Understanding Radiation Units. Radiation Protection in Paediatric Radiology L02. Understanding radiation units 22 Answer True or False 1.The same amount

Understanding Radiation Understanding Radiation UnitsUnits

Radiation Protection in Paediatric Radiology L02. Understanding radiation units 22

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.


Introduction

• Several quantities and units are used in the field of diagnostic radiology to measure and describe radiation dose.

• Dosimetry is the quantitative determination of radiation doses.

• Some can be measured directly while others can only be mathematically estimated.


Hot Coffee – Energy contained in a sip

Excess Temperature = 60º - 37 = 23º

1 sip = 3ml3x 23 = 69

calories


Radiation Dose

Lethal Dose= 4GyLD 50/60 = 4 GyFor 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 X-rays can be lethal

X-

rays


Dose of Radiation

Radiation energy absorbed by a body per unit mass.

Dosimetry is the quantitative determination of radiation doses.


Basic Radiation Quantities

• Used to quantify a beam of X or γ-rays

• There are:• Quantities to

express total amount of radiation.

• Quantities to express radiation at a specific point

Radiation at a specific point

•Photon fluence

•Absorbed dose

•Kerma

•Dose equivalent

Total radiation

•Total photons

•Integral dose


Exposure (X)

• Exposure is a dosimetric quantity for measuring ionizing electromagnetic radiation (X-rays & Ɣ-rays), 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

http://images.google.se/imgres?imgurl=http://www.hko.gov.hk/education/dbcp/radiation/eng/image/ionization.jpg&imgrefurl=http://www.hko.gov.hk/education/dbcp/radiation/eng/r7.htm&usg=__TnBtdWPl6MyaIL834IcXFfdTZMM=&h=166&w=353&sz=17&hl=sv&start=9&tbnid=S6KWNr4QQNIiVM:&tbnh=57&tbnw=121&prev=/images?q=ionization&gbv=2&hl=sv


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.


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


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).

In diagnostic radiology, KERMA and D are equal.

Absorbed dose: D

Harold Gray

http://www.spock.com/picture_frame/78c81wb451a51421

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 dose(rad).

• 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 rad.



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.


Now things get a little more complicated !

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

Radiation Protection in Paediatric Radiology L02. Understanding radiation units

Radiation Weighting Factors, wR

Radiation type 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

51020105(Source: ICRP 103) Radiation Protection in Paediatric Radiology L02. Understanding radiation units


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 : absorbed dose and equivalent dose have the same value Gy = Sv.

Rolph Sievert


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



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.



1. False -Different organs have different radio-sensitivity and tissue weighting factors.

2. False -It can be only calculated using different methods.

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?


a. 75.0 rad ( 1 Gy/ 100 rad) = 0.0750 Gyb. eq.dose = absorbed dose x Wr = (0.0750) ( 5)

= 0.375 Sveq.dose = 7.50 rad (5)

= 37.5 remc. Absorbed dose = eq.dose / Wr = 0.375

Sv/20= 0.0188 Gy


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?


Answer.

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 Gyabsorbed dose = E/mE = absorbed dose (m) = 0.25 Gy (60 kg) = 15 J



Mass Defect and Binding Energy

E = mc2


It tells of the huge amount of energy locked up in ordinary matter.


• 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.


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.


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


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.


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 amu2mp + 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


• 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


• 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


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. n __ Zr Ce U 1

09440

14058

23592

Given: m U-235 = 235.0439 amu m Zr-94 = 93.9065 amu mCe-140 = 139.9055 amu n = 1.00866 amu

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.


Calculate the energy involved in the reaction.


n 3 Kr Ba U n 10

9536

13856

23592

10

,

(1)

mass of the reactants = 235.0439

= 1.0087

Total mass of reactants = 236.0526

(2)

Mass of the products = 137.9050

= 94.9

3 = 3.0260

Total mass of products = 235.831


U23592

n10

Ba13856

Kr9536

n10

Total mass of reactants = 236.0526 amu

Total mass of products = 235.831 amu

Δ m = 0.2216 amu

0.222 amu


E = (0.222 ) (931 MeV/amu ) = 206 MeV


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• G-3: Radioactive Dating



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


• If the KAP is calculated by the system, you must know if the user added filtration you use is included or not !

Kerma-Area Product: KAP


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


Dosimetric Quantities for CT

• Computed Tomography Dose Index (CTDI)

• CT air kerma index

• Dose-Length Product (DLP)

• Air kerma-length product


ICRU 74 / IAEA TRS 457

• CT air kerma index•Free-in-air (Ck)• In phantom

(Ck,PMMA)

• Air kerma length product (PKA)



Principal dosimetric quantity in CT is CT air kerma index:

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.

50

50

100, )(1

dzzKNT

Ca



Weighted CT air kerma index, CW, combines values of CPMMA,100 measured at the centre and periphery of a standard CT dosimetry phantoms

pPMMAcPMMAw CCC ,100,,100, 23

1



Pitch (IEC, 2003):

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 rotationRadiographic, 2002, 22:949-62

NT

Ip


• 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):

• Average dose over x, y and z direction• Protocol-specific information


l

NTCC WVOL



• Kerma-length product (PKL):

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)

LCP VOLKL


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


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


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


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 about 4 Gy


DRLs for Complex Procedures

3rd level “Patient risk”

2nd level “Clinical protocol”

1st level“Equipment

performance”

Dose rate and dose/image(BSS, CDRH, AAPM)

Level 1 + No. images + fluoroscopy

time

Level 2 + DAP + Peak Skin Dose (MSD)

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)

Documents

Understanding Radiation Units. Radiation Protection in Paediatric Radiology L02. Understanding radiation units 22 Answer True or False 1.The same amount