Ver 2.13
Chapter XI
Half Value Thickness
&
Half Life
Joseph F. Buono RTT
Allied Health Science
Nassau Community College
1 Education Drive
Garden City, NY 11530-6793
phone: 516 - 572 - 9640 office
- ext: 26556
email: joseph.buono@ncc.edu
website: rtscanner.com
Title
MAIN MENU
- Objectives
- Introduction
- Half Value Layer
Problems
- linear attenuation coefficient
- Transmission Fraction ( TF )
- TF and inverse square problem
- Half Life
Problems
- decay constant
- Activity Fraction ( AF )
- AF and inverse square problem
Menu
Objectives
1 - Define Half-Value Layer.
2 - State the equation for half-value layer and define all the terms.
3 - Define Half-Life.
4 - State the equation for half-life and define all the terms.
5 - Graph intensity verses absorber thickness on normal and semi log graph paper.
6 - Graph radioactive decay verses time on normal and semi log graph paper.
7 - Calculate the final intensity using the half-value layer equation.
8 - Calculate the final activity using the half-life equation.
MENU
obj 1
Objectives
9 - Calculate linear attenuation coefficient.
10 - Calculate decay constant.
11 - Calculate the final intensity from the Transmission Fraction (TF) and the initial blocking.
12 - Calculate a final intensity using AF, TF, blocking, time and distance from a source.
MENU
obj 2
Half-Value Layer
&
Half-Life
In this section, the effects of blocking and radioactive decay are examined.
They are studied together because, for simple cases, they both follow the
same mathematical equation.
This equation has the form of If = Ii e-μx for Intensity with blocking (half
value thickness) and Af = Ai e-λt for Activity with time (half-life).
In the equations, If, Ii and Af, Ai are the final and initial intensity and activity
respectively, where 'e' is the natural number 2.718281828459... .
Looking at the intensity equation first, you have m which is the linear
attenuation coefficient and 'x' which is the thickness of the block or absorber.
Looking at the activity equation next, you have l
which is the decay constant and 't' which is time.
MENU
hvl 1
Half-Value Layer
Half Value Thickness (H.V.T):
This is that amount of absorber that, if place in the path of the beam, will
reduce the initial intensity by half, if measured at the same DISTANCE
FROM THE SOURCE.
The Half Value Thickness for Cobalt 60 is 1.1 cm of lead.
Co60
source
Except for electron contamination
(if the block is closer than 15 - 20
cm from the measuring point),
moving the block closer or further
from the source will have no effect
on the measured intensity.
1.1cm
lead
block
distance
80cm
detector
150cGy/min
75cGy/min
75cGy/min
No change in intensity as
block distance is varied.
MENU
hvl 2
Half-Value Layer
Repeating this process of measuring the intensity with different thickness of lead and plotting the
results on graph paper will yield the intensity curve shown in the next diagram.
150
125
cm of
lead
intensity
cGy/min
0
1.1
2.2
3.3
4.4
150
75
37.5
18.75
9.375
intensity
cGy/min
100
75
50
25
0
0
1
2 HVL 3
4
5
0
1.1
2.2
3.3
cm of lead
4.4
5.5
MENU
hvl 3
Half-Value Layer
Repeating this process of measuring the intensity with different thickness of lead and plotting the
results on graph paper will yield the intensity curve shown in the next diagram.
150
125
cm of
lead
intensity
cGy/min
0
1.1
2.2
3.3
4.4
150
75
37.5
18.75
9.375
intensity
cGy/min
100
75
Next draw the
curve that fits the
data points.
50
25
0
0
1
2 HVL 3
4
5
0
1.1
2.2
3.3
cm of lead
4.4
5.5
MENU
hvl 4
Half-Value Layer
Repeating this process of measuring the intensity with different thickness of lead and plotting the
results on graph paper will yield the intensity curve shown in the next diagram.
150
125
cm of
lead
intensity
cGy/min
0
1.1
2.2
3.3
4.4
150
75
37.5
18.75
9.375
intensity
cGy/min
100
75
Next draw the
curve that fits the
data points.
50
25
0
0
1
2 HVL 3
4
5
0
1.1
2.2
3.3
cm of lead
4.4
5.5
MENU
hvl 5
Half-Value Layer
Plotting this information on semi-log graph paper will give a straight line.
150
125
cm of
lead
intensity
cGy/min
0
1.1
2.2
3.3
4.4
150
75
37.5
18.75
9.375
1000
150
100
intensity
cGy/min
100
75
37.5
75
18.75
9.375 10
50
25
0
1
0
1
2 HVL 3
4
5
0
1
2 HVL 3
4
5
0
1.1
2.2
3.3
cm of lead
4.4
5.5
0
1.1
2.2
3.3
cm of lead
4.4
5.5
MENU
hvl 6
Half-Value Layer
Plotting this information on semi-log graph paper will give a straight line.
A number of real world problems follow this type of curve.
Examples:
1000
150
1) The time it takes a thermometer to indicate the
temperature after being placed in a bath of ice water.
125
150
2) The time it takes a capacitor to fully discharge.
intensity
cGy/min
100
3) The decay of Radioactive materials.
75
100
75
37.5
18.75
50
4) Simple cell survival curves.
9.375 10
25
0
1
0
1
2 HVL 3
4
5
0
1
2 HVL 3
4
5
0
1.1
2.2
3.3
cm of lead
4.4
5.5
0
1.1
2.2
3.3
cm of lead
4.4
5.5
MENU
hvl 7
Half-Value Layer
The reason for plotting this on semi-log paper is that we know the general equation of a straight line
curve on semi-log paper. We have been studding these types of curves in mathematics for centuries.
They all follow the same general equation.
If = Ii e-μ x
Which for Intensity is:
150
1000
125
150
100
intensity
cGy/min
100
75
37.5
75
18.75
9.375 10
50
25
0
1
0
1
2 HVL 3
4
5
0
1
2 HVL 3
4
5
0
1.1
2.2
3.3
cm of lead
4.4
5.5
0
1.1
2.2
3.3
cm of lead
4.4
5.5
MENU
hvl 8
Half-Value Layer
The reason for plotting this on semi-log paper is that we know the general equation of a straight line
curve on semi-log paper. We have been studding these types of curves in mathematics for centuries.
They all follow the same general equation.
Which for Intensity is:
If = Ii e-μ x
and for Radioactivity is Af
= Ai e-l x
The intensity equation is read:
I final equals I initial times e to the minus mu x.
Where:
I final ( If ) is the intensity after the block is placed in the path of the beam.
I initial ( Ii ) is the intensity without the block in the path in the beam.
e is number 2.718...
Mu ( m ) is the percent reduction in intensity for each unit thickness of absorber.
- The value of mu ( m ) is dependent on the material and the energy of the radiation.
- the technical name for mu ( m ) is: linear attenuation coefficient.
X is the thickness of the absorber.
MENU
hvl 9
Half-Value Layer
μ =
.693
HVL
Examining the intensity equation:
If = Ii e-μ x
If block thickness (x) is equal to
the half value layer (HVL) THEN:
BUT THEN:
If = Ii e-μ HVL
If = ½Ii
or
Ii = 2If
Substituting into equation for
Divide both sides by
If = 2If e-μ HVL
Ii
2If
If = 2If e-μ HVL
2If
2If
1
2
Take the natural log of
both sides of the equation:
rearrange equation:
MENU
.5
=
e-μ HVL
1
= e-μ HVL
ln (.5)
=
ln ( e-μ HVL )
-.693
=
-μ HVL
μ =
.693
HVL
hvl 10
Half-Value Layer
Example 1:
Find the linear attenuation coefficient if the HVT is 1.37 cm of lead.
μ =
=
=
=
μ =
.693
HVL
.693
HVL
.693
1.37cm
.506
cm
.506/cm
MENU
hvl 11
Half-Value Layer
Example 1:
Find the linear attenuation coefficient if the HVT is 1.37 cm of lead.
μ =
=
=
=
μ =
.693
HVL
.693
HVL
.693
1.37cm
.506
cm
.506/cm
Example 2:
Find the HVT if the linear attenuation coefficient is .86/cm of lead.
μ =
.693
HVL
Solve for HVL:
HVL =
.693
μ
.693
= .86/cm
= .806 cm
MENU
hvl 12
Half-Value Layer
There is one other equation, Transmission Fraction,
which is useful and will be derived at this time.
The Transmission Fraction ( TF ), is defined as the fraction of the beam that penetrates the block.
With the TF, a simple equation can be written to solve for the
final intensity given the initial intensity with blocking.
The mathematical equation would be:
Derivation of TF:
If = (TF) Ii
Thus:
If
TF = I
i
If = Ii e-μ x
MENU
hvl 13
Half-Value Layer
The mathematical equation would be:
Derivation of TF:
Divide by sides by Ii
If
BUT TF = I
i
BUT m = .693 / HVL
If = (TF) Ii
I
Thus:
f
-n
TF = 2TF
= I
i
If = Ii e-μ x
Ii
Ii
If
-μ x
Ii = e
TF = e-μ x
TF =
= e- ( .693 / HVL ) x
1
2n
TF = 2-n
= e- ( .693 ) ( x / HVL )
= (e- .693) ( x / HVL )
=
block thickness
Where: n = ——————————
Half-Value Thickness
( x / HVL )
1
e.693
= 1
( x / HVL )
2
Let n = x/HVL
Substituting into equation:
= 1 ( x / HVL )
2
MENU
hvl 14
Half-Value Layer
The mathematical equation would be:
If = (TF) Ii
TF = 2-n
Example:
If the initial intensity of an x-ray beam, which has a HVL of 1.5 cm of lead, is 175 cGy/min, find
the new intensity if a block of lead which is 4.25 cm thick is placed in the path of the beam.
Solution 1:
If = (TF) Ii
= (TF) ( 175 cGy/min )
where
TF = 2-n
= ( .14 ) ( 175 cGy/min )
where
block thickness
n = ——————————
Half-Value Thickness
= 2 - 2.38
4.25 cm
= ————
1.5 cm
= .14
= 2.38
= 24.6 cGy/min
MENU
hvl 15
Half-Value Layer
The mathematical equation would be:
TF = 2-n
If = (TF) Ii
Example:
If the initial intensity of an x-ray beam, which has a HVL of 1.5 cm of lead, is 175 cGy/min, find
the new intensity if a block of lead which is 4.25 cm thick is placed in the path of the beam.
Solution 1:
If = (TF) Ii
= (TF) ( 175 cGy/min )
= ( .14 ) ( 175 cGy/min )
=
24.6 cGy/min
Solution 2:
recall
.693
If = Ii e - m x
= Ii e – ( .693 / HVL
m = ——
HVL
)x
= 175 cGy/min e – ( .693 / HVL ) x
= 175 cGy/min e – ( .693 / 1.5cm )
= 175 cGy/min e – ( .462/cm )
4.25cm
4.25cm
= 175 cGy/min e – 1.96
= 175 cGy/min × .14
= 24.6 cGy/min
MENU
hvl 16
Half-Value Layer
Example:
The initial intensity of a cesium source is 27 cGy/hr at .1 m. The source is then placed in a lead
pig which is 3.15 cm thick. If the cesium's HVL is .55 cm of lead than find:
a) the intensity at .1 m from the source with the block in the path of the beam.
b) the intensity at 2.75 m from the source with and without the block.
c) the dose at the end of a 40 hr work week at BOTH points with and without BLOCK.
Solution :
(I find a diagram of the setup is very helpful in problem solving.)
lead pig
2.75 m
cesium
source
×
.1 m
Point 2
×
Point 1
Part a:
MENU
hvl 17
Half-Value Layer
Example:
The initial intensity of a cesium source is 27 cGy/hr at .1 m. The source is then placed in a lead
pig which is 3.15 cm thick. If the cesium's HVL is .55 cm of lead than find:
.51 cGy/hr
a) the intensity at .1 m from the source with the block in the path of the beam.
b) the intensity at 2.75 m from the source with and without the block.
c) the dose at the end of a 40 hr work week at BOTH points with and without BLOCK.
Solution :
(I find a diagram of the setup is very helpful in problem solving.)
lead pig
Point 2
2.75 m
cesium
source
×
.1 m
×
Point 1
Part a:
remember
If = Ii (TF)
block thickness
n = ——————————
= 27 cGy/hr × 2-n
Half-Value Thickness
3.15 cm
= 27 cGy/hr × 2- ————
.55 cm
= 27 cGy/hr × 2- 5.73
= 27 cGy/hr × .019
MENU
= .51 cGy/hr
hvl 18
Half-Value Layer
Example:
The initial intensity of a cesium source is 27 cGy/hr at .1 m. The source is then placed in a lead
pig which is 3.15 cm thick. If the cesium's HVL is .55 cm of lead than find:
.51 cGy/hr
a) the intensity at .1 m from the source with the block in the path of the beam.
b) the intensity at 2.75 m from the source with and without the block.
c) the dose at the end of a 40 hr work week at BOTH points with and without BLOCK.
Solution :
(I find a diagram of the setup is very helpful in problem solving.)
lead pig
2.75 m
cesium
source
×
.1 m
Point 2
×
Point 1
Part b: 1) intensity @ 2.75 meters
with block (Point 2)
recall inverse
square law
I1
I2
.51 cGy
I2
(d2)2
=
(d1)2
( 2.75 m)2
=
( .1 m)2
I2 =
MENU
.51 cGy × .01 m2
7.56 m2
= .00067 cGy/min
hvl 19
Half-Value Layer
Example:
The initial intensity of a cesium source is 27 cGy/hr at .1 m. The source is then placed in a lead
pig which is 3.15 cm thick. If the cesium's HVL is .55 cm of lead than find:
.51 cGy/hr
a) the intensity at .1 m from the source with the block in the path of the beam.
.00067 cGy/min b) the intensity at 2.75 m from the source with and without the block.
.036 cGy/min
c) the dose at the end of a 40 hr work week at BOTH points with and without BLOCK.
Solution :
(I find a diagram of the setup is very helpful in problem solving.)
lead pig
×
.1 m
×
Point 1
Part b: 1) intensity @ 2.75 meters
with block (Point 2)
recall inverse
square law
I1
I2
.51 cGy
I2
(d2)2
=
(d1)2
( 2.75 m)2
=
( .1 m)2
I2 =
MENU
Point 2
2.75 m
cesium
source
.51 cGy × .01 m2
7.56 m2
= .00067 cGy/min
2) intensity @ 2.75 meters
with NO block (Point 2)
I1
I2
27 cGy
I2
(d2)2
=
(d1)2
( 2.75 m)2
=
( .1 m)2
I2 =
27 cGy × .01 m2
7.56 m2
= .036 cGy/min
hvl 20
Half-Value Layer
Example:
The initial intensity of a cesium source is 27 cGy/hr at .1 m. The source is then placed in a lead
pig which is 3.15 cm thick. If the cesium's HVL is .55 cm of lead than find:
.51 cGy/hr
a) the intensity at .1 m from the source with the block in the path of the beam.
.00067 cGy/min b) the intensity at 2.75 m from the source with and without the block.
.036 cGy/min
c) the dose at the end of a 40 hr work week at BOTH points with and without BLOCK.
Solution :
(I find a diagram of the setup is very helpful in problem solving.)
lead pig
×
.1 m
×
Point 1
Part c: dose after 40 hours at .1 m
1) dose without block
Dose = Dose Rate × Time
dose after 40 hours at .1 m
1) dose without block
Dose = Dose Rate × Time
= 27 cGy/hr × 40 hr
= .036 cGy/hr × 40 hr
= 1080 cGy
= 1.44 cGy
2) dose with block
Dose = Dose Rate × Time
MENU
Point 2
2.75 m
cesium
source
2) dose with block
Dose = Dose Rate × Time
= .51 cGy/hr × 40 hr
= .00067 cGy/hr × 40 hr
= 20.4 cGy
= .027 cGy
hvl 21
Half Life
Half Life (T½ ): This is the amount of time that is required to reduce
the radioactivity to one half it present value.
Radioactive Source
Gold (Au 198)
As it decays it emits monenergetic
gamma ray photons of 0.412 MeV.
MENU
T½ - 1
Half Life
Half Life (T½ ): This is the amount of time that is required to reduce
the radioactivity to one half it present value.
Radioactive Source
Gold (Au 198)
As it decays it emits monenergetic
gamma ray photons of 0.412 MeV.
It also emits beta particles with a
maximium energy of 0.98 MeV.
But the beta particles are stopped by the 0.1 mm
thick platinum wall surrounding the seed.
MENU
T½ - 2
Half Life
Half Life (T½ ): This is the amount of time that is required to reduce
the radioactivity to one half it present value.
Radioactive Source
Gold (Au 198)
The Half Life for radioactive gold 198 is 2.7 days.
This means that after 2.7 days, the radioactivity is reduced
by one half its present value and therefore, its associated
dose rate is also reduced by the same amount.
Example: At time zero the initial activity is 30 milli curies
( 1 Ci = 3.7 × 1010 disintegrations per second ).
30 mCi
The process of measuring the activity every 2.7
days will result in the table seen to the right.
time
(days)
activity
(mCi)
0
2.7
5.4
8.1
10.8
30
15
7.5
3.75
1.875
Next plotting the results on graph paper will yield
the activity curve shown in the next diagram.
MENU
T½ - 3
Half Life
Radioactive Source
Gold (Au 198)
30
25
activity
( mCi )
20
15
10
5
time
(days)
activity
(mCi)
0
2.7
5.4
8.1
10.8
30
15
7.5
3.75
1.875
0
0
1
2
0
2.7
5.4
T½ 3
8.1
4
10.8
5
13.5
days
MENU
T½ - 4
Half Life
Radioactive Source
Gold (Au 198)
30
25
time
(days)
activity
(mCi)
0
2.7
5.4
8.1
10.8
30
15
7.5
3.75
1.875
activity
( mCi )
20
Next draw the
curve that fits the
data points.
15
10
5
time
(days)
activity
(mCi)
0
2.7
5.4
8.1
10.8
30
15
7.5
3.75
1.875
0
0
1
2
0
2.7
5.4
T½ 3
8.1
4
10.8
5
13.5
days
MENU
T½ - 5
Half Life
Just as in the intensity problem, the activity, if plotted
on semi log graph paper, will also follow a straight line.
Radioactive Source
Gold (Au 198)
30
25
time
(days)
activity
(mCi)
0
2.7
5.4
8.1
10.8
30
15
7.5
3.75
1.875
30
15
10
20
activity
( mCi )
100
7.5
3.75
15
1.875
1
10
5
0
.1
0
1
2
0
2.7
5.4
T½ 3
8.1
days
4
10.8
5
0
1
2 T½ 3
13.5
0
2.7
5.4
8.1
4
10.8
5
13.5
days
MENU
T½ - 6
Half Life
Just as in the intensity problem, the activity, if plotted
on semi log graph paper, will also follow a straight line.
The equation for Activity will be an exact copy of the
intensity equation with the appropriate symbols.
Radioactive Source
Gold (Au 198)
The equation is:
Af = Ai e- λ t
100
30
The equation is read:
A final equals A initial times
e to the minus lambda T.
25
15
10
20
activity
( mCi )
30
7.5
3.75
15
1.875
1
10
5
0
.1
0
1
2
0
2.7
5.4
T½ 3
8.1
days
4
10.8
5
0
1
2 T½ 3
13.5
0
1.1
2.2
3.3
4
5
4.4
5.5
days
MENU
T½ - 7
Half Life
The equation is:
Af = Ai e- λ t
The equation is read:
A final equals A initial times
e to the minus lambda T.
Where:
• A final ( Af ) is the activity at the end of time t.
• A initial ( Ai ) is the activity at start of the measurements.
• e is number 2.718...
• Lambda ( l ) is the percent reduction in intensity for each unit thickness of absorber.
- The value of lambda ( l ) is dependent on the material and the energy of the radiation.
- the technical name for lambda ( l ) is: decay constant.
• t is the amount of time from the initial reading.
MENU
T½ - 8
Half Life
λ =
.693
T½
Examining the intensity equation:
Af = Ai e- λ t
If time (t) is equal to
the half life (T½) THEN:
BUT THEN:
Af = ½ Ai
Af = Ai e- λ T½
or
Ai = 2Af
Ai
Af = 2Af e- λ T½
2Af
Af = 2Af e- λ T½
2Af
2Af
Substituting into equation for
Divide both sides by
1 =
2
Take the natural log of
both sides of the equation:
e- λ T½
1
- λ T½
.5 = e
- λ T½ )
ln (.5) = ln ( e
rearrange equation:
MENU
-.693 = - λ T½
λ =
.693
T½
T½ - 9
Half Life
λ =
Example 1:
.693
T½
The half-life of cobalt 60 is 5.26 years. Find the decay constant for this element.
λ =
.693
T½
.693
= 5.26 yr
= 0.132 / yr
This is the decay constant on a per year time frame.
Next look at the decay constant on a per month time frame.
λ = 0.132 / yr ×
1 yr
12 mth
0.132
= 12 mth
Change to a Percent:
MENU
= 0.011 / mth
= 1.1% / mth
NOTE:
This is the correction factor
for Cobalt 60. Which is
rounded off to 1% per month.
T½ - 10
Half Life
Activity Fraction (AF)
Again, as in the Intensity problems, there is one other
useful equation that will be derived at this time.
Activity Fraction (AF), which is defined as the fraction of the
activity that is remaining after a given amount of time.
With the AF, a simple equation can be written to
solve for the final activity given the initial activity.
The mathematical equation would be:
Derivation of AF:
Af = (AF) Ai
Af
=
Thus: AF
Ai
Af = Ai e- λ t
MENU
T½ - 11
Half Life
Activity Fraction (AF)
The mathematical equation would be:
Derivation of AF:
A
f
BUT AF = A
i
Af = Ai e- λ t
Ai
Ai
Af
Af = (AF) Ai
Thus: AF = A
i
Divide by sides by Ai
Af
= e- λ t
Ai
AF =
BUT λ = .693 / T½
AF = 2-n
AF = e- λ t
= e- ( .693 / T½ ) t
= e- ( .693 ) · ( t / T½ )
=
=
Let n = t / T½
Substituting into equation:
( t / T½ )
1
1
2n
AF = 2-n
Total time
Where: n = —————
Half-Life
e.693
1
2
( t / T½ )
= 1 ( t / T½ )
2
MENU
T½ - 12
Half Life
Activity Fraction (AF)
AF = 2-n
Af = (AF) Ai
Example 2:
Af = Ai e- λ t
If the initial activity of a radioactive palladium source is 15 mCi with a half life of
17.0 days, what is the activity after 47.5 days?
Solution 1:
Af = (AF) Ai
= (AF) 15mCi
where:
AF =
2-n
= (1.44 ) 15mCi
= 2- 2.79
= 2.2 mCi
= 1.44
Total time
n = —————
Half-Life
47.5 days
= —————
17 days
=
2.79
MENU
T½ - 13
Half Life
Activity Fraction (AF)
AF = 2-n
Af = (AF) Ai
Example 2:
Af = Ai e- λ t
If the initial activity of a radioactive palladium source is 15 mCi with a half life of
17.0 days, what is the activity after 47.5 days?
Solution 1:
Af = (AF) Ai
Solution 2:
where
λ = .693 / T½
Af = Ai e- λ t
= (AF) 15mCi
= Ai e- ( .693 / T½ ) t
= (1.44 ) 15mCi
= 15 mCi e- ( .693 / 17.0 dy
= 2.2 mCi
) 47.5 dy
= 15 mCi e- 1.936
= 15 mCi
×
.144
= 2.2 mCi
MENU
T½ - 14
Half Life
Example 3:
The initial activity of radioactive iridium (I-192) source is 135 mCi . The intensity for each mCi of iridium
is 4.69 R/hr at 1 cm from the source. The half life is 74.2 days and the HVL is 2.5 mm of lead find the:
a) activity after 175 days.
b) exposure rate at 75 cm from the sources
at the begging and the end of 175 days.
Solution :
c) exposure rate at 75 cm from the
sources if they are placed in a lead
box of .83 cm at the end of 175 days.
diagram of the setup:
75 cm
iridium
source
×
.01 m
Point 2
×
Point 1
Part a:
MENU
T½ - 15
Half Life
Example 3:
The initial activity of radioactive iridium (I-192) source is 135 mCi . The intensity for each mCi of iridium
is 4.69 R/hr at 1 cm from the source. The half life is 74.2 days and the HVL is 2.5 mm of lead find the:
a) activity after 175 days. 26.3 mCi
b) exposure rate at 75 cm from the sources
at the begging and the end of 175 days.
Solution :
c) exposure rate at 75 cm from the
sources if they are placed in a lead
box of .83 cm at the end of 175 days.
diagram of the setup:
Point 2
75 cm
iridium
source
×
.01 m
×
Point 1
Part a:
Af = (AF) Ai
recall
AF = 2-n
= .195 × 135 mCi
= 2- 2.36
= 26.3 mCi
= .195
where:
Total time
n = —————
Half-Life
175 days
= —————
74.2 days
=
2.36
MENU
T½ - 16
Half Life
Example 3:
The initial activity of radioactive iridium (I-192) source is 135 mCi . The intensity for each mCi of iridium
is 4.69 R/hr at 1 cm from the source. The half life is 74.2 days and the HVL is 2.5 mm of lead find the:
a) activity after 175 days. 26.3 mCi
b) exposure rate at 75 cm from the sources
at the begging and the end of 175 days.
Solution :
c) exposure rate at 75 cm from the
sources if they are placed in a lead
box of .83 cm at the end of 175 days.
diagram of the setup:
Point 2
75 cm
iridium
source
×
.01 m
×
Point 1
Part b: exposure 175 days at point 2
first exposure at 1 cm ( .01 m ) at 175 days
exposure = Af × G
= 26.3 mCi × 4.69 R/hr mCi
=
123.35 R/hr
this is the
Exposure Rate Constant
which gives the exposure
rate at 1 cm from a 1 mCi
radioactive substance
MENU
T½ - 17
Half Life
Example 3:
The initial activity of radioactive iridium (I-192) source is 135 mCi . The intensity for each mCi of iridium
is 4.69 R/hr at 1 cm from the source. The half life is 74.2 days and the HVL is 2.5 mm of lead find the:
a) activity after 175 days. 26.3 mCi
b) exposure rate at 75 cm from the sources
at the begging and the end of 175 days.
Solution :
c) exposure rate at 75 cm from the
sources if they are placed in a lead
box of .83 cm at the end of 175 days.
diagram of the setup:
75 cm
iridium
source
exposure at after 175 days = .022 R/hr Point 2
×
.01 m
×
Point 1
Part b: exposure 175 days at point 2
first exposure at 1 cm ( .01 m ) at 175 days
NEXT: exposure at 75 cm
I1
I2
exposure = Af × G
= 26.3 mCi × 4.69 R/hr mCi
123.35 R/hr
I2
=
MENU
(d2)2
=
(d1)2
( 75 cm )2
=
( 1 cm)2
123.35 R/hr
I2 =
123.35 R/hr × 1 cm2
5625 m2
= .022 R/hr
T½ - 18
Half Life
Example 3:
The initial activity of radioactive iridium (I-192) source is 135 mCi . The intensity for each mCi of iridium
is 4.69 R/hr at 1 cm from the source. The half life is 74.2 days and the HVL is 2.5 mm of lead find the:
a) activity after 175 days. 26.3 mCi
b) exposure rate at 75 cm from the sources
at the begging and the end of 175 days.
Solution :
c) exposure rate at 75 cm from the
sources if they are placed in a lead
box of .83 cm at the end of 175 days.
diagram of the setup:
lead box
iridium
source
75 cm
exposure at after 175 days = .022 R/hr Point 2
×
.01 m
×
Point 1
lead box: 0.83 cm thick
Part c:
recall
TF = 2-n
If = (TF) Ii
= .1 × .022 R/hr
=
.0022 R/hr
= 2- 3.32
= .1
END
where:
Block
n = ———
HVL
.83 cm
= —————
.25 cm
=
3.32
MENU
T½ - 19