G1.  D

G2. C

A is absorbed dose (gray), and D is dose equivalent (sievert). Exposure is defined only for photons below 3 MeV; in the SI system units of coulomb/kg replace the roentgen.

G3. E

The SI unit of dose equivalent is the sievert.   1 sievert is 100 rem.

G4. D

Q depends on the LET (linear energy transfer) of the radiation. Values used for radiation protection are:

Radiation type                                              Q

x-rays, electrons and gamma-rays             1

neutrons                                                         20

G5. A

Unlike charged particles attract.

G6. C

Like charged particles repel.

G7. A

Unlike charged particles attract.

G8. A

Unlike charged particles attract.

G9. B

There is no electric force between a charged and a neutral particle.

G10. A

This particle has negligible mass; A is the only choice.

G11. E

This particle has a charge of 2; E is the only choice.

G12. C

C is the only choice with a charge of +1.

G13. B

A, B and D have no charge: however, if C is a positron, the neutron must have a greater mass, so B is the coreect choice.

G14. D

Isotopes are forms of the same element, and thus have the same atomic number, Z, the number of protons, but, different numbers of neutrons, thus different A (neutrons plus protons). Isobars have the same A but different Z. Isomers have the same A and z, but different energy states. lsotones have the same number of neutrons but different Z.

G15. C

The mass number (A) is defined as the number of nucleons (protons and neutrons) in the atomic nucleus.

G16. C

The number of neutrons in a nucleus is the mass number (total no. of nucleons) minus the atomic number (no. of protons), i.e., 60-27=33.

G17. D

This is the definition of binding energy.

G18. B

G19. B

The binding energy per nucleon usually increases after radioactive decay, as the daughter nucleus is more stable than the parent. High binding energy implies stability.

G20. C

The maximum number in the outer shell is 8 (hence 8 groups in the periodic table) but in general the maximum no. of electrons in a shell of quantum number n is 2n*2.

G21. D

The average life is 1.44 x half-life. The half-life is inversely proportional to the decay constant, and is unaffected by T or P.

G22. C

In a time equal to half of a half-life, decay will be slightly more than one fourth, so 14 would be a good guess. Using the decay equation gives 14.14.

A = Aoe*-λt where λ = 0.693/T1/2 = 0.693/6 hr, and t = 3 hr.

G23. C

Combining physical and biological half-lives always gives an effective half-life shorter than either of the two componens.

Teff =  (Tbio x Tphys) / (Tbio + Tphys)

or

1/Teff = 1/Tphys + 1/Tbio

G24. D

G25. A

The mass number (A) does not change during isobaric or isomeric decay.

G26. B

Beta plus is possible because the atomic mass difference (M -> M - 2 MeV) is greater than 1.022 MeV.

Isobars have the same A.

An isomeric transition implies that Z does not change.

G27. A

Molybdenum decays to the metastable state of technetium by beta minus decay. The decay of technetium from the metastable to the ground state releases a 140 keV gamma-ray, which is used in nuclear medicine imaging.

G28. D

The rest masses of the two particles combine to yield two gamma-rays of 0.511 MeV each.

G29. D

Electron capture, example: 51/24 Cr to 51/23 V

G30. B

Adding neutrons to the nucleus may result in too many neutrons for stability, leading to beta minus decay.

031. B

In general, the higher the atomic number, the larger the neutron-proton ratio for stability. Thus when a uranium nucleus splits into two intermediate-Z nuclei, these have too many neutrons, and thus will decay by beta minus.

G32. B

Secular equilibrium occurs when the parent has a long half-life. After about four half-lives of the daughter, the apparent decay of the daughter and parent are equal.

G33. E

The output of a step-up transformer is higher voltage, but lower current and lower power than the input.

G34. D

5.0 Ci =5 x 3.7x 10*10 Bq = 18.5  x 10*10 Bq or 185 GBq.

G35. E

Exposure rate = (Exp. rate const.) x activity x (1/d)2 where d and the exposure rate constant are both expressed in cm.   Exp. rate = 3.3 x 10 x (1/100)*2 = R/hr = 3.3 mR/hr.

G36. C

Cobalt emits two gammas per disintegration (99.8% of the time). Half-life has no influence on gamma factor, and although cesium has a lower energy gamma than cobalt, this does not have a very marked effect on the gamma factor.

G37. A

The rotating anode spreads the heat load over a washer shaped disk rather than a single spot, thus increasing the heat load of the tube while maintaining a small effective focal spot.

G38. C

The process whereby a filament is heated to a sufficient temperature to emit electrons is called thermionic emission.

G39. E

Most of the energy (99%) is lost as heat. Of the energy which is converted into x-rays, approximately 90% is bremsstrahlung and 10% is characteristic radiation, although this varies with the Z of the target material.

G40.  A

B C-D are irrelevant. A is true only if the electron energy increases above the binding energy of one of the electron shells in the atom.

G41-43. BCB

Possible characteristic x-rays are found by taking energy differences between shells (30,29.3,26, 4, 3.3, 0.7) given in keV. Bremsstrahlung can have any energy up to the maximum incident electron energy of 50 keV.

G44. D

Filament current affects mA, hence beam intensity only.

G45. D

HVL = 0.693 / Linear attenuation coefficient.

G46. B

For a monochromatic beam, two half-value layers reduce the intensity to 25%, but for a broad spectrum the added absorbers harden the beam, increasing the penetrability; the second HVL is greater than the first.

047. D

048. A

049. C

G50. B

The radiation must have sufficient energy to ionize atoms in the film emulsion.

051. E

Using the inverse square law: I[100] = l[50] x (50/100)2 = 25 mR.

G52. D

Wavelength (m) x frequency (Hz) = velocity of light (m/sec).

Wavelength = 3x1O*8/ 3x10*6 = lOOm.

G53. D

The linear attenuation coefficient depends on beam energy and attenuating medium. The formula is used to calculate the intensity of a beam after it has passed through x cm of an attenuating medium.

G54. C

The mass attenuation coefficient is the linear attenuation coefficient divided by the density of the material. Whereas the probability of an interaction is proportional to Z cubed for the photoelectric effect, and Z for pair production, it is independent of Z for the Compton effect, and all materials have an equal probability, gram for gram (since all materials except H have almost equal numbers of electrons per gram). The mass attenuation coefficient represents the attenuation per unit mass of a material.

G55. C

G56. A

057. B

The primary radiation is reduced by 2 TVLs.  Half of the attenuating material will reduce the dose by one TVL (10%).   The total dose however, will include the contribution of scattered radiation which will depend on the energy of the radiation.

G58. C

The activity is increased by a factor of 4 (1 to 4 Ci). Therefore two additional HVLs are required.

G59. E

As the beam passes through the Al in which the HVL is measured, it is filtered and hardened. Therefore, more Al is required to reduce the beam's intensity from 50% to 25% than from 100% to 50%.

G60. D

The higher atomic number of calcium results in greater absorption by the photoelectric process.

G61. A

The photon is most likely to undergo a photoelectric interaction when its energy is just above the binding energy of the K shell electrons.

G62. C

The probability of a photoelectric interaction is proportional to Z cubed. Thus, 4:8 cubed is 1:2 cubed or 1:8.

G63. C

The probability decreases rapidly with energy.

G64. A

No energy is lost in classical scattering.

G65. B

In the Compton process, interaction at any angle other than grazing incidence leads to a significant transfer of energy to the electron. That is why scattered radiation from megavoltage beams is significantly lower in energy than the primary.

G66. A

Because the electron density of all materials (except hydrogen) is nearly constant, their mass attenuation coefficients, at energies. where Compton interactions predominate, are also nearly the same. (Note, however, that linear attenuation will depend on density.)

G67. B

The threshold energy for pair production is the energy necessary to create an electron pair, an electron, and a positron. (E = mc*2)        2 x me c*2 = 1.02 MeV. The rest mass (mec*2) of an electron or positron is 0.511 MeV.

G68. B

1.O2MeV is reemitted as two 0.511 MeV photons.

G69. A

G70. D

G71. E

G72. C

The probability of a Compton interaction decreases slowly with increasing photon energy.

G73. A

Coherent scatter, also known as Raleigh or classical scatter, occurs only for very low energy x-rays, and is of little concern in radiology.

G74. D

Pair production increases slowly with increasing energy, and may become the dominant process in high-Z materials at high MeV energies.

G75. C

Only 2 is false. Photoelectric effect is a photon interaction, not an electron interaction.

G76. E

The increase in LET is related to the mass and charge of the particles. The electron has a single charge and relatively low mass, the proton has a single charge and greater mass, and the alpha has a double charge and the greatest mass. The photon has no charge, and is less likely to lose energy along its path than an electron of equal energy.

G77. B

Neutrons may be captured into the target nucleus creating a new isotope. In interactions with water, the predominant interaction is with the hydrogen atoms which have mass similar to the neutron. Nuclear disintegration and the induction of radioactivity are a common result of neutron irradiation. Nuclear particles which can be ejected include alphas.

G78. D

In order to release a neutron from the nucleus, the photon must have an energy greater than the "binding energy" of the nuclear particles. This energy threshold is about 8 MeV.

G79. C

Collimating the beam decreases the area of exposure, using screens and a high kV technique both reduce patient dose. Grids improve contrast by "cleaning up" scatter, but require a somewhat higher dose to compensate for attenuation by the grid.

G80. D

A faster screen needs less radiation to produce the same amount of film darkening. It does this by using a thicker layer of more absorbent crystals. This increase in thickness sacrifices resolution. The screen converts liberated electrons into light.

G81. C

The "p value" represents the probability of making an error in accepting the conclusions of the statistical analysis, i.e., there is a   1 % chance that the results are not different.

082. B

G83. B

ROM is read only memory as opposed to RAM, random access memory. Programs may be put in or out of ROM and used but not revised. Programs may be put in and out of RAM and used and revised.

G84. D

The following table lists typical storage capacity in kilobyte (kB), megabyte, (MB) and/or gigabytes (GB) (1 byte = 8 bits)

O - Optical disk (1 to 2GB)

M - Multiplatter hard disk 300 to 700 MB

T - Magnetic tape  46MB

H - Hard disk 40MB to 2.5 GB

F - Floppy disk  1.5 MB

G85. D

Millions of diagnostic x-ray procedures are done every year. The next most significant source of radiation dose to the population as a whole is from nuclear medicine exams.

G86. C

Radon is a radioactive gas emitted primarily from rocky soils and/or building materials which contain trace amounts of uranium and radium ores. Radon concentrations in the air can reach hazardous levels in poorly ventilated basements and ground floors, particularly in certain parts of the country where soil and rocks contain large amounts of these ores.

G87. E

Although there are many valid arguments for other models, the simple linear no-threshold approach is used by the regulators.

G88. C

See BEIR V, Biological Effects of Ionizing Radiation Committee of the National Academy of Science.

G89. C

NCRP Report #33 recommends that "personnel monitoring shall be performed for occupationally exposed individuals for whom there is a reasonable possibility of receiving a dose exceeding one-fourth of the applicable MPD. It specifically excludes the use of monitors when the individual is "exposed as a patient for medical or dental reasons."

090. E

See NCRP Report#91.

G91. B 

Energy is inversely proportional to wavelength.

Thus Evisible = 1.2 x 10*-10/6 x 10*-5 MeV = 2eV.

G92. A

In a controlled area the occupancy factor is generally one for all locations. This reflects the fact that an employee will have freedom of movement throughout the entire area and can be anywhere within it for any portion of the work day.

G93. C

The current values of annual maximum permissible exposure are 100 mrem for non-controlled areas, 5 rem (5,000 mrem) for controlled areas. This factor of fifty difference requires approximately 6 more HVLs for the greater level of protection.

G94. C

Radiation produces ionization in the chamber gas which is measured as an electric current.  The measured current (or integrated charge) can be used to determine the number of roentgens or the dose, but neither of these quantities can be measured directly.

G95. A

Each ionization event in a Geiger counter generates a single large pulse, making it very sensitive to low levels of radiation. Geiger counters cannot, however, measure dose directly.

G96. D

3 years is 36 months, or approximately 18 half-lives.

The activity remaining is approximately 25 x [1/2]*18 = 9.5 x 10*-5 mCi

The exact value is: 25 exp -(0.693 x 3 x 365/60) =8 x lO*-5 mCi.

The exposure rate at 1cm is: 1.45 x 8 x 10*-5 R/hr = 0.12 mR/hr. The photon energy is about 35 keV. This exposure rate is so low that it is neither a hazard to staff, nor could it affect a radiograph.

G97. A

See 10 CFR Part 2O.

G98. C

0.1 R/mA-s x 60sec/min=6.0R/mA-min.

150mA - min/week x 6R/mA-min=900R/week at 1 meter.

By inverse square, at 3 meters WUT = 100 R/week.

To reduce this to 0.1 R: 100/0.1 = 1,000 times reduction. 103 = 1,000.

G99. A

See NRC Publication 10 CFR Part 20.

Dl. E

Approximately 1% of the energy imparted by electrons which bombard the anode is converted into x-rays. The remainder (99%) is converted to heat. Hence, one must be careful that the x-ray exposure does not deliver excessive amounts of heat to the anode which could damage it.

D2. E

The amount of x-ray intensity increases directly as the kVp squared and directly with the tube current (mA) and the exposure time (seconds). The anode material also affects the percentage of energy converted to x-rays by Bremsstrahlung and the energy of the characteristic x-rays. The collimation merely restricts the size of the area being irradiated.

D3. C

The atomic number of the material affects the photoelectric interactions and attenuation; and therefore, it influences the subject contrast of the objects. As the contrast level decreases, the object's size must be much larger to be easily visualized. Small focal spot sizes have an effect on the geometric unsharpness which primarily affects the visualization of high contrast objects which are small. Large density and thickness differences obviously also affect the subject contrast. The addition of contrast media will enhance the contrast of certain tissues.

D4. B

Fourier transforms are used to change line spread measurements into modulation transfer functions (MTF). The other terms listed are either mathematical functions that are used for other purposes by physicists, or in the case of back projection, it is an algorithm that is used to reconstruct CT images.

D5. D

The replenishment rates, developer concentration and developer temperature directly Influence the density and speed of the various film screen systems. Similarly, the developer immersion time is directly related to film density and the relative speed of the system. Bromine concentration in the developer does effect both the speed and contrast of the rnammograms. However, this question talks about nitrate depletion which has no effect on the development of mammography films.

D6. B

Step-down transformer

D7. D

Diode rectifiers

D8. E

Automatic exposure control (AEC)

D9. A

Step-up transformer

Dl0. C

Autotransformer

D11. A

Low kVp techniques are used to increase the contribution of photoelectric effect (absorption), hence reduce scattered radiation and increase subject contrast.

D12. E

Screen film mammography usually employs thick single emulsion films to reduce receptor blur by eliminating light cross-over through the emulsion. Parallax unsharpness, caused by double emulsion, will also be eliminated.

D13. D

Compressed breast generally produces greater sharpness, less scatter (better contrast) and decreased patient dose due to less tissue, hence less scattered radiation.

D14. C

Typical focal spot sizes for non-magnified mammography units are 0.3 mm to 0.4 mm. With magnification, 0.1 mm - 0.15 mm focal spot sizes are utilized to minimize geometric unsharpness.

Processor sensitometry is performed daily to check the processor stability.

The half-value layer is energy dependent.  Typically, the HVL for a 28 kVp beam is in the 0.3 mm Al -0.37 mm Al range.

Molybdenum is used as filter and target material. Molybdenum filters will selectively remove x-rays above and below the target characteristic x-rays, producing improved subject contrast.

D15. D

Skin dose to a 4.5-cm breast depends upon, among other factors, kVp selection and phototiming characteristics. At 28 kVp with the grid, skin exposures of 800 mrad could be obtained.

D16. B

Average glandular doses to a 4.5cm breast are also kVp and filtration dependent.  Typical values per view are in the 1.1 mGy -1.6 mGy (110-160 mrad) range. A value of 1.2 mGy is within that range.

D17. A

D18. E

Electron deceleration in the field of the nucleus gives Bremsstrahlung (continuous x-ray spectrum).  Electron interaction with orbital electrons causes ionization and gives characteristic x-rays. Compton scattering occurs when x-ray photons strike orbital electrons, not when electrons strike an atom. Electron capture is a nuclear decay process.

D19. B

D20. E

D21. A

D22. D

23. A

A thicker screen has a higher absorption efficiency, stops more x-rays and reduces patient dose.

D24. B

Noise is not decreased; it remains the same because the same number of x-ray photons is utilized to produce the required optical density.

D25. A

The limiting resolution is the highest number of line pairs per millimeter that can be obtained by an imaging system. Screen blur, caused by light diffusion in the screen, has inferior resolution.

D26. A

Rare earth screens have higher conversion efficiency. As conversion efficiency increases, more light per x-ray photon absorbed is used, hence fewer x-ray photons are required to produce the desired optical density, and noise increases.

D27. A

Rare earth screens are faster than conventional screens. In the diagnostic energy range, rare earth screen absorption and conversion efficiency factors are higher, hence they require fewer x-ray photons to produce the desired OD on the film. Examples of elements used in rare earth screens include lanthanum and gadolinium.

D28. E

Optical Density = Log Opacity = log (1/transmittance).   From simple mathematics, a film with an optical density greater than 1.0 must have an opacity greater than 10 and a transmittance less than 10%. If the optical density is I .0, opacity = 10, and transmittance is 10%. An increase in optical density of 0.3 (from 1-1.3) doubles the opacity to 20, and reduces the transmittance to 5%. A film with an OD of 1.3 has 50% lower transmittance. A film with an optical density of 1.3 is within the normal useful density range. Dark films have densities above 2.2.

D29. B

Average gradient is the slope of a straight line on the characteristic curve joining two points of specified densities Dl and D2 (usually 0.25 and 2.0 above base + fog). From the curve, B+F =0.2, D1= 0.25 + 0.2 = 045 and D2 = 2.0 + 0.2 = 2.2. Corresponding log exposures: log E1 = 1.1 and log E2 = 1.75. Average gradient = (D2-D1)/(logE2 - logE1) = (2.2 - 0.45)1(1.75 - 1.1) = 3

D30. E

From the previous question average gradient = 3

3= (D2-Dl)/(logE2 - logE1) = (D2-DI)/log E2/E1

if El produces a density of 0.5. and E2 = (2E1) produces a density D2, then 3 = D2 - 0.5)/log 2

3 Log 2 = D2-0.5, D2 = 1.5

D 31. C

Exposure under bone would be in the lower part of the characteristic curve at a value of about 0.8.

D32. C

One can assume that each emulsion receives an identical exposure from each screen. When the densities produced in the two emulsions are added, the double-emulsion film will produce twice the contrast of a single emulsion film.  Extended processing time or increasing developer temperature improves contrast. It may, however, cause other problems such as developer instability and film fog.   FiIm processing steps are developing, fixing, and washing.  Replemishment solutions are used to maintain the activity of the developer by restoring depleted preservative and developing agents. They contain alkaline agents and must be free of bromide.

D33. C

The lowering of the specific gravity of the developer solution is similar to diluting the chemistry with water. This results in a lowering of both the film speed and the film average contrast gradient.  The developer solution does not have an effect on the archival storage. The archival storage is affected by the fixer concentration and the amount of time the film spends in the fixer solution.  Dilution of the chemistry will also decrease the base + fog of the film down to its lowest level which is usually related to the color tint of the film and the age of the film.

D34. C

As in the previous questions, double-emulsion films produce twice the contrast from a single emulsion film. Processor caused artifacts such as scratch lines and developer variations may have the same effect on single and double emulsion films. In fact, single emulsion films are likely to shrink in area after processing (curl toward the emulsion side). An identical emulsion on each side of the base prevents this curling. Double emulsion / Double screen systems are faster, therefore cause reduction in patient dose. The use of screens is associated with reduced spatial resolution due to light diffusion in the screens.

D35. D

D36. B

The equation for the Larmor frequency says that it is equal to a constant 42.58 Megahertz per Tesla multiplied by the field strength of the magnet in Tesla. Therefore, the frequency is directly related to the magnetic field strength.

D37. E

All the types of pulse sequences listed in these questions have been employed in MRI studies with the exception of the last item which is fictitious.

D38. C

Thinner slices and smaller pixel sizes enclose fewer protons in the voxel space. Therefore, they produce smaller signal levels. More NEX results in averaging the signal which reduces the noise; and therefore, this change improves the signal-to-noise ratio. Short TR times results in a smaller signal being utilized to form the image. The signal-to-noise ratio improves as the field strength of the magnets increases.

D39. B

The T2* relaxation lime is influenced by both structural differences of the anatomy and by magnetic field inhomogeneities. The use of a 180 degree pulse to reverse the direction of the spins causes rephasing from which the effect of field homogeneities can be recovered.

D40. B

Fringe magnetic fields often interfere with electronic devices such as biostimulators and pacemakers. RF heating effects can occur if metal or wire structures are placed near the skin.  Magnetic surgical clips in patients have been known to move during MRI scans and cause  bleeding in the patients.  Large ferromagnetic objects brought near a large field strength MRI have been known to go flying through a room and sometimes injure staff in the room. For magnetic field strengths below 2 Tesla, the effect of magnetic fields inducing electrical potential in neurons has not been observed. However, at much higher field strengths, it has been speculated that such effects may occur.

D41. C

Computed radiography has a wide dynamic range, in effect a wide latitude. It produces acceptable films with sub-optimal x-ray technique, such as with portable x-ray producers. It uses image plates covered with photostimulable phosphor (Europium activated Barium Fluoride) as a receiver. Resolution with screen-film Systems (5-10 lp/mm) is better than that produced in computed radiography (2.5-5.0 lp/mm). The reject rate in CR is low due to post processing techniques available electronically. Contrary to the film system, all mentioned operations of image capture. storage and display are performed by different components of the CR system.

D42. A

D43. B

As the imaging performance improves, the ROC curve moves toward the upper left-hand corner.  An ideal system would give no false positives unless the observer insisted on calling everything positive. Its ROC curve would therefore hug the left hand edge.

D41. A

If the image conveyed no information at all to the observer and the observer simply guessed whether the object was present, the resulting ROC curve would be diagonal as in the dotted line.

D45. A

D46. B

As the imaging system signal-to-noise ratio increases, the ROC curve will move away from point 3, towards the upper left -hand corner.

D47. D

D48. E

All of the above are correct. Grids are placed between the patient and the film to cut down on scatter radiation produced from the interaction of x-rays with the patient.   A 12:1 grid will allow less scatter radiation through. This means better image contrast, but requires more x-rays, i.e., higher patient dose to compensate for the lost photons (lost density). 12:1 grids are more sensitive to distance and lateral decentering.

D49. B

Specificity is the true negative (TN) fraction = TN/(TN+FP) =20/40=0.5

D50. C

Sensitivity is the true positive (TP) fraction = TP/(TP+FN) = 30/40=0.75

D51. D

Accuracy is the fraction of correct diagnosis (TP+TN)/(total readings) = 50/80 =0.625

D52. C

Cerebral angiography with iodinated contrast uses magnification to improve fine vessels visualization. It results in lower contrast enhancement at higher kVp due to larger differences between beam mean energy and iodine K-edge absorption energy. A small focal spot size is recommended to minimize focal spot blur effect. Contrast resolution is mainly obtained from differences in linear attenuation coefficient of contrast medium (iodine) and vessel wall. The dominating interaction in iodine is the photoelectric effect.

D53. D

Each additional component in imaging chain degrades the image further.

The spot film cassette is the most direct view of the information contained in the x-ray beam. The spot film cassette typically has a spatial resolution of 4-8 line pairs per millimeter.

The next step on the imaging system would be the image intensifier. The image intensifier typically has a resolution of 4-5 line pairs per millimeter.

The next most direct step would be the 100 millimeter fluoroscopic spot film camera (FSFC). The 100 mm camera has essentially the same resolution as the image intensifier. 

The next step in the imaging chain is the television monitor. The television monitor degrades the spatial resolution due to the limited number of raster lines. For a 6-inch field of view, the television monitor typically has a resolution of 1.82 to 2 line pairs per millimeter.

The last step in the imaging chain would be the digital spot film device which has slightly less resolution than the television system.

D54. A

Magnification = Image size / Object size = (FFD) / (FFD-OFD)

32mm / (obj. size) = 120mm / (120-30 mm) = 120/90

Object size =90/120 x 32 mm = 24 mm.

D55. B

Due to distance increase, the exposure time must be increased by (D2/D1)*2 = (120/100)*2 = 1.44

D56. A

According to recent ACR standards for Teleradiology, small matrix systems require:

a) digitization systems with 0.5 K x 0.5 K x 8 bits array or better

b) display systems with 0.5 K x 0.48 K x 8 bits array or better,

while large matrix systems require:

a) digitization systems with 2K x 2K x 12 bits array or better.

b) display systems with 2K x 2K x 8 bits array or better.

There may be less-stringent guidelines for display systems when these display systems are not used to produce the official authenticated written interpretation.

D57. C

D58. A

With DSA contrast differences less than 1% in x-ray transmission can be visualized. Differences of 2-3% may be missed when using conventional film/screen receiving systems. The skin dose per frame with a 23-inch image intensifier system would be about 200 mR / frame. The noise level in the mask or live image will be lower than that in the final subtracted image. The subtracted image noise can be reduced by using frame averaging. Full or partial pixel shift is one method to correct for patient movement during DSA acquisition. The resolution obtained in DSA images is generally less than 2 lp/mm for 1000 line TV system.

D59. B

Scatter radiation from the patient is the major source of personnel exposure during fluoroscopy.  All other factors make insignificant contribution, if any. 

D60. C

As the fluoroscopic beam is positioned over thick or more dense areas of a patient, the penetration of x-rays decreases. The displayed technique factors of 120 kVp and 10 mA indicate that the image intensifier is not receiving enough transmitted radiation through the patient. At such high technique factors the patient skin entrance dose is extremely high. A malfunctioning 5 minute timer will not affect the technique factors. A broken timer, however, must be repaired. If the anti- scatter grid is not in the beam, the technique factors will be reduced.

D61. C

The image intensifier should be placed as close to the patient as possible in order to reduce patient dose. Improvement in image quality is also possible due to decreased distortion of anatomy and lower image blur. If procedure circumstances mandate that fluoroscopy be done with more than 25 cm distance between the patient and the image intensifier, then one should consider removing the anti-scatter grid.

D62. D

Many image intensifiers are of the multi field type. Some standard modes of operation are 12 inch, 9 inch, 6 inch or 4 inch. The smaller the field size, the more magnified the image appears on the TV monitor.  In the 9 inch II mode, a Cesium iodine (CsI) II tube can image 4 lp/mm (about 0.125 mm objects); in the 4 inch mode, the resolution is approximately 6 lp/mm (about 0.08 mm objects). The improved resolution with the smaller size Il input is associated with increased patient dose.

D63. E

Minification gain is the increase in image brightness that results from reduction in image size from the input phosphor to the output phosphor. The output phosphor of an image intensifier tube, usually made of zinc cadmium sulfide, is typically one inch in diameter. For a 9 inch image intensifier, the minification gain is (9/1)2 or 81.

D64. D

As the collimators open to expose a larger area of the patient, more scatter radiation is produced.  Scatter radiation reaching the image intensifier reduces image contrast.   Higher tube current increases patient and personnel exposure. It will, however, decrease image noise if all other factors are kept constant. As explained in an earlier question, decreased image intensifier to patient distance improves image quality. Higher kVp will result in lower image contrast. If the fluoroscopy image does not contain sufficient contrast for the procedure, then the kVp should be lowered slightly with appropriate increase in mA. This should be continued until a satisfactory combination of image contrast and image brightness is achieved.

D65. E

With pulsed fluoroscopy, the radiation is not continuously on during fluoroscopy. The radiation is pulsed on either 7 1/2, 15, or 30 times a second. Because of the short pulse duration, some savings in radiation dose of the order of 30-50% can be realized.

The automatic iris determines how much light is transmitted from the image intensifier to the television camera. The iris is opened when the radiation dose starts to increase which results in a reduction in radiation dose.

Similarly, the TV automatic gain control can electronically boost the measured signal from the television camera to compensate for studies in which few of the x-rays are penetrating the patient.  These images produce the required brightness on the television monitor without using excess radiation. The penalty is increased quantum mottle in the TV monitor image. 

Last frame capture or last frame hold is a digital system that maintains the last image on the television monitor when the fluoro is stopped. In this manner the total fluoro time can be reduced since the image can be studied without having the x-rays on continuously.

D66. C

D67 A

99mTC photons energy is 140 keV. The dominant interaction with soft-tissue at this energy is Compton scattering.

D68. C

Depending on filtration type and thickness, the 30 kVp x-ray beam spectrum is dominated by relatively low energy (soft) x-ray photons. The dominant interaction with breast tissue is the photoelectric effect.

D69. A

201Tl main photon energy is about 70 keV. At this energy, the dominant gamma photon interaction with bone is Compton scattering. Photoelectric effect makes less than 25% of the interactions at this energy.

D70. B

D71. B

D72. C

D73. D

D74. E Image contrast in nuclear medicine is generally high due to radioactivity localiaation in organ of

interest. In SPECT imaging helter Contrast is obtained due to the elimination of overlapping

structures. Septal penetration, scatter radiation and reduced target to background uptake ratio

reduce image contrast.

D75. E SPECT employs high-sensitivity collimators (low resolution) to maintain adequate photon

statistics per projection. With elliptical travel, the gamma camera will be at a minimum distance

from the patient at all limes during rotation. This will increLse sensitivity and rsssolution. Thinner

crystal minimizes light diffusion, hence increases resolution.

D76. E Anger camera field uniformity is the ability of the system to produce a uniform image from the

entire crystal surface and attached photomultiplier tubes. Uniformity is tested every day of use of

the system using any of the methods in A, B, or C. 99'DTc is inexpensive and easy to prepare, but a

new source must be prepared every day. 57Co has longer half-life so it doesn't have to be made

daily. It is, however more expensive and can't be used to test for the ~~Tc peak. 137~ has

relatively higher energy and is used in testing the constancy and acenracy of the dose ealibmtoi~

D77. D By definition the acoustic impedance is the product of the wave speed times the tissue density. The

unit of impedance is usually Rayls.

D78. A The equation for calculating the Doppler shift in color flow measurements depends on both

wavelength, ultrasound speed. angle of incidence and the frequency shift that occurs due to

Doppler effects.

D79. A The radiation dose calculations ftw adininistered radiopharrnaceuticals are given in the MIRD

pamphlets. These calculations nidicate that the radiation dose is dependent on all the faclors listed.

however, as the organ mass increases, the radiation dose decreases. Hence, the radiation dose is

inversely related 10 organ ln&L~.

RAPHF.X1997 0 Answers 0 Pa:e13

DSO. B The pulse repetilion rale (PRR) refers 10 the number of separate pulses of sound that are

transmitted per secoud. Its value is generally 1000 pulses/sec (1 kHz). High PRR limits the

penetration depth because the transducer receiving time decreases. Echoes from distant par~ may

still be out, or in the way when the next pulse starts. The spatial pulse length is the number of

sonic waves in the pulse tllnes the wavelength. Two objects will be resolved if the pulse length is

less than one-half the separation. This means that depth, or axial resolution is limited to half the

pulse length . If the frequency increases, wavelength decreases, and so does the pulse length but

the axial resolution increases. For a given transducer, as the wavelength decreases, the frequency

increases and tissue penetration will decrease.

D81. C

D82. C Lower frequencies, than B-mode imaging, are used in Doppler ultrasound to minimize attenuation.

Doppler imaging detects changes in the sound frequency emitted by a moving source. High C-

factor transducers are desirable because they produce a narrow range of frequencies. From the

Doppler shift formula, change in frequency is inversely proportional to the angle between

transducer and direction of flow. When 0 increases, cos 8 decreases, and change in frequency'

decreases. 1'wo transducers are employed in continuous Doppler, one sends and one receives.

Change in frequency is calcul~~icd fr(~m the difference between the two transducer signals.

D83. B 5 MHz transducers produce very good- axial resolution when used for large organs such as the

breast. Endorectal Ir,uisducers use 5-7 MHz with limited range to image the prostate. 2.5 - 5 MHz

transducers with good depth penetration are used for abdominal imaging. High C-factor

transducers have no backing material behind the crystal. The near zone (focal, or useful zone)

incre&se, with transducer size and frequency.

D84. A The Computerized Tomography Dose Index (C'IDl) is based upon a series of 15 C,l,, slices which

are adjacent to each other. The dose in the central slice is the sum of the diret exposure plus

scattering from adjacent slices. Ilence, this measurement already includes the fact that there are a

large number of CT slices in the series. The dose does not significantly increase as a(iditional CI'

slices are laken far away fr(ml the central slice. Higher kVp with the same mAs would result in an

increase in the radiation dose. I ~rge abdomen sizes place the patient tissue closer to the x-ray tube

which results in a slight iticTOL'e ill r,idiation dose. The radiation dose at the center of the body is

approximately 1/2 the value of the radiation dose at the surface due to tissue attenuation.

Obviouslyj, an increase in the mAs will incr~'se the total amount of radiation delivered to the body.

D85. C The window width is defuied as (lie r~ige of numbers from the lowest number to the highest

number that is being displayed. I lence the window width must range from -1000 for air to little

over +1000 for cortical bone. This range is a window width of 2000 Hounsfield units. The window

level is set in the middk' of the (~ number range that is being displayed. Hence the window level

should be set a~ 0.

D86. D CT number scale is equal to 1000 times (the ratio of the linear attenuation coefficient of the

substance divided by the linear attenuation of water minus one). The difference in the ratio of the 2

linearattenuationaiefficic'itsis 1.2-I x 1000=0.2x 1000=200.

D87. C It is important that the CT scanner detectors be small in size and packed closely together.

Furthermore, the detector should stop most of the incident radiation and respond quickly to the

changes in radiation which requires a fast temporal response. An isotropic response means that the

detector measures radiation uniformly from all directions. In the CT scanner the x-ray beams are

coming tlirough a small angle essentially perpendicular to the detector.

RAPHJ~X J~7 0 Answers 0 Page 14

DS~. E Volume averaging i%.' having sevcral dilTerent substances within a given voxel which produces

some average CT number represcntiitg the mixture of the materials found in that particular volume

element. In order to minitnize this elTect, it is important to use thin slice thicknesses.

D89. C The CT noise increases as 'lie paticnt attenuation increases. The CT noise decreases as the slice

thickness, patient d()se and pixel size are increased. For half the normal size thickness, the CT

noise increases by a factor of 42. By increasing the mAs by a factor of 2, the CT noise decreases

by a factor of 41/2. The net effect is no change.

1)90. E Magnetic tapes typically can store at a rate of 1200 bytes per inch ~PI). Therefore, the storage

capacity of magnetic tapes in bytes is 2400 x 12 x 1200. This value is approximately 34.5

megabytes (MB). Therefore, a 2 gigabyte (1)illion byte) optical disk can store approximately 50

times the amount of information as a magnetic tape.

1)91. B MB stands for megabytes and is the total amount of information that can be stored in millions of

bytes. The term MIPS stands for millions of instructions per second and is a measure of the speed

of the computer. The term RV~ means relative value unit and it is used in a financial sense to

weight the various diagnostic procedures. The term BAUD refers to the amount of bytes per

second which can be transmitted over some type of telephone line. The term BPI stands for bytes

per inch and is the storage rate on a magnetic tape.

1)92. C Convergence is used in mathe'n.'itics to indicate that a solution approaches a set answer and

divergence is a solution that docs not come to an answer but expands. Linearity says that items are

proportional to each odier. Compliance has to do with regulations and being within the speaf led

requirements of 'lie regulatory system. Parity is a check of l's and 0's to see whether they are odd

or even and this can identify problems in which one of the digits has been changed.

1)93. D NCRP publication #116 lists the risk of fatal cancer induction as a result of the radiation for

various organs. The stomach is most sensitive in this publication and is followed closely by the

lung and the GI system.

1)94. B The regulatory limit for radiation dose to the fetus of a pregnant worker is given as 500

microsieverts per month (50 millirem per month).

1)95. D The risk factors for radiation workers are found in NCRP Report #116. This risk is approximately

45 times 10-2 per sievert. One merely multiples the amount of Sieverts per year times the number

of years of exposure times the risk factor. (1/10:: (0.05 SvIy) x 40 y x 45:: (1~2/y)) = 0.09=0.1.

1)96-DIOO. The GSD is defined as the goziad dose that, if received by every member of the population, would be expected to produce the total genetic effect on the population as the sum of the individual doses actually received. It is a measure of the population's gonadal radiation exposure, taking into consideration certain physical and demographic characteristics such as the age of the population, their sex and probability of having cli ildren. Based on the above:

1)96. B The U.S. population ('i~I) is estimated at 0.2 mGy (20 niril).

1)97. A Some scatter radiation will reach the gonads.

1)98. A Won't have children.

1)99. B Won't have children.

AAPHFX 1997 0 Answer: 0 Page 15

DIOO B Ultrasound is non-ionizing radial ion.

DIOl B It takes 6 cm to reduce this bcam intensity from 32000 to 2000 photons. The 6 cm thickness is therefore equal to 4 llV1~. Onc IIVL is equal to 6 cmI4 = 1.5 crn. To reduce the number from 2000 to 500, two addilioi~I I IVi ~ ar~' needed at an additional thickness of 2 x 1.5 = 3 cm.

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