# Differences Between Gamma Radiation Biology Essay

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The main difference between the gamma radiation and x-ray is how they are produced. Gamma rays are generated in the nucleus of a radionuclide during decay. Whereas x-rays are produced when electrons strike a target.

## Does 1 mm of lead provide the same effective shielding for both types of radiation?

For effective Gamma rays require large amounts of either concrete or lead. So in this case 1mm of lead would not be sufficient, as the gamma rays would pass through without any problems. In the case of x-rays 1mm of lead provides effective shielding.

## Finally, provide values for the HVL and TVL for 140 kV X-rays and 140 kV gamma-rays.

X-rays 140 kV:

Gamma rays 140 kV:

## Question 2

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

Calculate:

## The mass of 99Mo loaded during production.

In order to find the mass first need to find the decay constant:

## Half life = ln2/Î»

Half-life = 2.214x10^5 seconds

Î»= decay constant

Rearrange the equation so:

Î»= (ln2)/ (2.214x10^5)

Î»=3.13x10^-6 secs^-1

Now want to find the number of atoms use the following equation:

## A=- Î»N

A=100x10^9 Bq

Î»=3.13x10^-6 secs^-1

N= Number of atoms

Put these values into the equation

N= (A)/ (Î»)

N= (100x10^9)/ (3.13x10^-6)

N=3.19X10^16 atoms

Now use to following equation to find the mass

## Mass= [(number of atoms) x (Molecular Weight)]/ (Avogadro's number)

Number of atoms=3.19X10^16

Molecular Weight=99 a.m.u

Mass= [(3.19X10^16) x (99)]/ (6.022x10^23)

Use the equation

## A(t)=A(0)e^- Î»t

Where

A(0)=100 GBq

Î»=3.13x10^-6 secs^-1

t=69 hrs = 2.484x10^5 secs

Put these values into the above equation

A(t)=(100)e^- (3.13x10^-6 )( 2.484x10^5)

A(t)=100(0.463)

## Activity of 99mTc obtained during this elution

Now use the Bateman equation to find the activity of 99mTc the daughter of 99Mo

## Ad (t) =AP (0) x {Î»d/ (Î»d-Î»p)} x {e^(-Î»p)(tp) -e^(- Î»d)(td)}+Ad(0)e^(-Î»d)(t)

The last part can this equation can be ignored as Ad(0) is equal to 0. So now just use

## Ad (t) =AP (0) x {Î»d/ (Î»d-Î»p)} x {e^(-Î»p)(tp) -e^(- Î»d)(td)}

AP (0) = 100 GBq

Î»d = 3.209x10^-5 sec-1

Î»p=3.13x10^-6 sec-1

tp=2.376x10^5 secs

td= 2.160x10^4 secs

Now put these numbers into the equation

Ad (t) =100 x {3.209x10^-5 / ((3.209x10^-5) -(3.13x10^-6))} x {e^(-3.13x10^-6)( 2.376x10^5) -e^(- 3.209x10^-5)( 2.160x10^4)}

## Question 3:

Explain the physiology of tissue accumulation of 18F-FDG

18F-FDG contains glucose and a radioactive emitting isotope with a half-life of 110 minutes. All cells use glucose with a high intake of glucose in the brain, kidneys and cancer cells. Glucose passes the blood-brain barrier and enters the brain easily, the amount of 18F-FDG that is found will give a measure of the metabolic activity in the brain. "FDG is transported from blood to tissues in a manner similar to glucose and competes with glucose for hexokinase phosphorylation to FDG-6-phosphate. However, since FDG-6-phosphate is not a substrate for subsequent glucose metabolic pathways and has very low membrane permeability, the FDG-6-phosphate becomes trapped in tissue in proportion to the rate of glycolysis or glucose utilization of that tissue."

## Question 4:

Briefly describe the construction of a PET scanner and the method of detection radiation used in it. Clearly outline the individual components and their role.

The PET scanner works by the detection of positron annihilation. Radioisotopes that are injected into the body work as tracers, as they decay by positron emission. Inside the body the radioisotopes can be followed due to the emission of the annihilation pairs which coincident gamma rays at 180 degrees. The emission can be viewed at different angles and gives the exact location of the radioisotopes. The ring of detectors is used to construct an image of a slice of the body.

Figure 4.1:

This image shows a PET scanner

During positron annihilation two gamma rays are emitted, both travel in opposite directions. The detection of these two gamma rays on two different detectors means the source is on a line between the two detectors.

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

## Question 5

What is the magnitude of natural background radiation in Ireland?

Compare the dose from natural sources with dose from nuclear medicine exams.

The magnitude of natural background radiation in Ireland is "3950 microsieverts (Î¼Sv)" . Background radiation comes from both man-made and natural sources. Natural sources account for 86% of the total annual radiation dose and man-made sources accounting for 14% of the total dose. The following are some of the sources of this background radiation:

Cosmic radiation: A person living at sea level receives a dose of 300 Î¼Sv of cosmic radiation (which is the high energy radiation from space). This dose varies depending on altitude and frequency of flights taken.

Natural radioactivity in soil : The soil and rock contain radioactive elements which provide an average dose of 300 Î¼Sv

Thoron : This is another naturally occurring radioactive gas which gives an average dose of 280 Î¼Sv

Medical exposure of patients: The dose from this depends on the procedure being given. The average dose for a person is 540 Î¼Sv

## Question 6

Describe a number of test procedures for the quality control of gamma camera s.

Peaking: In a spectrum of gamma rays, the Photopeak is the peak photon energy. Peaking is carried out in two ways, manually or automatically. Manual peaking is achieved by adjusting energy window setting and automatically by using the camera and which is checked to ensure the energy window is centred. Peaking is performed before use everyday and just before a different radionuclide is used.

## "peaking" the scintillation camera [centering the single-channel analyzer (SCA) window on the photopeak].

Uniformity: Uniformity tests highlight any malfunctions of the gamma camera. Testing can be either intrinsic using a Tc-99m source or systematic using a Co-57 planar source. For a large field 5 million counts is sufficient for daily testing. Uniformity should be evaluated daily.

## Right: The same camera, with an inoperative photomultiplier tube.

Spatial Resolution: Testing spatial resolution involves looking at the intrinsic and collimation resolution, while the system resolution is a combination of the intrinsic and collimator resolution. In order to determine the resolution a four-quadrant bar phantom is used. This is them imaged 4 times at 90 degree rotation. The resolution of the camera should be tested once a week.

## This image shows a Image of a bar phantom used for evaluating spatial resolution. The lead bars and spaces between the bars are of equal widths.

Linearity: The linearity of the system is checked to make sure that lines on a bar pattern are not wavy and there is no changes. For testing an orthogonal a whole pattern or parallel line with equal spacing phantoms are best suited for testing the linearity. This test should be performed at least once a week.

Fig 6.4

This image shows a bar phantom that is used to check the linearity of the gamma camera.