physics tutorial
DESCRIPTION
CORE PHYSICS TIPSTRANSCRIPT
PHYSICS & RADIATION BIOLOGYQUESTIONS
1.- Which is not a unit of Energy?a) Ergb) Joulec) Wattd) British thermal unite) Electron volt
Answer: c = Watt.
Charge: Coulomb (SI unit) Coulomb (CGS unit) Current: Amperes (Coulomb/sec) (SI & CGS) Potential: Volt (SI & CGS) Resistance: Ohm (SI & CGS) Energy: Joule (SI) Erg (CGS) Power: Watt (SI) Watt (CGS) Force: Newtons
Power is the rate of performing work. It is measured in Watts (Joules/sec)
Energy is measured in Joules Work is measured in Joules
The amp is a unit of electric current and it is equivalent to Coulombs/sec.
One proton has a charge of 1.6 x 10-19 .
2.- Which of the following descriptions about the electric charge of a positron is true?a) it can be annihilatedb) It changes in the magnetic fieldc) It is the same as the charge of an ionized hydrogen atomd) It has the same magnitude and sign as the charge of an
electron.
Answer: c
3.- Which of the following statements about the production of electromagnetic radiation is false?
a) It is generated by orbital motion of electrons within a specific shell
b) It may be produced by heat loss from a hot surfacec) The application of a high-frequency alternating current
to a wire generates electromagnetic radiation.
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d) It is produced by electrons slowing down in the proximity of the nucleus.
Answer: a
The application of a high frequency alternating current to a wire generates electromagnetic radiation. This is an example of electromagnetic induction.
4.- Which of the following statements about Planck’s constant is true?a) It is equal to the energy of a photon divided by its
frequencyb) It is measured in joulesc) It is proportional to the speed of lightd) It is proportional to the wavelength of a photon.
Answer: a
The energy of a photon can be expressed in terms of either frequency or wavelength. E = f E= energy , = Planck’s constant and f = frequency.
5.- Which of the following characteristics does not affect the quantity of x-rays produced by an x-ray tube.
a) target material of the anodeb) tube potentialc) voltage waveformd) field of view.
Answer: d
6.- Which of the following quantities will not increase if the x-ray tube potential is increased without changing any other parameter?a) Amount of heat produced at the anodeb) Quantity of emerging x-raysc) Energy of characteristic x-raysd) Maximum x-ray energy.
Answer: c = Energy of characteristic x-rays
7.- Which of the following statements about a general radiographic x-ray spectrum (tube potential, 120 kVp) is false?
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a) There are few 10-20 keV photons because they are preferentially absorbed by the x-ray tube envelope and housing.
b) Tungsten characteristic x-ray are presentc) The maximum x-ray wavelength is 120 armstrongsd) The spectrum will have an average energy of approximately
40-60 keV.
Answer: c The maximum x-ray wavelength is 1 nanometer
8.- Which of the following actions will not produce an x-ray spectrum with a higher effective energy?
a) Increasing filtrationb) Choosing an anode with a lower atomic numberc) Increasing the tube potentiald) Changing from a single to a three-phase generator.
Answer: b
9.- An x-ray beam is attenuated by three half-value layers. The x-ray beam intensity is reduced by a factor of :
a)2b)4c)8d)16e) 32
Answer: c
Each HVL reduces the intensity by ½, so the total intensity reduction is (½)3. The reduction will be a factor of 8 for this particular case.
10. Filters remove the following radiation from an x-ray beam:
a) secondaryb) low energyc) high energyd) leakagee) stray
Answer: b
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Filters are used to remove the low-energy photons that increase the patient dose but do not contribute to the image.
11.- The effective photon energy of an x-ray beam cannot be changed by:a) tube currentb) beam filtrationc) tube voltaged) voltage waveforme) passage through a patient
Answer: a Beam quality is independent of the tube current(mA),
which primarily determines the x-ray beam output (intensity).
12.- The x-ray beam filters used in the different radiographic examinations:
a) Routine radiography:b) Film/screen mammography:c) Chest radiography
Answers:
a) Aluminumb) Molybdenumc) Copper/aluminum
13.- The adequacy of the filtration of an x-ray tube may be determined by:
a) physical inspectionb) x-ray tube documentationc) kVp measurementd) x-ray output measuremente) HVL measurement
Answer: e
Adequacy of x-ray tube filtration is determined by measuring the HVL, usually at 80 kVp and ensuring it exceeds 2.5 mmAl.
14.- Indicate the HVL for each one of the following situations:
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a) Measured HVL on x-ray unit @ 80 kVb) Minimum HVL @ 80 kVc) Film/screen mammography unit HVLd) HVL for chest x-ray unit @ 140 kV.
Answers: a) 3.5 mm Alb) 2.5 mm Alc) 0.3 mm Ald) 6 mm Al
the higher the HVL the higher the beam quality and mean energy.
15.- The x-ray beam HVL does not depend on the:a) radiation intensityb) peak kVc) voltage waveformd) filtratione) x-ray spectrumf) anode material
Answer: a
The HVL expressed in mm Al only depends on the x-ray spectrum; it is thus independent of the intensity (quantity or mAs) of the beam as measured by exposure.
16.- Increasing the x-ray tube voltage (kVp) will NOT increase:
a) x-ray beam intensityb) patient penetrationc) beam HVLd) beam filtratione) heat produced in the anode.
Answer: d Beam filtration depends on the x-ray tube window and
added filters, and it is therefore independent of the x-ray tube kV/mA
17.- The soft tissue contrast in chest radiography performed at 140 kVp is primarily due to:
a) coherent scatter
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b) Compton scatterc) Photoelectric effectd) Pair productione) Photodesintegration
Answer: b
Compton scatter is the primary interaction for soft tissues at high photon energy levels (> 25 keV or > 75 kVp).
18.- The reasons for high subject contrast on a barium enema examinations is:
a) coherent scatterb) Compton scatterc) Photoelectric effectd) Pair productione) Photodesintegration
Answer: c
The high atomic number of barium results in strong absorption of the incident x-ray beam and results in very high subject contrast.
19.- By what process are electrons emitted from the x-ray tube filament?a) Bremsstrahlungb) Inductionc) Thermionic emissiond) Characteristic x-ray production
Answer : c
20.- Most of the kinetic energy of the accelerated electrons in the x-ray tube is converted into: a) Heatb) Gamma raysc) Leakage radiationd) X-rays
Answer: a
21.- The bremsstrahlung process is:
a) An interaction between two electronsb) Independent of the atomic number of the target material
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c) Identical to characteristic x-ray productiond) An interaction between an electron and a nucleus.
Answer: d
22.- Which phrase best completes the following sentence: <In the process of characteristic x-ray production>?
a) a nucleus interacts with an electronb) An outer shell orbital electron fills an inner shell
vacancyc) A continuous spectrum of x-ray energies is producedd) The energy of the accelerated electron is unimportant.
Answer: b
23.- A larger filament size may be selected instead of a smaller one for a particular imaging task when:
a) Motion blurring is a concernb) Low patient dose is requiredc) High patient dose is requiredd) Improved spatial resolution is needed.
Answer: a
24.- Which of the following pairs refers to the two types of x-ray anodes commonly in use?
a) Rotor and statorb) Cathode and anodec) Rhodium and berylliumd) Fixed and rotating
Answer: d
25.- Very low-energy x-rays contribute primarily to:
a) image contrastb) spatial resolutionc) patient dosed) x-ray tube cooling
Answer: c
26.- The effective size of the focal spot depends on:
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a) the anode angle and the actual focal spot sizeb) image contrastc) the anode stem materiald) the type of detector
Answer: a
27.- Which of the following characteristics is significantly affected by effective focal spot size?
a) Atomic number of the target materialb) Spatial resolutionc) Subject contrastd) Patient dose.
Answer: b
28.- For which of the following reasons is rhodium chosen as a target material for mammographic x-ray tubes?
a) It is cheaper and more plentifulb) It produces characteristic peaks at higher energies than
molybdenumc) It is necessary when a grid-based tube is used.d) It produces characteristic peaks at higher energies than
tungsten.
Answer: b
29.- The modulation transfer function:
a) Describes the system resolutionb) Compares image to object contrastc) Approaches 1.0 at low spatial frequenciesd) Decreases with increasing spatial frequency
Answer: all true.
30.- A detail screen compared to a regular screen has a higher:
a) Spatial resolution
31. GI contrast is increased by all of the following except:
a) Infusion of bariumb) Reduction of kVp
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c) Tight beam collimationd) Increase in grid ratioe) increasing the mA.
Answer: e
Increasing the mA normally will not increase image contrast.
32. Input devices in a computer include all except:
a) keyboardb) trackballc) light pend) monitore) touch screensf) joystick
Answer: d
Input devices in a computer include: keyboard, trackball, light pen, touch screens and joystick. Output devices: cathode ray tube, monitors, laser film, printers and paper printers.
33.- Compared to a Par speed film screen system, a 400 speed system will be:
a) Fasterb) slowerc) higher in contrastd) higher in latitude
Answer: a
The overall efficiency of a screen-film system is usually expressed by the speed of the system. So-called Par speed systems are usually assigned an arbitrary speed of 100.
Faster systems have higher numbers and screen-film system speeds of 50,200,400 and 600 are in common usage in radiology.
Faster films are more efficient at converting x-ray exposure to optical density, and thus they require less radiation and usually are noisier.
34.- The Bucky factor refers to:
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a) The artifacts that an antiscatter grid can useb) The factor by which the exposure needs to be turned up
when a grid is used.c) The inverse of the primary grid transmission factor.d) The increase in cost of the radiographic room when grids
are added to the base price.
Answer: b
There are many terms that have been used to describe the properties of an antiscatter grid: Bucky factor: Increase in incident exposure when a grid
is used. Selectivity: The relative transmission of the primary
radiation divided by the relative transmission of the scattered transmission.
Grid ratio: The height of the grid bars divided by the width of the interspace material.
Primary transmission The fraction of primary radiation that penetrates the grid.
Scatter transmission The fraction of scattered radiation that penetrates the grid
Contrast improvement ratio. The improvement in contrast when a grid is used. Equal to the ratio of the contrast with a grid over the contrast without a grid.
A radiograph that is produced without the use of a grid will require a certain radiographic technique (kV & mAS). When a grid is used to produce an image of the same object at the same kV, both the primary and scattered radiation will be reduced; Therefore, to produce the same level of film darkening, the mAS will have to be increased when the grid is used. The amount of increase that is needed. The amount of increase needed when a grid is used is called the Bucky factor.
35.- Where I refers to the light intensity coming through the film and Io is the intensity of the viewbox without the film on it, the definition of Optical Density is:
a) OD = - Log10 (I/Io)b) OD = Log e(Io/I)c) OD = Log10(I/Io)d) OD = Exp (I/Io)
Answer: aOD = - Log (T)
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The optical density of film describes the amount of light transmitted through the film base and developed emulsion at a given location on the film. If the film transmits a certain fraction of the light shone upon it, the OD will be: OD = - Log (T).
The relationship between transmission and OD is as follows:
Transmission=1.000 ; OD = 0.0 ; 100% transmission (blank film) (pp.148 Bushberg)
Transmission=0.741 OD=0.13 Typical base +fog Transmission=0.100 OD=1.0 Good exposure Transmission 0.001 OD=3.0 Very dark, need hot lamp Transmission 0.0003 ; OD=3.5 Maximal darkness in medical
radiography.
36.- The average gradient (AG) for OD1=2.0 and OD2=0.25, where E1 and E2 refer to the respective exposures is defined as:
a) AG=[OD1-OD2]/[LogE1-LogE2]b) AG=[OD1-OD2]/[E1-E2]c) AG=[LogE1-LogE2]/[OD1-OD2]d) AG=Log [OD1-OD2]/[LogE1-LogE2]
Answer: a
* See page 151; Bushberg.
37.- The biggest advantage that photostimulable phosphor systems have in terms of image quality over conventional screen-film radiography is:
a) much better spatial resolutionb) higher absorption efficiencyc) better signal to noise characteristicsd) wide dynamic range (exposure latitude)
Answer: d The dynamic range of photostimulable phosphor systems is
wider than the dynamic range of film. This gives PSP systems much broader exposure latitude than screen-film systems. This broad latitude is achieved through the use of sensitive elctronic components and by the use of pre-scan technique, which is a type of automatic gain control.
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38.- The scatter to primary ratio:
a) increases with patient thickness but decreases with field size
b) increases with patient thickness and field sizec) decreases with increasing kVd) is about 10-20 for mammography.
Answer: b
Scatter gets worse with the irradiated volume, and hence both when the object gets thicker and when the field of view increases.
In screen-film radiography, scattered radiation causes a reduction of contrast in the image.
Scattered radiation contributes an almost constant background fog level to the film image. For digital images, where contrast can be arbitrarily increased by using windowing and leveling techniques, the main effect of scattered radiation is to increase the noise in the image.
The amount of scattered radiation energy can be expressed relative to the amount of primary radiation energy deposited at the same point in the image. This ratio is called the scatter to primary ratio (SPR). The SPR increases with thicker patient anatomy and larger field sizes.(pp161 Bushberg)
Eventhough scatter increases with increased x-ray energy, so does the amount of transmitted primary radiation, and therefore the SPR only increases by a few percent as the kVp is increased from 80-120 for example.
High SPR (abdomen in a thick patient) are about 4-6. A ratio of 20 for mammography is too high.
39.- Radiographic contrast of bone will increase when:
a) the kV of the examination is loweredb) the average gradient of the film is increasedc) the scattered radiation is reduced by using a higher grid
ratio grid. d) All of the above.
Answer: d
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40.- The typical scatter fractions in mammography and abdominal radiography (respectively)
a) 0.5 & 1.2b) 0.1 & 0.8c) 0.5 & 0.75d) 0.8 & 0.05
Answer : c
The scatter fraction in mammo is 0.5 or so for larger breasts (scatter to primary ratio, SPR, ranges from 0.5 to 1.0), and in thicker patients with higher energy beams (abdominal) this can go to 0.75 (SPR=3 or even higher). Note that this values are without a grid. The grid reduces the scatter.
41.- The biggest practical effect that scattered radiation has on screen-film radiography is:
a) increased exposure time due to its presenceb) loss of signal to noise ratioc) loss of radiographic contrastd) effectively makes H&D curve steeper
Answer: c = loss of radiographic contrast
In screen-film radiography, scattered radiation causes a reduction of contrast in the image.
42.- The biggest overt effect that scattered radiation has on an appropriately windowed digital radiograph is:
a) increased exposure time due to its presenceb) loss of signal to noise ratioc) loss of radiographic contrastd) effectively makes H&D curve steeper.
Answer : b
* With digital images, the contrast can be arbitrarily increased mathematically by windowing more narrowly on the image. With this freedom, the limits of how tightly one can window are governed largely by the noise in the image.
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43.- Compared to the light emitted by calcium tungstate screens, the light emitted by Gd2O2S:Tb is:
a) longer in wavelengthb) shorter in wavelengthc) much more continuous than the line emission
characteristic of CaWO4
d) dimmer
answer: a
CaWO4 emission is down in the blue region and is continuous. Rare earth’s generally emit up in the green region of the visible spectrum, and tend to emit at several discrete wavelengths. Green light is longer in wavelengths than the blue light.
In the energy and wavelength spectrum from longer to shorter wavelengths: red, orange, yellow, green, blue and violet. Red has the longer wavelength and violet the shortest.
44.- The absorption efficiency of most screens:
a) increases rapidly just before the k-edge, then decreases at the k-edge
b) tends to increase with the color of light emitted by the screen
c) decreases in general with increasing x-ray beam energyd) is only about 2% in chest radiography.
Answer: c
The absorption goes down gradually with increasing energy. Keep in mind that in radiology we are dealing with very broad x-ray spectra, and the k-edge absorption peaks of the various screen phosphors are generally averaged out by the broad incident spectra.
45.- The visible light region of the electromagnetic spectrum spans the range of about:
a) 300 nm – 700 nmb) 3,000 nm to 7,000 nmc) 1.0 m 70 5.0 md) 4000 A to 5000 A
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Answer: a The wavelength in the visible range goes approx from 300-
700 nm
46.- The sequence of colors in the visible wavelengths running from short wavelengths to long wavelengths is:
a) red-orange-yellow-green-blue-violetb) violet-red-orange-yellow-green-bluec) violet-blue-magenta-red-yellow-greend) violet-blue-green-yellow-orange-red.
Answer: d
D is the correct answer from short wavelengths (violet) to long wavelengths (red). Remember the short wavelength edge of the visible spectrum abuts the ultraviolet region and the long wavelength edge meets the infrared region.
The other way to remember would be: ultraviolet rays have higher energies than infrared rays and thus, shorter wavelengths.
47.- Compared to thinner screens, thicker screens:
a) have comparable absorption efficiency, but much better spatial resolution
b) have better absorption efficiency but lower spatial resolution
c) usually absorb greater than 95% of the x-rays in chest radiography
d) last much longer in clinical use.
Answer: b
See pp. 143 Bushberg.
48.- A {detail} cassette used for hand radiography might have a speed rating of about.
a) 5b) 50c) 100d) 400
Answer: b
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PAR speed is 100 and detail cassettes are generally a bit slower than PAR systems.
49.- At the same mAS setting, increasing the kV will:
a) reduce the optical density due to reduced absorption efficiency of the screen
b) result in less radiation dose to the patientc) result in more radiation dose to the patientd) will not effect dose but will effect overall density of
the film.
Answer : c
The key here is “at the same mAS setting”. Increasing kV increases the energy in the beam (it also increases the output of the tube in terms of mR/mAS). Energy is the essence of dose (dose=energy/mass). In a clinical setting, increasing the kV would be accompanied by a dramatic decrease in the mAS and that is why we say that higher kV settings are generally lower in dose. The trick to this question is that the mAS is held constant.
True/False section
50.- The most commonly used film screen cassette for chest radiography uses a single screen and a single emulsion film.
Answer: false Chest radiography uses two sided casettes.
51.- The low kV used in mammography is used specifically to make it a lower dose examination.
Answer: False.
The low kV in mammo is to increase subject contrast. It actually increases the dose.
52.- As the grid ratio increases, the amount of unavoidable grid cutoff will also increase.
Answer: true
This is true because of geometrical considerations.
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53.- If an infinite focus grid is used at 36’’, the center of the image will be darker than the edges.
Answer: true.
This is an example of an off-focus grid. In the center of the field of view, the focus will be OK and radiation will penetrate the grid with little problem, but towards the periphery, the grid interspaces are pointing out compared to the divergent incoming x-rays. Hence, the grid cutoff will be severe at the periphery of the image, causing it to be lighter at the edges and darker in the middle.
54.- Focal spot blooming will be worse (i.e. larger effective spot), at higher mA settings compared to lower mA settings.
Answer: true.
Focal spot blooming occurs when a high tube current is used. As the electrons travel from the filament to anode, if there is a thick dense cloud of them (higher mA) they will tend to repel each other (like charges repel) and this will cause the beam to spread out (bloom).
55.- The Z for tungsten is lower than the Z of gadolinium
Answer: False
Z tungsten = 74/ Z gadolinium = 64.
56.- The Z of calcium is higher than that of iodine
Answer: false
Z Ca= 20 , Z of iodine = 53
57.- The intensification factor refers to the reduction of exposure when a screen is used versus when one is not used.
Answer: true.
58.- On dedicated geometrical tomography systems, the grid:
a) must be removed for all tomographic motions except linear
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b) is never usedc) must rotate to maintain alignment with the tubed) is usually a crossed grid.
Answer = c
If the grid does not rotate, then grid cut-off could be severe.
Antiscatter grids are used in geometrical tomography, and because the x-ray focal spot is moving with respect to the detector, the grid alignment is critical. Tomographic systems generally use linear grids, which need to be rotated in all but linear motion techniques to ensure alignment of the grid bars with primary radiation.
59.- Stereoradiography
a) requires two discrepant imagesb) makes use of a shift in position of the x-ray sourcec) allows true 3-D imaging of the patientd) all of the above
Answer: d = all of the above
Stereo is the only true 3-D imaging procedure
60.- In order to view a stereoradiographic pair of images without a special viewing device:
a) The radiologist might get a headacheb) Convergence and accomodation must be decoupledc) The images must be positioned correctly together on the
viewbox.d) All of the above.
Answer: d = all of the above
61.- Which of the following is not a motion produced by dedicated geometrical tomography equipment?
a) hypercycloidalb) trispiralc) arctangentiald) circular
Answer: c
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The motions possible with dedicated tomographic systems include: Linear, circular,trispiral and hypercycloidal.
62.- To create the subtraction mask from the mask image in preparation for film subtraction, the subtraction mask film should have the following properties:
a) gamma of 3.0 positive filmb) gamma of 3.0 negative film (whites copy to black, black
to white)c) gamma of 1.0 positive filmd) gamma of 1.0 negative film
Answer : d
A gamma of one is needed so that contrast is not enhanced during the production of the subtraction mask. If contrast were enhanced, optical subtraction would be unsatisfactory. The subtraction mask has to be of opposite polarity and therefore a negative image is needed
63.- Geometrical tomography with a tomographic angle of about 10 deg is called:a) thin slice tomographyb) zonographyc) the Bucky-Potter viewd) hybrid tomography
Answer: b
Narrow tomographic angles of 10 degrees result in a relatively thick plane of focus, and this technique is called zonography. It is sometimes used to delineate suspected pathology better in a thick section of the patient.
True or False section
64.- The slice thickness in geometrical tomography gets thinner with increasing tomographic angle.
True: Slice thickness gets thinner with increasing tomographic angle
65.- Linear tomography is best when the motion direction is perpendicular to long bones
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True
66.- The further that an object is away from the focal plane in geometrical tomography, the bigger the blur patter of that object on the film.TrueFor a small BB with a circular tomographic motion, if it is 2 cm out of plane it will be blurred over an anulus (a circle)
67.- With the patient supine on the table, conventional tomography results in a sagital slice in focus. Answer: False
Conventional tomography delivers coronal slices with the patient supine. Sagital slices can be achieved by placing the patient on his or her side.
68.- Automatic brightness control in a fluoroscopic system works by:
a) sensing the signal level into the TV monitor and controlling the gain on the TV camera
b) sensing the entrance exposure to the patient using a exposure meter in the table and modulating the fluoroscopic technique (kV and mA) in response
c) measuring the thickness of the patient using a laser and modulating the output of the x-ray tube accordingly
d) sensing the light output by the image intensifier and modulating the fluoroscopic technique (mA and kV) in response.
Answer: d
The purpose of the automatic brightness control is to keep the brightness of the image constant at the monitor .
It does this by regulating the x-ray exposure rate incident on the input phosphor of the II; the goal of the ABC is to keep the luminance at the output phosphor relatively constant.
The fluoroscopic ABC circuitry in the x-ray generator is capable of changing the mA and the kV.
69.- By changing the aperture in the lens system from f/5.6 to f8.0, the exposure rate to a patient in an automatic brightness controlled system:
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a) will increaseb) will decreasec) will not changed) will fluctuate with unknown consequences on exposure
rate.
Answer: a = will increase.
Changing from f/5.6 to f/8.0 results in a physically smaller aperture, allowing less light to pass between the image intensifier and TV camera.
This change, in effect, turns down the gain of the imaging system and to maintain constant brightness at the monitor the ABC system will call for more radiation (higher exposure rate).
70.- The radiologist pans the image intensifier from a thin part of the patient to a thicker part. The ABC senses the change and needs to adjust the fluoroscopic technique. Which of the following adjustment scenarios will result in the lowest dose to the patient?
a) increase mA, kV stays the sameb) increase kV, mA stays the samec) increase kV and mAd) increase mA, kV decreases.
Answer: increase kV, mA stays the same.
More radiation gets through the patient at higher kVs depositing dose. The highest dose tehcnique would be D, then A, C and B.
71.- A cardiologist is performing a cardiac catheterization procedure with balloon angioplasty and stent placement. He has been using exclusively the specially activated fluoroscopy in small field of view mode, where the exposure rate to the patient is 50 R/min. The procedure has been difficult due to tortuous vascular anatomy and already the tech has rrecorded 40 minutes of fluoroscopy time. What is the estimated entrance dose so far?
a) 1,200 radsb) 1.2 radsc) 2000 mRads ( 2.0 Rads)d) 2000 Rads.
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Answer: d = 2000 rads.
(50 R/min x 40 minutes) These numbers are realistic when specially activated fluoroscopy is used. Newer federal laws regulate specially activated fluoro to 20 R/min, but many systems are still around which exceed this. At 2000 Rads, the patient will definitely have erythema and even a 2nd degree radiation burn.
72.- The skin dose when the patient’s skin will turn red due to radiation is approximately:
a) 5 radsb) 50 radsc) 500 radsd) 5000 rads
Answer: 500 Rads
600 rads is often quoted as the erythema dose and so 500 Rads is the best answer.
73.- A radiologist is buying a new R/F suite for her practice, and wants to get a 14” diameter image intensifier and a digital photospot camera, and eliminate the spot film device. The salesman is suggesting that she get a high line TV camera (1023 vertical lines) due to the large image intensifier. To compare, she calculates that the limiting spatial resolution in the vertical direction for this II is 14” mode will be : __________with a 525 line TV system and ____ with a 1023 line system.
a) 1.4 & o.7 lp/mmb) 0.67 & 0.33 lp/mmc) 0.7 & 1.4 lp/mmd) 7 & 14 lp/mm
Answer: c = 0.7 & 1.4 lp/mm
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74.- A fluoroscopic study is to be performed on a newborn and the parents are very concerned about radiation dose to their new baby. Which of the following measures can the radiologist do to reduce the dose in this examination?
a) change the aperture setting from 5.6 to 3.4 (in the optics between the II and TV)
b) remove the antiscatter gridc) use of a higher kVd) All of the above
Answer: all of the above.
All these things will reduce the dose. The change in aperture setting (A) will increase the gain of the imaging system, reducing the exposure rate (the image quality will be worse also, but this is the low dose compromise). For a small newborn, scattered radiation will not be too bad and the grid could be removed. This could save a factor of 2 to 3 in terms of dose. Finally, increasing the kV means that more of the radiation will penetrate the newborn and hit the detector.
75.- An automatic brightness controlled fluoroscopic system is placed in a mode where only mA is changed to accommodate for different patient thicknesses (kV remains constant). Regardless of the different patient thicknesses, what parameter will remain the same?
a) signal to noise ratiob) entrance exposure to the patientc) number of x-ray photons entering the patient per secondd) x-ray tube current.
Answer: a = signal to noise ratio
With the kV fixed as stated, the mA adjusts the x-ray photons per unit time. It does this so that the same number of photons will be detected per unit time, regardless of patient thickness.
Since the SNR is related to the number of detected photons (SNR=N, where N=detected number of photons).
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Answers B, C and D are all affected by changing the mA. ( in D, the mA is the tube current).
76.- Which of the following sequences describes the signal chain in a fluoroscopic system from x-rays in to light out of the TV monitor?
a) X-rays-light-electrons-lightb) x-rays-electrons-light-electrons-lightc) x-rays-light-electrons-light-electrons-lightd) x-rays-light-electrons-soft x-rays-light-electrons-light.
Answer: c . The II does the x-ray-light-electron-light conversion and
the Tv camera then converts light to electrons and the TV monitor converts electrons to light.
77.- In an angiography fluoroscopic system where the x-ray tube is under the table, the II is over the patient, and there is no under table shielding other than what’s in the tube, where will scattered radiation striking the radiologist who is not wearing an apron be the most intense?
a) knee capsb) shouldersc) headd) hands.
Answer: knee caps.
The backscatter is much more intense than the forward scatter, and therefore the entrant surface of the patient is the most intense “emitter” of scattered radiation. This is because the beam is significantly attenuated (100X or more) before it comes out the distal side of patient. The knee caps are the only body part that is under the entrance surface.
True/False section
78.- The maximum exposure rate to the patient by law in non-specially activated fluoroscopy is 10 R/min
Answer: True
79.- After the maximum exposure rate is met for a large patient, modern fluoroscopy systemscan keep the image bright
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at the monitor by increasing the electronic gain of the TV camera.
Answer: true.
Once the 10 R/min rate is met, the brightness can be increased by more x-ray flux, so often the automatic gain control of the TV camera then kicks in. Modern systems might also start to open up an electronically controlled aperture in the lens system, to let more light from the II get through to the TV camera.
80.- Doubling the physical diameter of the aperture will result in a four-fold reduction in dose to the patient.
Answer: true
Doubling the diameter will quadruple the area, and thus the effective gain of the system will be quadrupled, leading to the need for 4 times less radiation exposure.
81.-The output phosphor of modern image intensifiers is usually made of ZnCdS
Answer: true.
82.- Smaller objects can be seen best in the 14 inch mode of a 14/10/5 trimode image intensifier.
Answer: false.
The best spatial resolution (to see smaller objects) is enjoyed at the magnification modes (5 inch mode for example).
83.- Pincushion distortion is worst in the 5 inch mode of a 14/10/5 inch trimode II.
Answer: false
Pincushion distortion is worse in the largest field of view mode-the 14 inch mode in this case. The smallest field of view covers a smaller part of the curved input phosphor, and therefore suffers less from pincushion distortion.
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84.- After a service visit on an R/F room, the radiologist notices a moire pattern on the fluoroscopic TV monitor in 5 inch mode. Changing the II to 7 inch mode changes the moire pattern completely, but it still does not go away. This suggests that the service man probably put the grid back on incorrectly.
Answer: True
A moire pattern can be generated when the grid bars are aligned almost parallel to the TV scan direction (horizontal). When the grid bars, which are quite small, are aligned perpendicular to the TV scan direction, the TV cannot resolve them and they do not produce the moire pattern.
Changing mag modes will change the pattern.
85.- A 16:1 grid ratio is very common for fluoroscopic systems
Answer: false.
A 16:1 grid ratio is way too high for fluoroscopy; more typical values may be 8:1, 10:1 or 12: 1 grid ratios. Fluoro systems often change their SID (source to image distance) and therefore high grid ratios would suffer too much from off-focus grid cutoff.
86.- With the ABC operational in an under table x-ray tube fluoroscopic configuration, moving the image intensifier up (further away from the patient) will cause a decrease in entrance exposure to the patient.
Answer: False
moving the II further away from the detector will reduce the primary radiation reaching it due to the inverse square law and therefore, more radiation will be called for by the ABC system. Also, less scatter from the patient will reach the II due to the air gap, and this loss will also cause the ABC to call for more entrance exposure.
87.- The output phosphor of an image intensifier has to have much better spatial resolution than the input phosphor.
Answer: True
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Let’s say that a 0.5 mm object is imaged at the input phosphor, with the II in 9 inch mode. At the input phosphor, the object size is close to its true size, and will be imaged using a 0.5 mm of the input phosphor. After the signal is propagated through the II, however, the whole 9 inch diameter image striking the input is now compressed onto a 1 inch diameter circle (the output phosphor). The 0.5 mm object at the input is now mapped down to a 55m shadow at the output phosphor. Thus, the resolution cell has to be much tinier at the output phosphor.
88.- Gamma 20 is:a) Exposure rate per unit activity at 20 inches from a point
radiation source b) Gamma exposure rate constant for a specified
radionuclide’s photons above 20 keVc) Beta particle exposure constant for a specified
radionuclide for all beta particles above 20 keV Emaxd) Dose rate constant for all photon emitting radionuclides
(above 20 keV) at 1 metere) The number after T 19.
Answer: b
Expressed as R-cm2/mCi-hr or R-m2/Ci-hr
89.- Detector balancing is much more critical in 4th generation CT compared to 3rd generation
Answer: False.
4th generation uses a detector fan instead of a source fan, and therefore is intrinsically less susceptible to intra-detector differences.
A detector problem in a 3rd generation scan will typically give a ring artifact.
89.- Xenon detector technology can only be used in 3rd generation scanners, not fourth
Answer: true
Xenon detectors are long thin ionization chambers with Xe in them. Because they are so long and thin, they are very
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directional and have to maintain precise alignment with the x-ray source. This alignment is maintained in 3rd generation CT where the tube and detectors ride the moving gantry together, but it is not in 4th generation, where hthe detectors are expected to detect x-rays efficiently while the tube makes a wide arc around it.
90.- If a material has lower than water, it will have a positive CT number.
Answer: false.
CT number = 1000 x [(-w)/ w)]. If <w (water) then the CT number will be negative.
91.- Partial volume artifacts can be reduced by decreasing the slice thickness.
Answer : true.
Partial volume averaging is made worse with thicker slices.
92.- CT has better contrast resolution than radiography.
Answer: true
Contrast resolution is what CT excells at
93.- CT has better spatial resolution than radiography
Answer: false
Screen film systems have spatial frequencies of up to 8-10 line pairs per mm, while CT is limited to perhaps 1.5 lp/mm or 15 lp/cm
94.- Mag mode fluoroscopy has better spatial resolution than CT
Answer: true.
Mag mode fluoro has limiting resolutions of about 2.5-3.5 lp/mm, much higher than the 1.5 lp/mm which CT can deliver.
95.- The dose in 2 view mammography is much higher than in a head CT.
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Answer: false.
A head CT is about 5-8 rads whereas the glandular dose in one mammogram is about 150 mrad, a 2 view would be 300 mRad.
96.- Electron density is related to N Z/A
Answer: true
NZ/A is the electron density in electrons per gram.
97.- Helical CT at the same slice thickness has a much better slice sensitivity profile than conventional CT.
Answer: false.
With helical acquisition, the table is moving and so the information in the Z dimension is blurred out more than with conventional, non-helical acquisition.
98.- The dose in body CT is about:
a) 3-5 radsb) 3-5 mradc) 3-5 Grayd) 3-5 rads (whole body)
Answer: A = 3-5 rads
Clearly, 3-5 mrad and 3-5 Gray (300-500 rads) are incorrect. D is not right because the dose in CT is delivered to the tissue in the beam, not the whole body. Background radiation is typically delivered whole body.
Special equipment is necessary to measure the exposure in CT because of the rotating nature of the x-ray tube. A homogeneous Lucite or plexiglas phantom is used that has holes bored to accept a pencil ionization chamber. Thermoluminiscent dosimeters are sometimes used as well.
Common values for CT doses range from 30 mGy (3Rads) to as high as 90 mGy (9Rads) for high mAs studies.
Typical doses are: Head CT: 65 mGy (6.5 Rads) Spine: 23 mGy (2.3 Rads) Body: 15-25 mGy (1.5-2.5 rads)
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Probably the best measure of risk (of cancer) is not the dose but the integral dose. Integral dose is just a measure of the amount of energy deposited in tissue, regardless of its distributions.
To calculate the integral dose, one calculates the average dose (energy/mass) for a mass of tissue and then multiplies by the mass of tissue.
The units of absorbed dose are the Rad (100ergs/gram) or preferably, in SI units the Gray (1 joule/kg).
The Rad and the Gray measure the energy deposited per unit mass, where 1 Gy=100 rad.
99.- The contrast resolution capabilities of CT are about:
a) 0.5%b) ten times better than screen filmc) 5 CT numbersd) all of the above.
Answer: d = all of the above.
Screen-film contrast resolution is about 5%, which is 10 times worse than CT.
5 CT scan numbers turns out to the typical noise you may find in CT
100.- Contrast resolution:
a) Depends on the mAs used in the CT scanb) Refers to the abiliaty to see very tiny objectsc) Improves as the SNR is decreasedd) All of the abovee) None of the aboce
Answer: a
When you turn up the mAs, the dose goes up, the number of photons used to make the image goes up, the signal to noise ratio goes up and the contrast resolution goes up.
101.- Spatial resolution depends upon:
a) detector size and spacingb) x-ray focal spot sizec) magnification factor of objectd) number of projection rays acquired
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e) all of the above
Answer: e all of the above.
* All these things have some influence on the spatial resolution of the scanner
102.- Iterative reconstruction techniques:
a) are required in helical CTb) are much faster than backprojectionc) make use of repeated guesses and updatesd) are only possible in parallel ray geometry.
Answer: c
Iterative reconstruction techniques were used in the early days of CT. They are also used for tomographic reconstruction in specialized applications.
Iterative reconstruction techniques repeat a series of updates to the reconstructed image and each update is called an iteration.
103.- Beam hardening artifacts:
a) are the primary cause of streaks from surgical clipsb) can result in lower CT numbersc) can result in higher CT numbersd) all of the abovee) none of the above.
Answer: b
Streaks from clips are primarily from motion Beam hardening causes to be reduced and thus will cause
the CT number to be lower High CT numbers result from scatter. A cupping artifact is a classic example of beam
hardening. Beam hardening artifacts usually occur in the soft tissue regions between large bony structures and result in a decreased image intensity not representative of the tissues
104.- Which of the following changes will result in the same primary dose per slice (before and after the change)?
a) halving slice thickness, doubling mAs
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b) halving slice thickness, halving mAsc) doubling slice thickness, no other changed) halving kV, doubling time, no change in mA.
Answer: c
Any time you change the mAs in CT you will change the dose. By increasing the collimator separation and doubling the slice thickness, twice the photons will be used but will be absorbed in twice the volume (and thus mass) of tissue.
Dose is energy/mass. More tissue will be irradiated, but the primary dose will be the same.
105.- The linear attenuation of water at typical CT beam energies is about 0.19 cm-1. A voxel with a material with = 0.38 cm-1 will have a CT number of:
a) 2b) 200c) 500d) 1000e) None of the above.
Answer: d = 1000
* CT # = 1000 x [(-w)/ w)]= 1000 x (.19/.19)= 1000 x 1
106.- A ramp filter in CT refers to:
a) the shape of the copper placed in the x-ray beam to reduce beam hardening
b) the shape of the kernel in frequency domainc) the shape of the kernel in the spatial domaind) the saw tooth artifact common in helical scanners.
Answer: b In the frequency domain, the filter enhances frequencies
linearly with increasing frequency. When plotted as amplitude of kernel (y axis) versus
frequency (x-axis), the plot looks like a ramp.
107.- For a given controlled area, the occupancy factor of each adjacent area:
a) depends on the type of x-ray device
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b) depends on the use factorc) would be the same if the controlled area was made
uncontrolledd) is usually chosen to reflect patient/exam workloade) none of the above
Answer: c
The occupancy factor indicates the fraction of time a single individual might spend in an adjacent room
The workload indicates the amount of x-rays that are produced
The radiation exposure level based upon the typical technique (kVp) used for the exam, the workload and the calculated scatter and leakage radiation levels
The use factor indicates the fraction of time the primary radiation will be directed toward a particular wall or barrier.
108.- For a given optical density on a film, all of the following will reduce the patient dose except:
a) beam collimationb) using screensc) using higher kVpd) using a high-grid ratioe) none of the above
Answer: d = using a high grid-ratio
* High grid ratio reduces the scatter on the film but requires a higher dose to achieve the same optical density on a film.
109.- Gradient coils are used to:
a) localize the plane to be imagedb) acquire the FID signal as it is produced in the
transverse planec) excite the protons in the bodyd) make sure the RF power into the patient is not excessivee) improve the homogeneity of the magnetic field
Answer: Localize the plane to be imaged
Gradient coils encode the spatial location of the protons within the patient volume
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Gradient coils are saddle coil shaped conductors that provide a linear proton localization based upon the known precessional frequency variations
Shim coils are active or passive magnetic field devices that externally affect the main magnetic field to improve the homogeneity in the sensitive central volume of the scanner.
RF transmitter and receiver body coils are also located within the magnet bore. After the gradients and transceiver equipment are installed, a 1 meter bore diameter reduces to about 60 cm.
Surface coils: specialized for imaging specific body areas are used in lieu of the body coil when high resolution or smaller FOV areas are desired. The control interface, computer system, power supplies, analog to digital converter and image display are crucial components of the MR system
110.- Calcium deposits in an acute hemorrhage can cancel T2 shortening effects of the blood degradation products because of _______properties.
a) diamagneticb) paramagneticc) magnetization transferd) hysteresise) solidification
Answer: a = diamagnetic
Calcium, because of its paired nuclear spins, exhibits a diamagnetic property that has the effect of diminishing the local magnetic field. Paramagnetic agents such as methemoglobin augment the local magnetic field. In certain situations, these properties offset, and the image does not exhibit the expected appearance of a hemorrhage.
111.- Match:
a) highest power consumptionb) requires cryogensc) heaviest systemd) strongest magnetic fielde) lowest operational costf) fringe fields extensive
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with
1) permanent magnet2) resistive magnet3) superconductive magnet
Answer: a) 2 b) 3 c) 1 d) 3 e) 1 f) 3
Permanent magnests have low operating costs and small fringe fields but are heavy and can only generate fields to approximately 0.35 T.
Resistive magnets can generate magnetic fields up to approximately 0.5 T
Resistive magnets can be turned on and off, but consume a large amount of power and need cooling because of the heat generated.
Fields strengths of 2T or higher can be generated by superconducting magnets. Superconductivity is the ability of certain materials to conduct electrical current without any resistance.
Superconducting magnets are generally kept cool using liquid helium surrounded by liquid nitrogen
112.- Which of the following has the shortest T1?
a) waterb) musclec) cerebrospinal fluidd) bonee) fat
Answer: fat.
At 0.5 T fat has a T1 of 200 ms and CSF 2000. From low to high, the distribution is as follows:
Fat 200 Liver 320 Kidney 500 White matter 530 Spleen 540 Muscle 550 Grey matter 650 CSF 2000
113.- Given a strong external magnetic field, the nuclear magnetic dipoles are:
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A) ALL ALIGNED PARALLEL TO THE APPLIED FIELDB) ALL ALIGNED ANTIPARALLEL TO THE APPLIED FIELDC) Aligned randomlyD) Aligned perpendicular to the applied fieldE) Aligned parallel and antiparallel to the applied field
Answer: e = alligned parallel and antiparallel to the applied field.
The individual magnetic dipoles are aligned parallel and antiparallel due to thermal energy of the sample. Transition from the parallel (low energy) to the anti-parallel (high energy) states result in the change of direction of the sample magnetic moment.
114.- For the spin-lattice relaxation time, the lattice is:
a) the x,y,z coordinate systemb) tissuec) an imaginary framed) a set of adjacent nucleie) the digital image matrix.
Answer: b
The lattice is the environment in which the spin exists.
115.- A spin-echo sequence used with a TE of 200 msec and TR of 3 seconds will result in an image that is primarily______weighted.
a) T1 weightedb) Proton density weightedc) T2 weightedd) T1/T2 weightede) Magnetization transfer.
Answer: c
This range of TE (relatively long) and TR (long) results in T2 weighting.
116.- Of the following nuclei, , which would not be of use for MRI?
a) 1Hb) 10 Ne
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c) 19Fd) 23Nae) 17O
Answer: b
Nuclei containing an even atomic mass number (paired spins) will not give rise to a net magnetic moment.
Magnetic moment is a vector that represents the stregth and orientation of a magnetic dipole.
Nuclei with an even number of protons and neutrons have no net magnetic moment. The protons and neutrons pair up with their magnetic moments aligned in opposite directions and cancerl each other.
The hydrogen nucleus has the largest magnetic moment and its abundance in the body makes it the basis of most clinical MR imaging.
117.- For image intensifiers, CRT’s and scintillation cameras, the ____ gauss line of the magnetic field is recommended as an exclusion zone.
a) 0.05b) 1c) 5d) 15e) 100
Answer: c = 5
Electronic instruments that depend on electron focusing, etc, are very sensitive to changes of even the earth’s magnetic field. Therefore, fringe fields greater than one gauss can cause severe problems. The “health” limit is 5 gauss (pacemakers, stc).
118.- The purpose of shim coils is to:
a) provide magnetic field gradients for selective excitationb) reduce the firnge fields of superconducting magnetsc) boost the amplitude of the RF pulse from the transmitterd) make the Bo field more homogeneouse) improve the acquisition time of the procedure.
Answer: d
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Shim coils adjust the homogeneity of the main magnetic field.
Shim coils are active or passive magnetic field devices that externally affect the main magnetic field to improve the homogeneity in the sensitive central volume of the scanner.
119.- When hydrogen nuclei are immersed in a 1.5 T magnet, the resonance frequency is closest to ____ MHz
a) 1.5b) 15c) 43d) 64e) 92
Answer: d= 64.
When magnetic moments are placed into a magnetic field, a torque causes the moments to perform a precession motion similar to a spinning top.
The Larmor frequency is the precession frequency of nuclei in a magnetic field. Fl= /2 x B. The gyromagnetic ration () is a constant for any nucleus.
Replecing the Larmor frequency (fL) with the angular frequency () where = 2 x fL, allows the Larmor equation to be expressed as = x B
Protons have a Larmor frequency of 21 MHz at 0.5 tesla (T), 42 MHz at 1 T and 63 MHz at 1.5 T.
119.- A gadolinium contrast agent will most likely cause one of the following:
a) decreased proton density signalb) increased proton density signalc) increased T2d) increased T1e) decreased T1
Answer: e = decreased T1
Gadolinium is a paramagnetic agent, and will cause a decrease in the T2 time because of magnetic inhomogeneities causing a more rapid phase dispersion of the transverse magnetization. It also interacts with the lattice of the tissues, causing a decrease in the T1 time.
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Paramagnetism is caused by the presence of unpaired atomic electrons or molecular electrons.
Gadolinium DTPA is an exampleof a paramagnetic contrast agent. Gadolinium has seven unpaired electrons with magnetic moments app 1000 times stronger than the proton magnetic moment
Gadolinium acts as a relaxation agent of nearby protons and reduces T1 significantly and T2 slightly.
120.- Gradient field stregth is measured in:
a) gauss/cmb) magnetic momentsc) megahertzd) MHz/Te) Ppm
Answer: a
Gradient strengths range from 1-10 mT/m (0.1-1 G/cm). The gradient field strength is on the order of 0.3-0.5
gauss/cm
121.- The dominaNT artifact affecting image quality, most often in the phase encoding direction is:
a) wrap aroundb) aliasingc) motiond) A & Be) RF zipper
Answer: c
The phase encoding gradient is applied with varying strengths throughout the acquisition; algorithms assume no motion. Since even a slight motion will cause a substantial amount of phase shift, most of the motion artifacts are manifested in the phase encode gradient direction.
Gibbs phenomenon (ring artifact) occurs near sharp boundaries, where high contrast transitions in the object occur. The ringing artifact appears as multiple, regularly spaced parallel bands of alternating bright and dark signal that slowly fade with distance. They are caused by the undersampling of the high spatial frequencies at sharp boundaries in the image. Ringing
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artifacts are more prominent for smaller digital matrix sizes. These artifacts can be reduced by increasing the matrix size for a given FOV.
Wrap around artifact-Aliasing. The wrap-around artifact is a result of mismapping of anatomy that lies outside the FOV but within the slice volume. The anatomy is usually displaced to the opposite side of the image.
Chemical shift artifacts are represented by a shift of fat to water mainly in the frequency encode gradient direction. Determination of fatty tissues can be made by swapping the phase encode gradient (PEG) and the frequency encode gradient (FEG) and examining the resultant shift of the tissues.
122.- Chemical shift artifact is most frequently manifested along:
a) the phase encode gradientb) the frequency encode gradientc) the slice select gradientd) A&Be) None of the above.
Answer: b
The chemical shift artifact is manifested due to the lower precessional frequency of protons in a lipid environment relative to a hydrated environment. The MR signal acquisition during the frequency readout gradient application cannot distinguish between protons resonating at a given frequency due to the gradient vewrsus those resonating at a slightly slower frequency due to chemical shift. The lipid part of the material shifts towards the weaker polarity of the applied gradient.
TRUE/FALSE;
123.- Ghost images are due to reinforced and cancelled partial copies of moving protons manifested in the frequency encode direction.
Answer: False
Ghost images are manifested in the phase encode gradient direction as partial copies of the anatomy. This occurs because the application of the PEG is divorced from the signal acquisition step (frequency encode gradient) and
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applied constantly throughout the signal acquisition process.
124.- Gibbs phenomenon refers to massive streaking across the image due to non-selective RF interference.
Answer: false.
Gibbs phenomenon (ring artifact) occurs near sharp boundaries, where high contrast transitions in the object occur. The ringing artifact appears as multiple, regularly spaced parallel bands of alternating bright and dark signal that slowly fade with distance. They are caused by the undersampling of the high spatial frequencies at sharp boundaries in the image. Ringing artifacts are more prominent for smaller digital matrix sizes. These artifacts can be reduced by increasing the matrix size for a given FOV.
125.- The time between the absorption of radiation energy and the expression of biological damage is referred to as the:
a) dominant periodb) latent periodc) G2 periodd) Interphase periode) Cell cycle period
Answer: latent period
126.- A barium enema was performed on a 25 year old female who was determined to be 3 weeks pregnant at the time of examination. As the consulting radiologist, you should:
a) recommend a therapeutic abortionb) counsel the patient that the embryo is at a significantly
high risk for gross malformations as a result of the radiation exposure; however, an abortion is not necessarily warranted.
c) Discuss the implications of the radiation exposure with the hospital’s legal department
d) Do not discuss any potential effects of the radiation exposure on the embryo because very little is known about in utero radiation exposure and your comment would be speculative
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e) Explain to the referring physician and patient that the radiation received by the embryo by this diagnostic procedure is relatively small and that the increase in risk is negligible compared to the spontaneous incidence of congenital abnormalities.
Answer: e
127.- Which of the following organ systems has the highest risk of a radiation induced fatal cancer?
a) thyroidb) female breastc) stomachd) bone marrowe) liver
Answer: stomach
The probability of fatal stomach cancer is 1.1 x 10-4 rem-1 compared to 0.08, 0.4, 0.5 and 0.15 x 10-4 rem –1 for the thyroid, female breast, bone marrow, and liver respectively. Not to be confused with detriment for which bone marrow (acute leukemia) has the highest risk coefficient owing largely to its short latency (avg of 10 years compared to 20 for most solid tumors) and low survival rate (< 5%).
128.- Which examination would cause the greatest radiation exposure to an embryo?
a) upper GIb) KUBc) IVPd) Pelvise) Hip
Answer: IVP
Radiation exposure to the embryo: IVP>KUB>Pelvis>Hip>Upper GI.
129.- The overall fatal cancer risk per rad of whole body low LET radiation of a population selected at random would be on the order of:
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a) 104
b) 102
c) 10-4
d) 10-6
e) 106
Answer: 10-4
130.- Match:
a) Non-stochastic (deterministic)b) Stochasticc) Relative riskd) Absolute riske) CNS syndromef) Cataract formation
With
1) Radiation induced genetic effects2) Gastrointestinal syndrome3) 5,000 rads whole body radiation exposure4) independent of age at time of exposure.
Answers: 1) b 2) a 3) a 4) d
Biological effects of radiation exposure can be classified as either stochastic or deterministic
A stochastic effect is one in which the probability of the effect, rather than its severity, increases with the dose. Radiation induced cancer and genetic effects are stochastic in nature. For example: the probability of radiation induced leukemia is greater after an exposure to 1 Gy (100 rad) than to 1 cGy (1rad) but there will be no difference in the severity of the disease if it occurs.
Stochastic effects are believed not to have a threshold because injury to a few cells or even a single cell could theoretically result in the production of the disease
If radiation exposure is very high, the predominant biological effect is cell killing. In this case, the severity of injury rather than its probability of occurrence, increases with dose.These so-called deterministic effects differ from stochastic effects in that they require much higher doses to produce an effect. There is also a threshold dose below which the effect is not seen. Cataracts, erythema, fibrosis and hematopoietic
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damage are some of the deterministic effects that can result from large radiation exposures.
Acute radiation syndrome: (Approximate dose that will produce prodromal symptoms in 50% of persons exposed 120 rads: anorexia 170 rads: nausea 210 rads : vomiting 240 rads: diarrhea hematopoietic syndrome: 50-1000 rad GI syndrome: above 1200 rad, usually 1,000-5,000 rads Neurovascular syndrome: > 5000 rads.
131.- Match:
a) High LETb) Low LETc) Half-lifed) Decay constante) Free radicals
With
1) 0.693/2) essentially independent of O2 tension3) neutron exposure4) produces direct effects from radiation exposure5) Dose from high energy beta particles, e.g. I-131.
Answer: 1)c 2) a 3) a 4) a 5) b
T1/2 = 0.693/ Interactions occur via the direct effect and are not
mediated by free radical production or oxygen enhancement Neutrons produce recoil protons which are high LET
particles The high ionization density produced by high LET
radiation has a higher probability of interacting via direct ionization of a target that via free radical production
Electrons, gamma and x-radiation all produce sparsely ionizing low LET radiation.
132.- Concerning irradiation in utero:
a) during the preimplantation period, gross congenital abnormalities are the expected response
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b) the number of days post-conception is important in predicting potential radiation injury because the radiation response will vary considerably with gestational age.
c) The fetal growth stage is the most sensitive period for radiation-induced gross congenital abnormalities
d) There is positive evidence of radiation induced genetic abnormalities in children born to A-bomb survivors.
Answers: a) F b) T c) F d) F
Organogenesis occurs between the 2nd and 6th week of pregnancy. Radiation exposure prior to this period typically results in either a spontaneous abortion or no observable abnormality.
Organogenesis is the most sensitive period for radiation induced congenital abnormalities. The organ systems which are most actively developing at the time of irradiation are at greatest risk.
While no evidence exists that indicates radiation exposure caused by any increase in genetic abnormalities, it is generally believed thay this is due to the number of people exposed and the low yield of abnormalities per unit radiation exposure compared to the naturally relatively high incidence.
133.- Which of the following statements on the linear attenuation coefficient for tissue in the diagnostic x-ray energy is true?
a) it decreases continuously with increasing energyb) it is expressed in units of grams per cm cubedc) it can be used to determine the mass attenuation
coefficient when its multiplied by the densityd) it is proportional to the tissue thickness.
Answer: a
Attenuation is the reduction of the intensity of an x-ray beam as it traverses matter.
An attenuation coefficient is a measure of the quantity of radiation attenuated by a given thickness of an absorbing material
The linear attenuation coefficient symbolized by is the fractional change in x-ray intensity per the thickness of the attenuating material because of interactions in a given material.
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= N/Nx The mass attenuation coefficient is obtained by dividing
the linear attenuation coefficient by the density of the material thfough which the photons pass and thus is represented by the symbol /.
As photon energy increases, the attenuation produced by a given thickness of absorber decreases.
134.- Which of the following statements best defines the HVL of an x-ray beam?a) it is equal to one half the linear attenuation
coefficientb) it is directly proportional to the mass absorption
coefficientc) it is inversely proportional to the linear attenuation
coefficientd) it is equal to twice the linear attenuation coefficient\
Answer: c
The HVL is defined as the thickness of a standard material that reduces the beam intensity to one half.
= 0.693 / HVL HVL = 0.693/ TVL=2.303/ Considering this equation, then HVL is inversely
proportional to the linear attenuation coefficient.
135.- Which of the following values represents the decrease in intensity that would result if 6 mm of aluminum filtration were added to a diagnostic x-ray beam with an HVL of 3 mm of aluminum?
a) Less than 25%b) Greater than 25%c) 25% d) 12.5%
Answer: b
If 3 mm is the HVL, this means that 3 mm reduces the intensity by 50%, an additional HVL will reduce it to 25% and a third one to 12.5% which means that the decrease in intensity will be greater than 25%.
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Remember that a decrease in intensity is a decrease in the number of photons of the beam.
136.- Which of the following statements best describes the effective energy of an x-ray beam?
a) it is proportional to the atomic number of the anode material
b) It is always less than the kilovolt peakc) It is not affected by added filtrationd) It is proportional to the mAs
Answer: b The effective energy of an x-ray beam is the energy of a
monoenergetic beam of photons that is attenuated at the same rate as the x-ray beam, in other words, that has the same HVL as the spectrum of photons in the beam. The effective energy is about 30-50% of peak energy.
137.- A diagnostic x-ray beam has 2mm of aluminum filtration. Which of the following statements best describes the x-ray beam that would result if 1 mm of the filtration were removed?
a) It would have a higher dose rate and greater HVLb) It would have a lower effective energyc) It would have an HVL of 1 mm of aluminumd) It would have the same HVL but a higher dose rate.
Answer: b
Diagnostic x-ray beams are polychromatic, and the mean energy is approximately 30-50% of the peak energy.
The increase in effective energy that occurs with increasing thickness of attenuating material is called beam hardening
Added filtration provides several advantages: It alters the shape of the x-ray specturm It causes a shift in the the effective energy of
the x-ray beam by selectively removing more low energy photons than higher energy photons
It reduces the intensity of the beam (decreases the total number of photons in the beam).
Increases the HVL of an x-ray beam
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Decreases patient exposure Improves image quality for a given dose A disadvantage of added filtration is that it
necessitates the increase of exposure factors to compensate for the reduction in intensity of the beam.
138. Which of the following values represents the fraction of the beam that is transmitted when a monoenergetic photon beam with a linear attenuation coefficient of 0.693/cm traverses 10 cm of attenuating material?
a) 0.50b) 0.10c) 0.37d) 0.69
Answer: a
139.- Which of the values for the HVL is correct if the linear attenuation coefficient is 0.55/cm?
a) 0.05b) 0.693c) 0.0347d) 13.86
Answer d
To solve this problem , the equation = 0.693 / HVL has to be applied. Solving for HVL, then HVL= 0.693/0.05 = 13.86
140.- Which of the following conditions will result in the second HVL of an x-ray beam being approximately equal to the first HVL?
a) the initial x-ray beam is monoenergeticb) the initial x-ray beam is polyenergeticc) the x-ray beam is less than 100 kVpd) the x-ray beam is less than 1.02 MeV
Answer: a
A monoenergetic beam is attenuated according to the exponential attenuation law.
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Thus, if the first HVL reduces the beam to one-half, a second HVL will reduce it by one half again to one-quarter.
With a monoenergetic beam, the first and second HVL’s are equal.
With a polychromatic beam, photons of low energy are attenuated more rapidly than photons of higher energy. The second HVL (the thickness required to reduce the penetration to one-quarter) is larger than the first HVL. The ratio of the two HVL’s ; first HVL/second HVL is called the homogeneity coefficient.
The homogeneity coefficient for a polychromatic beam is less than one.
141.- Which of the following expressions of the relationship between an HVL and a TVL is correct?
a) TVL = 3.3 HVLb) TVL = 2.3 HVLc) TVL = 0.693 HVLd) TVL = 0.5 HVL
Answer: a
142.- Which of the following factors does not control the HVL of an x-ray beam?
a) anode materialb) filtrationc) tube potentiald) mAs
Answer: d
143.- Compton scattering in soft tissues is the predominant mode of photon interaction for which of the following ranges of x-ray energy?
a) 10-100 keVb) 20-200 keVc) 30-300 keVd) 30-30000 keV
Answer: d
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X-ray photons undergoing Compton scattering do so by interacting with so-called free or valence shell electrons, in which the incident photon energy greatly exceeds the binding energy of the valence shell electron that is ejected.
Compton scattering predominates not only in the diagnostic energy range of x-rays in tissue (above 30 keV) but continues to predominate well beyond diagnostic energies of x-rays (to app. 30 MeV).
The probability of Compton scattering is proportional to the number of electrons per gram.
Once a Compton electron is ejected from the atom, it loses its kinetic energy through excitation and ionization of atoms in the surrounding material.
The Compton scattered photon, on the other hand, can traverse through the medium without interaction or may undergo any of a number of additional photon interactions, including a subsequent Compton scattering, photoelectric absorption or, if the photon energy is quite low, Rayleigh scattering.
144.- In a Compton scattering interaction, the wavelength of the scattered photon is ____ and its energy is ____ after large angle scatter compared with small angle scatter.
a) larger, lowerb) smaller , higherc) larger, higherd) unchanged, unchanged.
Answer: a In Compton interaction, the lower energy of the scattered
photon relative to the incident photon can alternatively be expressed as the increase in wavelength of the scattered photon relative to the wavelength of the incident photon.
The wavelength of the scattered photon depends on the angle of scatter relative to the trajectory of the incident photon.
The maximum energy transfer to the Compton electron (and thus the maximum reduction in incident photon energy) occurs with a 180 degree photon backscatter. In fact, the maximum energy of the scattered photon is limited to 511 keV, which occurs at a 90 degree scvattering angle and a maximum of 255 keV during a backscatter event.
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Even with maximal energy loss, the scattered photons still have a relatively high energy and thus a good probability of detection.
145.- For a given photon energy, the probability of photoelectric absorption in bone (effective Z=14) is ____ times greater than in fatty tissue (effective Z= 7)
a) 2b) 8c) 0.125d) 0.5
Answer: b
Photoelectric absorption is another mechanism of x-ray attenuation important to diagnostic imaging in which the incident photon interacts with a tightly bound electron (typically from the K or L shell). The x-ray photon is completely absorbed and the electron is ejected (now referred to as an ejected photoelectron) with a kinetic energy equal to the incident photon energy minus the binding energy of the ejected electron.
For photoelectric absorption to occur, the photon energy must be at least equal to or greater than the binding energy of the electron that is ejected.
146.- The abrupt increase in attenuation that is observed when the energy of photons just exceeds the binding energy of inner shell electrons is referred to as: _______ .
a) Compton plateaub) Absorption edgec) Lambda shiftd) Pair threshold.
Answer: b
The probability of photoelectric absorption increases dramatically with the atomic number of the absorber (i.e. proportional to Z3. Conversely, the probability of photoelectric absorption decreases dramatically with increasing incident photon energy( proportional to 1/E3).
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Thus, for a given absorbing material there is generally a rapid decrease in attenuation as photon energy is increased.
At photon energies equal to the binding energy of inner shell electrons, there is a rapid and dramatic increase in attenuation. This rapid increase is referred to as an absorption edge, at which point the number of electrons available for interaction dramatically increases, resulting in a rapoid rise in the attenuation cross section.
The phenomenon of the absorption edge is used in radiographic contrast agents such as iodine and barium. For these materials, the absorption edges of 33 and 37 keV create substantially increased values of x-ray attenuation relative to that surrounding tissues.
The high atomic number of these contrast agents also increases the probability of photoelectric absorption and decreases scattered radiation.
Photoelectric absorption is the primary mode of interaction of diagnostic x-rays with screen phosphors, contrast materials and bone.
147.- After the photoelectric absorption of an 80 keV photon with an atom whose K-,L- and M-shell binding energies are 60,10 and 2 keV respectively, one would expect K and Kcharacteristic x rays of ____ and ____ keV respectively.
a) 20 & 70b) 20 & 78c) 50 & 58d) 50 & 8
Answer: c
The nomenclature used to identify the characteristic x-rays is established such that the capital letter indicates the final destination of the cascading electron and subscripts Greek letter indicates whether the transition occurred from an adjacent or nonadjacent shell.
For example, an L characteristic x-ray indicates a transition from the M shell to the L shell, in which the subscript indicates that the origin of the cascading electron was the adjacent M shell. A K characteristic x-ray indicates an electron transition to the K shell from a nonadjacent shell.
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Taking this into consideration, the problem is solved as follows: the K characteristic x-ray has an energy equal to 60-10 (K & L electrons) and the K characteristic x-ray has an energy of 60-2.
148.- An alternative to characteristic x-ray emission from an atom with an unfilled inner electron orbital would be the emission of _____ which would have ___ energy than a characteristic x-ray.
a) Bremsstrahlung radiation, lessb) Bremsstrahlung radiation, morec) Auger electron, lessd) Auger electron, more
Answer: c Another form of energy dissipation is Auger electron
emission. In this process, the energy that otherwise would appear as a characteristic X-ray after an electron transition is transferred to a orbital electron whose binding energy is less than that of the characteristic x-ray and subsequently ejected.
The kinetic energy of the Auger electron is equal to that of the characteristic x ray minus the binding energy of the ejected electron.
149.- Which of the following relationships best describes the probability of photoelectric absorption?
a) Z/Eb) Z3/E3
c) E/Zd) E3/Z3
Answer: b
150.- Which of the following factors would not improve subject contrast in mammograms?
a) use of molybdenum targets in place of tungsten targetsb) breast compressionc) use of an aluminum filter in place of a molybdenum filterd) Lowering the tube voltage from 30 kVp to 26 kVp
Answer: c
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See page 464/Imaging and Therapeutic technology/ Radiographics.
151.- What is the ultimate fate of a positron produced during pair production?
a) it combines with an electron and converts their rest mass into energy
b) it combines with an electron, resulting in an electrically neutral atomic particle
c) it becomes part of the free electron poold) it is captured by the atomic nucleus of a nearby atom
Answer: a
In pair production, a high-energy photon, under the influence of the atomic nucleus is converted into a matter and antimatter pair, namely the elctron and positron.
The threshold photon energy required for this interaction is 1.02 MeV, which is equal to the rest mass energy equivalent of the positron-electron pair
The electron loses its kinetic energythrough excitation and ionization and becomes associated with another atom or is eventually absorbed into the free electron pool.
The positron (antimatter) also loses its kinetic energy by excitation and ionization, but its fate is much different than that of the elctron. The positron will eventually combine with an electron in annihilation reaction in which the rest mass energy of the positron-electron pair is completely converted into electromagnetic radiation in the form of two 511-keV photons.
These photons, emitted at approximately 180 deg to one another, are referred to as annihilation radiation.
152.- Which of the following x-ray interactions is predominant in CT of the liver performed at 120 kVp (62 effective energy)?
a) Rayleigh scatteringb) Compton scatteringc) Photoelectric absorptiond) Pair production
Answer: b
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153.- Which of the following techniques should be selected to maximize contrast in a radiograph?
a) the technique that uses the highest tube voltage possibleb) the technique that produces optical densities in the
shoulder portion of the characteristic curvec) the technique that produces optical densities in the toe
portion of the characteristic curved) the technique that produces optical densities in the
straight line portion of the characteristic curve
Answer: d
When the anatomic areas are properly exposed, the optical densities fall within the linear portion of the characteristic curve and the contrast is greatest.
If the anatomic areas are overexposed, the optical densities fall within the shoulder portion of the curve and contrast is reduced.
154.- Which of the following phrases best completes this sentence: ‘ A radiograph obtained with high tube voltage technique will ____ than a radiograph obtained with a low tube voltage technique’?
a) have lower contrastb) have higher contrastc) require a higher patient dosed) use a less penetrating x-ray beam.
Answer: a
155.- Which of the following factors has an effect on image blur?
a) Screen thicknessb) Magnificationc) Exposure timed) All of the above
Answer: d
All these factors are involved in the generation of image blur.
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156.- Which of the following statements is true when a radiograph is acquired with no magnification?
a) a large air gap is requiredb) there is no receptor blurc) the focal spot blur will be zerod) the total image blur will be zero
Answer: c
The focal spot blur in the image plane increases as the object is moved closer to the focal spot.
For the same magnification, the focal spot blur will increase as the focal spot size increases.
In addition to the focal spot size, the blur also depends on the magnification. When there is no magnification, the focal spot blur is zero.
Receptor blur is primarily caused by the spreading of light photons formed by x-rays interacting with the intensifying screen. Because the spreading of emitted light increases as the distance between the x-ray interaction and film increases, the amount of blur depends on the thickness of the screen phosphor layer. A thick, high speed screen has an inherent blur of app. 0.7 mm, whereas the blur of a thin, detail screen is 0.2-0.3 mm.
The receptor blur in the object plane will decrease as an object is magnified.
The total image blur in a radiograph is a composite of the focal spot blur and the receptor blur.
When little magnification is used, receptor blur dominates, when the magnification is large, the focal pot size becomes the major determining factor in the total image blur.
157.- Which of the following types of generator requires the longest exposure time?
a) single phaseb) three phasec) high frequencyd) constant potential
Answer: a
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The four basic types of generators are: single phase, three phase, high frequency and constant potential
The highest patioent exposure is needed when a single phase generator is used. This type of generator has 100% ripple.
Generator types with large voltage ripple require longer exposure times, which results in greater motion blur.
The total exposure time required when a single-phase generator used is longest of the four generator types. The single phase generator has 100% ripple Three phase six pulse generator: 13-25% Three phase 12 pulse: 3-10% High frequency generator: 4-15% Constant potential generator: No ripple
158.- Which of the following statements about use of a three-phase generator, compared with a single phase generator is true?
a) it requires greater patient exposureb) it exhibits lower voltage ripplec) it results in a greater likelihood of motion blurd) it requires a longer exposure time.
Answer: b The four basic types of generators are: single phase,
three phase, high frequency and constant potential The highest patioent exposure is needed when a single
phase generator is used. This type of generator has 100% ripple.
Generator types with large voltage ripple require longer exposure times, which results in greater motion blur.
The total exposure time required when a single-phase generator used is longest of the four generator types. The single phase generator has 100% ripple Three phase six pulse generator: 13-25% Three phase 12 pulse: 3-10% High frequency generator: 4-15% Constant potential generator: No ripple
159.- Which of the following statements about the results of grid use is false?
a) It increases image contrastb) It increases patient dose
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c) It reduces the number of scattered x-rays produced in the patient
d) It improves absorption of scattered radiation.
Answer: c
Grids absorb the scattered radiation coming from the patient. These are the most common method of reducing the level of scattered radiation reaching the image receptor.
The grid is placed between the patient and the image receptor.
An increase in contrast is achieved with the grids but at the expense of increased patient dose.
The ratio of exposure required with grid use and without grid use is called the Bucky factor
The Bucky factor is higher for higher ratio grids and higher energy exposures.
160.- Which of the following statements about the results of air gap technique is true?
a) it decreases image contrastb) it increases image contrastc) it does not change image contrastd) the effect of image contrast cannot be determined
Answer: b
Another method of reducing the level of scattered radiation that reaches the image receptor is tto place a gap between the patient and thereceptor.
The typical air-gap distance is 15-45 cm. Air gap increases image contrast.
161.- Which of the following phrases best completes this sentence: ‘To improve subject contrast when soft tissue is imaged, a low tube voltage technique should be selected to ________’
a) increase the number of photoelectric interactionsb) maximize the number of Compton interactionsc) better penetrate the tissued) decrease motion blur.
Answer: a
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The probability of photoelectric interactions increases as Z increases and decreases as energy increases.
Compton interactions are more likely to occur when a patient is imaged with a high energy x-ray beam (>40 keV). However, Compton interactions are not as effective as photoelectric interactions for distinguishing tissue types.
For soft tissue radiography, to maximize the photoelectric effect, a low-energy x-ray beam must be used.
162.- Which of the following factors does not affect subject contrast?
a) atomic numberb) magnificationc) tube voltaged) physical density.
Answer: b
Subject contrast is defined as the relative radiation intensities of the x-ray beam exiting the patient.
The subject contrast is larger is x-ray penetration through an object is much different from the penetration through adjacent background tissue.
The penetrability or penetrating power, is determined by the effective energy of the x-ray beam: Higher energy x-ray beams penetrate matter farther than lower energy beams do.
Because x-ray beam energy is directly affected by changing the tube voltage, the latter is a major factor in determining radiographic contrast.
163.- Atoms in unstable states because of a vacancy in an inner shell return to a more stable state by an outer shell-electron droping down to fill the vacancy, resulting in the emission of an:
a) x-ray or Auger electron b) Beta minus particlec) Gamma rayd) Neutrino
Answer: a
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164.- If the binding energies of the K,L and M shells of an atom are 80,10 and 2 keV, respectively, what are the energies of K-L shell transition characteristic x-rays and Auger electrons that result from de-excitation after a vacancy is created in the K-shell?
a) 78 and 76 keVb) 70 and 68 keVc) 70 and 8 keVd) 70 and 2 keV
Answer: b
165.- Isotopes have the same:
a) atomic numbersb) neutron numbersc) atomic mass numbersd) energy states
Answer: a
Isotopes have the same atomic number Isobars have the same atomic mass number Isotones have the same neutron number Isomers have identical characteristics except for the
energy states of the nuclei.
166.- A radionuclide with a high neutron to proton ratio will decay toward the line of stability by:
a) emitting a gamma rayb) emitting an alpha particlec) converting a neutron to a proton and emitting a beta
minus particle and an antineutrinod) converting a proton to a neutron and emitting a positron
and a neutrino.
Answer: c
If the proton number (i.e. atomic number Z) is plotted versus the neutron number (N) for the stable nuclides, the plot quickly deviates from the identity line, rapidly reaching a ratio of app. 1.5 neutrons for each proton.
The increased number of neutrons is necessary in nuclides of higher atomic number to separate the protons, which
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reduces the effects of the electromagnetic force of repulsion and permits the short-range strong forces to predominate.
Radionuclides are positioned on either side of this ‘line of stability’.
Neutron rich radionuclides, positioned on the left of the LOS, decay by Beta minus emission toward the line of stability by reducing the neutron to proton ratio.
Proton rich radionuclides, positioned on the right, decay by alpha or positron emission or electron capture toward the line of stability by increasing the neutron-proton ratio.
167.- Which of the following radiations will be emitted in an isobaric decay process involving only positron decay due to the annihilation of the positron with an electron?
a) Gamma raysb) Positronsc) Neutrinosd) 511 keV photons.
Answer: d
A radioactive atom may decay by one of seven basic decay processes and these may be grouped into three major categories. Alpha decay Isobaric processes Isomeric transitions or gamma emission.
In positron decay, a positively charged electron is emitted from the nucleus, resulting in a daughter with an anatomic number that is one less than that of the parent. In this transition, a proton is effectively converted to a neutron in the nucleus with the emission of matter (neutrino)-antimatter (positron) pair.
An example of positron decay is the decay of O-15 to N-15 The positron is an unstable particle; consequently, as
soon as it loses its kinetic energy, it immediately combines with a negatively charged electron and the masses of the two particles are converted into energy (1.022 meV), a reaction that is referred to as the annihilation process.
168.- In the electron capture process, the filling of the electron vacancy in the inner shell results in the
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production of x-rays or Auger electrons that are characteristic of the ___________.
a) parentb) daughterc) Both a and bd) Neither a or b
Answer: b
Electron capture is characterized by the absorption of an inner shell electron by the nucleus. The captured electron combines with a proton to form a neutron. Excess energy is carried off by the emission of a neutrino.
The result is a daughter with the same atomic mass number as the parent and an atomic number that is one less than that of the parent.
This process leaves the atom in an unstable state because of the vacancy created in an inner shell of the atom. The atom immediately returns to a more stable state by outer-shell electrons dropping down to fill the inner-shell vacancies, resulting in emission of an x-ray or Auger electron that is characteristic of the daughter.
An example of electron capture is the decay of thallium-201 to mercury-201.
169.- Nuclei in excited states usually return to the ground state by the emission of a:
a) x-ray or Auger electronb) Beta minus particlec) Gamma ray or internal conversion electrond) Neutrino.
Answer: c In decay processes, the parent may either decay directly
to the ground state of the daughter or to an excited state, which immediately de-excites by gamma emission.
In certain gamma ray emission, the gamma ray is not always emitted from the atom. Occasionally, the energy of the gamma ray is transferred to an inner shell electron, resulting in the ejection of the electron from the atom. This process is termed internal conversion and the ejected electron is known as an internal conversion electron.
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In certain gamma decay processes, the gamma ray is not emitted, but its energy is transferred to an inner shell electron and the electron is ejected from the atom. This process is called internal conversion. The atom is left in an unstable state and it immediately returns to a more stable state by an outer shell electron dropping down to fill the vacancy , resulting in characteristic x-ray or Auger electron emission.
In this way, internal conversion competes with gamma emission resulting in a decrease in the measured gamma ray intensity.
An example of internal conversion is the decay of I-123 to an excited state of tellurium-123
The decay scheme for Tc99m also demonstrates internal conversion. The excited state decays to the ground state by gamma emission with some of the gamma rays being internally converted, resulting in the emission of internal conversion electrons.
In medicine, radionuclides such as Tc99m that have a high ratio of gamma to electron emission (low internal conversion coefficients) are generally used because the internal conversion electrons only add to patient radiation dose.
169,- In an isomeric transition, the release of gamma rays causes the atomic mass number to: _____________.
a) decrease by 1b) Increase by 1c) Decrease by 4d) Remain unchanged
Answer: d
170.- In a radionuclide decaying by a single isomeric transition involving internal conversion, how does the presence of the internal conversion process affect the number of gamma rays detected?
a) It does not cause a changeb) It increases the number of gamma raysc) It decreases the number of gamma raysd) It does not cause a change;only the energy of gamma rays
is reduced.
Answer: c
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Internal conversion competes with gamma emission resulting in a decrease in the measured gamma ray intensity.
171.- Which of the following statements about the decay constant is true?
a) it gives the energy of the radioactive emissionsb) it is app. The same for all radionuclidesc) it is the probability of decay of a radionuclide per unit
timed) it is not related to half life
Answer: c
172.- Which of the following statements about radioactive decay is false?
a) One is not absolutely certain when a particular radioatom will decay
b) The behaviour of the ensemble of a large group of radioatoms is very predictable
c) All radioactive decay follows an exponential formd) All radioatoms decay at once
Answer: d
173.- Which of the following statements refers to specific activity ? a) It gives the energy of the radioactive emissionsb) It gives the activity per mass of a radionuclidec) It is the time when half of the radioatoms present will
have decayedd) It is not related to half life.
Answer: b
174.- What percentage of the initial radioisotope activity remains after three half lives?
a) 50%b) 33%c) 6.25%d) 12.5%
Answer: d
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175.- Which of the following statements about half-life is false?
a) it is inversely related to the probability of radionuclide decay
b) it can be used in the computation of specific activityc) it is the probability of decay of a radionuclide per unit
timed) it can be used to solve for activity remaining over time.
Answer: c
176.- Which of the following statements about average half-life of a radionuclide is true?
a) it is related to the half-life of the radionuclideb) app. 25% of the radionuclide remains after 2 average half
livesc) it is not related to the probability of decay of a
radionuclide per unit time. d) It is not related to the half-life of the radionuclide.
Answer: a
177.- The number of radioatoms per unit activity differs for radionuclides because of all the following factors except:
a) radionuclide half-life differsb) the atomic mass of the radionuclide differsc) the probability of decay of a radionuclide per unit time
differsd) Avogadro’s number is the same.
Answer: d
178.- Which of the following is a unit of activity?
a) Megabasketsb) Microcurdsc) Millikermasd) Gigabecquerels
Answer: d
179.- Which of the following statements about the secular equilibrium is false?
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a) The half life of the parent and daughter are nearly equalb) A constant supply of daughter radionuclide is producedc) It can only occur in natured) The half-life of the parent is much longer than that of
the daughter.
Answer: a
180.- Which of the following statements about transient equilibrium is true?
a) it is not related to the half-life of either the parent or the daughter
b) the half-life of the parent is shorter than that of the daughter
c) it will cause a nuclear explosiond) it can result in a slightly higher daughter activity when
equilibrium is reached.
Answer: d
CONCEPTS: Electrons flowing through a medium: Current Attraction or repulsion between two bodies: Force Mass or electromagnetic radiation: energy Restricts electric current: Resistance Energy expended per unit time: Power [Watts = Joules/sec] Mass on the atomic scale is measured in atomic mass units
(amu). One amu equals one twelfth the mass of a Carbon atom or 1.6 x 10-27 kg.
The charge of one electron is equal to -1.6 x 10 -19 C and the charge of proton is + 1.6 x 10-19 C .
An alpha particle has a charge of + 2. Electromagnetic radiation can travel in a vacuum. It
needs no medium to conduct or transport it . It travels at the speed of light (3 x 108m/sec).
The term force is used to describe the property of the environment that acts on an object. There are four types of force: Strong: holds nuclei together. Relative strength: 1 Electromagnetic/Electrostatic. Causes charged particles
to be attracted or repelled by one another. Binds electrons in atoms.
Weak: Interacts at subparticle level. Responsible for beta decay.
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Gravitational. Attraction between masses The weakest force.
Electrical charge can exist as a monopole: a positively charged object or a negatively charged object can exist independently. Magnetic materials exist in the form of a dipole and must have a north and south pole.
Electromagnetic radiation have the following frequencies: Radiowaves: 107
Microwaves: 1010
Visible light: 1014
Ultraviolet: 1016
X-rays:1018
Remember: The K-edge energy increases with increasing atomic number. For different materials: Hydrogen (z=1): 0.01 keV Oxygen (z=8): 0.5 keV Calcium (z=20): 4 keV Molybdenum (Z=42): 20 keV Iodine (Z=53) : 33 keV Barium (Z=56): 37.4 keV Tungsten (Z=74) : 69.5 keV Lead (Z=82) : 88.0 keV
An atom that has lost its outer shell is called an ion. Neutrons are not directly ionizing radiation. All
charged particles are directly ionizing; neutrons ionize by producing recoil protons. Photons ionize by generating electrons.
Ionizing particulate radiation: Beta rays Nonionizing particulate radiation: neutrinos Ionizing electromagnetic radiation: X-rays Nonionizing electromagnetic radiation: Radiofrequency. One curie is 3.7 x 1010 transformations per second. 1mCi= 37 MBq; 1 Ci = 37 kBq ATTENUATION OF RADIATION:
Linear attenuation coefficient () is the fraction of incident photons “lost” from the beam when traveling a unit distance. The attenuation coefficient accounts for all posible x-ray interactions including PE effect and Compton scatter.
The attenuation coefficient () is generally a function of photon energy (E), atomic number (Z) and density ().
The attenuation coefficient generally decreases with increasing photon energy and increases with atomic number and density.
An attenuation coefficient of 0.1/cm means that 10% of the incident photons are lost (absorbed or scattered)
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in traveling 1 cm, with the remaining 90% being transmitted.
The linear attenuation coefficient normally depends on the density of the absorbing material.
The mass attenuation coefficient is the linear attenuation coefficient divided by the density (), the mass attenuation coefficient is independent of the density of the material.
Filters preferentially absorb low-energy photons. Filtration does not affect the maximum energy of the x-ray beam spectrum.
Beam hardening: Beam hardening refers to the preferential loss of lower energy photons with filtration. The x-ray output is decreased with increased filtration but the average x-ray energy is increased. This is mainly a result of photoelectric effect.
Beam hardening does not occur with monochromatic x-ray beams because the HVL remains constant.
The quality of an x-ray beam is specified as the thickness of aluminum (mm) that reduces the x-ray beam intensity by 50%. At 80 kVp, the legal minimum x-ray beam HVL is 2.5 mm of aluminum.
Typical HVLs are: 0.3 mm aluminum in mammography (28kVp) 1.5 mm aluminum in xeromammography (45 kVp) 3.0 mm aluminum in conventional radiography (80 kVp)
Half value layers increase with increasing filtration. See page 41 Huda.
Remember: The photoelectric effect occurs between tightly bound (inner shell) electrons and incident x-ray photons.
The PE effect occurs when a photon is totally absorbed by an inner shell electron and a photoelectron is emitted. In addition to photoelectron emission, a positive atomic ion is formed. Scatter photons are not produced since the photon is completely absorbed. Characteristic radiation and Auger electrons may be produced.
The photoelectric effect is proportional to Z3.
Exposure is the ability of radiation to ionize air. Exposure is measured in Coulombs/kg (SI system) and Roentgens in the non- SI units. 1 R = 2.58 x 10-4 C/kg.
Kerma is the kinetic energy released in the medium. Kerma is specified in joules/kg.
Absorbed dose measures the amount of radiation energy absorbed per unit mass of absorbing medium. The absorbed dose is specified in gray in the SI system and rad in non-SI units. One gray is equal to 1 joule of energy deposited per kilogram. 1 Gy = 100 rad.
Absorbed dose is not source-related as is exposure ®.
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The f-factor is the conversion factor between exposure and absorbed dose. The relation between absorbed dose (D) and exposure (X) is: D=fX where f is the roentgen-to-rad conversion factor or f-factor.
LINEAR ENERGY TRANSFER. (LET) represents the energy absorbed by the medium per unit length of travel. For a given medium, LET is proportional to the square of the particle charge and is inversely related to particle kinetic energy. Thus, low speed particles with multiple charges such as slow-moving alpha particles have the highest LET values.
HIGH LET: Neutrons, protons, alpha particles and heavy ions. High LET radiations with values ranging from 3-200 keV/m
LOW LET: Photons, gamma rays, electrons and positrons, Low LET with values ranging from 0.2-3 keV/m.
High LET radiations are much more effective in producing biological damage.
DOSE EQUIVALENT: Attempts to quantify the biological damage arising from the deposition of ionizing radiation in tissues. The dose equivalent is the absorbed dose (D) multiplied by the quality factor (QF) of the radiation or
H= D x QF For low LET radiation sources (electrons, beta
partticles, x-rays, gamma rays) QF = 1. For high LET radiation sources (protons, neutrons, alpha particles, QF may be as high as 20. Dose equivalent is expressed in sievert (Sv) in the SI system and rem (radiation equivalent man) in non-SI units. 1 Sv = 100 rem; 1 rem = 10 mSv.
The maximum bremsstrahlung x-ray energy is equal to the energy of the incoming electron and is determined by the tube potential.
The maximum x-ray energy (90keV) is determined from the x-ray tube potential (90kVp).
The energetic electrons impinging on the target in an x-ray tube lose their energy by way of 3 basic processes: Radiation Ionization Excitation
For electrons in the diagnostic energy range (50-200 keV) and tungsten target, only 1% of the energy of the electron stream is converted to x-rays. This means that 99% of the energy goes into heating the anode.
Tungsten is an excellent target material due to its high melting point (3,370 C), and high atomic atomic number.
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The efficiency of x-ray production with the bremsstrahlung process increases dramatically with energy: At 4 MeV, app 40 % of the energy is converted to x-rays.
The energetic electrons are produced by thermionic emission from the negatively charged cathode and accelerated toward the positively charged anode. Modern x-ray tubes use rotating anodes to dissipate the heat over a wider area. The relatively thick targets produce x-rays isotropically (in all directions). A large number of x-rays are absorbed by the x-ray tube and casing.
High quality beams are more monoenergetic and have a greater percentage of higher-energy photons than beams of lower quality. Higher quality beams are better for imaging with x-rays. Higher quality beams produce a better image and impart a lower dose.
Parameters involved in the production of x-rays that affect the quantity and quality of the x-ray spectrum include: tube potential tube current exposure time distance anode material beam filtration generator type
The maximum photon energy is determined by the tube potential.
The x-ray quantity (area under the spectral curve) is directly proportional to the tube current and the exposure time or their product (mAs). Changing the tube current or exposure time has no effect on the shape (or quality) of the x-ray spectrum .
The x-ray quantity is directly proportional to the atomic number of the target material. The target material is fixed for most x-ray tubes and is typically an allow consistent of 90% tungsten (z=74) and 10% rhenium (z=75).
The target material determines the position (energy) of the characteristic x-rays and does influence the x-ray quality.
Beam hardening is the process by which an x-ray spectrum changes shape after passing through an attenuating material. Beam hardening increases the average energy of the beam, which increases the amount of material required
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to decrease the number of photons in the beam by a factor of two (HVL)
Beam hardening produces a more monoenergetic x-ray beam but it decreases the absolute number of photons in the beam.
The effect of beam hardening or filtration on the x-ray spectrum is to decrease the quantity while shifting the spectrum to higher energies. The maximum energy of the spectrum remains unchanged. Because the x-ray quantity increases with tube potential squared, the generator type (voltage waveform) affects the quantity and the quality of the x-ray spectrum.
RELATIONSHIP BETWEEN TRANSMITTANCE AND OPTICAL DENSITY. Transmittance: * 100% O.D.= 0.0
50% O.D = 0.3 25% O.D.= 0.6 10% O.D.= 1.0 5% O.D.= 1.3 1% O.D.= 2.0 0.1% O.D.= 3.0
Region of highest film density: OD between 1.0 and 2.0 Densities > 2.2 require the use of a hot light. High gradient (> 1.0) means that radiographic contrast is
amplified. Charged couple devices are used to detect light not x-
rays as in video camcorders. Devices to detect x-rays include ionization chambers, Geiger counters, thermoluminiscent dosimeter.
Ionization chambers are used to measure the total output of an x-ray tube. GM detectors are sensitive and are used to detect low levels of radioactive contamination. Scintillation crystal are used by gamma cameras in nuclear medicine. These detectors are used to determine the energy of incident gamma ray photons. The thermoluminiscent dosimeter measure the total energy absorption, not the energy of single photons.
Lithium fluoride (LiF) is the TLD used in diagnostic radiology because of its low atomic number (z=8.3). TLD can measure doses as low as 0.1 mGy (10 mrad) or as high as 10 Gy (1000 rad). TLD are frequently used to measure patient exposure during radiographic examinations and may be used for personnel dosimetry.
Radiation detectors that use photomultiplier tubes include: NaI crystal
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Thermoluminiscent dosimeter Photostimulable phosphor.
Quenching gases are used in Geiger-Muller counters to improve stability by minimizing the production of secondary discharges.
Scintillators are materials that emit light when exposed to radiation. The conversion efficiency (CE) is the percentage of absorbed energy that is converted to light.
SCINTILLATORS USED IN DIAGNOSTIC RADIOLOGY ZnCdS.Ag CE:18% Light: Blue & Green NaI 12% Blue (Nuc Med camera) CsI 10% Blue (Image
intensifier) La2O2S:Tb 12% Green (Rare earth screen)
Y2O2S:Tb 18% Blue (Rare earth screen) CaWo4 3.5% Blue (Old screen –pre-
70’s) Fast screen-film combinations allow for a decrease in
mAs, and this decrease in radiation exposure decreases patient dose.
Rare earth screens are faster than the calcium tungstate screens (CaWO4). They absorb more x-rays and are more efficient at providing light which allows the same film density to be obtained with less radiation. The patient dose decreases.
Grid ratio is h/D where h = height of the lead strip and D is the distance of the gap between the strips.
Scatter is undesirable in diagnostic radiology because it reduces subject contrast. Antiscatter grids are the most effective and practical method for removing scatter in diagnostic radiology.
High ratio grids will increase required mAs, image contrast, patient dose and they will remove scatter.
Grids primarily attenuate Compton scatter. Air gaps reduce scatter and thus improve image contrast.
Air gaps reduce scatter because the scattered photons are less likely to reach the screen/film receptor. With air gaps, magnification is introduced, a larger x-ray tube output is required and focal spot blurring is increased.
Typical grid usage: No grid: extremity radiography 4:1 mammography grid (60 lines/cm) 6:1 portable examinations (40 lines/cm)
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12:1 common Bucky grid. (30 lines/cm) Stationary grids with low ratios of 6:1 and about 45
lines/cm are used with mobile x-ray units because a low grid ratio tolerates beam misalignment. 12:1 grids are sensitive to misalignment problems. Grids used for portable examinations are 6:1.
Modulation transfer function is a curve that describes the resolution capability of an imaging system. The MTF is the ratio of output contrast to input contrast in an imaging system.
The modulation transfer function decreases if there is an increase in spatial frequency. Output contrast decreases as the spatial frequency increases.
At low spatial frequencies, the MTF is close to 1.0 At high spatial frequencies, the MTF falls to 0. Resolution is often expressed in line pairs per
millimeter which is a measure of spatial frequency. The modulation transfer function:
Describes the system resolution Compares image to object contrast Approaches 1.0 at low spatial frequencies Decreases with increasing spatial frequency
Image sharpness is affected by: Focal spot size Motion Screen thickness Screen-film contact
Geometric magnification can improve system spatial resolution. Geometric magnification does not require more radiation
at screen-film. Geometric magnification introduces an air-gap which
reduces the scatter. Geometric magnification increases focal spot blur but
reduces screen blur. Poor screen-film contact will result in a significant
loss of image detail. The gap between the screen and film will increase diffusion of light reading the film and increase image blur.
Thin screens have poor x-ray absorption efficiencies but excellent spatial resolution. A thin screen will most likely increase the spatial resolution of a screen-film combination.
Spatial resolution of different systems:
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Photospot film: 4-5 line pairs /mm 35 mm cine: 3.5 line pairs /mm DSA: 2.0 line pairs /mm CT scan: 0.7 line pairs /mm Screen-film: 5-10 line pairs /mm Mammography: 15-20 line pairs /mm Film alone: 25-100 line pairs /mm
When a rare-earth screen with 3 times the conversion efficiency and twice the absorption efficiency is substituted for a CaW04 screen, the patient dose would be expected to: be reduced by a factor of 6.
Remember that thicker screens (fast screens) have improved x-ray absorption efficiencies and require decreased exposure times (lower patient doses) but have inferior resolution.
The speed of an imaging system can be increased without increasing the noise by using: Phosphor with a higher absorption efficiency Thicker phosphor. ( A thicker screen has a higher
absorption efficiency, stops more of the incoming x-rays and reduces dose). Noise is not increased because the same number of x-rays are used to produce the image. If a more efficient phosphor is used (emits more light per x-ray photon absorbed), then fewer photons need to be absorbed in the screen to produce the desired optical density and then noise increases (fewer x-ray photons means more noise and vice-versa). As conversion efficiency increases so does quantum mottle.
Reducing the temperature of a film processor from 95-90 and keeping the film density constant, this will increase: Patient dose is increased because as temperature is
reduced, the film speed is reduced requiring greater exposure to produce the same amount of blackening.
Film contrast as opposed to subject contrast is primarily affected by: the film optical density level. Factors such as kVp, beam filtration, presence of contrast agents and tissue density differences affect subject contrast.
A detail screen compared to a regular screen has a higher: Spatial resolution.
Detail screens are thinner and therefore slower but have better spatial resolution.
Increase in kVp in screen-film radiography while maintaining a constant film density, will generally increase : the scatter. Scatter generally increases with kVp (Compton scatter increases).
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GI contrast is increased by: Infusion of barium Reduction of kVp Tight beam collimation Increase in grid ratio. (increasing the mA normally
will not increase image contrast). Image quality defects:
Abnormally dark films: Broken automatic exposure control (AEC)
Reduced image contrast: increased kVp Local area of decreased resolution: poor screen-film
contact Line artifacts: bad rollers
Malfunction and phototimed x-ray result: Blown x-ray tube: Film uniformly clear KVp too low: increased image contrast mA too high: film normal Broken phototimer: Film too dark. “When the phototimer
does not work, the x-ray exposure continues until the back-up timer switches the tube off and thus, the film will be over-exposed.
Image intensifier parts: Absorbs light and emits electrons: photocathode Absorbs x-rays and emits light: input phosphor Absorbs electrons and emits light: output screen Provides positive voltage: anode. The input image intensifier phosphor is made of: CsI The output phosphor is made of: ZnCdS
The brightness gain of an image intensifier tube does not depend on patient dose. Brightness gain is a measure of how much light you get out of the image intensifier for a given light level at the input phosphor and is the product of minification gain and flux gain. BG = MG x FG
Minification gain is the increase in image brightness that results from reduction in image size from the input phosphor to the output phosphor. MG = (di / do)2. di = input diameter do = output diameter
Flux gain is the increased number of light photons emitted from the output phosphor compared to the input phosphor. The FG is typically 50-100 so for each light photon emitted at theinput phosphor, there are 50-100 light photons emitted at the output phosphor.
Input phosphor: x-ray absorption and light generation (50 & 15% efficiency).
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Photocathode: electron emission: 10% Output phosphor: electron absorption (90%) Output phosphor: light generation: 10%. Typical values for modern image intensifiers:
Minification gain: 100 Flux gain: 50-100 Contrast ratio (ratio of periphery to central light
intensities when imaging a lead dic which is one-tenth the diameter of the input phosphor. A typical II contrast ratio is 20:1.
Brightness gain: 5000 Spatial resolution: 5 lp/mm Diameters: range from 23-57 cm
Operational parameters: Accelerating voltage: 25 kV Entrance exposure rate: 3 R/min Vignetting: 20% loss of light intensity at field edges
(typically less than 25%) Contrast resolution: 5 lp/mm
Remember: If an II diameter is reduced from 30-15 cm, the irradiated area is reduced by a factor of 4 and the exposure level must be increased fourfold to maintain the same brightness level at the II output phosphor.
The reason for interlacing two fields to form one frame in a conventional TV system is to reduce the flicker:
Vertical resolution of a standard north american TV is 0.7 x 262.5 line pairs. Only about 70% of the theoretical vertical resolution is achieved in TV systems. The ratio of measured to theoretical vertical resolution is called the Kell factor (normally 0.7).
Horizontal resolution is determined by the bandwidth. Vertical resolution of a TV system is primarily
determined by the number of TV lines. In 35 mm cardiac cine, for a constant film density,
patient entrance skin entrance skin exposure is reduced by increasing the: kV- increasing the kV increases patient penetration
and therefore reduces entrance skin dose for a constant II input level.
Resolution of viewing modes: 525-line TV = 1.2 lp/mm video-casette recorder = 1.0 lp/mm 1025-line TV = 2.5 lp/mm II output phosphor image = 5.0 lp/mm
Low contrast detectability in fluoroscopy images can be improved by increasing the grid ratio. Remember: the
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visibility of low contrast objects improves as scatter rejection improves, thereby increasing contrast.
Fluoroscopy and radiographic film/screen exposure use similar: kVp added x-ray beam filtration grids (grid ratios of about 10:1 for both) mA and exposure times are substantially different.
Input x-ray photons (quantum mottle) is the dominant noise.
TYPICAL EXPOSURES PER SINGLE FRAME: Conventional fluoroscopy: 2 R Screen-fil imaging: 300 R Photospot: 100 R Last image hold: 0 R
Digital subtraction angiography: Image intensifiers are used to create the images Low noise TV cameras should be used Analog to digital converters are used. Limiting resolution is about 2 lp/mm Image archival on disc is possible (DSA systems require
fast and high capacity disk drives up to about 50 Mbytes.
The most important component affecting spatial resolution in DSA is the digitation matrix. The digitization matrix is the primary determinant of DSA spatial resolution performance.
Comparing cine angio and DSA. Dose per frame: DSA is about 3 times higher than cine
(100 R vs 30 R). Spatial resolution: DSA has lower spatial resolution
(2.5 lp/mm vs 3.5 for cine.) Frame rates: lower for DSA (7.5 vs 30 frames/sec for
cine) Vessel visibility: DSA vessel visibility is better.
COMPUTERS 1 bit codes for 2 gray scale levels 8 bits = 1 byte 2 bytes = 1 word (16 bits) The shades of gray are given by 2n. For example: a total of 256 shades of gray are given by
1 byte (8 bits) 28. Computer hardware items and their performance:
RAM= storage device loses information when power goes out. Random Access Memory
Floppy disk= Can be erased if brought close to an MR scanner. It has a storage capacity of 1 MB
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CPU= performs logical and arithmetic operations. It is the central processing unit. It workjs under the control of software instruction.
Optical disk jukebox: The amount of image information generated by a radiology department in a year is about 1 terabyte and could be stored in an optical jukebox (20 GB-3TB).
ROM= Read only memory. Storage cannot be overwritten. It is for storage only.
Regarding digital computers: Hard disk: 20 MB-2 GB access time: 10 ms Magnetic tape: 600 MB- 5GB access time 10 sec—minutes. Optical disk: 600 mB- 10 GB access access time: 16 ms A byte is always 8 bits. A file is a collection of data treated as a unit. A microprocessor is a single integrated circuit and are
commonly known as chips. A modem is used to transmit information on telephone
lines. Input devices in a computer include: keyboard, trackball,
light pen, touch screens and joystick. Output devices: cathode ray tube, monitors, laser film, printers and paper printers.
Computers can communicate with each other using: Coaxial cables Telephone lines Fiber-optic cables Microwave links.
Components found in PACS are likely to include: Digital acquisition devices Digital archives Diagnostic display stations Patient data base Computers
PACS are also called: image management and communications (IMAC).
In CT scan, a projection is a profile of transmitted x-ray intensities through the patient at any given location of the x-ray tube, with up to 1000 projections acquired and used to reconstruct the CT image.
The image reconstruction algorithm that is most often used in current commercial CT scanners is: Filtered back projection.
In CT, the measured x-ray transmission values are called projections.
CT scan:
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1st generation: X-ray tube and detector were rotated 1 degree. It took aprox 5 minutes to generate an image.
2nd generation : translate-rotate 3rd generation: rotating fan beam and rotating detector 4th generation: rotating tube and fixed ring of
detectors. CT tubes have focal spots of approximately 1 mm The heavy filtration used with CT scanners typically
produces a beam with aluminum HVL of approximately 10 mm. Detectors.
The most common material used in solid state detectors is cadmium tungstenate (CdWO4).
Only solid state detectors are used for 4th generation scanners, which require thin detectors because of detector geometry.
Gas detectors: Xenon gas ionization detectors consist of a gas-filled chamber with anodes and cathodes maintained at a potential difference. Gas detectors can only be used in 3rd generation scanners. Gas detectors are more stable than solid-state detectors.
Most modern CT scanners make use of slip-ring technology in which high voltage is supplied to the tube through contact rings in the gantry. Most modern CT scanners use bow-tie filters to equalize the radiation level incident on the x-ray detectors, including helical scanners.
CT scanner spatial resolution improves with an increase of reconstruction matrix size. It is most likely to improve detectability of small high contrast objects.
MAs will have no effect on CT numbers, but will affect the precision with which they are measured.
Low contrast objects are difficult to see because of noise and increasing the mAs increases the number of photons used and hence decrease CT image noise.
CT noise: CT noise is determined primarily by the number of
photons to make an image (quantum mottle) Quantum mottle decreases as the number of photons
increases. CT noise is reduced by increasing kVp. Ct noise is also reduced by increasing voxel size
(decrease matrix size, increase field of view or increase section thickness.
Resolution: Typical CT scan resolution ranges from 0.7-1.5 lp/mm Axial resolution may be improved by operating in a high
resolution mode using a smaller FOV or a larger matrix size.
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Other factors that may improve CT spatial resolution include: smaller focal spot, smaller detectors and more projections.
Remember: Pixel size = Field of view / matrix size. A larger matrix size improves spatial resolution and
hence is likely to reduce volume averaging known as partial volume effect.
Nuclear Medicine
Radionuclide produced in a generator generally decay by a beta minus process:
Beta minus: A---A Z---Z+1 N---N - 1 Eliminate 1 electron and antineutrino. A neutron inside the nucleus is converted into a proton
and the excess energy is released as an energetic electron called Beta particle.
These are obtained by bombarding with neutrons in a reactor. Adding neutrons to nuclei results in excess neutrons which leads to a beta minus decay.
Cyclotron produced radionuclides generally decay by either a beta plus process or electron capture.
Bombarding with protons in a cyclotron results in a beta plus emission. Adding protons to nuclei requires them to lose their excess positive charge by positron emission or electron capture.
Beta plus: A ----- A Z ----- Z - 1 N ----- N + 1 Emits positrons and neutrinos.
A proton inside the nucleus is converted into a neutron. The excess energy is emitted as a positively charged electron called a positron and a neutrino. A positron is an electron with a unit positive charge. It interacts with matter like an electron.
Electron capture: A ------ A Z ------ Z- 1 N ------ N + 1 Emits neutrinos and x-rays. In electron capture a proton inside the nucleus is
converted into a neutron by capturing an electron from
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one of the atomic shells. A neutrino is emitted. Electron capture is more likely from the K-shell.
Both Beta plus and electron capture occur in nuclei with too few neutrons.
Electron capture radionuclides include: Co75, Ga 67, In 111, I 123, I125 and Tl 201.
Alpha emission: A ----- A-4 Z ----- Z-2 N ----- N-2 A radionuclide emits an alpha particle consisting of 2
neutrons and 2 protons. Alpha decay is most common in atoms with atomic numbers
greater than 82. Alpha particles have an energy of 4-7 MeV. The typical alpha emitter is 226 Ra which decays to form Radon.
Common radionuclides: 67Ga------cyclotron-----electron capture----78hr half
life 99mTc-----generator----isomeric transition--- 6 hr 111In-----cyclotron----electron capture----68 hr 123 I ----cyclotron----electron capture----13 hr 125 I-----reactor -----electron capture----60 days 131I------fission------beta decay ---------8 days 133 Xe-----fission-----beta decay----------5.3 days 201 Tl----cyclotron----electron capture----73 hr.
FORMULAS:1.- Velocity of a wave (v) is equal to the frequency (f) of the wave times its wavelength ()
v= f
2.- The decay constant is:
= 0.693/ T½3.- The relation between the linear attenuation coefficient () and HVL is :
HVL = 0.693/
4.- Mass attenutaion coefficient:
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Mass attenuation coefficient: /
The mass attenuation coefficient is the linear attenuation coefficient () divided by the density (). The mass attenuation coefficient is independent of density of material.
Chapter 4; Question 16. High LET radiations such as neutrons and alpha particles
produce tracks of densely clustered ionizations. The aim of a radiation protection standard is twofold: to
prevent deterministic (nonstochastic) health effects such as skin damage and cataracts and to limit stochastic health effects such as cancer and genetic effects.
High LET: neutrons, alpha particles Low LET: x ray, gamma ray, electorns STOCHASTIC EFFECT: No threshold. Genetic mutation, single
cell can be affected. Example: radiacion induced cancer, leukemia. It has three phases:
1. Initiation: change in the genome2. Promotion. The initiated cell undergoes the change
necessary for neoplastic transformation3. Progression: Successive changes in the cell give rise to
increasingly malignant subpopulations. NON-STOCHASTIC. Also called Deterministic effect. It is
the radiation induced changes associated with a certain dose. In other words, it does have a threshold. Examples: induction of cataracts, dermatitis, sterility, blood cell damage.
CHAPTER 5 : Non-medical exposure.
Radon equivalent dose in average per year in the US is: 24 mSv/y (2.4 rem/y).
Effective dose: 2 mSv. Radon is the largest significant contributor to human exposure to natural radiation. Radon produces highly ionizing alpha particles Rn-222.
An increased risk of lung cancer (small cell anaplastic) has been establsihed among uranium miners and other underground miners and is related to the inhalation of radon daughter products.
There is a synergistic effect between alpha particle radiation and cigarette smoking that enhances the lung cancer risk.
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Radon enters homes through the foundations = most common in basements.
Radon: The concentration should not exceed 300 Bq/m3 (8pCi/L).
This results in an annual dose of 14 mGy (1.4 rad) to bronchial epithelium. Assuming a radiation weighing factor of 20 for alpha particles, the equivalent dose to the bronchial epithelium would be 0.28 Sv/yr. The average indoor concentration of radon in US homes is 55 Bq/m3.
The US environmental protection agency recommends 150 Bq/m3.
The estimated lifetime risk of lung cancer death from chronic exposure to 300 Bq/m3 is about 2% based on a no-threshold linear risk model for high LET radiation.
The annual average effective dose to the US population from diagnostic medical radiation is estimated to be 0.53 mSv (53mrem).
The average annual effective dose to the global population from naturally occurring sources of ionizing radiation is 0.7-2.0 mSv (70-200 mrem).
The average annual whole body equivalent dose from radiations of cosmic origin range from about 0.2-0.6 mSv (20-60 mrem) with a gross average of about 0.3 mSv (30mrem).
Bananas are radioactive because they are a source of K-40. Radiation from these low LET sources account for 0.2 mSv/y (cosmogenic radionuclides).
Other than radon, the principal source of high LET radiation to humans is Po-210.
Chapter 6 Epidemiology.
Cancer mortality in the US exceeds 20% and is the second leading cause of death.
3 concepts of risk exist in epidemiology:1. Absolute risk2. Relative risk 3. Attributable risk
Attributable risk: fraction of affected individuals that is uniquely attributable to the exposure to the etiologic agent (ionizing radiation-cancer).
Relative risk is the ratio of incidence of disease or death in exposed and unexposed population.
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Absolute risk is the absolute occurrence of the disease that is directly attributable to the exposure to the etiologic agent. It is synonymous with excess risk.
Cohort study. Identifies the specific population by exposure to the causal agent (radiation). Studies groups of people exposed and not exposed to the stimulus.
The case-control study. Identifies the specific population by the disease (breast cancer in women) and studies groups of patients with and without the disease.
For most epidemiologic studies, a relative risk of greater than 1.5 would be necessary and may be sufficient to suggest a statistically significant association.
Atomic bomb survivors: The range of individual radiation absorbed doses extended
from 10 mGy (1 rad) to 6 gy (600 rads), with a mean whole body absorbed dose of 0.24 Gy (24 rad). The exposure was instantaneous.
Question:
What is the estimated excess number of fatal cancers induced in a population of 10,000 radiologists and technologists who have each received a whole-body dose of 10 mGy ( 1 rad) of low LET radiation?
Answer: 3 – 6.
The GI tract represents the principal organ system at risk for radiation exposure.
Radiation induced leukemia may have a latent period as short as 2 yr and runs its course in about 25 yr, although some risk may remain beyond that time.
Chapter 7.- Cancer mortality and radiation risk
22.5 % of all deaths in the USA are caused by cancer. The incidence in lung cancer has increased. On the other
hand, gastric and cervix cancer have decreased. In the USA, skin cancer is the only form of cancer for
which there is a higher incidence in whites than in blacks.
Factors that would be expected to influence the risk of cancer after a dose of 20 mGy (2 rads) of radiation. Age of the person at the time of exposure Linear energy transfer (LET) of the radiation Portion of the body exposed.
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The latent periods for solid cancers are in general, longer than for leukemia.
For cancers such as those of the breast, lung and probably stomach, the latent period appears to become longer the younger the age of exposure.
It has also become clear that exposure at a young age involves a greater risk of cancer than does exposure at older ages. For example: exposure at the age of about 20 years or below, carries the greatest risk of breast cancer, whereas the exposure at the age of about 45 or above appears to involve little or no risk.
The carcinogenic effect of high LET radiation is greater than that of low LET radiation.
Chapter 8 - Leukemias.
The most common acute leukemia in the pediatric population is the acute lymphoblastic leukemia, with a peak incidence of 2-6 yr of age.
Leukemias are much more common in the white population. ALL is more common in boys. Below 5 yr of age ANLL (acute
non-lymphocytic leukemia) is more common in girls. After 15 years of age, ANLL is more common in males.
Chronic myelogenous leukemia was the first human cancer shown to have a characteristic chromosomal aberration: a translocation resulting in the Philadelphia chromosome. This chromosome is found in about 90% of cases diagnosed as CML.
The translocation involves chromosomes 9 & 22 and the relocation of the c-abl oncogene.
Excess CML has been noted in populations exposed to radiation.
Radiation does not appear to increase the incidence of CLL.
Congenital disorders that have unstable chromosome patterns or show chromosome fragility are associated with an increased incidence of ANLL ( Fanconi’s anemia, Blooms syndrome [autosomal recessive w/ facial telangiectasia, photosensitivity and dwarfism], von Recklinghausen and others).
The genetic predisposition to leukemia is exemplified by the fact that an identical twin of a leukemia child has a 1 in 3 chance of developing leukemia. (this is 1000 times more likely than the iincidence in the general population of white children in the USA.
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The principal epidemiologic studies that demonstrate a positive relationship between radiation and incidence of leukemia include:
1. Atomic bomb survivors in Hiroshima and Nagasaki2. Women given radiotherapy for cervical cancer3. Individuals who received therapeutic radiation for
ankylosing spondylitis. In most worker populations exposed to radiation, the
death rates are lower than in the general population, this is a phenomenon known as the healthy-worker effect.
Exposure in utero during pelvimetry has been associated with an elevated incidence of childhood leukemias. Children exposed to less than 0.1 Gy (10rads) of atomic bomb radiation in utero, have not shown such increases in leukemia.
For leukemia, the minimum latent period is about 2-3 years. The probability of lifetime excess mortality from leukemia after exposure to 0.1 Gy (10rads) at a high dose rate is about 0.1%.
The excess incidence of all solid cancers is considerably greater than that of leukemias.
There are no dose-repsonse relationships for the individual types of leukemia but susceptibility to induction of CML appears to be the highest. There has been no increase in incidence of CLL following radiation exposure.
CHAPTER 9 THYROID CANCER.
I-131 was the original radioisotope used for medical scintigraphic imagign of the thyroid.
The isotope emits a high energy beta particle and has a half life of aprox. 8 days.
The agent of choice are now a shorter-lived radioisotope of iodine, I-123 which decays by electron capture and Tc-99m which decays by isomeric transition.
Data on medical applications of I-131 suggest the acute thyroid dose from shorter-lived radioisotopes may be more carcinogenic than the protracted dose form I-131.
Radiation-induced thyroid cancers have not been implicated in the group exposed to medically administered I-131.
Carcinomas are primarily of the papillary type, with follicular types second in prevalence and no induced anaplastic cancers.
Radiation induced benign adenomas and adenomatous nodules are about twice as likely as radiation induced cancers.
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Women have a 2-3 fold higher risk of radiation induced thyroid cancer than men do.
The relative risk for cancer is about 8.3 at 1Gy (100 rads) for the average population. The risk for adults is lower than for children by a factor of at least 2.
A threshold carcinogenic radiation dose cannot be established since doses as low as 0.90 mGy ( 9rad( to the thyroid have been estimated to be involved in radiation induced cancers.
Papillary carcinoma is the most common type of thyroid cancer and it is the most common radiation-induced thyroid cancer.
Evidence that either spontaneous or radiation induced benign adenomas undergo malignant transformation is lacking.
CHAPTER 10. LUNG CANCER More than 90% of lung cancers arise from the bronchial or
bronchoalveolar epithelium and the bronchial mucous glands. The major cancer types are squamous cell carcinoma, small cell carcinoma, adenocarcinoma and large cell anaplastic carcinoma.
Cigarette smoking is the principal cause of lung cancer, accounting for about 85% of all lung cancers in the US.
In the US, lung cancer is the leading cancer in men . Primary lung cancer represents about 15% of all cancer cases.
The lung cancer rates for both men and women in UTAH are lower than the overall rates in the US. The main reason for this is the proscription of the use of tobacco products by Mormons.
If a young patients (<10 yr) has received chest radiation up to 100 rads, the potential for developing lung cancer is:
Excess lung cancer has been observed in patients with ankylosing spondylitis who were exposed to x-ray therapy and in the Japanese atomic bomb survivors.
The excess lifetime mortality for lung cancer por 1000 persons instantaneously exposed to 1 Gy (100 rads) of low LET radiation is 5.0 based on an absolute risk model and 15.1 based on a relative risk model (UNSCEAR)
The latent period between radiation exposure and cancer development appears to be inversely related to the age at exposure and is a minimum of 5-10 yr.
Persons exposed as children appear to be less at risk than those exposed as young or middle-aged adults.
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Smoking habits are not expected to influence the radiation risk from X-rays.
Other high risk exposures include: Mustard gas, chloromethylethers, non-organic arsenical compounds, nickel dust and arsenic; chromate rubber and vinyl chloride industries.
CHAPTER 11: BREAST CANCER.
The risk factors for breast cancer include: Age at full-term pregnancy (early age has a protective
effect) Family history Race Estrogen levels in blood Early menarche Later than average age at cessation of the menses Breast feeding appears to have a protective effect. Increased breast cancer rates are associated with a
greater fat and animal protein consumption.
Radiation related risk in breast cancer has been associated with:1) women who had multiple fluoroscopic studies for pulmonary
TB2) women who were exposed to radiation from atomic bombs at
Hiroshima or Nagasaki3) women who received radiotherapy for post-partum mastitis,
For women who were irradiated, there is a latent period of about 15 yr, a linear dose-response relationship and a greater risk in women who were irradiated during adolescence.
No studies have implicated the low doses received during a mammographic examination with an increased risk for breast cancer.
Atomic bomb survivors showed an excess breast cancer rate. The latent period was at least 10 years. The increased risk occurred among women exposed to 10 rads or more. The greatest risk occurred at doses of about 100 rads (1 Gy) and for exposure during adolescence.
Latent periods for the appearance of breast cancer are variable. It appears to decrease with increasing age at exposure. The minimum is 5-10 years. There is no maximum latent period.
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CHAPTER 12. Mammography
A 35 yr old woman who undergoes a baseline mammogram and receives an absorbed dose of 2 mGy (200mrads) to the glandular tissue has a lifetime risk for radiation-induced breast cancer of 0.0004-0.002 %.
Atomic bomb data show a marked statistically significant increase in relative lifetime risk for women exposed to high doses (100rads) when younger than 40.
The baseline incidence of breast cancer in the US is about 3 times greater than in Japan due in part to certain types of fats in the diet.
In women who lack a family history of breast cancer, annual or biennial mammographic screenings are recommended for women 40 yr old and older. Biennical mommograms are not necessary for women under 40 if they lack a family history.
One of the reasons not to perform screening mammograms in younger women is that the risk of radiation induced cancer is highest for women under age 30. In addition, the spontaneous incidence of breast cancer is so low below this age that the detection rate would be too low to justify the screening.
Evidence that screening mammography is essential to reduce breast cancer mortality in women older than 40 has been demonstrated in the Health Insurance Plan of Greater New York (HIP) screening program.
This study demonstrated a 33% reduction in breast cancer mortality evident after 4 yr, which was attributed to early detection and surgery.
The ACR recommends that, for asymptomatic women, the first or baseline mammogram should be obtained by age 40. Initial screening at an earlier age is preferable when there is a personal history of breast cancer or a history of pre menopausal breast cancer in a patient’s mother or sister.
The American Cancer Society recommends that:1. A monthly breast self-examination should be performed
starting at age 20.2. Physical exam of the breast should be performed at 3 yr
intervals between ages 20-40 by a health professional3. A baseline mammogram should be obtained between 35-40
years of age.4. Annual or biennial mammograms should be obtained from
ages 40-49.5. Annual mammograms should be obtained from age 50 and on.
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The optimal combination of technical factors used for mammography of an average size breast include:
1. Molybdenum anode material2. 26 KVp3. HVL: 0.3 Al4. Receptor: Screen-film grid.
All other combinations are not well suited for the purpose of a good quality mammogram.
Screen-film systems differentiate soft tissues through by producing a high-contrast image.
Selenium plates (xeromammography) differentiate soft tissues through an edge-enhancement effect.
Geometric unsharpness and patient motion must be minimized to produce adequate spatial resolution. This requires the use of small focal spots and geometric distance between the x-ray source, the patient and the image receptor. The central axis of the x-ray beam should be positioned so that it is parallel to the chest wall.
Good mammographic technique also requires vigorous compression to reduce geometric blur, motion and patient dose.
Focal spots used for conventional mammographic systems typically range from 0.3-1.0 mm in actual size.
Screen-film systems require molybdenum targets with molybdenum filters to enhance the low energy characteristic x-rays produced from the molybbdenum target.
The characteristic x-rays range in the energy from 17-19 KeV. These low energies are necessary to produce sufficient soft tissue contrast on the film.
The breast gland dose typical of state of the art systems ranges from 1-5 mGy (100-500 mrads) for a 2 view breast study.
Adequate compression in mammography helps increase image sharpness (decreases geometric unsharpness)
NOTES ON BREAST COMPRESSION:1. The absorbed radiation dose is lower with breast
compression since the compressed breast is thinner, radiation penetrates more readily. This results in reduced radiation dose.
2. Compression does not change the physical density of breast structures.
3. Vigorous breast compression spreads the geometric configuration of fatty and glandular tissues and reduces the number of overlapping structures within the breast.
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4. Compression helps decrease the magnification by moving tissue closer to the image receptor.
5. For magnification imaging, where the breast is intentionally displaced from the image receptor, a very small focal spot must be used to decrease the geometric unsharpness.
6. By making the breast thinner and easier to penetrate, exposure time is reduced and thus, the likelihood of motion unsharpness is decreased.
7. Compression also holds breast in place and thus reduces motion unsharpness due to both patient motion and arterial rhythmic motion.
Chapter 13. Radiation Protection.
Q 45: When applying protection phylosophy to patients undergoing Dx radiographic examination:
1.- Exposures are to be restricted to patients considered to benefit medically from the study.2.- The radiologic examination must contribute to decisions on management.3.- the dose given should be the lowest practicable for satisfactory diagnosis. medical exposures are considered an important aspect of
population exposure, but for the individual patient they are excluded from dose limitation regulation for occupational or public exposures.
ALARA (as low as reasonable achievable).
Q. 46. Actions helpful in establishing an ALARA (as low as reasonably achievable program for the benefit of patients and personnel in a diagnostic radiology department include:
1. daily checks on the performance of dose calibrators.
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2. An investigation into the causes of abnormally high radiation badge reading
3. An ongoing analysis of the number of and reasons for films rejected as non-diagnostic.
Because cancer induction and genetic effects of radiation exposure are stochastic phenomena, it is assumed that some risk exists down to the lowest doses. Exposures to radiation must be justified on the basis of benefit and must be kept as low as reasonably achievable while remaining consistent with diagnostic accuracy. This is known as the ALARA principle.
Dual film badging is reasonably for fluoroscopists involved in interventional radiology procedures or female fluoroscopist of child bearing potential. To employ dual film badging in all workers is unnecessary, unproductive and not consistent with “reasonable” methods. Although increasing the kVp can reduce the entrance exposure to patients from diagnostic examinations, it may also degrade the quality of the image and the potential for reliable diagnosis. Use of a 90 kVp technique to perform a urographic study is inadequate because such radiation energies are not consistent with the attenuation characteristics of iodinated contrast agents.
Chapter 14. CATARACTS.
A 25 YEAR OLD WOMAN WITH NEUROLOGIC DISEASE REQUIRED A BRAIN ct SCAN INVOLVING 2 MM thick slices that inbcluded the orbits. This was followed by a cerebral angiogram. Her lifetime risk for radiation-induced cataracts is:
Answer: 0 %. A radiation induced cataract is considered a
deterministic effect. Deterministic effecects occur only after a threshold dose
is exceeded and the severity of the effect increases as the dose increases.
Acute exposures of patients undergoing radiotherapy to less than 2 Gy ( 200 rads) of X-rays generally cause no lens opacities. At the 2 Gy dose level, there may be a
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low incidence of small opacities that do not progress to impair vision.
Question 49: Regarding radiation-induced vision-impairing cataracts:
Answer: They occur after a single dose of 400 rads (4 Gy) of fast neutron irradiation.
The lens is avascular, nutrition depends on transport from the aqueous humor in the anterior chamber.
The term cataract describes a change in the translucency of the lens. To the patient, it appears as a barrier that covers the eye and reduces visual acuity but allows light through.
Radiation-induced cataracts develop in a characteristic sequence:
First changes occur in the posterior subcapsular area; a dot or a small number of dots or opacities can be seen with a slit lamp microscope.
With time, the dot enlarges and may become surrounded by vacuoles and granules.
Although a threshold of about 2 Gy (200 rads) of acute low LET radiation has been established for radiogenic cataracts, when the time over which radiation is given is increased, (fractionation regimens), the cataractogenic dose increases. i.e. if the exposure extends over a period of 3 weeks, the low LET is about 400 rad, if extended over more than 5 months, it is about 550 rads.
Once a radiation induced cataract is established, the probability that it will progress appears to be dose dependent. At doses above 600 rads, the probability that an opacity will progress with an eventual visual impairment is high.
Cataracts probably appear earlier and progress more rapidly when exposure is at a very young age.
The time between exposure and the detection of a cataract is inversely related to the dose. In the adult, it is usually 1 year or more.
The lens is unique among cell renewal systems in having no apparent mechanism for cell removal.
The risk of cataract from exposure to neutrons is greater than that for exposure to gamm or x –rays. In radiotherapy, patients treated with a total dose of 220 rads, radiation induced cataracts have been noted.
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Chapter 15. Occupational Exposure.
Q 49. A patient requires an emergency angiographic study of the abdomen. The only radiologist available to perform the procedure is pregnant (9weeks post-conception). Which is the most appropriate guideline for the radiologist?
Answer: She should perform the procedure in conformance with standard radiologic practice because the dose to the conceptus will be very small.
The national council on radiation protection and measurements (NCRP) currently recommends the following for pregnant women working with radiation:
1. The equivalent dose to the conceptus of the woman shall not exceed 5 mSv(0.5 rem) for the entire gestation.
2. Once pregnancy becomes known, the equivalent dose to the conceptus shall not exceed 0.5 mSv (0.05 rem) in any one month of gestation.
The sensitive period for radiation-induced malformation runs from approximately week 2 to app. Week 15 post-conception.
The radiation dose received by the conceptus of a pregnant angiographer can be kept well within the guidelines of the NRCP and ICRP as long as standard practices of radiation protection for angiographers are maintained.
Question 50. Regarding the dose to the unshielded thyroid of a radiologist during fluoroscopy and angiography:
Answer: The dose is at least five times greater than the dose to the shielded thyroid.
Doses reported in the literature are about 0.05-0.25 mGy ( 5-25 m rad) to the head and neck area per abdominal angiogram.
Thyroid shields will reduce the dose to 10% or less of the original level if they are 0.5 mm lead equivalent.
Question 52: Excess mortality from leukemia, aplastic anemia, oral cancer, lymphoma and skin cancer has been observed in radiologists in the United States who entered the profession before 1940. Excess mortality among radiologists who entered the profession after 1940 is still significant for:
Answer : Oral cancers.
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The only cancer that consistently shows a statistically significant excess in mortality for radiologists, is oral cancer.
The only other cancer from which there is excess mortality in radiologists who entered the field after 1940 is multiple myeloma.
Question 53.- The characteristics of all occupational radiation-induced cancers include:
Answer: There is a demonstrated dose-response relationship in the exposed populations.
For most radiation-induced tumors, the age at exposure appears to affect the risk. This is particularly true for young children.
No histologic types of tumors are pathognomonic of radiation exposure.
There appears to be a higher tumor incidence at higher absorbed dose levels of radiation.
Question 54. Regarding radiation-protective eye-wear used by radiologists:
Answer: Eye-wear with side-shields is more protective than conventional eye-wear designs.
The current ICRP and NCRP recommendations are that the lenses of the eyes not receive more than 0.15 Sv ( 15 rem) in any 1 year, which corresponds to 0.15 Gy ( 15 rads) of low LET radiation.
This is equivalent to a weekly schedule of 45 cardiac catheterizations, 15 intramedullary rod placements, 15 VCUG’s, 27 percutaneous nephrolithotomies or 90 neuro-angios.
Although the use of leaded apron is a requirement, the use of leaded glasses is voluntary.
There are no data to indicate that leaded eyewear reduced the incidence of cataracts in radiologists.
Chapter 16: X-ray exposure to technologists during fluoroscopy.
Question 55: During conventional fluoroscopy, the entrance dose rate to a patient is 30 mGy (3 rads) per minute. The x-ray tube is under the table and the radiologist has chosen
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not to use the leaded drapes. Which one BEST describes the dose rate in front of the lead apron to a technologist standing 1 m from the image intensifier?
Answer: 0.001-0.01 mGy/min ( 0.1-1.0 mrad/min).
When the leaded drapes are not used, the scattered radiation dose rate outside the lead apron of an individual standing 1 m from the irradiated volume of the patient would be only 0.01%-0.03% of the entrance dose to the patient.
In a patient receiving an entrance dose rate of 30 mGy (3 rads) per minute, the dose in front of the lead apron to a technologist standing 1 m from the image intensifier is 0.001-0.01 mGy/min (0.1-1.0 mrad/min).
Q 56. One of your fluoroscopic technologists is 3 months pregnant. Which one of the following actions is recommended by the National Council on radiation protection and measurements?
Answer: When assessing the total absorbed dose to the conceptus, take into consideration the shielding effects of protective aprons and the attenuation of overlying maternal tissues. In general, it has been recommended that the equivalent
dose to the conceptus BE KEPT BELOW 5 MsV (500 MREM) DURING THE ENTIRE GESTATION PERIOD.
When assessing the total absorbed dose to the conceptus, take into account the shielding effects of protective aprons and the attenuation by overlying maternal tissues.
Thermoluminiscent dosimeters often provide more accurate readings than do film badges.
Pocket dosimeter: major use: industrial radiography for accident management and in high radiation area where an instant readout is desirable.
A technologist whose conceptus is likely to receive more than 0.5 mSv (50 mrem) in any one month, should be transferred to another area.
Q 57 Your 3 month pregnant tech has had film-badge readings for the last 90 days showing a total equivalent dose of 6 mSv (600 mrem). The technologist wore her badge on her collar outside the lead apron at all times. Concerning this level of exposure:
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If a technologist has film-badge readings of 600 mrem and has wore her badge on the collar, outside the apron, we must understand that the dose to the product is 10-20 times smaller than the collar dose.
Conservatively assuming a factor of 10, the equivalent dose to the abdomen underneath the lead apron is 0.6 mSv ( 60 mrem). The overlying tissues would provide additional protection of at least a factor of 2 and therefore, the fetal dose would be less than 0.3 mSv (30mrem).
Chapter 17: Radiation exposure to the hands of Radiologists and technologists.
Q 58.- A radiologist who is often involved in percutaneous removal of renal calculi notes erythema on his dominant hand. Which of the comments is correct? A: If erythema due to chronic radiation exposure is present, epilation usually also occurs.
Typically, the entrance absorbed dose to the skin of a patient exposed to fluoroscopy are in the range of 30-40 mGy (3-4 rads/min.
Skin erythema may occur 24-48 hr after an absorbed dose of 2 Gy (200 rads) to the skin.
A more serious erythema may occur 8-9 days after an absorbed dose of 6-8 Gy (600-800 rads)
Chronic exposures to radiation usually have to be in excess of 50 Gy (5,000 rads) before vascular damage is identified.
The most recent recommendation of the NCRP and ICRP for the maximum permissible annual equivalent dose to the hands of radiation workers is 0.5 Sv (50 rem).
The most common types of radiogenic skin cancer are squamous cell and basal cell carcinoma.
Q 59. Which of the following observations on radiogenic skin cancer are correct?Answer: The radiocarcinogenic sensitivity of the hands is lower than that of the whole bone marrow or the lungs.
Ionizing radiation and ultraviolet radiation are synergistic carcinogens.
Induction of neoplasms is believed to be a stochastic effect; there is no well-defined threshold, and thus the lower the dose, the lower the risk.
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Occupational dose limits are related to risk. Exposure to the trunk of the body often exposes the thyroid, lungs, breasts and bone marrow, all of which are at a higher risk for radiogenic tumors than are the skin or bone. Thus, for the same level of risk for stochastic effects, the permissible dose to the hands is higher than that to the trunk.
Question 60: Concerning the absorbed radiation dose to the hands of a nuclear medicine technologist:
The use of unshielded thin-walled plastic syringes results in skin doses from Tc-99m ranging from about 0.15 to 0.30 mSv/MBq/hr (about 0.5 to 1 rem/mCi per hour).
The nuclear regulatory commission requires the use of a syringe shield except in special cases.
A thermoluminiscent hand dose monitor should be worn on the ring finger with the detector facing the palm of the hand. With this configuration, the badge will detect the maximum dose from a hand-held source, including doses due to low-energy x-rays and some high-energy beta rays. When the badge faces dorsally, much of the radiation can be attenuated by the finger, which results in too low a dose estimate.
Chapter 18 : Radionuclides and X-Rays outside Radiology Departments.
Question61.- A patient has been returned to the intensive care unit following pulmonary ventilation-perfusion scintigraphy. Xe-133 gas and Tc-99m macroaggregated albumin were the agents used. Concerning this procedure:
Answer: None of the options given are true. Maximum level for the public in one year: 5 mSv ( 500
mrem) Activities of radionuclides administered to patients for
diagnostic purposes are quite low. Exposures to medical personnel would typically be much less than 0.1 mSv ( 10 mrem) for patients undergoing diagnostic studies.
Since the average exposure rate at 1 m from the average patient is app. 0.2 mR/hr (0.002 mSv/hr), a nurse would have to stand within 1 m of 2500 patients for 1 hr immediately following injection to reach the curretn annual regulatory limit of 5 mSv (500 mrem) for members of the general public.
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Question 62. Concerning portable radionuclide imaging performed in the intensive care unit.
Answer: for radiation safety reasons, Tc-99m DTPA aerosol is
preferred over Xe –133 gas for pulmonary ventilation scintigraphy.
Tc-99m DTPA aerosols are prepared in a nebulizer. The patients inhales app. 25 MBq (700 microCi) during aerosol ventilation imaging performed before perfusion scintigraphy and about 100 MBq ( 3 mCi) for studies done after perfusion scintigraphy.
Xe-133 is an inert gas administered under positive pressure and the patient must inhale it and then exhale it. As it is exhaled it can be collected in a xenon trap which usually contains activated charcoal.
Although containment with a breathing apparatus sounds simple, it is actually very difficult to collect the xenon from patients who have difficulty breathing and who are coughing. Xenon gas permeates the environment and the lungs of personnel and patients will be exposed to the beta-emitting radionuclides.
In these cases (portable studies), it is preferable to use Tc-99 DTPA because this will minimize exposure to surrounding personnel and patients.
The amount of activity (radiation) received by a nurse standing within 1 m of a patient undergoing a portable study is usually in the range of 0.2 mR/hr. This does not represent a significant hazard.
Many radiopharmaceuticals are excreted in the urine: these include Tc-99m DTPA (whether given as aerosol or IV) and Tc-99m diphosphonates used for bone scintigraphy. There is no need to segregate the urine from these patients or to treat it ina manner different from tthat of other patients because exposure to personnel and patients from such radioactive waste is minimal.
Question 63:
A nurse is standing 1 m away from a patient getting a portable chest x-ray. The dose that the nurse received is:
Answer: 0.5 Sv (0.05 mrem).
in general, the radiation dose received by an individual standing 1 m from the body of a patient receiving a
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radiographic chest exam with a portable machine is about 0.1% of the entrance dose to the skin of the patient.
Entrance dose to patients are typically (0.3) mSv (.30 mrem) for portable chest radioigraphy. The absorbed dose to the individual standing 1 m from the patient is then about 0.3 Sv (0.03 mrem).
For the purposes of radiation protection, a rule of thumb is that individuals should remain 2 m from patients receiving mobile chest exams. Otherwisem they should be shielded with lead aprons.
Chapter 19: Absorbed doses from Radiography and CT.
Concerning skin entrance doses:
Chest ( 2 views) in average adult : < 2 mGy ( < 0.2 rad) Kidneys, average adult/urographic study: 20-100 mGy (2-10
rad) Heart in average adult cardiac catheterization 150-500
mGy (15-50 rads). Newborn abdomen (single view) < 2 mGy (< 0.2 rad)
Free in air exposure at skin entrance (ESE) is defined as the exposure measured at the position where the x-ray beam enters the patient. It does not include backscatter.
ESE is relevant to the potential for radiation-induced skin cancer at the site exposed, but they bear little relevance to the overall risk of cancer.
Skin cancer is generally not observed at cumulative doses of lower than 2 Gy (200 rads).
For radiography, the doses to organs at mid-depth in the average abdomen, are typically 10-25% of the skin dose.
For abdominal CT, the midline dose is typically 50-80% of the surface dose.
Skin entrance doses in the range of 0.60-1.0 Gy (60-100 rads) are expected during coronary angioplasty.
Cardiac catheterization is a high-dose exam. Technologically achievable floroscopic entrance dose rates are about 20-30 mGy/min of beam on time (2-3 rads/min). Typical entrance doses per cine frame range from 0.2-0.4 mGy/frame (20-40 mrads/frame). For a 7
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second single view run at 30 frames/sec, the dose would be about 60 mGy (6 rads).
For children and infants, the reported average skin dose during cardiac catheterization ranges from 3-30 rads.
Remember: Children are more susceptible to radiation-induced disease than are adults, perhaps by a factor of 2-10.
Question 68. A patient has a CT scan involving 20 contiguous 10 mm slices over the abdomen. The surface dose is 40 mGy (4 rads). It is decided to perform another study consisting of a seires of 5 mm contiguous scans over the kidneys, using twice the mA value as before. The radiation field width and the image field width are well matched. Which one of the following is closest to the total skin surface dose from both series of scans?
Answer : 120 mGy (12 rads)
In CT, the absorbed dose to the exposed tissues depends on the kVp, mAs (tube current and scan time). Geometry of the scanner, tube filtration, detector technology, collimation, spacing of slices, number of slices acquired and patient size.
KVp: the dose increases with the square of the kVp MAs: The dose increases linearly with increases in scan
time and tube current. Collimation: stringent collimation in CT is important to
eliminate unnecessary exposure to tissues outside the image slice volume of each scan. Optimal collimation also improves image quality by reducing scatter.
Patient size: Patient size is important since smaller patients receive higher doses for the same scanning techniques.
For pediatric CT studies, the scanning technique should be modified from that of adults to result in lower exposure.
Common CT doses: Abosrbed skin in CT typically range from 30-80 mGy (3-8 rads) for a series of 10 mm contiguous slices. Doses to internal organs are typically 50-80% of the skin dose and range from about 10-60 mGy (1-6 rads).
Overlapping slices produce higher doses than contiguopus slices.
As more slices are acquired, more tissue is exposed
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Question 69: During endoscopic retrgrade pancreatography, the entrance dose rate to a patient during a study confined to the biliary tree was ap.. 30 mGy (3 rads/min). The radiologist reset the timer 3 times during the procedure. Five spot radiographs, each involving an entrance dose of app. 5 mGy (500 mrads) were obtained during the study. Which one is the Closest to the total entrance dose received from this examination?
Answer: > 450 mGy (> 45 rads)
Entrance dose rates are typically about 20 mGy (2 rads) per mA/minute for conventional abdominal fluoroscopy.
Spot films add 25 mGy (2.5 rads) to the total dose, and so the total dose is in the range from 475-625 mGy (47.5-62.5 rads).
Question 70. The entrance dose rate to a patient undergoing gastrointestinal fluoroscopy is 20 mGy/minute (2 rads/min) in the 230 mm (9in) mode. The physician magnifies the fluoroscopic image by going to the 100 mm (4 in mode). Assuming that the brightness gain is adjusted solely by changing the tube current, which one of the following is closest to the new entrance dose rate?
Answer: 90-100 mGy/min (9-10 rads/min).
When magnification is used in fluoroscopy by converting to a field size of 100 mm (4 in diameter) there is a loss of brightness that must be recovered. This is done by increasing the radiation input to the image intensifier. Typically, this is achieved by changing the kVp, MA or both.
Changing the kVp instead of the mA helps minimize the entraNCE RADIATION DOSE TO THE PATIENT BUT RESULTS IN LOSS OF CONTRAST.
The loss in brightness due to loss of minification gain is proportional to the input area of the two field sizes.
The ratio of the 230 mm diameter field area to the 100 mm diameter field area is app. 5:1 and therefore, brightness is decreased by about a factor of 5. To compensate for this loss in brightness, the mA must be increased by the same factor.
The maximum regulatory free in air entrance exposure rate in the US is 10 R/min which is roughly 110 mGy/min or 11 rads/min.
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Question 71: For abdominal screen-film radiography, which one of the following is most likely to result in an increase in patient entrance exposure by a factor of about 4 in order to yield a properly exposed radiograph?
Answer: A patient with abdominal posteroanterior thickness of 300 mm is examined rather than a patient of 200 mm thickness. Higher beam energies produce more penetrating radiation.
Typically, for every increase of 10 kVp, the mAs must be reduced by a factor of 2. For a 20 kVp increase, the mAs would have to be reduced by a factor of 4 from that used for the lower kVp. Since higher kVp are more penetrating, this results in a lower entrance dose to the patient.
A faster film results in an image that has increased quantum mottle. By doubling the speed of the film, only half the radiation is required to produce a film of satisfactory optical density.
Patient size is one of the more critical factors in determining proper radiation output for radiography. For beam energies less than about 90 kVp, the amount of radiation doubles for every 40-50 mm of additional patient thickness. In this case, an increase of 100 mm would result in at least a factor of 4.
The interesting feature about reduced field size is that even though the entrance dose may be slightly higher, the potential patient risk is lower because much less tissue is exposed.
Collimation that is used to reduce the field size by a factor of 4 is likely to result in a slight increase in the entrance dose because the amount of scattered radiation is reduced in the image and a concomitant increase in output is required to maintain proper optical density.
Question 72: Effective means to reduce potential risks to the patient from ionizing radiation in a diagnostic study include:
1. use of PA projection for scoliosis films in female patients.
2. For pediatric studies, addition of a 0.2 mm copper filter to the x-ray beam and appropriate adjustment of contrast by changing the kVp.
3. Use of a fiber interspaced grid rather than an aluminum interspaced grid.
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4. Use of low-pulse-rate fluoroscopy with videodisc memory during pedaitric cardiac catheterization fluoroscopy.
The breasts and the thyroid are among the most radiosensitive organs for long-term cancer risk.
In scoliosis studies, obtaining films in PA position offers an advantage:
Since the breasts receive only an exit dose from the radiation during these studies, the breast doses are typically lower by nearly a factor of 10, than they would be if the studies were performed in the AP projection.
Further dose reductions and uniformly dark images may be obtained by using lead-acrylic shields, fast-screen-film combinations and app. Collimation and shielding. Gradient intensifying screens do not reduce the patient dose and are not preferred.
The use of a 0.2 mm copper filter with app. Adjustment of the kVp has been shown to be an effective means of reducing the entrance dose to pediatric patients.
Rare earth screen-film systems were introduced in the 1970’s and have resulted in patient dose reductions of 30-50% compared with calcium tungstate systems.
It is always important during fluoroscopy to keep the image intensifier as close to the patient as possible to improve the image quality and reduce the patient dose. A large air-gap between the patient and the image intensifier reduces the radiation intensity at the input of the image intensifier; the reduction is described by the inverse squre law.
Question 73: Taking into account the relative susceptibility for radiation-induced neoplasm and assuming a linear no-threshold response mode;, which one of the following examinations is associated with the HIGHEST risk of radiation-induced neoplasm ina 20 year old woman?
Answer: CT of the thorax (20 contiguous 10 mm slices from the sternal notch to the xiphoid, 120 kVp, 400 mAs, average-sized patient, third generation scanner).
Four of the organs that are most sensitive to radiation-induced neoplasm from low –LET radiations are the thyroid, female breast, bone marrow and lung. The GI tract is also considered sensitive. The one examination that delivers the highest dose to these organs is the CT examination (3-8 rads).
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Xeromammography on average, will deliver 3-25 mGy (0.3-2.5 rads) to the glandular tissue.
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