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28/05/2012 Yaser Gholami Radiation Physics Prac Assignment

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Part.1 (Q.1).Explain the need to determine recombination and polarity values for dose measurements using an ionization chamber.

Polarity effect : As x-ray beam or electron enters the chamber it will start ionizing the medium inside the chamber which could be gas or liquid . For the case when high energy photon is penetrating through the chamber, it will produce high energy Compton electron which constitutes a current called Compton current which is independent of the gas ionization. This effect collected. Also the external current such as the current collected outside the sensitive of the chamber could cause polarity effect. In addition the irradiation of the cable connecting the chamber with the electrometer could cause an external current.

The error due this effect could be minimized by reversing the polarity and taking the mean value of the collector current.

Recombination : some of the ions, on their way to the collection electrode might recombine with ions that have opposite charge and hence they are not collected.

Q(2) The Farmer ionization chamber has a maximum operating voltage of 400V. Explain the reasons why you would not use a higher voltage.

Basically when x-ray beam enters the chamber it will loses energy to gas by creating excited molecules, positive ions and negative ions (electrons). Without an external electrical field charges would recombine whereas in an applied electric field the positive and negative ions would drift in opposite directions and as a result an electric current will be produced that can be measured externally. This is the voltage at which the ionization chamber is operated. At voltage greater then voltage (400V) multiplication begins, where electron's energy would be greater then ionization energy and hence as a reasult it will result in charge multiplication (secondary charge will be generated). Knowing that Ionization chambers operated in direct current mode and it measures the average rate of ion formation in the chamber, it can only operate at a voltage lower than the multiplication region.

Q.3) Determine the recombination for the Farmer type ionization chamber that was used in the measurements with the 6MV x-ray beams. Note that the radiation output from the linear accelerator is pulsed radiation and not continuous.

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Part 2 Radiation dose output variation with beam size 1. Plot the relative dose output as a function of field size for the dose measurements at the different depths. Your doses should

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Relative dose output Vs field size

Relative dose output (nc)

0

D at 10 cm Datat points for D Max

200

400

D Max

600

800

1000

1200 1400 1600

Data points for Dat 10 cm

2. Explain why the relative doses are different for the measurements at Dmax and at 10 cm depth in the solid water phantom.

The dose rate at point depends on the field size A; as it was absorved from the expriment ,the larger the field size, the larger the dose. The relative dose factor (RDF) (referred to the machine output factor) is defined as the ratio of , the dose at P in a phantom for field A, to

Therefore this ndicates that the RDF(A) contains two factors : 1.scatter from the collimator 2. scatter from the phantom

3. In this particular task, dose measurements were performed in a Solid Water phantom. a) Describe the properties of Solid Water that allow it to be used as a substitute for water : The Solid water Phantom has a similar radiological properties to water. These properties are :

28/05/2012 1. physical density, 2.relative electron density,(because for megavoltage photon beams ,Compton effect is the main interaction in the clinical range),scattering of radiation 3.effective atomic number, 4.Similar absorption and Scattering of radiation b) Give some reasons why Solid Water is useful to use in a radiotherapy department.

Solid Water is easer to for transporting and setting up and also it eliminate some of the thcnical issues such as filling water tanks. Also For some cases it is not possible to put the radiation detector in water, therefore in these cases using soild water phantom is usful.In addition Solid water Phantom can be used for both electron and photon beam calibrations.

Questions: 1. Explain why you get a dose buildup region for megavoltage x-ray beams.

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Charge reading vs depth 12 11 10 9 Charge reading ( nc) 8 7 6 5 0

5

10 depth ( cm)

15

20

The dose region between the surface (depth = 0) and is called the dose buildup region. The incident photon interact with phantom and it produces secondary electrons which deposit thier energy downstream.The electron fluence and thus, absorbed dose increases until reach maximum (approximately range of electrons) at which electronic equilibrium occurs for photon beams; the point where dose reaches its maximum value. Same time-photon fluence decreases with depth at constant rate. Therefore, fewer photons to eject electrons as increase depth.The combination of all these factors yields the dose build up region (region before dmax).

2. Describe the differences in buildup region doses for the following beams: Cobalt60, 4MV, 6MV and 18 MV linac x-ray beams. the differences in buildup region doses depends on the energy of the photon beam. As the energy of the beam increases ,the depth of maximum ionization increases too . High energy photons beams can penetrate deeper in the meduim then lower energy , and they create more energetic electrons. hence , the maximum dose occurs at a deeper depth for high energy beams which crosspond to a larger build up region.

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2. From this database, determine and plot the mass-energy absorption coefficients for the following materials in the energy range from 10 keV to 10 MeV: Air, Aluminium, Cortical bone, Lung Tissue and Water.

Air, Dry (Near sea Level)

1

(c

/g)

1 Photon.Energy.(Mev)

5

10

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Z=13 , Aluminum

10 5

1 (c

/g)

1 Photon.Energy (Mev)

5

10

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Water

1

(c

/g )

1 Photon.Energy. (Mev)

5

10

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Bone, cortical, (ICRU-44)

10 5

1 (c

/g)

1 Photon.Energy. (Mev)

5

10

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Lung Tissue 5

1

(c

/g)

1 Photon.Energy. (Mev)

5

10

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Air, Aluminum,Water,Bone cortical, (ICRU-44),Lung tissue

10

5

1 (c

/g)

1 Photon.Energy. (Mev) Air

Water

Lung Tissue

Al

5

Bone

3. Comment on the differences in the mass-energy absorption coefficients at the different photon energies.

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The probability for a photon to take any interaction with an attenuator medium depends on the energy of the photon and on the atomic number Z of the attenuating Meduium. The photoelectric effect is tha main interation at low photon energies whereas the Compton effect is the main interaction at intermediate energies and pair production at high energies.

The mass energy-absorption coefficient is the average amount of incident photon energy transferred to kinetic energy of charged particles as a result of the interactions.

For Example, a 100 keV photon will interact with high Z materials (such as Aluminum, Cortical bone) mainly through the as it was shown in the above , they are similar).While a 100 keV photon will interact with soft tissue (Zeff = 7.5) and water predominantly through the Compton effect.

In addtion a 10 MeV photon, on the other hand interacts with high Z materials (such as Aluminum, Cortical bone) mainly through pair production and with tissue and water through the Compton effect.

4. Explain the differences in the doses between the measurement in air and the measurement at the surface of the solid water phantom.

The does rate in a scatter-free condition such as dose rate in air will be lower then that measured at surface of a Phantom.

This is due to the back sccatring facotr.

5. Can you explain the reason why backscatter is important for dosimetry measurements in the lower energy x-ray range from As it was mentioned above the does rate in a scatter-free condition such as dose rate in air will be lower then that measured at surface of a Phantom. This is is becuase of the secondary x-rays which are produced as a result of interaction between the incident x-ray and the phantom's meduim through compton scattering. These secoondary x-rays will be scattered in all directions so the radiation detector will be irraidted not only by the radiation coming directly from the radiation source but also by secondray radtion from within the phantom. So :

Or And as we mentioned the dominated interaction in low enerygies is compton scattering ( Also by looking at the plots ) we can see this scattring facter is an impportant factor for Doese Measrument.