Radiation Dosimetry
(Measurement of Absorbed
Dose)
Faiz Khan (Chapter 8)
Podgorsak (Chapter 2)
Introduction



Dosimetry attempts to quantitatively
relate specific radiation measurements to
chemical/biological changes that could be
produced
Essential for quantifying biological
changes as a function of radiation received
For comparing experiments
Introduction continued

Radiation interaction
• Produces ionized and excited atoms and
molecules
• Secondary electrons


Produce additional ionizations and excitations
Finally all energies are expended.
• Initial electronic transitions rapid (<10-15s)
• Represent the initial physical perturbations
from which all effects evolve.

So, ionization and energy absorption are
the starting point for radiation dosimetry
Quantities and Units
•
Absorbed dose is a measure of the
biologically significant effects
produced by ionizing radiation
• Definition of absorbed dose, dEavg/dm
•
dEavg : the mean energy imparted by
ionizing radiation to material of mass dm
• Unit
•
The old unit of dose is rad,

1 rad = 100 ergs/g = 10-2J/kg
• The SI unit of dose is the gray (Gy),

1 Gy = 1J/Kg
Quantities and Units

Exposure
• Defined for X and gamma radiation




In terms of ionization of air
Old unit called “roentgen” (R)
Initially defined in 1928, current definition
is:
1 R  2.58 x 10-4C kg-1 of air, exactly
• Applies only to electromagnetic
radiation; the charge and mass refer
only to air.
Roentgen - original definition

Amount of radiation that produced 1 esu
of charge in 1 cm3 of air at STP
• 1 esu = 3.335 x 10-10C
• At STP air has a density of 0.001293 g cm-3
• 1 kg of air has a volume of 7.734 x 105 cm3


1 R = 3.335 x 10-10 C cm-3 of air
How much energy is absorbed in air from
1R?
Absorbed Dose and Exposure

What is the absorbed dose in air when the
exposure is 1 R?
• Need to know W for air (energy to produce ion
pair in air)

33.7 eV/ip = 33.7 J/C.
• 1 R  2.58 x 10-4C/kg x 33.7 J/C = 8.8 x 10-3 J
kg-1

This equals 8.8 x 10-3 Gy (0.88 rad)



Similar calculations show that 1R would produce a dose
of 9.5 x10-3 Gy (=0.95 rad) in soft tissue.
Why is there a difference between air and tissue???
This is why one can say that 1 R ~ 1 rad in tissue.
Exposure Measurement Free Air Chamber




Feasible to measure exposure at radiation
energies between few keV and several
MeV
Definitive measure is by laboratory device
known as free air chamber
X-ray beam enters through a portal and
interacts with cylindrical column of air
defined by entry diaphragm
Ions created in defined space are
measured
Parallel Plate Free-air
Ionization Chambers
All the energy of the primary
HV
electrons expended in chamber
Wires
Diaphragm
scattered
photon
e4
Xray
beam
e3
e1
Monitor
Electrometer
e2
Free Air Chamber

Photons enter chamber and interact with
fixed quantity of air
• PE, CS



Ions from air collected by plates
Lead lined (shielded)
Electric fields are kept perpendicular to
plates by guard rings and guard wires
• Guard wires assure uniform potential drop
across plates
Free Air Chamber





Field intensity is ~ 100 V/cm
Collect ions prior to recombination
Voltage low enough so no secondary
ionizations
Current flow is measured
All energy of primary electrons must
be deposited in sensitive volume of
air for meter to work properly
Free Air Chamber


What if all primary electrons are not
collected in sensitive volume?
If equal number coming in from
outside of sensitive volume as is
going out: Electronic Equilibrium
Electronic Equilibrium
++++++++++++++
ee-
e----------------------
Electronic Equilibrium


For every electron which escapes the
sensitive volume, another electron of
equal energy enters the sensitive volume
and deposits energy in the detector
A layer of air between entrance port of
free air chamber and the sensitive volume
can provide enough air so that electronic
equilibrium is attained
Free-Air Ionization Chamber
http://physics.nist.gov/Divisions/Div846/Gp2/wafac.html
Measuring Exposure


Free air chamber practical only as
laboratory device
Portable instrument needed
Exposure Measurement:
Air Wall Chamber
Plastic
Charging
diaphragm
Anode


Practical alternative to free-air chamber
Built as a capacitor
• Central anode, insulated from rest of chamber
• Given an initial charge
• When exposed to photons, 20 electrons
neutralize charge & lower potential between
anode and wall
• Change in potential difference is proportional
to total ionization (and therefore exposure)
Air Wall Chamber


Better for field use than the Free Air
Chamber
Simulates compressing air into a small
volume by using ‘air equivalent’
material



X-ray absorption properties similar to that of air
Walls must be thick enough to generate
enough primary electrons
Walls must be thin enough so that
primary radiation is not shielded
Air Wall Chamber


Ideal air wall chambers have only
primary electrons ionizing the air
in the sensitive volume
Ideal wall thickness is almost
energy independent over a range
from 200 keV to almost 2 MeV
Air Wall Chamber


Greater than 3 MeV primary electrons
have long range
Impractical to build air wall chamber of
sufficient size
• When walls are made thick enough to
generate primary electrons, radiation is
attenuated significantly
• Radiation intensity will no longer be
constant
• Primary electrons not produced uniformly
• No electronic equilibrium
Exposure-Dose Relationship

Exposure
• measures charge produced in a mass of
air
• C/kg

Absorbed dose
• Measures energy absorbed per mass
• J/kg

How to relate measurement in air to
absorbed dose in something besides
air?
Exposure-Dose Relationship






Energy absorption in air  energy
absorption in tissue
Dose in air  dose in tissue
1 R = 87.7 ergs/gair = 95 ergs/gtissue
1 rad = 100 ergs/gtissue
For regulatory purposes, frequently 1
R is assumed to be equal to 1 rad
Conversion can be done if required
Exposure to Dose Conversion
mm = Energy absorption coefficient for
tissue
 ma = Energy absorption coefficient
for air
 rm = Tissue density
 ra = Air density
mm
rads 
rm
87.7

 roentgens
100 m a
ra
Bragg-Grey Theory


How to measure absorbed dose?
The best way would be
calorimetry...but not very
practical. Instead, absorbed dose
is measured by:
• measuring ionization
• use of correction factors
• calculating (approximating) dose

This is done with BRAGG-GREY
CAVITY THEORY
Measurement of Absorbed
Dose


Bragg-Gray principle relates
ionization measurements in a gas to
absorbed dose in some material.
Consider a gas in a walled enclosure
irradiated by photons:
Gas
e2
e1
Wall
Bragg-Gray,
cont’d
Gas
e1
Wall
e2

Photons interact in cavity and wall

Chose wall material that has similar radiation
absorption properties as tissue (e.g., Z)
• Cavity is very small

(doesn’t change angular and velocity
distributions of secondary electrons)
• “Electronic equilibrium” exists in
cavity


(# e- stopping = # e- starting in cavity)
requires wall thickness > range of secondary
e
Bragg-Gray cont’d



Ionizations in the gas
Can measure the charge liberated.
If you know the energy required to
ionize the gas,
W = 33.85

eV
J
= 33.85 ( for air )
ion pair
C
Then the dose to the gas is:
D gas
Q

W
m gas
Bragg-Gray, cont’d

where,
• Q = coulombs of charge liberated
• W = average ionization energy for the gas
• m = kg of gas in the cavity


Example: A cavity filled with (1 cm3) air
at STP is exposed to a radiation field
that liberates 3.336X10-10 C. What is
the dose to the air?
At STP:
3
kg
  -6 m 
3 
-6
=
(1
)
1.293
=
1.293X
kg



m gas
cm 
10
10
3
3

m 
cm 
Example, cont’d
J
 3.336X 10 - 10 C  
-3 J
  33.85  = 8.73X 10
D gas = 
-6
C
kg
 1.293X 10 kg  

Know the dose to the gas.
• What about the dose to the medium
surrounding it?

Assume our cavity is really
small...small enough that it does
not disrupt the electron
spectrum.
Gas
Bragg-Gray,
e
contd
Wall
1
e2


Wall thickness must be as great as range as
secondary charged particles (not too great to
attenuate beam)
Then energy absorbed per unit mass of wall is
related to that absorbed per unit mass of gas by:
Dw  Dg 

N gW
m
Note - this special case is where the wall
and gas are the same type of material
Gas
Bragg-Gray
e
, contd
Wall
1
e2





Dw is the dose to the wall
Dg is the dose to the gas
Nq is the number of ions produced in
the gas
W is eV required to produce ion pair
m is the mass of gas in the cavity
Gas
Bragg-Gray
e
, contd
Wall
1
e2

If gas and wall don’t have same atomic
composition, a slight modification is
required:
Dw 


Dg S w
Sg

N gWS w
mS g
where Sg,w are the mass stopping powers
of the wall and the gas
the cavity and gas pressure must be small
Example

1 cm3 of air in a block of carbon is
exposed to 60Co γ
• Q=3X10-8 C is produced.

What is the absorbed dose to the
carbon?
kg 

at STP m gas = ( 10 m )  1.293 3  = 1.293X 10 - 3 kg

m 
-6

3
Mean mass stopping power ratio for
60Co γ in carbon relative to air = 1.009
Example, continued
(3X 10 - 8 C) 
J
 33.85  (1.009)
D carbon =
-3
(1.293X 10 kg) 
C
J
= 0.792 = 0.792 Gy = 79.2 rad
kg

This equation allows us to measure
the ionizations in a gas and relate it
to dose to the medium.
Gas
Bragg-Gray
e
contd
Wall
1
e2


If neutrons are present, the wall
must be at least as thick as the
maximum energy range of any
secondary charged recoil nuclei
produced by the nuclear interactions.
Chambers that meet these
conditions can be used to measure
absorbed dose to the medium
Kerma
“Sum of the initial kinetic energies per
unit mass of all charged particles
produced by the radiation”
• This is regardless of where the energy is
deposited
• Bremsstrahlung photons are not counted,
whether they escape or not
• Annihilation radiation is not counted,
regardless of fate of annihilation photons

Initial positron, if primary ionizing particle, is
counted
Energy Transfer - A Two Stage
Process - Kerma and Absorbed Dose
Scattered photon
h
h’
Primary ionizing particle
(pe, cs electron, e+e- pairs,
scattered nuclei (neutrons)
h”
Quantity of transferred energy is called Kerma (j/kg)
Kerma

Etr is just the kinetic energy received by
charged particles in a specified volume
V, regardless of where or how they
spend the energy
d E tr d tr
K

dm dm

( ICRU , 1980)
Kerma is the expectation value of the
energy transferred to charged particles
per unit mass at a point of interest,
including radiative-loss energy, but
excluding energy passed from one
charged particle to another
Quantities to Describe a Radiation
Beam

Fluence

• # photons/area
•  = dN/da

• # photons/(time
area)
•  = d/dt
Energy fluence
• Energy / area
• = dN h/da
Fluence rate

Energy fluence rate
• Energy / (time area)
• = d /dt
Relationship of Kerma to
Photon Fluence

m
  
r
m
K =      E tr
r
• gives the number of photon interactions that
take place per unit mass of material.


m is attenuation coefficient
r is density
Relationship of Kerma to Photon
Fluence
For a spectrum of energies that can
be described by dΦ(hv) /d hv, then:

h max
K=

0
d ( h )  m ( h ) 

  E tr ( h )  dh
dh
 r 
Calculating Kerma


Given incident on a block of carbon
• 10 MeV photons
•  = 1014 m-2
What is kerma?
m
K =    E tr
r
2
m
m
  = 0.00196
kg
r
E tr = 7.3 MeV
Calculating Kerma, cont’d
14
2
10
m
J
- 13
K  2  0.00196
 7.30 MeV  1.602x10
m
kg
MeV
J
 0.229
kg

Kerma is easy to calculate - but
very difficult to measure!
Relating Kerma & Absorbed
Dose

Kerma
• a measure of kinetic energy transferred at a
point in space.

Absorbed dose is more “interesting”.
• Energy is transferred in the medium
• not all is retained there.
• absorbed dose is the energy retained in the
medium brought about by the ionizations along
the track of the charged particle.

Kerma and Absorbed Dose do not take
place at the same location
Calculating Absorbed Dose
d E ab
D
dm

dEab is the mean energy “imparted” by the
ionizing radiation into a mass, dm.
• Mass should be sufficiently small so that the
absorbed dose is defined at a point, but not so
small that statistical fluctuations become
important

From the previous example, dEtr = 7.3 MeV
• fraction of 10 MeV photon energy transferred to
the medium.

A smaller amount is absorbed along the
electron track: dEab = 7.06MeV
Kerma and Absorbed Dose,
cont’d

dEtr- dEab
• The difference, 7.30-7.06 = 0.24 MeV, is
bremsstrahlung.

What is the path length of the 7.3
MeV electron in C?
• Estimate from graphs or tables of
electron ranges from literature,
• ~ 4.2 g cm-2.
• Divide by the density of carbon
• Path length: 1.9 cm.
Dose and Kerma
m
K =    E tr
r
m
D    E ab
r
Important Relationship

Relate absorbed dose in air to
exposure:
• assuming CPE (electronic equilibrium)
Dair  K c(air )  X Wair
J/kg
J/kg
C/kg
J/C
Electronic (Charged-Particle)
Equilibrium

The transfer of energy (kerma)
occurs upstream from the absorbed
dose.
• Kerma can be easily calculated from
fluence
• Absorbed dose cannot. Why?
• Kerma remains constant
• Absorbed dose takes time to build up as
upstream electrons increase:
No Attenuation of Photon Beam,
Range R
Φ Constant
A
B
C
100
100
100
D
E
F
100
G
Number of electron tracks set in motion by photon
interaction
• Φ constant with depth (small # interactions)
• Same # electrons set in motion in each square
• i.e., interactions per volume constant through
target
Absorbed Dose and Kerma
A
B
C
100
100
100
D
E
F
kerma
Absorbed dose
Build up region
Electronic equilibrium
depth
100
G
Beam Unattenuated

Same number of photon tracks set in
motion in each square
• e.g., square D is traversed by 400 tracks
• ionization in D is the same as total
ionization started in A
• absorbed dose is proportional to ionization
produced in each square
• dose reaches a maximum at R (range of
2ndary electron)
• kerma constant with depth, equals
absorbed dose beyond R
Absorbed Dose and Kerma
100
95
90
86
kerma
Absorbed dose
In this region there is not
strict electronic equilibrium
Equilibrium thickness
Build up region
depth
82
78
Attenuation of Photon Beam



Beam attenuation,
Φ decreases with depth.
Dose increases to a maximum (at
maximum range of particle)
overshoots, then tracks kerma.
Attenuation of Photons in
Tissue
Isotope
137
Maximum
Beam
Dose Depth Attenuation
( mm in
(% of
Tissue)
original
beam)
Cs
2
1
Co
5
2
6 MV
15
6
60
CPE will generally exist in a uniform medium at a
point more than the maximum range for the
secondary charged particles from the
boundary of the medium
Relating Energy Fluence and
Exposure
p

Radioactive beam incident on an area
• What is relationship between energy fluence
and exposure at point p?


Assume small mass of air at p
The dose at p is: D= (m/r)Ēab=  (mab/r)
• Can relate to R as:



1 R = 0.00873 J/kg, then
/X = 0.00873 J/ ((mab/r)kg R)
Complicated variation of energy absorption
coefficient for air and energy of beam
Relating photon fluence to
exposure

Relationship between energy fluence
and photon fluence:
•  = dN/da
• = dN h/da
• So, =  h, and


X
0.00873 J
 m ab 
 kg R
h 
 r  air