Image Quality

Question 1

What is a point spread function?

Answer 1

It is a function that describes the response of an imaging system to a point source or point object. The PSF is convolved with an original image, which produces the blurring.

Question 2

What is a line spread function?

Answer 2

It is a line convolved with a point spread function, which results in a line spread function. Shows the effective spread of a perfect line due to system and detector limitations.

Question 3

What is an example of a fundamental theoretical limitation of the spatial resolution of an imaging modality.

Answer 3

The wavelength of the energy used to probe the object. In practice, many resolutions are larger than this limit. For example, an x-ray wavelength is 500nm, but the resolution of images are much bigger.

Question 4

Which imaging modalities typically have the smallest resolution?

Answer 4

Radiography, mammography, fluoroscopy, and ultrasound. Screen film radiography has a resolution of around 0.08mm, while digital has around 0.17mm. Screen film mammography has a resolution of 0.03mm while digital is about 0.05 - 0.1mm. Ultrasound has a resolution of 0.3mm. Fluoroscopy has a resolution of 0.125mm.

Question 5

Which imaging modalities typically have the largest resolution?

Answer 5

PET, with a resolution of 5mm and Single Photon Emission Computerized Tomography (SPECT) with 7mm.

Question 6

What physical mechanisms cause blurring?

Answer 6

The “camera” is not in focus (image not at focal point), optical diffusion (secondary particles from x-ray interaction scatter and spread out), digital averaging.

Question 7

What happens to the 2D Fourier transform of an object when the image is blurred?

Answer 7

The high frequencies decrease, so the high x and y points fade or disappear, making a more centralized plot than previous.

Question 8

What does MTF stand for and what is it?

Answer 8

Modulation Transfer Function is a very complete description of the resolution properties of an imaging system. The MTF illustrates the fraction (or percentage) of an object’s contrast that is recorded by the imaging system, as a function of the size (i.e., spatial frequency) of the object. It plots the fraction of the original amplitude that will be recorded as a function of frequency.

Question 9

What units are frequency measured in?

Answer 9

Cycles/mm.

Question 10

Are high spatial frequencies due to small or big objects?

Answer 10

Small objects.

Question 11

What physical property does the MTF describe?

Answer 11

The resolution of the imaging system.

Question 12

How are the line spread function and the modulation frequency functions related?

Answer 12

The MTF is typically calculated from a measurement of the line spread function (LSF). As the line spread function gets broader, the corresponding MTFs plummet to lower MTF values at the same spatial frequency, and the cutoff frequency (where the MTF curve meets the x-axis) is also reduced.

Question 13

What is the term “limiting resolution” used for?

Answer 13

It is the smallest frequency at which the MTF is at least 10%, it is used to quote the resolution of the imaging system.

Question 14

How can you use the display contrast to resolve difficult to see objects?

Answer 14

You can narrow the display window so that the range from maximum to minimum contrast is over the small signal range that is relevant to the object. Objects outside of this signal level will appear either black or white (high saturation).

Question 15

What is the contrast to noise ratio?

Answer 15

It is the difference between the average signal value (per pixel) and the average background value divided by the standard deviation of the background. Essentially, it says how many factors of the background error is the signal significance.

Question 16

What is the signal to noise ratio.

Answer 16

It is the sum of the difference between a signal and the average background output for every cell, divided by the standard deviation of the background. The size of the signal object (ie number of cells/pixels) thus increases the signal to noise ratio, as it is summed over every cell.

Question 17

Which one changes as a function of the size of the object: the contrast to noise ratio or the signal to noise ratio.

Answer 17

The signal to noise ratio because it is summed over every signal cell.

Question 18

Which one is invariant as a function of the size of the object: the contrast to noise ratio or the signal to noise ratio.

Answer 18

The contrast to noise ratio as the numerator is only the difference between the average signal size and the average background size, there is no sum.

Question 19

What is the true positive fraction and in what context is it used?

Answer 19

It is the sensitivity of a diagnostic decision parameter, defined as the number of true positive diagnoses to the total number of people with the disease (true positive plus true negative): TP/(TP + FP). It is used in conjunction with ROC (receiver operator curve) to define the accuracy of a diagnostic.

Question 20

What is the true negative fraction and in what context is it used?

Answer 20

It is the specificity of a diagnostic decision parameter, defined as the number of true negative diagnoses (people who truly do not have the disease and are diagnosed as not having it) divided by the total number of people who do not have the disease (true negative plus false positive): TNF=TN/(TN + FP). It is used in conjunction with ROC (receiver operator curve) to define the accuracy of a diagnostic.

Question 21

What is an ROC curve?

Answer 21

A receiver operator curve. It plots the Ture Positive Fraction vs the False Positive Fraction to characterize how much better than a coin toss is the diagnostic performing (want curve pushed to the top left of the graph to indicate high true positive and low false positive).

Question 22

What is the false positive fraction?

Answer 22

It is the fraction of people in the normal population who are told that they have the disease. It is defined as FPF=1-TNF (true negative fraction).

Question 23

What is the sensitivity?

Answer 23

The True Positive Fraction, the fraction of abnormal population that is correctly diagnosed: TP/(TP + FN).

Question 24

What is the specificity?

Answer 24

It is the True Negative Fraction, the fraction of the normal population that is correctly identified as not having the disease: TN/(TN + FP).

Question 25

What is meant by kVp?

Answer 25

kVP is the kilovoltage of the x-ray tube that accelerates the electrons. The electrons hitting the target have an energy equal to the kVP in KeV and the emerging photons have up to this maximum energy, but the mean photon energy is typically around 1/3 the kVP after filtering out the lowest energy photons.

Question 26

What are the two mechanisms that produce x-rays when a beam of electrons hit a tungsten target?

Answer 26

Bremsstrahlung and the production of characteristic x-rays (collide with an electron in the shell, eject it, and another atom falls into the open spot, emitting an x-ray with the energy difference).

Question 27

Into what form is the majority of energy converted when electrons hit a target?

Answer 27

Heat energy via small collisional energy exchanges with electrons in the target.

Question 28

How does bremsstrahlung occur in the context of x-ray production?

Answer 28

An electron is slowed down due to Coulomb interactions with the nucleus of an atom. This change in energy causes the electron to emit a photon equal to the energy difference.

Question 29

What are characteristic x-rays, in the context of x-ray production?

Answer 29

An incoming electron collides with a bound electron, ejecting it from an inner shell. This vacancy is filled by an electron in an outer shell, which emits a photon of energy equal to the energy difference between its previous position and the new shell position. They are called characteristic x-rays because their energies are equal to the energy differences between different shell energies for a particular medium. This causes large spikes in x-ray energy distributions at these energies.

Question 30

What causes large spikes in x-ray energy distributions at certain energies?

Answer 30

Characteristic x-rays caused by electrons falling from one shell energy to a lower one after a vacancy is produced.

Question 31

What is the purpose of the anode tube current in an x-ray tube?

Answer 31

It pulls the electrons off of the filaments towards the target.

Question 32

What is the purpose of the filament current in an x-ray tube?

Answer 32

It heats the filament, creating free electrons.

Question 33

What are typical voltage and current values for an x-ray tube filament circuit?

Answer 33

10 V and 3 to 7A.

Question 34

What is the purpose of rotating the anode disk?

Answer 34

It dissipates heat, as the electrons will collide on different parts of it as it rotates.

Question 35

What are the advantages/disadvantages of a large anode angle with a small filament length?

Answer 35

Large field coverage Small effective focal spot Poor power loading: a small spot on the target receives a lot of heat, it will overheat and turn off.

Question 36

How does the filament length control aspects of the x-ray production?

Answer 36

The longer the filament length, the wider a strip of electrons that will hit the tungsten target and emerge straight down at a right angle to the orginal electron beam. The larger the filament length, the larger the effective focal spot of the x-rays on the object.

Question 37

How does the anode angle control aspects of the x-ray production?

Answer 37

The anode angle is angle between the tungsten target surface and the electron beam. Increasing this angle results in a larger target surface area presented to the electron beam and, thus, a larger field of coverage (beam of x-rays emerging from the tube and spreading outwards).

Question 38

What does it mean for a cathode to be space charge limited?

Answer 38

It is producing more free electrons that can be pulled off the filament by the anode. Increasing the anode tube current can fix this.

Question 39

What is the heel effect in x-ray production?

Answer 39

In X-ray tubes, the heel effect or anode heel effect is a variation of the intensity of X-rays emitted by the anode depending on the direction of emission. Due to the geometry of the anode, X-rays emitted towards the cathode are in general more intense than those emitted perpendicular to the cathode–anode axis. This is because the anode is on an angle and attenuates, or absorbs, some of the x-rays traveling straight down perpendicular to the cathode-anode axis.

Question 40

What is off focal radiation in x-ray production?

Answer 40

It is radiation that originates from any part of the anode that is not the focal point (small target area that the filament is aimed at). This occurs when electrons striking the focal spot bounce off and are attracted back to the anode outside the focal spot region, hitting the target and producing an x-ray, which passes through the collimator.

Question 41

How does filtration affect the energy profile of x-rays emitted from an x-ray tube?

Answer 41

The unfiltered x-ray output is linear, with the most x-rays being emitted at very low energy. The filter preferentially attenuates low energy photons, removing almost all below 15 KeV, and creating a humped distribution peaking near the middle energy of the range (~40 KeV in class example of a 90 kVp voltage) and decreasing in a linear way at the high range.

Question 42

What is voltage ripple?

Answer 42

Small residual periodic variation of the direct current (dc) output of a power supply which has been derived from an alternating current (ac) source. The example showed how an ac current is turned into dc by removing the negative part of the dc curve and adding many pulses with slight shifts.

Question 43

What is meant by the quality of radiation?

Answer 43

Its penetrating ability.

Question 44

How does the Z of the x-ray tube anode tube affect the likelihood of interactions?

Answer 44

Interactions are more likely with high Z.

Question 45

How is the efficiency affected by the Z of the anode material in x-ray tubes?

Answer 45

It is roughly proportional to Z.

Question 46

What factors affect x-ray emission from an x-ray tube?

Answer 46

The anode target material, the tube voltage, and the tube current.

Question 47

What is the purpose of the x-ray tube voltage?

Answer 47

It creates a voltage difference between the filament and the target (anode), which accelerates the electrons across the tube to the target.

Question 48

How does the x-ray tube voltage affect the quality and quantity of the x-rays?

Answer 48

Increased voltage increases x-rays: # X-rays is proportional to kV^2.

Increased voltage increases the average energy, making the photons more penetrating.

Question 49

How does the tube current affect the emitted x-rays?

Answer 49

The number of x-rays is proportional to the tube current, since the number electrons accelerated across the tube is proportional to the current. More crashing electrons increases x-rays linearly.

Question 50

How does the beam filtration affect the outgoing x-rays from a tube?

Answer 50

It reduces the quantity of x-rays, but increases the quality by attenuating low energy photons, which increases the average energy of the beam. It can reduce dose to the patient, as low energy beams tend to be absorbed in the body and only contribute dose.

Question 51

What is a half value layer (HLV)?

Answer 51

It is the amount of material that must be put in front of the beam to reduce the beam by 1/2. In safety regulations, it is usually quoted as a minimum requirement because the low energy stuff just contributes to harmful dose. That is, if the HLV is less than the width of a person’s body, then half of that beam is going to give a dose to them. Longer HLV means more will just pass through.

Question 52

Why is the Half Value Layer (HLV) usually a minimum requirement in safety regulations?

Answer 52

In safety regulations, the HLV is usually quoted as a minimum requirement because the low energy stuff just contributes to harmful dose. That is, if the HLV is less than the width of a person’s body, then half of that beam is going to give a dose to them. Longer HLV means more will just pass throu

Question 53

How is magnification produced and defined for radiography.

Answer 53

Magnification occurs because the x-ray beam spreads wider over time. Thus, the width of the beam hitting the object grows wider before reacher the imager. It is defined as the ratio between the image length and the object length: M= L(image)/L(object).

Similarly, it can be defined as the distance between the x-ray source and the patient plus the distance from the patient to the imager, divided by the first distance: M = a+b/a

Question 54

What is the effect of having a large anode angle and a large filament length in an x-ray tube?

Answer 54

Large field coverage, large effective focal spot, good power loading.

Question 55

What is the effect of having a small anode angle and a long filament length in an x-ray tube?

Answer 55

Small field coverage, small effective focal spot, good power loading. Results in a sharper image, but smaller coverage.

Question 56

Why would we want to minimize the distance between the patient and the film for radiography?

Answer 56

To reduce blurring. X-rays hitting edges originate from slightly different sources in the beam, so between edge and film, they could separate in different directions. Want to minimize this effect.

Also reduce the magnification. It’s not a bad thing, but might be desired in some circumstances.

Question 57

How does the magnification change when considering a patient with measureable thickness (ie a real person)?

Answer 57

The magnification factors are adjusted to account for the surface to source measurement. M= (a’ + b’)/a’, where a’ is the distance from the patient to the source, and b’ is the distance from the opposite surface of the patient to the film.

Question 58

How does the thickness of the screen used in radiography affect the signal?

Answer 58

Thicker screens produce more signal (more interactions inside the screen), but they are more blurred because there is more space for diffusion of signal. The signal is stronger, but wider with a thick screen.

Question 59

Why are the parts of the radiography film that didn’t get any photons lighter than the areas that did?

Answer 59

The film is coated in positive silver ions. When photons hit the film, they knock electrons free, which fill the ion holes, creating regular silver. In the developing process the excess silver ions are washed away, leaving silver only in the spots that recieved a dose.

Question 60

When reading radiography films, how is the optical density defined?

Answer 60

The optical density is the log fraction of the intensity of light coming through the film, defined as

OD= - log_10(T), where T is the ratio of the intensity of light traveling through the area of the film from a light table to the intensity of the light table without the film in the way: T=I/I_o

Thus,

OD=1 is corresponds to T=10%

OD=2 is corresponds to T=1%

OD=3 is corresponds to T=0.1% of light getting through the film.

Question 61

How is the optical density of a radiography film related to the kerma?

Answer 61

The kerma is proportional to the number of photons that reached the film through the patient. A typical OD vs Kerma plot shows a linear relationship when the kerma is plotted on a log scale, where both the OD and Kerma increase together after an initial flat “toe” region and before a flattening at the end called the “shoulder”. The linear region is the part useful for measurements.

Question 62

What type of response does the light passing through the film have in relation to the kerma?

Answer 62

The light intensity increases logarithmically with kerma. This is why the Optical Density is defined as a log.

Question 63

How could one alter a radiography film to make it more sensitive?

Answer 63

Add more silver ions.

Question 64

What is meant by “high speed” radiography film?

Answer 64

High sensitivity.

Question 65

What is meant by a latitude measurement on an Optical Density plot of radiography screen film?

Answer 65

It is a measure of the horizontal distance between two points on a curve. As the horizontal axis represents kerma, and the y-axis the optical density, if two identical y points are compared on two differently sloped characteristic curves, a more narrow lattitude means that a smaller difference in kerma produced a larger difference in OD, corresponding to a more sensitive film. Likewise, the curve with a broader latitude is less sensitive.

Question 66

What is a characteristic curve for radiography screen film?

Answer 66

It is the optical density (OD) vs log Kerma plot.

Question 67

What is direct detection for a thin film transistor array?

Answer 67

The scintillators are grown in little towers like a semi conductor. A voltage difference is applied between the top and bottom so that ion pairs are drawn to the surface and depth onto detector elements for detection. These are used for digital radiography. The signal is a single spike, not a point spread function.

Question 68

What is indirect detection for a thin film transistor?

Answer 68

A scintillator is used to increase the signal, the photons make their way to the array of light sensitive detectors. Because the photon showers can difuse, the signal has a point spread function, not a single spike.

Question 69

What is the effect of increasing the conversion efficiency of a digital radiography screen?

Answer 69

A higher percent of photon energy is converted to light, you get a higher electronic gain for the same screen thickness. Thus, fewer incident x-rays are required and a lower dose is given to the patient.

Question 70

What is the effect of increasing the absorption efficiency of a digital radiography screen?

Answer 70

The thicker screen gives a higher electronic gain.

Question 71

What is a typical absoption efficiency for x-rays incident on a screen?

Answer 71

50%.

Question 72

What is a typical conversion efficiency (photons converted to light in the scintillator) for photons on a radiography screen?

Answer 72

10%. Sincethe light photons are about 2eV each and typical x-rays are around 140kPV, that’s about 2,000 light photons per x-ray.

Question 73

What is a scatter to primary ratio?

Answer 73

It is the ratio of the number of scattered photons reaching the film to the number of primary photons reaching the film:

SPR= S/P (scatter over primary)

The fraction of scattered to total is:

F=S/(S + P)

Question 74

What is an anti-scatter grid?

Answer 74

It is a grid above the TFT of little stoppers, or fences, that attenuates any photons that aren’t traveling on a path directly from the source. Gets rid of the scatter.

Question 75

What does a high anti-scatter grid ratio affect?

Answer 75

The grid ratio, which is the ratio of the height of the grid to the width between the fences H/W, reduces more scatter at higher ratio, but also reduces the number of primaries reaching the detector.

Question 76

What is the approximate Z value for soft tissue?

Answer 76

7.5

Question 77

What factors do we take into consideration when choosing a filter material for mammography?

Answer 77

We want to cut off higher energy regions and leave the nice spikes from the characteristic radiation of the target, as these spikes around 20 KeV fall into a region where the attenuation of the glandular and fat tissues are more separated from “infiltrating ductal carcinoma”, early stage cancer, allowing for them to be distinguished on the scans. So, look for a material that has a nice spike in attenuation after 20 KeV (ie some characteristic radiation). Molybdenum (Z=42) has a k-edge of 20 KeV and Rhosdium has k-edge of 23.2 KeV, so works well.

Question 78

What is the Z number of Molybdenum?

Answer 78

42

Question 79

What is the k-edge of Molybdenum and what are the energies of the characteristic x-rays?

Answer 79

k-edge of 20 KeV and characteristic x-rays of 17.9 and 19.5 KeV.

Question 80

What is the k-edge of Rhodium and what are the energies of the characteristic x-rays?

Answer 80

k-edge of 23.2 KeV and characteristic x-rays of 20.2 and 22.7 KeV.

Question 81

What is a normal focal spot size of mammography?

Answer 81

0.3 mm

Question 82

What are the advantages to flattening the breast for a mammography?

Answer 82

Flattening the breast reduces scatter and reduces dose, as the photons have a shorter path to travel.

Question 83

What is the ideal photon energy for mammography?

Answer 83

Around 19-20 KeV

Question 84

How many images does a typical fluoroscopy scan have?

Answer 84

18,000

Question 85

What are the dose requirements for fluoroscopy?

Answer 85

Extremely low. Because there are so many frames, each frame must have 1/1000 the dose of a radiographic image.

Question 86

What is the frame rate that our eye percieves as “real time”?

Answer 86

30 frames/sec

Question 87

How does the image intensifier work for fluoroscopy?

Answer 87

Just like a photomultiplier tube, but also with focusing, as the image output is much smaller in size than the incoming image.

Question 88

How is the x-ray signal converted to light in fluoroscopy?

Answer 88

The x-ray hits CsI tubes, grown in thin tall towers that act like a scintillator. They carry the light to the photocathode, which emits electrons when struck.

Question 89

What is pincushion distortion?

Answer 89

It is a distortion of the original fluoroscopy input image to be curved inwards at the edges. It is caused by the transition from a curved input surface to a flat output surface. The field of view must be shrunk to avoid the large distortions at the end.

Question 90

What is the brightness gain for fluoroscopy?

Answer 90

It is around 2500 to 7000, defined by the ratio of intensity in to intensity out.

Question 91

What is continuous fluoroscopy?

Answer 91

It is a mode of fluoroscopy where the beam is on continuously. The images are binned in 1/30th of a second for the display.

Question 92

What is variable frame rate pulsed fluoroscopy?

Answer 92

The beam has little pauses in it between frames, which reduces dose.

Question 93

How is the resolution affected when you increase the field of view on a fluoroscopy display?

Answer 93

The resolution gets worse.

Question 94

What are typical resolutions for a 27 cm flat panel system and image intensifier?

Answer 94

image intensifier: 1.6 and flat panel: 2.5 line pairs/mm

Question 95

What are typical resolutions for a 40 cm flat panel system and image intensifier?

Answer 95

image intensifier: 1.3 and flat panel: 1.8 line pairs/mm

Question 96

What are typical resolutions for a 14 cm flat panel system and image intensifier?

Answer 96

image intensifier: 2.7 and flat panel: 3.2 line pairs/mm

Question 97

Fallout in brightness at the periphery of a fluoroscopic image is called what?

Answer 97

Pincushion distortion.

Question 98

What is fluoroscopy used for at TBCC?

Answer 98

It is used for liver cancer. You can watch the organs moving and test breath hold and compression techniques in preparation for radiation therapy.

Question 99

Frames averaging in fluoroscopy likely decreases what?

Answer 99

The frame rate on the display.

Question 100

What is axial or sequential CT acquisition?

Answer 100

It is a scanning process where the beam is turned off after a cycle, the bed is moved, the beam is turned on again.

Question 101

Describe the view of a coronal cross section

Answer 101

It’s a cross section as if you are seeing a layer of the person facing forward.

Question 102

Describe the view of an axial scan.

Answer 102

It’s an image as if you are looking down from above.

750 × 783 - waent.org

Question 103

Describe the view of a sagittal cross section.

Answer 103

Viewing the person from the side.

1268 × 2028 - commons.wikimedia.org

Question 104

How is the heel effect reduced in CT scanners?

Answer 104

The anode-cathode axis is placed perpendicular to the imagine plane. The unsymmetric parts of the beam are then along the z-axis (length of the patient). This creates a uniform beam around the axis of rotation.

Question 105

What are slices in a CT scan? How many are typical?

Answer 105

Slices are individual rows of detectors along the z-axis (patient axis), which dictates the slice thickness of an image. Typically they have 64 to 320 slices.

Question 106

What is the typical thickness of a CT slice?

Answer 106

Typical CT slices are 0.5 mm for diagnostic and 3 mm thick for simulation.

Question 107

What physical property of the x-ray beam dictates how many slices a CT scanner can have?

Answer 107

The cone angle of the beam dictates the width of the beam along the z-axis (patient axis) and therefore controls how many slices will be covered by the beam.

Question 108

How does a CT scanner reduce scatter?

Answer 108

CT scanners use an anti-scatter grid.

Question 109

What are some advantages to having the anode rotate in the same plane as the gantry?

Answer 109

Rotating the anode in the same plane as the gantry results in:

• A uniform g-force, causing the structure to be more stable
• Causes the heel effect to be along z-axis - uniform along coronal axis.

Question 110

What is the fan angle of a CT scan?

Answer 110

BetaA and BetaB.

Question 111

What is the difference between a cone beam and a fan beam (CT scanners)?

Answer 111

Question 112

What is the energy of a typical CT scanner?

Answer 112

CT scanner x-ray tubes operate at a range from 80-140KVp.

Question 113

What photon interactions are important at CT x-ray energies?

Answer 113

CT x-ray energies range from about 20-140KeV, depending on the filters (x-ray tubes operate at 80-140KeV). At these energies, photoelectric and compton dominate in water, although by 140KeV, the photoelectric effect has dropped significantly.

Question 114

How is the linear attenuation coefficient for the photoelectric effect related to the Z of the material?

Answer 114

The linear attenuation coefficient for the photoelectric effect is proportional to Z^2.

Question 115

How is the linear attenuation coefficient for the compton effect related to the Z of the material?

Answer 115

It is not proportional to Z, it is proportional only to the physical density of the material. As the compton effect has a higher cross section at the energy ranges for CT scanners, the scanners are measuring differences in physical density over differences in Z value. Of course, they are really measuring the linear attenuation coefficients, but these depend on the density.

Question 116

What is a Hounsfield unit?

Answer 116

A Hounsfield unit is the unit displayed on the CT scanner. It is scales the linear attenuation coefficient of the measured medium by that of water and is defined as

HU = 1000*(mu - mu_w)/mu_w

Question 117

What is the HU value of

• water
• air

Answer 117

HU of

• water = 0
• air = -1000 as mu_(air) = 0

Question 118

What is the range of HU or CT numbers for organs and muscle?

Answer 118

The range of HU or CT numbers for organs and muscle is 30 to 220.

Question 119

What is the range of HU or CT numbers for fat?

Answer 119

The range of HU or CT numbers for fat is -80 to -30.

Question 120

What is the range of HU or CT numbers for bone?

Answer 120

The range of CT numbers for bone is 1500 1800.

Question 121

How are the linear attenuation coefficients for voxels handled.

Answer 121

The mus in a voxel are averaged together, the resolution controls the size of the voxel.

Question 122

What is the purpose of using a bow tie filter?

Answer 122

Bow tie filters reduce the dose to patients. Patients are generall thinner on their edges and thicker in the middle and would get a larger dose to their edges if the beam had the same intensity as in the middle.

Question 123

What would be the affect of having a bow tie filter that was too small?

Answer 123

If the bow tie is too small, the beams on the far left and right of the fan beam will pass through the full thickness of the filter (the elliptical hole in the middle will be smaller than the edge of the beam). The outer edges will be attenuated too much, resulting in the center of the body recieving a higher dose than the edges.

Question 124

Do CT detectors use a direct or indirect detection system?

Answer 124

Usually an indirect system. In an indirect system, x-ray photons are first converted into light photons by passing through a scintillator (in the case of CT, it is ceramic and made of rare earth metals).

Question 125

What parameters can you adjust in a CT to reduce the dose to the patient?

Answer 125

You can reduce the effective slices by reducing the beam and binning several slices together during reconstruction. You should use as few slices as possible.

Question 126

What is overscan?

Answer 126

Overscan occurs when the beam width is wider than the detector array that recieves the signal. It results in extra, unneccesary dose to the patient. It is quantized by

Epsilon = detected width/beam width

Question 127

What is the definition of a CT projection?

Answer 127

Projections are a superposition of linear attenuation coefficients.

Question 128

How does a single spot in a phantom look in a sinogram?

Answer 128

Like a sin wave. The y-axis of the sinogram is the attenuation coefficients along the width of the beam at that beam position. The x-axis is the angle of the anode when that particular image was taken (ie beam position).

Question 129

What information does a reference detector provide on a CT?

Answer 129

A reference detector sits outside of the field of view and meaures the brightness gain:

I_r = g_r I_0

Question 130

What is back projection?

Answer 130

Back projection is a way to reconstruct a CT image. It allocates the measured total attenuation equally to each pixel along the x-ray path through the patient.

Question 131

What is a weakness of back projection?

Answer 131

In filtered back projection the projection images are convolved with a mathematical filter to removed the blurring that resulted in the non-filtered back projection image. It adds negative values around the periphery of the image which was originally a 1/r function convolved with our image.

Question 132

Are sharper images more noisy or less noisy?

Answer 132

Sharper images are noisier.

Question 133

What is an advantage of using ultrasound imaging?

Answer 133

It does not use ionizing radiation and is safe for the patient.

Question 134

What is an ultrasound wave?

Answer 134

It is a mechanical perturbation that propagates as a longitudinal wave by the ompression and refaction of particles in the medium.

Question 135

In what direction is the displacement of the medium due to ultrasound waves?

Answer 135

Displacement of the medium is parallel to the propagatiom of the wave.

Question 136

What properties define a medium for ultrasound?

Answer 136

The medium is defined by its density and compressibility.

Question 137

How is the compressibility of an object defined?

Answer 137

It is a measure of the relative volume change of a medium as a response to isotropic pressure.

Question 138

How is the speed of sound in a medium related to the properties of the medium?

Answer 138

The speed of sound is equal to

1/sqrt(Beta*rho), the compressibility times the density.

Question 139

Is the speed of sound greater in mediums with great density?

Answer 139

The speed of sound is often greater for denser mediums, but this is not always true. The only reason that it is usually true even though c is proportional to 1/sqrt(rho) is because the bulk modulus also changes and

c= sqrt(K/rho), K=1/beta where beta is the compressibility.

Question 140

Compared to atmospheric pressure, how large is the typical peak pressure delievered by diagnostic ultrasound?

Answer 140

Diagnostic ultrasound typically delivers peak pressure levels that exceed ten times the earth’s atmospheric pressure.

Question 141

In the context of ultrasounds, what is a half-value thickness?

Answer 141

A half-value thickness is the medium thickness that is required to reduce the intensity of the beam by half.

Question 142

How are decibals defined?

Answer 142

Decibles:

dB = 10 log(I_2/I_1)

Question 143

For ultrasounds, how is the intensity related to the pressure amplitude?

Answer 143

Intensity is the power per unit area and is proportional to the square of the pressure amplitude.

Question 144

What do decibles measure?

Answer 144

The relative intsnsity of a wave as it is attenuated in a medium is measured in decibels. It is a logarithmic scale so that the number range is manageable. It can be calculated either using the fraction of intensity of the remaining wave or the fraction of pressure amplitude that remains:

db = 10 log (I2/I1)

db = 20 log (P2/P1)

As pressure squared is proportional to intensity.

Question 145

What is acoustic impedence and what does it control?

Answer 145

Acoustic impedance is like the stiffness of a compressible material. When two materials are connected, the energy transfer (and thus reflection and transmission) from one to another depends on the impedence.

Question 146

Does a large difference in acoustic impedance between materials result in a large or small perturbation to the acoustic wave.

Answer 146

A large perturbation.

Question 147

When does specular reflection occur in ultrasound imaging?

Answer 147

Specular reflection occurs from large smooth surfaces (significantly larger than the wavelength of the ultrasound). It is reflection that does not split the wave up into several directions (non-diffuse).

Question 148

When an ultrasound wave hits a surface, what three things can happen?

Answer 148

Specular reflection, transmission, and Rayleigh scattering.

Question 149

Why do we use gel between the face of the ultrasound transducer and the skin?

Answer 149

The gel has a high acoustic impedance, so that the waves actually get transmitted into the skin. When the impedance of the gel and skin are similar to each other, the reflection is smaller.

Reflection coefficient:

Rp = Z2 -Z1/(Z2 + Z1). Z is impedance

Question 150

For waves perpendicular to the medium boundary, what is the equation for the reflection coefficient?

Answer 150

Reflection coefficient:

Rp = Z2 -Z1/(Z2 + Z1). Z is impedance

Question 151

When does scatter occur for ultrasound waves?

Answer 151

Scattering occurs when the inhomogeneity dimension is on the order of the wavelength of the wave. This occurs at the small (relative to the wave length of the ultrasound) subtle boundaries that exist within tissue. At these, small amounts of energy are absorbed and retransmitted in all directions as if from a point source, in a manor that loosely resembles a pebble dropped into a pond.

Organs have a sginature “scatter pattern”.

Question 152

If the scattering object is very small compared to the ultrasound wavelength, how will the wave scatter?

Answer 152

It will scatter uniformly in all directions when the scattering object is very small compared to the ultrasound wavelength.

Question 153

In ultrasound, what is speckle?

Answer 153

Speckle is the interference pattern formed by a medium containing sub-resolution scattering objects.

Question 154

What law determines the reflection and refraction of waves not perpendicular to a new medium?

Answer 154

Snells law,

c1sin(theta1) = c2 sin(theta2)

c=speed of sound in medium

Question 155

What is a good rule of thumb for ultrasound attenuation rates in soft tissue?

Answer 155

Ultrasound attenuates at 1 dB per cm travelled per MHz.

Question 156

What mathematical relation governs the attenuation of waves in a medium.

Answer 156

It’s exponential.

Question 157

Does higher frequency ultrasound attenuate more rapidly or less rapidly?

Answer 157

Higher frequency ultrasound attenuates more rapidly.

Question 158

How is a wave mathematically characterized?

Answer 158

A wave is defined by the magnitude of the pressure change as a function of time.

Question 159

What physical property does the ultrasound detector measure?

Answer 159

We measure the change in acoustic impedence.

Question 160

What is the piezoelectric effect?

Answer 160

The Piezoelectric effect: a mechanical stress applied to a piezoelectric material induces a change in thickness that forces dipole realignment and a result electric field. The inverse effect causes piezoelectric materials to expand and contract as polarity is changed.

Question 161

What happens to a piezoelectric material when the electric field is removed?

Answer 161

When the electric field is removed, the piezoelectric material returns to equilibrium.

Question 162

What is A mode in ultrasound?

Answer 162

A-mode
(amplitude): display of digital signal proportional to echo amplitude as a function of time.

Question 163

What is B mode in ultrasound?

Answer 163

B mode: brightness mode. Converts A-mode inofmration into brightness modulated pixels. Results in the normal image we picture when we think of ultrasound.

640 × 480 - medison.ru

Question 164

What are the main components of a transducer?

Piezoelectric crystal, matching layer, acoustic backing (damping), and the electrodes.

570 × 215 - ndk.com

Question 165

What is the purpose of the matching layer in an ultrasound transducer?

Answer 165

Ultrasonic waves transmitted from the piezoelectric element are reflected off a target because there is a big difference in acoustic impedance between the piezoelectric element and the tissue. To avoid this phenomenon, an intermediate material is inserted between the two so that ultrasonic waves can efficiently enter the object. This is the role of the acoustic matching layer.
This acoustic matching for less reflection of ultrasonic waves makes it possible to implement highly sensitive probes.

The acoustic matching layer is designed to have adequate acoustic impedance value using a combination of different resin materials.

Question 166

What is the purpose of the damping material in an ultrasound transducer?

Answer 166

The backing material is located behind the piezoelectric element to prevent excessive vibration. Reducing excessive vibration will cause the element to generate ultrasonic waves with a shorter pulse length, which are the type needed for practical ultrasound signals. The damping then improves axial resolution in images.

Question 167

How is the thickness of the piezoelectric material related to the frequency of the generated ultrasound wave?

Answer 167

Q: How is the thickness of the piezoelectric material related to the frequency of the generated ultrasound wave?

A: They are proportional, t = lambda/2

Double the thickness, double the wavelength.

Question 168

What is a transducer, in general terms?

Answer 168

A transducer is a device that converts one form of energy to another.

• In the case of ultrasound it converts electrical energy into acoustic energy (sound waves) during transmission.
• Coverts acoustic energy to electrical energy during reception
• Conversion is accomplished through the piezoelectric effect

Question 169

What is the bandwidth of ultrasound transducers control?

Answer 169

Bandwidth is the useful range of contiguous frequencies over which
transducer can operate. Ultrasound transducer typically have multiple frequencies. It is usually defined as the full-width-at-half-maximum for pressure, ie 1/sqrt(2) in the intensity spectra.

Question 170

What is the Q factor for transducers?

Answer 170

Transducers are characterized by their centre frequency as well as the bandwidth range. The Q factor characterizes the significance of the damping used for the transducer. High Q factor = narrow bandwidth.

Q = f_0/bandwidth, f_0=centre frequency value

Figure 14-11 Effect of damping block on the frequency…

Question 171

Does a wide bandwidth have a high or low Q factor?

Answer 171

A wide bandwidth has a low Q factor as Q is inversely proportional to bandwidth:

Q = f_0/bandwidth

Question 172

What is time gain compensation, in the context of ultrasounds.

Answer 172

Time gain compensation is a reconstruction fix that applies an amplication to returning frequencies whose magnitude is dependent on the time it took for the signal to return (ie the amount of material it would have traversed and thus the amount of attenuation it experienced).

Question 173

How do transducer arrays work to create a collimated beam of ultrasound waves?

Answer 173

There are many piezoelectric elements in an array long the transducer. A time delay is placed on the elements in the middle so that the beam becomes collimated at a certain point.

Question 174

What is the near field zone of the ultrasound beam.

Answer 174

It is the distance along the beamline where the beam is converging prior to reaching the focal point. Near field length = d^2/(4 lambda), d=minimum beam diameter.

Question 175

What is the far field zone of an ultrasound beam?

Answer 175

It is the distance along the beamline where the beam is diverging after reaching the focal point. Near field length = d^2/(4 lambda), d=minimum beam diameter. After this length it is far field.

Question 176

How is beam steering achieved for ultrasound?

Answer 176

The arrays of the piezoelectric elements are activated several at a time in a delay to make a collimated beam. This pattern is iterated over time along the transducer from left to right using different delays to achieve the direction of the focal point desired.

Question 177

What is axial resolution in ultrasound and how is it determined?

Answer 177

Axial resolution is the ability to distinguish between two objects in a direction parallel to the beamline, but separated in depth. It has to have a pulse lying between the two objects. Can resolve these:

|||| x ||||| x

axial resolution = spatial pulse length/2 = N lambda/2, where N = number of cycles in the pulse.

Question 178

What is lateral resolution in ultrasound and how is it defined?

Answer 178

Lateral resolution refers to the ability to resolve two objects in the direction perpendicular to the beamline. It is determind by the width of the ultrasound beam and the depth of imaaging. Usually referenced by -3 dB beam width.

Lateral resolution = R lambda/d

Question 179

Lateral resolution is depth dependent since the beam’s width varis as a function of axial depth. Is it possible to maintain lateral resoultion with depth somehow?

Answer 179

Yes, lateral resolution can be maintained if multiple focal points are used. It affects the frame rate though.

Question 180

What is pulse repetition frequency in the context of ultrasound?

Answer 180

Pulse repetition frequency is the number of pulses per second.

Question 181

What is pulse repetition period in the context of ultrasound?

Answer 181

Pulse repetition period is the time between sequential pulses.

Question 182

What is the ultrasound duty factor?

Answer 182

Duty factor is

Pulse Duration/pulse repetition period.

Typical imaging factors are ~1%. Ie, most of the time is spent receiving echoes.

Question 183

What is the frame rate for an ultrasound?

Answer 183

Frame rate is

1/Tframe

Question 184

If you want to expand your field of view and maintain your ultrasound resolution, how can you achieve that?

Answer 184

You could increase the line density.

Question 185

What is M mode ultrasound?

Answer 185

M mode ultrasound is motion imaging. You don’t move the transducer, just watch one spot. Useful for vascular imaging.

Question 186

What are the basic assumptions that are made when forming an ultrasound object?

Answer 186

Assumptions:

• Sound waves travel directly to a reflector and back to the transducer
• Reflections occur only from structors along the beam’s central axis
• Intensity of reflection corresponds to the reflector scatter strength
• Sound travels at exactly 1540 m/s

Question 187

What are shadowing artifacts in ultrasound?

Answer 187

Shadowing occurs when there is a marked decrease in ultrasound intensity beyond a reflector or attenuator. There is no intensity in the wave left to image past the object.

Question 188

What is an enhancement artifact in ultrasound?

Answer 188

Enhancement occurs when the ultrasound pulse was attenuated less than expected after traversing an object with lower than expected density. Thus the time gain compensation correction was too high for the next object that reflects the wave after the cavity. Cysts and ducts cause enhancement artifacts.

Question 189

What is a reverberation artifact in ultrasound?

Answer 189

Reverberation artifacts arise when a pulse reflects repeatedly between interfaces.

Question 190

What is a mirror artifact?

Answer 190

Mirror artifacts arise from multiple beam relections between a mass and a strong reflector, such as the diaphragm.

The primary beam reflects from such a surface (e.g. diaphragm) but instead of directly being received by the transducer, it encounters another structure (e.g. a nodular lesion) in its path and is reflected back to the highly reflective surface (e.g. diaphragm). It then again reflects back towards the transducer.

There is a technical false presumption that the returning echo has been reflected once and hence the delayed echos are judged as if being returned from a deeper structure, thus giving a mirror artifact on the other side of the reflective surface.

Question 191

What is the relative noise for a bone CT reconstruction filter?

Answer 191

The relative noise for a bone CT reconstruction filter is 3.

Question 192

What is the relative noise for a smooth CT reconstruction filter?

Answer 192

The relative noise for a smooth reconstruction filter is 1 (ie no additional noise).

Question 193

What is iterative CT reconstruction?

Answer 193

You start the reconstruction with a back projection. Then you forward project that image (look at the attenuation coefficient plot it gives you) and compare it to your real signal. You adjust some of the numbers to correct that and check again, iteratively.

This wasn’t an option until recently because computing is now better and cheaper.

Question 194

What is a CT window width?

Answer 194

The CT window with is the range of HU that is covered between the minimum and maximum that will appear on the scale. Narrowing the window gives a greater contrast between individual HU numbers as greyscale has 256 shades. HU values outside of the window will appear as either complete black or complete white.

Question 195

What is a CT window level?

Answer 195

The CT window number controls which HU is in the center of the CT window. The window width will eminate in the positive and negative direction from this value.

Question 196

What is helical CT scanning?

Answer 196

The table is moved while the scanning happens, so the anode path around the patient is like a helix.

Question 197

What is pitch, in the context of CT?

Answer 197

Pitch expresses the tightness of the helix anode path around the patient.

pitch = L/W L is table movement per rotation, W is axial width of beam (z coverage of the fan)

Small pitch means a tight spiral, high pitch is a looser spiral.

Question 198

Why is overscan neccessary in CT scanning?

Answer 198

Every part of the image must be scanned by both the front and the back of the beam, as they have a different profile. So must overlap when making the helix.

Question 199

What is cardiac gating in CT?

Answer 199

Cardiac gating is when the heart beat is tracked and the scanner is turned on only at a certain phase in order to get a consistent scan. You can also do it retrospectively if you have a recording of the heart, can only use data points during certain phase.

Question 200

What is a dual source CT?

Answer 200

A CT scanner with two anodes, typically separated by 45 degrees, each with a corresponding detector array. Can speed up timing, or can set them at different energies to better distinguish different mediums.

Question 201

What is mA modulation in CT scanning?

Answer 201

The amps are reduced when above us and increased when beside us, as we are wider than we are thick (oval in shape). Saves dose on our thinner parts.

Question 202

How does the collimation of the CT beam increase noise?

Answer 202

Scatter is increased with a wider beam, increasing the noise.

Question 203

How do we determine spatial resolution of our CT scanners?

Answer 203

To determine spatial resolutions, we use phantoms with lines or wires close together. We determine how small of a separation we can distinguish by eye.

Question 204

What factors affect CT spatial resolution?

Answer 204

Factors that affect CT spatial resolution:

• XRay tube - focal spot distirbution
• Gantry motion - have to shoot back at an angle because it’s moving
• Detector size and sample
• Reconstruction filter

Question 205

How does sigma change with the mA of a CT scanner?

Answer 205

Sigma decreases by the square root of the factor that the mA increase. For example, if the mA increases by 4, the sigma goes down by 2.

Question 206

What causes streaks in a CT image?

Answer 206

Streaks are caused by bream hardening after going through something with very high Z. For example, gold and silver dental implants. This results from beam hardening and Compton scatter. Photons could end up in a different detector than they should be in. The biggest problem is when scattered photons end up in a detector that would usually have very few photons. For example, if a metal implant blocks all photons, then the corresponding detector element will only detect scattered photons.

Question 207

How can beam hardening affect a CT image?

Answer 207

It can cause streaks or shadows. Shadows occur when a very dense object hardens the beam. Then, the beam has a lower linear attenuation coefficient than expected and is not attenuated much afterwards.

Question 209

What does the mass number A stand for?

Answer 209

A is the number of protons and neutrons.

Question 210

What does the mass number A stand for?

Answer 210

A is the number of protons and neutrons.

Question 211

What does the atomic number Z stand for?

Answer 211

Z is the number of protons.

Question 212

Describe Isomeric transition.

Answer 212

An isomeric transition is a radioactive decay process that involves emission of a gamma ray from an atom where the nucleus is in an excited metastable state, referred to in its excited state, as a nuclear isomer.

The emission of a gamma ray from an excited nuclear state allows the nucleus to lose energy and reach a lower energy state, sometimes its ground state. In certain cases, the excited nuclear state following a nuclear reaction or other type of radioactive decay, has a half life that is more than 100 to 1000 times longer than the average 10−12 seconds, and this excited state is referred to as a metastable nuclear excited state. Some nuclei are able to stay in this metastable excited state for minutes, hours, days, or occasionally far longer, before undergoing gamma decay, in which they emit a gamma ray.

The process of isomeric transition (that is, the gamma decay of nuclear isomers), is therefore similar to any gamma emission from any excited nuclear state, but differs in that it involves excited metastable states of nuclei with longer half lives. These states are created, as in all nuclei that undergo gamma radioactive decay, following the emission of an alpha particle, beta particle, or occasionally other types of particles that leave the nucleus in an excited state.

The gamma ray may transfer its energy directly to one of the most tightly bound electrons causing that electron to be ejected from the atom, a process termed the photoelectric effect. This should not be confused with the internal conversion process, in which no gamma ray photon is produced as an intermediate particle.

Question 213

What energy does ^99mTc emit?

Answer 213

99mTc emits gamma rays of 140.5 keV.

Question 214

What is the half life of ^99mTc?

Answer 214

The half life of ^99mTc is 6 hours.

Question 215

What parent of ^99mTc is used for its production for nuclear medicine?

Answer 215

⁹⁹Mo is manufactured to, which decays to ^99mTc inside the generator at the clinics.

Question 216

After how many half-lives of ^99mTc should one “milk the generator” for the best efficiency?

Answer 216

4 half-lives = 24 hrs for ^99mTc.

Question 217

What does SPECT stand for?

Answer 217

Single photon emission computed tomography.

Question 218

What is used in conjuction with SPECT detectors to discriminate photons traveling in different directions?

Answer 218

Collimators are used to discriminate photons traveling in different directions.

Question 219

Why is the emission energy of ^99mTc nice?

Answer 219

140keV photons are nice because the photoelectric effect dominates at this energy, so the Compton continuum is not an issue. The energy is also nicely matched with the NaI(Ti) scintillator efficiency and spatial resolutions.

Question 220

What is an acceptance window used for in nuclear imaging?

Answer 220

An acceptance window is the allowable energy range outside of the expected energy of the isomer that is accepted by the detector. It is used to discriminate signal from background.

Question 221

How do you combine biological elimination and physical half-life to get an effective half life?

Answer 221

You add the activities:

lambda_e = lambda_b + lambda_p

where lambda_p=ln(2)/half-life

Question 222

How is SPECT data reconstructed?

Answer 222

SPECT is reconstructed using filtered back projection or iterative reconstruction.

Question 223

How big is the matrix for SPECT?

Answer 223

The matrix for SPECT is either 64² or 128².

Question 224

What type of filter is used for the FBP image reconstruction of SPECT?

Answer 224

Not a ramp filter because it causes too much noise.

Question 225

What does the attenuation correction corecct for in SPECT? What does the image look like before correction?

Answer 225

The attenuation correction accounts for the fact that you will get more counts from near the surface of the object (that contains the isomer) than from the centre because the stuff in the centre had to travel more and got attenuated. Before the correction, the centre looks lighter.

Question 226

How is spatial resolution measured for SPECT?

Answer 226

A phantom developed by NEMA (electrical engineering governing body) is used. Line sources of ⁵⁷Co are placed in phantom and a ramp filter is used and the FWHM is determined. Require 7-8mm periphery and 9.5 - 12 mm central.

Question 227

How many slices are typically used for PET?

Answer 227

PET uses ~109 slices.

Question 228

What is FDG?

Answer 228

Fludeoxyglucose (18F) (INN), or fludeoxyglucose F 18 (USAN and USP), also commonly called fluorodeoxyglucose and abbreviated [18F]FDG, 18F-FDG or FDG, is a radiopharmaceutical used in the medical imaging modality positron emission tomography (PET). The uptake of 18F-FDG by tissues is a marker for the tissue uptake of glucose, which in turn is closely correlated with certain types of tissue metabolism.

Question 229

What is the energy of photons from PET?

Answer 229

The photons originate from a positron anihilation, so there are two 511 keV photons.

Question 230

How do the scintillators in PET help guide the photons in a straight path?

Answer 230

The scintillator crystals have slits cut in them to create an effect like fiber optics to guide the light.

Question 231

What is the formula for the random rate of PET detectors?

Answer 231

R_random = tau S₁S₂

Where tau=coincidence window

S_i = count rate of detector i

Question 232

What is the random/true ratio for PET?

Answer 232

Random/true ratio is the ratio of random events (incorrectly matched events) to properly matched events. It increases with activity and decreases with the time window.

Question 233

What happens when you remove the septa in PET detectors?

Answer 233

The septa are the tungsten scatter grids. Removing them allow you detect in 3D instead of 2D, but it increases the random coincidence rate and the scatter fraction.

Question 234

Which uses attenuation correction before reconstruction, PET or SPECT?

Answer 234

PET uses attenuation correction before reconstruction, SPECT uses attenuation correction during reconstruction. PET achieves this by using a CT scanner at the same time.

Question 235

What radionuclide has been proposed to replace ⁹⁹Tc?

Answer 235

⁸²Rb has been proposed.

Question 236

What is the half-life of FDG?

Answer 236

FDG, is a radiopharmaceutical used in the medical imaging modality positron emission tomography (PET). Its half-life is 110 min.

Question 237

What is the principle of projection data collection for SPECT and PET?

Answer 237

SPECT uses collimation, PET uses annihilation coincidence detection.

Question 238

How does attenuation affect SPECT and PET?

Answer 238

SPECT: Attenuation is less severe. Radioactive attenuation correction sources or x-ray CT scan can correct for attenuation.

PET: Attenuation more severe. Radioactive attenuation correction sources or x-ray CT can correct for attenuation.

Question 239

What is meant by a Standardized Uptake Value (SUV)?

Answer 239

The SUV is the normalized uptake of FDG to:

• Administered activity
• radioactive decay from time of injection
• patient body mass

SUV = (activity concentration in a voxel or group of vodels)/(activity administered/body mass)

Question 240

What factors affect SUV?

Answer 240

Factors affecting Standardized Uptake Value:

• Accuracy of administered activity (material left in syringe)
• Leakage of activity during administration (??)
• PET/CT Calibration and attenuation correction
• Elapsed time before imaging (accuracy)
• Patient Physiological state (fasting, insulin etc)
• Body composition
• Size of tumor (partial volume effects)
• Motion (respiratory)
• Region of Interest Selection

Question 241

What is the half life of FLT?

Answer 241

The half life of FLT is 109 minutes and emits positrons with E_max = 1.65 MeV. It accululates in cells that are undergoing cell division (tumors divite a lot). This way you can look directly for dividing cells instead of metabolic uptake.

Question 242

What particle interaction dominates for PET?

Answer 242

Compton interactions dominate for PET, with 511 keV photons.

Question 243

What type of particle interactions dominate for CT scans?

Answer 243

At 70 keV, photoelectric effect dominates for CT scans.

Question 245

What is the emission energy of ⁶⁷Ga?

Answer 245

The emission energies of ⁶⁷Ga are: 93 keV, 185 keV, and 300 keV.

Question 246

Explain how a ⁹⁹Mo/^99mTc generator works.

Answer 246

A technetium-99m generator, or colloquially a technetium cow or moly cow, is a device used to extract the metastable isotope 99mTc of technetium from a source of decaying molybdenum-99. 99Mo has a half-life of 66 hours and can be easily transported over long distances to hospitals where its decay product technetium-99m (with a half-life of only 6 hours) is extracted.

The column, usually glass, containing a bed of aluminium oxide (alumina) as a support for the parent radionuclide. ⁹⁹Mo (molybdate) will bind strongly to this support media and is not washed off during the subsequent elution of the daughter radionuclide ^99mTc (pertechnetate).

Pouring normal saline solution through the column of immobilized ⁹⁹Mo elutes the soluble ^99mTc, resulting in a saline solution containing the ^99mTc as the pertechnetate, with sodium as the counterbalancing cation.

When the generator is left unused, 99Mo decays to 99mTc, which in turn decays to 99Tc. The half-life of 99Tc is far longer than its metastable isomer, so the ratio of 99Tc to 99mTc increases over time. Both isomers are carried out by the elution process and react equally well with the ligand, but the 99Tc is an impurity useless to imaging.