RAPHEX

Question 1

A CT image consists of 200 slices, each 512 × 512 pixels, each pixel having a 16-bit pixel depth. The size of the file is _

Answer 1

16 bits equals 2 bytes. 2 × 512 × 512 × 200 = 104857600 bytes = 100 MB

Question 2

highest to lowest energy radiation

Answer 2

gamma - xray- UV- infrared - microwave - radio

Question 3

isomeric transition

Answer 3

Isomeric transition is a radioactive decay process that occurs in an atom where the nucleus is in an excited meta state (e.g. following the emission of an alpha or beta particle). The nucleus emits a gamma ray which can sometimes interact with the orbital electrons and emit an internal conversion electron. If this occurs, it can be followed by a characteristic x-ray, or an Auger electron.

Question 4

secular equilibrium

Answer 4

Secular equilibrium defines a special case of transient equilibrium in which the half-life of the parent is very long compared with that of the product. In this case the product activity approaches but never truly achieves equilibrium.

Question 5

transient equilibrium

Answer 5

half-life of the daughter is shorter than the half-life of the parent. Contrary to secular equilibrium, the half-life of the daughter is not negligible compared to parent’s half-life

Question 6

variation of the focal spot size in the image occurs in which direction?

Answer 6

anode-cathode direction

Question 7

typical filament current for extremity radiation

Answer 7

5000 mA

tube current is 1-5 mA for fluoro and 200-1000 mA for radiography

Question 8

explain sensitometric strip

Answer 8

daily QA
exposes a film with a calibrated sensitometer similar to a screen phosphor
steps 1-8 show a greater response to increase in developer temperature
step 14 shows that almost all silver bormide grains were used and that increase in temperature will not increase the density as much as for the lower density steps
steps 21 shows very high density and is likely to be in shoulder part of characteristic curve

Question 9

most appropriate image receptor for fluoro

Answer 9

cesium iodide scintillator, photocathode

Question 10

is CsI used in screen-film radiography?

Answer 10

Cesium iodide is an excellent scintillator used in fluoroscopy and digital radiography; however it is too sensitive to moisture to be used for screen-film radiography

Question 11

what screens are used for screen-film mammo?

Answer 11

Gadolinium oxysulfide

Question 12

grid ratio

Answer 12

strip height/width of interspacing

higher grid ratio= better scatter cleanup

Question 13

linear blurring tomography

Answer 13

-high contrast for thin sections

Wide-angle tomography produces thin sections, which have low contrast. This is because all the anatomy above and below the section in focus is blurred across the image with an effect similar to fog. The patient exposure is higher because most of the time the x-ray beam is passing through the patient at an angle, i.e., the patient is effectively thicker.

The simplest method is linear tomography, in which the X-ray tube is moved in a straight line in one direction while the film moves in the opposite direction. As these shifts occur, the X-ray tube continues to emit radiation so that most structures in the part of the body under examination are blurred by motion. Only those objects lying in a plane coinciding with the pivot point of a line between the tube and the film are in focus.

Question 14

typical Pb equivalent in lead aprons in fluoro rooms

Answer 14

0.5 mm Pb

reduces x-ray intensity by 95-99% depending on kVp

Question 15

contrast resolution of CT

Answer 15

0.3-0.6%

Question 16

minimum mammo resolution

Answer 16

With a bar line-pair pattern on a 4.5 cm high phantom, the high contrast spatial resolution produced by a mammography unit should be at least 11 lp/mm with the bars perpendicular to the anode-cathode axis and 13 lp/mm with the bars parallel to the anode-cathode axis

Current commercial full field mammography units have image receptors which are capable of resolving 5 to 7 lp/mm, whereas screen-film receptors designed for mammography can often resolve 15 to 20 lp/mm.

Question 17

where is resolution in mammo lowest at?

Answer 17

at the top of the breast because geometric blur increases with magnification

Question 18

A technologist notices that a DR (or CR) spine image appears unacceptably noisy on the review workstation. The most appropriate action is

Answer 18

Increase the AEC density control and repeat exposure.

DR/CR systems cannot compensate for excessive noise due to quantum mottle, i.e., dose too low. If unacceptable, the view needs to be repeated with a higher exposure to the image receptor (and to the patient).

Question 19

high speed mode yields?

Answer 19

noisy images
speed is sensitivity

spatial resolution is independent of speed for digital imaging systems

Question 20

what will the appearance of an overexposed DR or CR image be?

Answer 20

Overexposed images are generally very good in DR/CR because they will be low in noise. The system will rescale the brightness and contrast so they will appear properly exposed.

Question 21

best estimate value for radiogenic fatal cancer

Answer 21

10%/Sv

Question 22

A patient is examined with an auto brightness, triple mode FOV, image intensifier x-ray fluoroscope. The patient’s entrance air kerma rate (EAKr) is 20 mGy/min when the 14 cm FOV is used and the x-ray beam is collimated to the full FOV, the patient’s EAKr

Answer 22

Increases more than 30% when the FOV is decreased to 10 cm (full field collimation).

Decreasing the FOV decreases the geometric light amplification factor of the image intensifier. More radiation is therefore needed to produce the required output light level. Also, less scatter is produced (and reaches the image intensifier) when the beam is collimated. A small increase in EAKr is needed to offset this effect.

Question 23

ratio of brightness gains equals the ratio of what in an image intensifer?

Answer 23

ratio of magnification gains

Question 24

major differences between fluoro and standard radiography

Answer 24

tube potentials are the same

The focal spots for fluoroscopy are typically 0.3 or 0.6 mm; those for standard radiography are usually 1.0 to 1.2 mm. The spatial resolution for fluoroscopy is usually limited by the TV system to 1.8 to 2.5 lp/mm, while radiography has resolutions of 4 to 8 lp/mm. The tube current for fluoroscopy is usually 1 to 3 mA in order to limit anode heating for the long exposure times of 3 to 10 minutes; because of the short exposure times (less than 1 second) of radiography, tube currents of 200 to 800 mA can be used. Tube potentials are the same for both procedures. SSDs in fluoroscopy are usually 18 to 20 inches, while the SSDs for radiographs are typically about 25 inches (except for chest radiographs).

Question 25

how is 18F produced

Answer 25

bombard 18O target with proton beam in cyclotron

Question 26

only accurate method for measuring activity of a pure beta source

Answer 26

A liquid scintillation counter, in which the sample is dissolved in a liquid scintillant is the only accurate method. The other methods are inaccurate, because the activity determination of a pure beta source is extremely position and geometry sensitive

Question 27

The window setting used on a gamma camera for 99mTc is set with the center at 140 keV with a width of ±14 keV, i.e., 20%. What is the reason for this?

Answer 27

Photons, which impinge upon the crystal, lose energy by Compton scattering and the photoelectric effect. Both processes convert the gamma-ray energy into electron energy. On average, approximately one electron-hole pair is produced per 30 eV of gamma-ray energy deposited in the crystal. These electrons result in the release of visible light when trapped in the crystal. These light quanta are collected and amplified by photomultiplier tubes. The statistical fluctuation in the number of light quanta collected and their amplification is what causes the spread in the detected energy peak, even when most of the 99mTc photons deposit exactly 140 keV in the NaI crystal.

Question 28

When counting an 125I source of unknown activity, the multi-channel analyzer shows a peak at 60 keV, in addition to the main peak at 30 keV. What is this due to?

Answer 28

A second peak is seen at twice the energy of the main peak. Pulse pile-up occurs when two photons impinge upon the detector at the same time, resulting in an event being recorded as if it had twice the energy of the initial photon.

Question 29

difference between 2D and 3D PET

Answer 29

2-D scans have septa in front of the detectors to reduce events from scattered photons. The septa reduce a large fraction of the cross-plane coincidences, in order to reduce coincidence events from scattered photons. Both 2-D and 3-D scans acquire all axial slice data simultaneously, and can reformat the data to display sagittal or coronal images

Question 30

In nuclear medicine imaging quality control, what is checked daily using standardized long half-life source.

Answer 30

dose calibrator constancy

Question 31

In nuclear medicine imaging quality control, what is checked daily by placing a small amount of a known source of radioisotope in front of the camera

Answer 31

Photopeak window of the pulse height analyzer

Question 32

Most PET scans consists of two parts: an emission scan and a transmission scan. What is the purpose of these two processes?

Answer 32

Emission provides an image of the activity distribution; transmission provides a map of the attenuation. The emission scan provides an image of the distribution of radioactivity within the patient. The intensity of the emission image depends upon the point-to-point thickness of the patient. For this reason a transmission image is performed with an external 511 keV 68Ge transmission source (or alternatively a 137Cs source or CT unit), so as to measure the attenuation profile of the patient (with couch).

Question 33

where is collimation necessary?

Answer 33

in all procedures except PET!

Question 34

The attenuation coefficient of US is strongest in lung or bone?

Answer 34

Lung | Air is a strong attenuator of ultrasound, particularly in alveolar structures

Question 35

dB

Answer 35

10log(I1/I2)

Question 36

doppler mode continuous or pulsed?

Answer 36

Continuous and pulsed Doppler are both available. Meaningful information is displayed by moving and non-moving interfaces (e.g., normal vasculature versus clot).

Question 37

what is duplex US?

Answer 37

Anatomical images and a selectable Doppler window gate simultaneously displayed. Duplex ultrasound refers to the simultaneous acquisition of dynamic B scan grayscale information and Doppler information within a predetermined “gate”.

Question 38

explain about harmonic frequencies

Answer 38

Harmonic frequencies, which are integral multiples of the transducer fundamental frequency, are generated in tissue near mid field along the beam. They are caused by wave distortion. No harmonics are generated in the near field because the wave has not traveled far enough to distort. With increased depth far field harmonics will be attenuated faster than they are being generated. Because they originate from reflecting boundaries, harmonics travel half the distance traveled by the fundamental pulse and would have a stronger signal from deeper structures. Harmonics appear closer to the center of the returning beam with minimal side lobes, hence improved lateral resolution and better gray-scale contrast.

Question 39

what is the lattice for spin-lattice relaxation time?

Answer 39

tissue

Question 40

gadolinium contrast does what?

Answer 40

decreases T1 Gadolinium, a paramagnetic agent, will cause a decrease in the T2 time because of magnetic inhomogeneities causing a more rapid phase dispersion of the transverse magnetization. It also interacts with the lattice of the tissues, causing a decrease in the T1 time

Question 41

A measurement of the cardiac dimensions is obtained from a chest film. The SID is 72" and the heart is midway in a 14" chest. The distance between the chest changer surface and the film is 3". The dimensions on the film are how much bigger or smaller than the actual anatomy

Answer 41

The heart is situated 10" from the film (14"/2+3"). The magnification factor is 72/62 = 1.16. 16% larger

Question 42

equation for magnification

Answer 42

SID/SOD

Question 43

does focal spot size affect dose?

Answer 43

No | affects resolution

Question 44

what contributes to average glandurlar dose in mammo?

Answer 44

The imaging system's target material and geometry, the HVL of the beam, the adiposity, and the degree of compression (which affects thickness) all contribute to the average glandular dose (AGD). The focal spot size affects resolution, especially in magnification views

Question 45

Interventional radiology procedures require significant fluoroscopy, and can deliver patient entrance doses of up to _______ mGy

Answer 45

2000

Question 46

In an anterior spot image of the thyroid, a starburst artifact may be seen. The cause of this artifact is?

Answer 46

septal penetration Septal penetration occurs when photons travel the shortest distance through the lead collimator holes. The star-like appearance is caused by the hexagonal arrangement of holes in the collimator.

Question 47

Parallel opposed lung fields, 13.0 × 17.5 cm, are treated at 100 cm SAD, d=10 cm, to a total dose of 180 cGy/fraction. If the beam energy is 6 MV, the maximum tissue dose in the treated volume will be about _____ % greater than the prescribed dose at the isocenter.

Answer 47

The maximum dose occurs at dmax. The only reasonable answer for this energy and separation is 106% of the midplane dose. To calculate the exact value, use the formula: Dose at depth2 = Dose at depth1 × [TMR(d2)/TMR(d1)] × [dist 1/dist2]2 Entrance dose = 90 × (1.0/0.804) × (100/91.6)2 = 133.4 cGy. Exit dose = 90 × (0.597/0.804) × (100/108.4)2 = 56.9 cGy. Total dose at dmax = 190.3 cGy = 1.057 × 180 cGy.

Question 48

Given a square field and an elongated rectangular field of the same area, which would you expect to have the greater percent depth dose for 4 MV photons?

Answer 48

Scatter from the corners of the rectangle has farther to travel to the field center than scatter from the periphery of the square and will contribute less to the depth dose.

Question 49

Advantage of physical wedge over dynamic wedge

Answer 49

With a dynamic wedge the wedge factor (measured on the beam axis) is a function of both the starting and ending moving jaw position, i.e., both field width and offset.

Question 50

what angle of wedge yields the most homogeneous dose?

Answer 50

For parallel-opposed fields, the most homogeneous distribution is achieved when the wedge turns the isodose lines so that they are perpendicular to the beam axis.

Question 51

what is purpose of beam spoiler?

Answer 51

increase dose in build-up region If only high-energy photons are available and superficial structures would be underdosed, spoilers may be used. The ideal is to maintain a low skin dose and increase dose in the build-up region, to emulate a lower energy beam. However, while it is impossible to exactly mimic a lower energy beam with a spoiler, the build-up characteristics may be preferable to using bolus, which could result in too high a skin dose.

Question 52

lower electron energy does what to skin dose?

Answer 52

decreases it

Question 53

In a single isocenter 4-field breast plan (i.e., using half-blocked fields for the tangents, supraclav, post axilla) the tangents are given a collimator rotation of 10° degrees to align with the chest wall. In order to match the supraclav field to the tangents, without creating a cold area at the match, the supraclav field requires _____ .

Answer 53

a couch rotation of 90° and a gantry rotation of 10°.

Question 54

Adjacent single direct fields of heights 25 cm and 28 cm are matched at 6.0 cm depth. The gap to be left on the skin between the light field edges is _____ cm

Answer 54

gap = (1/2)[coll1 + coll2] × (match depth/SAD) = 0.5 × 53 × 6/100 = 1.6 cm.

Question 55

The collimator rotation required to align 25 × 25 cm cranial fields with a direct spinal axis field of height 36 cm is _____ degrees.

Answer 55

The angle required to rotate the brain fields so that their inferior edge matches the divergence of the upper border of the spine field is: tan−1 [(36/2)/100] = 10°.

Question 56

is penumbra sharper for lower MV photons?

Answer 56

yes, except for Co60, because of large source diameter)

Question 57

A patient’s whole brain is treated isocentrically to 3000 cGy using 6 MV photons. If the calculation was done using 8.5 cm depth, but the patient’s separation was in fact 15.0 cm, the dose received was about _____ cGy.

Answer 57

There is 1 cm less attenuation in each beam, so the dose is increased by approx. 3%.

Question 58

The depth of dmax for an 18 MV photon beam is 3.5 cm. For parallel parallel-opposed 15 × 15 cm fields, with a separation of 20 cm, the minimum depth at which 95% of the midplane dose occurs is _____ cm.

Answer 58

For parallel-opposed fields the higher the photon energy, the more homogeneous the dose profile below dmax, and the lower the skin dose. A possible disadvantage is that dmax is deeper, so tissue in the build-up region can be underdosed. If tissue between the surface and dmax needs to be treated, the actual dose profile should be examined. The depth at which 95% of the midplane dose occurs, for parallel-opposed fields, is generally much less than the depth of dmax, but this depends on field size and separation, and should be calculated for each case.

Question 59

do superficial xrays or MV photons have a sharper penumbra?

Answer 59

superficial have sharper penumbra

Question 60

heterogeneity corrections are greater for 18 MV or 6 MV? Lung or bone?

Answer 60

In a lower energy beam, the attenuation per centimeter is greater, so the difference between soft tissue and lung will be greatest. The corrections for bone and lung are approximately the same, per centimeter, but in opposite directions. The path length in lung is generally greater than in dense bone, so lung corrections usually have the greatest impact.

Question 61

The dose under a 1.5 cm width cord block (5 HVL thickness) in a 6 MV photon beam at 5 cm depth is approximately _____ % of the dose in the open beam.

Answer 61

15%. Five HVLs is equivalent to 3% transmission. The additional dose is due to scatter from the surrounding tissue.

Question 62

A film with a magnification of 1.4 has a block drawn on it that measures 3.5 cm on the film. The actual size of this block on a tray at 65 cm, and the projected size on the patient’s skin at 100 cm SSD are _____ cm and _____ cm, respectively.

Answer 62

By similar triangle geometry: 3.5 cm at 140 is equivalent to 3.5 × (65/140) = 1.6 cm at 65, and 3.5 × (100/140) = 2.5 cm at 100.

Question 63

to improve homogeneity in TBI for parrallel-opposed fields, should you use longer or shorter SAD?

Answer 63

longer SAD = bigger PDD= more homogeneity for POP

Question 64

can a cerrobend block be more conformal than MLC?

Answer 64

yes, it can be cut to match any shape whereas MLCs have a steeped edge Blocks can also have edges to match the beam divergence

Question 65

A radiotherapy department has a choice of 6, 10, and 18 MV photons. For IMRT prostate plans, _____ MV photons are chosen, because:

Answer 65

10, this delivers a lower neutron dose than 18 MV, but acceptable dose to normal tissues.

Question 66

Reasons for fusing an MRI scan with a CT scan for treatment planning brain tumors include all of the following except:

Answer 66

The CT image is required for heterogeneity corrections. Although CT images are required for heterogeneity corrections, these are not necessary for partial brain plans, since the thickness of dense bone in the skull makes very little difference to the effective depth of the photon beams.

Question 67

Orthogonal films of a gynecological applicator are required for dosimetry planning. AP and lateral films are taken, but the lateral film has very poor contrast. All of the following solutions may provide films with acceptable contrast except:

Answer 67

Retake lateral with increased mAs. Increasing mAs will make the film darker, but have no effect on contrast. Contrast is improved by reducing scatter. Reducing the collimator setting to the minimum necessary field of view usually has the greatest effect. If different grids are available, the one with the greatest grid ratio will “clean up” the most scatter. If these two techniques still don’t work, for patients with very large lateral separations one solution is to take orthogonal R and L ant obliques, making use of the somewhat reduced separation. Increasing mAs would increase CNR but not contrast...

Question 68

A 10 × 10 cm 9 MeV electron field has the 90% depth dose at approximately _____ cm depth.

Answer 68

rule of thumb- D90 is at about E/3.. Khan disagrees with this... i.e. 2, 3, 4, 5 Rp, R50, R90, R100

Question 69

where does the tail of electron PDD come from?

Answer 69

Bremsstrahlung x-rays created by electron interactions with high Z components in the head. When electrons interact with high Z components in the head of the linac, bremsstrahlung and a smaller number of characteristic x-rays are produced. This bremsstrahlung “tail” increases with increasing energy, but is usually between 2% and 5%

Question 70

The dose rate in cGy/hr in air at 1 m from a radioactive source is closest numerically to the source strength expressed in units of:

Answer 70

Units of Ci and Bq are related to the number of disintegrations/second of the source. Air Kerma Rate (AKR) has units of dose rate. The recommended units of Air Kerma Strength are μGym2h−1, but this can be converted to cGy m2h−1, i.e., cGy/hr at 1 m.

Question 71

The exposure rate constant is:

Answer 71

8.25 R-cm2/mCi-hr for 1 mg Ra equivalent of any radionuclide. The exposure rate constant is different for each radionuclide, and is the exposure rate for a given activity at a given distance, usually per mCi at 1 cm, or per Ci at 1 m. One “mg Ra equivalent” is the quantity of a radionuclide which has the same exposure at 1 cm as 1 mg of radium.

Question 72

For a Fletcher applicator with loading of 15-10-10 mg-Ra equ in the tandem and 15 mg-Ra equ in each ovoid, a lateral displacement of 0.2 cm in the location of point A would cause about a __________ % change in the dose rate at that point.

Answer 72

This can be calculated roughly using the inverse square law: (2/2.2)2 = 0.83, i.e., 17%. If all sources were 2.0 cm from point A, this would be the exact answer; however, since most sources are further away, the answer is about 10%.

Question 73

The Patterson-Parker system cannot be used for a single plane breast implant, using equal strength 192Ir seeds placed on a 1 cm grid, because:

Answer 73

The system requires unequal strength sources at the periphery and in the center. The Patterson-Parker system is a set of rules which, if followed, delivers an even dose (within prescribed limits) to a plane at a given “treating distance” from the implant. It requires a higher activity (mCi/cm) at the periphery than in the center, to counteract the inverse square effect. For implants with equal activity per seed, the rules do not apply.

Question 74

Does the maximum dose rate on the surface of the container have to be stated on the label when Ir192 are sent to hospital?

Answer 74

No | TI must be there, along with radionuclide and its activity

Question 75

the dose in tissue at 1 cm depth from surface on vaginal cylinder, expressed as a percent of surface dose, is greatest for largest or smallest diameter cylinder?

Answer 75

Largest because IS less significant at greater distance

Question 76

the long term success of intra-vascular brachy depends on what?

Answer 76

- dose gradient across vessel - dose to innermost lumen must not be above or below therapeutic range - dose pattern should be reasonable symmetric along length of irradiated area, and should extend proximally and distally beyond balloon injured area - rate of dose delivery, 2-3 min for beta, 20 min for gamma, is not crucial

Question 77

A Co 60 source is stuck in "on" position, delivering 100 cGy/min at isocenter. The therapist enters the room to rescue the patient and spends 1 min at distance of 1 m from isocenter, Dose received by therapist is about X times weekly dose limit.

Answer 77

I don't agree with this answer. Says weekly dose limit is 50 mSv/yr / 50 = 1 mSv/wk says dose rate at 1 m lat is 1/1000 that at isocenter.