Safety

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

A

Which of the following body parts is (are) included in whole-body dose?GonadsBlood-forming organsExtremities A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

qThe Correct Answer is: BWhole-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. (Bushong, 8th ed., p. 555)

Skin erythema follows a ____________________ dose response relationship. A Stochastic, Non-threshold B Deterministic, Threshold C Stochastic, Threshold D Deterministic, Non-threshold

Dose response relationships manifest in several forms. Early effects, like skin erythema and epilation are deterministic, threshold relationships. One could say that the individual is determined to experience the effect after a threshold dose has been reached. Recognize that sunburn is a type of skin erythema. Sunburns occur after individuals stay out in the sun too long (exposure time). Sunburns do not manifest instantaneously, nor do they manifest years later; you must stay out in the sun beyond your threshold. Once you do, the sunburn effects get worse with more exposure. Stochastic effects are usually late effects, and the magnitude of the dose is related to the probability of the effect, not it's severity. Cancer is an example of a stochastic effect; you are not guaranteed to have it after receiving a dose of radiation. Cancer is not more or less awful; it progresses in the same stages regardless of the dose. (Sherer, 8th edition, pgs 160-162)

What two rules of electrostatics from the list of five below occur at the focusing cup and function to confine the space charge to a smaller area, thus focusing the beam of electrons onto the target anode? A Opposite charges attract while like charges repel. B An increase in the magnetic flux velocity increases the induced current. C The electrostatic force is proportional to the product of the charges and inversely proportional to the square of the distance between them. D Excess charges gather at the sharpest radius of curvature of an electrified object. E Voltage is the product of Current and Resistance.

Electrifying the focusing cup with a negative charge will cause a repulsive force, as described in choice A, between the surface of the cup and the electrons excited into the space charge. The magnitude of this force is given by Coulomb's law, which is choice C. If the cup surrounds the space charge, it collapses it to a smaller volume directly in the center of the cup, in front of the filament. Choices B and E pertain to Faraday's Law and Ohm's Law, both describing electrodynamics. While D is certainly a rule of electrostatics, the design of the focusing cup varies according to manufacturer, thus it cannot be said that the cup will feature the sharpest radius of curvature in the environment of the cathode assembly in all cases. (Bushong, 11th edition, pgs 63-64, 107)

Which of the following terms refers to the number of x-ray examinations performed weekly? A Use factor B Workload C Time of occupancy factor D Controlled area

Explanation The answer is B. Thicker lead barriers are needed in areas that perform a high number of x-ray examinations. The workload represents the quantity of examinations performed on a weekly basis. Workload is stated as milliampere-minutes per week (mAmin/week) (B). The use factor is a term used to define the percentage of time that the beam is directed to a certain protective barrier; the higher workload percentage present impacts the lead requirements of the protective barrier (A). Time of occupancy factor refers to the amount of time the area would be occupied by non-radiation workers and is set to 3 levels: full, frequent, and occasional (C). Full would be used for work areas, frequent would be hallways, restrooms, and exam rooms, while occasional would refer to areas such as stairwells, elevators, closets, and the like. Areas occupied by radiation workers are termed controlled areas (D). (Bushong, 11th ed., pp. 556-557)

Which of the following placements of an oscillating grid will decrease radiation dose to the patient during ABC equipped image intensified fluoroscopy? A Between the fluoroscopy tube and the patient. B Between the patient and the input phosphor of the Image Intensifier. C Between the output phosphor and the CCD. D None of these.

Grids are designed to filter scatter radiation from the remnant beam as it exits the patient. Grids are used to impact image quality by reducing scattered radiation noise. If an oscillating/moving grid is integrated into a fluoroscope, it will lie between the patient and the input phosphor of the Image Intensifier, reducing scatter radiation from being included in the final displayed image. Since some useful x-ray photons will also be absorbed, the output signal from the Image intensifier will dim. If the machine is ABC equipped, the dose to the patient will increase as a result. (Bushong, 11th edition, pgs 194-200, 405)

Increasing the voltage on the electrostatic lenses within the Image Intensifier of an ABS equipped fluoroscopic system will result in: A Increased brightness gain B Increased flux gain C Increased magnification D Decreased patient dose

Multifield image intensifiers produce different magnification of the image. In normal operation, electrons from the entire surface of the photocathode are focused onto the output phosphor. When voltage to the electrostatic lenses is increased, a smaller inner diameter of the photocathode is used, and electrons originating from the perimeter of the Image Intensifier are not directed to the output phosphor. With ABS equipped fluoroscopy, brightness, flux, and minification gain would dip initially, until the ABS software responded by increasing technique. With kV and mA increased, the patient dose necessarily increases in magnified mode as a result of this adjustment. (Bushong, 11th edition, p410)

Which of the following is considered the unit of exposure in air? A air kerma B Gy C Sv D ergs/gm

The Correct Answer is: A The unit kerma expresses kinetic energy released in matter. X-rays expend kinetic energy as they ionize the air. Joule/kilogram is used to measure air kerma and 1 J/kg = 1 Gya. The subscript a represents air as the absorber. The mGya is the SI unit of measure of radiation exposure/intensity. The SI unit used to describe absorbed dose is Gray (Gyt)—the subscript t represents tissue. It describes tissue in general; various organs and tissue types require assigned specific weighting factors in order to describe the effective dose. Absorbed dose and energy deposited are strongly related to chemical change and biologic damage. The SI unit of measurement to describe effective dose to biologic material is the Sv (Sievert). The Sievert is the unit of occupational radiation exposure. Ergs/gm refers to amount of energy absorbed per gram of any absorber.

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

The Correct Answer is: A 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 and genetic mutations, are random/stochastic effects and 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. (Bushong, 11th ed., p. 523)

What is the approximate ESE for the average upright PA chest radiograph? A 0.10 mGy B 1.0 mGy C 1.50 mGy D 2.75 mGy

The Correct Answer is: A Patients will occasionally 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 can also be referred to the medical physicist for more detailed information. The average PA chest delivers an ESE of about 0.1 mGy. The average AP supine lumbar spine delivers an ESE of about 2.5 - 3 mGy; the average AP thoracic spine about 1.8 mGy; the average AP cervical spine about 1.0 mGy. The average AP pelvis delivers about 1.5 mGy. The average lateral skull about 0.7 mGy, the shoulder about 0.90 mGy, and the extremity about 0.5 mGy. (Bushong 11th ed p569, 595)

If an individual receives an exposure of 4 mGy/hr at a distance of 2 feet from a radiation source, what will be their dose after 20 minutes at a distance of 5 feet from the source? A 0.21 mGy B 0.53 mGy C 0.64 mGy D 1.6 mGy

The Correct Answer is: A The relationship between x-ray intensity and distance from the source is expressed in the inverse square law of radiation. The formula is Substituting known values: 4 / x = 25 / 4 25 x = 16 x = 0.64 mGy/hr ; 0.21 mGy in 20 min Distance has a profound effect on dose received and therefore is one of the cardinal rules of radiation protection. As distance from the source increases, dose received decreases. (Bushong, pp 68-70)

An increase of 1.0 mm added aluminum filtration of the x-ray beam would have which of the following effects?Increase in average energy of the beamIncrease in patient skin doseIncrease in mGya output A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: AAluminum filters are used to decrease patient skin dose by absorbing the low-energy photons (therefore, decreased mGya output) that do not contribute to the image but do contribute to patient skin dose. HVL is defined as that thickness of any absorber that will decrease the intensity of a particular beam to one-half its original value. As filtration of an x-ray beam is increased, the overall average energy of the resulting beam is greater (because the low-energy photons have been removed) and, therefore, the HVL thickness required would be greater. (Bushong, 8th ed., p. 165)

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 Correct Answer is: AAs 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 (total/true absorption) 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 can 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. (Bushong, 11th ed., pp. 148-150)

Which of the following contributes most to patient dose? A The photoelectric effect B Compton scatter C Classical scatter D Thompson scatter

The Correct Answer is: AAs 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 (total/true absorption) 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 doing so, 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 doing so, 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 classical scatter, a low-energy photon interacts with an atom but causes no ionization; the incident photon disappears in the atom, then immediately reappears and is released as a photon of identical energy but changed direction. Thompson scatter is another name for classical scatter. (Bushong, 11th ed, pp 148-150)

A student radiographer who is under 18 years of age must not receive an annual occupational dose of greater than A 1 mSv B 5 mSv C 50 mSv D 100 mSv

The Correct Answer is: ABecause 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 1 mSv in any given year. It is important to note that this 1 mSv is included in the 0.5 mSv dose limit allowed for the student as a member of the general public. (Bushong, 9th ed., p. 620)

All of the following device(s) can be used to reduce patient dose, exceptgridcollimatorgonad shield A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: ACollimators or other kinds of beam restrictors limit the amount of tissue being irradiated and, therefore, can reduce patient dose significantly. The use of gonadal shielding protects the reproductive organs from unnecessary radiation exposure and should be employed whenever possible. Grids function to absorb scattered radiation before it reaches the image to cause fog. Grids improve the radiographic image considerably; however, their use requires a significant increase in milliampere-seconds, that is, patient dose. (Frank, Long, and Smith, 11th ed., vol. 1, pp. 32-33)

The interaction between x-ray photons and matter shown in the figure below is associated withtotal energy transfer from photon to electronan outer-shell electronCompton scatter A 1 only B 1 and 2 only C 1 and 3 only D 2 and 3 only

The Correct Answer is: ADiagnostic x-ray photons interact with tissue in a number of ways, but most frequently they are involved in the photoelectric effect or in the production of Compton scatter. The photoelectric effect is shown in Figure 3-9; it occurs when a relatively low-energy x-ray photon uses all its energy to eject an inner-shell electron. That electron is ejected (photoelectron) from the innermost (K) shell, leaving a "hole" in the K shell and producing a positive ion. An L-shell electron then drops down to fill the K vacancy and in so doing emits a characteristic ray whose energy is equal to the difference between the binding energies of the K and L shells. The photoelectric effect occurs with high-atomic-number absorbers such as bone and positive contrast media and is responsible for the production of contrast. Therefore, its occurrence is helpful for the production of the radiographic image, but it contributes significantly to the dose received by the patient (because it involves complete absorption of the incident photon). Scattered radiation, which produces a radiation hazard to the radiographer (as in fluoroscopy), is a product of the Compton scatter interaction occurring with higher energy x-ray photons. (Selman, 9th ed., pp. 125-126)

Occupational radiation monitoring is required when it is possible that the individual's annual dose might exceed A 5 mSv B 10 mSv C 20 mSv D 50 mSv

The Correct Answer is: ADifferent types of monitoring devices are available for the occupationally exposed, and anyone who might receive more than one-tenth the annual dose limit must be monitored. The annual dose limit is 50 mSv; one tenth of that is 5 mSv. 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 in proportion 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. (Bushong, 11th ed., p. 605)

The term effective dose refers to A whole-body dose B localized organ dose C genetic effects D somatic and genetic effects

The Correct Answer is: AEvery radiographic examination involves an ESE, which can be determined fairly easily. It also involves a gonadal dose and a marrow dose, which, if needed, can be calculated by the radiation physicist. If the ESE of a particular examination were calculated to determine the equivalent whole-body dose, this would be termed the effective dose. For example, the ESE of a PA chest radiograph is approximately 0.7 mGy, whereas the effective dose is 0.1 mGy. The effective (whole body) dose is much less because much of the body is not included in the primary beam. (Fosbinder and Kelsey, p. 390)

Which of the following contributes most to patient dose? A The photoelectric effect B Compton scatter C Classic scatter D Thompson scatter

The Correct Answer is: AIn the photoelectric effect, a relatively low-energy photon uses all its energy (total/true absorption) to eject an inner-shell electron, leaving a vacancy. An electron from the shell above drops down to fill the vacancy and in so doing emits a characteristic ray. This type of interaction is most harmful to the patient because all the photon energy is 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 energy and exits the body as an energetic scattered ray. 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 reappears immediately and is released as a photon of identical energy but with changed direction. Thompson scatter is another name for classic scatter. (Bushong, 11th ed., p. 148-150)

The interaction between x-ray photons and tissue that contributes to radiographic contrast and contributes significantly to patient dose is A the photoelectric effect B Compton scatter C coherent scatter D pair production

The Correct Answer is: AIn the photoelectric effect, the incident (low-energy) photon is completely absorbed by the part, contributing significantly to patient dose, and contributing to production of short scale contrast. In Compton scatter, only partial absorption occurs, and most energy emerges as scattered photons. In coherent scatter, no energy is absorbed by the part; it all emerges as scattered photons. Pair production occurs only at very high energy levels, at least 1.02 MeV. (Bushong, 8th ed., p. 176)

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 Correct Answer is: ALET 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. (Bushong, 8th ed., p. 496)

The correct way(s) to check for cracks in lead aprons is (are)to fluoroscope them once a yearto radiograph them at low kilovoltage twice a yearby visual inspection A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: ALead aprons require certain maintenance and care if they are to continue to provide protection from ionizing radiation. They can be kept clean with a damp cloth. It is very important that they be hung when not in use rather than being folded or left in a heap between examinations. A folded or crumpled position encourages the formation of cracks in the leaded vinyl. Lead aprons should be fluoroscoped (at about 120 kVp) at least once a year to check for development of any cracks. Radiographing the "lead" apron with low kV would be ineffective - high kV would be required if using the radiographic method. (Bushong, 8th ed., p. 464)

The annual dose limit for medical imaging personnel must be tracked and reported, and includes radiation from the following sourcesoccupational exposurebackground radiationmedical x-rays A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: AOccupationally 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)

Which of the following will increase patient dose during fluoroscopy?Decreasing the SSDUsing 2.5 mm Al filtrationRestricting tabletop intensity to less than 10 R/min A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: APatient dose during fluoroscopy can be considerable because the x-ray tube is in close proximity to the patient. Therefore, we can decrease patient dose by increasing the SSD as much as possible. The law states that the SSD must be at least 15 in. with fixed fluoroscopic equipment and at least 12 in. with mobile fluoroscopic equipment. The use of 2.5-mm-Al-equivalent filtration in equipment operated above 70 kVp is also required by law to reduce patient skin dose. Another requirement of fluoroscopic equipment is that the tabletop intensity not exceed 10 R/min. (Bushong, 10th ed., p. 551)

The term used to describe the gradual decrease in exposure rate as an x-ray beam passes through matter is A attenuation B absorption C scattered radiation D secondary radiation

The Correct Answer is: AThe gradual decrease in exposure rate as radiation passes through matter is called attenuation. Attenuation is attributed to the two major types of interactions that occur in tissue between x-ray photons and matter in the diagnostic x-ray range. In the photoelectric effect, absorption and secondary radiation occur. In the Compton effect, scattered radiation is produced. With each of these occurrences, there is a decrease in the exposure rate that is referred to as attenuation. (Bushong, 8th ed., p. 185)

Which of the following radiation-induced conditions is most likely to have the longest latent period? A Leukemia B Temporary infertility C Erythema D Acute radiation lethality

The Correct Answer is: ARadiation effects that appear days or weeks following exposure (early effects) are in response to relatively high radiation doses. These should never occur in diagnostic radiology today; they occur only in response to doses much greater than those used in diagnostic radiology. One of the effects that may be noted in such a circumstance is the hematologic effect—reduced numbers of white blood cells, red blood cells, and platelets in the circulating blood. Immediate local tissue effects can include effects on the gonads (i.e., temporary infertility) and on the skin (e.g., epilation and erythema). Acute radiation lethality, or radiation death, occurs after an acute exposure and results in death in weeks or days. Radiation-induced malignancy, leukemia, and genetic effects are late effects of radiation exposure. These can occur years after survival of an acute radiation dose or after exposure to low levels of radiation over a long period of time. Radiation workers need to be especially aware of the late effects of radiation because their exposure to radiation is usually low level over a long period of time. Occupational radiation protection guidelines, therefore, are based on late effects of radiation according to a linear, nonthreshold dose-response curve. (Sherer, 7th ed., p. 217)

What is the relationship between kV and HVL? A As kV increases, the HVL increases. B As kV increases, HVL decreases C If the kV is doubled, the HVL doubles. D If the kV is doubled, the HVL is squared.

The Correct Answer is: AThe 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)

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 image-processing area D used as an extra badge for new personnel

The Correct Answer is: AThe 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 degree of exposure is compared with the degrees of exposure of the monitors worn in radiation areas. Exposure greater than that of the radiation-free monitor are reported in millirem units. (Bushong, 8th ed., p. 596)

Which of the following expresses the gonadal dose that, if received by every member of the population, would be expected to produce the same total genetic effect on that population as the actual doses received by each of the individuals? A Genetically significant dose B Somatically significant dose C Maximum permissible dose D Lethal dose

The Correct Answer is: AThe genetically significant dose (GSD) illustrates that large exposures to a few people are cause for little concern when diluted by the total population. On the other hand, we all share the burden of the radiation that is received by the total population, and especially as the use of medical radiation increases, each individual's share of the total exposure increases. (Bushong, 8th ed., p. 589)

Which of the following groups of exposure factors will deliver the least amount of exposure to the patient? A 400 mA, 0.25 second, 100 kVp B 600 mA, 0.33 second, 90 kVp C 800 mA, 0.5 second, 80 kVp D 800 mA, 1.0 second, 70 kVp

The Correct Answer is: AThe mAs setting regulates the quantity of radiation delivered to the patient, and the kVp setting regulates the quality (i.e., penetration) of the radiation delivered to the patient. Therefore, higher energy (i.e., more penetrating) radiation (which is more likely to exit the patient), accompanied by lower milliampere-seconds (mAs), is the safest combination for the patient. (Thompson et al., p. 275)

Which of the following tissues is (are) considered to be particularly radiosensitive?Intestinal mucous membraneEpidermis of extremitiesOptic nerves A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: AThe 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. (Dowd and Tilson, 2nd ed., p. 155)

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 Correct Answer is: AThe 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. In pair production, the x-ray photon disappears, and is replaced by two oppositely charged electrons. (Bushong, 10th ed., pp. 151-153)

The primary source of scattered radiation is the A patient. B tabletop. C x-ray tube. D grid.

The Correct Answer is: AThe scatterer between the target and the image recorder is the patient. After the radiation has scattered once, it has been significantly attenuated. The intensity of scattered radiation 1 m from the patient is approximately 0.1% of the intensity of the primary beam. (Bushong, p 552)

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 Decreased milliamperage

The Correct Answer is: AThe 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)

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

The Correct Answer is: AThe tissue weighting factor (W t ) represents the relative tissue radiosensitivity of irradiated material (e.g., muscle vs. intestinal epithelium vs. bone) and is used to account for differences in tissue sensitivity to ionizing radiation when determining effective dose E. The radiation weighting factor (W r ) 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 W r of different ionizing radiations depends on the LET of that particular radiation. The following formula is used to determine effective dose E: (Bushong, 8th ed., p. 556)

Which of the following is (are) likely to improve image quality and decrease patient dose?Beam restrictionLow kilovolt and high milliampere- second factorsGrids A 1 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: AThe 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. (Bushong, 9th ed., pp. 229-233)

Biologic material is least sensitive to irradiation under which of the following conditions? A Anoxic B Hypoxic C Oxygenated D Hyperbaric

The Correct Answer is: ATissue is most sensitive to radiation when it is oxygenated and least sensitive when it is devoid of oxygen. Anoxic refers to tissue without oxygen; hypoxic refers to tissue with little oxygen. Anoxic and hypoxic tumors typically are avascular (with little or no blood supply) and, therefore, more radioresistant. Oxygen at a higher level than atmospheric pressure (hyperbaric) can be used to treat infections and promote healing, as well as treat carbon monoxide poisoning. (Bushong, 10th ed., p. 481)

Which of the following will produce the most significant increase in patient dose? A Decreased mAs B Decreased SID C Increased filtration D Increased kVp

The Correct Answer is: B One of the most important ways to decrease exposure to ionizing radiation is to increase the distance between the source and the individual exposed. As the distance between the radiation source and the irradiated individual decreases, tissue exposure increases dramatically according to the Inverse Square Law of radiation. An increase in mAs will increase patient dose proportionally. An increase in filtration will decrease patient skin dose, as low energy photons are removed from the x-ray beam. An increase in kVp will increase the number of high energy photons, thereby increasing patient dose. However, when accompanied by an appropriate decrease in mAs, increased kVp serves to decrease patient dose. (Bushong, 8th ed, p 552)The Correct Answer is: ACharacteristic 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)

The x-ray unit of exposure defines which of the following? A Sievert B air kerma C Gyt D Becqueral

The Correct Answer is: B The SI unit air kerma is used to express kinetic energy released in matter; it is described as the unit of exposure and identifies ionization in air. The SI unit used to describe absorbed dose is Gray (Gyt). The SI unit of measurement to describe dose to biologic material is the Sievert (Sv). Various organs and tissue types require assigned specific weighting factors in order to describe the effective dose. Absorbed dose and energy deposited is strongly related to chemical change and biologic damage. The SI unit of measurement to describe effective dose to biologic material is the Sievert (Sv). The Sievert is the unit of occupational radiation exposure. The Becquerel is used to measure radioactivity.

What is the approximate entrance skin exposure (ESE) for the average AP pelvis radiograph? A 0.1 mGy B 1.5 mGy C 3.0 mGy D 4.0 mGy

The Correct Answer is: B Patients will occasionally 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 can also be referred to the medical physicist for more detailed information. The average PA chest delivers an ESE of about 0.1 mGy. The average AP supine lumbar spine delivers an ESE of about 2.5 - 3 mGy; the average AP thoracic spine about 1.8 mGy; the average AP cervical spine about 1.0 mGy. The average AP pelvis delivers about 1.5 mGy. The average lateral skull about 0.7 mGy, the shoulder about 0.90 mGy, and the extremity about 0.5 mGy. (Bushong 11th ed p569, 595)

If the exposure rate to a body standing 3 feet from a radiation source is 2.5 mGy what will be the exposure rate to that body at a distance of 7 feet from the source? A 0.12 mGy B 0.45 mGy C 1.07 mGy D 1.30 mGy

The Correct Answer is: B The relationship between x-ray intensity and distance from the source is expressed in the inverse square law of radiation. The formula is Substituting known values,: 2.5 mGy / x mGy = 49 / 9 49 x = 22.5 x = 0.45 mGy Note the inverse relationship between distance and dose. As distance from the source of radiation increases, dose rate decreases significantly. (Bushong, p 67)

Which of the following cells is the least radiosensitive? A Myelocytes B Myocytes C Megakaryocytes D Erythroblasts

The Correct Answer is: BBergonié and Tribondeau theorized in 1906 that all precursor cells are particularly radiosensitive (e.g., stem cells found in bone marrow). There are several types of stem cells in bone marrow, and the different types differ in degree of radiosensitivity. Of these, red blood cell precursors, or erythroblasts, are the most radiosensitive. White blood cell precursors, or myelocytes, follow. Platelet precursor cells, or megakaryocytes, are even less radiosensitive. Myocytes are mature muscle cells and are fairly radioresistant. (Bushong, 8th ed., p. 495)

If an individual receives 1.8 mGy while standing 4 ft from a source of radiation for 2 minutes, which of the following option(s) will most effectively reduce his or her radiation exposure to that source of radiation? A Standing 3 ft from the source for 2 minutes B Standing 8 ft from the source for 2 minutes C Standing 5 ft from the source for 1 minute D Standing 6 ft from the source for 2 minutes

The Correct Answer is: BA quick survey of the distractors reveals that option (A) will increase exposure dose and thus is eliminated as a possible correct answer. Options (B), (C), and (D) will serve to reduce radiation exposure because in each case either time is decreased or distance is increased. It remains to be seen, then, which is the more effective. Using the inverse-square law of radiation, at a distance of 8 ft (choice B), the individual will receive 0.45 mGy in 2 min (double distance from source = one-fourth the original intensity). 1.8 mGy / x mGy = 64 (82) / 16 (42) 64 x = 28.8 x = 0.45 mGy at 8 ft in 2 min At 5 ft, the individual will receive 0.57 mGy in 1 min: 1.8 mGy / x mGy = 25 (52) / 16 (42)25 x = 28.8 x = 1.15 mGy at 5 ft in 2 min; 0.57 mGy in 1 min At 6 ft, the individual will receive 0.8 mGy in 2 min: 1.8 mGy / x mGy = 36 (62) / 16 (42) 36 x = 28.8 x - 0.8 mGy at 6 ft for 2 min. Thus, x = 22.22 mR in 2 minutes at 6 ft. Therefore, the most effective option is (B), 12.5 mR in 2 minutes at 8 ft. (Bushong, 8th ed., p. 66)

The unit of absorbed dose is the A air kerma B Gy C Sv D Bq

The Correct Answer is: BAir kerma (kinetic energy released in matter) is the SI unit of exposure. It's symbol is Gya. The Gray is the unit of absorbed dose; it's unit of measure is Gyt. The Sievert is the SI units of effective dose; it's symbol is Sv. The Becquerel is the SI unit of radioactivity; it's symbol is Bq. (Bushong 11th ed, p 22)

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

The Correct Answer is: BAlpha 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)

Which of the following has(have) been identified as source(s) of radon exposure?Indoors, in housesSmoking cigarettesRadiology departments A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BBecause minerals in rocks and the earth can emanate radioactivity, high levels of radon gas inside homes have been of recent concern. Another source of radon gas is from burning cigarettes, whether as a smoker or as passive exposure. Uranium miners have been identified with a much higher incidence of lung cancer; many of these individuals also were smokers. Radiology departments are not known as a source of radon gas exposure. (Dowd and Tilson, 2nd ed., p. 147)

With milliamperes (mA) increased to maintain fluoroscopic 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 Correct Answer is: BBecause of the divergent quality of the x-ray beam, as source-to-skin distance (SSD) increases, entrance skin exposure (ESE) decreases for the necessary exit dose. SSD must be at least 30 cm on mobile fluoroscopic equipment, and at least 38 cm on fixed fluoroscopic equipment. (Bushong, 11th ed., p. 553-554)

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 Correct Answer is: BBecause 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. (Selman, 9th ed., pp. 88-89)

If a patient received 20 mGy during a 10-minute fluoroscopic examination, what was the dose rate? A 0.02 mGy/min B 0.2 mGy/min C 2.0 mGy/min D 20 mGy/min

The Correct Answer is: C If 20 mGy were delivered in 10 minutes, then the dose rate is 20 ÷ 10, or 2.0 mGy/min.

Which of the following terms describes the amount of electric charge flowing per second? A Voltage B Current C Resistance D Capacitance

The Correct Answer is: BCurrent is defined as the amount of electric charge flowing per second. Voltage is the potential difference existing between two points. Resistance is the property of a circuit that opposes current flow. Capacitance describes a quantity of stored electricity. (Selman, 9th ed., pp. 46-47)

An optically stimulated luminescence dosimeter contains which of the following detectors? A Gadolinium B Aluminum oxide C Lithium fluoride D Photographic film

The Correct Answer is: BDifferent types of monitoring devices are available for the occupationally exposed. The film badge has photographic film; the pocket dosimeter contains an ionization chamber; TLDs use lithium fluoride crystals. OSL dosimeters are personnel 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. (Bushong, 8th ed., p. 594)

Patient dose increases as fluoroscopic A FOV increases B FOV decreases C FSS increases D FSS decreases

The Correct Answer is: BDuring fluoroscopic procedures, as field of view (FOV) decreases, magnification of the output screen image increases, and contrast and resolution improve. The focal point on an image intensifier's 6-in. 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 increased automatically to compensate for the loss in brightness with smaller FOVs. Focal spot size (FSS) is unrelated to patient dose. (Fosbinder, p. 285)

What should be the radiographer's main objective regarding personal radiation safety? A Not to exceed his or her dose limit B To keep personal exposure as far below the dose limit as possible C To avoid whole-body exposure D To wear protective apparel when "holding" patients for exposures

The Correct Answer is: BEven the smallest exposure to radiation can be harmful. It, therefore, must be every radiographer's objective to keep his or her occupational exposure as far below the dose limit as possible. Radiology personnel never should hold patients during an x-ray examination. (Bushong, 8th ed., pp. 9-10)

Which of the following dose-response curve characteristics represent genetic and some somatic responses to radiation?1.Linear2.Nonthreshold3.Sigmoidal A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: BGenetic 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 sigmoidal dose-response curve has a threshold and is thought to be generally correct for most somatic effects. (Sherer, Visconti, Ritenour, and Haynes, 8th ed p 161)

Which of the following contributes most to occupational exposure? A The photoelectric effect B Compton scatter C Classic scatter D Thompson scatter

The Correct Answer is: BIn the photoelectric effect, a relatively low-energy photon uses all its energy to eject an inner-shell electron, leaving a vacancy. An electron from the shell above drops down to fill the vacancy and in so doing gives up a characteristic ray. This type of interaction is most harmful to the patient because all the photon energy is 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 it usually retains most of its energy and exits the body as an energetic scattered ray. This scattered ray can contribute to image fog/noise or pose a radiation hazard to personnel depending on its direction of exit; thus, Compton scatter contributes the most to occupational exposure. In classic scatter, a low-energy photon interacts with an atom but causes no ionization; the incident photon disappears into the atom and then is released immediately as a photon of identical energy but with changed direction. Thompson scatter is another name for classic scatter. (Bushong, 11th ed, p 148)

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

The Correct Answer is: BIrradiation 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 200 mGy is exceedingly small and that most general diagnostic examinations are likely to deliver fetal doses of less than 100 to 200 mGy. (Bushong, 8th ed., pp. 544-546)

Isotopes are atoms that have the same A mass number but a different atomic number B atomic number but a different mass number C mass number but a different neutron number D atomic number and mass number

The Correct Answer is: BIsotopes are atoms of the same element (the same atomic number or number of protons) but a different mass number. They differ, therefore, in their number of neutrons. Atoms having the same mass number but different atomic number are isobars. Atoms having the same neutron number but different atomic number are isotones. Atoms with the same atomic number and mass number are isomers. (Bushong, 8th ed., p. 47)

The tabletop exposure rate during fluoroscopy shall not exceed A 10 mGya/min/mA B 21 mGya/min/mA C 10 mGya/hr D 5 R/h

The Correct Answer is: BIt is important to limit tabletop exposure during fluoroscopy because the SSD is so much less than in overhead radiography, so a much higher skin dose is delivered to the patient. For this reason, the tabletop exposure rate during fluoroscopy should not exceed 21 mGya/min per mA at 80 kV. Without optional HLC (High Level Control), table top intensity of 100 mGya/min is maximum permitted. When optional HLC is available, the maximum tabletop exposure permitted is 200 mGya/min. (Bushong 11th ed p555)

Which of the following factors will affect both the quality and the quantity of the primary beam?HVLkVmA A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: BKilovoltage (kV) and the half-value layer (HVL) effect a change in both the quantity and the quality of the primary beam. The principal qualitative factor of the primary beam is kilovoltage, but an increase in kilovoltage will also increase the number of photons produced at the target. HVL, defined as the amount of material necessary to decrease the intensity of the beam to one-half, therefore changes both beam quality and beam quantity. Milliamperage (mA) is directly proportional to x-ray intensity (quantity) but is unrelated to the quality of the beam. (Carlton & Adler 5th ed p179)

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 Correct Answer is: BLate 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. Tissue reactions, early or late (formerly referred to as deterministic effects) are those that are unlikely to occur below a particular threshold dose and that increase in severity as the dose increases. (Sherer 8th ed p 191)

All the following statements regarding mobile radiographic equipment are true except A the exposure cord must permit the operator to stand at least 6 ft from the patient, x-ray tube, and useful beam B exposure switches must be the two-stage type C a lead apron should be carried with the unit and worn by the radiographer during exposure D the radiographer must alert individuals in the area before making the exposure

The Correct Answer is: BNCRP Report No. 102 states that the exposure switch on mobile radiographic units shall be so arranged that the operator can stand at least 2 m (6 ft) from the patient, the x-ray tube, and the useful beam. An appropriately long exposure cord accomplishes this requirement. The fluoroscopic and/or radiographic exposure switch or switches must be of the "dead man" type; that is, the exposure will terminate should the switch be released. A lead apron should be carried with every mobile x-ray unit for the operator to wear during the exposure. Lastly, the radiographer must be certain to alert individuals in the area, enabling unnecessary occupants to move away, before making the exposure. (NCRP Report No. 102, p. 25)

Any wall that the useful x-ray beam can be directed toward is called a A secondary barrier B primary barrier C leakage barrier D scattered barrier

The Correct Answer is: BProtective barriers are classified as either primary or secondary. Primary barriers protect from the useful, or primary, x-ray beam and consist of a certain thickness of lead. They are located anywhere that the primary beam can possibly be directed, for example, the walls of the x-ray room. The walls of the x-ray room usually require a 1 / 16 -in. (1.5-mm) thickness of lead and should be 7 ft high. Secondary barriers protect from secondary (scattered and leakage) radiation. Secondary barriers are control booths, lead aprons, gloves, and the wall of the x-ray room above 7 ft. Secondary barriers require much less lead than primary barriers. (Bushong, 8th ed., p. 571)

Radiation dose to personnel is reduced by which of the following guidelines?Exposure cords on fixed equipment must be very short.Exposure cords on mobile equipment should be long.Exposure cords on fixed and mobile equipment should be of the coiled, expandable type. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BRadiographic and fluoroscopic equipment is designed to help decrease the exposure dose to patient and operator. One of the design features is the exposure cord. Exposure cords on fixed equipment must be short enough to prevent the exposure from being made outside the control booth. Exposure cords on mobile equipment must be long enough to permit the operator to stand at least 6 ft from the x-ray tube. (Bushong, 8th ed., p. 569)

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 Correct Answer is: BScattering occurs when there is partial transfer of the photon'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, 11th ed., p. 187)

The interaction between ionizing radiation and a cell that is most likely to occur within the human body is termed A direct action B indirect action C target action D random action

The Correct Answer is: BSince 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 essential molecules. The two major types of effects that occur are the direct action and the indirect action. Direct action usually occurs with high-LET radiations (like alpha particles) and when ionization occurs at the molecule itself. Indirect action, which occurs most frequently, happens when ionization takes place away from the 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). (Sherer, Visconti, Ritenour, Haynes, 8th ed., p. 121)

Which of the following is recommended for the pregnant radiographer? A Change dosimeters weekly. B Wear a second dosimeter under the lead apron. C Wear two dosimeters and switch their positions appropriately. D The pregnant radiographer must leave radiation areas for duration of the pregnancy.

The Correct Answer is: BSpecial arrangements are required for occupational monitoring of the pregnant radiographer. The pregnant radiographer will wear two dosimeters—one in its usual place at the collar and the other, a baby/fetal dosimeter, worn over the abdomen and under the lead apron during fluoroscopy. The baby/fetal dosimeter must be identified as such and always must be worn in the same place. Care must be taken not to mix the positions of the two dosimeters. The dosimeters are read monthly, as usual. The pregnant radiographer may not be made to leave the radiation area/department because of her pregnancy. (Bushong, 8th ed., p. 560)

Which of the following is (are) acceptable way(s) to monitor the radiation exposure of those who are occupationally employed?TLDOSL dosimeterQuarterly blood cell count A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: BThe OSL dosimeter and TLD are used frequently to measure the radiation exposure of radiographers. The pocket dosimeter may be employed by radiation workers who are exposed to higher doses of radiation and need a daily reading. A blood test is an unacceptable method of monitoring radiation dose effects because a very large dose would have to be received before blood changes would occur. (Sherer et al., 5th ed., pp. 257-261)

The effects of radiation on biologic material depend on several factors. If a quantity of radiation is delivered to a body over a long period of time, the effect A will be greater than if it is delivered all at one time B will be less than if it is delivered all at one time C has no relation to how it is delivered in time D solely depends on the radiation quality

The Correct Answer is: BThe effects of a quantity of radiation delivered to a body depend on several factors—the amount of radiation received, the size of the irradiated area, and how the radiation is delivered in time. If the radiation is delivered in portions over a period of time, it is said to be fractionated and has a less harmful effect than if the radiation were delivered all at once. With fractionation, cells have an opportunity to repair, so some recovery occurs between doses. (Bushong, 8th ed., p. 496)

Sources of natural background radiation exposure include1.the food we eat.2.air travel.3.medical and dental x-rays. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BThe entire population of the world is exposed to varying amounts of background (environmental) radiation. Sources of background radiation are either natural or man-made. Exposure to natural background radiation is a result of cosmic radiation from space, naturally radioactive elements within the earth's crust, and our own bodies. Naturally, the closer we are to the cosmic radiations from space, the greater our personal exposure will be; living at higher elevations and air travel expose us to greater amounts of radiation. Living or working in a building made of materials derived from the ground exposes us to some background radiation from the naturally radioactive elements found in the earth's crust. The food we eat, the water we drink, and the air we breathe, all contribute to the quantity of radiation we ingest and inhale. Man-made radiation, however, is the type of background radiation over which we have some control. Medical and dental x-rays contribute to our exposure to man-made background radiation. (Bushong 11th ed, p 5-6, Sherer 8th ed p24)

What is the intensity of scattered radiation perpendicular to and 1 m from a patient compared with the useful beam at the patient's surface? A 0.01% B 0.1% C 1.0% D 10.0%

The Correct Answer is: BThe patient is the most important radiation scatterer during both radiography and fluoroscopy. In general, at 1 m from the patient, the intensity is reduced by a factor of 1,000 to about 0.1% of the original intensity. Successive scatterings can reduce the intensity to unimportant levels. (Bushong, 8th ed., p. 572)

The interaction illustrated belowcan pose a safety hazard to personnelcan have a negative impact on image qualityoccurs with low-energy incident photons A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BThe 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. In the illustrated Compton scatter, a fairly high-energy x-ray photon ejects an outer-shell electron. Although the x-ray photon is deflected with reduced energy (modified scatter), it retains most of its original energy and exits the body as an energetic scattered photon. Because the scattered photon exits the body, it does not pose a radiation hazard to the patient. It can, however, contribute to image fog and pose a radiation hazard to personnel (as in fluoroscopic procedures). In the photoelectric effect, a low- energy x-ray photon uses all its energy to eject an inner-shell electron, leaving an orbital vacancy. An electron from the shell above fills the vacancy and in so doing gives up energy in the form of a characteristic ray. The photoelectric effect is more likely to occur in absorbers having high atomic number and contributes significantly to patient dose because all the photon energy is absorbed by the patient. Coherent (unmodified) scatter does not involve ionization. (Bushong, 8th ed., p. 174)

What quantity of radiation exposure to the reproductive organs is required to cause temporary infertility? A 1000 mGy B 2000 mGy C 3000 mGy D 4000 mGy

The Correct Answer is: BThe reproductive cells are considered among the most radiosensitive cells in the body. The immature female sex cells are the oogonia; they mature to ova. The immature male sex cells are the spermatogonia; they mature to sperm. Different amounts of ionizing radiation to these cells can cause differing levels/degrees of response. Doses as low as 100 mGy can cause menstrual changes in women and decrease the number of sperm in men. At 2000 mGy, temporary infertility is likely, and at 5000 mGy, sterility will result. (Bushong, 8th ed., p. 523)

The source of electrons within the x-ray tube is via A electrolysis. B thermionic emission. C rectification. D induction.

The Correct Answer is: BThe thoriated tungsten filament of the cathode is heated by its own filament circuit. The x-ray tube filament is made of thoriated tungsten and is part of the cathode assembly. Its circuit provides current and voltage to heat it to incandescence, at which time it undergoes thermionic emission—the liberation of valence electrons from the filament atoms. Electrolysis describes the chemical ionization effects of an electric current. Rectification is the process of changing AC to unidirectional current. (Bushong, p 118)

The amount of time that x-rays are being produced and directed toward a particular wall is referred to as the A workload B use factor C occupancy factor D controlling factor

The Correct Answer is: BUse factor describes the percentage of time that the primary beam is directed toward a particular wall. The use factor is one of the factors considered in determining protective barrier thickness. Another is workload, which is determined by the number of x-ray exposures made per week. Occupancy factor is a reflection of who occupies particular areas (radiation workers or nonradiation workers) and is another factor used in determining radiation barrier thickness. (Bushong, 8th ed., p. 573)

The advantages of beam restriction include which of the following?Less scattered radiation is produced.Less biologic material is irradiated.Less total filtration will be necessary. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BWith greater beam restriction, less biologic material is irradiated, thereby reducing the possibility of harmful effects. If less tissue is irradiated, less scattered radiation is produced, resulting in improved IR contrast. The total filtration is not a function of beam restriction but rather is a radiation protection guideline aimed at reducing patient skin dose. (Selman, 9th ed., p. 247)

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 Correct Answer is: BX-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. (Bushong, 8th ed., p. 176)

Which of the following features of fluoroscopic equipment is (are) designed to minimize radiation exposure to the patient or personnel?Bucky slot coverExposure switch/foot pedalCumulative exposure timer A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: C D The Bucky slot cover shields the opening at the side of the table because the Bucky tray is parked at the end of the table for the fluoroscopy procedure; this is important because the opening created otherwise would allow scattered radiation to emerge at approximately the level of the operator's gonads. The exposure switch (usually a foot pedal) must be of the "dead man" type; that is, when the foot is released from the switch, there is immediate termination of exposure. The cumulative exposure timer sounds or interrupts the exposure after 5 minutes of fluoro time, thus making the fluoroscopist aware of accumulated fluoro time. In addition, source-to-tabletop distance is restricted to at least 15 in. for stationary equipment and at least 12 in. for mobile equipment. Increased source-to-tabletop distance increases source-to-patient distance, thereby decreasing patient dose. (Bushong, 11th ed., p. 555)

The unit of measurement used to express occupational exposure is the A kerma B Gy C Sv D RBE

The Correct Answer is: C The unit kerma expresses kinetic energy released in matter. X-rays expend kinetic energy as they ionize the air. Joule/kilogram is used to measure air kerma and 1 J/kg = 1 Gya. The subscript represents air as the absorber. The mGya is the SI unit of measure of radiation exposure/intensity. The SI used to describe absorbed dose is Gray (Gyt)—the subscript t represents tissue. It describes tissue in general; various organs and tissue types require assigned specific weighting factors in order to describe the effective dose. Absorbed dose and energy deposited is strongly related to chemical change and biologic damage. The SI unit of measurement to describe effective dose to biologic material is the Sv (Sievert). The Sievert is the unit of occupational radiation exposure.

If the ESE for a particular exposure is 3.0 mGy, what will be the intensity of the scattered beam perpendicular to and 1 m from the patient? A 0.3 mGy B 0.03 mGy C 0.003 mGy D 3.0 mGy

The Correct Answer is: C The patient is the most important radiation scatterer during both radiography and fluoroscopy. In general, at 1 m from the patient, the intensity is reduced by a factor of 1,000 to about 0.1% of the original intensity. Successive scatterings can reduce the intensity to unimportant levels. Calculate that 0.1% of 3.0 mGy is 0.003 mGy. (Bushong 11th ed p 556)

If the exposure rate to an individual standing 7.0 feet from a source of radiation is 0.5 mGy/hr, what will be the dose received after 20 minutes at a distance of 3.0 feet from the source? A 0.07 mGy B 0.21 mGy C 0.90 mGy D 2.72 mGy

The Correct Answer is: C The relationship between x-ray intensity and distance from the source is expressed in the inverse-square law of radiation. The formula is Substituting known values: 0.5 mGy/hr / x = 9 / 49 9 x = 24.5 x = 2.72 mGy/hr at 3 ft from source / 0.9 mGy in 20 minDistance has a profound effect on dose received and, therefore, is one of the cardinal rules of radiation protection. As distance from the source increases, dose received decreases. (Bushong, 8th ed., pp. 68-70)

The dose equivalent limit for a radiography student under the age of 18 years is A 50 mSv B 150 mSv C 1.0 mSv D 500 mSv

The Correct Answer is: C Whole-body annual occupational DL (dose limit) is 50 mSv. One of the special DL considerations regards radiography students under the age of 18. Their annual whole-body annual occupational DL must not exceed 1.0 mSv. This is the same DL used for the general public who might be exposed to larger amounts of radiation. The usual annual DL for the general public is 5 mSv. The DL to the lens of the eyes is 150 mSv; the DL for skin, hands, and feet is 500 mSv (Bushong, 11th ed, p 603)

Which of the following functions to protect the x-ray tube and the patient from overexposure in the event that the AEC fails to terminate an exposure? A Circuit breaker B Fuse C Backup timer D Rheostat

The Correct Answer is: CA phototimer is one type of automatic exposure device; another is the ionization chamber. When either is installed in an x-ray unit, it is calibrated to produce receptor exposures as required by the radiologist. Once the part being radiographed has been exposed to produce the proper receptor exposure, the AEC automatically terminates the exposure. The manual timer should be used as a backup timer should the photo timer fail to terminate the exposure, thus protecting the patient from overexposure and the x-ray tube from excessive heat load. Circuit breakers and fuses are circuit devices used to protect circuit elements from overload. In case of current surge, the circuit will be broken, thus preventing equipment damage. A rheostat is a type of variable resistor. (Shephard, p. 274)

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

The Correct Answer is: CAccording to the NCRP, the annual occupational whole-body dose-equivalent limit is 50 mSv. The annual occupational whole-body dose-equivalent limit for students under the age of 18 years is 1 mSv. The annual occupational dose-equivalent limit for the lens of the eye, a particularly radiosensitive organ, is 150 mSv. The annual occupational dose-equivalent limit for the thyroid, skin, and extremities is 500 mSv. (Bushong, 8th ed., p. 557)

At least how many HVLs are required to reduce the intensity of a beam of monoenergetic photons to less than 10% of its original value? A 2 B 3 C 4 D 5

The Correct Answer is: CAn HVL may be defined as the amount and thickness of absorber necessary to reduce the radiation intensity to half its original value. Thus, the first HVL would reduce the intensity to 50% of its original value, the second to 25%, the third to 12.5%, and the fourth to 6.25% of its original value. (Bushong, p 54)

Radiation output from a diagnostic x-ray tube is measured in which of the following units of measurement? A Gyt B Sv C Gya D Bq

The Correct Answer is: CAs x-ray photons emerge from the x-ray tube they immediately encounter air—before being intercepted by any material. The unit kerma expresses kinetic energy released in matter. X-rays expend kinetic energy as they ionize the air. Joule/kilogram is used to measure air kerma and 1 J/kg = 1Gya. The subscript a represents air as the absorber. The mGya is the SI unit of measure of radiation exposure/intensity. The SI unit used to describe absorbed dose is Gray (Gyt)—the subscript t represents tissue. It describes tissue in general; various organs and tissue types require assigned specific weighting factors in order to describe the effective dose. Absorbed dose and energy deposited is strongly related to chemical change and biologic damage. The SI unit of measurement to describe effective dose to biologic material is the Sv (Sievert). The Sievert is the unit of occupational radiation exposure.

Which of the following types of adult tissue is (are) comparatively insensitive to effects of ionizing radiation?Epithelial tissueNerve tissueMuscle tissue A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: CBecause muscle and nerve tissues perform specific functions and do not divide, they are relatively insensitive to ionizing radiation exposure. Epithelial cells cover the outer surface of the body; they also line body cavities and tubes and passageways leading to the exterior. They contain very little intercellular substance and are devoid of blood vessels. Because epithelial cells constantly regenerate through mitosis they are very radiosensitive. (Dowd and Tilson, 2nd ed., pp. 121-122)

The interaction between x-ray photons and matter illustrated in the figure below is most likely to occur Reproduced with permission from Saia DA. Radiography: Program Review and Examination Preparation, 2nd ed. Stamford, CT: Appleton & Lange, 1999. 1.in structures having a high atomic number.2.during radiographic examination of the abdomen.3.using high kV and low mAs exposure factors. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: CDiagnostic x-ray photons interact with tissue in a number of ways, but mostly they are involved in the production of Compton scatter or in the photoelectric effect. Compton scatter is pictured; it occurs when a relatively high-energy (kV) photon uses some of its energy to eject an outer-shell electron. In doing so, the photon is deviated in direction and becomes a scattered photon. Compton scatter causes objectionable scattered radiation fog in large structures such as the abdomen and poses a radiation hazard to personnel during procedures such as fluoroscopy. In the photoelectric effect, a relatively low-energy x-ray photon uses all its energy to eject an inner-shell electron, leaving a "hole" in the K shell. An L-shell electron then drops down to fill the K vacancy, and in so doing emits a characteristic ray whose energy is equal to the difference between the binding energies of the K and L shells. The photoelectric effect occurs with high-atomic-number absorbers such as bone and positive contrast media, and is responsible for the production of radiographic contrast. It is helpful for the production of the radiographic image, but it contributes to the dose received by the patient (because it involves complete absorption of the incident photon). (Saia, p 224)

If the entrance dose for a particular exam is 1200 mGy, the radiation exposure at 1 m from the patient will be approximately A 120 mGy B 12.0 mGy C 1.20 mGy D 0.12 mGy

The Correct Answer is: CDuring radiography and fluoroscopy, radiation scatters from the patient in all directions. In fact, the patient is the single most important scattering object in both radiographic and fluoroscopic procedures. The approximate intensity (quantity) of scattered radiation at 1 m from the patient is 0.1% of the entrance dose. Therefore, if the entrance dose for this image is 1200 mGy, the intensity of radiation at 1 m from the patient is 0.1% of that, or 1.20 mGy (0.001 × 1200 = 1.20). (Bushong, 8th ed., p. 572)

Radiation-induced abnormalities to the fetus during the seventh or eighth week of pregnancy is likely to cause which of the following? A Spontaneous abortion B Skeletal anomalies C Neurologic anomalies D Organogenesis

The Correct Answer is: CDuring the first trimester, specifically the second through eighth weeks of pregnancy (during major organogenesis), if the radiation dose is high enough, fetal anomalies can be produced. Skeletal anomalies usually appear if irradiation occurs in the early part of this time period, and neurologic (CNS) anomalies are formed in the latter part; mental retardation and childhood malignant diseases, such as cancers or leukemia, 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 spontaneous abortion. None of these responses are likely to occur with exposures less than 250 mGyt. It must be emphasized that the likelihood of producing fetal anomalies at diagnostic imaging doses is exceedingly small and that most general diagnostic examinations are likely to deliver fetal doses of less than 10 mGy. (Bushong, 11th ed., p. 536)

The classifications of acute radiation syndrome include all the following except A central nervous system B gastrointestinal C neonatal D hematologic

The Correct Answer is: CEarly somatic effects are manifested within minutes, hours, days, or weeks of irradiation and occur only following a very large dose of ionizing radiation. It must be emphasized that doses received from diagnostic radiologic procedures are not sufficient to produce these early effects. An exceedingly high dose of radiation delivered to the whole body in a short period of time is required to produce early somatic effects. These whole-body responses are grouped into three categories—reflecting the system(s) affected and the resulting symptoms: hematologic, gastrointestinal, and central nervous system. (Bushong, 8th ed., p. 484)

Which of the dose-response curves shown in the figure below is representative of radiation-induced skin erythema?Dose-response curve ADose-response curve BDose-response curve C A 1 only B 1 and 2 only C 3 only D 2 and 3 only

The Correct Answer is: CFigure 3-7 illustrates three dose-response curves. Curve A begins at zero, indicating that there is no safe dose, that is, no threshold. Even one x-ray photon theoretically can cause a response. It is a straight (linear) line, indicating that response is directly related to dose; as dose increases, response increases. Radiation-induced cancer, leukemia, and genetic effects follow a linear nonthreshold dose-response relationship. Curve B is another linear curve (response is directly related to dose), but this one illustrates that a particular dose of radiation must be received before a response will occur. That is, there is a threshold dose; this is called a linear threshold curve. Curve C is another threshold curve, but this curve is nonlinear. It illustrates that once the minimum dose is received, a response occurs slowly initially and then increases sharply as exposure increases. This threshold, nonlinear (sigmoid) dose-response curve, illustrates the effect to skin from exposure to high levels of ionizing radiation. (Bushong, 8th ed., p. 499)

The focal spot-to-table distance, in mobile fluoroscopy, must be A a minimum of 38 cm. B a maximum of 38 cm. C a minimum of 30 cm. D a maximum of 30 cm.

The Correct Answer is: CLead and distance are the two most important ways to protect from radiation exposure. Fluoroscopy can be particularly hazardous because the SID is so much shorter than in overhead radiography. Therefore, for patient protection, it has been established that fixed (stationary) equipment must provide at least 38 cm source to tabletop/skin distance. Mobile fluoroscopic equipment must provide at least 30 cm source-to-tabletop/skin distance. (Bushong, 10th ed p. 551)

The focal spot-to-table distance in mobile fluoroscopy must A be at least 38 cm B not exceed 38 cm C not be less than 30 cm D not exceed 30 cm

The Correct Answer is: CLead and distance are the two most important ways to protect from radiation exposure. Fluoroscopy can be particularly hazardous because the SID is so much shorter than in overhead radiography. Therefore, it has been established that mobile fluoroscopic equipment must provide at least 30 cm source-to-tabletop/skin distance for the protection of the patient. Stationary fluoroscopic equipment must provide at least 38 cm source-to-tabletop/skin distance (Bushong, 11th ed p 554)

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 Correct Answer is: CLead aprons are worn by occupationally exposed individuals during fluoroscopic and mobile x-ray procedures. Lead aprons are available with various lead equivalents; 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 lead equivalent apron will attenuate about 99.9% of a 50-kVp beam, 88% of a 75-kVp beam, and 75% of a 100-kVp beam. (Bushong, p 597)

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 Correct Answer is: CLead 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)

Which of the following will reduce patient dose during fluoroscopy?1.Decreasing the source-skin distance (SSD)2.Using 2.5 mm Al filtration3.Restricting tabletop intensity to less than 21 mGya/min A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: CPatient dose during fluoroscopy can be considerable because the x-ray tube is in close proximity to the patient. We can therefore decrease patient dose by increasing the SSD as much as possible. The law states that the SSD must be at least 38 cm (15 inches) with fixed fluoroscopic equipment and at least 30 cm (12 inches) with mobile fluoroscopic equipment. The use of 2.5 mm Al equivalent filtration in equipment operated above 70 kVp is also required by law to reduce patient skin dose. Another requirement of fluoroscopic equipment is that the tabletop intensity not exceed 21 mGya/min. Using high level control fluoroscopy, the tabletop intensity must not exceed 200 mGya/min. (Bushong, 11th ed., p. 552, 554)

The reduction in the intensity of an x-ray beam as it passes through material is termed A absorption B scattering C attenuation D divergence

The Correct Answer is: CThe reduction in the intensity (quantity) of an x-ray beam as it passes through matter as a result of absorption and scatter is called attenuation. Absorption occurs when an x-ray photon interacts with matter and disappears, as in the photoelectric effect. Scattering occurs when there is partial transfer of energy to matter, as in the Compton effect. (Bushong, 11th ed., p. 139)

qWhich acute radiation syndrome requires the largest exposure before any effects become apparent? A Hematopoietic B Gastrointestinal C Central nervous system (CNS) D Skeletal

The Correct Answer is: CRadiation 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 (10 Gy -50 Gy) 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 50 Gy is required to affect the normally resilient CNS. (Bushong, 8th ed., pp. 517-518)

Primary radiation barriers must be at least how high? A 5 ft B 6 ft C 7 ft D 8 ft

The Correct Answer is: CRadiation 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. (Bushong, 8th ed., pp. 571-572)

Which type of dose-response relationship represents radiation-induced leukemia and genetic effects? A Linear, threshold B Nonlinear, threshold C Linear, nonthreshold D Nonlinear, nonthreshold

The Correct Answer is: CRadiation-induced malignancy, leukemia, and genetic effects are late effects of radiation exposure. These can occur years after survival of an acute radiation dose or after exposure to low levels of radiation over a long period of time. Radiation workers need to be especially aware of the late effects of radiation because their exposure to radiation is usually low level over a long period of time. Occupational radiation protection guidelines, therefore, are based on late effects of radiation according to a linear, nonthreshold dose-response curve. (Bushong, 8th ed., p. 499)

Which of the following is (are) possible long-term somatic effects of radiation exposure?Blood changesCataractogenesisEmbryologic effects A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: CSomatic effects are those induced in the irradiated body. Genetic effects of ionizing radiation are those that may not appear for many years (generations) following exposure. Formation of cataracts or cancer (such as leukemia) and embryologic damage are all possible long-term somatic effects of radiation exposure. A fourth is life-span shortening. Blood changes are generally early effects of exposure to large quantities of ionizing radiation. (Bushong, 9th ed., pp. 551-558)

Thermoluminescent dosimetry systems use which of the following crystals? A Silver halide B Sodium bicarbonate C Lithium fluoride D Aluminum oxide

The Correct Answer is: CTLDs are personnel 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. TLDs are very accurate personal monitors. Even more accurate are optically stimulated luminescence (OSL) dosimeters. OSL dosimeters use aluminum oxide as their sensitive crystal. Silver halide is in film emulsion used in film badge dosimeters. (Bushong, 8th ed., p. 593)

The Gray may be described as A disintegrations per second. B ions produced in air. C energy deposited in an absorber. D biologic effects.

The Correct Answer is: CThe Becquerel is the unit of radioactivity, describing disintegrations per second. The air kerma is the unit of exposure; it measures the quantity of ionization in air. Gray measures the energy deposited in any material. Sievert includes the relative biologic effectiveness (RBE) specific to the tissue irradiated, and therefore is a valid unit of measure for the dose to biologic material.

Which of the following tissues or organs is the most radiosensitive? A Rectum B Esophagus C Small bowel D Central nervous system (CNS)

The Correct Answer is: CThe 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 Lieberkühich 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 Lieberküeing 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. (Dowd & Tilson, p 155)

If the exposure rate to an individual standing 2.0 m from a source of radiation is 130 mGy/min, what will be the dose received after 2 minutes at a distance of 5.0 m from the source? A 21 mGy B 27 mGy C 42 mGy D 52 mGy

The Correct Answer is: CThe relationship between x-ray intensity and distance from the source is expressed in the inverse square law of radiation. The formula is Substituting known values: 130/x = 25/4 25 x = 520 x = 20.8 mGy/min = 42 mGy in 2 min. Distance has a profound effect on dose received and therefore is one of the cardinal rules of radiation protection. As distance from the source increases, dose received decreases. (Bushong, pp 68-70)

All the following function to reduce patient dose except A beam restriction B high kVp, low mAs factors C a high-ratio grid D gonadal shielding

The Correct Answer is: CThe use of a grid requires an increase in milliampere-seconds and, therefore, patient dose; the higher the grid ratio, the greater is the increase in milliampere-seconds required. Collimation (beam restriction) restricts the amount of tissue being irradiated and, therefore, reduces patient dose. High kilovoltage reduces the amount of radiation absorbed by the patient's tissues (recall the photoelectric effect), and low milliampere-seconds reduces the quantity of radiation delivered to the patient. Appropriate gonadal shielding will always reduce patient dose to reproductive cells. (Bushong, 8th ed., pp. 11, 248)

A minimum total amount of aluminum filtration (inherent plus added) of 2.5 mm is required in equipment operated A above 50 kVp B above 60 kVp C above 70 kVp D above 80 kVp

The Correct Answer is: CThe x-ray tube's glass envelope and oil coolant are considered inherent (built-in) filtration. Thin sheets of aluminum are added to make a total of at least 2.5-mm-Al-equivalent filtration in equipment operated above 70 kVp. The function of the filtration is to remove the low-energy photons that serve only to contribute to skin dose. (Bushong, 8th ed., p. 568)

Biologic material is most sensitive to radiation exposure under which of the following conditions? A Anoxic B Hypoxic C Oxygenated D Deoxygenated

The Correct Answer is: CTissue 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. (Bushong, 8th ed., p. 496)

What is the annual dose limit (DL) for the lens of the eye? A 50 mSv/yr B 100 mSv/yr C 150 mSv/yr D 500 mSv/yr

The Correct Answer is: CWhole-body annual DL dose of 50 mSv/yr is calculated to include the especially radiosensitive organs such as the gonads and the blood-forming organs. The annual dose limit to the lens is 150 mSv; the DL to less sensitive organs such as skin, hands, and feet (extremities) is 500 mSv/year. (Sherer, 8th ed., p. 197)

Examples of potential late effects of ionizing radiation on humans can includeleukemiagenetic defectsmalignant disease A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: D Late or long-term effects of radiation can occur in tissues as a result of chronic exposure or tissues that have survived a previous irradiation months or years earlier. The major types of late effects are carcinogenesis, embryological (genetic) effects and cataractogenesis. Of these late effects, carcinogenesis and embryological are considered stochastic, whereas cataractogenesis is considered a tissue reaction (formerly referred to as deterministic). 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, can induce an effect (thus, the importance of radiation protection). Increasing that dose will increase the likelihood of the occurrence but will not affect its severity; these effects are termed stochastic. Tissue reactions (formerly referred to as Deterministic effects) can be early or late and are defined as those not occurring below a particular threshold dose and that increase in severity as the dose increases. Early effects of radiation exposure are in response to relatively high radiation doses. These never should occur in diagnostic radiology; they occur only in response to doses much greater than those used in diagnostic radiology. One of the effects that may be noted in such a circumstance is the hematologic effect—reduced numbers of white blood cells, red blood cells, and platelets in the circulating blood. Immediate local tissue effects can include effects on the gonads (temporary infertility) and on the skin (e.g., epilation and erythema). Acute radiation lethality, or radiation death, occurs after an acute exposure and results in death in weeks or days. (Bushong, 11th ed., p. 472, Sherer 8th ed p 191)

The National Council on Radiation Protection and Measurements (NCRP) recommends an annual occupational effective (stochastic) dose equivalent limit of A 1 mSv. B 5 mSv. C 15 mSv. D 50 mSv.

The Correct Answer is: DA 1984 review of radiation exposure data revealed that the average annual dose equivalent for monitored radiation workers was approximately 2.3 mSv. The fact that this is approximately one-tenth the recommended limit indicates that the limit is adequate for radiation protection purposes. Therefore, the NCRP reiterates its 1971 recommended annual limit of 50 mSv. (Bushong, 8th ed., p. 557)

All the following statements regarding TLDs are true except A TLDs are reusable B a TLD is a personal radiation monitor C TLDs use a lithium fluoride phosphor D after x-ray exposure, TLDs emit heat in response to stimulation by light

The Correct Answer is: DA TLD is a sensitive and accurate device used in radiation dosimetry. It may be used as a personal dosimeter or to measure patient dose during radiographic examinations and therapeutic procedures. The TLD uses a thermoluminescent phosphor, usually lithium fluoride. When used as a personal monitor, the TLD is worn for 1 month. During this time, it stores information regarding the radiation to which it has been exposed. It is then returned to the commercial supplier. In the laboratory, the phosphors are heated. They respond by emitting a particular quantity of light (not heat) that is in proportion to the quantity of radiation delivered to it. After they are cleared of stored information, they are returned for reuse. (Bushong, 9th ed., p. 593)

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 Correct Answer is: DA 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. (Selman, 9th ed., p. 400)

A controlled area is defined as onethat is occupied by people trained in radiation safetythat is occupied by people who wear radiation monitorswhose occupancy factor is 1 A 1 and 2 only B 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: DA 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)

Classify the following tissues in order of increasing radiosensitivityLiver cellsIntestinal crypt cellsMuscle cells A 1, 3, 2 B 2, 3, 1 C 2, 1, 3 D 3, 1, 2

The Correct Answer is: DAccording 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)

According to the NCRP, the pregnant radiographer's gestational dose-equivalent limit for a 1-month period is A 1 mSv B 5 mSv C 0.1 mSv D 0.5 mSv

The Correct Answer is: DAccording to the NCRP, the annual occupational whole-body dose-equivalent limit is 50 mSv. The annual occupational whole-body dose-equivalent limit for students under the age of 18 years is 1 mSv. The annual occupational dose-equivalent limit for the lens of the eye, a particularly radiosensitive organ, is 150 mSv. The annual occupational dose-equivalent limit for the thyroid, skin, and extremities is 500 mSv. The total gestational dose-equivalent limit for the embryo/fetus of a pregnant radiographer is 5 mSv, not to exceed 0.5 mSv in 1 month. (Bushong, 8th ed., p. 557)

Patient exposure can be minimized by using which of the following?1.Accurate positioning2.High-kV, low-mAs factors3.AEC A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: DAccurate positioning helps decrease the number of retakes. The use of high-kV and low-mAs exposure factors limits the quantity of radiation delivered to the patient. Use of AEC can enable the technologist to reduce repeats. Patient shielding should be used whenever appropriate and possible. (Selman, p 413)

X-ray photon beam attenuation is influenced by1.tissue type.2.subject thickness.3.photon quality. A 1 only B 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DAttenuation (decreased intensity through scattering or absorption) of the x-ray beam is a result of its original energy and its interactions with different types and thicknesses of tissue. The greater the original energy/quality (the higher the kilovoltage) of the incident beam, the less the attenuation. The greater the effective atomic number of the tissues (tissue type determines absorbing properties), the greater the beam attenuation. The greater the volume of tissue (subject density and thickness), the greater the beam attenuation. (Bushong, p 185)

Methods of reducing radiation exposure to patients and/or personnel include1.beam restriction.2.shielding.3.high-kV, low-mAs factors. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DBeam restriction is probably the single best method of protecting your patient from excessive radiation. It is also an important factor in obtaining high-quality radiographs because there will be less fog from scattered radiation. Shielding areas not included in the radiograph, especially particularly radiosensitive areas, is another effective means of reducing patient dose. If the patient is subjected to less radiation exposure, then so is the operator. Shielding, distance, and time are the three cardinal rules of radiation protection. High-kV, low-mAs exposure factors employ the use of fewer and more penetrating x-rays. (Bontrager, p 58)

In which of the following examinations would an IR front having very low absorption properties be especially important? A Abdominal radiography B Extremity radiography C Angiography D Mammography

The Correct Answer is: DBecause mammography uses such low-kilovoltage levels, IR front material becomes especially important. Any attenuation of the beam by the IR front would be most undesirable. Low-attenuating carbon fibers or special plastics that resist impact and heat softening (e.g., polystyrene and polycarbonate) are used frequently as IR front material. (Carlton and Adler, 4th ed., p. 619)

What is (are) the major effect(s) of deoxyribonucleic acid (DNA) irradiation?Malignant diseaseChromosome aberrationCell death A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DChromosome aberration, cell death, and malignant disease are major effects of DNA irradiation, often as a result of abnormal metabolic activity. If the damage happens to the DNA of a germ cell, the radiation response may not occur until one or more generations later. (Bushong, 8th ed., pp. 404-405)

Which of the following is most likely to require the lowest patient exposure? A Decreasing kilovoltage by 15% and doubling the milliampere-seconds value B Increasing kilovoltage by 15% and cutting the milliampere-seconds value in half C Changing collimation from 10 × 12 to 14 × 17 D Changing from an 8:1 grid technique to nongrid

The Correct Answer is: DConverting 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. Doubling the mAs would double patient dose. Therefore, the largest decrease will occur with removal of a grid. (Bushong, 11th ed., pp. 197, 203)

Which of the following statements regarding dual x-ray absorptiometry is (are) true?It is a low-dose procedure.Two x-ray photon energies are used.Photon attenuation by bone is calculated. A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: DDXA imaging is used to evaluate bone mineral density (BMD). 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, their attenuation is calculated to represent the degree of bone density. Bone densitometry/DXA can be used to evaluate bone mineral content of the body or part of it, to diagnosis osteoporosis, or to evaluate the effectiveness of treatments for osteoporosis. (Frank, Long, and Smith, 11th ed., vol. 3, pp. 454-455)

Which of the following safeguards is (are) taken to prevent inadvertent irradiation in early pregnancy?Patient postingsPatient questionnaireElective booking A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DElective booking of a radiologic examination after inquiring about the patient's previous menstrual cycle is the most effective means of preventing accidental exposure of a recently fertilized ovum. Patient questionnaires obtain this information from the patient and are also used often in an informed consent form. Patient postings in waiting and changing areas alert patients to advise the radiographer if there is any chance of pregnancy. These three safeguards replace the earlier 10-day rule, which is now obsolete. (Bushong, 8th ed., p. 559)

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 Correct Answer is: DExamples 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)

Radiation safety requirements for fluoroscopic equipment include the following:1.SSD at least 38 cm on stationary (fixed) equipment.2.SSD at least 30 cm on mobile equipment.3.high level/boost mode must have continuous audible signal. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DFluoroscopy is potentially a high patient dose procedure. The principal reason is that the source of x-ray photons is much closer to the patient than in overhead radiography. There are recommendations that provide guidelines for minimum SSD (Source-Skin Distance), maximum tube output, collimation, timer, exposure switch specifications, and others. SSD must be at least 38 cm in stationary (fixed) fluoroscopic equipment, and at least 30 cm for mobile fluoroscopic equipment. The tabletop intensity of the fluoroscopic beam must be less than 10 R/min. A cumulative timing device must be available to signal the fluoroscopist (audibly, visibly, or both) when a maximum of 5 minutes of fluoroscopy time has elapsed. High-level mode requires continuous manual activation and audible signal. (Dowd and Tilson, p 175)

To be in compliance with radiation safety standards, the fluoroscopy exposure switch must A sound during fluoro-on time B be on a 6-ft-long cord C terminate fluoro after 5 minutes D be the "dead man" type

The Correct Answer is: DFor radiation safety, the fluoroscopy exposure switch must be of the "dead man" type. When pressure from the foot/hand is removed from the fluoro pedal/switch, the "dead man" switch will terminate the exposure immediately. There must also be a cumulative fluoroscopy timer (a separate device) that will sound and/or interrupt exposure after 5 minutes of fluoroscopy. There is no audible sound during fluoro-on time. (Bushong, 11th ed., p. 565; NCRP Report No 102 p 14,15)

Gonadal shielding should be provided for male patients in which of the following examinations?FemurAbdomenPelvis A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DGonadal shielding should be used when the gonads lie within 5 cm of the collimated primary beam, when the patient has reasonable reproductive potential, and when clinical objectives permit. Because their reproductive organs lie outside the abdominal cavity, male patients are more easily and effectively shielded than are female patients, whose reproductive organs lie within the abdominal cavity. Therefore, radiographic examinations of the male abdomen and pelvic structures should include evidence of gonadal shielding. (Sherer, 5th ed., p. 177)

All of the following have an effect on patient dose except A inherent filtration. B added filtration. C SID. D focal spot size.

The Correct Answer is: DInherent filtration is composed of materials that are a permanent part of the tube housing: the x-ray tube's glass envelope and the oil coolant. Added filtration, usually thin sheets of aluminum, is included to make a total of 2.5 mm Al equivalent for equipment operated at 70 kVp and higher. Filtration is used to decrease patient dose by removing the low-energy photons that do not contribute to image formation but simply contribute to skin dose. According to the inverse square law of radiation, exposure dose increases as distance from the source decreases and vice versa. The effect of focal spot size is principally on spatial resolution; it has no effect on patient dose. (Shephard, pp 17-18)

Possible responses to irradiation in utero includespontaneous abortioncongenital anomalieschildhood malignancies A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DIrradiation during pregnancy, especially in early pregnancy, is avoided because the fetus is particularly radiosensitive during the first trimester. Especially high-risk examinations include pelvis, hip, femur, lumbar spine, cystograms and urograms, upper GI series, and barium enema (BE) examinations. During the 2nd through 10th weeks of pregnancy (i.e., during major organogenesis), fetal anomalies can be produced. Skeletal and/or organ anomalies can appear if irradiation occurs early on, and neurologic anomalies can be formed in the latter part; mental retardation and childhood malignant diseases can also result from irradiation during the first trimester. Fetal irradiation during the second and third trimesters, with sufficient dose, can cause some type of childhood malignant disease. Fetal irradiation during the first 2 weeks of gestation most likely will result in embryonic resorption or spontaneous abortion. It must be emphasized that the likelihood of producing fetal anomalies at doses below 2 Gy is exceedingly small and that most general diagnostic examinations are likely to deliver fetal doses of less than 0.01 to 0.02 Gy. (Bushong, 8th ed., pp. 544-546)

Late radiation-induced somatic effects includethyroid cancerscataractogenesisskin cancers A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DLate 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)

From which of the following primary beam sizes, all other factors remaining constant, will the greatest radiation exposure result? A 8 × 10 B 10 × 12 C 11 × 14 D 14 × 17

The Correct Answer is: DLimiting the irradiated field size, through collimation or other beam restriction, is perhaps the most effective way of controlling patient exposure dose. The smaller the irradiated area, the smaller the patient dose; the larger the irradiated area, the larger the patient dose. Therefore, the 14 × 17 primary beam size will result in the greatest patient exposure dose. With greater beam restriction, less biologic material is irradiated, thereby reducing the possibility of harmful effects. Additionally, if less tissue is irradiated, less scattered radiation is produced, resulting in improved contrast and image quality. (Dowd & Tilson, p 231)

Which of the following cell types has the greatest radiosensitivity in the adult human? A Nerve cells B Muscle cells C Spermatids D Lymphocytes

The Correct Answer is: DLymphocytes, a type of white blood cell concerned with the immune system, have the greatest radiosensitivity of all body cells. Spermatids are also highly radiosensitive, although not to the same degree as lymphocytes. Muscle cells have a fairly low radiosensitivity, and nerve cells are the least radiosensitive in the body (in fetal life, however, nerve cells are highly radiosensitive). (Dowd and Tilson, 2nd ed., p. 135)

Patient dose during fluoroscopy is affected by thedistance between the patient and the input phosphoramount of magnificationtissue density A 1 only B 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DMoving the image intensifier closer to the patient during fluoroscopy decreases the SID and patient dose (as SID is reduced, the intensity of the x-ray photons at the image intensifier's input phosphor increases; the automatic brightness control then automatically decreases the milliamperage and, therefore, patient dose). Moving the image intensifier closer to the patient during fluoroscopy also decreases the OID and, therefore, magnification. As tissue density increases, a greater exposure dose is required. (Fosbinder and Kelsey, pp. 265-267)

Each time an x-ray photon scatters, A its intensity increases 4 times. B its intensity increases 1000 times. C its intensity decreases 4 times. D its intensity decreases 1000 times.

The Correct Answer is: DOne of the radiation protection guidelines for the occupationally exposed is that the x-ray beam must scatter twice before reaching the operator. Each time the x-ray beam scatters, its intensity at 1 m from the scattering object is approximately 0.1% of the intensity of the primary beam, that is, one thousandth of its original intensity. That is why, in terms of radiation protection, the patient is considered the most important source of scatter. Of course, the operator should be behind a shielded booth while making the exposure, but multiple scatterings further reduce danger of exposure from scatter radiation. Other scattering objects include the x-ray table, the Bucky-slot cover, and the control booth wall. (Dowd & Tilson, p 199)

All the following radiation-exposure responses exhibit a nonlinear threshold dose-response relationship except A skin erythema B hematologic depression C radiation lethality D leukemia

The Correct Answer is: DThe genetic effects of radiation and some somatic effects, such as 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, epilation, hematologic depression, and radiation lethality (death). (Bushong, 8th ed., pp. 498-499)

Examples of stochastic effects of radiation exposure includeradiation-induced malignancygenetic effectsleukemia A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DRadiation effects that appear days or weeks following exposure (early effects) are in response to relatively high radiation doses. These should never occur in diagnostic radiology today; they occur only in response to doses much greater than those used in diagnostic radiology. One of the effects that may be noted in such a circumstance is the hematologic effect—reduced numbers of white blood cells, red blood cells, and platelets in the circulating blood. Immediate local tissue effects can include effects on the gonads (i.e., temporary infertility) and on the skin (e.g., epilation and erythema). Acute radiation lethality, or radiation death, occurs after an acute exposure and results in death in weeks or days. Radiation-induced malignancy, leukemia, and genetic effects are late effects (or stochastic effects) of radiation exposure. These can occur years after survival of an acute radiation dose or after exposure to low levels of radiation over a long period of time. Radiation workers need to be especially aware of the late effects of radiation because their exposure to radiation is usually low level over a long period of time. Occupational radiation protection guidelines, therefore, are based on late effects of radiation according to a linear, nonthreshold dose-response curve. (Bushong, 8th ed., p. 537)

If 1 - 10 Gy or more is received as a whole-body dose in a short period of time, certain symptoms will occur; these are referred to as A short-term effects. B long-term effects. C lethal dose. D acute radiation syndrome.

The Correct Answer is: DRadiation is most hazardous when it is received in a large dose, all at one time, to the whole body. This is called Acute Radiation Syndrome; ARC can occur as hematpoietic, gastrointestinal, or cerebrovascular, depending on total dose received. Biologic effects can appear within minutes to weeks (depending on the dose received). (Sherer, Visconti, Ritenour, and Haynes 8th ed, p 145)

How will x-ray photon intensity be affected if the SID is doubled? A Its intensity increases two times. B Its intensity increases four times. C Its intensity decreases two times. D Its intensity decreases four times.

The Correct Answer is: DSource-to-image-receptor distance (SID) has a significant impact on x-ray beam intensity (other terms are quantity, exposure rate, dose). As the distance between the x-ray tube and image receptor increases, exposure rate/intensity/dose (and therefore receptor exposure) decreases according to the inverse square law. According to the inverse square law, the exposure rate is inversely proportional to the square of the distance; that is, if the SID is doubled, the resulting beam intensity will be one fourth the original intensity; if the SID is cut in half, the resulting beam intensity will be four times the original intensity.

How will x-ray photon intensity be affected if the source-to-image distance (SID) is doubled? A Its intensity increases two times. B Its intensity increases four times. C Its intensity decreases two times. D Its intensity decreases four times.

The Correct Answer is: DSource-to-image-receptor distance (SID) has a significant impact on x-ray beam intensity (other terms we could use are quantity, exposure rate and dose). As the distance between the x-ray tube and IR increases, exposure rate/intensity/dose (and receptor exposure) decreases according to the inverse square law. According to the inverse square law, the exposure rate is inversely proportional to the square of the distance; that is, if the SID is doubled, the resulting beam intensity will be one-fourth the original intensity; if the SID is cut in half, the resulting beam intensity will be 4 times the original intensity. (Dowd and Tilson, 2nd ed., p. 199)

Reducing the number of repeat images is an important way to decrease patient exposure and can be accomplished by1.good patient communication.2.accurate positioning skills.3.using AEC. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DThe best way to ensure patient cooperation is through effective communication. A patient who understands what the examination entails, who knows what to expect, and what will be expected of him or her is better able to cooperate with the radiographer. This patient is more likely to be able to maintain the required position and suspend their respiration when required—thereby avoiding a repeated image. Radiographers who use their knowledge along with patience and critical thinking skills are more apt to obtain good images the first time around, thus avoiding repeat examinations. The use of AEC also helps avoid repeat radiographs; AEC will adjust the exposure—compensating for position, habitus, or pathology, and reducing the likelihood of repeat radiographs. (Dowd & Tilson, p 243)

Which of the following is (are) important for patient protection during fluoroscopic procedures?1.Intermittent fluoroscopy2.Fluoroscopic field size3.Focus-to-table distance A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DThe fluoroscopist should release his or her foot from the exposure pedal at frequent intervals, thus reducing total patient exposure, as the image will fade from the screen only slowly. Field size plays an important role in controlling patient dose in fluoroscopy (as in radiography). Focus-to-table distance is extremely important in controlling patient exposure; the law states that this distance must be a minimum of 12 inches (preferably 15 inches) to decrease patient dose. (Bushong, p 569)

Which of the following statements regarding the pregnant radiographer is (are) true?She should declare her pregnancy to her supervisor.She should be assigned a second personnel monitor.Her radiation history should be reviewed. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DThe pregnant radiographer should declare her pregnancy to her supervisor (but she is not required to); at that time, her radiation exposure history can be reviewed and appropriate assignments made. Special arrangements are required for occupational monitoring of the pregnant radiographer. The pregnant radiographer will wear two dosimeters—one in its usual place at the collar and the other, a baby/fetal dosimeter, worn over the abdomen and under the lead apron during fluoroscopy. The baby/fetal dosimeter must be identified as such and always must be worn in the same place. Care must be taken not to mix the positions of the two dosimeters. The dosimeters are read monthly as usual. (Bushong, 8th ed., p. 560)

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 Correct Answer is: DThe 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 effect on receptor exposure, or on interactions (photoelectric, Compton) within the body. Filtration serves to increase the overall average energy of the beam; it "hardens" the x-ray beam, making it less heterogeneous. (Bushong, 8th ed, p 167)

The target theory applies to A spermatagonia B oocytes C lymphocytes D DNA molecules

The Correct Answer is: DThe 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). (Bushong, 11th ed., p. 499)

If the exposure rate to a body standing 8 feet from a radiation source is 70 mGy/min, what will be the dose to that body at a distance of 5 feet from the source? A 27 mGy/min B 43 mGy/min C 112 mGy/min D 179 mGy/min

The Correct Answer is: DThe relationship between x-ray intensity and distance from the source is expressed in the inverse square law of radiation. The formula is Substituting known values, 70 / x = 25 / 64 25 x = 4480 x = 179.2 mGy at 5 ft from the source Note the inverse relationship between distance and dose. As distance from the source of radiation increases, dose rate significantly decreases. (Bushong, pp 68-70)

Which of the following radiation exposure responses exhibit a nonlinear threshold dose-response relationship?1.Skin erythema2.Hematologic depression3.Lethality A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DThe stochastic effects of radiation 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 tissue reactions (formerly referred to as deterministic effects)—such as skin erythema, hematologic depression, and radiation lethality (death).

The major function of filtration is to reduce A image noise. B scattered radiation. C operator dose. D patient dose.

The Correct Answer is: DX-rays produced at the target make up a heterogeneous primary beam. There are many "soft," low-energy photons that, if not removed, would contribute only to greater patient (skin) dose. They do not have enough energy to penetrate the patient and expose the film; they just penetrate a small thickness of the patient's tissue and are absorbed. These photons are removed by aluminum filters. (Fauber, pp 32-33)

What is the single-most important way to reduce patient exposure? A Restriction of the useful beam B Opening up the collimation C Double exposure D AP vs. PA

The answer is A. Beam restrictorsare devices used to restrict the coverage of an x-ray beam (A). The restriction eliminates exposure to the patient. Keeping the collimator to the size of the film will result in a lesser patient dose of radiation. Some examples of beam restrictors include the following: cones, cylinders, and collimators. These devices confine the useful or primary beam before it enters the area of interest and reduce the amount of scattered radiation in the tissue, which then prevents unnecessary exposure to anatomy. Opening up the collimation would increase exposure to the patient, thereby making more scatter radiation (B). Double exposure would increase the exposure to the patient, making more scatter and doubling the amount of radiation (C). AP vs. PA is used for radiation protection and PA results in less radiation for the patient, but it is not the single most important way to reduce patient exposure (D).(Sherer, Visconti, Ritenour, & Haynes, 8thed., p. 204)

The annual upper effective dose limit for radiation personnel is50 mSv500 mSv50 mrem5 rem A 1 and 4 B 1 and 3 C 2 and 3 D 2 and 4

The answer is A. For individuals 18 years and older who receive occupational radiation exposure, the annual upper effective dose limit is 50 mSv (1), which is equivalent to 5 rem (4). To keep levels as low as reasonably achievable (ALARA), personnel should use time, distance, and shielding to their advantage. In addition to dosimeters to monitor their dose, personnel should also invest in personal shielding accessories, such as goggles or glasses, when working in fluoroscopy to prevent excess radiation exposure.(Coakes and Ehrlich, 9th ed., pp. 40-41)

The minimum source-to-skin distance (SSD) in mobile fluoroscopy must be A 30 cm (12 in.) B 38 cm (15 in.) C 50 cm (20 in.) D 25 cm (10 in.)

The answer is A. Patient skin dose can be reduced during fluoroscopic procedures by using a when the radiographer using a higher kVp and a lower mA setting. The radiographer can further limit entrance exposure by ensuring that the source-to-skin distance (SSD) is no less than30 cm (12 in.)during mobile fluoroscopy procedures (A). The minimum SSD for stationary (fixed) fluoroscopes is 38 cm (15 in.) (B). A38 cm (15 in.)minimum SSD ispreferred, however, for all fluoroscopic procedures. The remaining selections,50 cm (20 in.)and25 cm (10 in.), are beyond or less than the minimum SSD requirement (C and D). (Statkiewicz, Visconti, Ritenour & Haynes, 8th ed., p. 222)

Why are protective lead aprons and gloves worn during radiographic procedures? A To protect from scatter radiation B To protect the machines C So you can work longer D To keep your hands clean

The answer is A. Protective lead aprons and shield barriers function as gonadal shields for diagnostic imaging personnel. These devices protect personnel from both scatter and leakage radiation, which are types of secondary radiation (A). Protective gloves are used to cover the hands of radiologists and radiographers when they must be in or near a primary beam. Lead aprons are available in various thicknesses. Lead aprons do not protect the equipment (B). Lead aprons protect the staff from occupational exposure. Disposable gloves are worn for infection control purposes. The other answer choices are incorrect as per the previous explanation (C and D).(Sherer, Visconti, Ritenour, and Haynes, 8thed., pp. 276-277)

The dose required to cause erythema in 50% of those exposed is roughly A 5 Gy B 10 Gy C 15 Gy D 20 Gy

The answer is A. The average dose required to potentially produce erythema in 50% of those exposed is 5 Gy (A). Everyone has their own level of radiosensitivity, which will determine whether they have a skin reaction to this exposure level. Doses at much higher levels, however, will produce erythema in addition to other radiation reactions, such as GI syndrome and CNS syndrome. The other answer choices are incorrect as per the previous explanation (B, C, and D).(Bushong, 11th ed., p. 513)

What is the cumulative effective dose limit for a 37-year-old radiographer? A 370 mSv B 150 mSv C 50 mSv D 5 mSv

The answer is A. The formula used to calculate the cumulative occupational effective dose is 10 mSv × age. If the radiographer is 37 years of age, simply multiply 37 by 10 mSv, which equals 370 mSv (A). The annual occupational equivalent dose limit for the lens of the eye is 150 mSv (B). The annual occupational effective whole-body dose limit is 50 mSv (C). The fetal equivalent dose limit for the entire gestation period is 5 mSv (D). (Statkiewicz, Visconti, Ritenour, & Haynes, 8th ed., p. 197)

Which of the following statements are true with respect to the accompanying animation? (select the two that apply)0:120:05 A Involves no ionization B Likely occurs in tissues having high atomic number C Responsible for scattered radiation fog D Occurs with the tungsten atom nucleus E All the incoming photon energy is absorbed by tissue F Most likely to produce long scale contrast

The answer is B and E. In photoelectric effect, a relatively low energy (low kV) x-ray photon interacts with tissue and expends all of its energy (true/total absorption) to eject an inner shell electron. This creates a positive ion, leaving an inner shell orbital vacancy. An electron from the shell above drops down to fill the vacancy and, in doing so, gives up energy in the form of a characteristic ray. The photoelectric effect is more likely to occur in absorbers having high atomic numbers (e.g., bone, positive contrast media) and with low energy photons (B). The photoelectric effect contributes significantly to patient dose, as all the x-ray photon energy is absorbed by the tissue (and, therefore, contributes to the production of short-scale contrast) (E). Its probability is Z3/E3 in the diagnostic energy range. Characteristics of photoelectric effect include: A low energy x-ray photon uses all its energy to eject an inner-shell electron. It produces a characteristic ray (absorbed in tissue). It is a major contributor to patient dose. It occurs in absorbers having high atomic numbers. Because it involves total absorption, no photon energy reaches the IR and short-scale contrast is likely. (Bushong, 11th ed., pp. 149, 150)

The average energy of a Bremsstrahlung x-ray photon in the x-ray emission spectrum is ______ the kilovoltage-peak (kVp) selected by the radiographer. A one-half of B one-third of C two-thirds of D equal to

The answer is B. The energy of a Bremsstrahlung x-ray photon can be found by subtracting the energy that the filament electron has upon leaving the atom of the anode target atom from the energy it had upon entering the anode target atom. The energy the electron possesses upon leaving the anode target atom is dependent upon how close the electron encounters the nucleus of the anode target atom. The average energy of the Bremsstrahlung photon in the x-ray emission spectrum is one-third of the kVp selected by the radiographer (B). The other answer choices are incorrect as per the previous explanation (A, C, and D). (Johnston and Fauber, 2nd ed., p. 64)

Select the factors that lead to the production of a high-energy bremsstrahlung (brems) x-ray when a filament (projectile) electron interacts with an anode tungsten atom (select the four that apply) A The filament electron passes far away from the nucleus of a tungsten atom. B The filament electron travels close to the nucleus of a tungsten atom. C The nuclear electrostatic field creates a strong force of attraction on the passing filament electron. D The nuclear electrostatic field creates a weak force of attraction on the passing filament electron. E The filament electron arrives at the anode with high kinetic energy. F The filament electron arrives at the anode with low kinetic energy. G The atomic number of the tungsten atom is high. H The atomic number of the tungsten atom is low.

The answer is B, C, E, and G. A bremsstrahlung (brems) x-ray is produced when a filament (projectile) electron is slowed as it passes through the electrostatic field of an anode target atom nucleus. Because energy can neither be created nor destroyed, but only can change from one form to another, the nuclear force of attraction on the electron causes it to "brake" and turn in a new direction, transforming some of the electron's kinetic energy into electromagnetic radiation in the form of a brems x-ray. Higher energy brems x-rays are created when an incoming projectile electron has high kinetic energy and travels close to the target atom's nucleus. The target atoms are tungsten atoms, which have a high atomic number (74), meaning they have many protons in their nuclei. Since the protons have a positive charge, this creates a strong positive electrostatic attraction on the negatively charged projectile electron as it passes by, causing the electron to slow down as it is drawn toward the nucleus. This redirection of a high-energy projectile electron passing close to the nucleus yields a high-energy brems x-ray photon (B, C, E, G). When compared to a high kinetic energy electron passing close to the target atom nucleus, a projectile electron with low kinetic energy passing far from the nucleus would produce a low-energy brems x-ray (A, D, F, H). (Bushong, 11th ed., pp. 126-127)

Which of the following statements are true with respect to the accompanying animation? (select the three that apply)0:300:00 A Resulting scatter is unmodified B Scattered photon retains much of its original energy C Occurs with inner shell electrons D End products are recoil electron, scattered photon, and positive ion E Occurs with high-speed projectile electrons F Responsible for scattered radiation fog

The answer is B, D, and F. In Compton scatter, a fairly high energy (high kV) x-ray photon interacts with tissue atoms, giving up some of its energy to eject a loosely bound outer shell electron and producing a positive ion. The ejected electron is called a recoil electron. The scattered x-ray photon is deflected with somewhat reduced energy (modified scatter) (D). However, it retains much of its original energy and exits the body as an energetic scattered photon (B). Compton scatter predominates in diagnostic imaging. Because the scattered photon exits the body, it does not pose a radiation hazard to the patient. It can, however, contribute to scattered radiation image fog and pose a radiation hazard to personnel (as in fluoroscopic procedures) (F). Characteristics of Compton scatter include: It is the interaction that predominates in the diagnostic x-ray range. A high-energy x-ray photon uses a portion of its energy to eject an outer shell electron. It is responsible for scattered radiation fog to the image. It poses a radiation hazard to personnel. (Bushong, 11th ed., pp. 148, 149)

Which of the following is the term used to describe a skin dose that exceeds 15 Gy? A Genetically significant dose B Sentinel event C Threshold dose D Relative risk

The answer is B. A skin dose of 15 Gy is referred to as a sentinel event (B). Doses at this level cause skin damage and are to be reported to the Center for Devices and Radiological Health (CDRH). The genetically significant dose is the dose assigned to the population gene pool as a result of gonad exposure (A). The threshold dose is the dose at which an individual poses low risk of biological damage from irradiation, and relative risk is an estimation of late biological effects of radiation to a population where the dose is not precisely known (C and D).(Bushong, 11thed., pp. 516-575)

Which of the following cell types has (have) low radiosensitivity in the adult human? 1. Muscle cells 2. Lymphocytes 3. Nerve cells A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is B. According to thelaw of Bergonie' and Tribondeau, cell radiosensitivity is directly proportional to their reproductive activity and inversely proportional to their degree of differentiation (specialization). Therefore, the most radiosensitive cells in the adult human would be those with the least maturity and differentiation, the greatest reproductive activity, and the longest mitotic phases. Themostradiosensitive cells in the adult human are thelymphocytes(A, C, and D). Adult muscle cells and nerve cells areinsensitiveto radiation, as they are highly specialized and do not divide and reproduce (B). The aforementioned applies to theadulthuman, but in the embryo-fetus, all cells are considered radiosensitive, especially in the first trimester. This is due to their immaturity and non-specialization. Radiographers should therefore be ever-cognizant of this when performing radiographic procedures on females of reproductive age and be diligent toward adhering to the 10-day rule. (Statkiewicz, Visconti, Ritenour & Haynes, 8th ed., pp. 133-136)

During mobile fluoroscopic procedures, the radiographer must use a source-to-skin distance of at least A 15 inches. B 12 inches. C 72 inches. D 40 inches.

The answer is B. During mobile (portable) fluoroscopy, the radiographer must maintain a source-to-skin distance (SSD) of at least 12 inches to minimize patient entrance dose. A radiolucent spacer cone is commonly attached to the x-ray tube to ensure at least a 12-inch SSD. Although a 12-inch SSD is required, a 15-inch minimal distance is preferred for all fluoroscopy systems (B). A 15-inch SSD is required for stationary (fixed) fluoroscopes (A). For upright chest radiography, a 72-inch SID is preferred to minimize magnification distortion of the heart and surrounding mediastinal structures and minimize image blur (C). For most radiographic procedures other than upright chest radiography, a minimum 40-inch SID is typically used (D). (Statkiewicz, Visconti, Ritenour, & Haynes, 8th ed., p. 222)

As recommended by the NCRP, the occupational dose limit for extremities is ___ the limit for non-radiation workers/the public. A 5 times B 10 times C 15 times D 25 times

The answer is B. For radiation workers, the occupational dose limit for extremities is 500 mSv. For the public, the dose limit for extremities is 50 mSv. This equates to a dose limit 10 times greater for radiation workers than for the public (B). The other answer choices are incorrect as per the previous explanation (A, C, and D). (Bushong, 11th ed., p. 603)

NCRP 102 guidelines regarding leaking radiation state the following A Must not exceed 10 mGya/h at 1 m from the x-ray tube B Must not exceed 1 mGya/h at 1 m from the x-ray tube C Must be attenuated by 0.25 mm Pb equiv D Must not exceed 50 mSv/year

The answer is B. Leakage radiation is that which is emitted from the x-ray tube housing in directions other than that of the primary beam. NCRP guidelines state that any leakage radiation from lead-lined x-ray tubes must not exceed 1 mGya/h (100 mR/h) when measured at a distance of 1 m from the x-ray tube (B). Other NCRP recommendations include: Protective lead aprons must be at least 0.25-mm Pb equivalent and must be worn by the workers in the fluoroscopy department. NCRP recommends a 0.5-mm Pb equivalent. Protective lead gloves must be at least 0.25-mm Pb equivalent. The unshielded hand must not be placed in the unattenuated or useful beam. The control panel must somehow indicate when the x-ray tube is energized (i.e., "exposure-on" time) by means of an audible or visible signal. The x-ray exposure switch must be a "dead-man" switch and situated so that it cannot be operated outside the shielded area of the control booth. The other answer choices are incorrect as per the previous explanation (A, C, and D). (Saia, PREP 9th ed., pp. 282, 283)

Exposure cords for fixed equipment must be 1. Attached to the console 2. Short 3. Positioned to allow for the technologist to step outside of the console booth into the examination room during exposure A 2 only B 1 and 2 only C 1 and 3 only D 2 and 3 only

The answer is B. On fixed x-ray equipment, the exposure control cord should be attached to the console. The cord should be short in order to prevent the technologist from stepping outside of the control booth and into the examination room during exposure. Mobile equipment, on the other hand, requires the use of a long cord so that the technologist can step far enough away from the exposure for personal protection.(Bushong, 11thed., pp. 552-553)

The hypothesis that small amounts of ionizing radiation can actually be beneficial is called A ALARA B Radiation hormesis C Action limits D No threshold

The answer is B. The concept ofradiation hormesisis that there exists a beneficial aspect or result to groups of individuals from continuing exposure to small amounts of radiation (B). This concept is a positive consequence of radiation for populations continuously exposed to moderately higher levels of radiation than ordinary background levels.ALARAmeans "as low as reasonably achievable" (A).Action limits/levelin radiation is defined as the specific dose of radiation or other parameter that if reached may indicate a loss of control of part of a licensee's radiation protection (C).No thresholdis a model used in radiation protection to quantify radiation exposure and set regulatory limits (D). This means that radiation is always considered harmful with no safety threshold, and the sum of several very small exposures are considered to have the same effect as one larger exposure.(Sherer, Visconti, Ritenour, and Haynes, 8thed., p. 196)

Which of the following exposures would most likely deliver the least dose to the ovaries? A AP pelvis radiograph B PA thoracolumbar scoliosis radiograph C AP thoracolumbar scoliosis radiograph D AP bilateral (Cleaves) hips radiograph

The answer is B. The ovaries lie in the anterior portion of the pelvic cavity. Because of this anatomical location, a PA thoracolumbar scoliosis projection would absorb some of the x-rays in the lower back, thereby reducing exposure and ovarian dose. Likewise, breast dosage is reduced in the thoracic region due to absorption of some of the primary x-rays in the upper back. (B). An AP thoracolumbar scoliosis radiograph would enable more of the x-rays in the primary x-ray beam to reach the ovaries and increase their dose (C). An AP pelvis radiograph would produce the highest ovarian dose because of their more anterior location in the pelvic cavity. The primary x-ray beam also encompasses the entire pelvic area, which allows for more direct exposure and produces more internal scatter radiation that may reach the ovaries (A). An AP bilateral Cleaves projection of the hips would result in approximately the same ovarian dose as the AP pelvis radiograph, because of the centering location one-inch superior to the symphysis pubis (D). (Long, Rollins, & Smith, 13th ed., vol. 1, pp. 437-438)

The average energy of the x-ray beam is increased by which of the following? 1. Increased added filtration 2. Increased generator voltage ripple 3. Increased kilovoltage A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is B. Theaverage energy (quality)of the x-ray beam can be increased withincreased added filtration. Filtration exists in the form of bothinherentandadded filtrationand functions to absorb the lower-energy x-rays existing within the heterogenous x-ray beam. As the lower-energy x-rays are filtered, the average energy of the beam increases.Increased kilovoltageincreases acceleration of filament electrons in the x-ray tube traveling toward the anode. Because of the increased electron kinetic energy produced by this increased potential difference, bremsstrahlung x-rays of higher energy are produced (B).Increased generatorvoltage rippleproduces lower average kilovoltage, resulting in lower x-ray beam average energy (A, C, and D). (Bushong, 11thed., pp. 130-134)

At exposures utilizing 75 kVp, the attenuation percentage difference between shielding with 0.25 mm Pb equivalent versus 1.0 mm Pb equivalent is A 15% B 33% C 52% D 78%

The answer is B. X-ray attenuation varies as lead or lead equivalent thickness changes. At 75 kVp, a 0.25 mm apron attenuates 66% of the x-rays, while a 1.0 mm apron will attenuate 99%. This results in a 33% change between the two aprons (B). At 100 kVp, the 0.25 mm apron attenuates 51% of the x-rays, whilea 1.00 mm apron will attenuate 94%; while the 1.0 mm aprons are heavier by 4-7 kg, the protection is significantly increased. The other answer choices are incorrect as per the previous explanation (A, C, and D).(Bushong, 11th ed., p. 609)

Which of the following statements is/are true regarding shadow shields?They attach to the collimator box Shadow shields are more effective than contact shieldsPositioning is determined by adjusting the shield within the light field A 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is B. Shadow shields are shielding that attach to the collimator box to help protect patients from unnecessary radiation exposure (1). They work by limiting the x-ray beam, and the area they cover can be adjusted and determined by the shadow they cast in the light field (3). They are, however, less effective than contact shields (2). (Coakes and Ehrlich, 9th ed., pp. 40-41)

Cumulative occupational effective dose is calculated as follows: ____ × age in years. A 1 mSv B 5 mSv C 10 mSv D 20 mSv

The answer is C. According to the NCRP, the cumulative occupational effective dose limit for radiation must be measured by 10 mSv × age in years. Therefore, for a 25-year-old technologist, the cumulative occupational effective dose limit would be 250 mSv. The other answer choices are incorrect as per the previous explanation (A, B, and D). (Bushong, 11th ed., p. 603)

An increase in X-ray beam filtration has the following effect(s) 1. Increase in patient skin dose 2. Increase in beam intensity 3. Increase in average energy of the beam A 1 only B 1 and 2 only C 3 only D 1, 2, and 3

The answer is C. Aluminum filtration is used todecreasepatient skin dose by absorbing low-energy photons. X-ray beam intensitydecreasesbecause there are fewer photons. As the filtration of an x-ray beam is increased, the overall average energy of the resulting beam is greater due to the low-energy photons that have been removed (C). The other answer choices are incorrect as per the previous explanation (A, B, and D). (Sherer, Visconti, Ritenour, and Haynes, 8thed., p. 208)

To monitor radiation dose, pocket dosimeters utilize chambers filled with which of the following? A Lithium fluoride crystals B Light C Air D Processing chemicals

The answer is C. Each type of radiation monitoring device requires a specific method to determine the dose received. With pocket dosimeters, the ionization chamber is filled with air and charged before use (C). The exposure to ionizing radiation (if it occurs) causes the charge in the pocket dosimeter to decrease, thereby giving evidence of exposure. With thermoluminescence dosimeters, the lithium fluoride crystals absorb the energy from the x-rays and store the electrons until the badge is exposed to heat. The light that the crystals give off is related to exposure received (B). Therefore, pocket dosimeters utilize chambers filled with charged air to monitor dose received, while TLD dosimeters rely on the use of lithium fluoride crystals (D).(Bushong, 11thed., pp. 559-600)

Which of the following methods is used to focus the electrons on the anode target of the x-ray tube? A Emitting them separately B Focusing them through a narrow port C Electrostatic repulsion D Using a lens system

The answer is C. Electrons emitted from the x-ray tube cathode must be directed to the anode's target. Because negatively-charged electrons repel each other, the beam of electrons traveling from the cathode filament of the x-ray tube tend to spread out. To correct this, a focusing cup surrounding the cathode filament made of molybdenum or nickel has a negative charge placed on it. This creates an electrostatic repulsion on the electron cloud, formed at the cathode filament and thereby narrowing and constricting the electron beam (C). Because electrons are formed in a cloud in an area of the cathode filament they cannot be confined or restricted in any other manner. The other answer choices are incorrect as per the previous explanation (A, B, and D). (Carroll, 3rd ed., p. 149)

Benefits of filtration on the primary beam include 1. Increased penetrability 2. Increased quality 3. Decreased HVL A 1 only B 3 only C 1 and 2 only D 2 and 3 only

The answer is C. Filtration acts to attenuate low energy waves that would most often not reach the image receptor. In doing so, the filtration increases beam penetrability (1), image quality (2), and HVL (A, B, C, and D). Additionally, by eliminating the low energy waves, filtration reduces unnecessary patient dose. (Bushong, 11th ed., pp. 141-144)

According to NCRP Report No.116, the annual occupational equivalent dose limit to localized areas of the hands is A 50 mSv B 15 mSv C 500 mSv D 150 mSv

The answer is C. Ionizing radiation exposure to radiation workers must be limited to minimize the risk of harmful biologic effects. During some procedures, especially interventional fluoroscopy procedures (e.g., drainage, biopsy, alteration of vascular occlusions or malformations), the hands of the operator are oftentimes either in, or close to, the x-ray beam or close to the scattered x-rays emanating from the patient. Therefore, it is important to measure the amount of dose delivered to the hands to ensure that the annual equivalent dose of500 mSvis not exceeded. This is especially important when operating under high level control (HLC) fluoroscopy conditions. For this reason, an extremity dosimeter, which is most commonly a thermoluminescent dosimeter (TLD) ring, should be worn as a second monitor when performing procedures that require the hands to be near the x-ray beam or patient (C). Foroccupational workers, an annualwhole-bodyeffective dose limit of50 mSvis recommended (A). Fornon-occupationalworkers, or the public, an annual equivalent dose limit of15 mSvto thelenses of the eyesis recommended (B). Foroccupational workers, an annual equivalent dose limit of150 mSvto thelenses of the eyesis recommended (D). (Statkiewicz, Visconti, Ritenour & Haynes, 8th ed., pp. 77, 225)

According to the Bohr model, a neutral atom of oxygen (Z = 8), will have how many valence electrons? A 3 B 4 C 6 D 8

The answer is C. The Bohr model utilizes two rules, primarily, to place electrons in orbital shells: 1. The maximum per shell is equal to 2n2, where n is the shell number (K =1, L =2, etc.) 2. The valence shell cannot hold more than 8 (the octet rule). We have 8 electrons to place, which we deduce from the value given for the atomic number, 8, and the use of the phrase "neutral atom," meaning that the charges are balanced. Therefore, the number of electrons is equal to the number of protons. Applying the rules above, the K shell is filled with two electrons, and the L shell has ample capacity for the remaining 6. The valence shell is always the outer shell, and since we have no electrons left to place, it is the L shell, with 6 electrons (C). The other choices indicate that the student is not sufficiently familiar with the rules utilized for the Bohr model and should review (A, B, and D). The Bohr model is foundational to understanding both the means of x-ray production and x-ray interactions with matter. (Bushong, 11th ed., pp. 29-31)

Half value layer (HVL) is defined as A the thickness of a material needed to reduce the x-ray quantity to one-fourth based on the inverse square law B the thickness of a wedge filter needed to provide a uniform exposure to the anatomical area of interest C the thickness of a material needed to reduce the x-ray quantity to one-half D the amount of filtration needed to eliminate the low-energy x-rays from the primary beam before they expose the patient's skin

The answer is C. The half-value layer (HVL) is defined as thethickness of an absorbing material needed to reduce the x-ray quantity (intensity) in the beam to one-halfits original amount. Beams with higher energy x-rays (produced by higher kVp settings) require thicker material to reduce the beam's x-ray quantity to one-half (C). The amount of material needed to reduce the x-ray quantity to one-fourth based on the inverse square law is not the definition of the HVL (A). The thickness of a wedge filter is designed to provide uniform exposure to an anatomical part that is of unequal thickness. The thinner portion of the filter is placed over the thickest portion of the anatomy, whereas the thicker portion is placed over the thinnest portion of the anatomy (B). The inherent and added filtration of the x-ray tube and collimator combined filter out the lower-energy x-rays in the primary beam, thereby reducing the patient's entrance skin dose (ESD) (D).(Bushong, 11th ed., p. 139)

All women of childbearing age should be assessed for pregnancy before an examination. Risks to the fetus are 1. Minimal during first 2 weeks of pregnancy 2. Proportional to time and dose 3. Highest during the 2nd to 10th weeks A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is C. The risk that radiation can cause to a fetus is proportional to time and dose (2); higher doses at an increased frequency pose a greater risk than lower doses and infrequent rates. The risks are highest during the 2ndthrough 10thweeks of pregnancy (3), which are the most crucial for major organogenesis (C). This explains why radiation exposure during early pregnancy is more concerning than during the third trimester (1), when most organ formation has already been under way (A, B, and D). Nevertheless, there is always a risk. It is always important to minimize exposure to pregnant patients.(Bushong, 11thed., p. 592)

During x-ray production at the target anode, which of the following interactions yields a photon of discrete energy? A Compton production B Bremsstrahlung production C Characteristic production D Pair production

The answer is C. There are two phenomena involving the interaction of a high-speed electron with tungsten atoms at the anode that produce x-ray photons: bremsstrahlung and characteristic cascade (B). Only characteristic production yields a photon of approximately 69.5 keV, a single (discrete) energy (C). This is because of the law of conservation of energy, which says that the energy of the resultant photon has to come from somewhere; the k-shell binding energy of tungsten is 69.5 keV. Bremsstrahlung photons get their energy from the loss of kinetic energy by the high-speed electron as it slows. Some of the electrons slow down a lot, and others not as much, creating a bell curve distribution of energies. Pair production and Compton effect are both interactions of x-rays with matter, not production reactions (A and D).(Bushong, 11th ed., p. 128)

When an individual will receive more than 1/10 of the radiation dose limit, which steps must be taken? 1. They are issued a radiation dosimeter 2. They must wear a lead apron during work hours to reduce dose 3. Their annual dose will be monitored A 1 only B 2 and 3 only C 1 and 3 only D 1, 2, and 3

The answer is C. When an individual's dose will exceed 1/10 of the limit, it is necessary for them to wear a radiation dosimeter to monitor their dose (1) (A). The dosimeter will provide an annual report (3), and most companies require new badges to be issued every month or once a quarter (C). A lead apron (2) would only be necessary if the personnel would not be stationed behind a protective barrier during exposure times, most often with use of fluoroscopy and operating room rotations (B and D). (Bushong, 11thed., pp. 604-606)

Muscle tissue absorbs more radiation than fat tissue because muscle tissue has which of the following? A Vastly higher average atomic number B Vastly lower average atomic number C Higher tissue density (mass/volume) D Lower tissue density (mass/volume)

The answer is C. When determining the behavior of ionizing radiation and matter, including various tissue types, the physical properties of the target tissue influence the outcome measurably. Two such physical characteristics are average atomic number and density. For fat tissue, the average atomic number is around 6.3; for muscle, it's 7.4. While muscle is slightly higher, there is not a vast difference (A and B). However, when we consider tissue mass density, muscle has a density of 1,000 kg/m3, whereas fat tissue density is only 910 kg/m3. This difference means there are more target atoms in muscle than fat for equal volumes of tissue (C and D). This is the deciding factor that makes muscle absorb more radiation. (Bushong, 11thed., p. 150-156)

The frequency of an x-ray photon is 1. Directly proportional to its wavelength 2. Directly proportional to its energy 3. Inversely proportional to its wavelength 4. Inversely proportional to its energy A 3 only B 1 and 4 C 2 and 3 D 2 and 4

The answer is C. X-ray photons are part of a larger phenomenon called electromagnetic radiation. EMR exhibits some predictable properties: it travels as waves of constant speed, given by the wave equation c = λf, and each photon possesses energy directly proportional to its frequency via the Planck equation, E = hf. Once we understand c and h are constants, both relationships become clear. If c is always 3 × 108 m/s, then frequency should decrease as wavelength increases, and vice versa. If frequency decreases, the energy of the photon will also decrease (C and D). A clever student will then be able to reason the relationship between wavelength and energy: as wavelength increases, energy will decrease (A and B). How does this apply practically? Most students have operated portable radios without getting erythema. That's because radio waves have low energy. Ultraviolet waves and x-rays have much shorter wavelengths and much higher energies, sufficient to cause damage to skin and other tissues.(Bushong 11th ed., pp. 48-56)

All diagnostic x-ray photons produced in the x-ray tube have which of the following characteristics? 1. Monoenergetic 2. Ionizing 3. Travel at the speed of light A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is C. X-rays are invisible electromagnetic waves (photons) having no mass or charge, andtravel at the speed of light(3 x 108m/sec or 186,000 miles/sec) through space. They are capable ofionizingatoms with which they interact, meaning they can bombard the atom's orbital electrons and eject them from their orbits, leaving the otherwise balanced atom positively ionized (C). X-raysvary in their energydepending on the random interactions of projectile filament electrons with the anode target, the applied kVp, and generator type being used during the exposure. Therefore, the primary x-ray beam produced is considered a heterogenous, polyenergetic, or multienergetic beam. Inherent and added filtration absorbs the lower-energy x-rays to reduce the patient's entrance skin dose (ESD) (A, B, and D). (Bushong, 11thed., p.42).

For primary protective barriers, when the proper thickness is used, concrete or brick may be used to replace lead. Approximately how many inches of brick or concrete are needed to provide the same protection as 1/16 inch of lead? A 2 inches B 3 inches C 4 inches D 5 inches

The answer is C. Masonry materials such as concrete and brick can be good substitutes for lead in primary barriers. In order to provide enough protection, there is a conversion chart to determine the thickness of masonry material needed to replace lead. The conversions are as follows: 1/64 inch of lead is equivalent to 1 3/8 inches of concrete; 1/32 inch of lead is equivalent to 1 7/8 inches of concrete; 3/64 inch of lead is equivalent to 2 7/8 inches of concrete; 1/16 inch of lead is equivalent to 3 3/4 inches of concrete (C). The other answer choices are incorrect as per the previous explanation (A, B, and D). (Bushong, 11th ed., p. 556)

Which of the following techniques can reduce scattered radiation to the IR and improve image quality?Larger field sizeBody part compressionGrid usage A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is C. Scatter is the result of x-ray photon interactions with matter. Many factors contribute to scatter production: part thickness, field size, technique selection, and grid usage can all impact image quality as a result of scatter reduction. With part thickness, using compression (2) or methods that can decrease body part thickness, such as imaging a recumbent lumbar spine PA instead of AP, can help reduce scatter. Grids (3) also help to absorb scattered radiation and keep it from reaching the IP, therefore helping to produce higher quality images. Lastly, as the field size grows, scatter production increases; therefore, reducing the field size would decrease scatter production (1). (Bushong, 11th ed., pp. 186-199)

Which of the following methods could be used to reduce radiation exposure and patient dose?Provide clear patient instructionsApply the 15% ruleUse anatomical compression A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The answer is D. All the methods listed could be used to reduce patient exposure and subsequent dose. By providing clear patient instructions prior to the exam, the patient can better cooperate and avoid motion unsharpness, which would otherwise increase the possibility of a repeat exposure. By applying the 15% rule, the mAs can be halved and the kVp can be increased by 15% to provide the required exposure. By using a more penetrating beam, fewer x-ray photons are absorbed in the patient, thereby reducing dose. By halving the mAs using half of the original exposure time, the possibility of motion unsharpness can also be minimized. Anatomical compression such as that used in mammography reduces tissue thickness. Therefore, fewer x-ray photons are absorbed resulting in a smaller dose. Compression also reduces magnification distortion and image blur and improves spatial resolution and contrast resolution (D). (Statkiewicz, Visconti, Ritenour & Haynes, 8th ed., pp. 234-235; Bushong, 11th ed., pp. 131, 250, 293, 382-383).

Electrons from the cathode filament that collide with K-shell electrons within the Tungsten target yield which of the following? A Bremsstrahlung rays with an energy of 69.5 keV B Characteristic rays with a heterogeneous spectrum of energies C Bremsstrahlung rays with a heterogeneous spectrum of energies D Characteristic rays with a discrete emission of energy

The answer is D. Atomic interactions at the anode target that result in the production of x-radiation are of two types, bremsstrahlung and characteristic. Bremsstrahlung production occurs when an inbound electron decelerates rapidly in a tight arc around the tungsten nucleus, and produces photons with a broad spectrum of energies, often described as heterogeneous or polychromatic. Characteristic emission occurs when an outer shell electron in the tungsten atom cascades down to fill the vacancy created as described in the question. As the electron descends to this lower energy state, it sheds its energy as a photon with equivalent energy to the binding energy of the shell it fills. In the case of tungsten, that energy is a discrete (specific) value, 69.5 keV (D). The other answer choices are incorrect as per the previous explanation (A, B, and C). (Bushong, 11thed., p. 125-126)

Differential absorption of the x-ray beam is dependent on 1. Atomic number of the irradiated tissues 2. Collimation 3. Beam quality 4. Part density A 1 and 4 only B 2 and 3 only C 2 and 4 only D 1, 3, and 4 only

The answer is D. Differential absorption refers to variation in x-ray photon absorption and is the determining factor for image contrast. Factors that influence differential absorption include the atomic number of the irradiated tissue (1), the quality of the x-ray beam (3), and the density of the part being imaged (4). Higher atomic numbers translate to increased photon absorption, such as the high atomic number of bones compared to adipose tissue. Additionally, higher energy waves pass through more tissue than low energy waves, and lead to less absorption of the beam. Part thickness, or density, is also critical for absorption since thicker body parts absorb more of the beam. While collimation can reduce the size of the x-ray field, it is not a determining factor of differential absorption.(Bushong, 11th ed., pp. 158-159)

Inherent filtration can be composed of 1. Glass 2. Metal 3. Beryllium A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The answer is D. Inherent filtration is the filtration included in the x-ray tube that helps filter out low-energy waves from the x-ray beam. Inherent filtration can be composed of various types of glass (1), metal (2), and special materials such as beryllium (3) utilized in mammography tube windows (A, B, C, and D). In a general x-ray, the inherent filtration adds roughly 0.5 mm Al equivalent.(Bushong, 11th ed., pp. 141-142)

What type of filtration is the x-ray tube glass envelope and its insulating oil? A Added B Oil C Total D Inherent

The answer is D. Inherent filtrationis the reduction in radiation energy by materials that make up the x-ray tube and tube housing (D). This includes the oil or glass that make up the x-ray tube (B). The glass envelope window will equal about 0.5 mm Al.Added filtrationcomes from the collimator that is built onto the x-ray tube (A).Total filtrationis all of the filtration combined that makes up the entire assembly (C). Filtration refers to the use of a material, usually aluminum (AL) or aluminum equivalent, to absorb x-ray photons from the x-ray beam. This filtration may be in the form of an inherent, added, or compensating filter.(Johnston & Fauber, 2nded., p. 65)

Radiation doses of 5 Gy to the ovaries A have no effect on fertility B delay menstruation C cause temporary infertility D cause sterility

The answer is D. Irradiation to the ovaries poses a risk to fertility and genetic mutations. At 5 Gy, the dose is high enough for the potential of sterility (D). Doses at 100 mGy could cause a delay in menstruation, while doses of 2 Gy can cause temporary infertility (A, B, and C). Smaller doses of 250‒500 mGy can cause damage to oocytes, but some oocytes survive and repair genetic damage before maturation.(Bushong, 11thed., p. 514)

Which of the following radiographic factors will affect x-ray quantity?Kilovoltage-peak (kVp)Source-image-distance (SID)Beam filtration A 1 only B 2 only C 3 only D 1, 2, and 3

The answer is D. Kilovoltage-peak (kVp) will affect x-ray beam quantity and energy (quality) (1). The increase in x-ray quantity is caused by the increased number of electron interactions at the tube target by higher energy electrons from the cathode filament; this will also cause an increase in x-ray energy. Source-image receptor-distance (SID) changes will also affect x-ray quantity in accordance with the inverse-square law (2). As SID increases, x-ray quantity decreases to the image receptor (IR). As SID decreases, x-ray quantity to the IR increases. Beam filtration, if increased, will decrease x-ray quantity due to the increased absorption of x-ray photons in the beam filter (3). (Carlton/Adler/Balac, 6th ed., p. 159)

In thermoluminescence dosimeters (TLD), lithium fluoride (LiF) is the most commonly used material due to its atomic number and attenuation of x-rays. With an atomic number of 8.2, which of the following is lithium fluoride most similar to? A Bone B Fat C Calcium D Soft tissue

The answer is D. Lithium fluoride has an atomic number of 8.2, similar to soft tissue, which has an atomic number of 7.4 (D). Bone has an atomic number of 13.8, which absorbs more radiation than soft tissue and therefore is easier to image with x-rays (A). With an atomic number of 6.3, fat absorbs too few x-rays to be used as a comparable material (B). Lastly, calcium has an atomic number of 20 (C). (Bushong, 11th ed., pp. 156-157, 564-565)

In order to reduce exposure to the lens, the reproductive organs, and the breasts, it is helpful to perform these examinations in which of the following projection? A AP B RAO C Left lateral D PA

The answer is D. Posteroanterior (PA) projections are preferred for radiation safety (D). PA offers less radiation to the eyes, gonads, and thyroid when compared to anteroposterior (AP). AP projection would get the most radiation to the thyroid because the patient would be facing the x-ray tube (A). In an RAO projection the patient will be oblique, and the lens, reproductive organs, and breasts would still get some radiation (B). The left lateral would also not be a good projection when protecting from radiation because the patient would be exposed to radiation from the right side (C).(Sherer, Visconti, Ritenour, & Haynes, 8thed., p. 236)

During C-arm fluoroscopic procedures, what should the patient-to-image intensifier distance be in order to reduce patient entrance dose? A Long B 90° angle C Higher D Short

The answer is D. The patient-to-image intensifier distance should be asshort as possibleto limit patient entrance dose exposure (A and D). The C-arm should be positioned under the patient whenever possible, to reduce the quantity of scattered radiation (B). Occupational radiation exposure increases when the x-ray tube is positioned over the patient (C).(Sherer, Visconti, Ritenour, and Haynes, 8thed., p. 224)

The basic (cardinal) principle(s) of radiation protection areTimeDistanceShielding A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The answer is D. The radiographer must practice the basic principles of radiation protection to minimize radiation dose to their patients and themselves. For the patient, the radiographer must limit the amount of time the patient is exposed to the x-ray beam. Ensuring the production of diagnostic images using the minimum exposure necessary and avoiding repeat radiographs due to motion unsharpness will reduce patient exposure and dose. Using the proper SID will also minimize patient entrance dose from low-energy x-rays in the primary beam. Proper shielding and beam restriction will also minimize patient dose. By reducing the time spent in the radiographic room during exposures, the radiographer can minimize their dose. The radiographer should also stand as far as possible from the x-ray tube and patient if he or she must be in the room during the exposure. Wearing appropriate lead shields, standing behind the lead wall of the control booth or portable lead barrier, and standing behind the physician who is operating the fluoroscopic equipment reduces the radiographer's dose (D). (Statkiewicz, Visconti, Ritenour, & Haynes, 8th ed., p. 6)

Which of the following influence the average energy of the primary x-ray beam? kVp mAfiltrationbremstrahlung interaction A 1 only B 2 and 3 only C 1, 2, and 4 only D 1, 3, and 4 only

The answer is D. The technical factor that primarily controls x-ray beam energy is kVp. Higher kVp settings increase average beam energy, whereas lower kVp settings decrease average beam energy. Higher energy beams exhibit higher penetrability and therefore decrease patient dose. Beam filtration exists in the form of inherent and added filtration. Inherent filtration includes the x-ray tube window, oil in the x-ray tube housing, and the housing port. Added filtration includes an aluminum plate and the collimator mirror. The plastic cover on the bottom of the collimator can also filter some low-energy x-rays. Filtration removes the lower-energy x-rays that would only be absorbed within the patient and not contribute to the diagnostic image. By removing these low-energy x-rays, particularly patient skin dose is decreased. Bremstrahlung (brems) interaction takes place in the x-ray tube target. Projectile filament electrons randomly interact with the anode target tungsten atoms. The closer the projectile electrons pass by the nuclei of the tungsten atoms, the higher the energy will be in the brems x-rays produced because of the strong electrostatic attraction of the negatively charged electrons passing by the positively charged tungsten atom nuclei (D). The milliampere (mA) setting is one of two factors that primarily controls x-ray quantity. The other factor is exposure time. Higher mA settings increase the filament current, thereby increasing the number of filament electrons emitted through the process of thermionic emission. Therefore, higher mA settings increase the number of projectile electrons available to potentially produce x-rays in the anode target. Lower mA settings reduce the number of x-rays that can be potentially produced (A, B, C). (Bushong, 11th. ed., pp. 126-127; 141-142)

A metal added to the cathode filament of an x-ray tube to increase thermionic emission and extend filament life is A aluminum. B molybdenum. C rhenium. D thorium.

The answer is D. Thorium is a radioactive metallic element that is added to the tungsten filament to increase thermionic emission (boiling off of electrons) and extend filament life (D). Aluminum is used in beam filters, and molybdenum and rhenium are often added to the anode target of an x-ray tube (A, B, and C). (Johnston and Fauber, 2nd ed., p. 55)

During a diagnostic x-ray examination, who is the greatest source of scattered radiation? A Radiologist B Radiographer C Nurse D Patient

The answer is D. While performing any diagnostic x-ray examination, the patient becomes the greatest source of scattered radiation because of the Compton interaction process (D). Staff and ancillary personnel are not sources of scattered radiation (A, B, and C).(Sherer, Visconti, Ritenour, and Haynes, 8thed., p. 276)

The purpose of the x-ray tube housing is to 1. Shield radiation workers from leakage radiation 2. Support and protect the x-ray tube 3. Contain cooling and insulating oil A 1 only B 2 only C 1 and 3 only D 1, 2, and 3

The answer is D. X-rays are emitted isotropically from the x-ray tube anode. Only those x-rays directed toward the patient are of diagnostic use. The metal x-ray tube housing, lined with lead, absorbs much of the non-diagnostic radiation, calledleakage radiation, therebyshielding radiation workerswhen they are operating the equipment in the procedure room (A). Properly designed diagnostic x-ray tube housings must limit the level of leakage radiation to no more than 1 mGya/hr. at one meter. The tube housing alsosupportsthe x-ray tube andprotectsit from mechanical shock (B). Finally, the housing containscoolingandinsulating oilsurrounding the x-ray tube (C). The oil provides a thermal cushion to dissipate the heat generated when the tube is activated and serves as an insulator, protecting the operator from electrical shock. The radiographer, however, must never touch the tube housing during operation (D).(Bushong, 11th ed., p. 106)

Cone cutting, or cutoff from cylinder or cone extensions for beam restriction, is caused by misalignment of theTubeCone or cylinderImage receptor A 1 and 3 only B 2 and 3 only C 1 and 2 only D 1, 2, and 3

The answer is D. Cone cutting is one of the disadvantages of cone and cylinder use. Even though the cones and cylinders provide increased clarity and reduced patient exposure when used properly, if the alignment of the tube (1), cone or cylinder (2), and image receptor (3) is incorrect, cutoff will occur. This will cause a portion of the necessary anatomy to remain unexposed, which requires repeat imaging. Accurate positioning and centering helps prevent these errors, enhances image quality, and reduces patient dose. (Bushong, 11th ed., pp. 192-193)

All of the following statements regarding dual x-ray absorptiometry are true, except1.radiation dose is considerable.2.two x-ray photon energies are used.3.photon attenuation by bone is calculated. A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The correct answer is: (A) Dual x-ray absorptiometry (DXA) imaging is used to evaluate bone mineral density (BMD). 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 more dense and attenuates x-ray photons more readily, their attenuation is calculated to represent the degree of bone density. 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.

Diagnostic x-radiation, compared to fast neutrons and alpha particles, is generally considered to be A low energy, low LET B low energy, high LET C high energy, low LET D high energy, high LET

The correct answer is: (A) 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, 11th ed., p. 485)

The exposure rate to a body 4 ft from a source of radiation is 140 mGy/h. What distance from the source would best serve to decrease the exposure to 53 mGy/h? A 5 ft B 7 ft C 9 ft D 14 ft

The correct answer is: (B) The relationship between x-ray intensity and distance from the source is expressed by the inverse-square law of radiation. The formula is Substituting known values: Thus, x = 6.5 ft (necessary to decrease the exposure to 53 mGy/h). Note that in order for the exposure rate to decrease, the distance from the source of radiation must increase.

What is the annual dose-equivalent limit for the skin and hands of an occupationally exposed individual? A 30 mSv B 150 mSv C 500 mSv D 650 mSv

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

Protective devices such as lead aprons function to protect the occupationally exposed fromscattered radiationthe primary beamremnant radiation A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The correct answer is: AProtective apparel functions to protect the occupationally exposed person from scattered radiation only. Lead aprons and lead gloves do not protect from the primary beam. No one in the radiographic room except the patient must ever be exposed to the primary beam. The occupationally exposed and those (family and friends) who might assist a patient during an examination must wear protective apparel and keep out of the way of the primary beam. (Bushong, 9th ed., pp. 624-625)

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 correct answer is: BGonadal 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. (Bushong, 8th ed., pp. 598-599)

Using direct and indirect active matrix flat panel imaging systems, increasing exposure time with no other changes will: A Increase brightness of the displayed image. B Decrease displayed image contrast. C Decrease image receptor exposure. D Increase patient dose.

Within the realm of digital radiography, technical factors are no longer the controlling factor in the brightness and contrast of the displayed image. The software of the imaging system, particularly automatic rescaling, the look-up table (LUT), and the window width and window level post-processing controls determine the appearance of the image. A longer exposure time will increase mAs, which is directly proportional to patient dose. Image receptor exposure is a complex consideration of several technical factors including kVp, mAs, SID, grid selection, part thickness, and part composition. (Radiography in the Digital Age, Carroll, 3rd edition, p 213 and 240)

For x-ray filtration, inherent filtration provides _______ mm Al equivalent of the required _______ Al equivalent for total filtration. A 1.5/2.0 B 1.0/2.5 C 0.5/2.5 D 0.5/1.0

The answer is C. X-ray tubes are required to meet a 2.5 mm Al equivalent of total filtration. Inherent filtration provides a roughly 0.5 mm Al equivalent, while the collimator provides an additional 1.0 mm Al equivalent. Lastly, a removable aluminum sheet can be inserted between the tube housing and collimator, which can fulfill the remaining 1.0 mm Al equivalent required (C). (Bushong, 11th ed., pp. 240-241)

The Gray is the unit of A absorbed dose. B exposure. C effective dose. D ionization in air.

The Correct Answer is: A The SI unit air kerma is used to express kinetic energy released in matter, the unit of exposure and ionization in air. The SI unit used to describe absorbed dose is Gray (Gyt). The SI unit of measurement to describe dose to biologic material is the Sievert (Sv). Various organs and tissue types require assigned specific weighting factors in order to describe the effective dose. Absorbed dose and energy deposited is strongly related to chemical change and biologic damage. The SI unit of measurement to describe effective dose to biologic material is the Sievert (Sv). The Sievert is the unit of occupational radiation exposure.

The automatic exposure device that is located immediately under the x-ray table is the A ionization chamber B scintillation camera C photomultiplier D photocathode

The Correct Answer is: AAutomatic exposure control (AEC) devices are used to produce consistent and comparable radiographic results. The most commonly used AEC is the ionization chamber, located just beneath the tabletop above the IR. The part to be examined is centered on it (the sensor) and radiographed. When a predetermined quantity of ionization has occurred (equal to the correct receptor exposure), the exposure terminates automatically. The old type of AEC, the phototimer/photomultiplier, a small fluorescent screen is positioned beneath the IR and fluoresces when remnant radiation emerging from the patient exits the IR. Once a predetermined amount of fluorescent light is "seen" by the photocell sensor, the exposure is terminated. A scintillation camera is used in nuclear medicine. A photocathode is an integral part of the image intensification system.

Which of the following groups of exposure factors would deliver the lowest patient dose? A 2.5 mAs, 100 kVp B 5 mAs, 90 kVp C 10 mAs, 80 kVp D 20 mAs, 70 kVp

The Correct Answer is: ABecause 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. (Bushong, 8th ed., p. 162)

Which of the following are considered most radiosensitive? A Lymphocytes B Ova C Neurons D Myocytes

The Correct Answer is: AMature white blood cells (lymphocytes) are considered the most radiosensitive cells. Ova (female germ cells) are very radiosensitive, but not to the same degree as lymphocytes. Myocytes (muscle cells) and especially neurons (nerve cells) are actually radioresistant. (Bushong, 10th ed., p. 512)

Under what circumstances might a radiographer be required to wear two dosimeters?During pregnancyWhile performing vascular proceduresWhile performing mobile radiography A 1 and 2 only B 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: ARadiographers usually are required to wear one dosimeter, positioned at their collar and worn outside a lead apron. Special circumstances, however, warrant the use of a second monitor. During pregnancy, a second "baby monitor" is worn at the abdomen, under any lead apron. During special vascular procedures, the dose to the radiographer can increase significantly. This is so because the leaded protective curtain is often absent from the fluorotower and because of the extensive use of cineradiography. As a result, the radiographer's upper extremities can receive a greater exposure (e.g., when assisting during catheter introduction), and a ring or bracelet badge is often recommended. A second dosimeter is not required when performing mobile radiography. (Bushong, 8th ed., p. 525)

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 Correct Answer is: ASoft 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)

If a quantity of radiation is delivered to a body in a single dose, its effect A will be greater than if it were delivered in a number of doses over a long period of time. B will be less than if it were delivered a number of doses over a long period of time. C has no relation to how it is delivered in time. D is solely dependent on the radiation quality.

The Correct Answer is: AThe effects of a quantity of radiation delivered to a body are dependent on several factors: the amount of radiation received, the size of the irradiated area, and how the radiation is delivered in time. If the radiation is delivered in portions over a period of time, it is said to be fractionated and has a less harmful effect than if the radiation were delivered all at once. With fractionation, cells have an opportunity to repair, and so some recovery occurs between doses. (Bushong, p 496)

Which of the following defines the gonadal dose that, if received by every member of the population, would be expected to produce the same total genetic effect on that population as the actual doses received by each of the individuals? A Genetically significant dose B Somatically significant dose C Maximum permissible dose D Lethal dose

The Correct Answer is: AThe genetically significant dose (GSD) illustrates that large exposures to a few people are cause for little concern when diluted by the total population. On the other hand, we all share the burden of that radiation received by the total population, especially as the use of medical radiation increases, so each individual's share of the total exposure increases. (Selman, 9th ed., pp. 386-387)

If the exposure rate at 3 ft from the fluoroscopic table is 1.5 mGy/h, what will be the exposure rate for 30 minutes at a distance of 5 ft from the table? A 0.27 mGy B 0.75 mGy C 0.54 mGy D 1.25 mGy

The Correct Answer is: AThe intensity/exposure rate of radiation at a given distance from a point source is inversely proportional to the square of the distance. This is the inverse-square law of radiation, and it is expressed in the following equation: Substituting known values: 1.5 mGy / x = 25 / 9 25 x = 13.5 x = 0.54 mGy in 60 min ; 0.27 mGy in 30 min.

Occupational exposure received by the radiographer is mostly from A Compton scatter B the photoelectric effect C coherent scatter D pair production

The Correct Answer is: AThe photoelectric effect and Compton scattering are the two predominant interactions between x-ray photons and matter in diagnostic radiology. In the photoelectric effect, the low-energy-incident photon is absorbed by the tissues being radiographed. In Compton scatter, the high-energy-incident photon uses only part of its energy to eject an outer-shell electron. It retains much of its original energy in the form of a scattered x-ray. Radiologic personnel can be exposed to that high-energy scattered radiation, especially in fluoroscopy and mobile radiography. Lead aprons are used to protect us from exposure to scattered radiation during these procedures. (Bushong, 8th ed., p. 174)

Which of the following is (are) used to account for the differences in tissue characteristics when determining effective dose to biologic material?Tissue weighting factors (W t )Radiation weighting factors (W r )Absorbed dose A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: AThe tissue weighting factor (W t ) represents the relative tissue radiosensitivity of irradiated material (e.g., muscle vs. intestinal epithelium vs. bone). The radiation weighting factor (W r ) 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 W r of different ionizing radiations depends on the LET of that particular radiation. The following formula is used to determine effective dose E: (Bushong, 8th ed., p. 556)

If an exposure dose of 0.5 mGy/hr is delivered from a distance of 4 ft, what would be the dose delivered after 20 minutes at a distance of 3 ft from the source? A 0.15 mGy B 0.29 mGy C 0.88 mGy D 1.12 mGy

The Correct Answer is: B The relationship between x-ray intensity and distance from the source is expressed by the inverse-square law of radiation. The formula is Substituting known values: 0.5 mGy / x = 9 / 16 9 x = 8 x = 0.88 mGy in 60 min ; 0.29 mGy in 20 min Distance has a profound effect on dose received and, therefore, is one of the cardinal rules of radiation protection. As distance from the source increases, dose received decreases. (Bushong, 8th ed., pp. 68-70)

If the exposure rate at 3.0 m from a source of radiation is 0.8 mGy/hr, what distance would be required to decrease the exposure rate to 0.2 mGy/hr? A 3 m B 6 m C 7.2 m D 36 m

The Correct Answer is: B The relationship between x-ray intensity and distance from the source is expressed in the inverse-square law of radiation. The formula is Substituting known values: 0.8 mGy / 0.2 mGy = x2 / 90.2 x2 = 7.2 x2 = 36 x = 6 m distance to reduce intensity to 0.2 mGy/hr Distance has a profound effect on dose received and, therefore, is one of the cardinal factors considered in radiation protection. As distance from the source increases, dose received decreases. Doubling the distance will quarter the exposure rate.

Which of the following types of adult tissues is (are) relatively insensitive to radiation exposure?1.Muscle tissue2.Nerve tissue3.Epithelial tissue A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BBecause muscle and nerve tissues perform specific functions and do not divide, they are relatively insensitive to radiation exposure. Epithelial cells cover the outer surface of the body; they also line body cavities and tubes and passageways leading to the exterior. They contain very little intercellular substance and are devoid of blood vessels. Because epithelial cells constantly regenerate through mitosis, they are very radiosensitive. (Dowd and Tilson, 2nd ed., pp. 121-122)

The variation in photon distribution between the anode and cathode ends of the x-ray tube is known as A the line focus principle. B the anode heel effect. C the inverse square law. D Bohr's theory.

The Correct Answer is: BBecause the focal spot (track) of an x-ray tube is along the anode's beveled edge, photons produced at the target are able to diverge toward the cathode end of the tube, but are absorbed by the "heel" of the anode at the opposite anode end of the tube. This results in a greater number of x-ray photons distributed toward the cathode end and is known as the anode heel effect. The line focus principle is a geometric principle illustrating that the effective focal spot is always smaller than the actual focal spot. The inverse square law of radiation deals with the relationship between distance and radiation intensity. Bohr's theory refers to an atom's resemblance to the solar system. (Selman, p 253)

For radiographic examinations of the skull, it is generally preferred that the skull be examined in the A AP projection B PA projection C recumbent position D supine position

The Correct Answer is: BBecause the primary x-ray beam has a poly-energetic (heterogeneous) nature, the entrance or skin dose is significantly greater than the exit dose. This principle may be employed in radiation protection by placing particularly radiosensitive organs away from the primary beam. To place the gonads further from the primary beam and reduce gonadal dose, abdominal radiography should be performed in the posteroanterior (PA) position whenever possible. Dose to the lens is decreased significantly when skull radiographs are performed in the PA position.

Which of the following is (are) associated with Compton scattering?1.High-energy incident photons2.Outer-shell electrons3.Characteristic radiation A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BCompton scattering occurs when a relatively high-energy incident photon uses part of its energy to eject an outer-shell electron, and in doing so changes its direction (is scattered). The energy retained by the scattered photon depends on the angle formed by the ejected electron and the scattered photon: The greater the angle of deflection, the less the retained energy. Compton scatter is very energetic scatter. It emerges from the patient and is responsible for scattered radiation reaching the image in the form of fog. Characteristic radiation is associated with the photoelectric effect. (Bushong, pp 174-176)

How many half-value layers will it take to reduce an x-ray beam whose intensity is 78 Gy to an intensity of less than 10 Gy? A 2 B 3 C 4 D 8

The Correct Answer is: BHVL may be used to express the quality of an x-ray beam. The HVL of a particular beam is that thickness of an absorber that will decrease the intensity of the beam to one-half its original value. If the original intensity of the beam was 78 Gy, the first HVL will reduce the intensity to 39 Gy, the second HVL will reduce it to 19.5 Gy, and the third HVL will reduce it to 9.75 Gy. (Bushong, 11th ed., p. 40)

It is necessary to question a female patient of childbearing age regarding herdate of last menstrual periodpossibility of being pregnantage at her first pregnancy A 1 only B 1 and 2 only C 1 and 3 only D 2 and 3 only

The Correct Answer is: BIt is our ethical responsibility to minimize radiation exposure to ourselves and our patients, particularly during early pregnancy. One way to do this is to inquire about the possibility of our female patients being pregnant or for the date of their last menstrual period (to determine the possibility of irradiating a newly fertilized ovum). The safest time for a woman of childbearing age to have elective radiographic examinations is during the first 10 days following the onset of menstruation. (Thompson et al., p. 487)

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 Correct Answer is: BLead 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)

What is the approximate entrance skin exposure (ESE) for the average anteroposterior (AP) cervical spine radiograph? A 0.1 mGy B 1.0 mGy C 1.8 mGy D 2.5 mGy

The Correct Answer is: BPatients will occasionally 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 can also be referred to the medical physicist for more detailed information. The average PA chest delivers an ESE of about 0.1 mGy. The average AP supine lumbar spine delivers an ESE of about 2.5 - 3 mGy; the average AP thoracic spine about 1.8 mGy; the average AP cervical spine about 1.0 mGy. The average AP pelvis delivers about 1.5 mGy. The average lateral skull about 0.7 mGy, the shoulder about 0.90 mGy, and the extremity about 0.5 mGy. (Bushong 11th ed p569, 595)

Types of secondary radiation include 1. scattered. 2. leakage. 3. primary. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BRadiographers must avoid unnecessary radiation exposure to themselves and strive to keep patient dose to an absolute minimum. The sources of radiation exposure to the radiographer are the primary beam and secondary radiation (scatter and leakage). Radiographers must never be exposed to the primary, or useful, x-ray beam. The patient is the principal source of scatter radiation; protection guidelines address secondary radiation exposure. (Bushong, 8th ed, p 618)

epilation B radiolysis C proliferation D repopulation

The Correct Answer is: BRadiolysis has to do with the irradiation of water molecules and the formation of free radicals. Free radicals contain enough energy to damage other molecules some distance away. They can migrate to and damage a DNA molecule (indirect hit theory). (Bushong, 8th ed., pp. 506, 507)

Personnel present in the x-ray room during fluoroscopic examinations wear lead aprons to protect them primarily from A photoelectric scatter. B Compton scatter. C pair production. D magnetic fringe field.

The Correct Answer is: BSome radiations are energetic enough to rearrange atoms in materials through which they pass, and they can therefore be hazardous to living tissue. These radiations are called ionizing because they have the energetic potential to break apart electrically neutral atoms, resulting in the production of negative and/or positive ions. The gradual decrease in exposure rate as ionizing radiation passes through tissues is called attenuation. Attenuation is principally attributable to the two major types of interactions that occur between x-ray photons and tissue in the diagnostic x-ray range of energies: Compton scatter and the photoelectric effect. In Compton scatter, a fairly high energy (high kVp) x-ray photon ejects an outer shell electron. Even though the x-ray photon is deflected with somewhat reduced energy (modified scatter), it retains most of its original energy and exits the body as an energetic scattered photon. Therefore, Compton scatter can be a personnel radiation hazard, especially in fluoroscopic procedures. 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. Because the scattered photon exits the body, it does not pose a radiation hazard to the patient. It can, however, contribute to and pose a radiation hazard to personnel (as in fluoroscopic procedures). (Bushong, 8th ed, p 175)

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 Correct Answer is: BThe NCRP recommends a total equivalent dose limit to the embryo/fetus of 5 mSv. This dose limit is the total for the entire gestational period. The dose limit for 1 month during pregnancy is 0.5 mSv. (Bushong, 8th ed., p. 557)

The minimum NCRP requirement for lead-equivalent content in protective aprons is A 0.05 mm Pb. B 0.50 mm Pb. C 0.25 mm Pb. D 1.0 mm Pb.

The Correct Answer is: BThe NCRP requires at least 0.5 mm Pb equivalent in protective aprons. NCRP publishes national guidelines that, although not law, are generally followed by radiologic facilities. According to 21 CFR/federal law lead aprons must contain at least 0.25 (1/4) mm Pb equivalent. Many radiologic facilities routinely use lead aprons containing 0.5 mm Pb . Although heavier, the 0.5 mm Pb aprons attenuate a higher percentage of scattered radiation. (Bushong, 11th ed p. 609)

The effects of radiation on biologic material depend on several factors. If a quantity of radiation is delivered to a body over a long period of time, the effect A will be greater than if it were delivered all at one time B will be less than if it were delivered all at one time C has no relation to how it is delivered in time D depends solely on the radiation quality

The Correct Answer is: BThe effects of a quantity of radiation delivered to a body depend on the amount of radiation received, the size of the irradiated area, and how the radiation is delivered in time. If the radiation is delivered in portions over a period of time, it is said to be fractionated and has a less harmful effect than if it were delivered all at once because cells have an opportunity to repair, and some recovery occurs between doses. (Bushong, 8th ed., p. 496)

The dose-response curve that appears to be valid for genetic and some somatic effects is thelinearnonlinearnonthreshold A 1 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BThe genetic effects of radiation and some somatic effects, such as 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. (Bushong, 9th ed., pp. 516-517)

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 Correct Answer is: BThe 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, 11th ed., pp. 149-150)

Characteristics of x-ray photons includea penetrating effect on all matteran ionizing effect on airtravelling at the speed of sound A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BX-rays are energetic enough to rearrange atoms in materials through which they pass, and they can, therefore, be hazardous to living tissue. X-rays are called ionizing radiation because they have the energetic potential to break apart electrically neutral atoms, resulting in the production of negative and/or positive ions. X-rays are infinitesimal bundles of energy called photons that deposit some of their energy into matter as they travel through it. This deposition of energy and subsequent ionization has the potential to cause chemical and biologic damage. Several of the outstanding properties of x-ray photons are: X-rays are not perceptible by the senses. X-rays travel in straight lines. X-rays travel at the speed of light. X-rays are electrically neutral. X-rays have a penetrating effect on all matter. X-rays have a physiological effect on living tissue. X-rays have an ionizing effect on air. X-rays have a photographic effect on film emulsion. X-rays produce fluorescence in certain phosphors. X-rays cannot be focused. X-rays have a spectrum of energies. X-rays are unaffected by a magnetic field.

The measure of the rate at which energy is transferred from ionizing radiation to soft tissue is termed A the characteristic effect B Compton scatter C linear energy transfer D the photoelectric effect

The Correct Answer is: CAs radiation passes through tissue, different types of ionization processes can take place depending on the photon energy and the type of material being irradiated. The photoelectric effect (whose end products include a characteristic ray) and Compton scatter are the two major interactions that take place in the diagnostic x-ray peak kilovoltage range. The rate at which energy is deposited in (or transferred to) tissue during these interactions is termed linear energy transfer (LET). The greater the LET, the greater is the potential biologic effect. Diagnostic x-ray is considered low-LET radiation. (Bushong 11th ed p 484)

The law of Bergonié and Tribondeau states that cells are more radiosensitive if they arehighly proliferativehighly differentiatedimmature A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: CBergonié and Tribondeau were French scientists who, in 1906, theorized what has now become verified law. Cells are more radiosensitive if they are immature (undifferentiated or stem) cells, if they are highly mitotic (having a high rate of proliferation), and if the irradiated tissue is young. Cells and tissues that are still undergoing development are more radiosensitive than fully developed tissues. (Bushong, 8th ed., p. 495)

Occupational radiation monitoring is required when it is likely that an individual will receive more than what fraction of the annual dose limit? A ½ B ¼ C 1/10 D 1/40

The Correct Answer is: CDifferent types of monitoring devices are available for the occupationally exposed, and anyone who might receive more than one-tenth the annual dose limit must be monitored. 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. (Bushong, 8th ed., p. 593)

For exposure to 1 mGy of each of the following ionizing radiations, which would result in the greatest dose to the individual? A External source of 1-MeV x-rays B External source of diagnostic x-rays C Internal source of alpha particles D External source of beta particles

The Correct Answer is: CElectromagnetic radiations such as x-rays and gamma rays are considered low-LET radiations because they produce fewer ionizations than the highly ionizing particulate radiations such as alpha particles. Alpha particles are large and heavy (two protons and two neutrons), and although they possess a great deal of kinetic energy (approximately 5 MeV), their energy is lost rapidly through multiple ionizations (approximately 40,000 atoms/cm of air). As an external source, alpha particles are almost harmless because they ionize the air very quickly and never reach the individual. As internal sources, however, they ionize tissues and are potentially the most harmful. It may be stated that the alpha particle has one of the highest LETs of all ionizing radiations. (Bushong, 8th ed., p. 496)

Stochastic effects of radiation are those thathave a thresholdmay be described as "all-or-nothing" effectsare late effects A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: CLate effects of radiation can occur in cells 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. Tissue reactions (formerly referred to as Deterministic effects) are those that are unlikely to occur below a particular threshold dose and that increase in severity as the dose increases. (Sherer, 8th ed., p. 191)

What is the minimum requirement for lead aprons, according to 21 CFR? A 0.05 mm Pb B 0.50 mm Pb C 0.25 mm Pb D 1.0 mm Pb

The Correct Answer is: CLead 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 21 CFR). 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, 10th ed., p. 592)

Which of the dose-response curves shown in Figure 3-7 is representative of radiation-induced skin erythema? Dose-response curve A Dose-response curve B Dose-response curve C A A only B A and B only C C only D B and C only

The Correct Answer is: CThe Figure illustrates three dose-response curves. Curve A begins at zero, indicating that there is no safe dose, that is, no threshold. Even one x-ray photon theoretically can cause a response. It is a straight (linear) line, indicating that response is directly related to dose; as dose increases, response increases. Radiation-induced cancer, leukemia, and genetic effects follow a linear nonthreshold dose-response relationship. Curve B is another linear curve (response is directly related to dose), but this one illustrates that a particular dose of radiation must be received before a response will occur. That is, there is a threshold dose; this is called a linear threshold curve. Curve C is another threshold curve, but this curve is nonlinear. It illustrates that once the minimum dose is received, a response occurs slowly initially and then increases sharply as exposure increases. This threshold, nonlinear (sigmoid) dose-response curve, illustrates the effect to skin from exposure to high levels of ionizing radiation. (Bushong, 8th ed., p. 499)

The largest dose to the male gonads is most likely to result from which of the following exposures? A Lateral thoracic spine B Oblique lumbar spine C Cross-table lateral hip D AP axial skull

The Correct Answer is: CThe cross-table lateral hip will bring the primary beam in closest proximity to the male reproductive organs. A grid is required, and gonadal shielding is most likely not possible. Close, accurate collimation is recommended to keep exposure to a minimum, but the cross-table lateral hip will deliver the greatest dose to the male reproductive organs. (Bontrager, p 267)

The annual dose limit for occupationally exposed individuals is valid for A alpha, beta, and x-radiations. B x- and gamma radiations only. C beta, x-, and gamma radiations. D all ionizing radiations.

The Correct Answer is: CThe occupational dose limit is valid for beta, x-, and gamma radiations. Because alpha radiation is so rapidly ionizing, traditional personal monitors will not record alpha radiation. Because alpha particles are capable of penetrating only a few centimeters of air, they are practically harmless as an external source. (Selman, 6th ed., p. 395)

The annual dose limit for occupationally exposed individuals is valid for A alpha, beta, and x-radiations B x- and gamma radiations only C beta, x-, and gamma radiations D all ionizing radiations

The Correct Answer is: CThe occupational dose limit is valid for beta, x-, and gamma radiations. Because alpha radiation is so rapidly ionizing, traditional personnel monitors will not record alpha radiation. However, because alpha particles are capable of penetrating only a few centimeters of air, they are practically harmless as an external source. (Bushong, 8th ed., p. 596)

The exposure rate to a body 4 ft from a source of radiation is 2.5 mGy/h. What distance from the source would be necessary to decrease the exposure to 1 mGy/hr? A 4.6 ft B 5.3 ft C 6.3 ft D 7.2 ft

The Correct Answer is: CThe relationship between x-ray intensity and distance from the source is expressed by the inverse-square law of radiation. The formula is Substituting known values: 2.5 mGy / 1.0 mGy = x2 / 16 1 x2 = 40x = 6.32 ft Thus, x = 6.3 ft (necessary to decrease the exposure to 1.0 mGy/hr). Note that in order for the exposure rate to decrease, the distance from the source of radiation must increase.

All the following have an effect on patient dose except A kilovoltage B milliampere-seconds C focal spot size D inherent filtration

The Correct Answer is: CThe selected milliampere-seconds is directly related to patient dose. That is, if milliampere-seconds are doubled, patient dose is doubled. Similarly, if milliampere-seconds are cut in half, patient dose is cut in half. The selected kilovolts peak is inversely related to patient dose. That is, if the kilovolts peak is increased, patient dose can be decreased because more x-ray photons are transmitted through the patient rather than being absorbed. Inherent filtration is provided by materials that are a permanent part of the tube housing, that is, the glass envelope of the x-ray tube and the oil coolant. Added filtration, usually thin sheets of aluminum, is present to make a total of 2.5 mm Al equivalent for equipment operated above 70 kVp. Filtration is used to decrease patient dose by removing the weak x-rays that have no value but contribute to the skin dose. The effect of focal spot size is principally on radiographic sharpness; it has no effect on patient dose. (Bushong, 8th ed., pp. 137, 138)

Which of the dose-response curves shown in Figure 6-2 illustrate(s) illustrates a linear relationship between dose and response?Curve number 1Curve number 2Curve number 3 A 1 only B 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: CThree dose-response (dose-effect) curves are illustrated, representing the body's response to ionizing radiation exposure. Dose is indicated by the horizontal axis (increasing to the right); response is indicated by the vertical axis (increasing upward). Two of the curves (numbers 1 and 3) are linear; that is, response is directly proportional to dose. Curve 2 is not a straight line and is, therefore, nonlinear. Curves 2 and 3 show that a particular dose (threshold quantity) of radiation is required before any effect will occur; therefore, curve 2 is nonlinear threshold, and curve 3 is linear threshold. Curve 1, however, shows that any dose of radiation (theoretically, even a single x-ray photon; that is, there is no threshold) can result in a particular biologic effect; therefore, it is linear nonthreshold. (Bushong, 10th ed., p. 484)

Which of the following radiologic projections would deliver the greatest ESE? A PA Chest B Lateral Skull C AP Pelvis D AP Lumbar spine

The Correct Answer is: D Patients will occasionally 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 can also be referred to the medical physicist for more detailed information. The average PA chest delivers an ESE of about 0.1 mGy. The average AP supine lumbar spine delivers an ESE of about 2.5 - 3 mGy; the average AP thoracic spine about 1.8 mGy; the average AP cervical spine about 1.0 mGy. The average AP pelvis delivers about 1.5 mGy. The average lateral skull about 0.7 mGy, the shoulder about 0.90 mGy, and the extremity about 0.5 mGy. (Bushong 11th ed p569, 595)

Medical and dental radiation accounts for what percentage of the general public's exposure to human-made radiation? A 10% B 50% C 75% D 90%

The Correct Answer is: D Artificial/human-made sources of radiation include radioactive fallout, industrial radiation, and medical and dental x-rays. The majority of the general public's exposure to artificial radiation is from medical and dental x-rays, and it is growing every day. The general public's exposure to medical ionizing radiation has quadrupled in the last 25 years! It is our professional obligation, therefore, to keep our patient's radiation dose to a minimum. (Bushong, 11th ed., p. 596)

If a patient received 0.2 Gy during a 5 minute fluoroscopic examination, what was the dose rate? A 0.04 mGy/min B 1.0 mGy/min C 10.0 mGy/min D 40.0 mGy/min

The Correct Answer is: D0.2 Gy is equal to 200 mGy. If 200 mGy were delivered in 5 minutes, then the dose rate would be 40 mGy/min:200 mGy / 5 min = x / 1 5 x = 200 x = 40 mGy/min

Which of the following has (have) an effect on the amount and type of radiation-induced tissue damage?Quality of radiationType of tissue being irradiatedFractionation A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: DAll the factors listed influence the effect of radiation on tissue. Larger quantities, of course, increase radiation's effect on tissue. The energy (i.e., quality and penetration) of the radiation determines whether the effects will be superficial (erythema) or deep (organ dose). Certain tissues (such as blood-forming organs, the lens, and the gonads) are more radiosensitive than others (such as muscle and nerve). The length of time over which the exposure is spread (fractionation) is important; the longer the period of time, the less are the tissue effects. (Bushong, 8th ed., pp. 496-497)

In 1906, Bergonié and Tribondeau theorized that undifferentiated cells are highly radiosensitive. Which of the following is (are) characteristic(s) of undifferentiated cells?Young cellsHighly mitotic cellsPrecursor cells A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: DCells that are termed undifferentiated are immature or young. They have no specific function and/or structure. They are usually precursor cells; their most important function is to divide. Mitosis is the most radiosensitive part of the cell cycle. (Bushong, 8th ed., p. 495)

Which of the following is most likely to result in the greatest increase in patient exposure? A Changing from 100 mA to 200 mA B Increasing kV 15% and cutting mAs in half C Decreasing the kV 15% and doubling the mAs D Changing from nongrid technique to 8:1 grid technique

The Correct Answer is: DConverting from nongrid to an 8:1 grid requires about a fourfold increase in mAs. Increasing the kVp by 15% and cutting the mAs in half would reduce patient dose. Decreasing kV by 15% and doubling the mAs would increase dose. Changing from 100 mA to 200 mA will double the exposure dose. Therefore, the largest increase would be by the addition of a grid. (Bushong, p 252)

A fluoroscopic examination requires 3 minutes of exposure on time. If the exposure rate for the examination is 3 mGy/hr, what is the approximate exposure for the three minute procedure? A 3 mGy B 1.5 mGy C 0.75 mGy D 0.15 mGy

The Correct Answer is: DIf the exposure rate for the examination is 3 mGy/hr (60 minutes), then a 3-minute examination would be proportionally less—as the equation below illustrates: 3 (mGy) / 60 (min) = x (mGy) / 3 min 60 x = 9 x = 0.15 mGy dose in 3 minutes

What percentage of x-ray attenuation does a 0.5-mm lead-equivalent apron at 75 kVp provide? A 51% B 66% C 75% D 88%

The Correct Answer is: DLead aprons are worn by occupationally exposed individuals during fluoroscopic and mobile x-ray procedures. Lead aprons are available with various lead equivalents; 0.5- and 1.0-mm 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 used rarely 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 lead-equivalent apron will attenuate about 99.9% of a 50-kVp beam, 88% of a 75-kVp beam, and 75% of a 100-kVp beam. (Bushong, 8th ed., p. 597)

Which of the following cell types has the greatest radiosensitivity? A Nerve cells B Muscle cells C Spermatids D Lymphocytes

The Correct Answer is: DLymphocytes, a type of white blood cell concerned with the immune system, have the greatest radiosensitivity of all body cells. Spermatids are also highly radiosensitive, though not to the same degree as lymphocytes. Muscle cells have a fairly low radiosensitivity, and nerve cells are the least radiosensitive in the body (in fetal life, however, nerve cells are highly radiosensitive). (Dowd & Tilson, p 135)

The radiation dose to an individual depends on which of the following?Type of tissue interaction(s)Quantity of radiationBiologic differences A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: DPhotoelectric interaction in tissue involves complete absorption of the incident photon, whereas Compton interactions involve only partial transfer of energy. The larger the quantity of radiation and the greater the number of photoelectric interactions, the greater is the patient dose. Radiation dose to more radiosensitive tissues, such as gonadal tissue or blood-forming organs, is more harmful than the same dose to muscle tissue. (Bushong, 8th ed., p. 197)

Which of the following groups of exposure factors will deliver the least patient dose? A 300 mA, 250 ms, 70 kVp B 300 mA, 125 ms, 80 kVp C 400 mA, 90 ms, 80 kVp D 600 mA, 30 ms, 90 kVp

The Correct Answer is: DSelection of exposure factors has a significant impact on patient dose. Remember that milliampere-seconds (mAs) are used to regulate the quantity of radiation delivered to the patient and kilovolts peak (kVp) determines the penetrability of the x-ray beam. As kilovoltage is increased, more high-energy photons are produced, and the overall average energy of the beam is increased. An increase in milliampere-seconds increases the number of photons produced at the target, but milliampere-seconds are unrelated to photon energy. Generally speaking, then, in an effort to keep radiation dose to a minimum, it makes sense to use the lowest mAs setting and the highest kVp setting that will produce the desired radiographic results. An added benefit is that at high kilovolts peak and low milliampere-second values, the heat delivered to the x-ray tube is lower, and tube life is extended. In this example, (A) = 75 mAs, (B) = 37.5 mAs, (C) = 36 mAs, and (D) = 18 mAs. Decreasing the milliampere-seconds and increasing the kilovolts peak appropriately is the most effective combination for reducing patient dose. (Bushong, 8th ed., pp. 162-163)

If an individual receives an exposure of 150 mR/h at a distance of 2 feet from a radiation source, what will be their dose after 30 minutes at a distance of 5 feet from the source? A 60 mR B 30 mR C 24 mR D 12 mR

The Correct Answer is: DThe relationship between x-ray intensity and distance from the source is expressed in the inverse square law of radiation. The formula is Substituting known values: x = 24 mR in 60 minutes, therefore 12 mR in 30 minutes Distance has a profound effect on dose received and therefore is one of the cardinal rules of radiation protection. As distance from the source increases, dose received decreases. (Bushong, pp 68-70)

As the x-ray tube filament ages, it becomes progressively thinner because of evaporation/vaporization. The vaporized tungsten is frequently deposited on the window of the glass envelope. This may1.act as an additional filter.2.reduce tube output.3.result in arcing and tube puncture. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DThrough the action of thermionic emission, as the tungsten filament continually gives up electrons, it gradually becomes thinner with age. This evaporated tungsten is frequently 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 may actually attract electrons from the filament, creating a tube current and causing puncture of the glass envelope. (Selman, pp 137-138)

The positive electrode of the x-ray tube is the A capacitor B grid C cathode D anode

The Correct Answer is: DX-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. This 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 nickel 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)

Thickness of protective barriers depends on the following (select the four that apply): A Distance between radiation source and barrier B Beam alignment C Time of occupancy factor D Source-to-skin distance E Automatic exposure control F Occupational dose limit G Public dose limit

The answer is A, C, F, and G.There are a multitude of factors that determine barrier thickness, including the distance between the radiation source and the barrier (A); this distance is measured to include the occupied area on the opposite side of the barrier, which could be a physician office, technologist area, or waiting room. Time of occupancy factor (C) also measures how often the space on the other side of the barrier is occupied; the less it is occupied, the thinner the barrier can be. The occupational dose limit (F) utilizes dose limits for radiation employees to determine how thick control area barriers need to be, while the public dose limit (G) specifies thickness for barriers that protect areas in which non-radiation employees, such as patients, will be occupying. Beam alignment(B), source-to-skin distance (D), and automatic exposure (E) control do not affect barrier thickness. (Bushong, 11th ed., pp. 557-558)

Which of the following methods may be used to reduce the radiographer's dose? (select the three that apply) A Increasing distance from the patient during mobile radiographic exams B Holding a patient from the side of the table during fluoroscopic exams C Decreasing distance from the patient during fluoroscopic exams D Standing behind a lead barrier during an exposure E Holding a pediatric patient during an exposure to reduce the chance of having to repeat the exam F Standing at 90 degrees from the patient during fluoroscopy exams

The answer is A, D, and F. Based on the inverse square law, which states that the intensity of radiation varies inversely to the square of the distance, the radiographer shouldstand as far back as possible from the patientduring a mobile radiographic exposure (A). The radiographer shouldstand behind a lead barrierwhen performing radiographic exams, which reduces their exposure from scatter radiation emitted by the patient (D). If a radiographer must be in the room during a fluoroscopic exam, thesafest place to stand is 90 degrees from the patient, as less scatter radiation is emitted by the patient at this angle (F).Holding a patient at the head end of the tableexposes the radiographer to more scatter radiation than when standing 90 degrees to the patient (B). Based on the inverse square law,decreasing the distance from the patient during fluoroscopic exams increasesthe radiographer's exposure to scatter radiation (C). The radiographer should use immobilization devices as a first option when performing radiographic procedures on uncooperative pediatric patients. If this option is non-viable,a non-pregnant family member or other non-radiology personnelwearing a lead apron and gloves should be used to hold the patient (E). (Statkiewicz, Visconti, Ritenour & Haynes, 8th ed., pp. 280-283, 286-287)

Which of the following personnel radiation dosimetry devices typically are sent to a lab for dose measurements? 1. Thermoluminescent dosimeter 2. Optically stimulated luminescence dosimeter 3. Digital ionization dosimeter A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is A. All the devices listed in the selections are types of personnel dosimetry devices. Athermoluminescent dosimeter (TLD)uses a lithium fluoride radiation detector that stores radiation energy. This device is typically sent to a dosimetry lab, where the detector is placed in an analyzer and heated to release and measure the stored radiation energy. This energy is released in the form of light (luminescent) energy in proportion to the amount of radiation dose received . Anoptically stimulated luminescence (OSL) dosimeteris one that uses analuminum oxideradiation detector to store radiation energy. The device is sent to a dosimetry lab and is placed in an analyzer, where a laser light exposes it and causes the stored energy to release in the form of light (luminescent) energy, which varyies in intensity based on the amount of radiation energy absorbed. The light intensity is measured and translated to the quantity of the radiation dosage received (A). Adigital ionization dosimeterhas the appearance of a flash drive and contains a small ionization chamber. The air contained in this chamber is ionized in proportion to the amount of radiation received and an electrical charge is produced and stored in a built-in semiconductor. When the device is connected to an on-site computer's USB port, a small voltage delivered to a transistor "gate" provides a digital signal that represents the radiation exposure received by the dosimeter. An instant measurement and read-out of the dose quantity is then displayed on the computer (B, C, and D). (Statkiewicz, Visconti, Ritenour & Haynes, 8th ed., pp. 78-86).

Filtration in the x-ray tube, housing, and collimator provides which of the following? A Reduced patient skin dose B Increased patient skin dose C Decreased average primary beam energy D Increased primary beam intensity

The answer is A. Beam filtration exists in the form of inherent and added filtration. Inherent filtration includes the x-ray tube window, oil in the x-ray tube housing, and the housing port. Added filtration includes an aluminum plate and the collimator mirror. The plastic cover on the bottom of the collimator can also filter some low-energy x-rays. Filtration in the x-ray tube, housing, and collimator absorbs the low-energy x-rays in the primary beam, which would otherwise be absorbed in the patient's skin or shallow tissues and would not contribute to the radiographic image (A). The opposite would be true if there were no such filtration of the primary beam (B). Inherent and added filtration serve to filter the low-energy x-rays from the primary beam, thereby increasing the average beam energy (C). The quantity (intensity) of x-ray photons in the x-ray beam is controlled by the mAs setting and filtration of the primary beam decreases the quantity of x-ray photons present in the primary beam (D). (Bushong, 11th ed., pp. 131, 142)

As the radiographer increases kilovoltage to compensate for decreased mAs when performing a radiographic procedure, which of the following accompanying changes would be expected? Patient dose would decreaseScatter striking the image receptor would increaseImage noise would decrease A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is A. Higher kilovoltage techniques decrease the number of photoelectric interactions (absorption) of x-rays in the patient, but with a proportionately higher number of Compton scattered x-rays being produced. At higher kilovoltages, the scattered x-rays are more energetic and more likely to reach the IR. Image noise Image noise is more likely to increase in the form of mottle as mAs decreases. (Bushong, 11th ed., pp. 187-188, 313, 317)

RBE and LET have what type of relationship? A Direct relationship B Indirect relationship C Inverse relationship D No relationship

The answer is A. Relative biologic effectiveness (RBE) refers to the capabilities of radiation to cause biologic effects. It has adirect relationshipto the linear energy transfer (LET) of the radiation. As LET increases, RBE increases. Conversely, as LET decreases, RBE decreases. Mathematically, RBE is expressed as the ratio of the dose of a reference radiation (e.g., 250 kVp x-rays) to the dose of another type of test radiation in question, which is necessary to produce the same biologic response as that caused by the reference radiation (A). The opposite of the direct relationship described is anindirect, orinverse, relationship (B and C). Because there is a direct relationship between RBE and LET, the remaining answer,no relationship, is incorrect (D). (Statkiewicz, Visconti, Ritenour & Haynes, 8th ed., pp. 119-120, 138, 140, 349)

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

The answer is A. The adult nervous system is composed of non-dividing differentiated cells and is the most radioresistant system in adults (A). Non-dividing differentiated cells are specialized, mature cells that do not undergo mitosis. As a result, they are rendered radioresistant. Lymphocytes and epithelial tissue contain many stem cells and are the most radiosensitive cells in the human body (B, C, and D).(Sherer, Visconti, Ritenour, and Haynes, 8thed., p. 6)

The benefits of shadow gonad shields include 1. Useful during surgery due to easier manipulation around sterile field 2. Increased positioning accuracy and ease 3. More beneficial for pediatric applications A 1 only B 2 only C 1 and 3 only D 1, 2, and 3

The answer is A. The shadow shields are a type of gonadal shielding that do not require direct contact with the patient to offer protection. The main advantage of shadow shields is their ability to be utilized during surgical procedures with reduced interference with the sterile field compared to a contact shield (1) (A). The positioning of shadow shields needs to be more precise to offer protection without interfering with the exam, and therefore is not necessarily easier than the positioning accuracy and ease of a contact shield (2) (B and D). Lastly, shadow shields are better for use with adult patients rather than pediatric patients (3), as pediatric patients tend to move more, creating shielding positioning difficulty (C).(Bushong, 11th ed., pp. 590-592

Which of the following factor(s) affect the position of an x-ray emission spectrum curve on the x-axis? 1. Tube voltage 2. Target material 3. Tube current A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is A. The x-ray emission spectrum curve shifts to either the right or left on the x-axis when the average x-ray beam energy either increases or decreases, respectively. As thetube voltage(kVp) is increased, the filament electrons are accelerated at a higher kinetic energy and therefore, higher energy bremsstrahlung x-rays are produced in the anode target. These higher-energy x-rays increase the average x-ray beam energy, thus shifting the emission spectrum curve to the right. The converse would be true if the tube voltage were decreased. As the atomic number of thetarget materialincreases, so does the energy of the bremsstrahlung x-rays produced, and vice versa (A). Thetube current(mA) is the number of filament electrons traveling from the cathode to the anode (B, C, and D). As the tube current quantity increases, so does the number of x-rays produced in the anode target, and vice versa. Changes in tube current changes the height, or amplitude, of the x-ray emission spectrum curve. It does not affect the average energy of the beam, so therefore does not shift the x-ray emission spectrum curve to either the right or left. (Bushong, 11thed., pp. 131-134)

What is the feature that most x-ray imaging systems have that terminates the x-ray exposure once adequate exposure reaches the image receptor? A Automatic exposure control B mAs C kVp D Back-up timer

The answer is A. Theautomatic exposure controlis a system used to control the amount of radiation that reaches the image receptor (A). Though the automatic exposure control is used for all AEC systems, the radiographer must still use individual discretion to select an appropriate kVp, mA, image receptor, and grid. However, the AEC device determines the exposure time (and therefore total exposure).mAsis the milliamperage (mA), which is a major factor in determining the quantity of x-rays produced, and the (s), which is the length of time the radiation is going. The mAs factor together affects film density by governing the amount of x-ray photons that reach the image receptor (B).kVpis the kilovoltage peak applied to the x-ray tube, making the tube current in space charge. kVp is responsible for the acceleration of electrons from the cathode to the anode (C). Theback-up timeris used when using AEC; it is the electronic timer set to a maximum exposure length in case the AEC fails to terminate the exposure within that time (D).(Sherer, Visconti, Ritenour, & Haynes, 8thed., p. 263)

In addition to shielding, which of the following choices will best reduce patient dose during a radiology exam? (select the two that apply) A Increasing mAs B Decreasing mAs C Increasing collimation D Decreasing collimation E Decreasing SID

The answer is B and C. Patient dose should be a consideration as a practicing technologist selects techniques for radiographic exams. Decreasing mAs reduces the total number of photons contained in the primary beam and is directly proportional to dose (B). Collimation is described in inverse relation to FOV size. As collimation increases, FOV decreases, and less patient tissue is exposed to radiation, reducing dose (C). Close collimation also reduces scatter generated within the patient, which often would be absorbed internally, especially with thicker body parts (D). Decreasing SID would put less air molecules in between the patient and the tube; thus, entrance skin dose does increase, but not measurably (E). Air is a lousy shield. The significant change with reduced SID is the intensity (photons/cm2) of the beam increases, via the inverse square law. Increasing mAs increases the number of photons that enter the patient with no appreciable change to percent transmission, as we have not mentioned a compensating increase or decrease in kVp (A). It follows that the patient will therefore absorb more photons and dose will increase. (Bushong, 11thed., p. 590)

Which of the following conditions would make tissue more radiosensitive? (select the three that apply) A The tissue is composed of highly specialized cells B The tissue is composed of stem cells C The tissue has a high mitotic rate D The tissue is well oxygenated E The tissue is anerobic

The answer is B, C, and D. Radiosensitivity of a particular tissue is primarily governed by two factors: the Law of Bergonié and Tribondeau, and the oxygen effect. The Law of Bergonié and Tribondeau states that tissue becomes more radiosensitive when its degree of specialization is low and its mitotic rate is high (A and C). Stem cells are cells that are constantly dividing and making replacement tissue, like the inner lining of the stomach, which continually loses its inner cells to acid damage (B). The oxygen effect states that tissue that has an abundance of oxygen is more likely to undergo unwanted chemical reactions after ionization (D). This is primarily because oxygen is very reactive in all environments, not just within human tissue. Anerobic tissue, which means that it's lacking oxygen, would see a relative decrease in radiosensitivity (E). Tissue that is well oxygenated typically has a better chance of repairing damage caused by radiation as well. (Sherer, 7thed., pp. 133, 148)

Federal law requires that all occupationally exposed individuals are issued dosimeters for monitoring purposes. Occupationally exposed personnel describes any individual employed in a role in which they are likely to receive what percentage or more of the annual effective dose limit of 50 mSv in the performance of their duties? A 5% B 10% C 1% D 20%

The answer is B. Most healthcare facilities issue dosimeters when personnel could receive 1% of the annual effective dose limit (C). However, the requirement for issuing dosimetry monitors to occupationally exposed personnel uses 10% of the annual effective dose limit as the threshold for enforcement (B). Dosimetry monitors can only provide an accurate measurement of occupational exposure if they are worn according to recommended guidelines. The other answer choices are incorrect as per the previous explanation (A and D). (Sherer, 7thed., p. 84)

Which of the following dosimetry devices contains an aluminum oxide detector? A Thermoluminescent dosimeter B Optically stimulated luminescence dosimeter C Pocket ionization chamber D Digital ionization dosimeter

The answer is B. All the devices listed in the selections may be used for personnel dosimetry. An optically stimulated luminescence (OSL) dosimeter is one that uses an aluminum oxide radiation detector. The x-ray energy imparted to the detector is stored until the detector is placed in an analyzer, where a laser light exposes it and releases the stored energy in the form of light (luminescent) energy, which varies in intensity based on the amount of x-ray energy absorbed. Therefore, the degree of light energy emitted provides a measurement of the radiation dosage received (B). A thermoluminescent dosimeter (TLD) uses a lithium fluoride radiation detector that stores x-ray energy. When the detector is placed in an analyzer and subjected to heat, the stored energy is released in the form of light (luminescent) energy in proportion to the amount of radiation dose received (A). A pocket ionization chamber (also called a pen dosimeter) contains a chamber of air that is ionized by bombarding radiation. The ionization creates an electrical charge that causes a small quartz fiber in a built-in electrometer to move across a measurement gauge that provides an immediate readout in proportion to the dosage received (C). A digital ionization dosimeter has the appearance of a flash drive and contains a small ionization chamber. When the air is ionized in proportion to the amount of radiation received, an electrical charge is produced. This charge is stored in a semiconductor. When the device is connected to a computer's USB port, a small voltage delivered to a transistor "gate" provides a digital signal that represents the radiation exposure to the dosimeter. An instant readout of dose information may be obtained through proprietary dosimetry software that produces a digital graphical display of the dosage received (D). (Statkiewicz, Visconti, Ritenour & Haynes, 8th ed., p. 78) Mark

The annual dose limit for radiographers includes radiation from which of the following? 1. Background radiation 2. Occupational exposure 3. Medical x-rays A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The answer is B. Any occupationally-exposed individual is required to use devices that will record and provide documentation of the radiation they receive over a period of time. The most commonly used personal dosimeters are the TLD, OSL, and film badge. These must be worn at all times for documentation ofoccupational exposureonly (B).Background radiationcomes from a variety of sources and the entire population in the world is exposed to it (A and D). This type of radiation would not require radiographers to wear their dosimeters. Badges are not to be worn for any medical or dental x-rays one receives as a patient (C).(Sherer, Visconti, Ritenour, and Haynes, 8thed., p. 161)

Superficial reddening of the skin caused by radiation exposure is called A erythrocytosis B erythema C leukopenia D rubeola

The answer is B. Erythemais a superficial reddening of the superficial layers of skin in areas sustaining high x-radiation exposure. A single absorbed dose of 2 Gyt can cause erythema to the exposed area within approximately 24-48 hours, potentially followed by desquamation, or shedding, of skin cells. The energy deposited (absorbed dose) in the skin causes injury or irritation, which triggers the inflammatory response. During the inflammatory response there is dilatation of the blood capillaries, which causes the reddening seen on the skin surface. High exposures to x-radiation in diagnostic radiography are more likely to occur during interventional fluoroscopic procedures such as cardiovascular and interventional procedures, when long exposure times or high-level fluoroscopy (HLF) may be used. During cardiovascular or other interventional procedures, exposure rates may range from 100-200 mGya/min, and sometimes higher. Due to an increasing number of radiation-induced injuries reported following the use of high-level fluoroscopy, imposing strict controls on its use is essential (B).Erythrocytosisis a condition which causes an abnormal increase in the number of circulating red blood cells in the body. Having too many red blood cells can make the blood thicker than normal and lead to blood clots or other complications (A).Leukopeniais a condition of an abnormal decrease in the number of white blood cells. This condition can be caused by high whole-body dosages and lead to the hematopoietic syndrome (C).Rubeolais a condition caused by the measles virus, which is characterized by a red skin rash (D).(Statkiewicz, Visconti, Ritenour & Haynes, 8th ed., pp. 150, 152)

At which dose range does gastrointestinal syndrome from excess radiation exposure peak? A 5 Gy B greater than 10 Gy C greater than 50 Gy D 75-150 Gy

The answer is B. Gastrointestinal (GI) syndrome can appear at approximately 6 Gy. After an exposure of 10-50 Gy (B), the onset of GI syndrome takes only a matter of hours to cause vomiting and diarrhea. After the early onset, death occurs within 4-10 days of the exposure. At doses higher than 50 Gy, central nervous system (CNS) syndrome occurs, causing death within hours to a few days. CNS syndrome is sometimes also referred to as cerebrovascular syndrome. (Bushong, 11th ed., pp. 510-511; Sherer, 8th ed p145)

The energy of an L characteristic ray is equal to the A difference in energy between the K and L shells. B difference in energy between the L and M shells. C energy of the incoming electron. D energy of the incoming electron minus the energy of the L shell.

The answer is B. In production of characteristic radiation, a high-speed electron having an energy of at least 70 keV encounters a tungsten atom within the anode and ejects a K shell electron, leaving a vacancy in that shell. An electron from the adjacent L shell moves to the K shell to fill its vacancy, and in doing so emits a K characteristic ray. The energy of the characteristic ray is equal to the difference in energy between the K and L shell energy levels. An L characteristic ray is produced when an L electron is ejected and replaced by an M electron (B). The other answer choices are incorrect as per the previous explanation (A, C, and D). (Saia, PREP 9th ed., pp. 388, 389)

When considering various types of radiation in regard to linear energy transfer, which of the following statements is NOT true? A Electromagnetic waves have low LETs compared to alpha particles. B Beta particles have a higher LET than alpha particles. C Gamma and x-radiation have similar LETs. D All types of radiation can potentially deposit energy into living tissue.

The answer is B. Linear energy transfer is a method used to quantify the potential of various types of radiation to deposit their energy into matter (D). In general, radiations that travel as waves, for example, gamma and x-ray radiation, have low LETs as compared to particle radiations: alpha, protons, and neutrons (C). Alpha particles, composed of two protons and two neutrons, have a large mass and a +2 electrical charge. They have the potential to disrupt a target molecule via both a collision and an electron stealing capacity, both of which can break one or several chemical bonds. Beta particles, by contrast, are high speed electrons with very little mass. The probability of multiple bond breaks by a single beta is much lower, and thus, it has a lower LET than an alpha (B). The difference between betas and alphas is akin to the difference in damage of a single rifle bullet (the beta) versus a cannonball (the alpha).(Sherer 8th ed., p. 75)

Point lesions occur when there is disruption in A double chemical bonds B outer shell electrons C single chemical bonds D inner shell electrons

The answer is C. When ionizing radiation interacts with macromolecules, damage can occur. Point lesions are the result of disruption of single chemical bonds (C). The point lesions can then cause the cell to malfunction and cause damage, thus affecting functionality and cell life. (Bushong, 11th ed., pp. 492-493)

X-rays within the x-ray beam that do not reach the image receptor during a radiographic exposure of an anatomic part describes which of the following? A Filtration B Attenuation C Off-focus radiation D Leakage radiation

The answer is B. Some x-rays in the x-ray beam never reach the image receptor after they interact with the patient's tissues. Therefore, the original quantity of x-rays (intensity) that could have potentially exposed the image receptor is reduced. This reduction in quantity is calledattenuation(B). The amount of attenuation varies depending on the type, thickness, and atomic number of the tissues with which the x-rays traverse. Attenuation occurs due to two tissue interactions wherein the x-rays are either absorbed (photoelectric interaction) or scattered away from the image receptor (Compton scattering). A reduction in the quantity of low-energy x-rays in the beam prior to reaching the patient is achieved throughfiltration, both inherent and added (A).Off-focus radiationis produced when some of the filament electrons striking the anode focal spot bounce off and then land on other areas of the target, producing x-rays from outside of the focal spot (C).Leakage radiationrefers to x-rays that escape the protective tube housing and result in unnecessary exposure of the patient and personnel within the procedure room (D). (Bushong, 11th ed., pp. 139, 157)

By using an x-ray beam filter, patient _______ exposure is primarily reduced. A bone marrow B skin C gonadal D GI

The answer is B. The diagnostic x-ray beam is heterogeneous, or polyenergetic, due to the interactions of filament projectile electrons and the anode target material. The various x-ray energies produced result from both characteristic and bremsstrahlung x-ray interactions. Since both interactions are random events in the target material, varying x-ray energies are produced. The lower-energy (or softer) x-rays do not have the penetrating ability to expose the deeper body tissues nor the image receptor. They would likely be absorbed in the patient'sskin, thereby increasing the skin dose or entrance skin dose (ESD). By filtering these low-energy x-rays from the beam using inherent and added filtration, and therefore increasing the average energy of the primary (or useful) beam, skin dose is reduced (B). Since bone marrow, gonadal, and GI tissues lie deeper than the patient's skin, they are less likely to receive the dosage from the low-energy x-rays (A, C, and D).(Bushong, 11th ed., p. 241)

Where is the safest area for the radiographer to stand when a cross-table lateral C-arm projection is performed? A On the tube side B On the image intensifier side C In the path of the primary beam D On the tube side, but at an angle

The answer is B. X-ray exposure to the radiographer during C-arm procedures comes primarily from the patient. When cross-table lateral projections are performed, the radiographer must understand that the exposure rate caused by the entrance surface of the patient is greater than that near the exit surface of the patient, typically two to three times more. Therefore, the radiographer should stand on the side farthest from the x-ray tubeon the image intensifier side(B). The most scatter is producedon the tube sidewhere the primary beam enters the patient (A). The radiographer must never standin the path of the primary beam, as this would significantly increase their dosage and would superimpose the radiographer's anatomy over the patient's anatomy (C). Even if the radiographer were to stand atan angle to the tube sideof the patient, their exposure would be greater than if they were standing on the image intensifier side (D). (Statkiewicz, Visconti, Ritenour & Haynes, 8th ed., p. 287)

A radiographic exposure of 36 µC/kg is recorded at a source-image-distance (SID) of 36 inches. If the same technical factors are used, what will the exposure in µC/kg be if the source-image-distance is increased to 72 inches? A 5 B 9 C 18 D 72

The answer is B. The relationship of x-ray exposure quantity (µC/kg) to distance (SID) is described in the inverse square law, which states the intensity of radiation at a given distance from the point source is inversely proportional to the square of the distance. This is calculated by applying the following formula: I1/I2 = D22/D12, where I1 = original intensity (mR), I2 = new intensity (mR), D1 = original distance (SID), and D2 = new distance (SID). Applying this formula to find the answer to this problem is as follows: 36 µC/kg /I2 = 72"2/36"2; I2 = 36 × 362 / 722 ; I2 = 36 × 1,296 / 5,184; I2 = 9 µC/kg (B). Because the SID is increased in this problem, the exposure intensity will be less than the original intensity; therefore, choice (D), being higher than the original intensity of 36 µC/kg, is incorrect. Choices (A) and (C) are less than the original exposure intensity of 36 µC/kg but do not reduce the µC/kg by the inverse of the square of the distance (SID) change. (Carlton/Adler/Balac, 6th ed., p. 164)

What is the component of personal dosimeters that evaluates radiation quality? A Control panel B Look-up table C Filters D kVp

The answer is C. Dosimeters contain filters that attenuate radiation such that radiation types and energies can be differentiated.Filtersare the component of personal dosimeters that evaluate radiation quality (C). Thecontrol panelis where the technologist sets the technique for each position and is located outside of the x-ray room (A). Thelook-up tableis used in digital radiography and is data stored in the computer that is used to substitute new values for each pixel during processing (B).kVpis the kilovoltage peak applied to the x-ray tube and determines the quality of the x-ray beam; it is not a component of personal dosimeters (D).(Sherer, Visconti, Ritenour, & Haynes, 8thed., pp. 211-212)

Patient entrance skin exposure during fluoroscopic procedures can increase due to 1. Increased field size 2. Increased exposure time 3. Patient age 4. Use of intermittent exposures A 1 only B 3 only C 1 and 2 only D 1, 3, and 4 only

The answer is C. Entrance skin exposure can be complicated to calculate. The fluoroscopic unit is typically moving quite a bit, which changes the exact location of the radiation source/field. The field size (1) is always changing as well, which impacts ESE; larger field sizes increase skin exposure (A). Increased exposure time (2) also increases ESE, and each examination requires a different amount of fluoro time based on the patient needs (C). Patient age (3) does not necessarily increase ESE, and the use of intermittent exposures (4) will decrease ESE (B and D).(Bushong, 11thed., p. 574)

What are early symptoms of acute radiation syndrome? 1. Nausea and vomiting 2. Cataracts 3. leukopenia A 1 and 2 only B 2 and 3 only C 1 and 3 only D 1, 2, and 3

The answer is C. If a large amount of radiation is delivered to the whole body at one time, the short-term somatic effect must be considered. However, if the whole body receives over 600 rads at one time then acute radiation syndrome is likely to occur. The early stages of acute radiation syndrome are nausea, vomiting, fatigue, diarrhea, and leukopenia (C). During occupational exposure, individuals generally receive small amounts of low-energy radiation over a long period of time. These individuals are concerned with potential effects of radiation such as cataracts (A, B, and D).(Sherer, Visconti, Ritenour, & Haynes, 8thed., p. 6)

Which of the following factors increase the probability of scatter production within the patient? (select the three that apply) A Average atomic number of the tissue B Decreased mAs C Increased OID D Increased tissue thickness E Decreased collimation F Increased tissue density (kg/cm3)

The answer is D, E, and F. Compton scatter is a random phenomenon that is influenced by external factors. Increased tissue thickness, decreased collimation, and increased tissue density all contribute to the same result—more atoms in the path of the primary beam (D, E, and F). Students (and technologists) should remember thatcollimateis a verb that means to trim. Therefore, decreasing collimation is actually an increase in field size and therefore more tissue will be exposed to ionizing radiation, not less (B). Average atomic number has a direct relationship with the probability of the photoelectric effect, not Compton scatter (A). Increasing OID, while making scatter miss the image receptor via the air gap technique, does nothing to influence how much scatter is produced (C).(Bushong, 11thed., pp. 148-149, 189)

Which of the following statements are true with respect to the accompanying animation? (select the three that apply)0:130:06 A Occurs with outer shell electrons B Produces discrete energy photons C Likely occurs with tissues having high atomic numbers D Occurs when high speed electrons are decelerated E Occurs with tungsten atom nuclei F Produces a wide range of x-ray energies

The answer is D, E, and F. Two types of x-radiation are produced at the anode through energy conversion processes: Bremsstrahlung radiation and characteristic radiation; Bremsstrahlung radiation predominates. To produce Bremsstrahlung (Brems) or "braking" radiation, electrons are accelerated toward the tungsten atoms within the anode focal track. The high-speed negative electron is attracted by the positive nucleus of a tungsten atom and, as a result, is pulled off course and redirected toward the nucleus (E). The electron's deflection from its original course, caused by the "braking"/slowing down, results in a loss of energy (D). This energy loss is given up in the form of an x-ray photon called Bremsstrahlung (Brems/braking) radiation. The electron might not give up all of its kinetic energy in one such interaction; it can go on to have several more interactions with tungsten atoms deeper in the target, each time giving up an x-ray photon having less and less energy. This is one reason why the x-ray beam is polyenergetic (that is, has a spectrum of energies) (F). Brems radiation comprises 70%-90% of the primary x-ray beam. (Bushong, 11th ed., pp. 126, 127)

Damage to the DNA double helix ladder structure, which results in spurs that form unintended bonds that join the macromolecule into irregular loops, is called A single side rail scission B radiolysis C double side rail scission D cross-linking

The answer is D. DNA macromolecule damage are classified into three types of effects. Single side scissions, or point lesions, are the simplest type; a single bond is broken on the side of the ladder (A). Main-chain or double side scission occurs when both sides are broken, resulting in a free-floating fragment (C). Cross-links are unintended bonds that form after ion pairs are created in adjacent sections of the DNA strand (D). Radiolysis is the breakdown of water molecules by ionizing radiation (B).(Bushong 11th ed., p. 492)

Which of the following produces a potential difference within the x-ray tube during an exposure? A mA B mAs C Ohms D kVp

The answer is D. During a radiographic exposure, there are two stages. The first stage is called "boost and hold," when the radiographer is preparing the x-ray tube for the exposure. This first stage causes the rotor and anode disk to rotate 3,400 rpm for standard diagnostic x-ray tubes and 10,000 rpm for high-capacity specialty tubes. During the first stage, the cathode filament is also heated to incandescence, whereby through the process of thermionic emission electrons are boiled off and hover in an electron cloud (or space charge) in front of the filament. When the patient has complied with the radiographer's instructions, the exposure button is completely depressed to make the exposure. During the second stage, alarge electrical potential difference (kVp)produced by the x-ray circuit's high voltage step-up transformer (or high-tension transformer) is applied to the x-ray tube. This large electrical force causes the negatively charged electrons in the electron cloud to travel to the positively charged anode disk with high kinetic energy, where they strike the actual focal spot (target) of the anode. The filament electrons interact with the tungsten atoms of the anode disk to produce both bremsstrahlung (brems) and characteristic x-rays, with brems x-rays being the predominant type. Increased kVp increases the kinetic energy of the projectile filament electrons and, upon interaction with the anode target, creates higher-energy brems x-rays (D). Milliamperage (mA) refers to the electrical current, or quantity of projectile filament electrons, flowing from the cathode to the anode during an exposure (A). ThemAsis the electrical current flowing per unit of time from cathode to anode (B).Ohmsis a measure of electrical resistance that varies inversely with the cross-sectional diameter of a conductor (e.g., copper wire), and directly with the length of a conductor. Remember that resistance also results from the back-electromotive force created in the step-up transformer under the influence of mutual induction (C). (Bushong, 11thed., p. 124)

Filtration increases beam quality through 1. Increased attenuation of high energy waves 2. Decreased HVL 3. Increased attenuation of low energy waves 4. Increased HVL A 1 and 4 only B 2 and 3 only C 1 and 2 only D 3 and 4 only

The answer is D. Filtration within an x-ray tube, whether added or inherent, acts to attenuate low energy waves (C) that do not contribute to the diagnostic quality of an x-ray. The low energy waves increase skin exposure for patients, which makes the filtration an important piece of the tube assembly for decreased patient dose. By filtering out the low energy waves, the filtration increases the half value layer (D): the part thickness required to reduce the x-ray intensity by 50%. The other answer choices are incorrect as per the previous explanation (A and B).(Bushong, 11th ed., pp. 141-142)

For a given exposure, the amount of scatter radiation produced by a large patient compared to that of a smaller patient is A less than a smaller patient B approximately equal to a smaller patient C slightly less than a smaller patient D more than a smaller patient

The answer is D. Radiographic imaging of large (thicker) patients enables the primary x-rays to interact with a larger volume of tissue, therebyincreasing the number of Compton scatter interactions. This results in increasedscatter fog (noise)on the radiographic image, which decreases contrast resolution. It is particularly important for the radiographer to use high-ratio grids when performing radiographic procedures on large patients, to reduce the amount of scattered radiation striking the image receptor. Compression of the body part, such as performing a PA projection of the abdomen instead of an AP projection, also reduces the tissue volume with which the primary x-ray beam may interact (D). Smaller (thinner) patients inherently have less tissue volume; therefore,fewer Compton scattering interactionsoccur. Less scatter production and fog results in improved contrast resolution (A, B, and C). (Bushong, 11thed., pp. 188-191)

Which of the following factor(s) affect the amplitude of an x-ray emission spectrum curve? 1. Tube current 2. Exposure time 3. Added filtration A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is D. The height, oramplitude, of an x-ray emission spectrum curve is determined by the quantity of x-rays emitted by the x-ray tube. Factors that affect the amplitude includetube current(mA),exposure time, andadded filtration. Thetube currentis the number of filament electrons traveling from the cathode to the anode (B). As the tube current increases, so does the number of x-rays produced in the anode target. Conversely, as the tube current decreases, so does the number of x-rays produced in the anode target. Theexposure timedetermines the length of time the filament electrons travel from the cathode to the anode (A and C). Therefore, as exposure time increases, more x-rays are produced in the anode target, whereas as time decreases, fewer x-rays are produced.Added filtrationserves to remove the low-energy x-rays in the beam, and therefore reduces the quantity of x-rays exiting the tube (D). (Bushong, 11thed., pp. 130-132)

When performing a scoliosis series, which projection will minimize dosage to the patient's breasts? A LPO B RPO C AP axial D PA

The answer is D. The radiographer is responsible for practicing the principle of "as low as reasonably achievable" (ALARA) when performing radiographic procedures. Part of this responsibility when performing a scoliosis survey of the thoracolumbar spine is to limit the dosage delivered to sensitive organs such as the breasts by applying adequate collimation and shielding. Another method the radiographer should employ is to use a PA projection rather than an AP projection. Because of their anatomical location, a PA thoracolumbar scoliosis projection would attenuate some of the x-rays in the upper back, thereby reducing breast dosage. Likewise, gonadal dosage is reduced in the lumbosacral region due to attenuation of some of the primary x-rays in the lower back (D). The remaining selections would require an anterior to posterior projection, and the breasts (and gonads) would be subjected to unnecessary dosage. Furthermore, the routine projections used in a scoliosis series are PA and lateral (A, B, and C). (Long, Rollins, & Smith, 13th ed., vol. 1, p. 491)

Tungsten is commonly used in a diagnostic x-ray tube anode because of its 1. low atomic number 2. low melting point 3. high atomic number 4. ability to produce x-ray energies capable of producing a diagnostic image A 1 and 4 only B 1 and 2 only C 2 and 3 only D 3 and 4 only

The answer is D. Tungsten has ahigh atomic numberof 74, has a thermal conductivity like copper that dissipates heat easily, and has a high melting point of 3,400° C. Since 99 percent of the kinetic energy of the filament electrons bombarding the anode is converted to heat, this element is very capable of handling the thermal stress. The primary reason for using tungsten in the anode, however, is its high atomic number. A high atomic number means there are many protons within the atomic nuclei. Since the presence of this large number of protons produces a strong positive charge, accelerated filament electrons passing by the nuclei, which carry a negative charge, experience a strong electrostatic attraction toward the nuclei. Therefore, they brake (decelerate) strongly and their kinetic energy is transformed into high-energy bremsstrahlung x-rays of sufficient penetrating ability. The higher energy x-rays have sufficient energy to penetrate the inherent and added filtration of the tube window, oil, and housing port through which the primary beam must pass before traversing the patient toproduce a diagnostic image(D). Use of a lower atomic number element with a lower melting point would not produce high-energy x-rays and the anode would likely melt down and create surface defects. There would also likely be excess coating of evaporated metallic material on the x-ray tube window, inhibiting the transmission of x-rays toward the anatomical area of interest and increasing the chances of electrical arcing (A, B, and C).(Bushong, 11th ed., p. 110)

If the exposure rate to a body standing 7 ft from a radiation source is 1.21mGy/h, what will be the dose to that body at a distance of 4 ft from the source in 30 minutes? A 59.2 mGy B 3.7 mGy C 1.85 mGy D 0.75 mGy

The correct answer is (C). The relationship between x-ray intensity and distance from the source is expressed in the inverse-square law of radiation. The formula is Substituting known values, 1.21/x = 16/4916x = 59.29x = 3.70 Thus, x = 3.70 mGy in 1 hour (60 minutes) and, therefore, 1.85 mGy in 30 minutes. Note the inverse relationship between distance and dose. As distance from the source of radiation increases, dose rate decreases significantly.

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 ( W r )2.Tissue weighting factors ( W t )3.Absorbed dose A 1 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The correct answer is: (A )The Radiation Weighting Factor (W r ) 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 W r of different ionizing radiations is dependent on the LET of that particular radiation. The Tissue Weighting Factor (W t ) represents the relative tissue radiosensitivity of irradiated material (eg, muscle vs intestinal epithelium vs bone, etc). To determine Effective Dose (E) the following equation is used: Effective Dose (E) = Radiation Weighting Factor (W r ) × Tissue Weighting Factor (W t ) × Absorbed Dose (Bushong, p 556)

A dose of 250 mGy 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 correct answer is: (B) During the first trimester, specifically the 2 to 8 weeks of pregnancy (during major organogenesis), if the radiation dose is at least 200 mGy, 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/intellectual disability 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 mGy is exceedingly small and that most general diagnostic examinations are likely to deliver fetal doses of less than 10 to 20 mGy. (Bushong, p 546)

Organize the following items so they appear in the correct order in which the primary x-ray beam passes through filters in their path.

X-rays are produced and emitted isotropically from the focal spot of the anode disk of the x-ray tube. Only those x-rays emitted toward the patient make up the primary x-ray beam and are of potential diagnostic quality. As the x-rays travel toward the patient, they must pass through several structures and substances in their path. First, they must pass through the thin x-ray tube window. Then, the x-rays must travel through the oil surrounding the x-ray tube that dissipates heat through the process of conduction and convection and provides electrical insulation. Next, the x-rays pass through the x-ray tube housing port. The aforementioned structures and oil constitute what is called inherent filtration. After passing through the inherent filtration, the primary x-ray beam passes through an added aluminum (Al) plate of 1-mm thickness for diagnostic x-ray tubes. Beyond this, the x-rays pass through the collimator mirror. The aluminum plate and mirror constitute what is called added filtration. Both the inherent and added filtration make up 2.5 mm Al equivalent total filtration for diagnostic x-ray machines. This total filtration is beneficial and serves to absorb the low-energy x-rays produced by either bremsstrahlung or characteristic interactions of the filament projectile electrons bombarding the anode target during a radiographic exposure. By filtering out these low-energy x-rays, patient skin entrance dose (SED) is reduced. (Bushong, 11th ed., pp. 10, 140, 143, 195, 241-242, 253, 343-344)