# Understanding Radiation Units

```Understanding Radiation Units
Answer True or False
1. The same amount of radiation falling on the
person at level of breast, head or gonads will
have the same biological effects.
2. Effective dose can be easily measured.
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Introduction
• Several quantities and units are used in the field of
diagnostic radiology to measure and describe
• Dosimetry is the quantitative determination of
• Some can be measured directly while others can
only be mathematically estimated.
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Hot Coffee – Energy contained in a sip
Excess Temperature = 60º - 37 = 23º
1 sip = 3ml
3x 23 = 69 calories
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Lethal Dose= 4Gy
LD 50/60 = 4 Gy
For man of 70 kg
Energy absorbed = 4 x 70 = 280 J
= 280/418= 67 calories
= 1 sip
Energy content of a sip of coffee if derived in the form of Xrays can be lethal
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absorbed by a body
per unit mass.
 Dosimetry is the
quantitative
determination of
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• Used to quantify a beam
of X or γ-rays
• There are:
• Quantities to express
total amount of
• Quantities to express
radiation at a specific
point
•Total photons
•Integral dose
specific point
•Photon fluence
•Absorbed dose
•Kerma
•Dose equivalent
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Exposure (X)
• Exposure is a dosimetric quantity for measuring
ionizing electromagnetic radiation (X-rays & Ɣrays), based on the ability of the radiation to
produce ionization in air.
Units:
coulomb/kg (C/kg)
or
roentgen (R)
1 R = 0.000258 C/kg
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EXPOSURE (X)
• It is the ability of a radiation beam to ionize air.
• It is measured in roentgens ( R), where 1 R of radiation
exposure produces ions carrying 2.58 x 10-4 C of charge per
kg of dry air.
• The exposure is the charge Q in the air per unit mass m
exposure = Q/m
• 1 R = 2.58 x 10-4 C /kg
• The only information X can give is how much radiation is
present. It does not say anything about whether all this
ionization is absorbed by the material.
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KERMA
KERMA (Kinetic Energy Released in a Material):
• Is the sum of the initial kinetic energies of all charged
ionizing particles liberated by uncharged ionizing
particles in a material of unit mass
• For medical imaging use, KERMA is usually expressed in
air
SI unit = joule per kilogram (J/kg)
or gray (Gy)
1 J/kg = 1 Gy
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Absorbed dose: D
Absorbed dose, D, is the amount of (E)
energy imparted by ionizing radiation to
matter per unit mass (m).
D=E/m
SI unit = joule per kg (J/kg) or gray (Gy).
Harold Gray
In diagnostic radiology, KERMA and D
are equal.
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Absorbed Dose (D)
• One gray (Gy) is the amount of radiation
(regardless of the type) that will deposit 1 J
of energy in 1 kg of matter.
• An older unit for absorbed dose is the rad,
an acronym for radiation absorbed
• One rad is the quantity of ionizing radiation
that will deposit 0.01 J ( 100 ergs) of energy
in 1 g of absorbing material. 1 gray = 100
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Mean absorbed dose in a tissue or organ
The mean absorbed dose in a tissue or organ DT
is the energy deposited in the organ divided by
the mass of that organ.
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Now things get a little more complicated !
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Equivalent Dose
•
•
•
•
•
•
•
The biological effects of radiation depends on the type of radiation.
Equal dose of Alpha particles and Gamma radiation does not affect the tissue
the same way.
To account for the difference of the biological effects of radiation, we use the
quantity equivalent dose.
Equivalent Dose is the product of absorbed dose and the radiation weighting
factor wR for the type of radiation use.
Equivalent dose = absorbed dose x wR
The SI unit of equivalent dose is sievert (Sv)
An older unit for equivalent dose is the rem , short for roentgen equivalent
man.
1 Sv = J/kg
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Radiation Weighting Factors, wR
Radiation weighting factor, wR
Photons
1
Electrons and muons
1
Protons and charged pions
2
Alpha particle, fission fragments,
heavy ions
5
Neutrons,energy < 10keV
>10-100 keV
>100keV – 2MeV
> 2-20 MeV
> 20 MeV Radiation Protection in Paediatric Radiology
5
10
20
10
5ICRP 103)
L02. Understanding
(Source:
Radiation Quantities and Units
Equivalent dose (Unit = sievert, Sv )
• Compares the biological effects for
different types of radiation, X-rays,
Ɣ-rays, electrons, neutrons,
protons, α-particles etc.
• For X-rays, Ɣ-rays, electrons :
Rolph Sievert
absorbed dose and equivalent
dose have the same value Gy = Sv.
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Summary
• Dosimetric quantities are useful to know the
•
•
•
•
potential hazard from radiation and to determine
radiation protection measures to be taken
Physical quantities - Directly measurable
Protection quantities - Defined for dose limitation
purposes, but not directly measurable.
Application specific quantities - Measurable in
medical imaging.
Diagnostic Refernce Levels
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Answer True or False
1. The same amount of radiation falling on the
person at level of breast, head or gonads will
have same biological effects.
2. Effective dose can be easily measured.
3. Diagnostic reference levels are not
applicable to paediatric radiology.
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Answer True or False
1. False -Different organs have different
radio-sensitivity and tissue weighting
factors.
2. False -It can be only calculated using
different methods.
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Example 1
• A certain biological sample is given a dose
of 7.50 rad from alpha particles.
a. Calculate the absorbed dose in grays.
b. Calculate the equivalent dose in
sievert and rems.
c. If the same dosage is delivered using
fast neutrons with wR of 20, how much
dosage in grays will be needed?
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a. 75.0 rad ( 1 Gy/ 100 rad) = 0.0750 Gy
b. eq.dose = absorbed dose x Wr
= (0.0750) ( 5)
= 0.375 Sv
eq.dose = 7.50 rad (5)
= 37.5 rem
c. Absorbed dose = eq.dose / Wr = 0.375 Sv/20
= 0.0188 Gy
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Solve:
1.How does the biological damage of 100 rad of beta
particles compare with that of 100 rad of alpha
particles?
2.A 75.0 mg tissue sample is irradiated. If it abbsorbs
0.240 mJ of energy, what is the absorbed dose?
3.A person whose mass is 60.0 kg has been given a
full-body exposure to a dose of 25.0 rad. How
many joules of energy are deposited in the body?
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1. For beta:
Eq.dose = absobed dose x Wr
100 rad (1) = 100 rem
For alpha: Eq.dose = absobed dose x Wr
100 rad (5) =500 rem
2. Absorbed dose = E/ m = 0.240 mJ / 0.075 kg
= 32 Gy
3. Convert rad to Gy : 25 rad = 1 Gy/ 100 rad = 0.25 Gy
absorbed dose = E/m
E = absorbed dose (m)
= 0.25 Gy (60 kg)
= 15 J
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Mass Defect and Binding Energy
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E=
2
mc
It tells of the huge amount of energy locked
up in ordinary matter.
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• Protons and neutrons make up the nucleus.
• The mass of the nucleus is equal to the
combined masses of its protons and
neutrons.
• Nuclear reaction can be analyzed in terms of
the masses and energies of the nuclei and
the particle before and after reaction.
• E = mc2 is used in analyzing nuclear
reaction.
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Nuclear masses are measured using an instrument
called mass spectrometer .
The mass of the atoms is then compared against a
standard which is in the case is C-12 and is
expressed in atomic mass unit (amu). The mass of
C-12 is equal to 12 amu.
 1amu = 1.6600 x 10-27 kg
When compared to C-12, the mass of hydrogen is
equal to 1.007825 amu.
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Masses of the Proton, Neutron and
Electron
Mass of proton, mp = 1.0073 amu
Mass of neutron, mn = 1.0087 amu
Mass of electron, me = 0.0005 amu
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 How does the measured mass of hydrogen
compare with its nuclear mass?
 Compare the measured mass of the helium
nucleus to the combined masses of all the two
protons and two neutrons.
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 Compare the measured mass of the helium nucleus to the
combined masses of all the two protons and two neutrons.
Sol .
mHe = 4.0015 amu
 2mp = 1.0073 amu x 2 = 2.0146 amu
 2mn = 1.0087 amu x 2 = 2.0174 amu
2mp + 2mn = 2.0146 amu + 2.0174 amu = 4.0320 amu
Note: The nuclear mass of the Helium atom is less than the
total mass of its constituent parts.
The difference in mass is known as the mass defect, Δ m.
Δ m = 4.0320 amu – 4.0015 amu
= 0.0305 amu
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• Mass seems to disappear when protons and neutrons
combine to form a nucleus .
• Einstein’s principle of mass- energy equivalence says that
the missing mass ( mass defect) is converted into energy.
• The energy equivalent of the mass defect is known as the
binding energy (BE) . From E = mc2
BE = Δmc2
• The binding energy for helium can be calculated:
BEHe = Δmc2
= (0.0305 amu) ( 3x108 m/s) 2
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• 1 amu = 931 MeV
• For He,
.
BEHe = (0.0305 amu) ( 931MeV/ amu)
= 28.4 MeV
• 1 amu = 1.66 x 10 -27 kg
• 1eV = 1.602 x 10 -19 J
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Solve:
1. Calculate the binding energy of deuterium consists of one
proton and one neutron. Its nuclear mass is 2.0135 amu.
2. Compute the binding energy of C-12 and compare it with
the binding energy of C-14.
3. Calculate the disintegration energy of the reaction below.
235
92
U 
140
58
Ce 
94
40
Zr  __ n
Given: m U-235 = 235.0439 amu
mCe-140 = 139.9055 amu
1
0
m Zr-94 = 93.9065 amu
n = 1.00866 amu
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Nuclear Fission and Fusion
• The process by which a heavy nucleus split
into medium-sized nuclei. It is induced by
bombarding a heavy nucleus with neutrons.
• When two nuclei come very close to one
another at very high temperature, the strong
nuclear binding force predominates and
allows the nuclei to fuse, releasing large
amount of energy.
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,
Calculate the energy involved in the reaction.
1
0
n 
235
92
U 
138
56
Ba 
95
36
Kr  3 n
1
0
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
(1)
mass of the reactants 
235
92
1
0
U = 235.0439
n
= 1.0087
Total mass of reactants = 236.0526
(2)
Mass of the products 
138
56
95
36
Ba
= 137.9050
Kr
= 94.9
3 01 n =
3.0260
Total mass of products = 235.831
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


Total mass of reactants
= 236.0526 amu
Total mass of products
= 235.831 amu
Δm
=
0.2216 amu
0.222 amu
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E = (0.222 ) (931 MeV/amu )
= 206 MeV
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Assignment : Applications of Radioactivity and
Nuclear Energy
• Enumerate applications of radioactivity and
•
•
•
•
nuclear energy that you are aware of.
Group 1-3 will present a report on the
applications of radioactivity and nuclear
energy to.
G-1: Food and Agriculture
G-2: Diagnosis and Therapy
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Backscatter Factors (Water)
HVL
Field size (cm x cm)
mmAl
10 x 10
15 x 15
20 x 20
25 x 25
30 x 30
2.0
1.26
1.28
1.29
1.30
1.30
2.5
1.28
1.31
1.32
1.33
1.34
3.0
1.30
1.33
1.35
1.36
1.37
4.0
1.32
1.37
1.39
1.40
1.41
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Kerma-Area Product: KAP
• If the KAP is calculated by the system, you must
know if the user added filtration you use is included
or not !
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Kerma-Area Product: KAP
• It is always necessary to calibrate and to check the
transmission chamber for the X-ray installation in
use
• In some European countries, it is compulsory that
new equipment is equipped with an integrated
ionization transmission chamber or with automatic
calculation methods
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Dosimetric Quantities for CT
• Computed Tomography
Dose Index (CTDI)
• CT air kerma index
• Dose-Length Product
(DLP)
• Air kerma-length product
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ICRU 74 / IAEA TRS 457
• CT air kerma index
• Free-in-air (Ck)
• In phantom (Ck,PMMA)
• Air kerma length
product (PKA)
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Dosimetric Quantities for CT
Principal dosimetric quantity in CT is CT air kerma index:
Ca ,100
1

NT
50
 K ( z )dz
50
where K(z) is air kerma along a line parallel to the axis of
rotation of the scanner over a length of 100 mm.
N = Number of detectors in multi-slice CT
T = Individual detector dimension along z-dimension
The product NT defines the nominal scan beam
width/collimation for a given protocol.
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Dosimetric Quantities for CT
Weighted CT air kerma index, CW, combines
values of CPMMA,100 measured at the centre and
periphery of a standard CT dosimetry phantoms
1
Cw  C PMMA,100,c  2C PMMA,100, p 
3
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Dosimetric Quantities for CT
Pitch (IEC, 2003):
I
p
NT
T= Single detector dimension
along z-axis in mm.
N=Number of detectors used in a
given scan protocol (N>1 for
MDCT), N x T is total detector
acquisition width or collimation
I=table travel per rotation
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Dosimetric Quantities for CT
• Volume CTDI describes the average dose over the
total volume scanned in sequential or helical
sequence, taking into account gaps and overlaps
of dose profiles (IEC, 2003):
CVOL
NT
 CW
l
• Average dose over x, y and z direction
• Protocol-specific information
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Dosimetric Quantities for CT
• Kerma-length product (PKL):
PKL  CVOL  L
where L is scan length is limited by outer margins
of the exposed scan range (irrespective to pitch)
• PKL for different sequences are additive if refer to
the same type of phantom (head/body)
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Maximum Skin Dose (MSD)
• Measurement/evaluation of MSD
• Point or area detectors
• Cumulative dose at IRP (interventional radiology
point)
• Calculation from technical data
• Off line methods
• Area detectors: TLD array, slow films, radiochromic
films
• From KAP and Cumulative dose measurement
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Method for MSD Evaluation:
Radiochromic Large Area Detector
Example: Radiochromic films type Gafchromic XR R 14”x17”
• useful dose range: 0.1-15 Gy
• minimal photon energy dependence (60 - 120 keV)
• acquisition with a flatbed scanner:b/w image, 12-16 bit/pixel
or, measure of OD measurement with a reflection densitometer
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Benefits of Radiochromic Films
• The radiochromic film:
• displays the maximum dose and its location
• shows how the total dose is distributed
• provides a quantitative record for patient files
• provides physician with guidance to enable safe
planning of future fluoroscopically guided procedures
• improves fluoroscopic technique and patient safety
• possible rapid semi-quantitative evaluation
Example of an exposed
radiochromic film in a cardiac
interventional procedure
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Rapid Semi-Quantitative Evaluation: Example
• For each batch number (lot #) of gafchromic film a Comparison
Tablet is provided
• In the reported example we easily can recognise that the darkness
area of the film, corresponding to the skin area that has received the
maximum local dose, has an Optical Density that correspond at
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DRLs for Complex Procedures
Reference levels (indicative of the state of the practice): to help
operators to conduct optimized procedures with reference to patient
exposure
For complex procedures
reference levels should
include:
• more parameters
• and, must take into
account the complexity
of the procedures.
(European Dimond
Consortium
recommendations)
3rd level
“Patient risk”
2nd level
“Clinical protocol”
Level 2 + DAP
+ Peak Skin Dose (MSD)
Level 1
+ No. images + fluoroscopy time
1st level
“Equipment
performance”
Dose rate and dose/image
(BSS, CDRH, AAPM)
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