The chief difference between the gamma radiation and X ray is how they are produced. Gamma beams are generated in the karyon of a radionuclide during decay. Whereas X raies are produced when negatrons strike a mark.

Does 1 millimeter of lead provide the same effectual shielding for both types of radiation?

For effectual Gamma rays necessitate big sums of either concrete or lead. So in this instance 1mm of lead would non be sufficient, as the gamma beams would go through through without any jobs. In the instance of x-rays 1mm of lead provides effectual shielding.

Finally, supply values for the HVL and TVL for 140 kV X raies and 140 kV gamma-rays.

X-rays 140 kilovolt:

Half-Value Layer ( Lead ) : 0.32mm

Tenth-Value Layer ( Lead ) :1.04mm

Gamma rays 140 kilovolt:

Half-Value Layer ( Lead ) : 0.027 centimeter

Tenth-Value Layer ( Lead ) : 0.083 centimeter

Question 2

During production of a 99Mo/99mTc generator, 100 GBq of pure Mo 99Mo was loaded into the column on 7-th of January 2011 at 12:00. Immediately after generator ‘s readying the first elution was made. Subsequently, the generator was delivered to the infirmary and was used clinically for the first clip on January 10-th at 9:00.

Calculate:

The mass of 99Mo laden during production.

In order to happen the mass foremost necessitate to happen the decay invariable:

Half life = ln2/I»

Half-life = 2.214×10^5 seconds

I»= decay invariable

Rearrange the equation so:

I»= ( ln2 ) / ( 2.214×10^5 )

I»=3.13×10^-6 secs^-1

Now want to happen the figure of atoms use the undermentioned equation:

A=- I»N

A=100×10^9 Bq

I»=3.13×10^-6 secs^-1

N= Number of atoms

Put these values into the equation

N= ( A ) / ( I» )

N= ( 100×10^9 ) / ( 3.13×10^-6 )

N=3.19X10^16 atoms

Now use to following equation to happen the mass

Mass= [ ( figure of atoms ) x ( Molecular Weight ) ] / ( Avogadro ‘s figure )

Number of atoms=3.19X10^16

Molecular Weight=99 a.m.u

Avogadro number=6.022×10^23

Mass= [ ( 3.19X10^16 ) ten ( 99 ) ] / ( 6.022×10^23 )

Mass= 5.25x 10 ^-6 gms

Activity of 99Mo at clip of first usage in the infirmary

Use the equation

A ( T ) =A ( 0 ) e^- I»t

Where

A ( 0 ) =100 GBq

I»=3.13×10^-6 secs^-1

t=69 hour = 2.484×10^5 secs

Put these values into the above equation

A ( T ) = ( 100 ) e^- ( 3.13×10^-6 ) ( 2.484×10^5 )

A ( T ) =100 ( 0.463 )

A ( T ) =46.3GBq

Activity of 99mTc obtained during this elution

Now use the Bateman equation to happen the activity of 99mTc the girl of 99Mo

Ad ( T ) =AP ( 0 ) x { I»d/ ( I»d-I»p ) } x { e^ ( -I»p ) ( tp ) -e^ ( – I»d ) ( td ) } +Ad ( 0 ) e^ ( -I»d ) ( T )

The last portion can this equation can be ignored as Ad ( 0 ) is equal to 0. So now merely usage

Ad ( T ) =AP ( 0 ) x { I»d/ ( I»d-I»p ) } x { e^ ( -I»p ) ( tp ) -e^ ( – I»d ) ( td ) }

AP ( 0 ) = 100 GBq

I»d = 3.209×10^-5 sec-1

I»p=3.13×10^-6 sec-1

tp=2.376×10^5 secs

td= 2.160×10^4 secs

Now set these Numberss into the equation

Ad ( T ) =100 ten { 3.209×10^-5 / ( ( 3.209×10^-5 ) – ( 3.13×10^-6 ) ) } x { e^ ( -3.13×10^-6 ) ( 2.376×10^5 ) -e^ ( – 3.209×10^-5 ) ( 2.160×10^4 ) }

Ad ( T ) =100 x ( 1.108 ) ten ( ( e^-0.7437 ) – ( e^-0.693 ) )

Ad ( T ) =110.8x ( 0.030 )

Ad ( T ) =3.324GBq

Question 3:

Explain the physiology of tissue accretion of 18F-FDG

18F-FDG contains glucose and a radioactive breathing isotope with a half life of 110 proceedingss. All cells use glucose with a high consumption of glucose in the encephalon, kidneys and malignant neoplastic disease cells. Glucose passes the blood-brain barrier and enters the encephalon easy, the sum of 18F-FDG that is found will give a step of the metabolic activity in the encephalon. “ FDG is transported from blood to tissues in a mode similar to glucose and competes with glucose for hexokinase phosphorylation to FDG-6-phosphate. However, since FDG-6-phosphate is non a substrate for subsequent glucose metabolic tracts and has really low membrane permeableness, the FDG-6-phosphate becomes trapped in tissue in proportion to the rate of glycolysis or glucose use of that tissue. ”

Question 4:

Briefly describe the building of a PET scanner and the method of sensing radiation used in it. Clearly outline the single constituents and their function.

The PET scanner plants by the sensing of antielectron obliteration. Radioisotopes that are injected into the organic structure work as tracers, as they decay by antielectron emanation. Inside the organic structure the radioisotopes can be followed due to the emanation of the obliteration brace which coincident gamma beams at 180 grades. The emanation can be viewed at different angles and gives the exact location of the radioisotopes. The ring of sensors is used to build an image of a piece of the organic structure.

Figure 4.1:

This image shows a PET scanner

During positron obliteration two gamma beams are emitted, both travel in opposite waies. The sensing of these two gamma beams on two different sensors means the beginning is on a line between the two sensors.

Figure 4.2: This image shows that for a given location, you can sum the signal from all sensor pairs that correspond to a line traveling through that location.

Question 5

What is the magnitude of natural background radiation in Ireland?

And where does this radiation come from? Your reply should include at

Least four beginnings of radiation.

Compare the dosage from natural beginnings with dosage from atomic medical specialty tests.

The magnitude of natural background radiation in Ireland is “ 3950 microsieverts ( I?Sv ) ” . Background radiation comes from both semisynthetic and natural beginnings. Natural beginnings account for 86 % of the entire one-year radiation dosage and semisynthetic beginnings accounting for 14 % of the entire dosage. The followers are some of the beginnings of this background radiation:

Radon: The mean dosage of Rn is about 2050 I?Sv. It is a radioactive gas that occurs of course in the land.

Cosmic radiation: A individual life at sea degree receives a dosage of 300 I?Sv of cosmic radiation ( which is the high energy radiation from infinite ) . This dose varies depending on height and frequence of flights taken.

Natural radiation in dirt: The dirt and stone contain radioactive elements which provide an mean dosage of 300 I?Sv

Thoron: This is another of course happening radioactive gas which gives an mean dosage of 280 I?Sv

Medical exposure of patients: The dosage from this depends on the process being given. The mean dosage for a individual is 540 I?Sv

Question 6

Describe a figure of trial processs for the quality control of gamma camera s.

Top outing: In a spectrum of gamma beams, the Photopeak is the peak photon energy. Peaking is carried out in two ways, manually or automatically. Manual peaking is achieved by seting energy window scene and automatically by utilizing the camera and which is checked to guarantee the energy window is centred. Top outing is performed before usage mundane and merely before a different radionuclide is used.

Fig 6.1: In this image ; Z pulsation ( left ) and MCA type ( right ) shows of a pulse-height spectrum used for

“ peaking ” the scintillation camera [ focus oning the mono analyser ( SCA ) window on the photopeak ] .

Uniformity: Uniformity trials highlight any malfunctions of the gamma camera. Testing can be either intrinsic utilizing a Tc-99m beginning or systematic utilizing a Co-57 planar beginning. For a big field 5 million counts is sufficient for day-to-day proving. Uniformity should be evaluated daily.

Fig 6.2: The images displayed here are, Uniformity images. Left: Image from a modern high-performance camera with digital spacial one-dimensionality and Z pulse rectification circuitry.

Right: The same camera, with an inoperative photomultiplier tubing.

Spatial Resolution: Testing spacial declaration involves looking at the intrinsic and collimation declaration, while the system declaration is a combination of the intrinsic and collimator declaration. In order to find the declaration a four-quadrant saloon apparition is used. This is them imaged 4 times at 90 degree rotary motion. The declaration of the camera should be tested one time a hebdomad.

Fig 6.3

This image shows a Image of a saloon apparition used for measuring spacial declaration. The lead bars and infinites between the bars are of equal breadths.

One-dimensionality: The one-dimensionality of the system is checked to do certain that lines on a saloon form are non wavy and there is no alterations. For proving an extraneous a whole form or parallel line with equal spacing apparitions are best suited for proving the one-dimensionality. This trial should be performed at least one time a hebdomad.

Fig 6.4

This image shows a saloon apparition that is used to look into the one-dimensionality of the gamma camera.

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