Download chapter 7 health physics instrumentations

CHAPTER 7
7-1
HEALTH PHYSICS INSTRUMENTATIONS
Radiation Detectors
We should depend on instruments to detect the radiation and measure the size of
radiation. The radiation detectors are categorized as follows:
1) electrical type:
- ionization chamber
- proportional counter
- Geiger counter
- solid state detector (semi-conductor)
2) chemical type:
- film (photographic emulsion; film badges)
- chemical dosimeter (solid and liquid)
3) light type:
- scintillation counter (crystal and liquid)
- Cerenkov counter (crystal and liquid)
4) thermoluminescence type:
- TLD (crystal, CaF2(Mn), LiF)
5) heat type:
- calorimeter (solid or liquid)
For neutron detection, we use:
- product charged particles of neutron(n,), (n,p), (n,fission)
- decay of radioactive product nuclei produced by neutron-induced
transmutations
- recoiled particles by neutron elastic scattering (proton recoil)
1
),
Example 7-1.1:
ionization chamber:
The ion chamber is composed of an anode rod in the center and a cathode of cylindrical
hull, and filled with gas. The electric current will be produced by collecting the ions
produced by radiation around the anode in the chamber.
Let’s consider an ionization chamber as follows:
x- or - radiation
5 cm
1 cm
-
air condition in the chamber:
20oC, 750 mmHg
-
external current appeared in the ampere meter:
0.01 A = 10-8 C/sec
-
volume of the chamber:
V = (1/2)2 5  = 3.927 cm3
The exposure (# of ion pairs produced in the STP air by gamma- or X-radiation) rate is

X =
10 8 C / sec
 293  760 


3
6 
(3.927cm )(1.293x10 )  273  750 


= 2.14x10-3 Xunit/sec
= 8.30 R/sec
2
Example 7-1.2:
pocket dosimeter:
The pocket dosimeter is an ion chamber which is charged with electric potential
originally and will be discharged by the ions produced by radiation in the chamber.
Let’s consider a pocket dosimeter as follows:
air in STP
electric capacitance
= 5 pF
chamber volume
+
-
-
voltage across the chamber =
= 2 c.c.
180 V before exposure (original)
160 V after exposure (discharged)
-
exposure time = 1/2 hour
The average exposure rate is
(C )( V )
(5 x10 12 F )(180V  160V )

X=
(V )(  )( 12 hr ) (2cm 3 )(1.293x10 6 kg / cm 3 )( 12 hr )

= 7.734x10-5 Xunit/hr
= 0.3 R/hr
3
7-2
Particle-Counting Instruments
7-2.1
gas-filled particle counters:
The ionizing particles produced by radiation change to electrical pulses.
detector
output
C
R
The voltage of the pulse is expressed by:
V 
Q
C
Volt
(7-2.1)
where,
Q = collected charges, coulomb, and
C = capacitance of the circuit, F.
Depending on the applied voltage, the pulse height changes and the counter types are
characterized as follows:
pulse
Geiger
height
proportional
ionization chamber
discharge
region
recombination
voltage
before collection
4
1) recombination region: ions are recombined before collection due to low voltage
across the chamber:
2) ionization chamber: for low voltage, there exist feeble output pulses. In this case,
the collected electric charge is given by:
- Q = Ne
(N = number of ions produced)
3) proportional counter: secondary electrons are produced by the collision of
accelerated ions. In this case, the electric charge is proportional to the voltage applied:
-
Q = MNe (M = multiplication factor)
4) Geiger counter: avalanche of electrons extends along the entire length of the anode
once triggered by the originally produced ions. This can eliminate the pulse amplifier,
and the sizes of all pulses are the same.
5) discharge region: for very high voltage, the electric discharge occurs.
5
7-2.2
resolving time:
The pulse shape depends on RC constant of the electric circuit. For example as the RC
constant goes to infinity, the pulse shape reaches the plateau height with respect to time
as shown in the figure below.
RC = 
V
pulse
height
0
10
20
30
As can be seen in the figure, the narrow pulse is required to differentiate successive
pulses. Hence, the small time constant is necessary for better separation of the pulses
between.
Since the ionization primarily occurs near the anode, the collection time for positive
ions formed near the surface of the anode is given by
t
(b 2  a 2 ) p ln b a 
2V
sec
(7-2.2)
where,
b = radius of cathode, cm
a = radius of anode, cm
p = gas pressure in the counter, mmHg
V = potential difference across the counter, volts

-
= 1070 for air
-
= 1040 for argon
6
Example 7-2.1:
collection time:
How long will it take to collect all the positive ions
in an argon-filled Geiger counter?
data:
V = 1000 volts,
b = 1 cm,
a = 0.01 cm,
p = 100 mmHg
Solution:
From Eq.7-2.2,
t
(12  0.012 )(100) ln 1 / 0.01
= 221x10-6 sec
2 x100 x1040
resolving time
dead
recovery
voltage
0
100
200
300
400
500
600
700
800
time, 
Here, we define:
dead time: “time required to attain the electric field intensity to initiate an avalanche by
another ionizing particle”
recovery time: “time interval between the dead time and the time of full recovery”
resolving time: “sum of dead time and recovery time.”
Let,
7
r = true counting rate,
Ro = observed counting rate,
 = resolving time,
then, the loss of count is
r - Ro = r (Ro )
where,
Ro  = fraction of time for which the counter is insensitive.
Hence, the true counting rate is
r
Ro
1  Ro
(7-2.3)
The measurement of resolving time of a given instrument could be performed via twosource method.
Let,
r1, r2, r12, rb = true counting rates with source 1 plus background, source 2 plus
background, sources 1 and 2 plus background, and background only, respectively
R1, R2, R12, Rb = corresponding observed counting rates. Then
r1+ r2 = r12 + rb
Inserting Eq.7-2.3,
Rb
R1
R2
R12



1  R1 1  R2 1  R12 1  Rb
R  R   R
2
1

1
2
 

 R2   R12  R12   Rb 
2
2
R1  R2  R12  Rb
R12  R2  R2
2
2
(7-2.4)
2
8
7-2.3
miscellaneous detectors and counters:
1) scintillation counters:
The principle of scintillation is that ionizing particles produce flash or light in the
scintillation materials such as: NaI(Tl), CsI(Tl), KI(Tl), anthracene, etc. Using photomultiplier and a series of amplifiers, the light-pulse is amplified, analyzed and recorded
as follows:
detector
pre-
linear
amp
amp
pulse
scaler
analyzer
recorder
crystal
high
photo-multiplier voltage
2) Cerenkov detector:
High-speed charged particles produce the visible light when they move greater than light
speed in the dense medium. The detector is used in the high-energy physics lab in
detecting high-speed charged particles in the dense medium.
3) semi-conductor detector:
When the radiation produces ions in semi-conductors like Ge(Li) and Si(Li), electric
pulses are generated when the electric potential is supplied across the semi-conductors
as shown as below:
n-region
depletion layer
p-region
output pulses
9
7-3
Dose Measuring Instruments
Dose measuring instruments and particle counting instruments are basically different
each other. The dose measuring instruments are used for dose measurement, i.e., the
measurement of the total energy deposition of radiation rather than simple number of
particles. They include: pocket dosimeters, film badges, TLD, ion current chamber, etc.
Example 7-3.1:
dose measurement vs. particle counting: Find the difference between
dose measurement and particle counting.
Let’s assume two different cases as follows:
Case 1: 0.1 MeV - 2000 photons/cm2sec; and
Case 2: 2 MeV - 100 photons/cm2sec.
Solution:
The both cases have the same energy flux; 200 MeV/cm2sec.
From Eq.3-2.5, the absorbed dose rates are given by:

  (E ) 
D  (1.6 x10 13 )  e i  Eii Gy/sec
 m
i 
Hence,

D(case 1)  (1.6 x10 13 )(25.2cm2 / kg)(0.1MeV )(2000 / cm2 sec)  8.1x1010 Gy / sec

D(case 2)  (1.6 x10 13 )(25.7cm2 / kg)(2MeV )(100 / cm 2 sec)  8.2 x10 10 Gy / sec
The absorbed dose rates are almost same in both cases, but the particle counting or flux
measuring (i.e. use of GM-tube) is 20 times higher with 0.1 MeV photons.
10
Example 7-3.2:
Dose measurement using ionization chamber:
Let’s consider an air-filled ionization chamber with 1 mg/cm2 wall (tissue equivalent)
window.
a) What is the current produced from 1200 Po210-/minute entering the chamber?

( N particles / sec)( E eV / particle )(1.6 x10 19 C / ion )
w eV / ion
data:
energy of Po210-
E = 5.3 MeV
b) What is the current when the window thickness is increased to 3 mg/cm2 ?
Solution:
a) What is the current produced from 1200 Po210-/minute entering the chamber?
range in air (cm):
Rair = 1.24 E - 2.62 = 3.95 cm
range in tissue (mg/cm2): Rtissue = Rair
air
Atissue
Aair
-energy after passing through the window =
= (3.95)(1.293)
14 .4
14 .5
= 5.1 mg/cm2
5.1  1
5.3 = 4.25 MeV
5.1
1200
(
particles / sec)( 4.25 x10 6 eV / particle )(1.6 x10 19 C / ion )
  60
35 eV / ip
= 3.9x10-13 amps
b) What is the current when the window thickness is increased to 3 mg/cm2 ?
-energy after passing through the window =
  (3.9 x10 13 )
2.18
 2 x10 13 amps
4.25
11
5.1  3
5.3 = 2.18 MeV
5.1
7-4
Neutron Measurements
The neutron measurement is indirectly carried out by measuring the charged particles:
-
instantaneously produced by neutron-induced reactions (n,), (n,), (n,p),
(n,fission);
-
produced by decay of radioactive product nuclei produced by neutron-induced
activations; and
-
recoiled by neutron elastic scattering (ex. proton recoil).
Example 7-4.1:
BF3 counter:
BF3 counter is a typical neutron counter (thermal neutron) which is based upon
B10(n,)Li7 reaction using BF3 gas in the ion chamber. Alpha particles are counted
instead of neutrons. When a thermal neutron counting rate is 600 cpm, what is the
thermal neutron flux?
Data: Filled with 96% enriched B10F3 gas (p=20 cmHg, T=20oC)
2 cm
20 cm
Solution:
# of B-10 atoms in the counter,
N=
( 12  20)(10 3 )  20  273 
23
20
 
 6.02 x10 (0.96)  3.97 x10 atoms
22.4
 76  273  20 


The average absorption X-section of B-10 (1/v-absorber) is,
  293 
1/ 2
 a (20 C ) 



2  293 
(4010 b)  3553 b
The total reaction rate becomes,
12
Na = (3.97x1020)(3553x10-24


-count)
 = 7.1 neutrons/cm2sec
13
7-5
Counting
7-5.1
counting statistics:
The counting statistics are very important in order to draw a meaningful results of
counting.
The mean value of statistics is given by:
mn
n
i
i
(7-5.1)
N
where,
m = true value
n = mean value
ni = counts in the ith interval, and
N = total # of counting intervals.
The variance (square of standard deviation) of this statistics is given by:
 m  n 
2


i
2
 2  mn 
i
N
Since m is not known and the mean value n is computed from the same data set, the
number of independent sets of data (degree of freedom) should be reduced by 1 in
computing the variance, i.e.,
 n  n 
2
2 
i
i
(7-5.2)
N 1
14
Example 7-5-1:
Consider the data of 30 separate measurements taking for a 1-minute
interval as shown in the table below:
i
ni
n  ni
29
0.8
0.64
2
36
7.8
60.84
3
19
9.2
84.64
.
:
:
:
.
:
:
:
845
815.04
For the above example; the mean value is given by:
n
i
i
N

845
 28.2 .
30
The variance is given by:
 n  n 
2
2 
i
1
Sum
n
n  n 
i
i
N 1

815.04
 28.1
30  1
15
2
7-5.2
binomial distribution:
Let’s consider a large set of objects (No) with two classes; A and B. For example, in the
process of radioactive decay; A = transformation and B= remain in radioactive material.
A
B
The probability that exactly n of No objects selected from the set will be of class A can
be
w(n) 
No!
p n (1  p) No n
N o  n!n!
(7-5.3)
where,
p = probability that an object selected at random will be of class A
(1-p) = probability that an object selected at random will be of class B
This is called as the “binomial distribution.” For radioisotopes,
p = 1 - e-t (transformed or decayed)
1 - p = e-t (remained)
In this case, the true mean m becomes
No
m   nw(n)  pN o
(7-5.4)
n 0
Hence,
No
 2  m  n 2   (m  n) 2 w(n)
(7-5.5)
n 0
Expanding Eq.7-5.5,
No
No
No
n 0
n 0
n 0
 2   m 2 w(n)   2mnw(n)   n 2 w(n)
16
m
2
No
No
No
n 0
n 0
n 0
 w(n)  2m nw(n)   n w(n)
2
No
 w(n)  1
n 0
No
 nw(n)  m
n 0
No
 n w(n)  n
2
2
n 0
 n2  m2
(7-5.6)
Substituting,
n2 No!
p n (1  p) No n
N o  n!n!
No
No
n   n w(n)  
2
2
n 0
n 0
2
 No
  No
m    nw(n)    
 n 0
  n 0
2

nN o !
p n (1  p) No n 
N o  n !n!

2
The standard deviation becomes,
  N o p(1  p)1 / 2
(7-5.7)
 m(1  p)
1/ 2
Consider a radioactive transformation scheme in the time t of a system containing No
radioactive atoms. In this case the probability that n atoms will decay is
w(n) 


n
No!
1  e   t ( e  t ) N o  n
N o  n !n!
(7-5.8)
The true mean number of atoms decaying in time t is
m  N o (1  e  t )
(7-5.9)
The standard deviation becomes
  N o (1  e t )e t 
1/ 2

 me t

1/ 2
17
-life),
  m1 / 2  n
1/ 2
(7-5.10)
1. statistical error
The count rate (cpm: count per minute) including the standard deviation becomes
r  
n
n
r

r
t
t
t
Example 7-5.1:
(7-5.11)
statistical error: Let’s compare 10 minute counting with 1 minute
counting for a same counting rate.
a) 10,000 counts for 10 minute counting
r
n 10,000

 1,000 cpm
t
10

r
1,000

 10 cpm
t
10
 r  = (1,00010) cpm
b) 1,000 counts for 1 minute counting
r

n 1,000

 1,000 cpm
t
1
r
1,000

 32 cpm
t
1
 r  = (1,00032) cpm
2. background counting
Let
gross count rate (including the background count rate) = rg   g , and
18
background count rate = rb   b , then
net count rate = rn   n
where,
rn  rg  rb
 n   g 2  b2
(7-5.12)
Hence,
rn   n  (rg  rb ) 
Example 7-5.2:
Let
tg

rb
tb
(7-5.13)
background counting
gross count:
ng = 510 counts for 5 minutes
background count:
nb = 2,400 counts for 1 hour
rn  rg  rb 
n 
rg
rg
tg

ng
tg

nb 510 2400


 102  40  62 cpm
tb
5
60
rb
102 40


 4.6 cpm
tb
5
60
 rn   n  62  4.6 cpm
3. optimization of counting time
In order to minimize the standard deviation, d
n 
rg
tg

rb
tb
or
 n2 
rg
tg

Differentiating with respect to time,
19
n
rb
tb
=0 is required.
2 n d n  
rg
tg
2
dt g 
rb
tb
2
dtb
Letting dn =0,
rg
tg
2
dt g  
rb
dtb
2
tb
Since the total counting time t = tg + tb is constant, dt = dtg + dtb = 0.
Hence,
 rg 


t 2 
 g 
 rb  dt
 2   b
 1
 t 
dt g
 b 
tb
r
 b
tg
rg
(7-5.14)
Example 7-5-3:
Optimize the counting time: 1 hour is available for counting.
Preliminary measurements show that the background count rate is around 15 cpm and
the gross count rate is about 22 cpm. Optimize the counting time.
From Eq.7-5.14,
t b t  t g 60  t g
r
15


 b 
tg
tg
tg
rg
22
Hence, the optimized counting times for gross counting and background counting are:
tg = 33 minutes
tb = 27 minutes
20