AART study guide

333. Particulate matter entering the respiratory bronchi can cause a. emphysema. b. empyema. c. pneumothorax. d. pneumoconiosis.

The answer is D. ii. EXPLANATION: Pneumoconiosis is a condition of the lungs characterized by particulate matter having been deposited in lung tissue; it sometimes results in emphysema. Overdistension of the alveoli with air is emphysema. The condition is often a result of many years of smoking and is characterized by dyspnea, especially when recumbent. Empyema is pus in the thoracic cavity; pneumothorax is air or gas in the pleural cavity.

157. What is the single most important factor controlling size distortion? a. Tube, part, IR alignment b. IR dimensions c. SID d. OID

The answer is D. ii. EXPLANATION: Shape distortion (foreshortening, elongation) is caused by improper alignment of the tube, part, and image receptor. Size distortion, or magnification, is caused by too great an object-image distance or too short a source-image distance. OID is the primary factor influencing magnification, followed by SID. (Bushong, 8th ed, p 284)

16. An increase in kilovoltage will have which of the following effects? a. More scattered radiation will be produced. b. The exposure rate will increase. c. Radiographic contrast will increase. i. 1 only ii. 1 and 2 only iii. 2 and 3 only iv. 1, 2, and 3

The answer is ii 2. EXPLANATION: An increase in kilovoltage (photon energy) will result in a greater number (i.e., exposure rate) of scattered photons (Compton interaction). These scattered photons carry no useful information and contribute to radiation fog, thus decreasing radiographic contrast. (Selman, 9th ed., p. 117)

247. According to the line-focus principle, an anode with a small angle provides 1. improved recorded detail. 2. improved heat capacity. 3. less heel effect. A. 1 and 2 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: The line-focus principle illustrates that as the target angle decreases, the effective focal spot decreases (providing improved recorded detail), but the actual area of electron interaction remains much larger (allowing for greater heat capacity). It must be remembered, however, that a steep (small) target angle increases the heel effect, and part coverage may be compromised. (Shephard, p. 219)

24. Which of the following devices converts electrical energy to mechanical energy? a. Motor b. Generator c. Stator d. Rotor

The answer is A. ii. EXPLANATION: A motor is the device used to convert electrical energy to mechanical energy. The stator and rotor are the two principal parts of an induction motor. A generator converts mechanical energy into electrical energy. (Selman, p 78)

315. The photoelectric effect is an interaction between an x-ray photon and a. an inner-shell electron b. an outer-shell electron c. a nucleus d. another photon

The answer is A. ii. EXPLANATION: In the photoelectric effect, a relatively low-energy incident photon uses all its energy to eject an inner-shell electron, leaving a vacancy. An electron from the next shell will drop to fill the vacancy, and a characteristic ray is given up in the transition. This type of interaction is more harmful to the patient because all the photon energy is transferred to tissue.

75. How would the introduction of a 6-in. OID affect image contrast? a. Contrast would be increased b. Contrast would be decreased c. Contrast would not change d. The scale of contrast would not change

The answer is A. ii. EXPLANATION: OID can affect contrast when it is used as an air gap. If a 6-in. air gap (OID) is introduced between the part and IR, much of the scattered radiation emitted from the body will not reach the IR, as shown in Figure 7-20. The OID thus is acting as a low-ratio grid and increasing image contrast. (Shephard, p. 205)

80. Cells concerned with the formation and repair of bone are a. Osteoblasts b. Osteoclasts c. Osteomas d. Osteons

The answer is A. ii. EXPLANATION: Osteoblasts are cells of mesodermal origin that are concerned with formation and repair of bone. Osteoclasts are cells concerned with the breakdown and resorption of old or dead bone. An osteoma is a benign bony tumor. An osteon is the microscopic unit of compact bone, consisting of a haversian canal and its surrounding lamellae. (Bontrager, p 745)

25. All the following are related to recorded detail except a. Milliamperage b. focal-spot size c. screen speed d. OID

The answer is A. ii. EXPLANATION: The focal-spot size selected will determine the amount of focal-spot, or geometric, blur produced in the image. Different screen speeds will create differing degrees of fluorescent light diffusion, affecting recorded detail. OID is responsible for image magnification and hence recorded detail. The milliamperage is unrelated to recorded detail; it affects only the quantity of x-ray photons produced and thus the radiographic density. (Selman, 9th ed., pp. 206-210)

331. The risk of inoculation with HIV is considered high for which of the following entry sites? 1. Broken skin 2. Perinatal exposure 3. Accidental needle stick A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: The overall chance that a person will become infected with HIV is high with entry sites such as the anus, broken skin, shared needles, infected blood products, and perinatal exposure. Low-risk entry methods include oral and nasal, conjunctiva, and accidental needle stick.

275. Which of the following body parts is (are) included in whole-body dose? 1. Gonads 2. Blood-forming organs 3. Extremities A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: Whole-body dose is calculated to include all the especially radiosensitive organs. The gonads, the lens of the eye, and the blood-forming organs are particularly radiosensitive. Some body parts, such as the skin and extremities, have a higher annual dose limit.

18. How should the wheelchair footrests be positioned as a patient is assisted into or out of a wheelchair? a. Accessible to the foot b. Moved aside c. Parallel to the floor d. Available for support

The answer is B. ii. EXPLANATION: When helping a patient into or out of a wheelchair, it must first be locked. Then, the footrests must be moved up and aside to prevent the patient from tripping over them or tilting the wheelchair forward. The wheelchair should be placed at a 45-degree angle with the x-ray table or bed, with the patient's stronger side closest to the x-ray table or bed. Once the patient is seated, the footrests should be lowered into place for the patient's comfort. (Adler and Carlton, 4th ed., p. 169)

204. Which projection(s) of the abdomen would be used to demonstrate pneumoperitoneum? 1. Right lateral Decubitus 2. Left lateral Decubitus 3. Upright A. 2 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: An erect abdomen or left lateral decubitus should be performed for demonstration of air-fluid levels in the abdomen. The right lateral decubitus position is used to demonstrate the layering of gallstones. It will not show free air within the peritoneum because of the overlying gastric bubble on the elevated left side of the body. (Bontrager, pp 107, 111)

126. Which of the following statements regarding film badges is (are) correct? 1. Film badges should be read quarterly. 2. Film badges must not leave the workplace. 3. Film badges measure quantity and quality of radiation exposure. A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: Film badges are supplied by a dosimetry service. They contain pieces of dental film held within a holder containing filters. When used properly, film badges measure the quantity and quality of radiation exposure. Film within the badges is usually changed monthly. The sensitive film emulsion is susceptible to deterioration and false readings if the badges are worn for longer periods, or if they are damaged by water, heat, light, and so on. To avoid the possibility of damage or exposure, film badges should not leave the workplace. (Ballinger & Frank, vol 1, p 49)

183. Shoulder arthrography is performed to 1. evaluate humeral luxation 2. demonstrate complete or partial rotator cuff tear 3. evaluate the glenoid labrum A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: Shoulder arthrograms (Figure 2-64) are used to evaluate rotator cuff tear, glenoid labrum (a ring of fibrocartilaginous tissue around the glenoid fossa), and frozen shoulder. Routine radiographs demonstrate arthritis, and the addition of a transthoracic humerus or scapular Y projection would be used to demonstrate luxation (dislocation). (Frank, Long, and Smith, 11th ed., vol. 2, p. 16)

37. The AP oblique projection (medial rotation) of the elbow demonstrates which of the following? 1. Radial head free of superimposition 2. Olecranon process within the olecranon fossa 3. Coronoid process free of superimposition A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: The AP oblique projection (medial rotation) of the elbow superimposes the radial head and neck on the proximal ulna. It demonstrates the olecranon process within the olecranon fossa, and it projects the coronoid process free of superimposition. The radial head is projected free of superimposition in the AP oblique projection (lateral rotation) of the elbow. (Saia, p 10)

103. All the following statements regarding CR IPs are true except a. IPs do not contain radiographic film. b. IPs use no intensifying screens. c. IPs must exclude all white light. d. IPs function to protect the PSP

The answer is C. ii. EXPLANATION: Externally, IPs (Image Plates) appear very much like traditional film-screen cassettes. However, the main function of an IP is to support and protect the PSP (SPS) that lies within the IP. IPs do not contain intensifying screens or film and, therefore, do not need to be lighttight. The photostimulable PSP is not affected by light. (Shephard, p. 51)

10. What is the minimum requirement for lead aprons, according to CFR 20? a. 0.05 mm Pb b. 0.50 mm Pb c. 0.25 mm Pb d. 1.0 mm Pb

The answer is C. ii. EXPLANATION: Lead aprons are secondary radiation barriers and must contain at least 0.25-mm Pb equivalent, usually in the form of lead-impregnated vinyl (according to CFR 20). Many radiology departments routinely use lead aprons containing 0.5 mm Pb (the NCRP recommends 0.5-mm Pb equivalent minimum). These aprons are heavier, but they attenuate a higher percentage of scattered radiation. (Bushong, 8th ed., p. 560)

279. Primary radiation barriers must be at least how high? a. 5 ft b. 6 ft c. 7 ft d. 8 ft

The answer is C. ii. EXPLANATION: Radiation protection guidelines have established that primary radiation barriers must be 7 ft high. Primary radiation barriers are walls that the primary beam might be directed toward. They usually contain 1.5 mm of lead (1/16 in.), but this may vary depending on use factor, and so on.

124. Below-diaphragm ribs are better demonstrated when a. respiration is suspended at the end of full inhalation. b. exposed using shallow breathing technique. c. the patient is in the recumbent position. d. the patient is in the AP erect position.

The answer is C. ii. EXPLANATION: The ribs below the diaphragm are best demonstrated with the diaphragm elevated. This is accomplished by placing the patient in a recumbent position and by taking the exposure at the end of exhalation. Conversely, the ribs above the diaphragm are best demonstrated with the diaphragm depressed. Placing the patient in the erect position and taking the exposure at the end of deep inspiration accomplishes this. (Frank, Long, and Smith, 11th ed., vol. 1, p. 490)

91. Which of the following factors impact(s) recorded detail? 1. Focal-spot size 2. Subject motion 3. SOD A. 1 and 2 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: Focal-spot size affects recorded detail by its effect on focal-spot blur: The larger the focal-spot size, the greater is the blur produced. Recorded detail is affected significantly by distance changes because of their effect on magnification. As SID increases and as OID decreases, magnification decreases and recorded detail increases. SOD is determined by subtracting OID from SID.

2. Which of the following is most useful for bone age evaluation? a. Lateral skull b. PA chest c. AP pelvis d. PA hand

The answer is D. ii. EXPLANATION: A PA projection of the left hand and wrist is obtained most often to evaluate skeletal maturation. These images are compared with standard normal images for the age and sex of the child. Additional supplemental images may be requested. (Bontrager and Lampignano, 6th ed., p. 654)

110. A film emulsion having wide latitude is likely to exhibit a. high density b. low density c. high contrast d. low contrast

The answer is D. ii. EXPLANATION: Every film emulsion has a characteristic curve representative of that film's speed, contrast, and latitude. A gentle curve (as opposed to a steep curve) usually indicates a film with slow speed, low contrast, and more latitude. (Shephard, p. 105)

227. A positive contrast agent 1. absorbs x-ray photons 2. results in a dark area on the radiograph 3. is composed of elements having low atomic number A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: Radiopaque contrast agents appear white on the finished image because many x-ray photons are absorbed. These are referred to positive contrast agents—composed of dense (i.e., high atomic number) material through which x-rays will not pass easily. Radiolucent contrast agents appear black on the finished image because x-ray photons pass through easily. An example of a radiolucent contrast agent is air. (Shephard, pp. 200-202)

146. Which of the following is (are) well demonstrated in the oblique position of the cervical vertebrae? 1. Intervertebral foramina 2. Disk spaces 3. Apophyseal joints A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: The cervical intervertebral foramina form a 45-degree angle with the MSP and, therefore, are well visualized in a 45-degree oblique position. Apophyseal joints are formed by articulating surfaces of the inferior articular facet of one vertebra with the superior articular facet of the vertebra below; they are well demonstrated in the lateral position of the cervical spine. The intervertebral disk spaces are best demonstrated in the lateral position. (Bontrager and Lampignano, 6th ed., p. 298)

32. The radiographer must perform which of the following procedures prior to entering a contact isolation room with a mobile x-ray unit? 1. Put on gown and gloves only. 2. Put on gown, gloves, mask, and cap. 3. Clean the mobile x-ray unit. A. 1 only B. 2 only C. 1 and 3 only D. 2 and 3 only

The answer is A. 2. EXPLANATION: When performing bedside radiography in a contact isolation room, the radiographer should wear a gown and gloves. The cassettes are prepared for the examination by placing a pillowcase over them to protect them from contamination. Whenever possible, one person should manipulate the mobile unit and remain "clean," whereas the other handles the patient. The mobile unit should be cleaned with a disinfectant on exiting the patient's room, not prior to entering.

214. Which of the following conditions would require an increase in exposure factors? 1. Congestive heart failure 2. Pleural effusion 3. Emphysema A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: Emphysema is abnormal distention of the alveoli (or tissue spaces) with air. The presence of abnormal amounts of air makes it necessary to decrease from normal exposure factors. Congestive heart failure and pleural effusion involve abnormal amounts of fluid in the chest and thus require an increase in exposure factors. (Carlton & Adler, p 258)

201. Which of the following affects both radiographic density and intensifying screen speed? 1. Thickness of phosphor layer 2. Type of phosphors used 3. Thickness of spongy screen support A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: Factors that affect screen speed also will affect radiographic density. Rare earth-type phosphors absorb x-rays more efficiently and convert their energy into fluorescent light; therefore, they affect both screen speed and radiographic density. The thickness of the phosphor layer affects speed and density similarly. As the thickness of the phosphor layer increases, speed and density increase. The spongy layer behind each intensifying screen helps to ensure good screen-film contact and, therefore, good recorded detail. The spongy layer is unrelated to radiographic density. (Shephard, p. 68)

329. Advantages of battery-powered mobile x-ray units include their 1. ability to store a large quantity of energy 2. ability to store energy for extended periods of time 3. lightness and ease of maneuverability A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: There are two main types of mobile x-ray equipment—capacitor-discharge and battery-powered. Although capacitor-discharge units are light, and therefore fairly easy to maneuver, the battery-powered mobile unit is very heavy (largely because it carries its heavy-duty power source). It is, however, capable of storing a large milliampere-seconds capacity for extended periods of time. These units frequently have a capacity of 10,000 mAs, with 12 hours required for a full charge.

228. Exposed silver halide crystals are changed to black metallic silver by the a. preservative b. reducers c. activators d. hardener

The answer is B. ii. EXPLANATION: As the film emulsion is exposed to light or x-rays, latent image formation takes place. The exposed silver halide crystals are reduced to black metallic silver in the developer solution. Automatic processor developer agents are hydroquinone and phenidone. The preservative—sodium sulfite—helps to prevent oxidation. The activator provides the necessary alkalinity for the developer solution, and hardener is added to the developer in automatic processing to keep emulsion swelling to a minimum. (Fauber, p. 164)

79. An unexposed and processed film will have a density of about a. Zero b. 0.1 c. 1.0 d. 2.5

The answer is B. ii. EXPLANATION: Film that is unexposed and has been processed will not be completely clear. The blue-tinted base contributes a small measure of density. A small but measurable amount of exposure from background radiation also can be present, and processing itself produces a small amount of density from chemical fog. Together, this is expressed as base-plus fog and should never exceed a density of 0.2. (Fauber, p. 198)

125. Which of the following will be demonstrated best in the 45-degree right anterior oblique (RAO) position? a. Right axillary ribs b. Left axillary ribs c. Sternum in the heart shadow d. Left scapular Y

The answer is B. ii. EXPLANATION: The axillary portions of ribs are demonstrated in a 45-degree oblique position. In order to place the axillary portions parallel to the image receptor (IR), the affected side is away from the IR in the PA oblique (RAO and LAO) positions and toward the IR in the AP oblique (RPO and LPO) positions. Radiography of the sternum, in the slight RAO position, requires greater obliquity for thinner patients and lesser obliquity for thicker patients. The scapular Y position of the shoulder is performed to demonstrate dislocation and requires a rotation of 45-60 degrees, with the affected side closest to the IR. (Frank, Long, and Smith, vol. 1, pp. 494-495)

13. The floor of the cranium includes all the following bones except a. the temporal bones b. the occipital bone c. the ethmoid bone d. the sphenoid bone

The answer is B. ii. EXPLANATION: The skull is divided into two parts—the cranial bones and the facial bones. There are eight cranial bones. Four of them comprise the calvarium—the frontal, the two parietals, and the occipital. The bones that comprise the floor of the cranium are the two temporals, the ethmoid, and the sphenoid. (Bontrager and Lampignano, 6th ed., p. 368)

221. Myelography is a diagnostic examination used to demonstrate a. internal disk lesions. b. posttraumatic swelling of the spinal cord. c. posterior disk herniation. i. 1 only ii. 2 only iii. 2 and 3 only iv. 1, 2, and 3

The answer is C. 2. EXPLANATION: Myelography is used to demonstrate encroachment on and compression of the spinal cord as a result of disk herniation, tumor growth, or posttraumatic swelling of the cord. This is accomplished by placing positive or negative contrast medium into the subarachnoid space. Myelography will demonstrate posterior protrusion of herniated intervertebral disks or spinal cord tumors. Anterior protrusion of a herniated intervertebral disk does not impinge on the spinal cord and is not demonstrated in myelography. Internal disk lesions can be demonstrated only by injecting contrast medium into the individual disks (diskography). (Bontrager and Lampignano, 6th ed., p. 762)

179. All the following x-ray circuit devices are located between the incoming power supply and the primary coil of the high-voltage transformer except a. the circuit breaker. b. the kilovoltage selector. c. the rectifiers. d. the autotransformer.

The answer is C. ii. EXPLANATION: All circuit devices located before the primary coil of the high-voltage transformer are said to be on the primary, or low-voltage, side of the x-ray circuit. The timer, circuit breaker, autotransformer, kilovoltage selector switch, and (prereading) kilovoltage meter are all located in the low-voltage circuit. The rectifiers, however, are placed after the secondary coil of the high-voltage transformer and before the x-ray tube. (Selman, pp. 150-151)

223. Radiation output from a diagnostic x-ray tube is measured in which of the following units of measurement? a. Rad b. Rem c. Roentgen d. Becqueral

The answer is C. ii. EXPLANATION: As x-ray photons emerge from the x-ray tube they immediately encounter air—before being intercepted by any material. The roentgen is the unit of exposure; it measures the quantity of ionizations in air. The roentgen is, therefore, the unit of choice for measuring x-ray tube output—and an ion-chamber dosimeter instrument is used for this purpose. Rad is an acronym for radiation absorbed dose; it measures the energy deposited in any material. Rem is an acronym for radiation-equivalent man; it includes the relative biologic effectiveness. Becqueral is the SI unit of measurement for radioactivity. (Bushong, 8th ed., p. 588)

191. Which of the following positions would best demonstrate the proximal tibiofibular articulation? a. AP b. 90 degrees mediolateral c. 45-degree internal rotation d. 45-degree external rotation

The answer is C. ii. EXPLANATION: In the AP projection, the proximal fibula is at least partially superimposed on the lateral tibial condyle. Medial rotation of 45 degrees will "open" the proximal tibiofibular articulation. Lateral rotation will obscure the articulation even more. (Frank, Long, and Smith, 11th ed., vol. 1, p. 289)

66. How much protection is provided from a 75-kVp x-ray beam when using a 0.50-mm lead equivalent apron? a. 51% b. 66% c. 88% d. 99%

The answer is C. ii. EXPLANATION: Lead aprons are worn by occupationally exposed individuals during fluoroscopic procedures. Lead aprons are available with various lead equivalents; 0.25, 0.5, and 1.0 mm are the most common. The 1.0-mm lead equivalent apron will provide close to 100% protection at most kVp levels, but it is rarely used because it weighs anywhere from 12 to 24 lb. A 0.25-mm lead equivalent apron will attenuate about 97% of a 50-kVp x-ray beam, 66% of a 75-kVp beam, and 51% of a 100-kVp beam. A 0.5-mm apron will attenuate about 99% of a 50-kVp beam, 88% of a 75-kVp beam, and 75% of a 100-kVp beam. (Thompson et al, p 457)

299. Which of the following pathologic conditions require(s) a decrease in exposure factors? 1. Pneumothorax 2. Emphysema 3. Multiple myeloma A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: All three pathologic conditions involve processes that render tissues more easily penetrated by the x-ray beam. Pneumothorax is a collection of air or gas in the pleural cavity. Emphysema is a chronic pulmonary disease characterized by an increase in the size of the air-containing terminal bronchioles. These two conditions add air to the tissues, making them more easily penetrated. Multiple myeloma is a condition characterized by infiltration and destruction of bone and marrow. Each of these conditions requires that factors be decreased from the normal to avoid overexposure.

290. The cardiopulmonary resuscitation (CPR) procedure for infants differs from that of adults with respect to 1. hand placement. 2. number of compressions. 3. volume of air delivered. A. 1 and 2 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: In CPR for infants, the rescuer places his or her index finger on the sternum just below the level of the intermammary line. The second and third fingers are used to compress the sternum 1/2 to 1 in. during compressions. The recommended rate is 100 compressions per minute. The volume of air delivered during ventilation should be just enough to make the infants chest rise and fall.

38. Using a 48-in. SID, how much OID must be introduced to magnify an object two times? a. 8-in. OID b. 12-in. OID c. 16-in. OID d. 24-in. OID

The answer is D. ii. EXPLANATION: Magnification radiography may be used to delineate a suspected hairline fracture or to enlarge tiny, contrast-filled blood vessels. It also has application in mammography. To magnify an object to twice its actual size, the part must be placed midway between the focal spot and the IR. (Selman, pp. 223-225; Shephard, pp. 229-231)

298. Which of the following waveforms has the lowest percentage voltage ripple? a. Single-phase b. Three-phase, six-pulse c. Three-phase, 12-pulse d. High-frequency

The answer is D. ii. EXPLANATION: Single-phase current has a 100% voltage drop between peak voltages. Three-phase current decreases this voltage drop considerably. Three-phase, six-pulse current has about a 13% voltage drop between peak voltages, and three-phase, 12-pulse current has only about a 4% drop between peak voltages. However, high-frequency current is almost constant potential, having less than 1% voltage ripple

300. At what level do the carotid arteries bifurcate? a. Foramen magnum b. Trachea c. Pharynx d. C4

The answer is D. ii. EXPLANATION: The common carotid arteries function to supply oxygenated blood to the head and neck. Major branches of the common carotid arteries (internal carotids) function to supply the anterior brain, whereas the posterior brain is supplied by the vertebral arteries (branches of the subclavian). The carotid arteries bifurcate into internal and external carotid arteries at the level of C4. The foramen magnum and pharynx are superior to the level of bifurcation, and the larynx is inferior to the level of bifurcation.

241. If a radiograph exposed using a 12:1 ratio grid exhibits a loss of density at its lateral edges, it is probably because the a. SID was too great b. grid failed to move during the exposure c. x-ray tube was angled in the direction of the lead strips d. central ray was off-center

The answer is A. ii. EXPLANATION: If the SID is above or below the recommended focusing distance, the primary beam will not coincide with the angled lead strips at the lateral edges. Consequently, there will be absorption of the primary beam, termed grid cutoff. If the grid failed to move during the exposure, there would be grid lines throughout. Central ray angulation in the direction of the lead strips is appropriate and will not cause grid cutoff. If the central ray were off-center, there would be uniform loss of density. (Carlton and Adler, 4th ed., p. 260)

212. Which of the following groups of exposure factors will produce the shortest scale of contrast? a. 200 mA, 0.25 s, 70 kVp, 12:1 grid b. 500 mA, 0.10 s, 90 kVp, 8:1 grid c. 400 mA, 0.125 s, 80 kVp, 12:1 grid d. 300 mA, 0.16 s, 70 kVp, 8:1 grid

The answer is A. ii. EXPLANATION: Of the given factors, kilovoltage and grid ratio will have a significant effect on the scale of radiographic contrast. The milliampere-seconds values are almost identical. Because decreased kilovoltage and high-ratio grid combination would allow the least amount of scattered radiation to reach the IR, thereby producing fewer gray tones, (A) is the best answer. Group (D) also uses low kilovoltage, but the grid ratio is lower, thereby allowing more scatter to reach the IR and producing more gray tones. (Shephard, p. 308)

52. Which blood vessels are best suited for determination of pulse rate? a. Superficial arteries b. Deep arteries c. Superficial veins d. Deep veins

The answer is A. ii. EXPLANATION: Superficial arteries are best suited for determination of pulse rate. The five most easily palpated pulse points are the radial, carotid, temporal, femoral, and popliteal pulses. The radial pulse is used most frequently. The apical pulse, at the apex of the heart, is most accurate and can be determined with the use of a stethoscope. (Adler and Carlton, 4th ed., p. 198)

335. For which of the following conditions is operative cholangiography a useful tool? 1. Patency of the biliary ducts 2. Biliary tract calculi 3. Gallbladder calculi A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: Operative cholangiography plays a vital role in biliary tract surgery. The contrast medium is injected, and imaging occurs following a cholecystectomy so that gallbladder calculi will not be visualized. This procedure is used to investigate the patency of the bile ducts, the function of the hepatopancreatic sphincter (of Oddi), and the presence of previously undetected biliary tract calculi.

83. An advantage of coupling the image intensifier to the TV camera or CCD via a fiber-optic coupling device is its 1. compact size 2. durability 3. ability to accommodate auxilary imaging devices A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: The image intensifier can be coupled to the TV camera via a fiber-optic bundle or via a lens coupling device. The fiber-optic connection offers less fragility, more compactness, and ease of maneuverability. The big advantage of the objective lens is that it allows the use of auxiliary imaging devices such as a cine camera or spot-film camera.(Bushong, 8th ed., p. 366)

246. X-ray film emulsion is most sensitive to safelight fog a. before exposure and development b. after exposure c. during development d. at low humidity

The answer is B. ii. EXPLANATION: X-ray film emulsion becomes more sensitive to safelight fog following exposure to fluorescent light from intensifying screens. Care must be taken not to leave exposed film on the darkroom workbench for any length of time because its sensitivity to safelight fog is now greatly heightened. (Shephard, p. 123)

186. Which of the following will occur as a result of a decrease in the anode target angle? 1. Less pronounced anode heel effect 2. Decreased effective focal spot size 3. Greater photon intensity toward the cathode side of the x-ray tube A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: Target angle has a pronounced geometric effect on the effective, or projected, focal spot size. As the target angle decreases, the effective (projected) focal spot becomes smaller. This is advantageous because it will improve radiographic detail without creating a heat-loading crisis at the anode (as would occur if the actual focal spot size were reduced to produce a similar detail improvement). There are disadvantages, however. With a smaller target angle, the anode heel effect increases; photons are more noticeably absorbed by the "heel" of the anode, resulting in a smaller percentage of x-ray photons at the anode end of the x-ray beam and a concentration of x-ray photons at the cathode end of the radiograph. (Shephard, p 221)

257. Which of the following is (are) characteristic(s) of a 16:1 grid? 1. It absorbs a high percentage of scattered radiation. 2. It has little positioning latitude. 3. It is used with high-kVp exposures. A. 1 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: High-kilovoltage exposures produce large amounts of scattered radiation, and therefore high-ratio grids are used in an effort to trap more of this scattered radiation. However, accurate centering and positioning become more critical to avoid grid cutoff. (Selman, p 243)

169. What is the established annual occupational dose-equivalent limit for the lens of the eye? a. 10 mSv b. 50 mSv c. 150 mSv d. 250 mSv

he answer is C. ii. EXPLANATION: According to the NCRP, the annual occupational whole-body dose-equivalent limit is 50 mSv (5 rem or 5,000 mrem). The annual occupational whole-body dose-equivalent limit for students under the age of 18 years is 1 mSv (100 mrem or 0.1 rem). The annual occupational dose-equivalent limit for the lens of the eye, a particularly radiosensitive organ, is 150 mSv (15 rem). The annual occupational dose-equivalent limit for the thyroid, skin, and extremities is 500 mSv (50 rem). (Bushong, 8th ed., p. 557)

64. Possible side effects of an iodinated contrast medium that is administered intravenously include all the following except 1. warm, flushed feeling. 2. altered taste. 3. rash and hives. A. 1 only B. 3 only C. 2 and 3 only D. 1, 2, and 3

. The answer is B. 2. EXPLANATION: Nonionic, low-osmolality iodinated contrast agents are associated with far fewer side effects and reactions than ionic, higher osmolality contrast agents. A side effect is an effect that is unintended but possibly expected and fundamentally not harmful. An adverse reaction is a harmful unintended effect. Possible side effects of iodinated contrast agents include a warm, flushed feeling, a metallic taste in the mouth, nausea, headache, and pain at the injection site. Adverse reactions include itching, anxiety, rash or hives, vomiting, sneezing, dyspnea, and hypotension. (Adler and Carlton, 4th ed., p. 279)

33. A radiolucent sponge can be placed under the patient's waist for a lateral projection of the lumbosacral spine to 1. make the vertebral column parallel with the IR 2. place the intervertebral disk spaces perpendicular to the IR 3. decrease the amount of SR reaching the IR A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

. The answer is B. 2. EXPLANATION: When placed in the recumbent lateral position, the average adult's lumbar spine will not be parallel to the x-ray tabletop. Because the shoulders and hips generally are wider than the waist, the vertebral column slopes downward in the central areas—making the lower thoracic and upper lumbar spine closer to the tabletop than the upper thoracic and lower lumbar spine. One solution is to place a radiolucent sponge under the patient's waist. This will elevate the sagging spinal area and make the vertebral column parallel to the x-ray tabletop and IR. It will also open the intervertebral disks better, placing more of them parallel to the path of the x-ray photons and perpendicular to the IR. This position also places the intervertebral foramina parallel with the path of the CR. The radiolucent sponge is strictly a positioning aid and has no impact on the amount of SR reaching the IR. (Bontrager and Lampignano, 6th ed., p. 335)

94. The drug acetaminophen is classified as a(n) a. diuretic b. antipyretic c. antihistamine d. emetic

. The answer is B. ii. EXPLANATION: An antipyretic is used to reduce fever. Tylenol (acetaminophen) is an example of an antipyretic. An antihistamine is used to relieve allergic effects. Benadryl (diphenhydramine hydrochloride) is an example of an antihistamine that is often on hand in radiology departments in the event of a minor reaction to contrast media. Ipecac is a medication used to induce vomiting and is classified as an emetic. This is easy to remember if you think of what an emesis basin is for. A diuretic is a medication that stimulates the production of urine. Lasix (furosemide) is an example of a diuretic. (Adler and Carlton, 4th ed., p. 265)

70. Which of the following is the factor of choice for the regulation of radiographic (optical) density? a. kVp b. mAs c. SID d. Filtration

. The answer is B. ii. EXPLANATION: The principal quantitative factor regulating radiographic (or optical) density is mAs. The mAs selected is directly proportional to radiographic density (ie, if the mAs is cut in half, radiographic density will be halved). Although SID affects exposure rate (according to the inverse square law of radiation), and therefore affects density (according to the density maintenance formula), it is not used to regulate radiographic density. According to the 15% rule, kVp may be used to change radiographic density, but kVp is not a major quantitative factor. The principal use of filtration is to decrease patient skin dose; filtration of that sort will not affect the radiographic image. (Shephard, p 170)

5. Congruence of the x-ray beam with the light field is tested using a. a pinhole camera b. a star pattern c. radiopaque objects d. a slit camera

. The answer is C. ii. EXPLANATION: Radiographic results should be consistent and predictable with respect to positioning accuracy, exposure factors, and equipment operation. X-ray equipment should be tested and calibrated periodically as part of an ongoing quality assurance (QA) program. The focal spot should be tested periodically to evaluate its size and its impact on recorded detail; this is accomplished using a slit camera, a pinhole camera, or a star pattern. To test the congruence of the light and x-ray fields, a radiopaque object such as a paper clip or a penny is placed at each corner of the light field before the test exposure is made. After processing, the corners of the x-ray field should be exactly delineated by the radioopaque objects. (Carlton and Adler, 4th ed., p. 484)

9. Moving the image intensifier closer to the patient during traditional fluoroscopy 1. decreases the SID 2. decreases patient dose 3. improves image quality A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

. The answer is D 2. EXPLANATION: Moving the image intensifier closer to the patient during traditional fluoroscopy reduces the distance between the x-ray tube (source) and the image intensifier (IR), that is, the SID. It follows that the distance between the part being imaged (object) and the image intensifier (IR), that is, the object-to-image distance (OID), is also reduced. The shorter OID produces less magnification and better image quality. As the SID is reduced, the intensity of the x-ray photons at the image intensifier's input phosphor increases, stimulating the automatic brightness control (ABC) to decrease the milliamperage and thereby decreasing patient dose (Figure 5-15). (Fosbinder and Kelsey, pp. 265-267)

73. What should be done to correct for magnification when using air-gap technique? a. Decrease OID b. Increase OID c. Decrease SID d. Increase SID

. The answer is D. ii. EXPLANATION: OID is used to effect an increase in contrast in the absence of a grid, usually in chest radiography. If a 6-in. air gap (OID) is introduced between the part and the IR, much of the scattered radiation emitted from the body will not reach the IR; thus, the OID acts as a low-ratio grid and increases image contrast. However, the 6-in. OID air gap will make a very noticeable increase in magnification. To correct for this, the SID must be increased. Generally speaking, the SID needs to be increased 7 in. for every 1 in. of OID. With a 6-in. OID, the SID usually is increased from 6 to 10 ft (120 in.). (Shephard, pp. 263, 264)

43. Use your mouse to drag the following tissues into order (A-D) from lowest weighting factor (Wt) to highest weighting factor. a. Skin b. Breast c. Lung d. Bone marrow

Skin, Breast, Lung, Bone marrow ii. The tissue weighting factor (Wt) represents the relative tissue radiosensitivity of irradiated material (e.g., muscle versus intestinal epithelium versus bone, etc.). The tissue weighting factor of the gonads is 0.20. The tissue weighting factor of bone marrow, colon, lung, and stomach is 0.12. The bladder, breast, esophagus, liver, and thyroid have a tissue weighting factor of 0.05. The skin and surface of bone weighting factor is 0.01. The radiation weighting factor (Wr) is a number assigned to different types of ionizing radiations in order to better determine their effect on tissue (e.g., x-rays versus alpha particles). The Wr of different ionizing radiations depends on the LET of that particular radiation. The radiation weighting factor of x and gamma radiation, as well as electrons, is 1. The radiation weighting factor of protons is 2 and of alpha particles is 20. The radiation weighting factor of neutrons can be anywhere between 5 and 20 depending on their energy. The following formula is used to determine effective dose (E): E = Wr x Wt x absorbed dose (Bushong, pp. 556, 558)

185. In the lateral projection of the scapula, the 1. vertebral and axillary borders are superimposed. 2. acromion and coracoid processes are superimposed. 3. inferior angle is superimposed on the ribs. A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: A lateral projection of the scapula superimposes its medial and lateral borders (vertebral and axillary, respectively). The coracoid and acromion processes should be readily identified separately (not superimposed) in the lateral projection. The entire scapula should be free of superimposition with the ribs. The erect position is probably the most comfortable position for a patient with scapular pain. (Frank, Long, and Smith, 11th ed., vol. 1, p. 214)

star pattern is used to measure 1. Focal spot resolution. 2. intensifying-screen resolution. 3. SID resolution. A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: A quality control program requires the use of a number of devices to test the efficiency of various components of the imaging system. A star pattern is a resolution testing device that is used to test the effect of focal spot size. A parallel-line-type resolution test pattern is used to test the resolving capability of intensifying screens.

211. A slit camera is used to measure 1. focal spot size. 2. intensifying-screen resolution. 3. SID resolution. A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: A quality control program requires the use of a number of devices to test the efficiency of various parts of the imaging system (see the figure below). A slit camera, as well as a star pattern or pinhole camera, is used to test focal spot size. A parallel line-type resolution test pattern is used to test the resolution capability of intensifying screens. (Bushong, p 462)

208. Which of the following is/are components of the secondary, or high voltage, side of the x-ray circuit? 1. Rectification system 2. Autotransformer 3. kV meter A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: All circuit devices located before the primary coil of the high-voltage transformer are said to be on the primary or low-voltage side of the x-ray circuit. The timer, autotransformer, and (prereading) kilovoltage meter are all located in the low-voltage circuit.The secondary/high-voltage side of the circuit begins with the secondary coil of the high-voltage transformer. The mA meter is connected at the midpoint of the secondary coil of the high-voltage transformer. Following the secondary coil is the rectification system, and the x-ray tube. (Selman, 9th ed., pp. 150-151). Transformers are used to change the value of alternating current (AC). They operate on the principle of mutual induction. The secondary coil of the step-up transformer is located in the high-voltage (secondary) side of the x-ray circuit. The step-down transformer, or filament transformer, is located in the filament circuit and serves to regulate the voltage and current provided to heat the x-ray tube filament. The rectification system is also located on the high-voltage, or secondary, side of the x-ray circuit. (Selman, 9th ed., pp. 155-156)

44. Recorded detail is directly related to 1. source-image distance (SID). 2. tube current. 3. focal spot size. A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: As SID increases, so does recorded detail, because magnification is decreased. Therefore, SID is directly related to recorded detail. As focal spot size increases, recorded detail decreases because more penumbra is produced. Focal spot size is thus inversely related to radiographic sharpness or recorded detail. Tube current affects radiographic density and is unrelated to recorded detail.

3. X-ray photon energy is inversely related to i. photon wavelength ii. applied milliamperes (mA) iii. applied kilovoltage (kV) A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: As kilovoltage is increased, more high-energy photons are produced, and the overall energy of the primary beam is increased. Photon energy is inversely related to wavelength; that is, as photon energy increases, wavelength decreases. An increase in milliamperage serves to increase the number of photons produced at the target but is unrelated to their energy.

271. Terms that refer to size distortion include 1. magnification 2. attenuation 3. elongation A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: Distortion is misrepresentation of the actual size or shape of the object being imaged. Size distortion is magnification. Shape distortion is a result of improper alignment of the x-ray tube, the part being radiographed, and the IR; the two types of shape distortion are foreshortening and elongation. The shapes of various structures can be misrepresented radiographically as a result of their position in the body, when the part is out of the central axis of the x-ray beam, or when the CR is angled (Figure 7-19). Parts sometimes are elongated intentionally for better visualization (e.g., sigmoid colon). Some body parts, because of their position in the body, are foreshortened, such as the carpal scaphoid. Attenuation refers to decreasing beam intensity and is unrelated to distortion.

81. In general, as the intensification factor increases, 1. radiographic density increases 2. screen resolution increases 3. recorded detail increases A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: Factors that contribute to an increase in the intensification factor generally function to reduce resolution. Slow-speed (detail or "extremity") screens resolve more line pairs per millimeter (lp/mm) than much faster screens. The use of fast screens results in some loss of recorded detail. As intensification factor increases, radiographic density generally increases.

182. Geometric unsharpness is directly influenced by 1. OID 2. SOD 3. SID A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: Geometric unsharpness is affected by all three factors listed. As OID increases, so does magnification—therefore, OID is directly related to magnification. As SOD and SID decrease, magnification increases—therefore, SOD and SID are inversely related to magnification. (Carlton and Adler, 4th ed., pp. 444-445)

141. Drugs that may be used to prolong blood clotting time include 1. heparin 2. diphenhydramine (Benadryl) 3. lidocaine A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: Heparin is produced by the body (especially in the liver) and functions to prevent intravascular clotting. Heparin is also produced artificially and used to treat thromboembolic disorders. Lidocaine and Benadryl are drugs that are usually available on crash carts for emergency use. Lidocaine is used to treat ventricular arrhythmias, and Benadryl is used to treat allergic reactions and acute anaphylaxis. (Torres et al., 6th ed., p. 284)

29. In which type of equipment does kilovoltage decrease during the actual length of the exposure? 1. Condenser-discharge mobile equipment 2. Battery-operated mobile equipment 3. Fixed x-ray equipment A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: Mobile x-ray machines are compact and cordless and are either the battery-operated type or the condenser-discharge type. Condenser-discharge mobile x-ray units do not use batteries; this type of mobile unit requires that it be charged before each exposure. A condenser (or capacitor) is a device that stores electrical energy. The stored energy is used to operate the x-ray tube only. Because this machine does not carry many batteries, it is much lighter and does not need a motor to drive or brake it. The major disadvantage of the capacitor/condenser-discharge unit is that as the capacitor discharges its electrical charge, the kilovoltage gradually decreases throughout the length of the exposure—therefore limiting tube output and requiring recharging between exposures. (Frank, Long, and Smith, 11th ed., vol. 3, p. 235)

295. Capacitor-discharge mobile x-ray units use capacitors to power the 1. x-ray tube 2. machine locomotion 3. braking mechanism A. 1 only B. 2 only C. 1 and 2 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: Mobile x-ray machines are smaller and more compact than their fixed counterparts in the radiology department. It is important that they be relatively easy to move, that their size allows entry into patient rooms, and that their locks enable securing of the x-ray tube into the required positions. Mobile x-ray machines are cordless and are either the battery-operated type or the condenser-discharge type. Condenser-discharge mobile x-ray units do not use batteries; this type of mobile unit requires that it be charged before each exposure. A condenser (or capacitor) is a device that stores electrical energy. The stored energy is used to operate the x-ray tube only. Because this machine does not carry many batteries, it is much lighter and does not need a motor to drive or brake it. The major disadvantage of the capacitor/condenser-discharge unit is that as the capacitor discharges its electrical charge, the kilovoltage gradually decreases throughout the length of the exposure—hence, the need for recharging between exposures.

131. Which of the following is (are) composed of nondividing, differentiated cells? 1. Neurons and neuroglia 2. Epithelial tissue 3. Lymphocytes A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: Nondividing, differentiated cells are specialized, mature cells that do not undergo mitosis. Having these qualities, they are rendered radioresistant, according to the theory proposed by Bergonié and Tribondeau. The adult nervous system is composed of nondividing, differentiated cells and thus is the most radioresistant system in the adult. Epithelial tissue and lymphocytes contain many precursor stem cells and hence are among the most radiosensitive cells in the body. (Bushong, 8th ed., p. 492)

71. The annual dose limit for medical imaging personnel includes radiation from 1. occupational exposure 2. background radiation 3. medical x-rays A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: Occupationally exposed individuals are required to use devices that will record and provide documentation of the radiation they receive over a given period of time, traditionally 1 month. The most commonly used personal dosimeters are the OSL, the TLD, and the film badge. These devices must be worn only for documentation of occupational exposure. They must not be worn for any medical or dental x-rays one receives as a patient, and they are not used to measure naturally occurring background radiation. (Thompson et al., p. 459)

200. Early symptoms of acute radiation syndrome include 1. leukopenia 2. nausea and vomiting 3. cataracts A. 1 and 2 only B. 2 only C. 1 and 3 only D. 2 and 3 only

The answer is A. 2. EXPLANATION: Occupationally exposed individuals generally receive small amount of low-energy radiation over a long period of time. These individuals, therefore, are concerned with the potential long-term effects of radiation, such as carcinogenesis (including leukemia) and cataractogenesis. However, if a large amount of radiation is delivered to the whole body at one time, the short-term early somatic effects must be considered. If the whole body receives 600 rad at one time, acute radiation syndrome is likely to occur. Early signs of acute radiation syndrome include nausea, vomiting, diarrhea, fatigue, and leukopenia (decreased white blood cells count); these occur in the first (prodromal) stage of acute radiation syndrome. (Bushong, 8th ed., p. 524)

84. In radiography of a large abdomen, which of the following is (are) effective way(s) to minimize the amount of scattered radiation reaching the image receptor (IR)? 1. Use of close collimation 2. Use of low mAs 3. Use of a low-ratio grid A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: One way to minimize scattered radiation reaching the IR is to use optimal kilovoltage; excessive kilovoltage increases the production of scattered radiation. Close collimation is exceedingly important because the smaller the volume of irradiated material, the less scattered radiation will be produced. The mAs selection has no impact on scattered radiation production or cleanup. Low-ratio grids allow a greater percentage of scattered radiation to reach the IR. Use of a high-ratio grid will clean up a greater amount of scattered radiation before it reaches the IR. Use of a compression band, or the prone position, in a large abdomen has the effect of making the abdomen "thinner"; it will, therefore, generate less scattered radiation. (Shephard, p. 203)

39. What are the effects of scattered radiation on a radiographic image? a. It produces fog. b. It increases contrast. c. It increases grid cutoff. i. 1 only ii. 2 only iii. 1 and 2 only iv. 1, 2, and 3

The answer is A. 2. EXPLANATION: Scattered radiation is produced as x-ray photons travel through matter, interact with atoms, and are scattered (change direction). If these scattered rays are energetic enough to exit the body, they will strike the IR from all different angles. They, therefore, do not carry useful information and merely produce a flat, gray (low-contrast) fog over the image. Grid cutoff increases contrast and is caused by an improper relationship between the x-ray tube and the grid, resulting in absorption of some of the useful/primary beam. (Bushong, 8th ed., p. 248)

272. What is used to account for the differences in ionizing characteristics of various radiations, when determining their effect on biologic material? 1. Radiation weighting factors (Wr) 2. Tissue weighting factors (Wt) 3. Absorbed dose A. 1 only B1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: The Tissue Weighting Factor (Wt) represents the relative tissue radiosensitivity of irradiated material (eg, muscle vs intestinal epithelium vs bone, etc). The Radiation Weighting Factor (Wr) is a number assigned to different types of ionizing radiations in order to better determine their effect on tissue (eg, x-ray vs alpha particles). The Wr of different ionizing radiations is dependent on the LET of that particular radiation. The following formula is used to determine Effective Dose (E): Effective Dose (E) = Radiation Weighting Factor (Wr ) x Tissue Weighting Factor (Wt) x Absorbed Dose

143. Which of the following bones participate(s) in the formation of the knee joint? 1. Femur 2. Tibia 3. Patella A. 1 and 2 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: The knee (tibiofemoral joint) is the largest joint of the body, formed by the articulation of the femur and tibia. However, it actually consists of three articulations—the patellofemoral joint, the lateral tibiofemoral joint (lateral femoral condyle with tibial plateau), and the medial tibiofemoral joint (medial femoral condyle with tibial plateau). Although the knee is classified as a synovial (diarthrotic) hinge-type joint, the patellofemoral joint actually is a gliding joint, and the medial and lateral tibiofemoral joints are hinge type. (Frank, Long, and Smith, 11th ed., vol. 1, p. 238)

337. Which of the following tissues is (are) considered to be particularly radiosensitive? 1. Intestinal mucous membrane 2. Epidermis of extremities 3. Optic nerves A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: The most radiosensitive portion of the GI tract is the small bowel. Projecting from the lining of the small bowel are villi, from the crypts of Lieberkühn, which are responsible for the absorption of nutrients into the bloodstream. Because the cells of the villi are continually being cast off, new cells must continually arise from the crypts of Lieberkühn. Being highly mitotic, undifferentiated stem cells, they are very radiosensitive. Thus, the small bowel is the most radiosensitive portion of the GI tract. In the adult, the CNS is the most radioresistant system, and the epidermis is composed of radioresistant-mature, postmitotic cells.

137. What is used to account for the differences in tissue sensitivity to ionizing radiation when determining effective dose E? 1. Tissue weighting factors (Wt) 2. Radiation weighting factors (Wr) 3. Absorbed dose A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: The tissue weighting factor (Wt) represents the relative tissue radiosensitivity of irradiated material (e.g., muscle vs. intestinal epithelium vs. bone). The radiation weighting factor (Wr) is a number assigned to different types of ionizing radiations in order to better determine their effect on tissue (e.g., x-ray vs. alpha particles). The Wr of different ionizing radiations depends on the LET of that particular radiation. The following formula is used to determine effective dose E: effective does E=Wr x Wt x absorbed dose (Bushong, 8th ed., p. 556)

282. Which of the following is (are) likely to improve image quality and decrease patient dose? 1. Beam restriction 2. Low kilovolt and high microampere- second factors 3. Grids A. 1 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: The use of beam restrictors limits the amount of tissue being irradiated, thus decreasing patient dose and decreasing the production of scattered radiation. High milliampere-second factors increase patient dose. Patient dose is reduced by using high-kilovolt and low-milliampere-second combinations. Although the use of a grid improves image quality by decreasing the amount of scattered radiation reaching the IR, it always requires an increase in exposure factor (usually milliampere-seconds) and, therefore, results in increased patient dose

236. Deposition of vaporized tungsten on the inner surface of the x-ray tube glass window 1. acts as additional filtration 2. results in increased tube output 3. results in anode pitting A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: Through the action of thermionic emission, as the tungsten filament continually gives up electrons, it gradually becomes thinner with age. This evaporated tungsten frequently is deposited on the inner surface of the glass envelope at the tube window. When this happens, it acts as an additional filter of the x-ray beam, thereby reducing tube output. Also, the tungsten deposit actually may attract electrons from the filament, creating a tube current and causing puncture of the glass envelope.

181. Which of the following voltage ripples is (are) produced by single-phase equipment 1. 100% voltage ripple 2. 13% voltage ripple 3. 3.5% voltage ripple A. 1 only B. 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is A. 2. EXPLANATION: With single-phase, full-wave-rectified equipment, the voltage drops to zero every 180° (of the AC waveform); that is, there is 100% voltage ripple. With three-phase equipment, the voltage ripple is significantly smaller. Three-phase, 6-pulse equipment has a 13% voltage ripple, and three-phase, 12-pulse equipment has only a 3.5% ripple. Three-phase, 12-pulse equipment comes closest to constant potential, as the voltage never falls below 96.5% of maximum value. (Selman, p 96)

63. Radiographs from a particular three-phase, full-wave-rectified x-ray unit, made using known correct exposures, were underexposed. A synchronous spinning top test was performed using 200 mA, 1/12 second, and 70 kVp, and a 20° arc is observed on the test film. Which of the following is most likely the problem? a. The 1/12-second time station is inaccurate b. The 200-mA station is inaccurate c. A rectifier is not functioning d. The processor needs servicing

The answer is A. ii. EXPLANATION: A synchronous spinning top test is used to test timer accuracy or rectifier function in three-phase equipment. Because three-phase, full-wave-rectified current would expose a 360° arc each second, a 1/12-second exposure should expose a 30° arc. Anything more or less indicates timer inaccuracy. If exactly one half of the expected arc appears, one should suspect rectifier failure. (Saia, p 434)

248. Which of the following is most likely to produce a radiograph with a long scale of contrast? a. Increased photon energy b. Increased screen speed c. Increased mAs d. Increased SID

The answer is A. ii. EXPLANATION: An increase in photon energy accompanies an increase in kilovoltage. Kilovoltage regulates the penetrability of x-ray photons; it regulates their wavelength—the amount of energy with which they are associated. The higher the related energy of an x-ray beam, the greater its penetrability (kilovoltage and photon energy are directly related; kilovoltage and wavelength are inversely related). Adjustments in kilovoltage have a big impact on radiographic contrast: As kilovoltage (photon energy) is increased, the number of grays increases, thereby producing a longer scale of contrast. In general, as screen speed increases, so does contrast (resulting in a shorter scale of contrast). An increase in mAs is frequently accompanied by an appropriate decrease in kilovoltage, which would also shorten the contrast scale. SID and radiographic contrast are unrelated. (Shephard, p 204)

305. A decrease in kilovoltage will result in a. a decrease in optical density b. a decrease in contrast c. a decrease in recorded detail d. a decrease in image resolution

The answer is A. ii. EXPLANATION: As kilovoltage is increased, more electrons are driven to the anode with greater speed and energy. More high-energy electrons will result in production of more high-energy x-rays. Thus, kilovoltage affects both quantity and quality (energy) of the x-ray beam. However, although kilovoltage and radiographic density are directly related, they are not directly proportional; that is, twice the radiographic density does not result from doubling the kilovoltage. With respect to the effect of kilovoltage on image density, if it is desired to double the radiographic density yet impossible to adjust the milliampere-seconds, a similar effect can be achieved by increasing the kilovoltage by 15%. Conversely, the density may be cut in half by decreasing the kilovoltage by 15%. Therefore, a decrease in kilovoltage will produce fewer x-ray photons, resulting in decreased density. Additionally, a decrease in kilovoltage will produce fewer shades of gray, that is, a shorter-scale, or higher/increased, contrast. Kilovoltage is unrelated to recorded detail and resolution.

166. Which of the following interactions between x-ray photons and matter is most responsible for patient dose? a. The photoelectric effect b. Compton scatter c. Classic scatter d. Thompson scatter

The answer is A. ii. EXPLANATION: As radiation passes through tissue, different types of ionization processes can take place depending on the photon energy and the type of material being irradiated. In the photoelectric effect, a relatively low-energy photon uses all its energy to eject an inner-shell electron from the target atom, leaving a vacancy in that shell. An electron from the shell beyond drops down to fill the vacancy and in so doing emits a characteristic ray. This type of interaction contributes most to patient dose because all the x-ray photon energy is being transferred to tissue. In Compton scatter, a high-energy-incident photon uses some of its energy to eject an outer-shell electron. In so doing the incident photon is deflected with reduced energy but usually retains most of its original energy and exits the body as an energetic scattered photon. In Compton scatter, the scattered radiation will either contribute to image fog or pose a radiation hazard to personnel depending on its direction of exit. In classic scatter, a low-energy photon interacts with an atom but causes no ionization; the incident photon disappears in the atom and then immediately reappears and is released as a photon of identical energy but with changed direction. Thompson scatter is another name for classic scatter. (Selman, 9th ed., pp. 125-128)

264. A part whose width is 6 inches will be imaged at 44 inches SID. The part to be imaged lies 9 inches from the IR. What will be the magnification factor? a. 1.25 b. 1.86 c. 4.9 d. 7.3

The answer is A. ii. EXPLANATION: As the object-to-image receptor distance (OID) increases, magnification of that object increases. Depending upon the information provided, we can determine the magnification factor, the percentage magnification, and image width. In the stated scenario, we are looking for image width. The formula used to determine magnification factor is: MF = SID/SOD. Substituting known factors the equation becomes: MF = 44/35, MF = 1.257. The "1" in the answer represents the actual object, while the ".257" represents the degree of magnification. The percent magnification can be determined by moving the decimal two places to the right. Thus, the percent magnification is 25.7%. (Shephard, p 230)

69. The condition that results from a persistent fetal foramen ovale is a. an atrial septal defect b. a ventricular septal defect c. a patent ductus arteriosus d. coarctation of the aorta

The answer is A. ii. EXPLANATION: Atrial septal defect is a small hole (the remnant of the fetal foramen ovale) in the interatrial septum. It usually closes spontaneously in the first months of life; if it persists or is unusually large, surgical repair is necessary. The ductus arteriosus is a short fetal blood vessel connecting the aorta and pulmonary artery that usually closes within 10 to 15 hours after birth. A patent ductus arteriosus is one that persists and requires surgical closure. Ventricular septal defect is a congenital heart condition characterized by a hole in the interventricular septum that allows oxygenated and unoxygenated blood to mix. Some interventricular septal defects are small and close spontaneously; others require surgery. Coarctation of the aorta is a narrowing or constriction of the aorta. (Tortora and Derrickson, 11th ed., p. 794)

53. The symbols 130/56Ba and 138/56Ba are examples of which of the following? a. Isotopes b. Isobars c. Isotones d. Isomers

The answer is A. ii. EXPLANATION: Ba and Ba are isotopes of the same element, barium (Ba), because they have the same atomic number but different mass numbers (numbers of neutrons). Isobars are atoms with the same mass number but different atomic numbers. Isotones have the same number of neutrons but different atomic numbers. Isomers have the same atomic number and mass number; they are identical atoms existing at different energy states. (Bushong, 8th ed., p. 49)

238. As grid ratio is decreased, a. the scale of contrast becomes longer b. the scale of contrast becomes shorter c. radiographic density decreases d. radiographic distortion decreases

The answer is A. ii. EXPLANATION: Because lead content decreases when grid ratio decreases, a smaller amount of scattered radiation is trapped before reaching the IR. More grays, therefore, are recorded, and a longer scale of contrast results. Radiographic density would increase with a decrease in grid ratio. Grid ratio is unrelated to distortion. (Carlton and Adler, 4th ed., p. 432)

218. Which of the following groups of exposure factors would deliver the lowest patient dose? a. 2.5 mAs, 100 kVp, 400-speed screens b. 10 mAs, 90 kVp, 200-speed screens c. 10 mAs, 70 kVp, 800-speed screens d. 10 mAs, 80 kVp, 400-speed screens

The answer is A. ii. EXPLANATION: Because patient dose is regulated by the quantity of x-ray photons delivered to the patient, the milliampere-seconds value regulates patient dose. Highly energetic x-ray photons (high kilovoltage) are more likely to penetrate the patient rather than be absorbed by biologic tissue. Consequently, the use of high-kilovoltage and low-milliampere-seconds exposure factors is preferred in an effort to reduce patient dose. The use of high-speed screens can assist in the reduction of exposure factors. (Bushong, 8th ed., p. 162)

188. A student radiographer who is under 18 years of age must not receive an annual occupational dose of greater than a. 0.1 rem (1 mSv) b. 0.5 rem (5 mSv) c. 5 rem (50 mSv) d. 10 rem (100 mSv)

The answer is A. ii. EXPLANATION: Because the established dose-limit formula guideline is used for occupationally exposed persons 18 years of age and older, guidelines had to be established to cover the event that a student entered the clinical component of a radiography educational program prior to age 18. The guideline states that the occupational dose limit for students under 18 years of age is 0.1 rem (100 mrem or 1 mSv) in any given year. It is important to note that this 0.1 rem is included in the 0.5-rem dose limit allowed for the student as a member of the general public. (Bushong, 9th ed., p. 620)

35. Aspirated foreign bodies in older children and adults are most likely to lodge in the a. right main stem bronchus b. left main stem bronchus c. esophagus d. proximal stomach

The answer is A. ii. EXPLANATION: Because the right main stem bronchus is wider and more vertical, aspirated foreign bodies are more likely to enter it than the left main stem bronchus, which is narrower and angles more sharply from the trachea. An aspirated foreign body does not enter the esophagus or the stomach because they are not respiratory structures. The esophagus and stomach are digestive structures; a foreign body would most likely be swallowed to enter these structures. (Tortora and Derrickson, 11th ed., p. 857)

36. The AP axial projection of the chest for pulmonary apices a. requires 15 to 20 degrees of cephalad angulation b. projects the apices above the clavicles c. should demonstrate the medial ends of the clavicles equidistant from the vertebral column i. 1 only ii. 1 and 2 only iii. 1 and 3 only iv. 1, 2, and 3

The answer is A. ii. EXPLANATION: Because the right main stem bronchus is wider and more vertical, aspirated foreign bodies are more likely to enter it than the left main stem bronchus, which is narrower and angles more sharply from the trachea. An aspirated foreign body does not enter the esophagus or the stomach because they are not respiratory structures. The esophagus and stomach are digestive structures; a foreign body would most likely be swallowed to enter these structures. (Tortora and Derrickson, 11th ed., p. 857)

154. In the production of characteristic radiation at the tungsten target, the incident electron a. ejects an inner-shell tungsten electron b. ejects an outer-shell tungsten electron c. is deflected, with resulting energy loss d. is deflected, with resulting energy gain

The answer is A. ii. EXPLANATION: Characteristic radiation is one of two kinds of x-rays produced at the tungsten target of the x-ray tube. The incident, or incoming, high-speed electron ejects a K-shell tungsten electron. This leaves a hole in the K shell, and an L-shell electron drops down to fill the K vacancy. Because L electrons are at a higher energy level than K-shell electrons, the L-shell electron gives up the difference in binding energy in the form of a photon, a characteristic x-ray (characteristic of the K shell). (Selman, 9th ed., p. 115)

113. The ability of an x-ray unit to produce constant radiation output, at a given mA, using various combinations of mAs and time is called a. linearity. b. reproducibility. c. densitometry. d. sensitometry.

The answer is A. ii. EXPLANATION: Each of the four factors are used as part of a complete quality assurance (QA) program. Linearity means that a given mA, using different mA stations with appropriate exposure time adjustments, will provide consistent intensity. Reproducibility means that repeated exposures at a given technique must provide consistent intensity. Sensitometry and densitometry are used in evaluation of the film processor, part of a complete QA program. (Bushong, 8th ed, p 569)

33. A type of cancerous bone tumor occurring in children and young adults and arising from bone marrow is a. Ewing sarcoma b. multiple myeloma c. Enchondroma d. osteochondroma

The answer is A. ii. EXPLANATION: Ewing sarcoma is a (primary) malignant bone tumor that arises from bone marrow and occurs in children and young adults. The disease is characterized by new bone formation in a layering effect—giving the bone the characteristic "onion peel" appearance radiographically. Multiple myeloma is also a cancerous bone tumor usually affecting adults between the ages of 40 and 70 years. Bone undergoes osteolytic changes, and radiographic demonstration appears as circular areas of bone loss. As their name implies (chondr), enchondroma and osteochondroma involve cartilage—they are both benign conditions

97. Grid cutoff due to off-centering would result in a. overall loss of density b. both sides of the image being underexposed c. overexposure under the anode end d. underexposure under the anode end

The answer is A. ii. EXPLANATION: Grids are composed of alternate strips of lead and interspace material and are used to trap scattered radiation after it emerges from the patient and before it reaches the IR. Accurate centering of the x-ray tube is required. If the x-ray tube is off-center but within the recommended focusing distance, there usually will be an overall loss of density. Over- or under-exposure under the anode is usually the result of exceeding the focusing distance limits in addition to being off-center. (Carlton and Adler, 4th ed., p. 257)

285. The most frequent site of hospital-acquired infection is the a. urinary tract. b. blood. c. respiratory tract. d. digestive tract.

The answer is A. ii. EXPLANATION: Hospital-acquired infections (HAIs) are also referred to as nosocomial. Despite the efforts of infectious disease departments, HAIs continue to be a problem in hospitals today. This is at least partly due to there being a greater number of older, more vulnerable patients and an increase in the number of invasive procedures performed today (i.e., needles and catheters). The most frequent site of HAI is the urinary tract, followed by wounds, the respiratory tract, and blood.

216. Which of the following groups of exposure factors would be most appropriate for a sthenic adult IVU? a. 300 mA, 0.02 s, 72 kVp b. 300 mA, 0.01 s, 82 kVp c. 150 mA, 0.01 s, 94 kVp d. 100 mA, 0.03 s, 82 kVp

The answer is A. ii. EXPLANATION: IVU requires the use of iodinated contrast media. Low kilovoltage (about 70 kVp) is usually used to enhance the photoelectric effect and, in turn, to better visualize the renal collecting system. High kilovoltage will produce excessive scattered radiation and obviate the effect of the contrast agent. A higher milliamperage with a short exposure time generally is preferable. (Fauber, 2nd ed., p. 264)

321. To produce a just perceptible increase in radiographic density, the radiographer must increase the a. mAs by 30% b. mAs by 15% c. kV by 15% d. kV by 30%

The answer is A. ii. EXPLANATION: If a radiograph lacks sufficient blackening, an increase in milliampere-seconds is required. The milliampere-seconds value regulates the number of x-ray photons produced at the target. An increase or decrease in milliampere-seconds of at least 30% is necessary to produce a perceptible effect. Increasing the kilovoltage by 15% will have about the same effect as doubling the milliampere-seconds.

202. The presence of dust or scratches on intensifying screens will cause a. decreased density in those areas of the image b. increased density in those areas of the image c. decreased density in all areas of the image d. increased density in all areas of the image

The answer is A. ii. EXPLANATION: If intensifying-screen phosphors are covered with dust, either they will not fluoresce or their fluorescence will not reach the IR emulsion. Similarly, if the screen is scratched and phosphors are removed, there will be no fluorescence to expose the IR. In both cases, then, those areas will exhibit little or no density. (Shephard, p. 75)

30. If a radiograph exposed using a 12:1 ratio grid exhibits a loss of density at its lateral edges, it's probably because the a. SID was too great b. grid failed to move during the exposure c. x-ray tube was angled in the direction of the lead strips d. central ray was off-center

The answer is A. ii. EXPLANATION: If the SID is above or below the recommended focusing distance, the primary beam will not coincide with the angled lead strips at the lateral edges. Consequently, there will be absorption of the primary beam, termed grid cutoff. If the grid failed to move during the exposure, there would be grid lines throughout. Central ray angulation in the direction of the lead strips is appropriate and will not cause grid cutoff. If the central ray were off-center, there would be uniform loss of density. (Carlton and Adler, 4th ed., p. 260)

27. The CR should be directed to the center of the part of greatest interest to avoid a. rotation distortion b. magnification c. foreshortening d. elongation

The answer is A. ii. EXPLANATION: Image details placed away from the path of the CR will be exposed by more divergent rays, resulting in rotation distortion. This is why the CR must be directed to the midpoint of the part of greatest interest. For example, if bilateral hands are requested, they should be examined individually; if imaged simultaneously, the CR will be directed to no anatomic part (between the two hands) and rotation distortion will occur. Magnification occurs when an OID is introduced, or with a decrease in SID. Foreshortening and elongation are the two types of shape distortion—caused by nonalignment of the x-ray tube, part/subject, and IR.

240. What is the effect on RBE as LET increases? a. As LET increases, RBE increases. b. As LET increases, RBE decreases. c. As LET increases, RBE stabilizes. d. LET has no effect on RBE.

The answer is A. ii. EXPLANATION: LET expresses the rate at which photon or particulate energy is transferred to (absorbed by) biologic material (through ionization processes); it depends on the type of radiation and the characteristics of the absorber. RBE describes the degree of response or amount of biologic change one can expect of the irradiated material. As the amount of transferred energy (LET) increases (from interactions occurring between radiation and biologic material), the amount of biologic effect/damage will also increase. (Thompson et al., pp. 419-420)

273. What is the relationship between LET and RBE? a. As LET increases, RBE increases. b. As LET increases, RBE decreases. c. As LET decreases, RBE increases. d. There is no direct relationship between LET and RBE.

The answer is A. ii. EXPLANATION: LET increases with the ionizing potential of the radiation; for example, alpha particles are more ionizing than x-radiation; therefore, they have a higher LET. As ionizations and LET increase, there is greater possibility of an effect on living tissue; therefore, the RBE increases. The RBE (sometimes called the quality factor) of diagnostic x-rays is 1, the RBE of fast neutrons is 10, and the RBE of 5-MeV alpha particles is 20.

58. For the average patient, the CR for a lateral projection of a barium-filled stomach should enter a. midway between the midcoronal line and the anterior abdominal surface b. midway between the vertebral column and the lateral border of the abdomen c. at the midcoronal line at the level of the iliac crest d. perpendicular to the level of L2

The answer is A. ii. EXPLANATION: Lateral projections of the barium-filled stomach (Figure 2-61) may be performed recumbent or upright for demonstration of the retrogastric space. With the patient in the (usually right) lateral position, the CR is directed to a point midway between the midcoronal line and the anterior surface of the abdomen at the level of L1. When the patient is in the LPO or RAO position, the CR should be directed midway between the vertebral column and the lateral border of the abdomen. For the PA projection, the CR is directed perpendicular to the IR at the level of L2. (Frank, Long, and Smith, 11th ed., vol. 2, pp. 152-153)

174. Which of the following systems function(s) to compensate for changing patient/part thicknesses during fluoroscopic procedures? a. Automatic brightness control b. Minification gain c. Automatic resolution control d. Flux gain

The answer is A. ii. EXPLANATION: Parts being examined during fluoroscopic procedures change in thickness and density as the patient is required to change positions and as the fluoroscope is moved to examine different regions of the body that have varying thickness and tissue densities. The automatic brightness control functions to vary the required milliampere-seconds and/or kilovoltage as necessary. With this method, patient dose varies, and image quality is maintained. Minification and flux gain contribute to total brightness gain. (Bushong, 8th ed., p. 360)

48. Which of the following combinations would pose the most hazard to a particular anode? a. 0.6 mm focal spot, 75 kVp, 30 mAs b. 0.6 mm focal spot, 85 kVp, 15 mAs c. 1.2 mm focal spot, 75 kVp, 30 mAs d. 1.2 mm focal spot, 85 kVp, 15 mAs

The answer is A. ii. EXPLANATION: Radiographic rating charts enable the operator to determine the maximum safe mA, exposure time, and kVp for a particular exposure using a particular x-ray tube. An exposure that can be made safely with the large focal spot may not be safe for use with the small focal spot of the same x-ray tube. The total number of HU that an exposure generates also influences the amount of stress (in the form of heat) imparted to the anode. The product of mAs and kVp determines HU. Groups A and C produce 2250 HU; groups B and D produce 1275 HU. Groups B and D deliver less heat load, but group D delivers it to a larger area (actual focal spot) making this the least hazardous group of technical factors.

22. Which of the following quantities of filtration is most likely to be used in mammography? a. 0.5 mm Mo b. 1.5 mm Al c. 1.5 mm Cu d. 2.0 mm Cu

The answer is A. ii. EXPLANATION: Soft tissue radiography requires the use of long-wavelength, low-energy x-ray photons. Very little filtration is used in mammography. Certainly, anything more than 1.0 mm of aluminum would remove the useful soft photons, and the desired high contrast could not be achieved. Dedicated mammographic units usually have molybdenum targets (for the production of soft radiation) and a small amount of molybdenum filtration. (Carlton & Adler, p 581)

45. Which of the following quantities of filtration is most likely to be used in mammography? a. 0.5 mm Mo b. 1.5 mm Al c. 1.5 mm Cu d. 2.0 mm Cu

The answer is A. ii. EXPLANATION: Soft tissue radiography requires the use of long-wavelength, low-energy x-ray photons. Very little filtration is used in mammography. Certainly, anything more than 1.0 mm of aluminum would remove the useful soft photons, and the desired high contrast could not be achieved. Dedicated mammographic units usually have molybdenum targets (for the production of soft radiation) and a small amount of molybdenum filtration. (Carlton & Adler, p 581)

116. In the AP axial projection (Towne method) of the skull, with the CR directed 30 degrees caudad to the orbitomeatal line (OML) and passing midway between the external auditory meati, which of the following is best demonstrated? a. Occipital bone b. Frontal bone c. Facial bones d. Basal foramina

The answer is A. ii. EXPLANATION: The AP axial projects the anterior structures (frontal and facial bones) downward, thus permitting visualization of the occipital bone without superimposition (Towne method). The dorsum sella and posterior clinoid processes of the sphenoid bone should be visualized within the foramen magnum. This projection may also be obtained by angling the CR 30 degrees caudad to the OML (Figure 2-46). The frontal bone is best shown with the patient PA and with a perpendicular CR. The parietoacanthal projection is the single best position for facial bones. Basal foramina are well demonstrated in the submentovertical projection. (Frank, Long, and Smith, 11th ed., vol. 2, p. 316)

89. What is the relationship between kV and HVL? a. As kV increases, the HVL increases. b. As kV decreases, the HVL decreases. c. If the kV is doubled, the HVL doubles. d. If the kV is doubled, the HVL is squared.

The answer is A. ii. EXPLANATION: The HVL of a particular beam is defined as that thickness of a material that will reduce the exposure rate to one-half of its original value. The more energetic the beam (the higher the kilovoltage), the greater is the HVL thickness required to cut its intensity in half. Therefore, it may be stated that kilovoltage and HVL have a direct relationship: As kilovoltage increases, HVL increases. (Selman, 9th ed., pp. 122-123)

109. The control dosimeter that comes from the monitoring company should be a. stored in a radiation-free area b. kept in a designated control booth c. kept in the film-processing area d. used as an extra badge for new personnel

The answer is A. ii. EXPLANATION: The control badge that comes with the month's supply of dosimeters is used as a standard for comparison with the used personal badges. The control badge should be stored in a radiation-free area, away from the radiographic rooms. When it has been processed, its density is compared with the densities of the monitors worn in radiation areas. Densities greater than the density of the radiation-free monitor are reported in millirem units. (Bushong, 8th ed., p. 596)

23. The image intensifier's input phosphor differs from the output phosphor in that the input phosphor a. is much larger than the output phosphor b. emits electrons, whereas the output phosphor emits light photons c. absorbs electrons, whereas the output phosphor absorbs light photons d. is a fixed size, and the size of the output phosphor can vary

The answer is A. ii. EXPLANATION: The image intensifier's input phosphor is 6 to 9 times larger than the output phosphor. It receives the remnant radiation emerging from the patient and converts it into a fluorescent light image. Very close to the input phosphor, separated only by a thin, transparent layer, is the photocathode. The photocathode is made of a photoemissive alloy, usually a cesium and antimony compound. The fluorescent light image strikes the photocathode and is converted to an electron image, which is focused by the electrostatic lenses to the small output phosphor. (Bushong, 8th ed., pp. 360-363)

62. The image intensifier's input phosphor differs from the output phosphor in that the input phosphor a. is much larger than the output phosphor b. emits electrons, whereas the output phosphor emits light photons c. absorbs electrons, whereas the output phosphor absorbs light photons d. is a fixed size, and the size of the output phosphor can vary

The answer is A. ii. EXPLANATION: The image intensifier's input phosphor is 6 to 9 times larger than the output phosphor. It receives the remnant radiation emerging from the patient and converts it into a fluorescent light image. Very close to the input phosphor, separated only by a thin, transparent layer, is the photocathode. The photocathode is made of a photoemissive alloy, usually a cesium and antimony compound. The fluorescent light image strikes the photocathode and is converted to an electron image, which is focused by the electrostatic lenses to the small output phosphor. (Bushong, 8th ed., pp. 360-363)

25. The submentovertical (SMV) oblique axial projection of the zygomatic arches requires that the skull be rotated a. 15 degrees toward the affected side. b. 15 degrees away from the affected side. c. 45 degrees toward the affected side. d. 45 degrees away from the affected side.

The answer is A. ii. EXPLANATION: The oblique axial projection is valuable when the zygomatic arches cannot be demonstrated bilaterally with the submentovertical projection because they are not prominent enough or because of a depressed fracture. The patient still may be positioned as for an SMV projection, but the head is obliqued 15 degrees toward the side being examined. This serves to move the zygomatic arch away from superimposed structures and provides a slightly oblique axial projection of the arch. (Bontrager and Lampignano, 6th ed., p. 426)

253. Which interaction between x-ray photons and matter results in total absorption of the incident photon? a. Photoelectric effect b. Compton scattering c. Coherent scattering d. Pair production

The answer is A. ii. EXPLANATION: The photoelectric effect and Compton scattering are the two predominant interactions between x-ray photons and matter in diagnostic x-ray. In the photoelectric effect, the low-energy incident photon uses all its energy to eject an atom's inner-shell electron. That photon ceases to exist—it has used all its energy to ionize the atom. The part has absorbed the x-ray photon. This interaction contributes to patient dose and produces short-scale contrast. In Compton scatter, the high-energy incident photon uses only part of its energy to eject an outer-shell electron. It retains most of its energy in the form of a scattered x-ray. (Bushong, 8th ed., pp. 176-177)

309. Tracheotomy is an effective technique used to restore breathing when there is a. respiratory pathway obstruction above the larynx b. crushed tracheal rings owing to trauma. c. respiratory pathway closure owing to inflammation and swelling d. all the above

The answer is A. ii. EXPLANATION: The respiratory passageways include the nose, pharynx, larynx (upper respiratory structures), trachea, bronchi, and lungs (lower structures). If obstruction of the breathing passageways occurs in the upper respiratory tract, above the larynx (i.e., in the nose or pharynx), tracheotomy may be performed to restore breathing. Intubation can be done into the lower structures, larynx, and trachea, moving aside any soft obstruction and restoring the breathing passageway.

152. Which of the following would be most likely to cause the greatest skin dose (ESE)? a. Short SID b. High kilovoltage c. Increased filtration d. Increased milliamperage

The answer is A. ii. EXPLANATION: The shorter the SID, the greater is the skin dose (ESE). This is why there are specific SSD restrictions in fluoroscopy. X-ray beam quality has a significant effect on patient skin dose. The use of high kilovoltage produces more high-energy penetrating photons, thereby decreasing skin dose. Filtration is used to remove the low-energy photons that contribute to skin dose from the primary beam. Although milliamperage regulates the number of x-ray photons produced, it does not affect photon quality. (Bushong, 8th ed., pp. 300-303)

330. Of the following groups of exposure factors, which will produce the shortest scale of radiographic contrast? a. 500 mA, 0.040 second, 70 kV b. 100 mA, 0.100 second, 80 kV c. 200 mA, 0.025 second, 92 kV d. 700 mA, 0.014 second, 80 kV

The answer is A. ii. EXPLANATION: The single most important factor regulating radiographic contrast is kilovoltage. The lower the kilovoltage, the shorter is the scale of contrast. All the milliampere-seconds values in this problem have been adjusted for kilovoltage changes to maintain density, but just a glance at each of the kilovoltages is often a good indicator of which will produce the longest scale or shortest scale contrast.

190. The device used to give film emulsion a predetermined exposure in order to test its response to processing is called the a. sensitometer b. densitometer c. step wedge d. spinning top

The answer is A. ii. EXPLANATION: To test a film's response to processing, the film first must be given a predetermined exposure with a sensitometer. The film then is processed, and the densities are read using a densitometer. Any significant variation from the expected densities is further investigated. A step wedge is used to evaluate the effect of kilovoltage on contrast, and a spinningtop test is used to check timer accuracy. (Shephard, pp. 99-100)

312. All the following are forms of mechanical obstruction seen in neonates or infants except a. paralytic ileus. b. meconium ileus. c. volvulus. d. intussusception.

The answer is A. ii. EXPLANATION: Volvulus and intussusception both involve a mechanical "closure" or obstruction of the intestinal lumen by a change in the continuous pathway of the GI tract—volvulus by a twisting of the bowel on itself causing obstruction and intussusception by "telescoping" of the bowel causing obstruction. Meconium ileus is another form of mechanical obstruction where meconium (first feces of a newborn) becomes hardened and impacted, causing obstruction. Paralytic (or adynamic) ileus, however, is an obstruction caused by loss of peristaltic movement of the intestine.

209. When examining a patient whose elbow is in partial flexion, a. the AP projection requires two separate positions and exposures. b. the AP projection is made through the partially flexed elbow, resting on the olecranon process, CR perpendicular to IR. c. the AP projection is made through the partially flexed elbow, resting on the olecranon process, CR parallel to the humerus. d. the AP projection is eliminated from the routine.

The answer is A. ii. EXPLANATION: When a patient's elbow needs to be examined in partial flexion, the lateral projection offers little difficulty, but the AP projection requires special attention. If the AP projection is made with a perpendicular CR and the olecranon process resting on the table-top, the articulating surfaces are obscured. With the elbow in partial flexion, two exposures are necessary to achieve an AP projection of the elbow joint articular surfaces. One is made with the forearm parallel to the IR (humerus elevated), which demonstrates the proximal forearm. The other is made with the humerus parallel to the IR (forearm elevated), which demonstrates the distal humerus. In both cases, the CR is perpendicular if the degree of flexion is not too great or angled slightly into the joint space with greater degrees of flexion. (Bontrager and Lampignano, 6th ed., p. 169)

6. When a slow screen-film system is used with a fast screen-film AEC system, the resulting images a. are too light b. are too dark c. have improved detail d. have poor detail

The answer is A. ii. EXPLANATION: When an AEC (phototimer or ionization chamber) is used, the system is programmed for the use of a particular screen-film speed (e.g., 400 speed). If a slower-speed screen IR is placed in the Bucky tray, the AEC has no way of recognizing it as different and will time the exposure for the system for which it is programmed. For example, if the AEC is programmed for a 400-speed screen-film combination, and if a 200-speed screen IR is placed in the Bucky tray, the resulting radiograph will have half the required radiographic density. (Shephard, pp. 65-66)

225. To obtain an AP projection of the right ilium, the patient's a. left side is elevated 40°. b. right side is elevated 40°. c. left side is elevated 15°. d. right side is elevated 15°.

The answer is A. ii. EXPLANATION: When the pelvis is observed in the anatomic position, the ilia are seen to oblique forward, giving the pelvis a "basin-like" appearance. To view the right iliac bone, the radiographer must place it parallel to the IR by elevating the left side about 40° (RPO). The left iliac bone is radiographed in the 40° LPO oblique position. (Ballinger & Frank, vol 1, p 308)

26. To demonstrate the pulmonary apices with the patient in the AP position, the a. central ray is directed 15° to 20° cephalad. b. central ray is directed 15° to 20° caudad. c. exposure is made on full exhalation. d. patient's shoulders are rolled forward.

The answer is A. ii. EXPLANATION: When the shoulders are relaxed, the clavicles are usually carried below the pulmonary apices. To examine the portions of the lungs lying behind the clavicles, the central ray is directed cephalad 15° to 20° to project the clavicles above the apices when the patient is examined in the AP position. (Ballinger & Frank, vol 1, p 472)

31. Using fixed milliampere-seconds and variable kilovoltage technical factors, each centimeter increase in patient thickness requires what adjustment in kilovoltage? a. Increase 2 kV b. Decrease 2 kV c. Increase 4 kV d. Decrease 4 kV

The answer is A. ii. EXPLANATION: When the variable-kilovoltage method is used, a particular milliampere-seconds value is assigned to each body part. As part thickness increases, the kilovoltage (i.e., penetration) is also increased. The body part being radiographed must be measured carefully, and for each centimeter of increase in thickness, 2 kV is added to the exposure. (Shephard, pp. 299-300)

161. Diagnostic x-radiation may be correctly described as a. low energy, low LET b. low energy, high LET c. high energy, low LET d. high energy, high LET

The answer is A. ii. EXPLANATION: X-radiation used for diagnostic purposes is of relatively low energy. Kilovoltages of up to 150 kV are used, as compared with radiations having energies of up to several million volts. Linear energy transfer (LET) refers to the rate at which energy is transferred from ionizing radiation to soft tissue. Particulate radiations, such as alpha particles, have mass and charge and, therefore, lose energy rapidly as they penetrate only a few centimeters of air. X- and gamma radiations, having no mass or charge, are low-LET radiations. (Bushong, 8th ed., p. 495)

138. Aluminum filtration has its greatest effect on a. low-energy x-ray photons b. high-energy x-ray photons c. low-energy scattered photons d. high-energy scattered photons

The answer is A. ii. EXPLANATION: X-ray photons emerging from the focal spot comprise a heterogeneous primary beam. There are many low-energy x-rays that, if not removed, would contribute significantly to patient skin dose. These low-energy photons are too weak to penetrate the patient and expose the IR; they simply penetrate a small thickness of tissue before being absorbed. Filters, usually made of aluminum, are used in radiography to reduce patient dose by removing this low-energy radiation (i.e., decreased beam intensity) and resulting in an x-ray beam of higher average energy. Total filtration is composed of inherent filtration plus added filtration. X-ray photons scatter only after they have interacted with the absorber/patient; scatter is unrelated to filtration. (Bushong, 8th ed., p. 165)

210. A slit camera is used to measure 1. focal-spot size 2. spatial resolution 3. intensifying-screen resolution A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: A quality-control (QC) program requires the use of a number of devices to test the efficiency of various parts of the imaging system. Spatial resolution is most significantly affected by the focal spot size. A slit camera, as well as a star pattern (Figure 5-12), or pinhole camera, is used to test focal-spot size. The slit camera is considered the standard for (annual) measurement of the effective focal spot size. A parallel line-type resolution test pattern (Figure 5-12) can be used to test the resolution capability of intensifying screens. (Bushong, 9th ed., pp. 307-308)

199. Which of the following types of radiation is (are) considered electromagnetic? 1. X-ray 2. Gamma 3. Beta A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: Alpha and beta radiation are particulate radiations; alpha is composed of two protons and two neutrons, and beta is identical to an electron. Gamma and x-radiation are electromagnetic, having wave-like fluctuations like other radiations of the electromagnetic spectrum (e.g., visible light and radio waves). (Bushong, 8th ed., p. 60)

205. Which of the following will improve the spatial resolution of image-intensified images? 1. A very thin coating of cesium iodide on the input phosphor 2. A smaller-diameter input screen 3. Increased total brightness gain A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: An image's spatial resolution refers to its recorded detail. The effect of the input screen's phosphor layer is similar to the effect of the phosphor layer thickness in intensifying screens; that is, as the phosphor layer is made thinner, recorded detail increases. Also, the smaller the input phosphor diameter, the greater is the spatial resolution. A brighter image is easier to see but does not affect resolution.

276. Structures involved in blowout fractures include the 1. orbital floor 2. inferior rectus muscle 3. zygoma A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: Blowout fractures of the orbital floor are caused by a direct blow to the eye. The orbital floor is caused to collapse; this carries the inferior rectus muscle through the fracture site and into the maxillary sinus. Diplopia (double vision) often results. Blowout fractures are well demonstrated with the Waters method (parietoacanthal projection) and by using tomographic studies. A parietoacanthal projection with the OML perpendicular and the CR angled 30 degrees caudad also will demonstrate the orbital floor in profile. The zygoma usually is not involved with a blowout fracture but rather with a tripod fracture. (Bontrager and Lampignano, 6th ed., p. 417)

44. Bone densitometry is often performed to I. measure degree of bone (de)mineralization II. evaluate results of osteoporosis treatment/therapy III. evaluate condition of soft tissue adjacent to bone A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: Dual x-ray absorptiometry (DXA) imaging is used to evaluate bone mineral density (BMD). Bone densitometry/DXA can be used to evaluate bone mineral content of the body, or part of it, to diagnose osteoporosis, or to evaluate the effectiveness of treatments for osteoporosis. It is the most widely used method of bone densitometry—it is low dose, precise, and uncomplicated to use/perform. DXA uses two photon energies—one for soft tissue and one for bone. Since bone is denser and attenuates x-ray photons more readily, the attenuation is calculated to represent the degree of bone density. Soft tissue attenuation information is not used to measure bone density. (Frank, Long, and Smith, 11th ed., vol. 3, pp. 454-455)

21. The late effects of radiation are considered to 1. have no threshold dose. 2. be directly related to dose. 3. occur within hours of exposure. A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: Exposure to high doses of radiation results in early effects. Examples of early effects are blood changes and erythema. If the exposed individual survives, then late, or long-term, effects must be considered. Individuals who receive small amounts of low-level radiation (such as those who are occupationally exposed) are concerned with the late effects of radiation exposure—effects that can occur many years after the initial exposure. Late effects of radiation exposure, such as carcinogenesis, are considered to be related to the linear nonthreshold dose-response curve. That is, there is no safe dose; theoretically, even one x-ray photon can induce a later response.

90. Greater latitude is available to the radiographer when using 1. high-kilovoltage factors. 2. a slow film-screen combination. 3. a high-ratio grid. A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: In the low-kilovoltage ranges, a difference of just a few kilovolts makes a very noticeable radiographic difference; that is, there is little latitude. High-kilovoltage techniques offer a much greater margin for error, as do slower film-screen combinations. Grid ratio is unrelated to exposure latitude, but higher-ratio grids offer less tube-centering latitude (i.e., leeway, margin for error) than low-ratio grids. (Carlton and Adler, 4th ed., p. 185)

105. Which of the following anomalies is (are) possible if an exposure dose of 40 rad (400 mGy) were delivered to a pregnant uterus in the third week of pregnancy? 1. Skeletal anomaly 2. Organ anomaly 3. Neurologic anomaly A. 1 only B. 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: Irradiation during pregnancy, especially in early pregnancy, must be avoided. The fetus is particularly radiosensitive during the first trimester, during much of which time pregnancy may not even be suspected. High-risk examinations include pelvis, hip, femur, lumbar spine, cystograms and urograms, and upper and lower gastrointestinal (GI) series. During the first trimester, specifically the 2nd to 10th weeks of pregnancy (i.e., during major organogenesis), if the radiation dose is sufficient, fetal anomalies can be produced. Skeletal and/or organ anomalies can appear if irradiation occurs in the early part of this time period, and neurologic anomalies can be formed in the latter part; mental retardation and childhood malignant diseases, such as cancers or leukemia, and retarded growth/development also can result from irradiation during the first trimester. Fetal irradiation during the second and third trimesters is not likely to produce anomalies but rather, with sufficient dose, some type of childhood malignant disease. Fetal irradiation during the first 2 weeks of gestation can result in embryonic resorption or spontaneous abortion. It must be emphasized, however, that the likelihood of producing fetal anomalies at doses below 20 rad is exceedingly small and that most general diagnostic examinations are likely to deliver fetal doses of less than 1 to 2 rad. (Bushong, 8th ed., pp. 544-546)

102. The use of which of the following is (are) essential in magnification radiography? 1. High-ratio grid 2. Fractional focal spot 3. Direct exposure technique A. 1 only B. 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: Magnification radiography is used to enlarge details to a more perceptible degree. Hairline fractures and minute blood vessels are candidates for magnification radiography. The problem of magnification unsharpness is overcome by using a fractional focal spot; larger focal-spot sizes will produce excessive blurring unsharpness. Grids are usually unnecessary in magnification radiography because of the air-gap effect produced by the OID. Direct-exposure technique probably would not be used because of the excessive exposure required. (Selman, 9th ed., pp. 226-228)

184. The advantages of capacitor-discharge mobile x-ray equipment include 1. compact size 2. light weight 3. high kilovoltage capability A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: Mobile x-ray machines are compact and cordless and are either the battery-operated type or the condenser-discharge type. Condenser-discharge mobile x-ray units do not use batteries; this type of mobile unit requires that it be charged before each exposure. A condenser (or capacitor) is a device that stores electrical energy. The stored energy is used to operate the x-ray tube only. Because this machine does not carry many batteries, it is much lighter and does not need a motor to drive or brake it. The major disadvantage of the capacitor/condenser-discharge unit is that as the capacitor discharges its electrical charge the kilovoltage gradually decreases throughout the length of the exposure—therefore limiting tube output and requiring recharging between exposures. (Frank, Long, and Smith, 11th ed., vol. 3, p. 235)

215. When blue-emitting rare earth screens are matched properly with the correct film emulsion, what type of safelight should be used in the darkroom? 1. Red 2. GBX-2 3. Ultraviolet A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: Most radiographic film (i.e., green- and blue-sensitive) is orthochromatic, that is, sensitive to all colors but red. The GBX-2 is a red filter that is safe with orthochromatic film emulsion. Blue-sensitive film is also sensitive to violet and ultraviolet as well as blue. Green-sensitive film is also sensitive to blue, violet, and ultraviolet, as well as green. (Shephard, p. 90)

207. The operation of personal radiation monitoring can be based on stimulated luminescence. Which of the following personal radiation monitors function(s) in that manner? 1. OSL dosimeter 2. TLD 3. Pocket dosimeter A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: Occupationally exposed individuals are required to use devices to record and document the radiation they receive over a given period of time, traditionally 1 month. The most commonly used personal dosimeters are the OSL dosimeter, the TLD, and the film badge. These devices are worn only for documentation of occupational exposure, not for any medical or dental x-rays received as a patient. TLDs are radiation monitors that use lithium fluoride crystals. Once exposed to ionizing radiation and then heated, these crystals give off light proportional to the amount of radiation received. OSL dosimeters are radiation monitors that use aluminum oxide crystals. These crystals, once exposed to ionizing radiation and then subjected to a laser, give off luminescence proportional to the amount of radiation received. The pocket dosimeter contains an ionization chamber (containing air), and the number of ions formed (of either sign) is equated to exposure dose. (Bushong, 9th ed., p. 594)

222. Radiographic contrast is a result of 1. differential tissue absorption 2. emulsion characteristics 3. proper regulation of milliampere-seconds A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: Radiographic contrast is defined as the degree of difference between adjacent densities. These density differences represent sometimes very subtle differences in the absorbing properties of adjacent body tissues. The type of film emulsion used also brings with it its own contrast characteristics. Different types of film emulsions have different degrees of contrast "built into" them chemically. The technical factor used to regulate contrast is kilovoltage. Radiographic contrast is unrelated to milliampere-seconds. (Selman, 9th ed., pp. 218-220)

127. Which of the following units is (are) used to express resolution? 1. Line-spread function 2. Line pairs per millimeter 3. Line-focus principle A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: Resolution describes how closely fine details may be associated and still be recognized as separate details before seeming to blend into each other and appear "as one." The degree of resolution transferred to the IR is a function of the resolving power of each of the system components and can be expressed in line pairs per millimeter (lp/mm), line-spread function (LSP), or modulation transfer function (MTF). Lp/mm can be measured using a resolution test pattern; a number of resolution test tools are available. LSP is measured using a 10-m x-ray beam; MTF measures the amount of information lost between the object and the IR. The effective focal spot is the foreshortened size of the actual focal spot as it is projected down toward the IR, that is, as it would be seen looking up into the emerging x-ray beam. This is called the line-focus principle and is not a unit used to express resolution.

219. The advantages of digital subtraction angiography over film angiography include 1. greater sensitivity to contrast medium 2. immediately available images 3. increased resolution A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: Superimposition of bony details frequently makes angiographic demonstration of blood vessels less than optimal. The method used to remove these superimposed bony details is called subtraction. Digital subtraction angiography (DSA) accomplishes this through a computer. The advantages of DSA over film angiography include greater sensitivity to contrast medium, immediate availability of images, and lower total cost. Although DSA applications are increasing, film angiography may be preferred in cases in which resolution is critical. (Frank, Long, and Smith, 11th ed., vol. 3, p. 28)

195. The AP projection of the coccyx requires that the CR be directed 1. 15 degrees cephalad 2. 2 inches superior to the pubic symphysis 3. to a level midway between the ASIS and pubic symphysis A. 1 only B. 2 only C. 1 and 2 only D. 1 and 3 only

The answer is B. 2. EXPLANATION: The AP projection of the coccyx requires the CR to be directed 10 degrees caudally and centered 2 inches superior to the pubic symphysis. The AP projection of the sacrum requires a 15-degree cephalad angle centered at a point midway between the pubic symphysis and the ASIS. (Bontrager and Lampignano, 6th ed., p. 344)

47. Which of the following could be used to improve recorded detail? 1. Slower screen/film combination 2. Smaller focal-spot size 3. Correct photocell selection A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: The factors that affect the recorded detail of traditional screen/film imaging are focal spot size, source-to-image distance (SID), object-to-image distance (OID), film/screen speed, and motion. Recorded detail is improved using small focal-spot size, largest practical SID, shortest possible OID, and slowest practical screen/film combination and avoiding motion of the part being imaged. Other imaging factors such as milliampere-seconds and kilovoltage selection and correct photocell selection influence the visibility of recorded detail by affecting density and contrast.

230. Exposures less than the minimum response time of an AEC may be required when 1. using high milliamperage 2. using fast film-screen combinations 3. examining large patients or body parts A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: The minimum response time, or minimum reaction time, is the length of the shortest exposure possible with a particular AEC. If less than the minimum response time is required for a particular exposure, the radiograph will exhibit excessive density. This problem becomes apparent when making exposures that require very short exposure times, such as when using high-milliamperage and fast film-screen combinations. To resolve this problem, the radiographer should decrease the milliamperage rather than the kilovoltage in order to leave contrast unaffected. (Carlton and Adler, 4th ed., p. 102)

118. The output phosphor can be coupled with the Vidicon TV camera or charge-coupled device (CCD) via 1. fiber optics. 2. an image distributor or lens. 3. closed-circuit TV. A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: The output phosphor of the image intensifier displays the brighter, minified, and inverted image. From the output phosphor, the light image is conveyed to its destination by some kind of image distributor—either a series of lenses and a mirror or via fiber optics. Fiber optics is often the method of choice where equipment size is of concern (e.g., mobile equipment). The image distributor, that is, the lens or fiber optics, then sends the majority of light to the TV monitor for direct viewing and the remaining light (about 10%) to the IR (e.g., photospot camera). (Bushong, 8th ed., p. 366)

287. Which of the following positions demonstrates the sphenoid sinuses? 1. Modified Waters (mouth open) 2. Lateral 3. PA axial A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: The parietoacanthal (Waters method) projection demonstrates the maxillary sinuses. The modified Waters position, with the CR directed through the open mouth, will demonstrate the sphenoid sinuses through the open mouth. The PA axial projection demonstrates the frontal and ethmoidal sinus groups. The lateral projection, with the CR entering 1 inch posterior to the outer canthus, demonstrates all the paranasal sinuses. X-ray examinations of the sinuses always should be performed erect to demonstrate leveling of any fluid present.

266. Which of the following circuit devices operate(s) on the principle of self-induction? 1. Autotransformer 2. Choke coil 3. High-voltage transformer A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: The principle of self-induction is an example of the second law of electromagnetics (Lenz's law), which states that an induced current within a conductive coil will oppose the direction of the current that induced it. It is important to note that self-induction is a characteristic of AC only. The fact that AC is constantly changing direction accounts for the opposing current set up in the coil. Two x-ray circuit devices operate on the principle of self-induction. The autotransformer operates on the principle of self-induction and enables the radiographer to vary the kilovoltage. The choke coil also operates on the principle of self-induction; it is a type of variable resistor that may be used to regulate filament current. The high-voltage transformer operates on the principle of mutual induction.

76. An AP oblique (lateral rotation) of the elbow demonstrates which of the following? 1. Radial head free of superimposition 2. Capitulum of the humerus 3. Olecranon process within the olecranon fossa A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: The radial head and neck are projected free of superimposition in the AP oblique projection (lateral rotation) of the elbow. The humeral capitulum is also well demonstrated in this external oblique position. The AP oblique projection (medial rotation) of the elbow superimposes the radial head and neck on the proximal ulna. It demonstrates the olecranon process within the olecranon fossa, and it projects the coronoid process free of superimposition. (Bontrager and Lampignano, 6th ed., p. 171)

78. The sensitometric curve may be used to 1. identify automatic processing problems 2. determine film sensitivity 3. illustrate screen speed A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: The sensitometric, or characteristic, curve is used to illustrate the relationship between the exposure, given the film and the resulting film density. It can be used to predict a particular film emulsion's response (i.e., speed and sensitivity) by determining how long it takes to record a particular density. The sensitometric curve is used in sensitometry to monitor automatic processing efficiency and consistency. A film is given a series of predetermined exposures and processed. The resulting densities are plotted, and the resulting curve is compared with a known correct curve. Any deviation between the two may indicate processing difficulties. The sensitometric curve illustrates the effects of exposure and processing on radiographic film emulsion; it is unrelated to film speed. (Shephard, pp. 104-108)

49. The steeper the straight-line portion of a characteristic curve for a particular film, the 1. slower is the film speed 2. higher is the film contrast 3. greater is the exposure latitude A. 1 only B. 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: The steepness of the characteristic curve is representative of image contrast. The steeper the curve, the greater is the density difference and the higher is the contrast. The speed of the film is determined by the curve's position on the log-relative scale: When comparing two or more characteristic curves, the faster film lies farthest to the left. The faster the film speed, the less is the exposure latitude. (Shephard, p. 105)

268. Which of the following statements is (are) true regarding lower-extremity venography? 1. The patient is often examined in the semierect position. 2. Contrast medium is injected through a vein in the foot. 3. Imaging begins at the hip and proceeds inferiorly. A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: To increase the concentration of contrast media in the deep veins of the leg, a Fowler's position is used with the x-ray table angled at least 45°. Tourniquets can also be used to force the contrast into the deep veins of the leg, especially when the patient is examined in the recumbent position. Contrast medium is injected through a superficial vein in the foot. Imaging may be performed with or without fluoroscopy, and may include AP, lateral, and 30° obliques of the lower leg in internal rotation. Imaging begins at the ankle and proceeds superiorly, usually including the inferior vena cava. (Ballinger & Frank, vol 2, p 542)

42. Which of the following precautions should be observed when radiographing a patient who has sustained a traumatic injury to the hip? 1. When a fracture is suspected, manipulation of the affected extremity should be performed by a physician. 2. The axiolateral projection should be avoided. 3. To evaluate the entire region, the pelvis typically is included in the initial examination. A. 1 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The answer is B. 2. EXPLANATION: Typically, traumatic injury to the hip requires a cross-table (axiolateral) lateral projection, as well as an AP projection of the entire pelvis. Both of these are performed using minimal manipulation of the affected extremity, reducing the possibility of further injury. A physician should perform any required manipulation of the traumatized hip.

34. What percentage of x-ray attenuation does a 0.5-mm lead equivalent apron at 100 kVp provide? a. 51% b. 66% c. 75% d. 94%

The answer is B. 2. EXPLANATION: When placed in the recumbent lateral position, the average adult's lumbar spine will not be parallel to the x-ray tabletop. Because the shoulders and hips generally are wider than the waist, the vertebral column slopes downward in the central areas—making the lower thoracic and upper lumbar spine closer to the tabletop than the upper thoracic and lower lumbar spine. One solution is to place a radiolucent sponge under the patient's waist. This will elevate the sagging spinal area and make the vertebral column parallel to the x-ray tabletop and IR. It will also open the intervertebral disks better, placing more of them parallel to the path of the x-ray photons and perpendicular to the IR. This position also places the intervertebral foramina parallel with the path of the CR. The radiolucent sponge is strictly a positioning aid and has no impact on the amount of SR reaching the IR. (Bontrager and Lampignano, 6th ed., p. 335)

77. A 15% increase in kVp accompanied by a 50% decrease in mAs will result in a(n) a. shorter scale of contrast b. increase in exposure latitude c. increase in radiographic density d. decrease in recorded detail

The answer is B. ii. EXPLANATION: A 15% increase in kVp with a 50% decrease in mAs serves to produce a radiograph similar to the original, but with some obvious differences. The overall blackness (radiographic density) is cut in half because of the decrease in mAs. However, the loss of blackness is compensated for by the addition of grays (therefore, longer-scale contrast) from the increased kVp. The increase in kVp also increases exposure latitude; there is a greater margin for error in higher kVp ranges. Recorded detail is unaffected by changes in kVp. (Fauber, pp 59-60)

245. Which portion of the characteristic curve would most likely represent a density of 1.0? a. Toe b. Straight-line portion c. Shoulder d. Dmax

The answer is B. ii. EXPLANATION: A characteristic curve is used to predict the speed, contrast, and exposure latitude of a particular film emulsion. It compares the exposure, given the film, with the resulting density. It has three portions—the toe, the straight-line portion (region of correct exposure), and the shoulder. The toe occurs immediately after base-plus fog, whose density must not exceed 0.2. The ascending straight-line portion follows the toe; the straight-line portion is the portion of correct exposure and extends from about 0.25 to 2.5. The curve then bends and levels off at the shoulder (Dmax) portion of the curve. (Bushong, 8th ed., p. 277)

324. A lesion with a stalk projecting from the intestinal mucosa into the lumen is a(n) a. fistula b. polyp c. diverticulum d. abscess

The answer is B. ii. EXPLANATION: A polyp is a tumor with a pedicle (stalk) that is found commonly in vascular organs projecting inward from its mucosal wall. Polyps usually are removed surgically because, although usually benign, they can become malignant. A diverticulum is an outpouching from the wall of an organ, such as the colon. A fistula is an abnormal tube-like passageway between organs or between an organ and the surface. An abscess is a localized collection of pus as a result of inflammation.

135. Verbal disclosure of confidential information that is detrimental to the patient is referred to as a. invasion of privacy b. slander c. libel d. assault

The answer is B. ii. EXPLANATION: A radiographer who discloses confidential information to unauthorized individuals may be found guilty of invasion of privacy. If the disclosure is in some way detrimental or otherwise harmful to the patient, the radiographer may be accused of defamation. Spoken defamation is slander; written defamation is libel. Assault is to threaten harm; battery is to carry out the threat. (Torres et al., 6th ed., p. 11)

104. During a double-contrast BE, which of the following positions would afford the best double-contrast visualization of the lateral wall of the descending colon and the medial wall of the ascending colon? a. AP or PA erect b. Right lateral decubitus c. Left lateral decubitus d. Ventral decubitus

The answer is B. ii. EXPLANATION: A right lateral decubitus position will demonstrate a double-contrast visualization of left-sided bowel structures, that is, the lateral side of the descending colon and the medial side of the ascending colon. A left lateral decubitus position will demonstrate a double-contrast visualization of right-sided bowel structures, that is, the lateral side of the ascending colon and the medial side of the descending colon. With the patient in the erect position, barium moves inferiorly and air rises to provide double-contrast visualization of the hepatic and splenic flexures. (Frank, Long, and Smith, 11th ed., vol. 2, p. 190)

54. During measurement of blood pressure, which of the following occurs as the radiographer controls arterial tension with the sphygmomanometer? a. The brachial vein is collapsed b. The brachial artery is temporarily collapsed c. The antecubital vein is monitored d. Oxygen saturation of arterial blood is monitored

The answer is B. ii. EXPLANATION: A stethoscope and a sphygmomanometer are used together to measure blood pressure. The sphygmomanometer's cuff is placed around the midportion of the upper arm. The cuff is inflated to a value higher than the patient's systolic pressure to temporarily collapse the brachial artery. As the inflation is gradually released, the first sound heard is the systolic pressure; the normal range is 110 to 140 mmHg. When no more sound is heard, the diastolic pressure is recorded. The normal diastolic range is 60 to 90 mm Hg. Elevated blood pressure is called hypertension. Hypotension, low blood pressure, is not of concern unless it is caused by injury or disease; in that case, it can result in shock. (Adler and Carlton, 4th ed., pp. 200-201)

254. A wire-mesh test is performed to evaluate screen a. lag b. contact c. resolution d. intensification

The answer is B. ii. EXPLANATION: A wire mesh supported between two rigid pieces of clear plastic is used to evaluate screen-film contact. The mesh is placed on an IR and radiographed. On viewing, any areas that appear unsharp or blurry are indicative of poor screen-film contact. A screen lag test is performed by radiographing a phantom using an empty cassette/IR and then loading it with film and leaving it for a few minutes. If, after processing, there is any indication of an image, there is most probably screen lag. (Shephard, p. 54)

316. Which of the following medications commonly found on emergency carts functions to raise blood pressure? a. Heparin b. Norepinephrine c. Nitroglycerin d. Lidocaine

The answer is B. ii. EXPLANATION: All four medications are found routinely on the typical emergency cart. Heparin is used to decrease coagulation and often used in the cardiovascular imaging suite to inhibit coagulation on catheters. Norepinephrine functions to raise the blood pressure, whereas nitroglycerin functions as a vasodilator, relaxing the walls of blood vessels and increasing circulation. Lidocaine is used as a local anesthetic or antidysrhythmic.

297. Which of the following artifacts is occasionally associated with the use of grids in digital imaging? a. Incomplete erasure b. Aliasing c. Image fading d. Vignetting

The answer is B. ii. EXPLANATION: An artifact associated with digital imaging and grids is "aliasing" or the "moiré effect." If the direction of the lead strips and the grid lines per inch (i.e., grid frequency) matches the scan frequency of the scanner/reader, this artifact can occur. Aliasing (or Moiré effect) appears as superimposed images slightly out of alignment, an image "wrapping" effect. This most commonly occurs in mobile radiography with stationary grids and can be a problem with DR flat panel detectors.

176. As the CR laser scanner/reader recognizes the phosphostimulated luminescence (PSL) released by the PSP storage plate, it constructs a graphic representation of pixel value distribution called a a. processing algorithm b. histogram c. lookup table d. exposure index

The answer is B. ii. EXPLANATION: As the CR laser scanner/reader recognizes the phosphostimulated luminescence (PSL) released by the PSP storage plate, it constructs a graphic representation of pixel value distribution called a histogram. The photostimulable storage phosphor (PSP) within the IP is the image receptor (IR). The PSP is a europium-doped barium fluorohalide coated storage plate. When the PSP is exposed by x-ray photons, the x-ray energy interacts with the crystals and a small amount of visible light is emitted, but most of the x-ray energy is stored (hence, the term storage plate). This stored energy represents the latent image. The IP is placed in the CR scanner/reader where a helium-neon, or solid-state, laser beam scans the PSP and its stored energy is released as blue-violet light (phosphostimulated luminescence [PSL]). This light signal represents varying tissue densities and the latent image that is then transferred to an analog-to-digital converter (ADC)—converting the signal to a digital (electrical) one to be displayed on a monitor. The PSL values will result in numerous image density/brightness values that represent various tissue densities (i.e., x-ray attenuation properties), for example, bone, muscle, blood-filled organs, air/gas, pathologic processes, and so on. The CR scanner/reader recognizes all these values and constructs a representative gray-scale histogram of them corresponding to the anatomical characteristics of the imaged part. Thus, all PA chest histograms will be similar, all lateral chest histograms will be similar, all pelvis histograms will be similar, and so on. A histogram is a graphic representation of pixel-value distribution. The histogram analyzes all the densities from the PSP and represents them graphically—demonstrating the quantity of exposure, the number of pixels, and their value. Histograms are generated that are unique to each body part that can be imaged. After a part is exposed/imaged, its PSP is read/scanned and its own histogram is developed and analyzed. The resulting analysis, and histogram of the actual imaged part, is compared to the programmed representative histogram for that part. Over time, if required diagnostic image characteristics change, a histogram can be updated to reflect the latest required characteristics. (Carlton and Adler, 4th ed., pp. 361-362)

265. All the following statements regarding beam restriction are true except a. beam restriction improves contrast resolution b. beam restriction improves spatial resolution c. field size should never exceed IR dimensions d. beam restriction reduces patient dose

The answer is B. ii. EXPLANATION: Beam restriction is used to determine the size of the x-ray field. This size never should be larger than the IR size. Because the size of the irradiated area can be made smaller, patient dose is reduced. Beam restriction reduces the production of scattered radiation that leads to fog and, therefore, improves contrast resolution. Spatial resolution is related to factors affecting recorded detail, not contrast resolution

45. With milliamperes (mA) increased to maintain output intensity, how is the ESE affected as the source-to-skin distance (SSD) is increased? a. The ESE increases b. The ESE decreases c. The ESE remains unchanged d. ESE is unrelated to SSD

The answer is B. ii. EXPLANATION: Because of the divergent quality of the x-ray beam, as source-to-image receptor distance (SSD) increases, entrance skin exposure (ESE) decreases. SSD must be at least 12 inches on mobile fluoroscopic equipment, and at least 15 inches on fixed fluoroscopic equipment. (Bushong, 9th ed., p. 582)

338. When the radiographer selects kilovoltage on the control panel, which device is adjusted? a. Step-up transformer b. Autotransformer c. Filament circuit d. Rectifier circuit

The answer is B. ii. EXPLANATION: Because the high-voltage transformer has a fixed ratio, there must be a means of changing the voltage sent to its primary coil; otherwise, there would be a fixed kilovoltage. The autotransformer makes these changes possible. When kilovoltage is selected on the control panel, the radiographer actually is adjusting the autotransformer and selecting the amount of voltage to send to the high-voltage transformer to be stepped up (to kilovoltage). The filament circuit supplies the proper current and voltage to the x-ray tube filament for proper thermionic emission. The rectifier circuit is responsible for changing AC to unidirectional current.

133. Incorrect spectral matching between rare earth intensifying screens and film emulsion results in a. longer-scale contrast. b. insufficient density. c. decreased recorded detail. d. excessive density.

The answer is B. ii. EXPLANATION: Calcium tungstate intensifying screens have a broad range of emitted light, and it is more likely that somewhat different film emulsions still could be compatible with them. However, rare earth phosphors emit light over a relatively short range, usually in the green portion of the spectrum. The film emulsion must be sensitive and responsive to that particular color, or the expected results will not occur. If, for example, a blue-sensitive emulsion were matched with green-emitting screens, the resulting radiograph would be underexposed because the blue-sensitive film emulsion would not be responsive to the green-emitting phosphors. (Shephard, pp. 65-66)

220. Impingement on the wrist's median nerve causing pain and disability of the affected hand and wrist is known as a. carpal boss syndrome b. carpal tunnel syndrome c. carpopedal syndrome d. radioulnar syndrome

The answer is B. ii. EXPLANATION: Carpal tunnel syndrome involves pain and numbness to some parts of the median nerve distribution (i.e., palmar surface of the thumb, index finger, and radial half of the fourth finger and palm). Carpal tunnel syndrome occurs frequently in those who continually use vibrating tools or machinery. Carpopedal spasm is spasm of the hands and feet, commonly encountered during hyperventilation. Carpal boss is a bony growth on the dorsal surface of the third metacarpophalangeal joint. (Bontrager and Lampignano, 6th ed., p. 142)

19. What type of precautions prevent the spread of infectious agents in droplet form? a. Contact precautions b. Airborne precautions c. Protective isolation d. Strict isolation

The answer is B. ii. EXPLANATION: Category-specific isolations have been replaced by transmission-based precautions: airborne, droplet, and contact. Under these guidelines, some conditions or diseases can fall into more than one category. Airborne precautions are employed with patients suspected or known to be infected with tubercle bacillus (TB), chickenpox (varicella), or measles (rubeola). Airborne precautions require that the patient wear a mask to avoid the spread of bronchial secretions or other pathogens during coughing. If the patient is unable or unwilling to wear a mask, the radiographer must wear one. The radiographer should wear gloves, but a gown is required only if flagrant contamination is likely. Patients under airborne precautions require a private, specially ventilated (negative-pressure) room. A private room is also indicated for all patients on droplet precautions, that is, with diseases transmitted via large droplets expelled from the patient while speaking, sneezing, or coughing. The pathogenic droplets can infect others when they come in contact with mouth or nasal mucosa or conjunctiva. Rubella ("German measles"), mumps, and influenza are among the diseases spread by droplet contact; a private room is required for the patient, and health care practitioners should use gowns and gloves. Any diseases spread by direct or close contact, such as MRSA, conjunctivitis, and hepatitis A, require contact precautions. Contact precautions require a private patient room and the use of gloves, masks, and gowns for anyone coming in direct contact with the infected individual or his or her environment. (Adler and Carlton, 4th ed., pp. 232-233)

27. Widening of the intercostal spaces is characteristic of which of the following conditions? a. Pneumothorax b. Emphysema c. Pleural effusion d. Pneumonia

The answer is B. ii. EXPLANATION: Chest radiographs demonstrating emphysema will show the characteristic irreversible trapping of air that increases gradually and overexpands the lungs. This produces the characteristic "flattening" of the hemidiaphragms and widening of the intercostal spaces. The increased air content of the lungs requires a compensating decrease in technical factors. Pneumonia is inflammation of the lungs, usually caused by bacteria, virus, or chemical irritant. Pneumothorax is a collection of air or gas in the pleural cavity (outside the lungs), with an accompanying collapse of the lung. Pleural effusion is excessive fluid between the parietal and visceral layers of pleura. (Bontrager and Lampignano, 6th ed., p. 92)

30. An exposed image plate will retain its image for about a. 2 hours b. 8 hours c. 24 hours d. 48 hours

The answer is B. ii. EXPLANATION: Computed radiography (CR) cassettes use no intensifying screens or film—hence, the term filmless radiography. The cassettes have a protective function (for the image plate within) and can be used in the Bucky tray or directly under the anatomic part; they need not be light-tight because the image plate is not light sensitive. The cassette has a thin lead-foil backing (similar to traditional cassettes) to absorb backscatter. Inside the cassette is the photostimulable phosphor (PSP) image plate, sometimes referred to simply as an image plate (IP). This PSP or IP within the cassette has a layer of europium-activated barium fluorohalide that serves as the IR as it is exposed in the traditional manner and receives the latent image. The PSP can store the latent image for several hours; after about 8 hours, noticeable image fading will occur. (Carlton and Adler, 4th ed., p. 358)

153. A dose of 25 rad to the fetus during the fourth or fifth week of pregnancy is more likely to cause which of the following: a. Spontaneous abortion b. skeletal anomalies c. neurologic anomalies d. organogenesis

The answer is B. ii. EXPLANATION: During the first trimester, specifically the 2 to 8 weeks of pregnancy (during major organogenesis), if the radiation dose is at least 20 rad, fetal anomalies can be produced. Skeletal anomalies usually appear if irradiation occurs in the early part of this time period, and neurologic anomalies are formed in the latter part; mental retardation and childhood malignant diseases, such as cancers or leukemia, can also result from irradiation during the first trimester. Fetal irradiation during the second and third trimester is not likely to produce anomalies, but rather, with sufficient dose, some type of childhood malignant disease. Fetal irradiation during the first 2 weeks of gestation can result in spontaneous abortion. It must be emphasized that the likelihood of producing fetal anomalies at doses below 20 rad is exceedingly small and that most general diagnostic examinations are likely to deliver fetal doses of less than 1 to 2 rad. (Bushong, p 546)

151. Which of the following projections is most likely to deliver the largest dose to the ovaries? a. AP lumbar spine, 7 x 17 in. cassette, 80 kVp b. AP lumbar spine, 14 x 17 in. cassette, 80 kVp c. AP abdomen, 80 kVp d. AP abdomen, 70 kVp

The answer is B. ii. EXPLANATION: Exposure dose to patients can be expressed as entrance skin exposure (ESE), sometimes referred to as skin entrance exposure (SEE). Exposure can also be expressed in terms of organ dose. Organ doses to the gonads, bone marrow, breast, thyroid, lens, and lung can be determined. Patient position and beam restriction often make a significant difference in patient dose. Examinations performed PA rather than AP often decrease exposure to sensitive organs. This is so because the lower energy x-ray photons will be absorbed by the anatomic structures closer to the x-ray source, and the higher energy photons will penetrate and exit the part (penetrating the sensitive part rather than being absorbed by it). PA abdomen radiographs deliver less quantity dose to the reproductive organs than AP abdomen radiographs do. An AP lumbar spine radiograph, 7 x 17 in. cassette, 80 kVp delivers about 74 mrad to the ovaries, whereas the same projection using a 14 x 17 in. cassette delivers 92 mrad. An AP abdomen radiograph with 70 kVp delivers 80 mrad, whereas at 80 kVp, the ovarian dose is 68 mrad. (Bontrager and Lampignano, 6th ed., p. 67)

256. What should be done to better demonstrate the mandibular rami seen in PA projection in Figure A? a. use a perpendicular CR b. angle the CR cephalad c. angle the CR caudad d. oblique the head 15° medial

The answer is B. ii. EXPLANATION: Figure A shows a PA projection of the mandible. The head is positioned PA with the OML perpendicular to the IR. The mandibular body is well demonstrated in this position. With the patient in the PA position, the rami can be better demonstrated with 20° to 25° cephalad angulation. A caudal angle could be employed if the skull was positioned in the AP position.

325. The most efficient type of male gonadal shielding for use during fluoroscopy is a. flat contact b. shaped contact (contour) c. shadow d. cylindrical

The answer is B. ii. EXPLANATION: Gonadal shielding should be used whenever appropriate and possible during radiographic and fluoroscopic examinations. Flat contact shields are useful for simple recumbent studies, but when the examination necessitates obtaining oblique, lateral, or erect projections, flat contact shields are easily displaced and become less efficient. Shaped contact (contour) shields are best because they enclose the male reproductive organs and remain in position for oblique, lateral, and erect projections. Shadow shields that attach to the tube head are particularly useful for surgical sterile fields.

92. Graves disease is associated with a. thyroid underactivity b. thyroid overactivity c. adrenal underactivity d. adrenal overactivity

The answer is B. ii. EXPLANATION: Graves disease is the most frequently occurring form of hyperthyroidism. Graves disease is an autoimmune disorder whose symptoms include enlargement of the thyroid gland and exophthalmos (protrusion of the eyes resulting from fluid buildup behind them). Hypothyroidism can result in cretinism in the child and myxedema in the adult. Adrenal overactivity produces Cushing syndrome; underactivity causes Addison disease. (Tortora and Derrickson, 11th ed., p. 659)

56. Continuous rotation of the CT x-ray tube and detector array, with simultaneous movement of the CT couch, has been accomplished through implementation of a. additional cables b. slip rings c. multiple rows of detectors d. electron beam CT

The answer is B. ii. EXPLANATION: In the 1990s, the implementation of slip ring technology allowed continuous rotation of the x-ray tube (through elimination of cables) and simultaneous couch movement. Sixth-generation CT scanning is termed helical (or spiral) CT—permitting acquisition of volume multislice scanning. Today's helical multislice scanners, employing thousands of detectors (up to 60+ rows), can obtain uninterrupted data acquisition of 128 "slices" per tube rotation and can perform 3D multiplanar reformation (MPR). Fifth-generation CT is electron beam; ultra high-speed CT is used specifically for cardiac imaging. (Bushong, 9th ed., p. 375; Romans, pp. 50-51)

196. A small container holding several doses of medication is termed a. an ampule. b. a vial. c. a bolus. d. a carafe.

The answer is B. ii. EXPLANATION: Injectable medications are available in two different kinds of containers. An ampule is a small container that usually holds a single dose of medication. A vial is a somewhat larger container that holds a number of doses of medication. The term bolus is used to describe an amount of fluid to be injected. A carafe is a narrow-mouthed container; it is not likely to be used for medical purposes. (Adler and Carlton, 4th ed., p. 309)

74. Which of the following is a major cause of bowel obstruction in children? a. Appendicitis b. Intussusception c. Regional enteritis d. Ulcerative colitis

The answer is B. ii. EXPLANATION: Intussusception is the telescoping of one part of the intestinal tract into another. It is a major cause of bowel obstruction in children, usually in the region of the ileocecal valve, and is much less common in adults. Radiographically, intussusception appears as the classic "coil spring," with barium trapped between folds of the telescoped bowel. The diagnostic BE procedure occasionally can reduce the intussusception, although care must be taken to avoid perforation of the bowel. Appendicitis occurs when an obstructed appendix becomes inflamed. Distension of the appendix occurs, and if the appendix is left untended, gangrene and perforation can result. Regional enteritis (Crohn disease) is a chronic granulomatous inflammatory disorder that can affect any part of the GI tract but generally involves the area of the terminal ilium. Ulceration and formation of fistulous tracts often occur. Ulcerative colitis occurs most often in young adults; its etiology is unknown, although psychogenic or autoimmune factors seem to be involved. (Bontrager and Lampignano, 6th ed., p. 119)

101. Late or long-term effects of radiation exposure are generally represented by which of the following dose-response curves? a. Linear threshold b. Linear nonthreshold c. Nonlinear threshold d. Nonlinear nonthreshold

The answer is B. ii. EXPLANATION: Late or long-term effects of radiation can occur in tissues that have survived a previous irradiation months or years earlier. These late effects, such as carcinogenesis and genetic effects, are "all-or-nothing" effects—either the organism develops cancer or it does not. Most late effects do not have a threshold dose; that is, any dose, however small, theoretically can induce an effect. Increasing that dose will increase the likelihood of the occurrence, but will not affect its severity; these effects are termed stochastic. Nonstochastic effects are those that will not occur below a particular threshold dose and that increase in severity as the dose increases.

35. How much protection is provided from a 100-kVp x-ray beam when using a 0.50-mm lead-equivalent apron? a. 40% b. 75% c. 88% d. 99%

The answer is B. ii. EXPLANATION: Lead aprons are worn by occupationally exposed individuals during fluoroscopic procedures. Lead aprons are available with various lead equivalents; 0.25, 0.5, and 1.0 mm of lead are the most common. The 1.0-mm lead equivalent apron will provide close to 100% protection at most kilovoltage levels, but it is rarely used because it weighs anywhere from 12 to 24 lb. A 0.25-mm lead-equivalent apron will attenuate about 97% of a 50-kVp x-ray beam, 66% of a 75-kVp beam, and 51% of a 100-kVp beam. A 0.5-mm apron will attenuate about 99% of a 50-kVp beam, 88% of a 75-kVp beam, and 75% of a 100-kVp beam. (Thompson et al., p. 457)

239. Which of the following drugs is used to treat dysrhythmias? a. Epinephrine b. Lidocaine c. Nitroglycerin d. Verapamil

The answer is B. ii. EXPLANATION: Lidocaine (Xylocaine) is an antiarrhythmic used to prevent or treat cardiac arrhythmias (dysrhythmia). Epinephrine (Adrenalin) is a bronchodilator. Bronchodilators may be administered in a spray mister, such as for asthma, or by injection to relieve severe bronchospasm. Nitroglycerin and verapamil are vasodilators. Vasodilators permit increased blood flow by relaxing the walls of the blood vessels. (Adler and Carlton, 4th ed., p. 304)

288. What is the approximate ESE for the average upright PA chest radiograph using 115 kVp and a grid? a. 20 rad b. 20 mrad c. 200 rad d. 200 mrad

The answer is B. ii. EXPLANATION: Patients occasionally will question the radiographer about the amount of radiation they are receiving during their examination. Most of these patients are merely curious because they have heard a recent news report about x-rays or have perhaps studied about x-rays in school recently. It is a good idea for radiographers to have some knowledge of average exposure doses for patients who desire this information. The curious patient can also be referred to the medical physicist for more detailed information. The average high-kilovolt-age chest radiograph with grid delivers an ESE of about 20 mrad (0.020 rad). The same chest radiograph done without a grid at 80 kVp would deliver an ESE of about 12 mrad (0.012 rad). The average AP supine lumbar spine radiograph delivers an ESE of about 350 mrad (0.35 rad). The average AP supine abdomen radiograph delivers about 300 mrad; the average AP cervical spine radiograph delivers about 80 mrad.

67. If a radiograph were made of an average-size knee using automatic exposure control (AEC) and all three photocells were selected, the resulting radiograph would demonstrate a. excessive density b. insufficient density c. poor detail d. adequate exposure

The answer is B. ii. EXPLANATION: Proper functioning of the photo timer depends on accurate positioning by the radiographer. The correct photocell(s) must be selected, and the anatomic part of interest must completely cover the photocell(s) to achieve the desired density. If a photocell is left uncovered, scattered radiation from the part being examined will cause premature termination of exposure and an underexposed radiograph.

68. If a radiograph were made of an average-size knee using automatic exposure control (AEC) and all three photocells were selected, the resulting radiograph would demonstrate a. excessive density b. insufficient density c. poor detail d. adequate exposure

The answer is B. ii. EXPLANATION: Proper functioning of the photo timer depends on accurate positioning by the radiographer. The correct photocell(s) must be selected, and the anatomic part of interest must completely cover the photocell(s) to achieve the desired density. If a photocell is left uncovered, scattered radiation from the part being examined will cause premature termination of exposure and an underexposed radiograph.

93. Which unit of exposure is described as 100 ergs of energy per gram of irradiated absorber? a. roentgen b. rad c. rem d. curie

The answer is B. ii. EXPLANATION: Rad is an acronym for radiation absorbed dose; it measures the energy deposited in any material; that is, it is equal to 100 ergs of energy per gram of any absorber. Roentgen is the unit of exposure; it measures the quantity of ionizations in air. Rem is an acronym for radiation equivalent man; it includes the RBE specific to the tissue irradiated and therefore is a valid unit of measurement for the dose to biologic tissue.

12. How are mAs and radiographic density related in the process of image formation? a. mAs and radiographic density are inversely proportional b. mAs and radiographic density are directly proportional c. mAs and radiographic density are related to image unsharpness d. mAs and radiographic density are unrelated

The answer is B. ii. EXPLANATION: Radiographic density is described as the overall degree of blackening of a radiograph or a part of it. The milliampere-seconds value regulates the number of x-ray photons produced at the target and thus regulates radiographic density. If it is desired to double the radiographic density, one simply doubles the milliampere-seconds; therefore, milliampere-seconds and radiographic density are directly proportional. (Selman, 9th ed., p. 214)

231. In radiation protection, the product of absorbed dose and the correct modifying factor (rad x QF) is used to determine a. roentgen (C/kg) b. rem (Sv) c. curie (Cu) d. radiation quality

The answer is B. ii. EXPLANATION: Rem (dose-equivalent) is the only unit of measurement that expresses the dose-effect relationship. The product of rad (absorbed dose) and the quality factor appropriate for the radiation type is expressed as rem or dose equivalent (DE) and may be used to predict the type and extent of response to radiation. (Sherer et al., 5th ed., p. 11)

156. What is the term used to describe x-ray photon interaction with matter and the transference of part of the photon's energy to matter? a. Absorption b. Scattering c. Attenuation d. Divergence

The answer is B. ii. EXPLANATION: Scattering occurs when there is partial transfer of the proton's energy to matter, as in the Compton effect. Absorption occurs when an x-ray photon interacts with matter and disappears, as in the photoelectric effect. The reduction in the intensity (quantity) of an x-ray beam as it passes through matter is termed attenuation. Divergence refers to a directional characteristic of the x-ray beam as it is emitted from the focal spot. (Bushong, 8th ed., p. 241)

65. Which of the following is most likely to occur as a result of using a 30-in. SID with a 14 x 17 in. IR to radiograph a fairly homogeneous structure? a. Production of quantum mottle b. Density variation between opposite ends of the IR c. Production of scatter radiation fog d. Excessively short-scale contrast

The answer is B. ii. EXPLANATION: Since x-ray photons are produced at the tungsten target, they more readily diverge toward the cathode end of the x-ray tube. As they try to diverge toward the anode, they interact with and are absorbed by the anode "heel." Consequently, there is a greater intensity of x-ray photons at the cathode end of the x-ray beam. This phenomenon is known as the anode heel effect. Because shorter SIDs and larger IR sizes require greater divergence of the x-ray beam to provide coverage, the anode heel effect will be accentuated.

111. Somatic effects of radiation refer to effects that are manifested a. in the descendants of the exposed individual b. during the life of the exposed individual c. in the exposed individual and his or her descendants d. in the reproductive cells of the exposed individual

The answer is B. ii. EXPLANATION: Somatic effects of radiation refer to those effects experienced directly by the exposed individual, such as erythema, epilation, and cataracts. Genetic effects of radiation exposure are caused by irradiation of the reproductive cells of the exposed individual and are transmitted from one generation to the next. (Selman, 9th ed., p. 382)

37. The device used to change alternating current to unidirectional current is a. a capacitor b. a solid-state diode c. a transformer d. a generator

The answer is B. ii. EXPLANATION: Some x-ray circuit devices, such as the transformer and autotransformer, will operate only on AC. The efficient operation of the x-ray tube, however, requires the use of unidirectional current, so current must be rectified before it gets to the x-ray tube. The process of full-wave rectification changes the negative half-cycle to a useful positive half-cycle. An x-ray circuit rectification system is located between the secondary coil of the high-voltage transformer and the x-ray tube. Rectifiers are solid-state diodes made of semiconductive materials such as silicon, selenium, or germanium that conduct electricity in only one direction. Thus, a series of rectifiers placed between the transformer and x-ray tube function to change AC to a more useful unidirectional current. (Bushong, 8th ed., p. 119)

99. The Centers for Disease Control and Prevention (CDC) suggests that health care workers protect themselves and their patients from blood and body fluid contamination by using a. strict isolation precautions. b. standard precautions. c. respiratory precautions. d. sterilization.

The answer is B. ii. EXPLANATION: Standard blood and body fluid precautions serve to protect health care workers and patients from the spread of diseases such as AIDS and AIDS-related complex. Although the precautions are indicated for all patients, special care must be emphasized when working with patients whose infectious status is unknown (e.g., the emergency trauma patient). Gloves must be worn if the radiographer may come in contact with blood or body fluids. A gown should be worn if the clothing may become contaminated. Blood spills should be cleaned with a solution of 1 part bleach to 10 parts water. (Torres et al., 6th ed., pp. 62-65)

48. An RT (ARRT) is supervising manager of a short-staffed imaging facility in a state requiring certification. An applicant arrives whose ARRT certification has lapsed. The manager hires him to fill a 20-hour position doing chest and extremity radiography. The supervisor is guilty of a. breaking the ARRT Code of Ethics b. breaking the ARRT Rules of Ethics c. malpractice d. nothing, because position responsibilities are limited

The answer is B. ii. EXPLANATION: The ARRT Standards of Ethics apply to those Registered Technologists holding ARRT certification and Candidates for ARRT certification. The Standards consist of two parts: the Code of Ethics (aspirational) and the Rules of Ethics (mandatory). The ARRT Rules of Ethics are mandatory minimum professional standards for all RTs and candidate RTs. Violators, and individuals who permit violation, of these Rules are subject to sanctions. Rules of Ethics numbers 15 and 21 specifically refer to those who knowingly assist another without proper certification to engage in the practice of radiologic technology, and/or those who fail to promptly report such activity to the ARRT—as being subject to sanction. (ARRT Standards of Ethics)

87. The National Council on Radiation Protection and Measurements (NCRP) has recommended what total equivalent dose limit to the embryo/fetus? a. 0.5 mSv b. 5.0 mSv c. 50 mSv d. 500 mSv

The answer is B. ii. EXPLANATION: The NCRP recommends a total equivalent dose limit to the embryo/fetus of 5 mSv (500 mrem, 0.5 rem). This dose limit is the total for the entire gestational period. The dose limit for 1 month during pregnancy is 0.5 mSv (50 mrem, 0.05 rem). (Bushong, 8th ed., p. 557)

322. The auditory, or eustachian, tube extends from the nasopharynx to the a. external ear. b. middle ear. c. inner ear. d. oropharynx.

The answer is B. ii. EXPLANATION: The auditory, or eustachian, tube extends from the middle ear to the nasopharynx. It is 3-4 cm in length and is lined with mucous membrane. Otitis media can result when the auditory tube becomes occluded during inflammatory processes. The middle ear contains the auditory ossicles (i.e., malleus, incus, and stapes). The inner ear contains the cochlea, semicircular canals, and vestibule.

303. The most effective method of sterilization is a. dry heat. b. moist heat. c. pasteurization. d. freezing.

The answer is B. ii. EXPLANATION: The most effective method of sterilization is moist heat, using steam under pressure. This is known as autoclaving. Sterilization with dry heat requires higher temperatures for longer periods of time than sterilization with moist heat. Pasteurization is moderate heating with rapid cooling; it is used frequently in the commercial preparation of milk and alcoholic beverages such as wine and beer. It is not a form of sterilization. Freezing also can kill some microbes, but it is not a form of sterilization.

36. Which interaction between x-ray photons and matter involves partial transfer of the incident photon energy to the involved atom? a. Photoelectric effect b. Compton scattering c. Coherent scattering d. Pair production

The answer is B. ii. EXPLANATION: The photoelectric effect and Compton scattering are the two predominant interactions between x-ray photons and matter in diagnostic x-ray. In Compton scatter, the high-energy incident photon uses only part of its energy to eject an outer-shell electron. It retains most of its original energy in the form of a scattered x-ray. The outer-shell electron leaves the atom and is called a recoil electron. Compton scatter is the interaction between x-ray photons and matter that occurs most frequently in diagnostic x-ray and is the major contributor of scattered radiation fog. In the photoelectric effect, the low-energy incident photon uses all its energy to eject an atom's inner-shell electron. When photon ceases to exist, it means it has used all its energy to ionize the atom. The part has absorbed the x-ray photon. This interaction contributes to patient dose and produces short-scale contrast. (Bushong, 8th ed., pp. 176-177)

32. According to the National Council on Radiation Protection and Measurements (NCRP), the monthly gestational dose-equivalent limit for embryo/fetus of a pregnant radiographer is a. 1 mSv b. 5 mSv c. 15 mSv d. 50 mSv

The answer is B. ii. EXPLANATION: The pregnant radiographer poses a special radiation protection consideration, for the safety of the unborn individual. It must be remembered that the developing fetus is particularly sensitive to radiation exposure. Therefore, established guidelines state that the occupational gestational dose-equivalent limit for embryo/fetus of a pregnant radiographer is 5 mSv (500 mrem), not to exceed 0.5 mSv in 1 month. According to the NCRP, the annual occupational whole-body dose-equivalent limit is 50 mSv (5 rem or 5,000 mrem). The annual occupational whole-body dose-equivalent limit for students under the age of 18 years is 1 mSv (100 mrem or 0.1 rem). The annual occupational dose-equivalent limit for the lens of the eye, a particularly radiosensitive organ, is 150 mSv (15 rem). The annual occupational dose-equivalent limit for the thyroid, skin, and extremities is 500 mSv (50 rem). (Bushong, 8th ed., p. 557)

332. Star and wye configurations are related to a. autotransformers b. three-phase transformers c. rectification systems d. AECs

The answer is B. ii. EXPLANATION: The terms star and wye (or delta) refer to the configuration of transformer windings in three-phase equipment. Instead of having a single primary coil and a single secondary coil, the high-voltage transformer has three primary and three secondary windings—one winding for each phase (Figure 5-13). Autotransformers operate on the principle of self-induction and have only one winding. Three-phase x-ray equipment often has three autotransformers

148. Which of the following functions to increase the milliamperage? a. Increase in charge of anode b. Increase in heat of the filament c. Increase in kilovoltage d. Increase in focal spot size

The answer is B. ii. EXPLANATION: The x-ray tube filament is made of thoriated tungsten. When heated to incandescence (white hot), the filament liberates electrons—a process called thermionic emission. It is these electrons that will become the tube current (mA). As heat is increased, more electrons are released, and milliamperage increases. (Bushong, 8th ed., p. 132)

260. What is the name of the device that exposes a film emulsion with an optical step wedge having a number of densities ranging from white to black? a. Densitometer b. Sensitometer c. Penetrometer d. Potentiometer

The answer is B. ii. EXPLANATION: Two devices that are required for QA evaluation purposes are a sensitometer and a densitometer. A sensitometer is a device that functions to expose a film; when developed, that film will illustrate an optical step wedge with a number of optical densities ranging from white through several shades of gray to black. A densitometer is a device that reads the various densities on the film. A penetrometer, or (usually aluminum) step wedge, can be used to demonstrate the effect of peak kilovoltage on radiographic contrast. A potentiometer is another name for a variable resistor. (Carlton and Adler, 4th ed., pp. 304-305)

145. The relationship between the ends of fractured long bones is referred to as a. angulation b. apposition c. luxation d. sprain

The answer is B. ii. EXPLANATION: Various terms are used to describe the position of fractured ends of long bones. The term apposition is used to describe the alignment, or misalignment, between the ends of fractured long bones. The term angulation describes the direction of misalignment. The term luxation refers to a dislocation. A sprain refers to a wrenched articulation with ligament injury. (Bontrager and Lampignano, 6th ed., p. 600)

274. Which of the following vessels does not carry oxygenated blood? a. Pulmonary vein b. Pulmonary artery c. Coronary artery d. Chordae tendineae

The answer is B. ii. EXPLANATION: Venous blood is returned to the right atrium of the heart via the superior (from the upper part of the body) and inferior (from the lower body) venae cavae and the coronary sinus (from the heart substance). During atrial systole, the blood passes through the tricuspid valve into the right ventricle. During ventricular systole, the blood is pumped through the pulmonary semilunar valve into the pulmonary artery (the only artery to carry unoxygenated blood) to the lungs for oxygenation. Blood is returned via the pulmonary veins (the only veins to carry oxygenated blood) to the left atrium. During atrial systole, blood passes through the mitral (bicuspid) valve into the left ventricle. During ventricular systole, the oxygenated blood is pumped through the aortic semilunar valve into the aorta. The coronary arteries supply oxygenated blood to the myocardium. The chordae tendineae are connective tissue fibers that help to limit the movement of valve flaps, preventing backflow of blood.

19. A three-phase timer can be tested for accuracy using a synchronous spinning top. The resulting image looks like a a. series of dots or dashes, each representative of a radiation pulse b. solid arc, with the angle (in degrees) representative of the exposure time c. series of gray tones, from white to black d. multitude of small, mesh-like squares of uniform sharpness

The answer is B. ii. EXPLANATION: When a spinning top is used to test the efficiency of a single-phase timer, the result is a series of dots or dashes, with each representing a pulse of radiation. With full-wave-rectified current and a possible 120 dots (pulses) available per second, one should visualize 12 dots at 1/10 s, 24 dots at 1/5 s, 6 dots at 1/20 s, and so on. However, because three-phase equipment is at almost constant potential, a synchronous spinning top must be used, and the result is a solid arc (rather than dots). The number of degrees formed by the arc is measured and equated to a particular exposure time. A multitude of small, mesh-like squares describes a screen contact test. An aluminum step wedge (penetrometer) may be used to demonstrate the effect of kilovoltage on contrast (demonstrating a series of gray tones from white to black), with a greater number of grays demonstrated at higher kilovoltage levels. (Selman, 9th ed., p. 106)

147. The radiograph illustrated in the figure below was made using a single-phase, full-wave-rectified unit with a timer and rectifiers that are known to be accurate and functioning correctly. What exposure time was used to produce this image? a. 1/10 second b. 0.05 second c. 1/12 second d. 0.025 second

The answer is B. ii. EXPLANATION: When a spinning top is used to test the timer efficiency of full-wave-rectified single-phase equipment, the result is a series of dots or dashes, with each dot representing a pulse of radiation. With full-wave-rectified current and a possible 120 dots (pulses) available per second, one should visualize 12 dots at 1/10 second, 6 dots at 0.05 second, 10 dots at 1/12 second, and 3 dots at 0.025 second. Because three-phase equipment is at almost constant potential, a synchronous spinning top must be used for timer testing, and the result is a solid arc (rather than dots). The number of degrees covered by the arc is measured and equated to a particular exposure time. (Selman, 9th ed., p. 106)

7. To demonstrate the glenoid fossa in profile, the patient is positioned a. 45 degrees oblique, affected side up. b. 45 degrees oblique, affected side down. c. 25 degrees oblique, affected side up. d. 25 degrees oblique, affected side down.

The answer is B. ii. EXPLANATION: When viewing the glenoid fossa from the anterior, it is seen to angle posteriorly and laterally approximately 45 degrees. To view it in profile, then, it must be placed so that its surface is perpendicular to the IR. The patient is positioned in a 45-degree oblique, affected-side-down position, which places the glenoid fossa approximately perpendicular to the IR. The arm is abducted slightly, the elbow is flexed, and the hand and forearm are placed over the abdomen. The CR is directed perpendicular to the glenohumeral joint. (Frank, Long, and Smith, 11th ed., vol. 1, pp. 192-193)

304. An axial projection of the clavicle is often helpful in demonstrating a fracture that is not visualized using a perpendicular CR. When examining the clavicle in the PA position, how is the CR directed for the axial projection? a. Cephalad b. Caudad c. Medially d. Laterally

The answer is B. ii. EXPLANATION: With the patient in the AP position, the CR is directed cephalad 25 to 30 degrees. This serves to project the clavicle away from the pulmonary apices and ribs, projecting most of the clavicle above the thorax. The reverse is true when the patient is examined in the PA position. The PA projection can be useful to obtain better recorded detail because of reduced OID.

277. To eject a K-shell electron from a tungsten atom, the incoming electron must have an energy of at least a. 60 keV. b. 70 keV. c. 80 keV. d. 90 keV.

The answer is B. ii. EXPLANATION: X-ray photons are produced in two ways as high-speed electrons interact with target tungsten atoms. First, if the high-speed electron is attracted by the nucleus of a tungsten atom and changes its course, as the electron is "braked," energy is given up in the form of an x-ray photon. This is called Bremsstrahlung ("braking") radiation, and it is responsible for most of the x-ray photons produced at the conventional tungsten target. Second, a high-speed electron having an energy of at least 70 keV may eject a tungsten K-shell electron, leaving a vacancy in the shell. An electron from the next energy level, the L shell, drops down to fill the vacancy, emitting the difference in energy as a K-characteristic ray. Characteristic radiation makes up only about 15% of the primary beam

311. According to NCRP regulations, leakage radiation from the x-ray tube must not exceed a. 10 mR/h b. 100 mR/h c. 10 mR/min d. 100 mR/min

The answer is B. ii. EXPLANATION: X-ray photons produced in the x-ray tube can radiate in directions other than the one desired. The tube housing, therefore, is constructed so that very little of this leakage radiation is permitted to escape. The regulation states that leakage radiation must not exceed 100 mR/h at 1 m while the tube is operated at maximum potential

123. The type of x-ray tube designed to turn on and off rapidly, providing multiple short, precise exposures, is a. high speed b. grid-controlled c. diode d. electrode

The answer is B. ii. EXPLANATION: X-ray tubes are diode tubes; that is, they have two electrodes—a positive electrode called the anode and a negative electrode called the cathode. The cathode filament is heated to incandescence and releases electrons—a process called thermionic emission. During the exposure, these electrons are driven by thousands of volts toward the anode, where they are suddenly decelerated. That deceleration is what produces x-rays. Some x-ray tubes, such as those used in fluoroscopy and in capacitor-discharge mobile units, are required to make short, precise—sometimes multiple—exposures. This need is met by using a grid-controlled tube. A grid-controlled tube uses the molybdenum focusing cup as the switch, permitting very precise control of the tube current (flow of electrons between cathode and anode). (Bushong, 8th ed., p. 132)

232. The true lateral position of the skull uses which of the following principles? 1. Interpupillary line perpendicular to the IR 2. MSP perpendicular to the IR 3. Infraorbitomeatal line (IOML) parallel to the transverse axis of the IR A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: A lateral projection generally is included in a routine skull series. The patient is placed in a PA oblique position. The MSP is positioned parallel to the IR, and the IOML is adjusted so as to be parallel to the long axis of the cassette. The interpupillary line must be perpendicular to the IR. In a routine lateral projection of the skull, the CR should enter approximately 2 inches superior to the EAM. (Frank, Long, and Smith, 11th ed., vol. 2, p. 306)

20. The Bucky slot cover is in place to protect the 1. patient 2. fluoroscopist 3. technologist A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: All fluoroscopic equipment has protective devices and protocols to protect the patient and user. Fluoroscopic equipment must provide at least 12 in. (30 cm) and preferably 15 in. (38 cm) between the x-ray source (focal spot) and the x-ray tabletop (patient), according to NCRP Report No. 102. The tabletop intensity of the fluoroscopic beam must not exceed 10 R/min or 2.1 R/min/mA. With under-table fluoroscopic tubes, a Bucky slot closer/cover having at least the equivalent of 0.25 mm Pb must be available to attenuate scattered radiation coming from the patient, posing a radiation hazard to the fluoroscopist and radiographer. Fluoroscopic milliamperes must not exceed 5 mA. Because the image intensifier functions as a primary barrier, it must have a lead equivalent of at least 2.0 mm. A cumulative timing device must be available to signal the fluoroscopist when a maximum of 5 minutes of fluoroscopy time has elapsed. Because occupational exposure to scattered radiation is of considerable importance in fluoroscopy, a protective curtain/drape of at least 0.25 mm Pb equivalent must be placed between the patient and fluoroscopist. (Bushong, 8th ed., p. 570)

11. The squeegee assembly in an automatic processor i. functions to remove excess solution from films ii. is located near the crossover rollers iii. helps establish the film's rate of travel A. 1 only B. 2 only C. 1 and 2 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: An exposed radiographic film contains an invisible (latent) image. Only through processing can this image be converted to a permanent, visible (manifest) image. As the film exits the developer section, it passes through the crossover assembly, and before it enters the fixer section, it passes through the squeegee assembly. The squeegee assembly rollers function to remove excess developer solution from the emulsion before the film enters the fixer. This process helps to maintain fixer strength/activity. The rate of travel through the processor is determined by the transport mechanism, that is, the speed of the rollers as established at time of manufacture. (Shephard, p. 143)

117. Which of the following will serve to increase the effective energy of the x-ray beam? 1. Increase in added filtration 2. Increase in kilovoltage 3. Increase in milliamperage A. 1 only B. 2 only C. 1 and 2 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: As filtration is added to the x-ray beam, the lower-energy photons are removed, and the overall energy or wavelength of the beam is greater. As kilovoltage is increased, more high-energy photons are produced, and again, the overall, or average, energy of the beam is greater. An increase in milliamperage serves to increase the number of photons produced at the target but is unrelated to their energy. (Selman, 9th ed., p. 171)

61. If a radiograph exhibits insufficient density, this might be attributed to 1. inadequate kVp. 2. inadequate SID. 3. grid cutoff. A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: As kVp is reduced, the number of high-energy photons produced at the target is reduced; therefore, a decrease in radiographic density occurs. If a grid has been used improperly (off-centered or out of focal range), the lead strips will absorb excessive amounts of primary radiation, resulting in grid cutoff and loss of radiographic density. If the SID is inadequate (too short), an increase in radiographic density will occur. (Selman, pp 214, 240-242)

197. When an image intensifier's magnification mode is used, 1. output screen gain is increased. 2. resolution increases. 3. patient dose increases. A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: During fluoroscopic procedures, as FOV decreases, magnification of the output screen image increases and contrast and resolution improve. The focal point on an image intensifier's 6-inch field/mode, is further away from the output phosphor than the focal point on the normal mode; therefore, the output image is magnified. Because less minification takes place, the image is not as bright. Exposure factors are automatically increased to compensate for the loss in brightness that occurs with smaller FOVs used in magnification mode. (Fosbinder, p 285

47. Which of the following are methods of limiting the production of scattered radiation? 1. Using moderate ratio grids 2. Using the prone position for abdominal examinations 3. Restricting the field size to the smallest practical size A. 1 and 2 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: If a fairly large patient is turned prone, the abdominal measurement will be significantly different from the AP measurement as a result of the effect of compression. Thus, the part is essentially "thinner," and less scattered radiation will be produced. If the patient remains supine and a compression band is applied, a similar effect will be produced. Beam restriction is probably the single most effective means of reducing the production of scattered radiation. Grid ratio affects the cleanup of scattered radiation; it has no effect on the production of scattered radiation.

140. The CR will parallel the intervertebral foramina in which of the following projections? 1. Lateral cervical spine 2. Lateral thoracic spine 3. Lateral lumbar spine A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: If the x-ray photons can pass through/between structures such as joint spaces and foramina, these joint spaces and foramina must be situated perpendicular to the IR and parallel to the CR. The intervertebral foramina of the thoracic and lumbar vertebrae are perpendicular to the IR and, therefore, parallel to the CR in the lateral projection. The cervical intervertebral foramina are well demonstrated when placed 45 degrees to the IR and CR. (Bontrager and Lampignano, 6th ed., pp. 292, 295, 322)

162. Which of the following affect(s) both the quantity and the quality of the primary beam? 1. Half-value layer (HVL) 2. Kilovoltage (kV) 3. Milliamperage (mA) A. 1 only B. 2 only C. 1 and 2 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: Kilovoltage and the HVL affect both the quantity and the quality of the primary beam. The principal qualitative factor for the primary beam is kilovoltage, but an increase in kilovoltage will also create an increase in the number of photons produced at the target. HVL is defined as the amount of material necessary to decrease the intensity of the beam to one-half its original value, thereby effecting a change in both beam quality and quantity. The milliampere-seconds value is adjusted to regulate the number of x-ray photons produced at the target. X-ray-beam quality is unaffected by changes in milliampere-seconds. (Carlton and Adler, 4th ed., p. 181)

178. Use of a portion of the input phosphor during fluoroscopy, rather than the entire input phosphor, will result in 1. a larger field of view (FOV). 2. a magnified image. 3. improved spatial resolution. A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: Multifield image intensifier tubes are usually either dual-field or trifield and are designed this way in order to permit magnification imaging. As voltage is applied to the electrostatic focusing lenses, the focal point moves back—closer to the input phosphor—and a smaller portion of the input phosphor is used. As a result, the FOV decreases and magnification increases, producing better spatial resolution. At the same time, brightness is decreased, requiring an increase in milliamperage (therefore increased patient dose). This increase in milliamperage increases image quality. It can be likened to an increase in signal-to-noise ratio (SNR), with milliamperes being the signal. (Fosbinder and Kelsey, p. 262)

294. The quantity of scattered radiation reaching the IR can be reduced through the use of 1. a fast imaging system 2. an air gap 3. a stationary grid A. 1 and 2 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: Scattered radiation adds unwanted degrading densities to the x-ray image. The single most important way to reduce the production of scattered radiation is to collimate. Although collimation, optimal kilovoltage, and compression can be used, a large amount of scattered radiation is still generated within the part being imaged, and because it adds unwanted non-information-carrying densities, it can have a severely degrading effect on image quality. A grid is a device interposed between the patient and IR that functions to absorb a large percentage of scattered radiation before it reaches the IR. It is constructed of alternating strips of lead foil and radiolucent filler material. X-ray photons traveling in the same direction as the primary beam pass between the lead strips. X-ray photons, having undergone interactions within the body and deviated in various directions, are absorbed by the lead strips; this is referred to as cleanup of scattered radiation. An air gap introduced between the object and IR can have an effect similar to that of a grid. As energetic scattered radiation emerges from the body, it continues to travel in its divergent fashion and much of the time will bypass the IR. It is usually necessary to increase the SID to reduce magnification caused by increased OID. Imaging system speed is unrelated to scattered radiation.

281. Resolution in CR increases as 1. laser beam size decreases 2. monitor matrix size decreases 3. PSP crystal size decreases A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: Spatial resolution in CR is impacted by the size of the PSP, the size of the scanning laser beam, and monitor matrix size. High-resolution monitors (2-4 MP, megapixels) are required for high-quality, high-resolution image display. The larger the matrix size, the better is the image resolution. Typical image matrix size (rows and columns) used in chest radiography is 2,048 x 2,048. As in traditional radiography, spatial resolution is measured in line pairs per millimeter. As matrix size is increased, there are more and smaller pixels in the matrix and, therefore, improved spatial resolution. Other factors contributing to image resolution are the size of the laser beam and the size of the PSP/SPS phosphors. Smaller phosphor size improves resolution in ways similar to that of intensifying screens—anything that causes an increase in light diffusion will result in a decrease in resolution. Smaller phosphors in the PSP (SPS) plate allow less light diffusion. Additionally, the scanning laser light must be of the correct intensity and size. A narrow laser beam is required for optimal resolution

29. Accurate operation of the AEC device depends on 1. the thickness and density of the object 2. positioning of the object with respect to the photocell 3. beam restriction A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: The AEC automatically terminates the exposure when the proper density has been recorded on the film. The important advantage of the phototimer, then, is that it can accurately duplicate radiographic densities. It is very useful in providing accurate comparison in follow-up examinations and in decreasing patient exposure dose by reducing the number of "retakes" needed because of improper exposure. The AEC automatically adjusts the exposure required for body parts with different thicknesses and densities. However, proper functioning of the phototimer depends on accurate positioning by the radiographer. The correct photocell(s) must be selected, and the anatomic part of interest must completely cover the photocell to achieve the desired density. If collimation is inadequate and a field size larger than the part is used, excessive scattered radiation from the body or tabletop can cause the AEC to terminate the exposure prematurely, resulting in an underexposed radiograph.

213. In the 45-degree medial oblique projection of the ankle, the 1. talotibial joint is visualized 2. tibiofibular joint is visualized 3. plantar surface should be vertical A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: The medial oblique projection of the ankle can be performed either as a 15- to 20-degree oblique or as a 45-degree oblique. The 15- to 20-degree oblique projection demonstrates the ankle mortise, that is, the articulations between the talus, tibia, and fibula. The 45-degree oblique opens the distal tibiofibular joint. In all three cases, although the MSP can change the plantar surface must be vertical. (Frank, Long, and Smith, 11th ed., vol. 1, p. 291)

20. Which of the following statements is (are) correct regarding the parietoacanthial projection (Waters' method) of the skull? i. The head is rested on the extended chin. ii. The orbitomeatal line (OML) is perpendicular to the (IR). iii. The maxillary antra should be projected above the petrosa. A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: The parietoacanthial projection (Waters' position) of the skull is valuable for the demonstration of facial bones or maxillary sinuses. The head is rested on the extended chin so that the OML forms a 37° angle with the IR. This projects the petrous pyramids below the floor of the maxillary sinuses and provides an oblique frontal view of the facial bones.

243. The automatic film processor's recirculation system functions to 1. add quantities of solution as required. 2. maintain uniform temperatures. 3. mix and agitate solutions. A. 1 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: The processor's pumping mechanisms transport/recirculate the solution through heating devices to maintain the proper temperature. The solution then is returned under pressure for further recirculation. The added pressure functions to agitate the solution and keep it in close contact with the film emulsion. Adding additional solutions at intervals is the function of the replenishment system.

194. Which of the following function(s) to reduce the amount of scattered radiation reaching the IR? 1. Grid devices 2. Restricted focal spot size 3. Beam restrictors A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: There are several ways to reduce the amount of scattered radiation reaching the IR. First, the use of optimum kVp is essential; excessive kVp will increase the production of scattered radiation. Second, conscientious use of the beam restrictor (collimator) will reduce scattered radiation; the smaller the volume of irradiated tissue, the less scattered radiation is produced. The use of grids helps clean up scattered radiation before it reaches the IR. The size of the tube focus has an impact on image geometry and recorded detail, but it has no effect on scattered radiation. (Shephard, p 203)

15. Which of the following radiographic examinations require(s) the patient to be NPO 8-10 hours prior to examination for proper patient preparation? i. Abdominal survey ii. Upper GI series iii. BE A. 1 and 2 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: There is no preparation required for an abdominal survey. For an upper GI series and a lower GI series (BE), the patient should be NPO, or have nothing by mouth, for 8 to 10 hours prior to the examination. In addition, a low-residue diet may be imposed, fluid intake may be increased, and cleansing enemas and laxatives may be prescribed to rid the colon of fecal matter.

261. During an upper gastrointestinal (GI) examination, the AP recumbent projection of a stomach of average shape will usually demonstrate 1. anterior and posterior aspects of the stomach. 2. barium-filled fundus. 3. double-contrast body and antral portions. A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is C. 2. EXPLANATION: With the body in the AP recumbent position, barium flows easily into the fundus of the stomach, displacing the stomach somewhat superiorly. The fundus, then, is filled with barium, while the air that had been in the fundus is displaced into the gastric body, pylorus, and duodenum, illustrating them in double-contrast fashion. Air-contrast delineation of these structures allows us to see through the stomach to the retrogastric areas and structures. Anterior and posterior aspects of the stomach are visualized in the lateral position; medial and lateral aspects of the stomach are visualized in the AP projection. (Ballinger & Frank, vol 2, p 110)

15. To radiograph an infant for suspected free air within the abdominal cavity, which of the following projections of the abdomen will demonstrate the condition with the least patient exposure? a. PA erect with grid b. Left lateral decubitus with grid c. Left lateral decubitus without grid d. Recumbent AP without grid

The answer is C. ii. EXPLANATION: Air-fluid levels are demonstrated in the erect or decubitus position. Grid radiography requires about a 3 to 4 times greater dose than nongrid radiography. A left lateral decubitus projection without a grid, then, would demonstrate fluid levels with a considerably smaller dose to the infant. PA erect with grid would be the next-best choice - the PA position reduces dose to radiosensitive future reproductive cells. A recumbent AP projection would not demonstrate air-fluid levels. Left lateral decubitus is preferred to right lateral decubitus because the liver moves down slightly in this position, creating a space between the liver and right hemidiaphragm (right subphrenic space). Any abdominal free air would then be visualized below the right hemidiaphragm in the subphrenic space along the lateral aspect of the liver. (Bontrager and Lampignano, 6th ed., p. 676)

51. Which section of the automatic processor shown in Figure 6-9 is associated with preservation of the x-ray image? a. Section 1 b. Section 2 c. Section 3 d. Section 4

The answer is C. ii. EXPLANATION: As the exposed film enters the processor from the feed tray, it first enters the developer section (number 1), where exposed silver bromide crystals are reduced to black metallic silver. The film then enters the fixer (number 2), where the unexposed silver grains are removed from the film by the clearing agent. The film then enters the wash section (number 3), where chemicals are removed from the film to preserve the image, improving archival quality. From the wash, the film enters the dryer section (number 4). (Selman, 9th ed., p. 194)

16. Which of the following combinations will result in the most scattered radiation reaching the image receptor? a. Using more mAs and compressing the part b. Using more mAs and a higher ratio grid c. Using fewer mAs and more kVp d. Using more mAs and less kVp

The answer is C. ii. EXPLANATION: As x-ray photons travel through a part, they either pass all the way through to expose the film/image receptor, or they undergo interaction(s) that may result in their being absorbed by the part or deviated in direction. It is those that change direction (scattered radiation) that undermine the image. With respect to the radiographic image, it is responsible for the scattered radiation that reaches the film/image receptor. Scattered radiation adds unwanted, degrading densities to the radiographic image. The single most important way to reduce the production of scattered radiation is to collimate. Although collimation, use of lower kVp (with appropriately higher mAs), and compression can be used, a large amount of scattered radiation is still generated within the part being radiographed and, because it adds unwanted noninformation-carrying densities, it can have a severely degrading effect on image quality, thus the need for grids. (Bushong, 8th ed, p 236)

46. According to the CDC, all the following precaution guidelines are true except a. airborne precautions require that the patient wear a mask b. masks are indicated when caring for patients on MRSA precautions c. patients under MRSA precautions require a negative-pressure room d. gloves are indicated when caring for a patient on droplet precautions

The answer is C. ii. EXPLANATION: Category-specific isolations have been replaced by transmission-based precautions: airborne, droplet, and contact. Under these guidelines, some conditions or diseases can fall into more than one category. Airborne precautions are employed with patients suspected or known to be infected with tubercle bacillus (TB), chickenpox (varicella), or measles (rubeola). Airborne precautions require that the patient wear a mask to avoid the spread of bronchial secretions or other pathogens during coughing. If the patient is unable or unwilling to wear a mask, the radiographer must wear one. The radiographer should wear gloves, but a gown is required only if flagrant contamination is likely. Patients under airborne precautions require a private, specially ventilated (negative-pressure) room. A private room is also indicated for all patients on droplet precautions, that is, with diseases transmitted via large droplets expelled from the patient while speaking, sneezing, or coughing. The pathogenic droplets can infect others when they come in contact with mouth or nasal mucosa or conjunctiva. Rubella ("German measles"), mumps, and influenza are among the diseases spread by droplet contact; a private room is required for the patient, and health care practitioners should use gowns and gloves. Any diseases spread by direct or close contact, such as methicillin-resistant Staphylococcus aureus (MRSA), conjunctivitis, and hepatitis A, require contact precautions. Contact precautions require a private patient room and the use of gloves, masks, and gowns for anyone coming in direct contact with the infected individual or his or her environment.

8. Europium-activated barium fluorohalide is associated with a. rare earth intensifying screens b. image intensifiers c. PSP storage plates d. filament material

The answer is C. ii. EXPLANATION: Computed radiography (CR) cassettes use no intensifying screens or film—hence, the term filmless radiography. The Image Plates (IPs) have a protective function (for the PSP/storage plate within) and can be used in the Bucky tray or directly under the anatomic part; they need not be light-tight because the PSP is not light sensitive. The IP has a thin lead-foil backing (similar to traditional cassettes) to absorb backscatter. Inside the IP is the photostimulable phosphor (PSP) storage plate. This PSP storage plate within the IP has a layer of europium-activated barium fluorohalide that serves as the IR as it is exposed in the traditional manner and receives the latent image. The PSP can store the latent image for several hours; after about 8 hours, noticeable image fading will occur. (Carlton and Adler, 4th ed., p. 358)

334. Which cholangiographic procedure uses an indwelling drainage tube for contrast medium administration? a. Endoscopic retrograde cholangiographic pancreatography (ERCP) b. Operative cholangiography c. T-tube cholangiography d. Percutaneous transhepatic cholangiography

The answer is C. ii. EXPLANATION: Contrast media may be administered in a variety of manners in cholangiography, including (1) an endoscope with a cannula placed in the hepatopancreatic ampulla (of Vater) for an ERCP, (2) a needle or small catheter placed directly in the common bile duct for an operative cholangiogram, (3) a very fine needle through the patient's side and into the liver for a percutaneous transhepatic cholangiogram, and (4) via an indwelling T-tube for a postoperative or T-tube cholangiogram.

82. In which type of monitoring device do photons release electrons by their interaction with air? a. Film badge b. TLD c. Pocket dosimeter d. OSL dosimeter

The answer is C. ii. EXPLANATION: Different types of monitoring devices are available for the occupationally exposed. Ionization is the fundamental principle of operation of both the film badge and the pocket dosimeter. In the film badge, the film's silver halide emulsion is ionized by x-ray photons. The pocket dosimeter contains an ionization chamber (containing air), and the number of ions formed (of either sign) is equated to exposure dose. TLDs are radiation monitors that use lithium fluoride crystals. Once exposed to ionizing radiation and then heated, these crystals give off light proportional to the amount of radiation received. OSL dosimeters are radiation monitors that use aluminum oxide crystals. These crystals, once exposed to ionizing radiation and then subjected to a laser, give off luminescence proportional to the amount of radiation received. (Selman, 9th ed., p. 400)

22. The dose of radiation that will cause a noticeable skin reaction is referred to as the a. LET b. SSD c. SED d. SID

The answer is C. ii. EXPLANATION: Erythema is the reddening of skin as a result of exposure to large quantities of ionizing radiation. It was one of the first somatic responses to irradiation demonstrated to the early radiology pioneers. The effects of radiation exposure to the skin follow a nonlinear, threshold dose-response relationship. An individual's response to skin irradiation depends on the dose received, the period of time over which it was received, the size of the area irradiated, and the individual's sensitivity. The dose that it takes to bring about a noticeable erythema is referred to as the SED. (Bushong, 8th ed., p. 521)

4. The instrument that is used frequently in quality-control programs to measure varying degrees of x-ray exposure is the a. aluminum step wedge. b. spinning top. c. densitometer. d. sensitometer.

The answer is C. ii. EXPLANATION: Every radiographic image is composed of a number of different densities. These densities may be measured and given a numeric value with a device called a densitometer. A sensitometer is another device used in QA programs; it is used to give a precise exposure to a film emulsion. An aluminum step wedge (penetrometer) may be used to show the effect of kilovoltage on contrast. A spinning top is used to test the accuracy of the x-ray machine's timer or rectifiers. (Bushong, 8th ed., p. 275)

327. Primary radiation barriers usually require which thickness of shielding? a. 1/4-in. lead b. 1/8-in. lead c. 1/16-in. lead d. 1/32-in. lead

The answer is C. ii. EXPLANATION: Examples of primary barriers are the lead walls and doors of a radiographic room, that is, any surface that could be struck by the useful beam. Primary protective barriers of typical installations generally consist of walls with 1/16 in. (1.5 mm) of lead and 7 ft high. Secondary radiation is defined as leakage and/or scattered radiation. The x-ray tube housing protects from leakage radiation, as stated earlier. The patient is the source of most scattered radiation. Secondary radiation barriers include that portion of the walls above 7 ft in height; this area requires only 1/32 in. of lead. The control booth is also a secondary barrier, toward which the primary beam must never be directed

255. The most common cause of x-ray tube failure is a. a cracked anode. b. a pitted anode. c. vaporized tungsten on glass envelope. d. insufficient heat production.

The answer is C. ii. EXPLANATION: Excessive heat production is a major problem in x-ray production. Of the energy required to produce x-rays, 0.2% is transformed to x-rays, and 99.8% is transformed to heat. The copper anode stem and the oil surrounding the x-ray tube help to move heat away from the face of the anode. Excessive heat can cause pitting of the anode (resulting in decreased output) or actual cracking of the anode or damage to the rotor bearings (resulting in tube failure). As the cathode filament is heated for exposure after exposure, some of its tungsten is vaporized and deposited on the inner surface of the glass envelope near the tube window. After a time, this can cause electric arcing and tube failure. This is the most common cause of tube failure because it can occur even with normal use.

167. Film base is currently made of which of the following materials? a. Cellulose nitrate b. Cellulose acetate c. Polyester d. Glass

The answer is C. ii. EXPLANATION: Film base functions to support the silver halide emulsion. Today's film base is made of tough, nonflammable polyester. Cellulose nitrate was used in the past, but it was highly flammable. Cellulose acetate, also used in the past, was not flammable, but it was not as durable as polyester. The earliest supports for emulsion were plates of glass (hence, the term flat plate). (Selman, 9th ed., p. 194)

177. Geometric blur can be evaluated using all the following devices except a. star pattern b. slit camera c. penetrometer d. pinhole camera

The answer is C. ii. EXPLANATION: Focal-spot size accuracy is related to the degree of geometric blur, that is, edge gradient or penumbra. Manufacturer tolerance for new focal spots is 50%; that is, a 0.3-mm focal spot actually may be 0.45 mm. Additionally, the focal spot can increase in size as the x-ray tube ages—hence the importance of testing newly arrived focal spots and periodic testing to monitor focal-spot changes. Focal-spot size can be measured with a pinhole camera, slit camera, or star-pattern-type resolution device. The pinhole camera is rather difficult to use accurately and requires the use of excessive tube (heat) loading. With a slit camera, two exposures are made; one measures the length of the focal spot, and the other measures the width. The star pattern, or similar resolution device such as the bar pattern, can measure focal-spot size as a function of geometric blur and is readily adaptable in a QA program to monitor focal-spot changes over a period of time. It is recommended that focal-spot size be checked on installation of a new x-ray tube and annually thereafter. (Bushong, 8th ed., p. 462)

262. An exposure was made of a part using 300 mA and 0.06 second with a 200-speed film-screen combination. An additional radiograph is requested using a 400-speed system to reduce motion unsharpness. Using 400 mA, all other factors remaining constant, what should be the new exposure time? a. 5 ms b. 11 ms c. 22 ms d. 44 ms

The answer is C. ii. EXPLANATION: High-speed imaging systems are valuable for reducing patient exposure and patient motion. However, some detail will be sacrificed, and quantum mottle can cause further image impairment. In general, doubling the film-screen speed doubles the radiographic density, thereby requiring that the milliampere-seconds value be halved to maintain the original radiographic density. Changing from 200 to 400 screens requires halving the milliampere-seconds value to 9 mAs. The new exposure time, using 400 mA, is 400x = 9. Thus, x = 0.0225-s exposure using 400 mA and 400-speed screens (0.0225 = 22.5 ms). (Selman, 9th ed., p. 181)

39. The appearance of underexposure on an image created using a high-speed film-screen system can be caused by all the following except a. insufficient mAs b. insufficient kV c. insufficient SID d. insufficient development

The answer is C. ii. EXPLANATION: Higher-speed film/screen imaging systems are used often to produce more density with less exposure. Since the milliampere-seconds value is the factor controlling image density, an insufficient amount would result in underexposure. Since kilovoltage has a definite effect on image density, an insufficient amount would result in underexposure. Underdevelopment of the latent film emulsion image also would result in insufficient density. However, insufficient SID (i.e., SID too low) would cause an increase in image density/overexposure. (Shephard p. 179)

26. Another name for Hirschsprung's disease, the most common cause of lower GI obstruction in neonates, is a. Intussusception b. Volvulus c. congenital megacolon d. pyloric stenosis

The answer is C. ii. EXPLANATION: Hirschsprung disease, or congenital megacolon, is caused by the absence of some or all of the bowel ganglion cells—usually in the rectosigmoid area but occasionally more extensively. Hirschsprung disease is the most common cause of lower GI obstruction in neonates and is treated surgically by excision of the affected area followed by reanastomosis with the normal, healthy bowel. Hirschsprung disease is diagnosed by BE or, in mild cares, by rectal biopsy. Intussusception is "telescoping" of the bowel, causing (mechanical) obstruction. Volvulus is twisting of the bowel on itself causing (mechanical) obstruction. Pyloric stenosis is a condition of the upper GI tract. (Bontrager and Lampignano, 6th ed., p. 659)

46. If the quantity of black metallic silver on a particular x-ray film is such that it allows 1% of the illuminator light to pass through the film, that film has a density of a. 0.1 b. 1.0 c. 2.0 d. 3.0

The answer is C. ii. EXPLANATION: If a film is placed on an illuminator and 100% of the illuminator's light is transmitted through the film, that film must have a density of 0. According to the equation density = log10(incident intensity/transmitted light intensity) if 10% of the illuminator's light passes through the film, that film has a density of 1. If 1% of the light passes through the film, that film has a density of 2; if 0.1% of the illuminator's light passes through the film the density is 3, and so on. (Shephard, p. 102)

173. If the quantity of black metallic silver on a particular radiograph is such that it allows 10% of the illuminator light to pass through the x-ray image, that image has a density of a. 0.01. b. 0.1. c. 1.0. d. 2.0.

The answer is C. ii. EXPLANATION: If an x-ray image were placed on an illuminator and 100% of the illuminator's light was transmitted through the image, that image must have a density of 0. According to the equation, density = log10(incident light intensity/transmitted light intensity) If 10% of the illuminator's light passes through the image, that image has a density of 1. If 1% of the light passes through the image, that image has a density of 2. (Fauber, 2nd ed., p. 197)

42. The cause of films coming from the automatic processor still damp can be a. air velocity too high b. unbalanced processing temperatures c. insufficient hardening action d. underreplenishment

The answer is C. ii. EXPLANATION: If the fixer fails to harden the gelatin emulsion sufficiently, water will remain within the still-swollen emulsion. The dryer mechanism will be unable to completely rid the emulsion of wash water, and the film will emerge from the processor damp and tacky. On the other hand, excessive hardening action may produce brittle radiographs. High air velocity usually encourages more complete drying. Unbalanced processing temperatures can result in blistering of the emulsion. Developer underreplenishment results in "light" images and can be the cause of transport problems as a result of insufficient hardener. (Carlton and Adler, 4th ed., p. 289)

112. What angle is formed by the median sagittal plane and the IR in the parieto-orbital projection (Rhese method) of the optic canal? a. 90° b. 37° c. 53° d. 45°

The answer is C. ii. EXPLANATION: In the parieto-orbital projection, the patient is PA with the acanthomeatal line perpendicular to the IR. The head rests on the zygoma, nose, and chin, and the MSP should form a 53° angle with the IR (37° with the central ray). Radiographically, the optic canal should appear in the lower outer quadrant of the orbit. Incorrect rotation of the MSP results in lateral displacement, and incorrect positioning of the baseline results in longitudinal displacement. (Ballinger & Frank, vol 2, pp 290-291)

132. Which of the following personnel monitoring devices used in diagnostic radiography is considered to be the most sensitive and accurate? a. TLD b. Film badge c. OSL dosimeter d. Pocket dosimeter

The answer is C. ii. EXPLANATION: Ionization is the fundamental principle of operation of both the film badge and the pocket dosimeter. In the film badge, the film's silver halide emulsion is ionized by x-ray photons. The pocket dosimeter contains an ionization chamber, and the number of ionizations taking place may be equated to exposure dose; it is accurate, but it is used only to detect larger amounts of radiation exposure. The TLD can measure exposures as low as 5 mrem, whereas film badges will measure a minimum exposure only as low as 10 mrem. TLDs contain lithium fluoride crystals that undergo characteristic changes on irradiation. When the crystals are subsequently heated, they emit a quantity of visible (thermo) luminescence/light in proportion to the amount of radiation absorbed. The relatively new OSL dosimeters contain aluminum oxide crystals that also undergo characteristic changes on irradiation. When the Al2O3 crystals are stimulated by a laser, they emit (optically stimulated) luminescence/light in proportion to the amount of radiation absorbed. OSL dosimeters can measure exposures as low as 1 mrem.

160. Which of the following ionizing radiations is described as having an RBE of 1.0? a. 10 MeV protons b. 5 MeV alpha particles c. Diagnostic x-rays d. Fast neutrons

The answer is C. ii. EXPLANATION: LET increases with the ionizing potential of the radiation; for example, alpha particles are more ionizing than x-radiation, and, therefore, they have a higher LET. As ionizations and LET increase, there is greater possibility of an effect on living tissue; therefore, the RBE increases. The RBE [sometimes called quality factor (QF)] of diagnostic x-rays is 1, the RBE of fast neutrons is 10, the RBE of 5-MeV alpha particles is 20, and the RBE of 10-MeV protons is 5.0. (Bushong, 8th ed., p. 496)

100. How much protection is provided from a 75-kVp x-ray beam when using a 0.50-mm lead equivalent apron? a. 51% b. 66% c. 88% d. 99%

The answer is C. ii. EXPLANATION: Lead aprons are worn by occupationally exposed individuals during fluoroscopic procedures. Lead aprons are available with various lead equivalents; 0.25, 0.5, and 1.0 mm are the most common. The 1.0-mm lead equivalent apron will provide close to 100% protection at most kVp levels, but it is rarely used because it weighs anywhere from 12 to 24 lb. A 0.25-mm lead equivalent apron will attenuate about 97% of a 50-kVp x-ray beam, 66% of a 75-kVp beam, and 51% of a 100-kVp beam. A 0.5-mm apron will attenuate about 99% of a 50-kVp beam, 88% of a 75-kVp beam, and 75% of a 100-kVp beam. (Thompson et al, p 457)

98. An exposure was made at a 36-in. SID using 12 mAs and 75 kVp with a 400-speed imaging system and an 8:1 grid. A second radiograph is requested with improved recorded detail. Which of the following groups of technical factors will best accomplish this task? a. 15 mAs, 12:1 grid, 75 kVp, 400-speed system, 36-in. SID b. 15 mAs, 12:1 grid, 75 kVp, 400-speed system, 40-in. SID c. 30 mAs, 12:1 grid, 75 kVp, 200-speed system, 40-in. SID d. 12 mAs, 8:1 grid, 86 kVp, 200-speed system, 36-in. SID

The answer is C. ii. EXPLANATION: Look over the choices again, keeping in mind the factors that affect recorded detail. Looking first at SID, the options may be reduced to (B) and (C) because the increase to a 40-in. SID certainly will improve recorded detail. There is one other factor that will affect detail—the speed of the system (intensifying screens). Because a slower system will render better recorded detail, the best answer is (C). The technical factors such as milliampere-seconds, kilovoltage, and grid ratio have no effect on recorded detail.

10. All the following statements regarding the use of iodinated contrast agents with patients taking metformin hydrochloride are true except a. metformin is used to help lower blood sugar levels in type 2 diabetic patients b. patients on metformin who have intravenous (IV) iodinated contrast agent administration are at risk for renal failure c. metformin should be withheld for 48 hours before IV iodinated contrast studies d. metformin should be withheld for 48 hours after IV iodinated contrast studies

The answer is C. ii. EXPLANATION: Metformin hydrochloride (Glucophage) is used as an adjunct to appropriate diet to lower blood glucose levels in patients who have type 2 diabetes and whose hyperglycemia is not being managed satisfactorily with diet alone. Patients on Glucophage who are having intravascular iodinated contrast studies can develop an acute alteration of renal function or acute acidosis. It is recommended that patients on Metformin hydrochloride (Glucophage) have it withheld 48 hours after the examination

278. Which of the following modes of a trifield image intensifier will result in the highest patient dose? a. Its 25-in. mode b. Its 17-in. mode c. Its 12-in. mode d. Diameter does not affect patient dose

The answer is C. ii. EXPLANATION: Most image-intensifier tubes are either dual-field or trifield, indicating the diameter of the input phosphor. When a change to a smaller-diameter mode is made, the voltage on the electrostatic focusing lenses is increased, and the result is a magnified but dimmer image. The milliamperage will be increased automatically to compensate for the loss in brightness with a magnified image, resulting in higher patient dose in the smaller-diameter modes.

3. Which of the following projections of the elbow should demonstrate the coronoid process free of superimposition and the olecranon process within the olecranon fossa? a. AP b. Lateral c. Medial oblique d. Lateral oblique

The answer is C. ii. EXPLANATION: On the AP projection of the elbow, the radial head and ulna are normally somewhat superimposed. The lateral oblique demonstrates the radial head free of ulnar superimposition. The lateral projection demonstrates the olecranon process in profile. The medial oblique demonstrates considerable overlap of the proximal radius and ulna, but should clearly demonstrate the coronoid process free of superimposition and the olecranon process within the olecranon fossa. (Saia, p 96)

43. Which of the following projections of the elbow should demonstrate the coronoid process free of superimposition and the olecranon process within the olecranon fossa? a. AP b. Lateral c. Medial oblique d. Lateral oblique

The answer is C. ii. EXPLANATION: On the AP projection of the elbow, the radial head and ulna are normally somewhat superimposed. The lateral oblique demonstrates the radial head free of ulnar superimposition. The lateral projection demonstrates the olecranon process in profile. The medial oblique demonstrates considerable overlap of the proximal radius and ulna, but should clearly demonstrate the coronoid process free of superimposition and the olecranon process within the olecranon fossa. (Saia, p 96)

259. Which of the following is a violation of correct sterile technique? a. Gowns are considered sterile in the front down to the waist, including the arms. b. Sterile gloves must be kept above the waist level. c. Persons in sterile dress should pass each other face to face. d. A sterile field should not be left unattended.

The answer is C. ii. EXPLANATION: Persons in sterile dress should not pass each other face to face. Rather, they should pass each other back to back to avoid contaminating each other. Gowns are considered sterile in the front down to the waist, including the arms. Sterile gloves must be kept above the waist level. If the hands are accidentally lowered or placed behind the back, they are no longer sterile. A sterile field should not be left unattended. Sterile fields should be set up immediately prior to a procedure and should be covered with a sterile drape if a few moments are to elapse before the procedure can begin. A sterile field should be monitored constantly to be certain that it has not been contaminated. (Adler and Carlton, 4th ed., p. 247)

128. Which acute radiation syndrome requires the largest exposure before any effects become apparent? a. Hematopoietic b. Gastrointestinal c. Central nervous system (CNS) d. Skeletal

The answer is C. ii. EXPLANATION: Radiation effects that appear days or weeks following exposure (early effects) are in response to high radiation doses; this is called acute radiation syndrome. These effects should never occur in diagnostic radiology; they occur only in response to much greater doses. Sufficient exposure of the hematologic system to ionizing radiation can result in nausea, vomiting, diarrhea, decreased blood cells count, and infection. Very large exposure of the GI system (1,000-5,000 rad) causes severe damage to the (stem) cells lining the GI tract. This can result in nausea, vomiting, diarrhea, blood changes, and hemorrhage. Exposure greater than 5,000 rad is required to affect the normally resilient CNS. (Bushong, 8th ed., pp. 517-518)

119. Which of the following combinations would deliver the least amount of heat to the anode of a three-phase, 12-pulse x-ray unit? a. 400 mA, 0.12 s, 90 kVp b. 300 mA, 1/2 s, 70 kVp c. 500 mA, 1/30 s, 85 kVp d. 700 mA, 0.06 s, 120 kVp

The answer is C. ii. EXPLANATION: Radiographic rating charts enable the operator to determine the maximum safe milliamperage, exposure time, and kilovoltage for a particular exposure using a particular x-ray tube. An exposure that can be made using the large focal spot may not be safe when the small focal spot of the same x-ray tube is used. The total number of heat units an exposure generates also influences the amount of stress (in the form of heat) imparted to the anode. Single-phase heat units are determined by the product of milliampere-seconds and kilovoltage. Three-phase, six-pulse heat units are determined from the product of milliamperage x time x kilovoltage x 1.35. Three-phase, 12-pulse heat units are determined from the product of milliamperage x time x kilovoltage x 1.41. In the examples given, then, group (A) produces 6,091 HU, group (B) produces 14,805 HU, group (C) produces 1,997 HU, and group (D) produces 7,106 HU. Therefore, group (A) exposure factors will deliver the least amount of heat to the anode. (Fauber, 2nd ed., pp. 36-37)

326. If your patient is unable to stay erect for a paranasal sinus examination, which of the following alternatives should be chosen? a. Recumbent AP b. Lateral recumbent c. Lateral cross-table recumbent d. Recumbent Waters'

The answer is C. ii. EXPLANATION: Radiography of the paranasal sinuses should be performed in the erect position whenever possible to demonstrate the presence of an air-fluid level. The only way air-fluid levels can be demonstrated is to have the central ray parallel the floor, as in erect, decubitus, and cross-table projections. Therefore, of the choices provided, the cross-table lateral is the only one that will demonstrate air-fluid levels.

122. A vasomotor effect experienced after injection of a contrast agent is characterized by all of the following symptoms except a. nausea. b. syncope. c. hypotension. d. anxiety

The answer is C. ii. EXPLANATION: Reactions to contrast agents are named and categorized according to the body system(s) affected, the nature of the reaction (i.e., allergic vs. nonallergic), and its severity (i.e., mild, moderate, or severe). These reactions are categorized as vasomotor (a nonallergic reaction), anaphylactic (allergic reaction), vasovagal (life-threatening), and acute renal failure (renal shutdown). Vasomotor effects are principally emotional and anxiety-based. They are characterized by anxiety, syncope, nausea, lightheadedness, and sometimes a few hives. The patient usually just requires reassurance and not medical attention. An anaphylactic reaction is a true allergic reaction to e.g. iodinated media and can lead to a life-threatening situation. Immediate medical attention is required. Symptoms of anaphylactic reaction include laryngo/bronchospasm, hypotension, moderate to severe urticaria, angioedema, and tachycardia. A vasovagal reaction is life-threatening and requires a declared emergency ("code"). Symptoms of a vasovagal reaction include bradycardia, hypotension, and no detectable pulse. The fourth type of reaction, acute renal failure, may not manifest for up to 48 hours following injection of the contrast agent. Patients should notify their physician if they experience any changes in their urinary habits or any other atypical symptoms. Treatment would include hydration, dispensation of a diuretic (e.g., Lasix), and possibly even renal dialysis. (Bontrager and Lampignano, 6th ed., p. 558)

234. All the following procedures demonstrate renal function except a. IVP. b. descending urography. c. retrograde urography. d. infusion nephrotomography.

The answer is C. ii. EXPLANATION: Retrograde urography is not considered a functional study of the urinary system. IVP, descending urography, and infusion nephrotomography are all considered functional urinary tract studies because the contrast medium is introduced intravenously and excreted by the kidneys. Retrograde urography involves introduction of contrast medium into the kidneys via catheter, thereby demonstrating their structure but not their function. (Frank, Long, and Smith, 11th ed., vol. 2, p. 207)

11. Which of the following is a measurement of dose to biologic tissue? a. Roentgen (C/kg) b. Rad (Gy) c. Rem (Sv) d. RBE

The answer is C. ii. EXPLANATION: Roentgen is the unit of exposure; it measures the quantity of ionization in air. Rad is an acronym for radiation absorbed dose; it measures the energy deposited in any material. Rem is an acronym for radiation equivalent man; it includes the RBE specific to the tissue irradiated, and therefore is a valid unit of measurement for the dose to biologic material. (Bushong, p 24)

283. In which of the following projections is the talofibular joint best demonstrated? a. AP b. Lateral oblique c. Medial oblique d. Lateral

The answer is C. ii. EXPLANATION: The AP projection demonstrates superimposition of the distal fibula on the talus; the joint space is not well seen. The 15- to 20-degree medical oblique position shows the entire mortise joint; the talofibular joint is well visualized, as well as the talotibial joint. There is considerable superimposition of the talus and fibula in the lateral and lateral oblique projections.

187. With the patient seated at the end of the x-ray table, elbow flexed 80 degrees, and the CR directed 45 degrees laterally from the shoulder to the elbow joint, which of the following structures will be demonstrated best? a. Radial head b. Ulnar head c. Coronoid process d. Olecranon process

The answer is C. ii. EXPLANATION: The axial trauma lateral (Coyle) position is described. If routine elbow projections in extension are not possible because of limited part movement, these positions can be used to demonstrate the coronoid process and/or radial head. With the elbow flexed 90 degrees and the CR directed to the elbow joint at an angle of 45 degrees medially (i.e., toward the shoulder), the joint space between the radial head and capitulum should be revealed. With the elbow flexed 80 degrees and the CR directed to the elbow joint at an angle of 45 degrees laterally (i.e., from the shoulder toward the elbow), the elongated coronoid process will be visualized. (Bontrager and Lampignano, 6th ed., p. 174)

171. Which of the following is a functional study used to demonstrate the degree of AP motion present in the cervical spine? a. Moving mandible position b. AP open-mouth projection c. Flexion and extension laterals d. AP right and left bending

The answer is C. ii. EXPLANATION: The degree of anterior and posterior motion is occasionally diminished with a whiplash-type injury. Anterior (forward, flexion) and posterior (backward, extension) motion is evaluated in the lateral position with the patient assuming flexion and extension as best as he or she can. Left and right bending images of the thoracic and lumbar vertebrae are obtained frequently when evaluating scoliosis. The AP open-mouth projection is used to evaluate the first two cervical vertebrae. The moving mandible AP projection is used to demonstrate the entire cervical spine while blurring out the superimposed mandible. (Frank, Long, and Smith, 11th ed., vol. 1, p. 402)

106. What is the annual dose-equivalent limit for the skin and hands of an occupationally exposed individual? a. 5 rem b. 25 rem c. 50 rem d. 100 rem

The answer is C. ii. EXPLANATION: The dose-equivalent limit for the hands and skin of an occupationally exposed individual is 50 rem (500 mSv). The dose-equivalent limit for the lens of the eye is 15 rem (150 mSv). An occupationally exposed individual may receive up to 3 rem (30 mSv) in a given calendar quarter, or 13-week period. However, that individual may not exceed 5 rem (50 mSv) in that particular year. If, for example, one received 3 rem (30 mSv) during the first 3 months of a year, that individual must not receive more than 2 rem (20 mSv) in the remaining 9 months. (Bushong, 9th ed, p 619)

55. During CR imaging, the latent image present on the PSP is changed to a computerized image by the a. PSP b. Scanner-reader c. ADC d. helium-neon laser

The answer is C. ii. EXPLANATION: The exposed CR cassette is placed into the CR scanner/reader, where the PSP (SPS) is removed automatically. The latent image appears as the PSP is scanned by a narrow, high-intensity helium-neon laser to obtain the pixel data. As the plate is scanned in the CR reader, it releases a violet light—a process referred to as photostimulated luminescence (PSL). The luminescent light is converted to electrical energy representing the analog image. The electrical energy is sent to an analog-to-digital converter (ADC), where it is digitized and becomes the digital image that is displayed eventually (after a short delay) on a high-resolution monitor and/or printed out by a laser printer. The digitized images can also be manipulated in postprocessing, transmitted electronically, and stored/archived. (Carlton and Adler, 4th ed., p. 358)

108. Which of the following can affect the amount of developer replenisher delivered per film in an automatic processor? a. Developer temperature b. Amount of film exposure c. Film size d. Processor capacity

The answer is C. ii. EXPLANATION: The film processor is automated, and replenishment quantities are preset. A microswitch is activated as a film enters the processor at the entrance rollers. Replenisher is added according to the length of the x-ray film—for as long as the detector senses the presence of film. Once the back end of the film passes the entrance roller sensor, replenishment stops. When films are fed into the processor the "long way," too much replenishment occurs, and the image can exhibit excessive density. (Shephard, p. 146)

150. Fluoroscopic imaging of the ileocecal valve is generally part of a(n) a. esophagram. b. upper GI series. c. small-bowel series. d. ERCP.

The answer is C. ii. EXPLANATION: The ileocecal valve is located at the terminal ileum, where it meets the first portion of the large bowel, the cecum. Most small-bowel examinations are performed following oral administration of barium sulfate suspension. The first small-bowel radiograph is taken 15 minutes after the first swallow of barium, with subsequent radiographs made every 15 to 30 minutes, depending on how quickly the barium is moving through the small bowel. Each image is shown to the radiologist, and a decision is made regarding the time of the next image. When the barium reaches the terminal ileum, fluoroscopy may be performed and compression spot images taken of the ileocecal valve. (Ballinger & Frank, vol 2, p 116)

14. The function of the developer solution chemicals is to a. reduce the manifest image to a latent image b. increase production of silver halide crystals c. reduce the latent image to a manifest image d. remove the unexposed crystals from the film

The answer is C. ii. EXPLANATION: The latent image is the invisible image produced within the film emulsion as a result of exposure to radiation. The developer solution converts this to a visible manifest image. The exposed silver halide grains in the emulsion undergo chemical change in the developer solution, and the unexposed crystals are removed from the film during the fixing process. (Fauber, 2nd ed., p. 163)

292. Which of the following positions demonstrates all the paranasal sinuses? a. Parietoacanthial b. PA axial c. Lateral d. True PA

The answer is C. ii. EXPLANATION: The parietoacanthial (Waters' method) projection demonstrates the maxillary sinuses. The PA axial with a caudal central ray (Caldwell) demonstrates the frontal and ethmoidal sinus groups. The lateral projection, with the central ray entering 1 inch posterior to the outer canthus, demonstrates all the paranasal sinuses. X-ray examinations of the sinuses should always be performed erect, to demonstrate leveling of any fluid present.

14. A particular milliampere-seconds value, regardless of the combination of milliamperes and time, will reproduce the same radiographic density. This is a statement of the a. line-focus principle b. inverse-square law c. reciprocity law d. law of conservation of energy

The answer is C. ii. EXPLANATION: The reciprocity law states that a particular milliampere-seconds value, regardless of the milliamperage and exposure time used, will provide identical radiographic density. This holds true with direct exposure techniques, but it does fail somewhat with the use of intensifying screens. However, the fault is so slight as to be unimportant in most radiographic procedures. (Shephard, p. 193)

142. The functions of which body system include mineral homeostasis, protection, and triglyceride storage? a. Endocrine b. Integumentary c. Skeletal d. Muscular

The answer is C. ii. EXPLANATION: The skeleton's design functions to protect vital internal organs such as the heart and lungs. Bone stores important minerals (e.g., calcium and phosphorus) and releases them into the blood as needed. Yellow bone marrow is composed mainly of fat cells and stores triglycerides for use as an energy reserve. The endocrine system is associated with hormone production; the integumentary system includes the skin that is important in protection and excretion; the muscular system is responsible for movement and heat production. (Tortora and Derrickson, 11th ed., p. 172)

12. Biologic material is most sensitive to radiation exposure under which of the following conditions? a. Anoxic b. Hypoxic c. Oxygenated d. Deoxygenated

The answer is C. ii. EXPLANATION: Tissue is most sensitive to radiation exposure when it is in an oxygenated condition. Anoxic refers to a general lack of oxygen in tissue; hypoxic refers to tissue with little oxygen. Anoxic and hypoxic tumors typically are avascular (with little or no blood supply) and, therefore, more radioresistant.

18. Which of the following radiologic examinations requires preparation consisting of a low-residue diet, cathartics, and enemas? a. Upper GI series b. Small bowel series c. Barium enema (BE) d. Intravenous (IV) cystogram

The answer is C. ii. EXPLANATION: To have high diagnostic quality, a barium enema (BE) examination requires rigorous and complete patient preparation. This usually consists of a modified low-residue diet for a few days before the examination, cathartics the day before, and cleansing enemas the morning of the examination. Instructions for a upper GI series, small bowel series, and IV cystogram are usually to be NPO after midnight. (Frank, Long, and Smith, 11th ed., vol. 2, p. 159)

57. Which of the following should be performed to rule out subluxation or fracture of the cervical spine? a. Oblique cervical spine, seated b. AP cervical spine, recumbent c. Horizontal beam lateral d. Laterals in flexion and extension

The answer is C. ii. EXPLANATION: When a cervical spine radiograph is requested to rule out subluxation or fracture, the patient will arrive in the radiology area on a stretcher. The patient should not be moved before a subluxation is ruled out. Any movement of the head and neck could cause serious damage to the spinal cord. A horizontal beam lateral projection is performed and evaluated. The physician then will decide what further images are required. (Frank, Long, and Smith, 11th ed., vol. 2, p. 35)

40. Which of the following technical changes would best serve to remedy the effect of very dissimilar tissue densities? a. Use of high-speed screens b. Use of a high-ratio grid c. High-kilovoltage exposure factors d. High milliampere-seconds exposure factors

The answer is C. ii. EXPLANATION: When tissue densities within a part are very dissimilar (e.g., chest x-ray), the radiographic result can be unacceptably high contrast. To "even out" these densities and produce a more appropriate scale of grays, exposure factors using high kilovoltage should be employed. Radiographic contrast generally increases with an increase in screen speed. The higher the grid ratio, the higher is the contrast. Exposure factors using high milliampere-seconds generally result in excessive image density, frequently obliterating much of the gray scale. (Bushong, 8th ed., p. 273; Shephard, p. 200)

175. Although the stated focal-spot size is measured directly under the actual focal spot, focal-spot size actually varies along the length of the x-ray beam. At which portion of the x-ray beam is the effective focal spot the largest? a. At its outer edge b. Along the path of the central ray c. At the cathode end d. At the anode end

The answer is C. ii. EXPLANATION: X-ray tube targets are constructed according to the line-focus principle—the focal spot is angled (usually 12-17 degrees) to the vertical (Figure 4-34). As the actual focal spot is projected downward, it is foreshortened; thus, the effective focal spot is always smaller than the actual focal spot. As it is projected toward the cathode end of the x-ray beam, the effective focal spot becomes larger and approaches the actual size. As it is projected toward the anode end, it gets smaller because of the anode heel effect. (Selman, 9th ed., p. 139)

34. A controlled area is defined as one 1. that is occupied by people trained in radiation safety 2. that is occupied by people who wear radiation monitors 3. whose occupancy factor is 1 A. 1 and 2 only B. 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: A controlled area is one that is occupied by radiation workers who are trained in radiation safety and who wear radiation monitors. The exposure rate in a controlled area must not exceed 100 mR/week; its occupancy factor is considered to be 1, indicating that the area may always be occupied and, therefore, requires maximum shielding. An uncontrolled area is one occupied by the general population; the exposure rate there must not exceed 10 mR/week. Shielding requirements vary according to several factors, one being occupancy factor. (Bushong, 9th ed., p. 586)

250. Classify the following tissues in order of increasing radiosensitivity 1. Liver cells 2. Intestinal crypt cells 3. Muscle cells A. 1, 3, 2 B. 2, 3, 1 C. 2, 1, 3 D. 3, 1, 2

The answer is D. 2. EXPLANATION: According to Bergonié and Tribondeau, the most radiosensitive cells are undifferentiated, rapidly dividing cells, such as lymphocytes, intestinal crypt (of Lieberkühn) cells, and spermatogonia. Liver cells are among the types of cells that are somewhat differentiated and capable of mitosis. These characteristics render them somewhat radiosensitive. Muscle cells, as well as nerve cells and red blood cells, are highly differentiated and do not divide. Therefore, in order of increasing sensitivity (from least to greatest sensitivity), the cells are muscle, liver, and then intestinal crypt cells. (Selman, 9th ed., pp. 374-375)

88. Indirect modes of disease transmission include 1. airborne 2. fomite 3. vector A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: Airborne, fomite, and vector are all indirect modes of transmitting microorganisms. Direct contact involves actual touching of the infected person. A fomite is an inanimate object that has been in contact with an infectious microorganism (e.g., doorknobs or x-ray tables). Although an inanimate object may serve as a temporary host for microbes, microbes flourish on and in the human host, where plenty of body fluids and tissues nourish and feed them. A vector is an animal host of an infectious organism that transmits the infection via a bite or sting, such as the mosquito or deer tick. Airborne contamination occurs via droplets (sneeze) or dust. (Adler and Carlton, 4th ed., pp. 225-226)

317. The decision as to whether to deliver ionic or nonionic contrast medium should include a preliminary patient history including, but not limited to 1. patient age. 2. history of respiratory disease. 3. history of cardiac disease. A. 1 and 2 B. 1 and 3 C. 2 and 3 D. 1, 2, and 3

The answer is D. 2. EXPLANATION: All the choices listed in the question should be part of a preliminary patient history before deciding to inject ionic or nonionic contrast media. As patients age, their general health decreases, and they are, therefore, more likely to suffer from adverse reactions. Patients with a history of respiratory disease, such as asthma or emphysema and COPD, are more likely to have a reaction and to suffer greater distress in the event of a reaction. Patients with cardiac disease run an increased risk of changes in heart rate and myocardial infarction. Patients also should be screened for decreased renal or hepatic function, sickle-cell disease, diabetes, and pregnancy.

129. To demonstrate the entire circumference of the radial head, exposure(s) must be made with the 1. epicondyles perpendicular to the cassette 2. hand pronated and supinated as much as possible 3. hand lateral and in internal rotation A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: Although routine elbow projections may be essentially negative, conditions may exist (such as an elevated fat pad) that seem to indicate the presence of a small fracture of the radial head. To demonstrate the entire circumference of the radial head, four exposures are made with the elbow flexed 90 degrees and with the humeral epicondyles superimposed and perpendicular to the cassette—one with the hand supinated as much as possible, one with the hand lateral, one with the hand pronated, and one with the hand in internal rotation, thumb down. Each maneuver changes the position of the radial head, and a different surface is presented for inspection.

168. Which of the following is (are) characteristics of the x-ray tube? 1. The target material should have a high atomic number and a high melting point. 2. The useful beam emerges from the port window. 3. The cathode assembly receives both low and high voltages. A. 1 only B. 2 only C. 1 and 2 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: Anode target material with a high atomic number produces higher-energy x-rays more efficiently. Because a great deal of heat is produced at the target, the material should have a high melting point so as to avoid damage to the target surface. Most of the x-rays generated at the focal spot are directed downward and pass through the x-ray tube's port window. The cathode filament receives low-voltage current to heat it to the point of thermionic emission. Then, high voltage is applied to drive the electrons across to the focal track. (Selman, 9th ed., p. 111)

7. Which of the following factors influence(s) the production of scattered radiation? i. Kilovoltage level ii. Tissue density iii. Size of field A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: As photon energy (kV) increases, so does the production of scattered radiation. The greater the density of the irradiated tissues, the greater is the production of scattered radiation. As the size of the irradiated field increases, there is an increase in the volume of tissue irradiated, and the percentage of scatter again increases. Beam restriction is the single most important way to limit the amount of scattered radiation produced. (Carlton and Adler, 4th ed., p. 228)

8. Which of the following contribute(s) to base-plus fog? i. Chemical fog ii. Base tint iii. Background radiation A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: Base-plus fog is the small amount of measurable density on unexposed and processed x-ray film. This fog is a result of environmental background radiation that is present during film manufacture, transportation, and storage. The (usually blue) tint, given the base to enhance contrast, adds more density. Finally, the emulsion receives further fog as the film is chemically processed. Base-plus fog should not exceed 0.2D. (Carlton and Adler, 4th ed., p. 307)

293. Which of the following is (are) essential to high-quality mammographic examinations? 1. Small-focal-spot x-ray tube 2. Short-scale contrast 3. Use of a compression device A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: Breast tissue has very low subject contrast, but it is imperative to visualize micro-calcifications and subtle density differences. Fine detail is necessary to visualize any micro-calcifications; therefore, a small-focal-spot tube is essential. High, short-scale contrast (and, therefore, low kilovoltage) is needed to accentuate minute differences in tissue density. A compression device serves to even out differences in tissue thickness (thicker at the chest wall, thinner at the nipple) and decrease OID and helps to decrease the production of scattered radiation.

296. Factor(s) that can be used to regulate radiographic density is (are) 1. milliamperage. 2. exposure time. 3. kilovoltage. A. 1 only B. 2 only C. 1 and 2 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: Factors that regulate the number of x-ray photons produced at the target can be used to control radiographic density, namely milliamperage and exposure time (mAs). Radiographic density is directly proportional to mAs; if the mAs is cut in half, the radiographic density will decrease by one-half. Although kilovoltage is used primarily to regulate radiographic contrast, it may also be used to regulate radiographic density in variable-kVp techniques, according to the 15% rule.

180. For which of the following can a radiographer be found liable for a negligent tort? 1. Radiographer images the wrong forearm. 2. Patient is injured while being positioned on the x-ray table. 3. Radiographer fails to question patient about possible pregnancy before performing x-ray examination. A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: For negligent tort liability, four elements must be present—duty (what should have been done), breach (deviation from duty), injury sustained, and cause (as a result of breach). The assessment of duty is determined by the professional standard of care. Examples of negligent torts include patient injury as a result of a fall while unattended on an x-ray table, in a radiographic room, or on a stretcher without side rails or safety belt. Radiographing the wrong patient and radiographing the opposite limb are other examples of negligence. If patient injury results from misperformance of a duty in the routine scope of practice of the radiographer, most courts will apply res ipsa loquitur; that is, "the thing speaks for itself." If the patient is obviously injured as a result of the radiographer's actions, it becomes the radiographer's burden to disprove negligence. In many instances, the hospital and/or radiologist also will be held responsible according to respondeat superior, or "the master speaks for the servant." (Adler and Carlton, 4th ed., p. 374)

134. In which of the following ways can higher radiographic contrast be obtained in abdominal radiography? 1. By using lower kilovoltage 2. By using a contrast medium 3. By limiting the field size A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: Higher contrast is shorter scale contrast; it is present in an image that has few shades of gray between white and black. High radiographic contrast is, in part, a result of lower energy photons (lower kVp). High radiographic contrast also results when radiographing anatomic parts that have high subject contrast, such as the chest. The abdomen has low subject contrast, and therefore abdominal radiographs will tend to have very low contrast unless technical factors are selected to increase contrast. To produce high radiographic contrast in abdominal radiography, lower kVp should be used. To better demonstrate high contrast within a viscus, a contrast medium such as barium, iodine, or air can be used. Restricting the size of the field will also function to increase contrast because less scattered radiation will be generated. (Carlton & Adler, p 397)

130. Late radiation-induced somatic effects include 1. thyroid cancers 2. cataractogenesis 3. genetic mutations A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: Late somatic effects are those that can occur years after initial exposure and are caused by low, chronic exposures. Occupationally exposed personnel are concerned with the late effects of radiation exposure. Bone malignancies, thyroid cancers, leukemia, and skin cancers are examples of carcinogenic somatic effects of radiation. Another example of somatic effects of radiation is cataract formation to the lenses of eyes of individuals accidentally exposed to sufficient quantities of radiation. The lives of many of the early radiation workers were several years shorter than the lives of the general population. Statistics revealed that radiologists, for example, had a shorter life span than physicians of other specialties. Life-span shortening, then, was another somatic effect of radiation. Certainly, these effects should never be experienced today. The human reproductive organs are particularly radiosensitive. Fertility and heredity can be greatly affected by the germ cells produced within the testes (spermatogonia) and ovaries (oogonia). Excessive radiation exposure to the gonads can cause temporary or permanent sterility and/or genetic mutations. (Bushong, 8th ed., pp. 532-534)

336. Which of the following is (are) associated with subject contrast? 1. Patient thickness 2. Tissue density 3. Kilovoltage A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: Radiographic contrast is the sum of film emulsion contrast and subject contrast. Subject contrast has by far the greatest influence on radiographic contrast. Several factors influence subject contrast, each as a result of beam-attenuation differences in the irradiated tissues. As patient thickness and tissue density increase, attenuation increases, and subject contrast is increased. As kilovoltage increases, higher-energy photons are produced, beam attenuation is decreased, and subject contrast decreases.

121. Shape distortion is influenced by the relationship between the 1. x-ray tube and the part to be imaged. 2. body part to be imaged and the IR. 3. IR and the x-ray tube. A. 1 and 2 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: Shape distortion is caused by misalignment of the x-ray tube, the body part to be radiographed, and the IR. An object can be falsely imaged (foreshortened or elongated) as a result of incorrect placement of the tube, the part, or the IR. Only one of the three need be misaligned for distortion to occur. (Selman, 9th ed., pp. 225-226)

155. Abdominal viscera located in the retroperitoneum include the 1. kidneys. 2. duodenum. 3. ascending and descending colon. A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: Structures located behind the parietal peritoneum are referred to as retroperitoneal. Retroperitoneal structures include the kidneys, adrenal glands, pancreas, duodenum, ascending and descending colon, portions of the aorta, and the inferior vena cava. (Tortora and Derrickson, 11th ed., p. 19)

59. Which of the following will influence recorded detail? 1. Screen speed 2. Screen-film contact 3. Focal spot A. 1 and 2 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: The faster the imaging system, the greater is the sacrifice of image clarity (recorded detail). As intensifying-screen speed increases, recorded detail decreases. Perfect screen-film contact is essential for good detail. Any areas of poor contact result in considerable blurriness in the radiographic image. Focal-spot blur is related to focal-spot size; smaller focal spots produce less blur and thus better recorded detail. (Selman, 9th ed., pp. 206-210)

198. Which of the following radiation exposure responses exhibit a nonlinear threshold dose-response relationship? 1. Skin erythema 2. Hematologic depression 3. Lethality A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: The genetic effects of radiation and some somatic effects, like leukemia, are plotted on a linear dose-response curve. The linear dose-response curve has no threshold; that is, there is no dose below which radiation is absolutely safe. The nonlinear/sigmoidal dose-response curve has a threshold and is thought to be generally correct for most somatic effects—such as skin erythema, hematologic depression, and radiation lethality (death). (Ballinger & Frank, vol 1, p 45)

96. The total number of x-ray photons produced at the target is contingent on the 1. tube current 2. target material 3. square of the kilovoltage A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: The greater the number of electrons making up the electron stream and bombarding the target, the greater is the number of x-ray photons produced. Although kilovoltage usually is associated with the energy of the x-ray photons because a greater number of more energetic electrons will produce more x-ray photons, an increase in kilovoltage will also increase the number of photons produced. Specifically, the quantity of radiation produced increases as the square of the kilovoltage. The material composition of the tube target also plays an important role in the number of x-ray photons produced. The higher the atomic number, the denser and more closely packed are the atoms making up the material, and therefore, the greater is the chance of an interaction between a high-speed electron and the target material. (Selman, 9th ed., pp. 112-115)

308. Inspiration and expiration projections of the chest are performed to demonstrate 1. partial or complete collapse of pulmonary lobe(s) 2. air in the pleural cavity 3. foreign body A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: The phase of respiration is exceedingly important in thoracic radiography because lung expansion and the position of the diaphragm strongly influence the appearance of the finished radiograph. Inspiration and expiration radiographs of the chest are taken to demonstrate air in the pleural cavity (pneumothorax), to demonstrate atelectasis (partial or complete collapse of one or more pulmonary lobes) or the degree of diaphragm excursion, or to detect the presence of a foreign body. The expiration image will require a somewhat greater exposure (6-8 kV more) to compensate for the diminished quantity of air in the lungs.

23. Tangential axial projections of the patella can be obtained in which of the following positions? 1. supine flexion 45° (Merchant) 2. prone flexion 90° (Settegast) 3. prone flexion 55° (Hughston) A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3 only

The answer is D. 2. EXPLANATION: The tangential axial projections of the patella are also often referred to as "sunrise" or "skyline" views. The supine flexion 45° (Merchant) position requires a special apparatus, and the patellae can be examined bilaterally. This position also requires patient comfort without muscle tension—muscle tension can cause a subluxed patella to be pulled into the intercondyler sulcus, giving the appearance of a normal patella. The two prone positions differ according to the degree of flexion employed. The 90° flexion (Settegast) position must not be employed with suspected patellar fracture.

313. Geometric unsharpness will be least obvious 1. at long SIDs. 2. with small focal spots. 3. at the anode end of the image. A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: The x-ray tube anode is designed according to the line-focus principle, that is, with the focal track beveled (Figure 6-24). This allows a larger actual focal spot to project a smaller effective focal spot, resulting in improved recorded detail with less blur. However, because of the target angle, penumbral blur varies along the longitudinal tube axis, being greater at the cathode end of the image and less at the anode end of the image. Therefore, better detail will be appreciated using small focal spots at the anode end of the x-ray beam and at longer SIDs.

6. Demonstration of the posterior fat pad on the lateral projection of the adult elbow can be caused by i. trauma or other pathology ii. greater than 90-degree flexion iii. less than 90-degree flexion A. 1 only B. 3 only C. 1 and 2 only D. 1 and 3 only

The answer is D. 2. EXPLANATION: There are three important fat pads associated with the elbow, best demonstrated in the true lateral projection. They cannot be demonstrated in the AP projection because of their superimposition on bony structures. The anterior fat pad is located just anterior to the distal humerus. The posterior fat pad is located within the olecranon fossa at the distal posterior humerus. The supinator fat pad/stripe is located at the proximal radius just anterior to the head, neck, and tuberosity. The posterior fat pad is not visible radiographically in the normal elbow. The posterior fat pad is visible in cases or trauma or other pathology and when the elbow is insufficiently flexed.

21. Which of the following criteria are used to evaluate a PA projection of the chest? i. Ten posterior ribs should be visualized. ii. Sternoclavicular joints should be symmetrical. iii. The scapulae should be lateral to the lung fields. A. 1 and 2 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: To evaluate sufficient inspiration and lung expansion, 10 posterior ribs should be visualized. The sternoclavicular joints should be symmetrical; any loss of symmetry indicates rotation. To visualize maximum lung area, the shoulders are rolled forward to move the scapulae laterally from the lung fields. (Ballinger & Frank, vol 1, p 527)

149. Routine excretory urography usually includes a postmicturition radiograph of the bladder. This is done to demonstrate 1. tumor masses. 2. residual urine. 3. prostatic enlargement. A. 2 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The answer is D. 2. EXPLANATION: Variance from the normal bladder contour will be noted while the bladder is full of contrast medium. However, a postmicturition (postvoiding) radiograph is also an essential part of an IVU/IVP. The presence of residual urine may be an indication of small tumor masses or, in male patients, enlargement of the prostate gland. (Frank, Long, and Smith, 11th ed., vol. 2, p. 217)

114. In what order should the following examinations be scheduled? 1. Upper GI 2. Intravenous pyelogram (IVP) 3. Barium enema (BE) A. 3, 1, 2 B. 1, 3, 2 C. 2, 1, 3 D. 2, 3, 1

The answer is D. 2. EXPLANATION: When scheduling patient examinations, it is important to avoid the possibility of residual contrast medium covering areas of interest on later examinations. The IVP should be scheduled first because the contrast medium used is excreted rapidly. The BE should be scheduled next. Finally, the upper GI series is scheduled. Any barium remaining from the previous BE should not be enough to interfere with the stomach or duodenum (a preliminary scout image should be taken in each case). (Torres et al., 6th ed., pp. 233-234)

280. Which of the following personnel radiation monitors will provide an immediate reading? a. TLD b. Film badge c. Lithium fluoride chips d. Pocket dosimeter

The answer is D. ii. EXPLANATION: A TLD is used to measure monthly exposure to radiation, as is the film badge. Lithium fluoride chips are the thermoluminescent material used in TLDs. A pocket dosimeter (a small personal ionization chamber) measures the quantity of ionizations occurring during the period worn and reads out in millirem; it is used primarily when working with large quantities of radiation.

159. QA was being performed on a three-phase, full-wave-rectified x-ray unit. A synchronous spinning-top test was performed using 300 mA, 60 ms, and 70 kVp, and a 22-degree arc is observed on the test film. Which of the following statements regarding these results is most correct? a. The timer is inaccurate. b. The milliamperage station is inaccurate. c. One rectifier is malfunctioning. d. The test results are satisfactory.

The answer is D. ii. EXPLANATION: A synchronous spinning-top test is used to test timer accuracy or rectifier function in three-phase equipment. Because three-phase, full-wave-rectified current would expose a 360-degree arc each second, a 60-ms (0.06-s) exposure should expose a 21.6-degree arc (360 degrees x 0.06 = 21.6 degrees). Anything more or less indicates timer inaccuracy. If exactly one-half the expected arc appears, rectifier failure should be suspected. (Selman, 9th ed., p. 106)

319. An x-ray image of the ankle was made at 40-SID, 200 mA, 50 ms, 70 kV, 0.6 mm focal spot, and minimal OID. Which of the following modifications would result in the greatest increase in magnification? a. 1.2 mm focal spot b. 36-in. SID c. 44-in. SID d. 4-in. OID

The answer is D. ii. EXPLANATION: All the factor changes affect recorded detail, but focal spot size does not affect magnification. An increase in SID would decrease magnification. Although a decrease in SID will increase magnification, it does not have as significant an effect as an increase in OID. In general, it requires an increase of 7 in. SID to compensate for every inch of OID.

28. All the following have an impact on radiographic contrast except a. photon energy b. grid ratio c. OID d. focal-spot size

The answer is D. ii. EXPLANATION: As photon energy increases, more penetration and greater production of scattered radiation occur, producing a longer scale of contrast. As grid ratio increases, more scattered radiation is absorbed, producing a shorter scale of contrast. As OID increases, the distance between the part and the IR acts as a grid, and consequently, less scattered radiation reaches the IR, producing a shorter scale of contrast. Focal-spot size is related only to recorded detail. (Shephard, p. 203)

41. Which of the following causes pitting, or many small surface melts, of the anode's focal track? a. Vaporized tungsten on the glass envelope b. Loss of anode rotation c. A large amount of heat to a cold anode d. Repeated, frequent overloading

The answer is D. ii. EXPLANATION: As the filament ages, vaporized tungsten may be deposited on the port window and act as an additional filter. Tungsten may also vaporize as a result of anode abuse. Exposures in excess of safe values deliver sufficient heat to cause surface melts, or pits, on the focal track. This results in roughening of the anode surface and decreased tube output. Delivery of a large amount of heat to a cold anode can cause cracking if the anode does not have sufficient time to disperse the heat. Loss of anode rotation would cause one large melt on the focal track because the electrons would bombard only one small area. If the anode is not heard to be rotating, the radiographer should not make an exposure. (Selman, 9th ed., pp. 137-138)

289. Using an AEC system, how will the mAs be adjusted as the film-screen speed combination is decreased? a. The mAs will increase as film-screen speed decreases. b. Both the mAs and the kV will increase as film-screen speed decreases. c. The milliamperage will decrease as film-screen speed decreases. d. The mAs remains unchanged as film-screen speed decreases.

The answer is D. ii. EXPLANATION: As the speed of the film-screen system decreases, an increase in milliampere-seconds usually is required to maintain radiographic density. However, when an AEC (phototimer or ionization chamber) is used, the system is programmed for the use of a particular film-screen speed. If a slower speed screen cassette/IR is placed in the Bucky tray, the AEC has no way of recognizing it as different and will result in the exposure required for the system for which it is programmed. For example, if the system were programmed for a 400-speed film-screen combination, and if a 200-speed screen cassette/IR were placed in the Bucky tray, the AEC would produce an exposure appropriate for the 400 speed system, and the resulting image would have half the required radiographic density.

270. Which of the following is most likely to permit the greatest decrease in patient exposure? a. Changing from a 200-speed system to a 400-speed system b. Increasing kilovoltage by 15% and cutting the milliampere-seconds value in half c. Changing collimation from 10 x 12 to 14 x 17 d. Changing from an 8:1 grid technique to nongrid

The answer is D. ii. EXPLANATION: Converting from an 8:1 grid to nongrid requires about a fourfold decrease in milliampere-seconds. Increasing the kilovoltage by 15% and cutting the milliampere-seconds in half would reduce patient dose by half. Changing from a 200-speed system to a 400-speed system also would require the milliampere-seconds to be halved. Therefore, the largest decrease will occur with removal of a grid.

164. Secondary radiation barriers usually require the following thickness of shielding: a. 1/4-inch lead b. 1/8-inch lead c. 1/16-inch lead d. 1/32-inch lead

The answer is D. ii. EXPLANATION: Examples of primary barriers are the lead walls and doors of a radiographic room, that is, any surface that could be struck by the useful beam. Primary protective barriers of typical installations generally consist of walls with 1/16 inch (1.5 mm) lead thickness and 7 feet high. Secondary radiation is defined as leakage and/or scattered radiation. The x-ray tube housing protects from leakage radiation as stated previously. The patient is the source of most scattered radiation. Secondary radiation barriers include that portion of the walls above 7 feet in height; this area requires only 1/32-inch lead. The control booth is also a secondary barrier, toward which the primary beam must never be directed. (Bushong, p 572)

172. Which of the following most effectively minimizes radiation exposure to the patient? a. Small focal spot b. Low-ratio grids c. Increased SID d. High-speed intensifying screens

The answer is D. ii. EXPLANATION: Focal spot size affects recorded detail and x-ray tube heat limits—it has no effect on patient dose. Low-ratio grids, although they require fewer milliampere-seconds than high-ratio grids, are not a means of patient protection. Long SIDs usually require the use of higher milliampere-seconds and so would not be an effective means of patient protection. The use of high-speed intensifying screens enables the use of lower milliampere-second values and, therefore, is an important consideration in limiting patient dose. Limiting the irradiated field size, through collimation or other beam restriction, is perhaps the most effective way of controlling patient exposure dose. (Bushong, 8th ed., p. 12)

31. Focusing distance is associated with which of the following? a. Computed tomography b. Chest radiography c. Magnification radiography d. Grids

The answer is D. ii. EXPLANATION: Focusing distance is the term used to specify the optimal SID used with a particular focused grid. It is usually expressed as focal range, indicating the minimum and maximum SID workable with that grid. Lesser or greater distances can result in grid cutoff. Although proper distance is important in computed tomography and chest and magnification radiography, focusing distance is unrelated to them. (Selman, 9th ed., pp. 239-240)

314. Which of the following positions will separate the radial head, neck, and tuberosity from superimposition on the ulna? a. AP b. Lateral c. Medial oblique d. Lateral oblique

The answer is D. ii. EXPLANATION: In the AP projection of the elbow, the proximal radius and ulna are partially superimposed. In the lateral projection, the radial head is partially superimposed on the coronoid process, facing anteriorly. In the medial oblique projection, there is even greater superimposition. The lateral oblique projection completely separates the proximal radius and ulna, projecting the radial head, neck, and tuberosity free of superimposition with the proximal ulna.

38. Changes in milliampere-seconds can affect all the following except a. quantity of x-ray photons produced b. exposure rate c. optical density d. recorded detail

The answer is D. ii. EXPLANATION: Milliampere-seconds (mAs) are the product of milliamperes (mA) and exposure time (seconds). Any combinations of milliamperes and time that will produce a given milliampere-seconds value (i.e., a particular quantity of x-ray photons) will produce identical optical density. This is known as the reciprocity law. Density is a quantitative factor because it describes the amount of image blackening. The milliampere-seconds value is also a quantitative factor because it regulates x-ray-beam intensity, exposure rate, quantity, or number of x-ray photons produced (the milliampere-seconds value is the single most important technical factor associated with image density and is the factor of choice for regulating radiographic/optical density). The milliampere-seconds value is directly proportional to the intensity (i.e., exposure rate, number, and quantity) of x-ray photons produced and the resulting radiographic density. If the milliampere-seconds value is doubled, twice the exposure rate and twice the density occur. If the milliampere-seconds value is cut in half, the exposure rate and resulting density are cut in half. The milliampere-seconds value has no effect on recorded detail. (Shephard, p. 170)

306. What is the approximate ESE for the average AP lumbar spine radiograph? a. 30 rad b. 30 mrad c. 300 rad d. 350 mrad

The answer is D. ii. EXPLANATION: Patients occasionally will question the radiographer regarding the amount of radiation they are receiving during their examination. Most of these patients are merely curious because they have heard a recent news report about x-rays or have perhaps studied about x-rays in school recently. It is a good idea for radiographers to have some knowledge of average exposure doses for patients who desire this information. The curious patient also can be referred to the medical physicist for more detailed information. The average AP cervical spine radiograph delivers about 80 mrad (0.080 rad). The average AP supine lumbar spine radiograph delivers an ESE of about 350 mrad (0.35 rad). The average AP supine abdomen radiograph delivers about 300 mrad.

226. All the following statements regarding pediatric positioning are true except a. for radiography of the kidneys, the CR should be directed midway between the diaphragm and the symphysis pubis. b. if a pediatric patient is in respiratory distress, a chest radiograph should be obtained in the AP projection rather than in the standard PA projection. c. chest radiography on a neonate should be performed in the supine position. d. radiography of pediatric patients with a myelomeningocele defect should be performed in the supine position.

The answer is D. ii. EXPLANATION: Radiography of pediatric patients with a myelomeningocele defect should be performed in the prone position rather than in the routine supine position. The supine position would put unnecessary pressure on the protrusion of the meninges and spinal cord. All the other statements in the question are true. The anatomic dimensions of children are different from those of adults, and this must be kept in mind when performing pediatric radiography. The liver occupies a larger area of the abdominal cavity in a child than in an adult. This causes the kidneys to be in a lower position. Generally, the kidneys will be midway between the diaphragm and the symphysis pubis. Chest radiography for the pediatric patient varies depending on the age of the child. Neonates are routinely radiographed in the supine position. Although infants also may be examined in the supine position, it is preferable to examine them by placing the infant securely in a support device to obtain a good PA erect radiograph. Exceptions to this rule are made if the infant is in respiratory distress. To avoid aggravating the respiratory distress, an erect AP radiograph usually is obtained. (Dowd and Tilson, 2nd ed., vol. 2, pp. 1004-1005, 1013)

242. Which combination of exposure factors most likely will contribute to producing the shortest-scale contrast? a. mAs: 10; kV: 70; Film-screen system: 400; Grid ratio: 5:1; Field size: 14 x 17 in. b. mAs: 12; kV: 90; Film-screen system: 200; Grid ratio: 8:1; Field size: 14 x 17 in. c. mAs: 15; kV: 90; Film-screen system: 200; Grid ratio: 12:1; Field size: 11 x 14 in. d. mAs: 20; kV: 80; Film-screen system: 400; Grid ratio: 10:1; Field size: 8 x 10 in.

The answer is D. ii. EXPLANATION: Review the groups of factors. First, because the milliampere-seconds value has no effect on the scale of contrast produced, eliminate milliampere-seconds from consideration by drawing a line through the column. Then check the two entries in each column that are likely to produce shorter-scale contrast. For example, in the kilovoltage column, because lower kilovoltage will produce shorter-scale contrast, place checkmarks next to the 70 and 80 kV. In the film-screen column, the faster screens (400) will produce higher (shorter-scale) contrast than the slower screens; place a checkmark next to each. Because higher-ratio grids permit less scattered radiation to reach the IR, the 10:1 and 12:1 grids will produce a shorter scale of contrast than the lower-ratio grids; check them. As the volume of irradiated tissue decreases, so does the amount of scattered radiation produced, and consequently, the shorter is the scale of radiographic contrast; therefore, check the 11 x 14 and 8 x 10 in. field sizes. An overview shows that the factors in groups (A) and (C) have two checkmarks, whereas the factors in group (D) have four checkmarks, indicating that group (D) will produce the shortest-scale contrast. (Shephard, pp. 306-308)

144. All the following are part of the Patient's Bill of Rights except 1 a. the right to participate in proposed research studies. b. the right to continuity of care. c. the right to considerate and respectful care. d. the right to review any institutional records.

The answer is D. ii. EXPLANATION: The AHA publishes a Patient's Bill of Rights, which details 12 specific areas of patients' rights and the health care professional's ethical (and often legal) responsibility to adhere to these rights. The Patient's Bill of Rights is summarized as follows: 1. The right to considerate and respectful care 2. The right to be informed completely and understandably 3. The right to refuse treatment 4. The right to have an advance directive (e.g., a living will or health care proxy) 5. The right to privacy 6. The right to confidentiality 7. The right to review one's records 8. The right to request appropriate and medically indicated care and services 9. The right to know about institutional business relationships that could influence treatment and care 10. The right to be informed of, consent to, or decline participation in proposed research studies 11. The right to continuity of care 12. The right to be informed of hospital policies and procedures relating to patient care, treatment, and responsibilities iii. Although it is the patient's right to appropriate and medically indicated care and services, only a physician may order an x-ray procedure—just as it is only the physician who may order prescription drugs. Although it is the patient's right to review his or her records and the patient's right to know about institutional business relationships that could influence treatment and care, the patient does not have the right to review just any hospital/institutional records. The AHA recently replaced the Patient's Bill of Rights with The Patient Care Partnership—Understanding Expectations, Rights and Responsibilities. This plain-language brochure includes the essentials of the Patient's Bill of Rights and reviews what patients can and should expect during a hospital stay. (Ehrlich et al., 6th ed., pp. 61-62)

165. Which of the following conditions will require an increase in x-ray photon energy/penetration? a. Fibrosarcoma b. Osteomalacia c. Paralytic ileus d. Ascites

The answer is D. ii. EXPLANATION: The ability of x-ray photons to penetrate a body part has a great deal to do with the composition of that part (e.g., bone vs. soft tissue vs. air) and the presence of any pathologic condition. Pathologic conditions can alter the normal nature of the anatomic part. Some conditions, such as osteomalacia, fibrosarcoma, and paralytic ileus (obstruction), result in a decrease in body tissue density. When body tissue density decreases, x-rays will penetrate the tissues more readily; that is, there is more x-ray penetrability. In conditions such as ascites, where body tissue density increases as a result of the accumulation of fluid, x-rays will not readily penetrate the body tissues; that is, there is less x-ray penetrability. (Carlton and Adler, 4th ed., p. 250)

24. To obtain an exact axial projection of the clavicle, place the patient a. supine and angle the central ray 30° caudally. b. prone and angle the central ray 30° cephalad. c. supine and angle the central ray 15° cephalad. d. in a lordotic position and direct the central ray at right angles to the coronal plane of the clavicle.

The answer is D. ii. EXPLANATION: The exact axial projection is performed by placing the patient in a lordotic position, leaning against the vertical grid device. This places the clavicle at right angles, or nearly so, to the plane of the IR. The central ray is directed to enter the inferior border of the clavicle, at right angles to its coronal plane. Other axial projections may include a prone position with a 25° to 30° caudal angle. However, none of these produce an exact axial projection of the clavicle. (Ballinger & Frank, vol 1, p 159)

328. The stomach of an asthenic patient is most likely to be located a. high, transverse, and lateral. b. low, transverse, and lateral. c. high, vertical, and toward the midline. d. low, vertical, and toward the midline.

The answer is D. ii. EXPLANATION: The four body types (from largest to smallest) are hypersthenic, sthenic, hyposthenic, and asthenic. The abdominal viscera of the asthenic person are generally located quite low, vertical, and toward the midline. The opposite is true of the hypersthenic individual: Organs are located high, transverse, and lateral

163. A radiograph made with a parallel grid demonstrates decreased density on its lateral edges. This is most likely due to a. static electrical discharge b. the grid being off-centered c. improper tube angle d. decreased SID

The answer is D. ii. EXPLANATION: The lead strips in a parallel grid are parallel to one another and, therefore, are not parallel to the x-ray beam. The more divergent the x-ray beam, the more likely there is to be cutoff/decreased density at the lateral edges of the radiograph. This problem becomes more pronounced at short SIDs. If there were a centering or tube angle problem, there would be more likely to be a noticeable density loss on one side or the other. (Carlton and Adler, 4th ed., p. 260)

192. The AP axial projection, or "frog leg" position, of the femoral neck places the patient in a supine position with the affected thigh a. adducted 25 degrees from the horizontal b. abducted 25 degrees from the vertical c. adducted 40 degrees from the horizontal d. abducted 40 degrees from the vertical

The answer is D. ii. EXPLANATION: The patient is supine with the leg abducted (drawn away from the midline) approximately 40 degrees. This 40-degree abduction from the vertical places the long axis of the femoral neck parallel to the IR. Adduction is drawing the extremity closer to the midline of the body. (Frank, Long, and Smith, 11th ed., vol. 1, p. 350)

310. Which type of personnel radiation monitor can provide an immediate reading? a. Thermoluminescent dosimeter (TLD) b. Optically stimulated luminescence (OSL) c. Film badge d. Ionization chamber

The answer is D. ii. EXPLANATION: The pocket dosimeter, or pocket ionization chamber, resembles a penlight. Within the dosimeter is a thimble ionization chamber. In the presence of ionizing radiation, a particular quantity of air will be ionized and cause the fiber indicator to register radiation quantity in milliroentgen (mR). The self-reading type may be "read" by holding the dosimeter up to the light and, looking through the eyepiece, observing the fiber indicator, which indicates a quantity of 0 to 200 mR. Although it provides an immediate reading while other personnel monitors require "processing," the disadvantage of the pocket dosimeter is that it does not provide a permanent legal record of exposure. (Adler and Carlton, 4th ed., p. 124)

323. Filtration is added to the x-ray beam to a. decrease photoelectric interaction. b. remove the "hard" x-rays. c. produce a more heterogeneous x-ray beam. d. produce an x-ray beam with higher average energy.

The answer is D. ii. EXPLANATION: The primary beam generally has a total filtration of 2.5 mm Al equivalent for patient protection purposes. In general-purpose radiographic tubes, the glass envelope usually accounts for about 0.5 mm Al equivalent and the collimator provides about 1.0 mm Al equivalent. These are considered inherent filtration. The manufacturer adds another 1.0 mm Al (added filtration) to meet the minimum requirements of 2.5 mm Al equivalent total filtration for radiographic tubes operated above 70 kilovolt peaks (kVp). This type of filter serves to remove the diagnostically useless x-ray photons that only contribute to patient (skin) dose. Because this radiation is "soft" (low energy) and would not reach the image receptor anyway, the x-ray tube total filtration has no real effect on radiographic density. Filtration serves to increase the overall average energy of the beam; it "hardens" the x-ray beam.

284. The target theory applies to a. spermatagonia b. oocytes c. lymphocytes d. DNA molecules

The answer is D. ii. EXPLANATION: The principal interactions that occur between x-ray photons and body tissues in the diagnostic x-ray range, the photoelectric effect and Compton scatter, are ionization processes producing photoelectrons and recoil electrons that traverse tissue and subsequently ionize molecules. These interactions occur randomly but can lead to molecular damage in the form of impaired function or cell death. The target theory specifies that DNA molecules are the targets of greatest importance and sensitivity; that is, DNA is the key sensitive molecule. However, since the body is 65% to 80% water, most interactions between ionizing radiation and body cells will involve radiolysis of water rather than direct interaction with DNA. The two major types of effects that occur are the direct effect and the indirect effect. The direct effect usually occurs with high-LET radiations and when ionization occurs at the DNA molecule itself. The indirect effect, which occurs most frequently, happens when ionization takes place away from the DNA molecule in cellular water. However, the energy from the interaction can be transferred to the molecule via a free radical (formed by radiolysis of cellular water).

244. Which of the following structures will be filled with barium in the AP recumbent position of a sthenic patient during an upper GI examination? a. Duodenal bulb b. Descending duodenum c. Pyloric vestibule d. Gastric fundus

The answer is D. ii. EXPLANATION: The stomach is normally angled with the fundus lying posteriorly and the body, pylorus, and duodenum inferior to the fundus and angling anteriorly. Therefore, when the patient ingests barium and lies AP recumbent, the heavy barium gravitates easily to the fundus and fills it. With the patient PA recumbent, barium gravitates inferiorly to the body, pylorus, and duodenum, displacing air into the fundus.

13. Which of the following structures will usually contain air, in the PA position on a sthenic patient, during a double-contrast upper GI (UGI) examination? a. Duodenal bulb b. Descending duodenum c. Pyloric vestibule d. Gastric fundus

The answer is D. ii. EXPLANATION: The stomach is normally angled with the fundus lying posteriorly and the body, pylorus, and duodenum inferior to the fundus and angling anteriorly. Therefore, when the patient ingests barium and lies AP recumbent, the heavy barium gravitates easily to the fundus and fills it. With the patient PA recumbent, barium gravitates inferiorly to the body, pylorus, and duodenum, displacing air into the fundus. (Frank, Long, and Smith, 11th ed., vol. 2, pp. 144-145)

60. The roentgen is the unit of a. radiation dose b. biologic dose c. dose equivalent d. ionization in air

The answer is D. ii. EXPLANATION: There are several radiation units that are used to express quantity and effects of radiation. Rad (radiation absorbed dose) expresses energy deposited (as a result of ionizations) in any kind of absorber. The unit of exposure, the roentgen, is used to express the quantity of ionization in air. The unit of dose equivalent is the rem (radiation equivalent man), which expresses dose to biologic material. (Selman, 9th ed., p. 131)

72. Which of the following sinus groups is demonstrated with the patient positioned as for a parietoacanthal projection (Waters method) with the CR directed through the patient's open mouth? a. Frontal b. Ethmoidal c. Maxillary d. Sphenoidal

The answer is D. ii. EXPLANATION: This is a modification of the parietoacanthal projection (Waters method) in which the patient is requested to open the mouth, and then the skull is positioned so that the OML forms a 37-degree angle with the IR. The CR is directed through the sphenoidal sinuses and exits the open mouth. The routine parietoacanthal projection (with mouth closed) is used to demonstrate the maxillary sinuses projected above the petrous pyramids. The frontal and ethmoidal sinuses are best visualized in the PA axial position (modified Caldwell method). (Bontrager and Lampignano, 6th ed., p. 442)

206. With the patient in the PA position, which of the following tube angle and direction combinations is correct for an axial projection of the clavicle? a. 5 to 15 degrees caudad b. 5 to 15 degrees cephalad c. 15 to 30 degrees cephalad d. 15 to 30 degrees caudad

The answer is D. ii. EXPLANATION: When the clavicle is examined in the PA recumbent position, the CR must be directed 15 to 30 degrees caudad to project most of the clavicle's length above the ribs. The direction of the CR is reversed when examining the patient in the AP position. (Bontrager and Lampignano, 6th ed., p. 202)

9. During an upper gastrointestinal (GI) examination, a stomach of average shape demonstrates a barium-filled fundus and double contrast of the pylorus and duodenal bulb. The position used is most likely a. AP erect b. PA c. RAO d. LPO

The answer is D. ii. EXPLANATION: With the body in the AP recumbent position (or LPO position), barium flows easily into the fundus of the stomach (from the more distal portions of the stomach), displacing/drawing the stomach somewhat superiorly. The fundus, then, is filled with barium, whereas the air that had been in the fundus is now displaced into the gastric body, pylorus, and duodenum, illustrating them in double contrast. Double-contrast delineation of these structures allows us to see through the stomach to the retrogastric areas and structures. The RAO position demonstrates a barium-filled pylorus and duodenum. Anterior and posterior aspects of the stomach are visualized in the lateral position; medial and lateral aspects of the stomach are visualized in the AP projection. (Frank, Long, and Smith, 11th ed., vol. 2, p. 142)

2. During IV urography, the prone position generally is recommended to demonstrate the filling of a. the ureters b. the renal pelvis c. the superior calyces i. 1 only ii. 1 and 2 only iii. 1 and 3 only iv. 1, 2, and 3

The answer is II 2. EXPLANATION: The kidneys lie obliquely in the posterior portion of the trunk with their superior portion angled posteriorly and their inferior portion and ureters angled anteriorly. Therefore, to facilitate filling of the most anteriorly placed structures, the patient is examined in the prone position. Opacified urine then flows to the most dependent part of the kidney and ureter—the ureteropelvic region, inferior calyces, and ureters. (Saia, 4th ed., p. 191)

1. The term latitude describes I. an emulsion's ability to record a range of densities II. the degree of error tolerated with given exposure factors III. the conversion efficiency of a given intensifying screen A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The answer is b 2. EXPLANATION: The term latitude may refer to either film emulsion latitude or exposure latitude. Exposure latitude refers to the margin of error inherent in a particular group of exposure factors. Selection of high-kilovoltage and low-milliampere-seconds factors will allow greater exposure latitude than low-kilovoltage and high-milliampere-seconds factors. Film emulsion latitude is chemically built into the film emulsion and refers to the emulsion's ability to record a long range of densities from black to white (long-scale contrast).

17. Component part(s) of x-ray film include which of the following? a. Phosphor layer b. Gelatin emulsion c. Adhesive layer i. 1 only ii. 1 and 3 only iii. 2 and 3 only iv. 1, 2, and 3

The answer is iii 2. EXPLANATION: The manufacture of x-ray film starts with a clear polyester base that serves as support for the emulsion. Applied next is an adhesive layer, which functions to hold the emulsion to the base. Next is the emulsion, consisting of silver halide grains suspended in gelatin. Finally, a supercoat of clear, hard gelatin is applied as an antiabrasive layer. A phosphor layer is used in the construction of intensifying screens. (Shephard, pp. 85-88)

320. In which of the following positions can the sesamoid bones of the foot be demonstrated to be free of superimposition with the metatarsals or phalanges? a. Dorsoplantar metatarsals/toes b. Tangential metatarsals/toes c. 30-degree medial oblique foot d. 30-degree lateral oblique foot

answer is B. ii. EXPLANATION: The tangential projection projects the sesamoid bones separate from adjacent structures. The patient is best examined in the prone position because this places the parts of interest closest to the IR. The affected foot is dorsiflexed so as to place its plantar surface 15 to 20 degrees with the vertical. The CR is directed perpendicular to the posterior surface of the foot (near the metatarsophalangeal joints). The dorsoplantar and oblique projections of the foot will demonstrate the sesamoid bones superimposed on adjacent bony structures.

49. In fluoroscopy, the automatic brightness control is used to adjust the a. kilovoltage (kVp) and milliamperage (mA) b. backup timer c. milliamperage (mA) and time d. kilovoltage (kV) and time

i. Answer is A ii. EXPLANATION: As body areas of different thicknesses and densities are scanned with the image intensifier, image brightness and contrast require adjustment. The ABC functions to maintain constant brightness and contrast of the output screen image, correcting for fluctuations in x-ray beam attenuation with adjustments in kilovoltage and/or milliamperage. There are also brightness and contrast controls on the monitor that the radiographer can regulate. (Bushong, 8th ed., p. 358)

40. Using a short (25-30 in.) SID with a large (14 x 17 in.) IR is likely to a. increase the scale of contrast b. increase the anode heel effect c. cause malfunction of the AEC d. cause premature termination of the exposure

i. Correct. The answer is B. ii. EXPLANATION: Use of a short SID with a large-size IR (and also with anode angles of 10 degrees or less) causes the anode heel effect to be much more apparent. The x-ray beam needs to diverge more to cover a large-size IR, and it needs to diverge even more for coverage as the SID decreases. The x-ray beam has no problem diverging toward the cathode end of the beam, but as it tries to diverge toward the anode end of the beam, it is eventually stopped by the anode (x-ray photons are absorbed by the anode). This causes a decrease in beam intensity at the anode end of the beam and is characteristic of the anode heel effect. (Carlton and Adler, 4th ed., p. 407)