3 m, what must be the
minimum ventilation rate if the' steady state concentration is not to exceed 1110
5
5
of the 10 CPR 20 limit of 3.7 x 10 Bq/m3 (1 x 10- /-lCilmL)?
The steady state concentration in a ventilated room is given by equation 11.26
SOLUTIONS FOR CHAPTER 11
Generation rate
311
G
c= Ventilation rate =Q
4rnL
Molar volume of gas at 273 K = 2.24 x 10 -1moe
Calculate G, the rate of generation;
1
x 1.3 x 1011 ｾ＠
x
273 K
= 5.32 x 105 Bq
G =0.1 rnL x
min 2.24xl04 mL
mole (273+25)K
min
mole
Calculate the maximum concentration allowed (10% of 10 CFR 20 limit);
5 Bq
4 Bq
C = 0.1 x 3.7 x 10 - 3 = 3.7 x 10
m
5 Bq
5.32 x 10 - .
Q=-=
ｭ］ＱＴＮｾ＠
C
3.7 x 104 B;
m
G
- 0
m.)
3
min
Solutions for Chapter 12
CRITICALITY
Cooling water circulates through a water boiler reactor core at a rate of 4 L/
min through a coiled stainless steel tube of 6.4 rom inside diameter and 213 cm
in length. The concentration of Na and CI in the water is 5 atoms each per
million molecules H 20. What is the concentration of induced Na and CI radioactivity in the cooling water after a single passage through the reactor core if the
11
2
mean thermal flux is 10 neutrons per cm /s and the mean temperature in the
core is 80°C?
12.1
The number of radioactive atomslL produced by thermal neutron irradiation of N
target atomslL during an irradiation time t seconds is:
n= N
targets
L
2
x cr
'" neutrons
cm
X 'I'
X t sec
target
cm 2 ·s
Since the irradiation time of one passage through the core is much less than the
half life of the activated isotopes, the resulting activity is :
A = A sec-1 x n
To correct for the cross section at a different temperature, the following equation
is applied to the cross sections (Ethrington):
1
0' = 0'0 1.128
293K
(273+80)K =0'0 x 0.808
The following reactions and their associated parameters are used for this prob-
313
12.1
314
THE HEALTH PHYSICS SOLUTIONS MANUAL
lem. Also listed is the number of target atoms per liter for each element, which
were calculated as shown below.
Reaction
Cross Sections
cr, b
cro,b
N
targets/L
Activated Isotopes
AS-I
,
T1f2
23N a (n,-o3 6CI
0.53
0.43
1.625 x 1020
35CI(n,y?6CI
43.6
35.2
1.23 x 1020 3.08 X 105 y 7.13 X 10- 14
3SCl(n,p?SS
0.489
0.40
1.23 x 1020
87 d
9.22 x 10-8
37CI(n,y)38CI
0.423
0.34
4.0 x 1019
37.5 m
3.08 x 10-4
15 h
1.28 x 10-5
3
Calculate the number of atoms of each nuclide present per cm .
H 20, with density of 0.97 g/cm3 (CRC) at 80°C, and a MW of 18 g/mole;
6.03 X 10 23 molecules 1 mole 970 g
-------- x
x----=mole
18 g
L
025 molecules H 20
·
3.25 xl
L
Since there are 5 atoms 23Na and CI for every million molecules of water;
25 molecules H2
3.25 x 10
L
°
5 atoms 23Na
6
X 10 molecules H
2
°
= l.625 X 1020 atoms 23Na and CI
L
There are two isotopes of CI; 75.4% 35CI and 24.6% 37CI , whose concentrations
are
20 atoms CI 123 10 20 atoms 35CI
0.754 x 1.625 x 10
L
=. x
L
and
20 atoms CI 40 10 19 atoms 37 CI
0.246 x1.625 x 10
L
=. x
L
Calculate the irradiation time that is, the time that the water spends in the reactor;
distance = velocity x time
SOLUTIONS FOR CHAPTER 12
315
volumetric flow rate
and velocity =- - - - - - - area
.
volumetric flow rate
xt
therefore: d Istance =
area
ＴＰ｣ｾ＠
213 cm=
3
mm
11:/4(0.64
x t
cmt 60 ｳｾ｣＠
x
IDln
t
= 1.03 seconds
For. the case of 24Na, the induced atomic. concentration after a single passage
through the reactor core:
2
cm
'" neutrons
targets
n=N
x cr
x 'I'
2
X t sec
L
target
cm· sec
23N
20 targets
a 043 10.24
n = 1.625 x 10
LX. x
2
cm
1011 neutrons
23N x
2
x 1.03 sec
cm . sec
atom
a
n = 7.2 x 106 atoms 24Na/L
and the activity is
Activity = A x n
Activity
e4Na) = 1.28 x 10. sec· x 7.2 x 106 = 92 Bq/L
5
l
By similar calculations, we find that for the other activations:
14
36CI: In = 7.13 x 10- sec- 1 x 1.23 x 1020 x 35.2 X 10.24 x 1011 x 1.03
= 3.2 x 10-5 ｾｱ＠
316
35
THE HEALTII PHYSICS SOLUTIONS MANUAL
S: In = 9.22 x 10-8 x 1.23 X 1020 x 0.40 X 10-24 x 1011 x 1.03
Bq
=0.47T
CI: In = 3.08 x 10-4 x 4 X 1019 x 0.34 X 10-24 x lOll x 1.03
38
Bq
=431L
12.2
12.2 If the cooling water in problem 1 circulates through a heat exchange
reservoir containing 400 L (including the water in pipes between the core and the
reservoir), what will be the concentration of induced activity in the reservoir
after 7 days operation of the reactor?
The rate of change of activity in the reservoir is:
dQ A . .
.
..
= CtIVIty rate III - aCtIVIty rate out
dt
-
For the i tb isotope, let:
Bq = th e actIVatIOn
. . rate 0 f tel
h ·tb Isotope.
•
K .-.'ffiln
Q.I Bq = quantity of the i isotope in the system at any time t
tb
·A. min- I =radionuclide decay constant for the ith isotope
I
.
QjBq
400 L = concentration at any time t
L)
.
.
Bq + ( Q Bq x 4
rate mto reserVOIr = K min
400 L min
4L) + A min- x QBq
(QBq
rate out of reservoir = 400 L x min
I
SOLUTIONS FOR CHAPTER 12
317
Replacing words with the above equations:
dQ ={K ｂｾ＠
+(Q Bq x ＴｾＩｽ｟ｻＨｑｂｱ＠
400L mm
dt
mm
x
400L
ＴｾＩＫａｭｩｮＭｬ＠
mm
xQ Bq}
Combining terms;
dQ
Bq
__ I .
- = K - . -Amm x QBq
dt
mm
dQ
-=K-AXQ
dt
Q = K (1- e-A./) , the activity for each individual isotope after irradiation time t.
A
.
Calculate the activity of 24Na. A value for K is calculated using the information
in problem 12.1;
Bq 4 L
Bq
K=92- x - . =368-.
L
mm
mm
x = _ _0._69_3_ _
15 hr x 60 min
1 hr
Putting values into the equation and solving for t = 1 X 104 min (7 days):
K
Q =A
(1 -e -A.X/) =
368 Bq
.
(1- e-7 .7x W- 4xlxlO4 ) = 4.78 x 105 Bq 24Na
mm
l
7.7 x 10-4 min-
THE HEALTII PHYSICS SOLUTIONS MANUAL
318
Since Q was defined as the activity in the total system, the concentration is:
5
C
4.78 X 10 Bq
1 liter
400 L
x 1000 mL
1.2 x 103 Bq = 32 pCi
L
mL
the activity concentrations for the other activated radioisotopes are calculated in
a similar manner, and the results are tabulated below. The values for C. and A are
taken from problem 12.1.
I
Isotope
I
24Na
36Cl
35S
38CI
12.3
L
Bq
K,-. =4Cj
mm
92
368
C, Bq
"-,min' 1
Q,Bq
7.7 x ＱＰｾＴ＠
4.78 x lOS
12
3.2 x 10-5 1.28 X 10.4 4.3 X 10.
5.5 x10· 6
1.88
0.47
431
0.0185
1724
1.3
1.83 x 104
9.32 x 104
C Bq
C pCi
L
mL
32
1.2 X 103
32 xl0· 3 8.6 x 10.5
46
1.2
233
6.3
12.3 If the tank of problem 12.2 is spherical, what will be the surface dose rate
due to the induced radioactivity one week after the start of operation?
R·m 2
.
4
re Na) is listed in Table 6.3 as 1.84 Ci. h ,which becomes, using 0.95 radlR
2
2
rC4Na) = 1.84 R· m x 103 mR X 10 4 cm x 10-12 Ci x 0.95 mad
Ci.h
R
m2
pCi
mR
2
rC4Na) = 175x 10-5 mrad·cm
, pCi·h
re 8CI) is not listed. It may be estimated by a variant of equation 6.18.
'-6
mrad ·cm
r = 4.75 x 10 'Lf;E;
PC'l' h r
2
-6
.6 mrad· cm
r=4.75x 10 (0.38 x 1.6+0.47x2.17)=7.73x 10
PC'l' hr
2
Only 24Na and 38CI are gamma emitters, and thus contribute to the surface dose.
Use equation 10.37 to fmd the dose rate at the surface of the sphere;
. -x
1 C x r x-x
4n (1 -e -f1.r )
D=
2
Jl'
SoumoNS FOR CHAPTER 12
319
activity
c=--volume
Calculate the radius of a 400 liter spherical tanle
4
3
3
)
V= -nr
3
r= ( - V
4n
3
ｾ＠ = ( --x400Lx
3
1000 cm ) ｾ＠
=45.7cm
4xn
L
The values in the table below for concentration, C, are from problem 12.2, and the
mean linear absorption coefficients in water, ;::t, are from the data in Table 5.3.
Y
r mrad ·cm
Isotope C pCi
cm 3
MeV
24Na
1.37
1
2.75
1
1.6
0.38
2.17
0.47
32
3sCI
6.3
f
2
/J.,
. mrad
D--
h
f-lGy
Dh
o
pCi· hr
cm-
1.75 x 10-5
0.026
9.409 x 10-2
0.94
7.73 x 10-6
0.027
8.033 x 10-3
0.08
1
Substituting the values from the table into the surface dose rate equation gives,
.[: 24
lor Na
2
iJ- .!x32 pCi3 X 1.75 x 10-5 mrad·cm x
- 2
cm
. 24
pCi·h
4n
x (1_e-00026X45o7)
0.026 cm- 1
-2 mrad
h
D( Na) = 9.409 x 10 - -
and for 3sCI , we find
iJ eSCl) = 8.033 x 10-3 mrad
h
e
iJ = iJ 4Na) + iJ eSCl) = (9.409 + 0.8033) x 10-2 = 0.1 mrad = 1 /J.Gy
h
h
320
THE HEALTH PHYSICS SOLUTIONS MANUAL
The resulting dose at the surface at the tank is 0.1 mradslhr = 1 fJGylhr. Note that
this is also the maximum (equilibrium) dose rate at a NaCI concentration of 5
6
NaCl atoms per 10 H20 molecules.
12.4
12.4 A research reactor, after going critical for the fIrst time, operates at a power
level of 100W for 4 h. How much fission product activity does the reactor
contain at the following times after shutdown of the reactor?
Using equation 12.23b;
A = 1.46 x P x [( 't - t ) -002 - 't -02]
0
t = total time
t = operation time in days = 4 hr = 4124 day = 0.167 day
't - t = cooling time
p= 100W
(a) 1 h
't = 1 hr + 4 hr = 5/24 day = 0.208 day
A = 1.46 x 100 W x [(0.208 d -0.167 d)-{)02 -(0.208 d)-{)02]= 75.4 Ci present
after 1 hr
(b) 8 h
't = 8 hr + 4 hr = 0.5 day
A = 1.46 x 100 W x [(0.5 d -0.167 dfo02 -(0.5 d)-{)02] = 14.2 Ci after 8 hr
(c) 7 days
't = 7 days + 4 hr = 7 day + 4124 day = 7.167 days
A = 1.46 x 100 W x [(7.167 d -0..167 d)-{)02 -(7.167 d)-{)02] = 0.46 Ci after 7
days
(d) 30. days
't = 30 days + 4 hr = 30 day + 0.167 day = 30..167 day
A = 1.46.x 100 W x [(30.167 d -0..167 d)-{).2 - (30.167 d)-{)02] = 0.08 Ci after 30
days
SOLUTIONS FOR CHAP1ER 12
321
12.5 A research reactor. after going critical for the first time, operates at a power
level of 100 W for 4 hr. How many curies of fission product activity does that
core contain?
Equation 12.23
A = 1.46 x P x
[( trO.2- -02] Ci,
'r -
'r
.
is valid only for times greater than or equal to 10 seconds after shutdown.
Therefore, the activity in the reactor can be approximated by graphing the
activity from times 10 - 60 seconds after shutdown, and then extrapolating the
curve back to t = O. Substituting
t = 4 hours = 116 day
'r - t = 10 seconds
=1.16 x 10-4 day
into the equation, we have
I
4 2
,A = 1.46 x 100 x [(1.16 X10- rO.
The calculated activity for the other times, until 60 seconds after shutdown, are
tabulated and plotted below. The extrapolated curve intersects the t = 0 axis at
1000 Ci.
Sec
Ci
10
686
15
616
2o
570
25
536
3o
509
35
488
4o
469
45
453
5o
440
55
427
6o
416
-1--------1
-1--------1
-+--------1
-I-----------f
-+----------f
-+----------f
-+-----------f
ｾｲｯｮ＠
ＫＭｾ＠
the following graph an activity of 1000 Ci is extrapolated.
12.5
322
THE HEALTH PHYSICS SOLUTIONS MANUAL
Ｑｾ｜＠
••
•••
••••
••••
ＱＰＫｾｈＴ＠
M
ｾ＠
ID
ｾ＠
m
••••••
N
N
ｾ＠
N
••••••••
•••••••••••••
ｾ＠
N
ｾ＠
M
Seconds
v
M
ｾ＠
M
0
V
M
V
ID
V
•••••••••
••••
m
V
N
ｾ＠
ｾ＠
ｾ＠
ｾ＠
ｾ＠
SoumoNS FOR CHAPTER 12
323
12.6 An accidental criticality occurred in an aqueous solution in a half filled
mixing tank 25 em diameter by 100 em high. The energy released during the
burst was estimated as 1800 J. Assuming that, on the average, each disintegration
of a fission product is accompanied by a 1 MeV "I-ray, estimate the "I-ray dose
rate at the surface of the tank (which maintained its integrity during the criticality) and at a distance of 25 feet from the tank at 1 min, 1 hr, 1 day, and 1 week
after the criticality.
According to Chapter 12, approximately 190 MeV per fission is released.
-13
J
.
190 MeV
- - - x 1.6 x 10
x X fissIOns = 1800 J
fission
MeV
13
X = 5.92 x 10 fissions required to produce 1800 J energy.
Equation 12.3b can be used to estimate the activity at a given time after a
criticality. Combine this information with the number of fissions required to
produce 1800 J energy.
A = 1.03 x 10-16 T-1. 2 = 1.03 x 10-16 T-1. 2
C.i x 5.92 x 1013 fissions
fiSSIOn
A = 6.1 X 10-3 T-1. 2 Ci
Calculate the activity at each time: ｾｔＬ＠
after criticality
-4
.
.
1 day
1 mm = 1 mm x 1440 min = 6.94 x 10 day
ｾｔ］＠
Ci = 6.1 x 10-3 (6.94 X 10-4 day)-1.2 Ci = 37.6 Ci is the
activity present 1 minute after criticality.
A = 6.1 X 10- ｾｔＭＱＮＲ＠
3
The average dose rate within the tank is calculated with equation 6.68
D(t) = C(t) x
rxg
Since the surface dose rate is one half the average within the tank, the surface
dose rate is
.
1
D (t) =-C(t) x r x gs
2
12.6
324
THE HEALTH PHYSICS SOLUTlONS MANUAL
C(t), the concentration at time t, is
C(t) =. A(t) =
vol.
A(t)
= _ _A_(_t)_ _
2:.(25 cm)2 x 50 cm
2.45 x 10 cm
4
3
4
r = OSf. f.E. R . m = 0.5 x 1 x 1
2
I
I
Ci. hr
r = 4.75 x 103 rad· cm
R
2
2
d
. m x 10 4 cm x 0.95 ra
Ci . hr
m2
R
2
Ci· hr
g , the mean geometry factor = 132.5 cm for a 25 cm diameter by 50 cm tall
cylinder of water (Table 6.4).
The unshielded surface dose rate at time t after criticality is:
b (t) =! x
$
2
2
A(t)
2.45 x 10 4 cm 3
x 4.75 x 10 3 rad . cm
.
x 132.5 cm
Ci·hr
.
rad
D (t) = 12.4
x A(t) Ci
S
Ci. hr
At I1T = 1 minute,
b (1 min) = 12.4 rad
G.hr
S
x 37.6 Ci =466 rads
hr
At a distance of 25 feet (762 cm), the 25 cm diameter by 50 cm tall tank can be
considered a "point" source. The dose rate may therefore be calculated with
equation 10.1
4.75 x 10 3 ｲ｡ｾＮ＠
r
b(t) = - 2 x A(t) =
d
b(t) = 8.18 x 10-3
cm
C1 · hr x A(t)
(762 cm) 2
rad x A Ci
Ci·hr
SOLUTIONS FOR CHAPTER 12
325
For 1 minute after criticality, the dose rate at 25 feet is
•
3
rad
.
rad
3 Gy
x 37.6 Cl =0.31- = 3.1 x 10- D 25 (l min) = 8.18 x 10Ci· hr
hr
hr
,
If we repeat these calculations for the dose rates at the later times post criticality,
we obtain
D (surface)
D (25 ft)
D.T
ACt), Ci
radlhr
Gylhr
radlhr
Gylhr
1 m (111440 d)
37.6
466
4.66
0.31
,3.1 x 10-3
1 h (1/24 d)
0.276
3.4
0.034
2.3 x 10-3
2.3 X 10-5
1d
6.1 X 10- 3
7.6x 10-2
7.6 x 10-4 .
5 X 10-5
5 X 10-7
1 wk (7 d)
5.9 x 10-4
7.3 X 10-3
7.3 X 10-5
4.8 X 10-6
4.8 X 10-8
12.7 A slab of pure natural uranium metal weighing 1 kg is irradiated in a
12
2
thermal neutron flux of 10 neutrons per cm /sec for 24 days at a temperature of
150°F. If the fission yield for 131 1 is 2.8%, how many millicuries of 1311 will be
extracted 5 days after the end of the irradiation?
The activity of a fission product whose yield isfper fission is given by a variant
of equation 5.59. If the fission product decays for a time td after the end of
neutron activation, then the activity is
A=
ｦｾｃｊｮ＠
3.7 x 10
7
(1- e- Ati )( e- Atd ) mCi
a, the fission cross section at 65.6°C (150°F) is calculated from a o (235U, 20°C) =
586 b using a variant of equation 5.53. A "not one over v" factor, 0.974, must be
used in calculating a (65.6°C), as well as the 111.128 factor to account for the
fact that the mean a in a Maxwellian distribution is greater than the most probable a.
o
a (1) = -CJ x not
1.128
#0
Yvv x -T
12.7
326
cr (65.6°C)
THE HEALTII PHYSICS SOLUTIONS MANUAL
0
0
273 + 20
273 0 + 65.60 = 471 b
586 b x 0.974 x
1.128
n, the number of 235U target atoms is
0.72 g 235 U 6.02 x 1023 atoms 235 U/mole
n =1000 g U x 100 g U x
235 g 235 U/mole
n = 1.84 x 1022 atoms 235U
0.693
A (1311) = T (1311) = ｾ＠
0.693
= 0.087 d-1
Substituting these values into the activity equation for 24 days irradiation and 5
days decay, we have
A=
n
2.8 x 10-2 x 10 12
x 4.71 x 10-:-22 cm 2 x 1.84 x 10 22 atoms
2
cm . s
(1- e
e
-O.08x24 )( -O.087X5).
7
3.7 X 10 dP%Ci
A = 3.72 x 103 mCi
12.8
12.8 What is the uranium concentration of uranyl sulfate U0 2S04 aqueous
solution that can go critical if the uranium is enriched to
(a) 10%
For criticility, k = ll£p/ = 1
Eta (11), the mean number of fission neutrons per neutron absorbed in uranium, is
calculated from equation 12.7 and from the data in Example 12.2 in the text.
These data can also be found in various nuclear references.
SOLUTIONS FOR CHAPTER 12
where
327
.
. 235
= 549 b fiISSlOn
cross sectIOn U
0'[5
0' as
. 23S
·
= 650 b ab sorptIOn
cross sectIOn U
0'
= 2.8 b absorption cross section 23S U
a8
N S = 90 = 9 = molar ratio of U-8 to
Ns
10
D-5 atoms in 10% enriched U
v = 2.5 for 235U = average number of neutrons per fission
11 =
549
650+(0.9) x 2.8
x 2.5 = 2.03
The fast fission factor, £, can be assumed to be 1, since this is a homogeneous
assembly. Additionally, p, the resonance escape probability, is also assumed to be
1, since the molar ratio of moderator to feul is very high (Example 12.2 in the
text book discusses each of these factors in more detail).
The value off, the thermal utilization factor that will lead to criticality, koo = 1 is
koo = ll£pj
1 = 2.03 x 1 x 1 xj
j= 0.4926
The dependency of jon the composition of the aqueous U02S04 solution is given
by equation 12.14
j=
cr a5 M 5
+ cragMs
cr a5 M 5 + cr as M 8 + cr aH20 M H2 0 + cr aOzS04 M OzS04
where
M5 = moles U-5 per mole water
Ms = moles U-8 per mole water
Mms04 = moles 02S04 per mole water
0'H20 = absorption cross section of H 0 = 0.664 b (Text, p. 526)
2
O'mso4 = absorption cross section of 02S04 = 0.491 b (Text, p. 526)
M H20r = mole water per mole water = 1
If M = the number of moles of U02S04 per mole H 20 to attain criticality, then for
10% enrichment,
U-
Yu = 0.1, U - Yu = 0.9 ,the number of moles per mole of H 0 for each of
2
the components of the U0 2S04 solution is given in the table below.
328
THE HEALTIl PHYSICS SOLUTIONS MANUAL
cra b
moles
cra x moles
ｾｏ＠
0.664
1
0.664
°2S 04
0.491
M
0.491M
U-5
650
O.lM
65M
U-8
2.8
0.9M
2.52M
Component
Substituting these values into equation 12.14 gives
65M +2.52 M
j- 65 M +252 M +0.664+0.491 M
0.4296
M = 0.00745 mole 10%"enriched U02S04 per mole of water for criticality.
The molecular weight of the enriched uranium is
(235 x 0.1) + (238 x 0.9) = 237.7 g/mole of enriched uranium.
and the molecular weight of the UO 2S04 solution is
237.7 + (2 x 16) + 32 + (4 x 16) = 365.7 g/mole U02S04
The concentration of 10% enriched U02S04 required to attain criticality:
365.7
g
103 g
x 745
. x 10-3 mol U0 2 SO 4 x 1 mol H 2 0 x
mol H 2 0
18 g I L
1 mol U0 2 SO 4
g U0 2 S0 4
= 151.4 LH 2
°
(b) The calculations for 90% enriched U02S04 are made in a similar manner,
using the values for 11 and for j appropriate to 90% enrichment
11 = "
"., +
ＨｾＺＩ＠
cr /5
V =
X".,
ＶＵＰＫＨＡｾＩ＠
"549
2.5 = 2.11
x 2.8
The value ofjneeded for criticality (koo =1) is
SOLUTIONS FOR CHAPTER 12
329
1 = ll Epf
1 = 2.11 x 1 x 1 xf
If we substitute equation 12.14 for f into the criticality equation, we have
By letting M =the number of moles of 90% enriched U02S04, we obtain the
following valueS' to substitute into the equation above:
Component
crb
a
moles
cra x moles
H2O
0.664
1
0.664
S 04
0.491
M.
0.491 M
U5
650
0.9M
585
U8
2.8
0.1 M
0.28M
°2
Substituting these values into the equation, we have
585M +0.28M
585M +0.28M +0.664+0.491M
1 = 2.1 1 x - - - - - - - - - - - -
The molecular weight of 90% enriched uranium is
(235 x 0.9) + (238 x 0.1) = 235.3
and the molecular weight of 90% enriched U02S04 is
235.3 + (2 x 16) + 32 + (4 x 16) = 363.3
The concentration of 90% enriched U02 S04 required in the aqueous solution to
attain criticality is
C = 363.3
g
x 1.023 X 10-3 mol U0 2 S0 4 x 1 mol H 2 0 x 1000 g
18 g I L
1 mol H 2 0
mol U0 2 S0 4
330
12.9
THE REALTII PHYSICS SOLUTIONS MANUAL
12.9 Calculate 11 for 239pU, given that the fission cross section is 664 barns and
the non-fission absorption cross section is 361 barns.
Eta (11), the number of fission neutrons per neutron, is given by equation 12.7:
11 =
fission cross section
.
total cross sectIOn
(J
.
x number of neutrons per fissIOn =_ f _ x v
(J total
For Pu, v = 3.0 neutrons per fission (p. 522 in text)
11( 239 Pu) =
12.10
664
x 3.0 = 1.94 fission neutrons
fission
664 + 361
12.10 The "blood plasma from a worker who was overexposed during a criticality
accident had a 2-Na activity of 37 Bq (0.001 JlCi) per mL 15 hr after the accident.
The accidental excursion lasted 10 msec. What was the absorbed dose due to (a)
the 14N(n,p/4C reaction and (b) the autointegral gamma ray dose due to the n,Y
reaction of the hydrogen. All the Na in nature is 23Na . The thermal ｮ･ｵｴｲｾ＠
activation cross section at 20°C" is 0.53 barns, and T 1I2C4Na) = ＱＵｨｲｾ＠
The neutron flux during the criticality is calculated from the initial induced
activity. The number, n, of the 24Na atoms per rnL that were made radioactive by
the 23Na(n, y)24Na reaction during the 10 millisecond exposure is given by
n= N
23Na atoms
rnL
x ｾ＠
neutrons
cm 2
"
2
X (J - - x t sec
cm . s
atom
The initial activity of these 24Na atoms is given by
Since we had 37 Bq/mL at exactly 1 half life (T = 15 h) after the criticality, the
initial activity was 74 Bq.
SOLUTIONS FOR CHAPTER 12
74 S-l =
0.693
331
xn
15 hx ＳＶＰｾ＠
h
n = 5.77 x 10 atoms 24Na
6
N, the number of 23Na target atoms, is detennined by the NaCI concentration in
the blood, 9 mg/mL (ICRP 23):
6.02 X 10 23 atoms Na
23
N =
mol
x 9 x 10-3 g NaCI = 9.3x 1019 atoms Na
mL blood
mL blood
58.5 g NaCI
mol
The activation cross section for 2200 mlsec neutrons, (Jo (20°C) is corrected to
37°C by a variant of equation 5.53
cr 0
cr(37 C) = 1.128
o
rr: 0.53 b
273 + 20
'IT = 1.128 x 273+37 =.0.46 b
Substituting these values into the activation equation, we have (note equation is
in split into two lines)
24N
5.77 x 10 6 atoms
a=
mL
23N
t
9.3 x 10 19 a oms
｡ｸｾ＠
mL
cm 2
23
n2 x 0.46 x 10-24
x10x10-3 S
cm . s
atom Na
ｾ＠ = 1.35 X 1013 __n__
cm 2 ·sec
a) The absorbed dose due to the Ｑｾ＠
(n,p) 14C reaction is the product of the dose
rate, equation 6.105, and the lOx 10-3 second exposure time
332
THE HEALTII PHYSICS SOLUTIONS MANUAL
J
<\>Ncr X Q X 1.6 X 10-13 - D = iJ X f =
MeV X t
l}{g
Gy
where
<\> = 1.35 X 10 13 _ _n__
cm 2 ·sec
24
N = 1.49 X 10 nitrogen atoms/kg tissue (Table 6.12)
cr = 1.75 X 10-24 cm2 (page 225 in text)
Q = 0.63 Me V = energy released per (n,p) reaction (page 616 in text)
3
t = lOx 10- sec
D(n,p) =
1.35 x 10 13
ｾ＠
cm ·s
X
t
.
2
J
1.49 X 10 24 a oms X 1.75 x 10-24 cm xO.63MeVx1.6xl0-13-kg
atom
MeV
D(n, p) = 35.5 X 10-3 Gy
(b) The absorbed dose rate from the IH(n, y)2H reaction is calculated by combining the (n, y) reaction rate, equation 6.106
where
<l> = neutron flux
25
NH =5.98 X 10 hydrogen atoms per kg (Table 6.12)
crH =0.33 X 10-24 cm2 (text, p. 252)
with the equation for a uniformly distributed gamma emitter, equation 6.82
.
A
D=-x<llxi1
m
SOLUTIONS FOR CHAP1ER 12
333
where
Aim = Bq/kg
<p = 0.278 (Table 6.8) = absorbed fraction for 2.23 MeV gamma unifonnly
distributed throughout the body
/). = dose rate in an infmite mass whose specific activity is 1 Bq/kg
!J.. = 2.23 MeV x 1.6 x 10-13 _1_ xl dps x 1 Gy
d
MeV
Bq
1 1/kg
!J.. = 3.57 X 10- 13 Gy/s
Bq/kg
The absorbed dose is the product of the absorbed dose rate and the exposure time
D(n, y) = D(n, y) x t
"Bq"
Gy/s
kg
Bq/kg
D( n, y) = (<I> N H cr H ) - - x <p x !J..
x t sec
Note that the equation is split into two lines:
D(n,y) =1.35x1013
2
atoms
24 cm
X 0.33 x 10--x
kg
atom
n
2
ｸＵＮＹＸＱＰｾ＠
cm ·s
x0.278x3.57x10- 13 Gy/s x 0.01 sec
Bq/kg
D(n, y) = 0.264 Gy
Note that this problem only considers the dose from the thermal portion of the
neutron spectrum.
12.11 (a) For the case where k
= 1.0025, and an initial number of 1000 neutrons,
how many neutrons will be present after 10 generations?
Equation 12.5 gives the multiplication factor
12.11
334
THE HEALTH PHYSICS SOLUTIONS MANUAL
N J+I
ｮ･ｵｴｲｯｳｾ＠
- = 10025
.
k eff = N
neutron
generatIOn
To find the number of neutrons after 10 generations (n = 10);
Nfn = (kefft- No = (1.0025)9 X (l000) = 1023 neutrons
I
(b) After how many generations will the neutron flux be doubled?
N = (k )n-l N
n
eff
0
1
2000 = (1.0025t X (l000)
Solve for n;
2 = (1.0025t-
1
log 2
1
n = log 1.0025 + = 279 generations
--
12.12
12.12 At 20 minutes after a criticality accident the dose rate in a laboratory from
the fission products was 15 Gy/hr (1500 radlhr). If the laboratory ventilation
system was shut down at the time of the criticality, how long would it take before
a person could enter the laboratory if his dose equivalent during a 15 minute
exposure time is not to exceed 50 mGy (5 rad)?
1hr
Tl = 20 min x. 60 min = 0.33 hr
.
Gy
D =151
hr
iJ = 50 mGy = 50 mGy x 60 min x
15 min
2
15 min
hr
1 Gy
= 0.2 Gy
1000 mGy
hr
Equation 12.4 is used to determine the time:
iJ
T-l.'l:
Dl
ｾＭｌＲ＠
- .2 - =2- -
_I
SOLUTIONS FOR CHAPTER 12
335
. . . ( Ii Jl.2
D 2 =D1 -T
2
Inserting values and solving for T2 :
5Gy (0.33hrJI.2
Gy
0.2-=1 - x
hr
hr
T2
T2 = 12.17 hrs. = 730 minutes after the accident, OR 710 minutes after the first
measurement (730 - 20 = 710)
15
12.13 A transient burst of 1 X 10 fissions in an unshielded accumulation of
fissile materials causes a total dose equivalent of 0.25 Sv (25 rem) at a distance
of 2 meters. If the neutron to gamma dose equivalent ratio is 9, what were the
absorbed doses from the gammas and from the neutrons?
The dose equivalent is found using 10CFR20 quality factors. The quality factor
for gamma rays is 1. The quality factor for fast neutrons (all neutrons are born
fast, Chapter 5) is 10. The equation representing the dose equivalent is:
H=QxD
H(n)
H(y) = 9
H (y) + H (n) =0.25 Sv
Replacing H(n),
9H(y) + H(y) =0.25 Sv
H(g) = 0.025 Sv
H(n) = 0.250 Sv - 0.025 Sv = 0.225 Sv
Gy
D(g) = 0.025 Sv x 1 Sv
= 25 mGy
1 Gy
D(n) = 0.225 Sv x 10 Sv = 0.0225 Gy = 22.5 mGy
12.13
336
12.14
THE HEALTII PHYSICS SOLUTIONS MANUAL
12.14 The composition, by weight percent, of a concrete mix used in reactor
shielding consists of oxygen, 52.17%, Si: 34.0%, Ca: 4.4%, AI: 3.5%, Na: 1.6%,
Fe: 1.5%, K: 1.3%, H: 1.0%. The density of the concrete is 2.35 g/cm3•
(a) Find and tabulate the thermal (2200 m/s) absorption cross section for each
element.
(b) Calculate the linear attenuation coefficient (macroscopic cross section) of the
concrete.
The microscopic cross section (cra in barns), for each element is tabulated in the
table below (CRC).
To calculate the total macroscopic cross section for the concrete, the macroscopic
cross section for each isotope of each element in the concrete is calculated, then
the total summed. The values were calculated in the table below in the following
manner;
From equation 5.23, the macroscopic cross section, L, is
For each element, N, the number of atoms present per cm3, must be calculated,
and then mUltiplied by the microscopic cross section. Using 0 as an example
6.02 X 10 23 atoms.
3
g- = 1.42 x 10 24 x ｾ＠
=m=o=le:.... x f. x 2.35_g_
cm 3
Ai
Ai mole
.____
N j , atoms/cm =
I
N(O) = 1.42 X 10 24 x 0522 = 4.63 x 1022 atoms
16.
cm 3
28
cr(O) x N(O) = 2.8 X 10- x 4.63 X 1022 = 1.29 X 10-5 cm- l
I
I
I
i.
!
j
J
SOLUTIONS FOR CHAPTER 12
j, wt.
337
Fraction,
abundance
in concrete
a.cM
x 10- 24
Cross
section, b
Oxygen
5.22E-01
2.80E-04
16
4.63E22
1.29E-05
Silicon
3.40E-01
1.70E-01
28
1. 72E22
2.91E-03
Calcium
4.40E-02
4.30E-01
40.1
1.56E22
6.67E-04
Aluminum
3.50E-02
2.30E-01
27
1.84E21
4.22E-04
Sodium
1.60E-02
5.25E-01
23
9.88E20
5. 17E-04
Iron
1.50E-02
2.56E+00
55.9
3.81E20
9.72E-04
Potassium
1.30E-02
2. 1OE+OO
39.1
4.72E20
9.88E-04
Hydrogen
1.00E-02
3.32E-01
1
1.42E22
4.70E-03
Sum
1.12E-02
Elerrent
1
AI'
N,J at/cc
ar X N j , cm- 1
Cross Section
g
mole
The macroscopic cross section for this concrete is 1.1 x 10-2 cm-1
12.15 The 1311 fission yield is 2.77%. What is the 1311 activity in the core of a
power reactor that has been operating at a power level of 3000 MW (t) for
(a) 8 days
(b) 30 days
(c) 60 days
(d) 180 days?
The fission product activity at time t after start of neutron irradiation is given by
a variant of equation 4.38.
where K is the production rate of the fission product.
fissions
131
1
13l 1
=2.74x10IsK = 3.3x 10 10 sec x3x10 9 Wx2.77x10-2
W
fission
sec
131
0.693 0.693
1
A( I) =T
= 8d = 0.087 d-
Substituting these values into the inventory equation, we have
12.15
338
THE HEALTIl PHYSICS SOLUTIONS MANUAL
131
1
ACt) = 2.74 x 10 x
sec
18
(1 -e
-0.087 x
I) dps (or Bq)
and in mCi, the activity is
18
(1
-0.087 XI)
2.74 x 10 S-1
A(t) x -e .
- 3.7 X 10 10 S-1 . Ci· 1
131 I
A(t) = 7.4 x 107_x(1_e-O·087 x')Ci
sec
For t = 8 days
131
I
A(8) = 7.4 x 107_x(1_e-0.087 x 8)=3.7 x 107 Ci = 37 MCi
sec
By similar calculations for the other operating times, we have
12.16
t, days
Bq
MCi
8
1.37E+18
37.01
30
2.54E+18
68.54
60
2.72E+18
73.64
180
2.74E+18
74.05
12.16 Tritium is produced in a nuclear reactor in ternary fission, in which one 3H
4
nucleus is produced in every 10 fissions. What is the tritium activity in a reactor
that had been operating at a mean power level of 3000 MW(t) for 2 years?
The activity of a fission product in the core of a reactor that had been operating
at a power level of P watts for a time t is given by
,
A(t) =K(1 - e-AI) dps (or Bq)
where K = production rate of the fission product, atoms/second.
SOLUTIONS FOR CHAPTER 12
339
fiss
atoms
K= Pwattsx 3.3x 10 10 - - x f - s· W
fiss
K = 3 X 10 9 W x 3.3 X 10 10 fiss x 1 X 10-4 atoms = 9.9 x 1015 atoms
s·W
fiss
0.693 _ 0.693
_I
A( H) = T
- 12.3 yr - 0.056 Y
3
In terms of curies,
1.1 X 10 15 Bq
4
A(2 y) = 3.7 X 10 10 Bq =2.9 X 10 Ci
Ci
s
Solutions for Chapter 13
EVALUATION OF PROTECTIVE MEASURES
.13.1
A series of measurements with threshold detectors showed the following
spectral distribution of neutrons:
13.1
Percent neutrons
Energy
40
20
10
10
10
10
Thermal
1000 eV
10,000 eV
0.1 MeV
1 MeV
10 MeV
When 500 mg 32S was irradiated for 2 h in this field and then counted in a 2n
counter 24 h after the end of irradiation, the result was 500 counts/min. What is
the dose rate in the neutron field?
The neutron dose rate depends on the neutron flux, <\>, which is determined by
measuring the induced activity. Fast neutron irradiation of 32S produces 32p,
whose Tlfl = 14.3 days:
The induced activity's relation to the flux after an irradiation time tj is given by
equation 5.59:
AN = ｾ｣ｲｮＨ＠
1- e-J..t)
The induced activity, A, determined with a 50% counting efficiency (2n geometry)
is
341
342
THE REALTII PHYSICS SOLUTIONS MANUAL
500 counts
2 dis
1 min
6 67 dis
x
=1.A ----x
mm
1 count 60 sec
sec
Correcting for 24 hours decay, we have
Ao = 17.5 dps
(j
e2S, 10 MeV) = 0.4 bam (Garber, D.l. and Kinsey, R.R.:Neutron Cross Sec-
tions, BNL 325, 1976). Only the 10 MeV neutrons contribute to the activation·
reaction because the reaction has a neutron energy threshold of 1.5 MeV.
11, the number of 32S target atoms is
11
= 0.5 g x
"A( 32 p) =
ｾＯＲ＠
23
6.02 X 10 atoms/mole
/
32 g mole
ep) -
0.693
_
2
21
= 9.4 x 10 atoms
0.693
-3 -I
14.3 d x 24 hid - 2 x 10 h
Substituting these values into the activation equation, we have
2
175 S-1 = <I> neut x 0.4 x 10-24 cm x 9.4 x 10 21 atoms(1- e-O.002h-lx2 h)
cm 2 . s
atom
<I> = 1.17 X 10 6
neut
cm 2 ·s
Since 10 Me V neutrons constitute only 10% of the total neutron flux, the fluxes
for the other neutrons are:
343
SOLUTIONS FOR CHAPTER 13
. mSv
H.,h
1
<I>.xl0 6 neut
<1>( msv)
cm2.s
40h
Energy
I
I
Thermal
4.68
270
433
1,000 eV
2.34
280
209
10,000 eV
1.17
290
101
0.1 MeV
1.17
58
504
1 MeV
1.17
10
2925
10 MeV
1.17
8.5
3441
iJ = ｾｩｊＮ＠
= 7613 mSv = 7.6 Sv
I
Column 3 in the table above, <1>(
ＱＴｾｶＩ＠
h
h
, which is taken directly from Table 9.5,
gives the neutron fluence rates, or fluxes, for a 1 mSv dose from a 40 hour
exposure. These fluxes thus lead to a dose rate of 1140 of 1 mSv/h. The dose rate,
if;, from each Qf these neutron fluxes is calculated by:
. mSv
<l>i
---;------'-:=
<1>(1 msv)
40h
H.h
I
1 mSv
40 h
For the thermal neutrons, we have
iJ(thermal) = ｾ＠
6
x 4.68 X 10
40
270
= 433 mSv
h
The dose rates from each of the other groups of neutrons were calculated similarly, and are listed in the table above. The dose rate in the neutron beam is
iJ = ｾｈｩ＠
i=!
= 7613 mSv = 7.6 Sv
h
h
344
13.2
THE HEAL11I PHYSICS SOLUTIONS MANUAL
13.2 A sealed 90Sr source is leak tested. The wipe, counted in a 2n gas-flow
counter, gave 155 counts in 5 min. The background was 130 counts in 5 min. At
the 95% confidence level, is the source contaminated?
ng = 155 counts
tg
=5 min
155 counts
.
= 31 cpm .= Mg
5mm
rg =
nb = 130 counts
tb
=5 min
130 counts
= 26 cpm = Mb
5 min
r =
b
Using equation 9.49 to find the "t" value:
t =
Mr==g=-=M=-,"b
.:.-1
1
-ll
_
\31- 26\
1.48
+ 26
5 5
Since this is a one tail test, a "t" value of 1.645 is required to determine whether
_Mg > Mb at the 95% confidence level (as listed in chapter 9). There is no difference between the two counts at the 95% confidence level, and therefore the
source is not contaminated.
13.3
13.3 An air sample on a filter paper was counted in a 2n gas flow counter, and
gave 800 counts in 5 min. A background count gave 260 counts in 10 min. What
was the standard deviation of the net counting rate?
n g = 800 counts
·t=5min
g
rg=
800 counts
5 min =160cpm
nb = 260 counts
tb
= 10 min
260 counts
r = - - - - 26cpm
b
10 min
SoumoNS FOR CHAPTER 13
345
Using equation 9.33 to find the standard deviation:
160
cr n =
26
5+ 10 = 5.9 cpm
Rnet = 160 cpm - 26 cpm = 134 cpm
134 ± 5.9 cpm
13.4 A radioisotope worker weighing 70 kg inadvertently drinks water contain-
ing 3.7 MBq (100 JlCi) 22Na. Following this accidental exposure, his body
burden was measured by whole body counts made over a period of 2 months.
The following retention function was fitted to the whole body counting data:
Q(t) = 1.8exp(-0.082t) + 1.gexp(0.052t) MBq
Calculate:
(a) The cumulative activity, in Bq days.
Equation 6.91
AEl = 0.082 d-
1
AE2 = 0.052 d-
1
ASI
(0) = 1.8 MBq
AS2
(0) = 1.9 MBq
Ii = 1.8 MBq + 1.9 MBq
0.082 d- 1
.
0.052 d- 1 - 58.5 MBq d
(b) The initial dose rate, assuming the 22Na to be uniformly distributed throughout the body
According to Fig. 4.8, 22Na emits a 0.544 MeV positron in 89.8% of the decays,
13.4
346
THE HEALTIl PHYSICS SOLUTIONS MANUAL
and a 1.277 MeV gamma in every decay. The dose rate from an internally
deposited radioisotope is given by equation 6.47:
D=
1 tps
- MeV
J
sec
Bq
t
MeV
hr
q Bq x - - x E - - x 1.6 x 10- 13 - - x 3600mkg x
Ｑｾ＠
/Gy
kg/'
In the case of 22N a, 3 different radiations contribute to the dose: the 0.544 MeV
positron (E = 0.216 MeV, NRHH), the 0.51 MeV annihilation photons (2
photons per positron), and the 1.277 MeV gamma. The effective energy per
transformation, E e, is the sum of these 3 contributions. In the calculation below
for E,e
1; = the number of particles per transformation
<l> (E ) = The fraction of the gamma ray energy absorbed in the body (Table 6.8,
Total body to Total body)
-()
-
J3+
MeV
t
J3 +
E J3+ = f x E = 0.898- x 0.216-- = 0.194 MeV/t
E(annih) = f x 0.511 x <I> = 0.898£ x 2 l x 0.511 MeV x 0.34 =0.312 MeV/t
y
MeV
t
Y
Y
J3+
t
.
.
E(Y) = f x E x <I> = 1- x 1.277-- x 0.31 = 0.396 MeV/t
Y
E e = 0.194 MeV/t + 0.312 MeV/t + 0.396 MeV/t
Ee = 0.902 MeV
t
When we substitute 0.902 MeV/t for Ee and 70 kg for m in the equation for iJ,
we have
3.7 X 10 6 Bq x 1 tps x 0.902 MeV x 1.6 x 10-13 _J_ x 3600 sec
D=
Bq
t
MeV
hr
70kg x
D = 2.74 X 10-6 Gy = 27.4 flGy
hr
hr
Ｑｾ＠
/Gy
kg/,
SOLUTIONS FOR CHAPTER 13
347
13.5 The maximum permissible skeletal burden of 90Sr is 74 kBq (2 /-lCi). Calcu-
late the number of transformations per minute per 24 hr urine sample that may be
expected from one fourth of this skeletal burden if 0.05% per day is eliminated in
the urine.
1 trans
74 kBq x 1000 Bq x sec x 60 sec = 4.44 x 106 trans
1 kBq
1 Bq
min
min
One quarter of the burden would then be:
6
6 trans
4.44 x 10 - . - x 0.25 = 1.11 x 10 tpm
mm
Only 0.05% is eliminated in the urine per day:
0.05
6
1.11 x 10 tpm x 100 = 555 tpm
13.6 Using the ICRP three compartment lung model and the data for the refer-
ence man, calculate the ratio of concentration of soluble 1 /-lm AMAD uranium
particles in the air to uranium in the urine, Bq/m3 air per Bq/L urine, for the case
where a steady state has been attained through continuous inhalation of the
uranium.
Figure 8.3 and Table 8.5 gives the details of deposition of material in the respiratory tract;
Region
Percent
Deposition
Deposited fraction
absorbed into
body
Fraction inhaled
absorbed into
body
N-P
30%
0.5
0.15
T-B
8%
0.95
0.076
P
25%
1
0.25
Total absorbed
0.476
ａｳｾｵｭ･＠
The total fraction of the inhaled uranium absorbed into the body is 0.476.
3
1 Bqlm exposure, and from Appendix C, a reference person breaths 20
3
m /day.
13.6
348
THE HEALTI-I PHYSICS SOLUTIONS MANUAL
3
1 Bq x 20 m = 20 Bq
day
day
m3
So a person would take in 20 Bq/day. Since only 0.476 is the fraction absorbed in
the body;
9.6 Bq x 0.476 = 4.6 Bq is absorbed. Reference man excretes 1.4 liters per day
day
day
(appendix C).
Under steady state conditions, intake = output
Bq
d
Bq
4.6-x
=3.3 -----"---d
1.4 L urine
L urine
For an exposure of 1 /lm AMAD at 1 Bq/m3, 3.3 Bq/liter of urine would be
expected if we neglect the small fraction of uranium deposited for long term
storage in the bones.
It is interesting to compare this with the older ICRP 2 two compartment model:
According to ICRP 2, only 25% of the material is absorbed into the body fluids.
The total fraction of the uranium absorbed into the body is 0.25. As calculated
earlier, a person would take in 20 Bq/day. Since only 0.25 if the fraction absorbed in the body;
20
Bq
d
x 0.25 = 5
Bq
d
is absorbed. Reference man excretes 1.4 liters per day
(appendix C).
= 3.6
Bq
5 Bq x
d
d
1.4 L urine
L urine
.
3
For an exposure of 1 Jlm AMAD at 1 Bq/m , 3.6 Bq/L of urine would be expected using the ICRP 2 two compartment model. Compare this with the three
compartment model result of 6.8 BqIL and note the difference is about a factor of 2.
SOLUTIONS FOR CHAPTER 13
349
13.7 The body burden of l37Cs at time t days following a single intake Q(O) is
.
by
gIVen
13 7
•
Q(t) = Q(O) (0.1 e-D·693t + 0.ge-D·Ollt).
If the ratio of urinary to fecal excretion is 9: 1, calculate the activity per 24 hr
urine sample 1 day and 10 days after ingestion of 50,000 Bq (1.35 f.!Ci) l37Cs.
Solve for day 1
Q(O) = 50,000 Bq
t = 1 day
Q(t) = Q(O) x (0.1 x e-D·693t + 0.9 x e-D·Ollt)
Q(1) = 50,000 x (0.1 x e-O· 693 xl + 0.9 x e-O·0I1 Xl) = 47008
The quantity still retained in the body is 47008 Bq, so the total quantity excreted
(in both the urine and feces) is:
3
50000 - 47008 = 3.0 x 10 Bq is the total activity excreted on day one after
exposure.
Since the urinary to fecal activity ratio is 9:1 (the total excretion would be 10,
the sum of both excretion pathways), the quantity in the urine would be 90% of
the total.
9
3
3
3.0 X 10 Bq x 10 = 2.7 X 10 Bq is the expected activity in the urine on day 1.
The activity excreted during day 10, A (10), is the difference between the body
burdens on day 9 and day 10:
A(10) = Q (9) - Q (10)
Q(9) = 50000 x (0.1 x e-O.693x9 + 0.9 x e-O· OllX9 ) = 4.08 x 104 Bq
Q(10) = 50000 x (0.1 x e-O.693 x 10 + 0.9 x e-O·OllXlO) = 4.03 x 104 Bq
Subtract the activity in the body on day 9 from day 10 to determine the excretion.
4
4.08 X 10 Bq - 4.03 X 10 Bq = 500 Bq
4
Since 90% of the total is excreted in the urine, the activity in the urine would be:
350
THE HEALTH PHYSICS SOLUTIONS MANUAL
9
500 Bq x 10 = 450 Bq is the activity expected in the urine on day ten
13.8
13.8 A chemist accidentally inhaled a 14C tagged organic solvent that is readily
absorbed from the lungs. The solvent is known to concentrate in the liver. That
part of the solvent which is eliminated before deposition in the liver leaves in the
urine; the detoxifIcation products are eliminated from the liver into the G.!. tract
and into the urinary tract; 25% is eliminated in the urine and 75% in the feces.
Following the inhalation, 24 hr urine samples were collected over a 2 week
period and the following data were obtained:
Days after inhalation
1
2
3
4
5
6
8
10
1-2
14
kBg/sample
98
57
39
265
20
18
12
10
7.4
5.9
(a) How much activity was absorbed into the body?
The activity absorbed into the body is equal to the amount excreted; and the
amount excreted is represented by the area under the excretion curve.
Graphing the data:
1000
100
•
•
•
•
•
•
•
10
•
o
2
4
6
8
Days
10
12
•
14
16
SOLUTIONS FOR CHAPTER 13
351
Since we know that clearance from the body follows first order kinetics, we plot
the data on semi-log paper. Since the data fall on a curved line on semi-log .
paper, we know that several different compartments are being cleared at different
rates. We note that the curve, after some time, eventually becomes a straight line.
This means that all the shorter lived compartments have been cleared, and only
the slowest clearing compartment is left. The intercept on the time = 0 axis of the
extrapolated line represents the activity initially deposited in the longest lived
compartment.
1000
•
100
•
>-
ｾ＠
'"
C.
E
.,'"
0al
---
""
10
o
4
2
6
8
10
12
14
Days
The extrapolated line intersects the t = 0 axis at 32 kBq. The slope of this line
can be determined from its half clearance time, that is, the time until the urine
activity reaches 16 kBq per sample. From the curve, we find the half clearance
time to be 5.7 days. The slope of this line is computed:
1
f\"
IL
0.693
= 5.7 d =0.12 d
-1
16
352
THE HEALTH PHYSICS SOLUTIONS MANUAL
·The equation representing the long term compartment of the clearance curve
therefore is:
ACt).
= 32 x e-O.l2xr kBq/day
unne
Since the clearance curve represents the sum of several compartments, the
activity in the remaining compartments is determined by subtracting the contribution of the long term compartments from the total, and plotting these differences on semi-log paper; as shown in the table and graph below. In this case, the
differences fall on a straight line, thus showing that there is only one rapidly
cleared compartment. By extrapolating this rapid clearance component to zero,
we find the intercept to be 128 kBq; the half clearance time for this compartment
is 1.04 days, which yields a slope of 0.67 per day. The equation for the urinary
clearance curve, which is the sum of 2 compartments is:
A(t).
= 128 x e-O·67xr + 32 x e-O· 12xr Bq/day
unne
1000
J,
•
100
>-
•
III
:g
a.
E
III
1:1
ｾ＠III
i.
II
.><
•
•
•
III
&
•
•
•
10
o
2
4
6
8
Days
•
10
•
12
•
14
16
SOLUTIONS FOR CHAPTER 13
353
Day
Total activity in urine,
Kbq/sample
Long term
kBq/sample
Difference
kBq/sample
1.
98
28
128
2
57
25
32
3
39
22
17
4
26
20
6
The urinary activity of the slowly clearing compartment represents only 114 of
the activity cleared from the liver. The total activity cleared from the liver is
therefore four times that found in the urine. Thus, the intercept for the total
"cleared" from the liver is 4 x 32, or 128 Bq. Therefore the curve representing
the total clearance from the body is
AT = 128 x e-O.67xt + 128 x e-D.l2xt
The total activity absorbed into the body and eliminated by both the urine and
feces is' given by the area under the curve described by the above equation.
Integrate to find the area:
ro
00
A = f 128 x e-O.67 x t dt + f 128 x e-Q·12 x t dt
TOO
AT =
128 kBq 128 kBq
0.67 + 0.12
191 kBq + 1067 kBq = 1258 kBq = 1.3 MBq is the
total activity absorbed into the body.
(b) What was the dose to the body during the 13 weeks after inhalation?
The absorbed dose during a time interval t following a single acute intake is
given by equation 6.57
Where
A = effective clearance rate
Do = initial dose rate
354
D· Ｍｾ＠
THE HEALTII PHYSICS SOLUTIONS MANUAL
q kBq X 10 3 tps X E MeV x 1.6 x 10-13 _1_ x 8.64 x 104 sec
kBq
t
MeV
day
m kg x _11_ /Gy
kg/'
o-
e
For the first compartment, in which E C) =0.049 MeV/t and whose
A = 0.67 d- I , liver weight = 1.8 kg, we have
4
ｄｏｉ］ＭｾＨＷＰＱＮＸＩｫｧｘ［ｇｹ＠
·
191 kBq x 10 3 tps x 0.049 MeV x 1.6 x 10-13 _1_ x 8.64 X 104 sec
MeV
day
kBq
t
·
-6 Gy
DOl = 1.92 x 10 - d
For compartment 2 (the liver weight = 1.8 kg)
｢ＰＲ］Ｍｾ［
1067 kBq x 10 3 tps x 0.049 MeV x 1.6 x 10-13 _1_ x 8.64 x 104 sec
Bq
t
MeV
d
1.8 kg x _11 Gy
.
kg
·
-4 Gy
D02
= 4.02 x 10 - d
The effective half life times of the 14C in both compartments, 1 day and 5.8 days,
are «<·13 weeks (91 days). Therefore the body dose is
(b)
b
D(body) = ｾ＠
Al
1.92 X 10-6 Gy
=
0.67 d-
I
d
= 2.87 x 10-6 Gy
(c) The dose to the liver (compartment 2) is
' )
D(llver
= ｾ＠
.
D02
4.0 X 10-4 Gy
=
0.12 d-I
d
= 3.35 X 10-3 Gy
SOLUTIONS FOR CHAPTER 13
355
(d) The committed effective dose equivalent is calculated with equation 8.2,
using wTof 0.05 for·the liver and 0.95 for the rest of the body, and 1 Gy ｾ＠ 1
Sv for beta radiation.
HE=LWTHT
HE = 0.95 x 2.87 x 10-6 + 0.05 x 3.35 x 10-3
HE = 1.7 X 10-4 Sv
13.9 A health physicist samples waste water to ascertain that the water may be
safely discharged into the environment. The water analysis is made by chemically separating the 90Sr, allowing the 90y daughter to accumulate, then extracting
and counting the 9Dy activity. The volume of the sample was 1 liter, the 90y
ingrowth time was 7 days, and the 90y activity was determined 15 hr after
extraction in an internal gas flow counter having an overall efficiency of 50%.
The background counting rate, determined bya 60 minute count was 35 counts/
min. The sample (including background) gave 2766 counts in 60 min. What was
the 90Sr concentration, at the 90% confidence level?
Only counting error is considered.
Calculating the 90% confidence interval:
n g = 2766 counts
tg = 60 min
2766 counts
rg =
60 min = 46.1 cpm
tb
= 60 min
rb = 35 cpm
rn = 46.1 - 35 = 11.1 cpm
U sing equation 9.43 to find the standard deviation of the net count rate
cr n =
46.1 35
60 + 60 = 1.16 cpm
The 90% confidence interval corresponds to 1.645 standard deviations, so the
90% confidence interval is:
1.16 x (1.645) = ±1.91 cpm is the interval associated with the average so the
-13.9
356
THE HEALTII PHYSICS SOLUTIONS MANUAL
percent errOf, or coefficient of variation, is: Equation 9.39a
1.91 cpm
% error = CV = 111
. cpm = 0.17 = 17%
The counter is only 50% efficient, so only half of the activity is counted, thus,
multiplying the net number of counts by 2 will produce the number of decays;
2
dis
11.1 counts 1 min
1 Bq 037 B
x--=.
q
x
x
count
min
60 sec
dps
Placing this value into equation 4.18 to account for the 15 hour delay before
. the 90y:
countmg
A - A e-J.I
- °
(
0.693
15)
0.37 Bq = AD x e - 64 h x
A o = 0.435 is the activity of the 90Y at the start of the 15 hour extraction process.
Now find the activity of the 90Sr at the start of the 7 day (168 h) growth period
using equation 4.40;
QA = activity of 90Sr at the start of ingrowth period
QB = activity of 90y at end of 7 day-ingrowth period (start of 15 hr period) =
0.435 Bq
0.693
-I
AY- 90 = 64 hr = 0.01083 hr
t = 7 days = 168 hr
I
0.435 Bq = Q (1- e -0.01083 hr- xI68 hr )
A
QA = 0.52 Bq/L is the initial activity of 90Sr isolated. Multiply this by the error
calculated earlier (0.17) to obtain; 0.52 x 0.17 = 0.09.
0.52 0.09 BqIL is the 90% confidence interval of the 90Sr concentration.
±
SOLUTIONS FOR CHAPTER 13
357
13.10 An air sample that was counted 4 hr after collection gave 1450 counts in
10 min. The background was counted for 30 min, and gave a rate of 45 counts/min.
The sample was counted again 20 hr later, and gave 990 counts in 10 min; a 60
-min background gave 2940 counts. If the volume of the .air sample was 1.0 m3, and
if the counting geometry was 50%, calculate the atmospheric concentration of the
3
long lived contaminant, Bq/m3 and I-lCi/cm , and the 95% confidence limits.
Calculating the 95% confidence limits:
At the 4 hour point;
n g1 = 1450 counts
tgl = 10 min
rgl = 145 cpm
tbl = 30 ｾｮ＠
r bl = 45 cpm
rnl = rg1 - r h1 = 145 cp'm - 45 cpm = 100 cpm at 4 hours
Twenty hours later;
ng2 = 990 counts
tg2 = 10 min
rg2 = 99 cpm
tb2 = 60 min
n b2 = 2940 counts
rb2 =
2940 counts
60 min = 49 cpm .
rn2 = rg2 - rb2 = 99 cpm - 49 cpm" = 50 cpm at 20 hours later
Using equation 9.43 to find the standard deviations and equation 9.39a to find
the coefficient of variation, CV:
CV( C1 ) 4 cpm = 0.04
100 cpm
358
THE BEAL1E PHYSICS SOLUTIONS MANUAL
After I1t = 20 hours, 10.6 hour ThB would have decreased, but the long lived
contaminant would still be there. The count rate of the long lived contaminant,
Cu is given by equation 13.34:
C _Ce-JJ:,.z
2
1-e
1
-A!J.t
where
C1 = the first net count rate = 100 cpm
C2 = the second count rate = 50 cpm
A = 0.693 = 0.693 _
-I
T(ThB)
10.6 hr - 0.0654 hr
I1t = time between C1 and C2 = 20 h
Substituting these values gives
50 - 100 x e -O.0654x(20)
Cu = -1-----e -O.0654x(20)
31.5 cpm
cr
Note that CV = - - (Equation 9.39a)
mean
cr( Cu) = .Jl 0.76 + 1.17 = 3.45 cpm
Therefore
Cu = 31.5 ± 3.5 cpm
SOLUTIONS FOR CHAPTER 13
cv=
359
35
=-=0.11
mean 31.5
°a
Since we have 50% geometry, the mean atmosphere concentration is
tpm
1 Bq
31.5 cpm x 2 - - x ----='-cpm 60 tpm = 1.05 B;
CLL1 m3
m
Since the CV = 0.11, cr(conc) = 0.11 X 1.05 = 0.12, and the 95% confidence
interval = ±1.96cr = 0.23
Therefore, CLL = 1.05 ± 0.23 Bq/m3
In traditional units, we have
a( C = CV x C
a )
LL
= 0.11 x 2.84 x 10-
11
= 3.12 X 10-
12
ｾ＠
Ci
12
The 95% confidence interval = 1.96a = 1.96 x 3.12 x 10- f.l
rnL
Ci
rnL
= 6.1 x 10-12 f.l
-12
CLL = (28.4 ± 6.1) x 10
f.lCi
rnL
13.11 A film badge worn by a worker in a fast neutron field showed the following
distribution of proton recoil tracks among 100 random microscopic fields of 2 x
-4
2
10 cm each:
13.11
360
THE REALTII PHYSICS SOLUTIONS MANUAL
Observed Tracks per Field
Frequency
o
40
40
18
2
1
2
3
(a) If 2600 tracks per square centimeter correspond to 1 mSv (100 mrems), what
was the fast neutron dose?
Observed Tracks per Fiekl,n
Frequency, f
fxn
0
40
0
1
40
40
2
18
36
3
2
6
Surn,Lfn.
I
I
82
82 tracks
= 4100 tracks
2
cm 2
100 fields x 2 x 10-4 cm
Since 2600 tracks per square centimeter corresponds to 100 mrems,
H = 4100
tracks
cm
2
x
H = 158 mrems x
100 mrems
2600 tracks
cm 2
1 mSv
100 mrems
=158 mrems
= 1.58 mSv
(b) What is the 95% confidence limit of this measurement?
Equation 9.36 is used to fmd the standard deviation of the measurement;
(J
=·frz = J82 = 9.06
The 95% confidence limit requires 1.96 standard deviations;
1.96 x 9.06 = 17.75
361
SOLUTIONS FOR CHAPTER 13
The 95% coefficient of variation would be found using equation 9.39a:
17.75
1.960'
95%=CV= - - = -82 =0.21
mean
Giving a 95% confidence interval of: ± 0.21 x 158 rnrems = ± 34.2 rnrems
H = 158 ± 34 mrems
1 mSv
95%s = 34 mrems x 100
= 0.34 mSv
mrems
H = 1.58 ± 0.34 mSv
13.12 Using the three compartment ICRP lung model and the physiologic data
for the reference person, compute the dose to the lungs and to the bone following
5
a single acute exposure of 1 Bq·s (2.7 x 10- ｾｃｩﾷｳＩ＠
per cubic meter of respirable
of (a) strontium titinate, (b) strontium chloride.
aerosol, MMAD = 2 ｾｭＬ＠
(a) strontium titinate
Converting to Bq using a breathing rate of 10 liters per minute (rounded from
Table 2 in Appendix C);
Bq . sec 1 min
L
1 m3
-4
1 m 3 x 60 sec x 10 mm x 1000 L = 1.67 x 10 Bq is the inhaled quantity of
activity.
The ICRP three compartment model distributes activity with MMAD = 2 /lm
into the following compartments (Fig. 8.4):
Region of
Deposition
Inhaled Activity,
Bq
Fraction activity
deposited in
region
deposited actvity, Bq
A, (0)
N-P
l.67 x 10-4
0.50
8.35 x 10-5
T-B
1.67 x 10-4
0.08
l.34 x 10- 5
P
l.67 X 10-4
0.18
3.01 x 10-5
N-P region deposition is cleared, and does not contribute significantly to the lung
13.12
362
THE HEALTII PHYSICS SOLUTIONS MANUAL
dose. The acti vity transferred to the blood and GI tract from the N-P region
deposition are calculated later.
The dose to the lung is calculated from equation 6.97
H (lung) = A (lung)Bq . d x S (lung ｾ＠
Sv
lung) - Bq·d
Clearance rates of particulates in the lung are given in Fig. 8.3. For 90SrTi02'
which is class w (moderately soluble), the biological clearance rate = the effective clearance rate because TI12 (90Sr) = 28 years. The activity deposited in each
region, and for each clearance pathway, the fraction of the deposit, its retention
half time T and clearance rate A (0.693fT), are listed in the table below:
Region Bq x 10-5 .
f
T, d
A , d- l
To
N-P
8.35
0.1
0.9
0.01
0.4
69.3
1.73
Blood
GI
T-B
1.34
0.5
0.5
0.01
0.2
69.3
3.47
Blood
GI
3.01
0.4
0.4
. 0.2
1
50
50
0.693
0.0139
0.0139
GI
GI
Blood
P
The cumulated activity, 11, for several compartments is given by a variant of
equation 6.91
-
A. (0)
A = L-,--li
Ai
If we substituted the values from the table above, we have
5
ｓＨｬｵｮｧｾＩ＠
- 05 x 1.34 x 10- Bq 05 x 1.34 x 10-5 Bq 0.4 x 3.01 x 10-5 Bq 0.6 x 3.01 x 10-5 Bq
A=
+
+
+ _ _ _ _---:--I
0.0139 d- I
0.693 d-1
69.3 d3.47 d- I
-4
=4.2 x 10
ｾ＠
rad
C· h
1·
r
(MIRD Pamphlet 11)
SOLUTIONS FOR CHAPTER 13
S = 4.2 X 10-4
363
rad x 1 Gy x
l/lCi
x 1 Sv = 272 X 10-9 Sv
4
/lCi·hr 100rads 3.7xl0 Bq 1Gy
Bq·d
H =; A x S(lung ｾ＠
lung)
H = 1.33 X 10-3 Bq . d x 272 X 10-9 Sv = 3.6 X 10- 12 Sv
Bq·d
According to the ICRP lung model, some of the inhaled 90Sr is transferred from
the lung to the blood, and some is transferred to the GI tract. Then,
30% from the blood goes to the bone
9% from the GI tract goes to the bone
From the N-P region, where 10% goes to the blood and 90% to the GI tract, the
activity deposited in the bone is:
via blood: 8.35 x 10-5 Bq x 0.1 x 0.3 = 2_51 x 10-6 Bq
via GI: 8.35 x 10-5 Bq x 0.9 x 0.09 = 6.77 x 10-6 Bq
In a similar manner, we can calculate the contributions of the T-B and P regions
to the bone, with the results that are tabulated below: The total activity deposited
in the bone is 1.59 x 10-5 Bq
From
VIa
Activity transferred, Bq
Deposition in bone, Bq,x 10- 6
N-P
Blood
GI
8.35 x 10-5 x 0.1 x 0.3
8.35 x 10- 5 x 0.9 x 0.09
2.51
6.77
T-B
Blood
GI
1.34 x 10-SBq x 0.5 x 0.3
1.34 x 10-SBq x 0.5 x 0.09
2.01
0.60
P
Blood
GI
3.01 x 10-SBq x 0.2 x 0.3
3.01 x 10-SBq x 0.8 x 0.09
1.81
2.17
Total transferred to bone
15.9 x 10- 6 Bq
ｾＭＫ
ｾＭｲ
ｾＭｲ
ｾＭＫ
The dose equivalent to the bone is
H(skeleton) = A(bone)Bq.d x S(skeleton ｾ＠
Sv
bone)-Bq·d
364
THE HEALTH PHYSICS SOLUTIONS MANUAL
The cumulated activity, A, is calculated using the ICRP value of 6400 days for
the effective half life of 90Sr in the bone.
-(
). As(O)
A bone =
=
AE
As (0)
0.693/TE
=
15.9xI0-6Bqx6400d
0.693
=0.147 Bq·d
s( skeleton ｾ＠
rad
0.01 Gy
1 /lCi
x ＲＴｾｸ＠
bone) = 8.45 x 10-5
X
X
4
. /lCi· h
rad
3.7 x 10 Bq
s( skeleton ｾ＠
bone) = 5.48 X 10- 10
Sv
Bq·d
d
1 Sv
Gy
(MIRD 11)
The dose to the skeleton is
H( skeleton) = 0.147 Bq· d x 5.48 X 10-10
Sv
Bq·d
= 8 X 10-11 Sv
(b) strontium chloride
Since the MMAD is the same for each compound, the activity deposition is the
same as for strontium titinate.
From Figure 8.4:
Activity
deposited in the
region in Bq
As(O)
Region of
Deposition
Total Activity in
Bq
Fraction activity
deposited in
region
N-P
1.67 x 10-4
0.50
8.35 x 10- 5
T-B
1.67 x 10-4
0.08
1.34 x 10-5
P
1.67 X 10-4
0.18
3.01 x 10-5
The dose to the lung from the inhaled 90SrCl2 is calculated with equation 6.97
H (lung) = A (lung)Bq . d x S (lung ｾ＠
..
Sv
lung) - Bq·d
The calculations for strontium chloride are very similar to part (a), except that it is
a class D compound, and therefore has different retention half times in the lung.
Since the radiological half life is 10ng compared to the biological half life, only the
SOLUTIONS FOR CHAP1ER 13
365
N-P region deposition is cleared, and does not contribute significantly to the
lung dose in the N-P region, so the deposition dose is not calculated. The activity
transferred to the blood and GI tract from the N-P region deposition are calculated later.
T-B region has 2 compartments, 95% clears at 0.01 days (1 = 69.3 d-\ Figure
1
8.3), and 5% clears with a half time of 0.2 d (1 = 3.47 d- ).
The dose to the lung from the inhaled 90SrCl2 is calculated with equation 6.97
H(lung) = A(lung)
x S (lung ｾ＠
lung) -SvBq·d
The cumulated activity is calculated as in part (a), using the table below:
-(l
Region
Bq x 10- 5
f
T,d
A, d- 1
To
N-P
8.35
0.5
0.5
0.01
0.01
69.3
1.73
Blood
GI
T-B
1.34
0.95
0.05
0.01
0.2
69.3
3.47
Blood
GI
P
3.01
1.0
0.5
1.39
Blood
)_ Ai(O)
A ung -
Ai
0.95x1.34x10-5 Bq 0.05x1.34x10-5 Bq 3.01x10-5 Bq
+-----.::.
=
+
69.3d-1
3.47d-1
1.39d-1
A (lung) = 2.19 x 10-5 Bq·d
9
Sv
= 2.72 x 10- Bq.d (Adopted from MIRD 11)
ｓＨｬｵｮｧｾＩ＠
-
H(lung) = A(lung) x S(lung ｾ＠
H(lung) = 6 x 10- 14 Sv
5
9 Sv
lung) = 2.19 x 10-. Bq·d x 2.72 x 10- Bq.d
366
THE HEALTII PHYSICS SOLUTIONS MANUAL
The activity transferred from the lung to the bone too is calculated as in part (a),
as shown in the table below:
From
VIa
Activity transferred, Bq
Deposition in bone, Bq, x 10-6
N-P
Blood
GI
8.35 x 10. 5 x 0.5 x 0.3
8.35 x 10-5 x 0.5 x 0.09
12.53
3.76
T-B
Blood
GI
1,.34 x 10- 5 Bq x 0.95 x 0.3
1.34 x 10- 5 Bq x 0.05 x 0.09
3.82
0.06
P
Blood
3.01 x 10-5 Bq x 0.05 x 0.3
9.03
Total transferred to bone
29.2 x 10-6 Bq
Sv
H(skeleton) = A (bone)Bq ·d x S(skeleton +-- bone)-Bq·d
6
29.2 X 10- Bq x 6400 d x.
548 x 10- 10 Sv = 15
. x 10- 10 S v
k I
H( seeton
)=
0.693
Bq·d
3
Summary of doses in Sv per Bq·s per m exposure
Lungs
Skeleton
Sr-titinate
3.6 x 10- 12
8 X 10- 11
Sr-chloride
6 x 10- 14
1.5 X 10- 10
13.13 13.13 The following size distribution was obtained on a sample of an aerosol:
Percent by number
Class intervaLl-lm
10
0.5-1.0
15
1.0-1.5
15
1.5-2.0
10
2.0-2.5
10
2.5-3.0
10
3.0-3.5
10
3.5--4.5
10
4.5-6.0
5
6.0-8.0
5.
8.0-10.0
SOLlJI10NS FOR CHAPTER 13
367
(a) Plot the cumulative frequency distributions on linear graph paper, on linear
probability paper, and on log probability paper, by number, surface area (assume
the particles to be spherical), and by mass (assume the particles to have a density
of 2.7 g/cm\
Tabulate the cumulative frequency for the diameter, surface area, and mass of the
particles below, in order to graph each:
Less than
% by number
cmnulative %
diameter
surfuce area, ( m)2
mass, g
10
10
1
3.14
1.41
15
25
1.5
7.07
4.77
15
40
2
12.57
11.31
10
50
2.5
19.63
22.09
10
60
3
28.27
38.17
10
70
3.5
38.48
60.61
10
80
4.5
63.62·
128.82
10
90
6
113.09
305.35
5
95
8
201.06
723.80
5
100
10
314.15
1413.68
Linear Graph
8
6
4
2
o
ｾＭｌ＠
0.0
__ｌＭｾ＠
__ｌＭｾ＠
20.0
__ｌＭｾ＠
40.0
__L - - L__ｌＭｾ＠
60.0
Cumulative percent particles
80.0
100.0
368
THE HEALlli PHYSICS SOLUTIONS MANUAL
Linear Probability Paper
10.0 r---.,-----.-,- - - . , - - - . , - - - - - ' ,- - - - " r - - - ' - - - - - r ,-----,
-------+-··-·············-------t----·············t---·-..---.-.+.-.....-..- ..-........+...........+--------.........-{-.....-.
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1.-...................... J1.... -..,
.---------1--.-.....
,
' I
1
1
1
1
1
I
I
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I
I
..Ｍﾷｾｉ＠
- - i - - - - - - f -I - - - - } I-
Ｍﾷｴｩｾ＠
8.0
J
-·-···--1-···
... --........-.•...
1
1
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1
1
}--------+----------1--1
1
1
1
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,
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1
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1
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1
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Ｍｾ］Ｚﾷｴ｟ｬＮ＠
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+---
4.0
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,
1
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,
ﾷＺ］ｊｾｴｦｬＱｅ＠
2.0
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1
1
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····--·----······t--·--·-·······--·········t·············1
.. ····--··-------·------1------·1
1
1
1
1
1
1
1
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1
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1
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1
.......•....••.....1 ...•...........
0.0 L...-_....I...-_ _ _ _--'-_ _---1_ _ _- ' - -_ _ _ _-'--_---L.._ _ _ _-'----'
5.0
10.0
30.0
50.0
70.0
90.0
99.0 99.5
95.0
Cumulative percent particles
Log Probability Paper
ＱＰｲＭｾ｟ＮＬ＠
----.1-...···_------·_·--1-.-..·----1-----_..-._-----1._---______.___. __.. . . ___._.__..j.----..-..-...-......-..•• ..•••._.
1
ｾ］｝ＺＭｉ｟ｲﾷｦｌ＠
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------. I····----;---··--------·-1-·----
］ｾｴｬＺｲＭＱ＠
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ｾＭＫ
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1
1
!I
II
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1
1
1
1
1
1
1
I
1
Ｍｾｔ｜Ｑ
I
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____
J__1----L
I
I I
I
1
1
1
1
1
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1
1
1
1
' I
1
1
1
1
II
I
50.0
70.0
I
I
90.0
95.0
I
!
I
1
5.0
10.0
30.0
Cumulative percent particles
99.0 99.5
SOLUTIONS FOR CHAPTER 13
369
Linear Graph
350.0
300.0
250.0
t<j
0
..... 200.0
--<0
U
ｾ＠ ....
150.0
;::l
U)
100.0
50.0
0.0
0.0
40.0
20.0
60.0
100.0
80.0
Cumulative percent particles
Linear Probability Paper
Ｍﾷｾｉ
201
ＭｔＱｾｉＬ＠
_._. __
1;3 151
ｾ＠
ＭｾＮＫ｟＠
u
ｾ＠ .....
a 101
ＭＫｾ
ＭＮＫｾ｟＠
1
1
1
1
1
1
'1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
----------11---
-.1---______________. . 111_____ _
ｾ＠
1
1
1
1
I1
I1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
I
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
30.0
50.0
1
1
1
:... -- -"-"-1'-" ---- --.. ------1
5.0
1
1
1
1
1
1
1
1
. I··-------------+--------.--------..
--.1-.-.---.-•.
-.--.-.-... -....·,.···-·11..·--··1
1
1
I1
I1
1
10.0
1
1
I1
I1
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1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1------·· .
1
90.0
95.0
99.0 99.5
---r----------;---1
1
1
1
1
1
-------+-_._--_._-------_._{-_
. _. _1
1
ｾＭ 1
.. _------+-------+------------------{------1
1
1
---+------------.1---------4--------4-Ｍｾｉ｟
1
1
1
1
51
ｬｌＭｾ］＠
1
1
1
1
1
1
1
1__ • __ • ______________ 1________ _l_
1 ___________ ._.l.
1 _____________________1-______
1
.. 1
I1
Ｍｾ［
o
Ｎ｟ｾ＠
1
1
------r·-·---·---------1'
---- ----., ---1'11 -.. --- --.--- -- ----- -----r--------·t
-----..-------..-- -1
1
1
70.0
Cumulative percent particles
370
THE HEALTH PHYSICS SOLUTIONS MANUAL
Log Probability Paper
················1······ .. -.......................{................... .J....
I
.,
I
..........•.•.•••... "........
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100
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f
1
10.0
5.0
30.0
50.0
70.0
90.0
95.0
99.0 99.5
Cumulative percent particles
Linear Graph Paper
1500.0
I
1000.0
500.0
0.0
0.0
ＭＬｾｬｪｉｊｲｩＮ＠
"j'fi r ,1/)!11,.jJ 1/
20.0
40.0
60.0
Cumulative percent particles
80.0
100.0
SOLUTIONS FOR CHAPTER 13
371
Linear Probability Paper
1500.0.--------,-----------,----,---------.------,
I
I
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1000.0
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•••••••••••••••••••••••• 1•••••••••• _•••••.••.• _•••••••••••••••.!. •.•••.•••••••••.. L ••••••••• _•••••••••••••••••••••• .1 ••• _....... .
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1.. •..••••.••••.•••..••••.•••••••• -:- •••..•••••.•
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0.0
50.0
70.0
90.0
99.0
95.0
99.5
Cumulative percent particles
Log Probability Paper
I
1000
I
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Ｚｾ｝ｉｅｔ＠
. . . . . v ...........
I
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..................
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l·······A .......................... {."........
.......... ...'ff' ....••••···•••
.. •• .... ••• ........ {········
• ................ ·1" .......... • ......................... • .. • .... ·t .. • .......................... -t ........................ ,..•...... ·t ....
.... , .. " ................. ••• ....
' .... , .. ..
........ -. ·l,......ｾ＠ .... ················ .--..}......
+. ._-_.............. -} ..,.......-... -.............. ·1·.... _...... -:- .......... -................... ｾ＠ ........A . A . . . . • .. • • . . . . . . . . . . . . . .
n
........ • ............... ｾａ＠
.... '
ｾＢＱＧ＠
Ｎｾ＠
ｾ＠
Ｑｾ､＠
P ...............
... A ........ • ..
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100
ＭｾＫ
• .................
'I· .. • .... • .. ••••• ................. •••• .... -r .. •.. •••• ..... · .......... •.. T .. • ...... • .. •••• .............. r . . ·....
ＧＢａｾｉ＠
ｾ＠
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Ｍｾ
•...........
" ..... ａﾷｾ＠
I
I
. 1...... ••••••
.. ········1' .. "· . ·· ........ · .. ····· .. · .. "·1.. • .. ••• .. •
r·· ............ -.......... ·r .... · ...... ·• ...... -.... ·r·.. . ............... ,,-_ ....... ·1·
· .... ·r-·· .. ····· .... ····•· ..... l· .... •··•
.AA ............ • • . . . . . .
ｾ＠
. ....................................... r ...... · .. ··· ...... T ........................ ·· ..... · .. 1····· ... ...
d ..
............ · .... ·t· .......... "..... ·· ........... ....... · .... T ...... ••· ............. ·T ................. ,.............
ｾ＠
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.....· . ··1····
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n
ｾ＠
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1 ...... •• ..... ..
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······'f'''· . ·· . ··..........· ...... ·· . ·· ..
r........ ....... ................ ............... ........ ........ ".. e................. ..... _.. _-_ ...............
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10
1
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____ ｾ＠ I _______________ ｾ｟＠ J
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_ _____ LI __________ IL ______________ ｾ＠
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r _____________ II ___ _
ｾ＠
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...........
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...................................
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30.0
50.0
70.0
90.0
95.0
99.0 99.5
1
5.0
10.0
Cumulative percent particles
..
372
THE HEALTII PHYSICS SOLUTIONS MANUAL
(b) Are the size distributions normally or log-normally distributed?
The size distributions are log-normally distributed, as can be seen by the straight
line formed when the size, surface area, and mass are plotted on log-probability
graphs.
(c) Compute the geometric mean and standard deviations for each of the three
types of distributions.
First determining CMD:
Log Probability Paper
10,---,---------,------.-----,---------.--__, -______, - - .
I
........ 1J... . ............... Il ............ 1
......I
•·•· .••..-.1 ...
I
I
I
I
,.
...,...,..
. ...,,
•••••••••• _ •.. f
J
J
J
I
I
. •,
I
I
......_.... I, . . --........ -_ ........ ｾＭﾷｴ＠
I
I
I
..... 1
ｉＭｾ＠
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..... ｾ＠ ..... j
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:
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,
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,
,
.,.
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........... _••••••1•• _....
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1
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. ..•... - •• 1
"""--"1'
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70.0
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.......................__ .,.I......... .
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50.0
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J
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30.0
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10.0
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.................
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,
..............., .......I
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···1··· .. ·
II
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I
I
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II
.... {........... __ .......... _-_.: ........ .
............... {
I
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5.0
1
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..... __.... .............. _-_....... --.}.
I
.
I
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I
I
I
I
I
I
I
____L -______Ｍｌｾ＠
90.0
95.0
99.0 99.5
Cumulative percent particles
Since we do not know the number of particles in the sample, we must determine
the CMD from the graph. On the log-probability plot of cumulative frequency vs.
Particle size, we find that the ordinate at the 50% point intersects the plot at" 2.5
fl· Next, we determine the size of the particles at one standard deviation above
and below the mean to determine the geometric standard deviation.
The geometric standard deviation,
=
cr
g
84% size
50% size
a, is given by
g
SOLUTIONS FOR CHAPTER 13
373
From the graph, we find the 84% size to be 5.111·
() = 5.1 = 2
g
25
Similarly from the graph, we find the 16% size to be 5 J.l.
25
() =-=2
g
1.25
Thus, the CMD = 2.5J.l: 2
If the number of particles is known,
-1
CMD = Iog
I log d
N
j
= Iog IIi Nlog di
-1
where
N = number of particles
th
f = number of particles in the i size interval
di = average diameter of particles in the i1h size interval
d = size of the i particle
th
j
The standard deviation can also be calculated as
0"=
I
(logd.)2
N
I
-
(-)2
.
log d
where log d is the average logarithm of the
particle size.
The mass median diameter, MMD, and the surface median diameter, SMD, can
be read from the respective graphs, or they can be calculated if we know the
CMD and 0".
g
Log(MMD) = Log(CMD) +6.9 x (Log(O"CMD))2
Log(MMD) = Log(2.5) +6.908 x (Log(2))2 = 1.024
MMD = 10.6 J.l
The surface median diameter, SMD, is given by:
Log(SMD) = Log(CMD) +5.757 x (Log(O"CMD))2
Log(SMD) = Log(2.5) +5.757 x (Log(2))2= 0.919
SMD = 8.2 J.l
374
13.14
THE HEALTH PHYSICS SOLUTIONS MANUAL
13.14 An instrument repairman suffered an accidental exposure to J3I 1 while
working in a customer's laboratory. Two days later his thyroid gland was found
to contain 2 x 104Bq (0.54 J.!Ci) 1311. Assuming he is a normal healthy man who
weighs 70 kg, calculate
(a) the amount of 1311 activity originally deposited in the thyroid,
The activity at any time t after an initial deposit of activity, A(O), is given by the
product of the retained fraction and A(O).
A(t) =R(t) x A(O)
R(t) = 07 exp ( -0.693)
- - t +0.3 exp ( -0.693)
--t
.
0.35
100
0.693) +O.3xA(O)xexp(0.693)
2xl04 =07xA(0)xexp ( ---x(2)
---x(2)
.
0.35
100
4
A(O) = 6.5 X 10 Bq was the activity initially deposited in the thyroid.
(b) the dose commitment to the thyroid as a result of the accident.
Equation 6.97;
D(thyroid) = A(thyroid) x S(thyroid ｾ＠
thyroid)
Find the cumulated activity, A, of the deposited radioactivity using equation
6.91;
Since this particular isotope is in two compartments,
6
_ As( 0) 0.7 x (6.5 X 104Bq) 0.3 x (6.5 X 104Bq)
A= A =
(0.693)
+
(0.693)
= 2.84 x 10 Bq·d
0.35d
100 d
ｓＨｴｨｹｲＰＱ､ｾＩ＠
Next, look in MIRD Pamphlet 11, p. 185, for ｓＨｴｨｹｲｯｩ､ｾＩ＠
•
.
-2
= 2.2 x 10 ｾｃｩＮ＠
rad
hr
x 0.01 Gy x
1 ｾｃｩ＠
x 24 h
rad
3.7 x 10 4 Bq
d
SOLUTIONS FOR CHAPTER 13
375
7
S(thyroidf-thyroid) = 1.43 x 10- Gy
Bq·d
Gy
Gy
S(thyroid-,thyroid) = 1.65 x 10-12 B q. sec = 1.43 X 10-7 B q. d
Replacing values into equation 6.97:
D(thyroid) = A(thyroid) xS(thyroid-,thyroid)
6
7 Gy
D(thyroid) = 2.84 x 10 Bq·d x 1.43 x 10- Bq. d - 0.4 Gy
13.15 A 20 liter breath sample was collected over 2 minutes. Analysis for 222Rn
showed the radon concentration to be 1 x 10-7 Bq per liter. Estimate the body
burden of 226Ra f rom th ese data.
The rate of exhalation of 222Rn is related to the body burden of 226Ra by equation
13.10;
Ae = 0.7 x q Bq xl minLet:
1
C = 1 x 10-7 Bq/liter
V = 20 litersl2min
AC26Ra) = 8.1 x 10-10 min-1 (chapter 13, p. 550)
The exhalation rate,'equation 13.11 is;
A = 1 x 10-7Bq x 20L =1x10-6 Bq
e
L
2 min
min
1x10-6 Bq
min
3
q = -0-.7-x-8-.1-x-1-0'=-I=O=Ill1-'n---1 = 2 x 10 Bq
13.15
376
13.16
THE HEALTH PHYSICS SOLUTIONS MANUAL
13.16 A lab worker accidentally ingested 2lDpo by using a contaminated cup for
his coffee. Twenty four hour urine samples were taken over a 60 day period and
analyzed. The following data were obtained:
Days after ingestion
1
5
10
15
20
25
30
40
50
60
Bq per sample
25
23
21
19
18
16
15
12
11
9
(a) plot the data on semi-log paper and fit an equation to the elimination data.
100
ｾ＠
:g. 10
III
o
10
20
30
40
50
60
70
Days
The data, when plotted on semi-log paper, fallon a straight line. Extrapolation of
the line intercepts the time axis at 25.4 Bq/d. The graph shows that this value
falls by 50% after a time interval of 40 days. The slope of the line, therefore, is
AE =0.693/40 d =0.0173 per day. The equation for the urinary clearance data,
therefore, is
SOLUTIONS FOR CHAPTER 13
V(t)
377
= 25.4 x e-O.017xt
(b) If 10% of ingested Po is known to be eliminated in the urine, and 90% is
eliminated in the feces, how much 21O pO was ingested?
The given data represent the effective clearance data, and the slope of the urinary
clearance rate. Since the radiological half life of 21OPO is 138 days, the biological
half life can be calculated. The effective half life, from the data, is 40 days.
(Equation 6.54)
T _ TR X TB
ETR +TB
40 d = 138 d x TB
138 d + TB
TB = 56.3 days
A = 0.693 = 0.0123 d-I
B
56.3 d
Since the slope of the biological clearance curve is slightly less than that for the
effective clearance curve, the intercept will be different. Since T elOpo) = 138 d,
it is reasonable to assume that the 1 day datum is not significantly affected by the
1 day's decay. The new intercept can thus be calculated
V(1) = Vo e-O.0123t
25 = V e-0.0123 x 1
o
VO= 25.3 Bq
The biological urinary clearance curve is
V(t)
= 25.3 x e-0.0123 x
t
Since only 10% of the ingested 21OpO is eliminated in the urine, the quantity
originally deposited = 10 x total activity eliminated in the urine.
A(O)
25.3 Bq/d
4
A(O) = lOS VU)dt = 10-- = 10 X
-I = 2.06 X 10 Bq
00
.
0
AB
0.0123 d
(c) If 13% of the 21OPO was deposited in the kidneys, what was the committed dose
378
THE HEALTH PHYSICS SOLUTIONS MANUAL
equivalent to the kidneys from this accidental ingestion?
Since we have a simple compartment clearance, A(kid) =AE, and the dose
commitment to the kidneys is
D= D(O)
AE
The activity deposited in the kidney is:
4
3
AS(O)K = 0.13 x 2.06 x 10 Bq = 2.67 X 10 Bq
The dose rate from this activity is
qBq x 1 tpS/ x E MeV x 1.6 x 10-13 _1_ x 8.64 x 104 sec
/Bq
t
MeV
mkg x
ｬｾＩｇｙ＠
d
kg
E elOpo) alpha = 5.4 MeV (including the KE of the recoil nucleus)
w of kidneys = 0.31 kg
Substituting these values into the equation for dose rate, we have:
ｄｯ］ＭｾＩ
2.67 X 10 3 Bq x 1 tps/ x 5.4 MeV x 1.6 x 10-13 _1- x 8.64 X 104 sec
B
/Bq
t
MeV
d
.
0.31 kg x Ｑｾ＠
kg
Gy
Do = 6.4 X 10-4 Gy/d
.
6.4 x 10-4 Gy
D =-:;: =
0.017 d-1
Do
13.17
d
0.038 Gy
13.17 What is the dose commitment to the skeleton due to the ingestion of 100
Bq/day, for 1 year, of 90Sr dissolved in drinking water?
Sr, a pure beta emitter, rapidly reaches secular equilibrium with its short lived
(64.1 h) 90y daughter. Both isotopes are pure beta emitters. The average ･ｮｾｲｧｹ＠
per 90Sr transformation is the sum of the average beta energies:
90
Ee = 0.1958 MeV + 0.9348 MeV =1.13 MeV/transformation
1------------------------------ ____ _
SOLUTIONS FOR CHAPTER 13
379
The dose commitment includes the dose during intake of the 9OSr_9CJy, and the
dose during the washout from the body following the end of the intake. The dose
during the 1 year intake period is given by equation 11.17
H(t) = H(SS{t; + AlE
(e-
AE
';
-1)]
Where fI (SS) is the steady state dose rate, at which time the daily intake and
daily elimination of 90Sr_90 y are equal. This steady state body burden is given by
an analog of equation 4.38
q(SS)
K
="i
where
K = the rate of deposition of the activity = intake rate x fraction.deposited
A =the effective turnover rate of the activity. The effective half life of 90Sr in the
skeleton is listed by the ICRP, and the ICRP indicates 9% of the ingested 90Sr is
deposited in the bone.
TE (90Sr) = 6400 d in the skeleton (ICRP)
A =0.693 = 0.693 = 1.08 x 10-4 d-l
E
TE
6400 d
The dose rate to the skeleton, whose weight is 7 kg, is given by:
1 tps E J x 864 x 104 secx
1 Sv
MeV
q Bqx--x
- x 16 x 10-13 - H=
Bq
e
t
.
MeV'
day
Gy
mkgxl Jig
Gy
8.33x10 4 Bqx 1 tps x 1.13 MeV x1.6x10-13_J-x8.64x104 sec xl SV
H(SS) =
Bq
t
MeV
day
Gy
7 kgxl Jig
Gy
iI(ss) = 1.86 x 10-4 Sv/d
380
THE HEALTII PHYSICS SOLUTIONS MANUAL
Substituting this value into the dose equation gives:
1
(e-l.08XIO-4d-1365 d -1)]
H(365 d) = l.86 X 10-4 Sv [365 d +
d
l.08 x 10-4 d- I
H(365 d) = 1.3 X 10-3 Sv
The dose during the washout is given by
H(washout) = iJ( 0)
.
AE
where iJ( 0) is the dose rate at the beginning of the washout period. Since the
steady state has not yet been reached at the end of 1 year (The effective half life
= 6400 d = 17.5 years), the dose rate after 1 year is given by a variant of equation
4.40.
I
.---t---------i . - - - - -
Equilibriu
Activity
in bone
1 year
Time in years
iJ(O) = 1.86 x 10-4 Sdv (1- e-l.08XIO-4X365) = 7.19 X 10-6 SdV
7.19 X 10-6 Sv .
H(washout) =
d1 = 6657 X 10-3 Sv
.
l.08xl0-4dDose commitment = 1.32 x 10-3 Sv + 66.57 X 10-3 Sv = 68 X 10-3 Sv
SOLUTIONS FOR CHAPTER 13
381
I
13.18 An accidental release of 21'1'0° 2 from a glove box leads to an atmospheric
From a recording air monitor
concentration of 1500 Bq/m3 (4.05 x 10- ｾｃｩｬ｣ｭ｜＠
whose alarm had failed it was later learned that a worker "had been exposed to the
airborne 21'1'0°2 for 1 hour. Measurements made with a cascade impactor showed
Using the
the mass median aerodynamic particle size (MMAD) to be 0.5 ｾｭＮ＠
data for the reference person, calculate
(a) the amount of activity deposited in the lung,
8
Since 10 CPR 20 is based upon ICRP 30, the ICRP three compartment model is
utilized.
The quantity inhaled is based on approximately 10m3 air inhaled per 8 hour day
(Appendix C).
3
1500 Bq xl hr x 10 m
m3
8 hr
1875 Bq is the quantity of activity inhaled.
For a MMAD of 0.5 mm particle, the following deposition occurs (Figure 8.4);
C
Fraction inhaled 21OPO
deposited in region
Total Activity in
Bq inhaled
Deposited activity in regions,
Bq
N-P
0.18
1875
Not in hmg
T-B
0.08
1875
150
P
0.34
1875
637.5
Sum
787.5
The total activity deposited in the lung is 787.5 Bq.
(b) the dose commitment to the lung from this accidental exposure.
The dose to the lung is calculated from equation 6.97
-
H(lung) = A(lung)Bq . d x S(lung ｾ＠
ｃｬ･｡ｲｮ｣ｾ＠
Sv
lung)-Bq·d
ｾＭ
rates of particles in the lung are given in Fig. 8.3. For 21Op002 , which is
class W (moderately soluble), the effective clearance half times = the biological
half times given in Fig. 8.3 for clearance from the T-B region (since TR = 138
days). From the P region, 40% of the deposit is cleared with a half time of 1 day.
However, 60% of the particles are cleared from the P region with a biological
half time of 50 days. The effective half time, therefore, is given by equation 6.54.
13.18
I
382
T
E
THE HEALTH PHYSICS SOLUTIONS MANUAL
= TR X TB = 138 d x 50 d = 37 d
TR + TB
138 d + 50 d
The activity deposited in each region, and for each clearance pathway. The
fraction of the deposit, its retention half time, and its clearance rate A (= 0.693/
TE ) are listed in the table below.
Region
Bq
f
N-P
337.5
Not in hmg
T-B
150
P
637.5
TE, d
A ,d- I
0.5
0.5
0.01
0.2
69.3
3.47
0.4
0.6
1
37
0.693
0.0187
The cumulated activity, A, for several compartments is given by a variant of
equation 6.91,
A. (0)
A=L-'-
Ai
If we substitute the values from the table above, we have
05 x 150 Bq . 05 x 150 Bq. 0.4 x 637.5 Bq 0.6 x 637.5 Bq
------=+
+
+ - - - - I0.0139 d0.693 d- 1
69.3 d- 1
3.47 d- 1
A =2.1 X 10 Bq·d
4
The dose rate to the lung, ｓＨｬｵｮｧｾＩＬ＠
S(lung ｾ＠
SvlBq·d is calculated
lung) =
1 Bq x 1 tps x E MeV x 1.6 x 10-13 _._1_ x 8.64 X 104 sec x Q Sv
Bq
t
MeV
day
Gy
ｭｫｧｸＱｾ＠
.
/Gy
kg/'
The lung has a mass of 1 kg, a 21Dpo alpha decay releases 5.4 MeV, and the quality
factor Q for alpha particles is 20. Substituting these values into the equation
above gives
---------Ｍｾ＠
-
SOLUTIONS FOR CHAPTER 13
383
Sv
S(lung +-lung) = 1.49 x ＱＰｾ＠
Bq·d
The lung dose is
H = 2.1 X 104 Bq . d x 1.49 X 10-6
Sv = 0.031 Sv
Bq·d
13.19 A demineralizer 20 cm in diameter x 20 cm high processes 200 liters per
minute contaminated water, and removes the following long lived isotopes:
Isotope
ｾｃｩｉｌ＠
Bqlliter
1048 x 104
004
1.11 x 105
3.0
1.85 x 106
50.0
The demineralizer operates for 180 days. Thirty days later,
(a) What is the activity of each of these isotopes in the demineralizer?
The activity buildup in the de mineralizer follows the same kinetics as serial
transformation. The activity at any time tb , A(t) after the start of operation is
given by a variant of equation 4.38:
and after a decay time of td after the end of buildup, the activity on the resins is
reduced to
A = Ｈｾ＠
(1- e-
AX
tb
))e-
AX
td
where
K=C ｾｃｩ＠
x ＲＰｾ＠
L
A = decay rate
x 1440 min =2.88 x 105 C
min
d
L
d
13.19
384
THE HEALTH PHYSICS SOLUTIONS MANUAL
For 60Co, T = 5.3 Y
A
= 0.693 = 0.693 x 1 yr = 358 x 10-9 d-I
Co-60· T.1/2
5.27 yr 365 d
004 /-lCi x 2.88 x 105 L
A( 60 Co) =
L
d x (1 _ e -3.58)(10-4)( 180 d )
358 X 10-4d- 1
[
j
X e -3.58)(10-4)( 30 d
A( 60 Co) = 1.99 x 10 /-lCi =19.9 Ci
7
A(
137
Cs) = 155 x 108/-lCi =155 Ci
For I44Ce, T = 285 d, A = 2043 X 10-3 d- I
j
50 /-lCi x 2.88 x 105 L
A( 144 Ce) =
L
·d x (1- e -2.43)(10-3x180 d)
1
358
X
10-4d[
Xe -2.43xlO-3x30 d
.
A( 144Ce) = 1.95 x 10 /-lCi =1950 Ci
9
(b) If the demineralizer approximates a point source at 4 meters, estimate the
gamma ray dose rate there.
Equation 10.1
2
R·mx radx A C·1
r.
Dose rate = _C=.i. . .:h:.::r'----=R=----'--_
(d, m) 2
60
R·m 2
(Table 6.3)
For Co, r CO -60 = 1.32.
CI·hr
SOLUTIONS FOR CHAPTER 13
.
D=
R.m2)
rad
.
x 19.9 CI
( 1.32 Ci . hr x 0.95R
(4
mr
6 rads
=1.5 - .
hr
2
For
137
0 33 R . m
Cs, lCS-137 - .
.
(Table 6.3)
CI·hr
. (
D=
_
0.33 R· m2) x 0.95 rad x 155 Ci
Ci . hr
R
(4 m)
2
rads
=3.04hr
2
For
144
. (
D=
R·m
Ce, r Ce_ l44 =0.04 Ci. hr (RHH)
0.04 R· m2) x 0.95 rad x 1950 Ci
Ci· hr
R
(4 m)2
rads
=4.63hr
The total dose rate = 1.56 + 3.04 + 4.63 = 9 radslh
(c) Estimate the gamma ray dose rate at the surface of the demineralizer.
The surface dose rate is estimated with equation 6.68,
D=crg
where D.is the mean dose rate from an isotope of concentration C and g is a
geometry factor (Table 6.4). The volume of the ion exchanger is
V = 1tr2 h = 1t (20) 2 20 = 6283 cm3
19.9 Ci
2
4 R ·cm
rad
3
3
b = 6.283 x 10 cm x 1.32 x 10
x 0.95- x 68.9 cm
Ci· hr
R
.
3
D= 2.74 x 10 radslhr
385
386
·
D =
•
THE HEALTII PHYSICS SOu.mONS MANUAL
155Ci
6.283 x 10 3 cm 3
4R·cm2
rad
x 0.95- X 68.9 cm
Ci· hr
R
X 0.33 X 10
3
D= 5.33 x 10 radslhr
·
D =
1950Ci
6.283 x 10 3 cm 3
4R·cm2
rad
x 0.95- X 68.9 cm
Ci·hr
R
X 0.04 X 10
b= 8.13 X 103 radslhr
Adding the contributions of the 3 radionuclides we have a total of
3
16.2 X 10 radslh
Since this is the mean dose rate, the surface dose rate is
3
b(surface) = 0.5 x 16.2 x 10 = 8.1 X 10 radslhr
13.20
3
13.20 In accidental releases to the air in a fuel reprocessing plant, the following
mixture of isotopes is usually found. Using DAC values for the air given in
10CFR20, calculate the atmospheric DAC for the total activity that must be
applied during cleanup of the contamination.
% of total activity
7
1
10
15
25
13
2
The DAC of a mixture of contaminants is given by
1
DAC(mixture) =- - p.
LDA'C.
1
SOLUTIONS FOR CHAPTER 13
387
th
where Pi is the proportion of the i contaminant, and DAC i is the DAC of the i th
contaminant.
For the radioisotopes in the mixture, we have (10CFR20, Appendix B, Table 2;
column 1)
DA C =[
]-1
0.01
0.1
0.15
0.25
0.13
0.02
0.07
+
+
+ - - -8
8 +
9 +
8 +
8
7
9
6xl0·
2xlO5x105x105x106xlO6x10-
DAC = 3 x 10-8 /lCilmL
Note that the DAC applies only to occupational radiation workers.
13.21 Tritiated water vapor was unknowingly released in a laboratory. An air
sample was taken using a freeze-out technique (100% freeze out) when the leak
was discovered. Further investigation revealed that the system had been leaking
for 24 hours prior to the discovery. Five hundred liters air were drawn through
the cold trap, and the collected moisture was diluted to 50 mL. One mL of the
dilution was counted for tritium betas in a liquid scintillation counter whose
background was 12 cpm and whose counting efficiency was 30%. The 1 mL
sample gave 3200 cpm.
(a) What was the tritium concentration in the air?
3200 cpm - 12 cpm (background) = 3188 net cpm
Since the counting technique was 30% efficient:
counts 1 min 100 decays
1 Bq
x
x
x
= 177.1 Bq is the actual activity in
min . 60 sec 30 counts 1 decay
sec
1 mL sample collected from the 50 mL dilution.
3188
Convert this to the activity in each cubic meter of air (50 mL sample represented
500 liters of air);
4 Bq
177.1 Bq 50 mL 103L
1 mL x 500 L x 1 m3 = 1.77 x 10 m 3
13.21
388
THE HEALlli PHYSICS SOLUTIONS MANUAL
(b) A technician who had been working in the lab for 8 hours left for a vacation
without leaving a urine sample. If the principal route of intake was inhalation,
and if all the inhaled tritium was taken up by the technician, estimate her dose
commitment. (Use the biological data given for reference person.)
First find how much air the technician takes in during 8 hours. From Appendix
C, it is found that 9100 liters (female) are taken during 8 hrs.
4
1.77 X10 Bq
q=
m3
entering her body.
m3
5
X 1000 LX 9100 L= 1.61 x 10 Bq is the amount of tritium
The dose rate, iJ, from this activity entering her body is given by equation 6.47:
q Bq xl tps x E MeV x 1.6 x 10-13 _1_ x 8.64 x 10 4 sec
Bq
e
t
MeV
dayｾ＠
____________________________
ｄ］Ｍｾ
mkgx Ｑｾ＠
/Gy
kg/'
where the mass of the reference person is 70 kg (Appendix C) and the average
energy of the tritium beta is 0.0057 MeV.
tps
MeV
.J
s
5
1.61 X10 Bq x 1 - x 0.0057 - - x 1.6 X 10-13 - - x 8.64 X 104 _
D=--__________
ｂｾｱ＠ ________t____ｾＭｍ･＠
__
V ________､｟｡ｹｾ＠
ＷＰｫｧｘｬｾ＠
/Gy
kg/'
iJ = 1.81 X 10-7 Gy
.
d
ICRP 30, part 1, p.66, gives the effective half life of tritium in the body as 10
days. Calculating the effective decay constant using equation 4.21;
A = 0.693 = 0.693 _
E
T
-I
10 d - 0.0693 d .
The dose cOmmitment can now be calculated using equation 6.58;
SOLlJI10NS FOR CHAPTER 13
389
Since the dose is only due to betas, a quality factor of 1 is used, and thus the dose
commitment is 2.61 x 10-6 Sv
(c)The technician submitted a urine sample 21 days later. What concentration of
tritium would be expected in the urine?
ICRP 30, part 1, p.65, gives the water content of the body as 42000 g =42 L.
Thus, the activity in the technician is distributed over the 42 liters of water in the
body, giving an initial concentration of tritium in the body fluids of:
5
1.61 x 10 Bq
3 Bq
.
C H =
= 3.83 x 10 is the initial concentration of tritium in
42L
L
the body.
3
)
(
Since the urine is a bodily fluid, it will have the same concentration as the rest of
. the bodily fluids. ICRP 30, part 1, p.65, shows that the retention of tritium
follows the following function, with t in days;
B
-0.693t
R(t) = A(O)e-
lO-
= 3.83 x 10
3
-.5! x e
L
-0.693 x 21
10
B
= 8942
L
13.22 A worker accidentally ingested an unknown amount of 60CO activity. His
body burden was measured by whole-body counting from day 1 until day 14
after the ingestion with the following results:
Day
1
2
3
4
7
14
kBq
75.3
63.4
54.3
49.2
42.0
31.0
The whole body retention function for ingested 60CO is given as
Q(t) = 0.5e-1.386t + 0.3e-O·1155t + 0.le-O.01155t + 0.1 e-S.663XI0-4t •
13.22
390
THE HEAL1H PHYSICS SOLUTIONS MANUAL
(a) Estimate the amount of ingested activity.
The intake retention function, Q(t), gives the fraction of the intake remaining at
time t days after intake. Therefore, the estimated intake, A51.(0), from a whole
body measurement is:
A . (0) = measured activity = ASi (t)
SI
Q(t)
Q(t)
Our best estimate of the intake is the mean of the intakes calculated from the
successive body burden measurements and the value of Q(t) at the time of the
measurement.
A (0) si
measured activity
-I
0.5 x e -L386xl + 0.3 x e -O.l155xl + 0.1 x e -O.OI155xl + 0.1 x e -8.663xlO XI
For 1 day, we have
A (0) =
sJ
75.3 kBq
.
0.5 x e-L386xl + 0.3 x e-O.1l55xl + 0.1 x e-O·Oll55xl + 0.1 x e-8.663xlO
_I
xl
As/0) = 127.39 kBq
Similarly, we can calculate the estimated intake from each of the whole body
measurements and then take the mean value as our best estimate of the intake.
The results, together with the calculated values for Q(t), are tabulated below:
Day
Q(t)
1
0.591
75.3
127.39
2
0.467
63.4
135.78
3
0.416
54.3
130.43
4
0.386
49.2
127.42
7
0.325
42
129.10
14
0.243
31
127.32
Ait), kBq AiO), kBq
Mean = 129.6 kBq
(b) What was the committed dose from the ingested radiocobalt?
SOLUTIONS FOR CHAPTER 13
391
The committed dose equation is given by equation 6.97.
_
Sv
H = A Bq·d x SC0-60Cbodyf-body) Bq. d
The cumulated activity, A, is given by equation 6.91.
A = 129.6 x 103B (
0.5
+
0.3
+
0.1
+
0.1
)
q 1.386 d -1 0.1155 d -1 0.01155 d -1 8.663 X 10-4 d- 1
A = 1.647 X 107 Bq·d
SC0-60 Cbod yf-body) =
_
-5
-2.7xl0 ｾｃｩＮｨｲ＠
rad
x 0.01 Gy x
1 ｾｃｩ＠
x 24 h x 1 Sv
rad
3.7xl04Bq
d
Gy
-10
SC0-60Cbodyf-body) = 1.75 x 10
Sv
Bq. d
Dose = 1.647 x 10 7 Bq . d x 1.75 x 10-
10
Sv
3
Bq. d = 2.9 x 10- Sv = 2.9 mSv
13.23 A rotameter is calibrated to read directly at 25°C and 760 mm Hg.1t is
used at an altitude of 5000 ft (1500 m), where the pressure is 633 mm Hg and the
temperature is 15°C. What was the actual flow rate when the rotameter reading
was 2.5 L/min?
Equation 13.17;
Po= 760mm
P a =633mm
To = 25°C = 273 + 25 = 298 K
Ta = 15°C = 273 + 15 = 288 K
13.23
392
THE HEALTH PHYSICS SOLUTIONS MANUAL
Q, =QoJPo T" ］ＲＮＵｾｸ＠
Pa To
13.24
760mm x 288K
ｾ＠ 633 mm 298 K
min
］ＲＮＷｾ＠
min
13.24 A rotameter that was calibrated for air at 25°C and 760 mm was used in a
. room at a temperature of 25°C and 760 mm to measure helium flowing into a gas
chromatograph. The rotameter reading was 28.3 L/min (1 cfm). What was the
actual flow rate of the helium?
The density of He, MW = 4, differs greatly from that of air, MW = 29. Equation
13.17 relates flow to molecular weight.
L f/9
L
Qactual - Qcalibrated 11rP calibrated - Qcalibrated 11 I MWcalibrated
MW
= 28.3 - - X - = 76.2-V
13.25
P He
V
He
min
4
min
13.25 The air in a lab 10m x 8 m x 5 m high has airborne 239pu at a concentra-
12
tion of 0.1 DAC (DAC = 7 x 10- 1lCilrnL = 3 x 10-1 Bq/m\ If all the airborne
activity were to settle out, what would be the areal concentration on the floor,
2
tpm, per 100 cm ?
5 m x 10m x 8 m = 400 m
3
The 239Pu activity in a column of air 100 cm2 in an area 500 cm (5 m) high, when
all the Pu settled, gives an areal concentration of
2
ｾｃｩ＠
(
)
6 tpm
Co =0.1 x 7 x 10-12 --3
X 100 cm x 500 cm x 2.22 x 10 - - .
.
cm
ｾｃｬ＠
2
Ca = 0.078 tpmll 00 cm
13.26
13.26 The continuous air monitor (CAM) of a radiochemical manufacturing
laboratory that synthesizes various 14C labeled compounds is set to alarm at 0.1
DAC. Long term data show that the mix of 14C in the air is:
SOLUTIONS FOR CHAPTER 13
393
Compound
%
DAC, flCilmL
14CO
20
7 X 10-4
2
70
9 X 10-5
Labeled
10
1 x 10--6
14CO
At what 14C concentration should the alarm be set? .
The DAC of the mixture is given by:
DAC.
1
= - - - ］ＭｾＬ
mIXture
I
p.
I
- - - - -
DAC.
I
DAC.
1
fraction 14CO fraction 14C02 fraction labeled
+
+-----DAC 14 CO
DAC 14 CO 2
DAC labeled
1
0.7
--6 !lCi
mL
= - - - - - - - - - - - - - = 9.25 x 10
0.2
nnxlUre
ＷｸｬＰＭＴｾ＠
C· +
mL
ＹｸｬＰＭＵｾ＠
0.1
C· +
mL
ｬｸＰＭＶｾ＠
C·
mL
Set the CAM for one tenth the DAC,
!lCi
7 !lCi
9.25 x 10--6 mL x 0.1 = 9 x 10- mL (since DAC's are normally rounded to the
nearest whole number)