Midterm Review Sheet

Question 1

OID on Magnification

Answer 1

Increased OID increases image magnification

Question 2

SID on magnification

Answer 2

Increased SID decreases magnification

Question 3

Image sharpness is affected by

Answer 3

Focal spot size
OID - Increases most rapidly with OID
SID

Question 4

Penumbra

Answer 4

Image unsharpness created because photon are being released from multiple sources and not just one single source

Question 5

Inverse Square law

Answer 5

I(1) D(2)2
— = —-
I(2) D(1)2

The outcome of the equation is the intensity change of the beam from the original
To know a new technique based on this you would divide the original mAS by the answer to the inverse Square Law. This would give a new mAS that worked for the new distance.

Question 6

mAs rule

Answer 6

mAs(1) D(1)2
— = —-
mAs(2) D(2)2

Answer to this formula will give the answer used to determine a new technique

Question 7

Precalculated estimates for changes in SID

Answer 7

40 to 56 = multiply mAs by 2
40 to 72 = multiply mAs by 3
40 to 80 = multiply mAs by 4

Question 8

Size distortion

Answer 8

Always magnification
Due to divergent beam
Fecreasing SID or increasing OID will be most effective with OID being prime

Question 9

Shape distortion

Answer 9

A result of unequal magnification of the shape of the structure
Occurs with a different OID of one end or the other and will appear fatter and foreshortened
Occurs with When the x-ray tube is angled and everything else is parallel this results in elongation

Question 10

Safe Lights inside darkrooms must be

Answer 10

At least 4 feet from work area
Must not exceed 15 watts
Covered by a filter that only allows light of a spectral emission that is not within a damaging range (only allows light from the red end of the visual spectrum)
Films should not be exposed to light for three minutes

Question 11

Safe light test

Answer 11

Expose an underdeveloped film to the light of the room and every thirty seconds move the film so that more light is exposed

First stripe will be most exposed and give an indication of what time length is safe

Question 12

The three T’s of processing

Answer 12

Time, Temperature, Titration
Increased TTT will overdevelop film and produce a fog whivh results from the development of unexposed silver halide processing

Question 13

Protocol for manual processing

Answer 13

68 degrees; development for 5 minutes; fixer for 10 minutes; Washing for 20 minutes; drying 20-40 minutes

estimated overall time is one hour per film

Question 14

Image exposure

Answer 14

Film is exposed to light
Exposure causes silver bromide crystals to seperate into a silver ion and a bromide ion
Sulfur in the crystal collects electrons and holds them away from the silver ion
Creates latent image

Question 15

Image development

Answer 15

Developer donates electrons that creates elemental silver (which is black in color)
Latent image becomes the visible image

Question 16

Image fixing

Answer 16

Fixer removes all unexposed silver halide from the emulsion and permenantly stops any possibility of exposure beyond what is already done

Question 17

Washing

Answer 17

Final wash throughly clears all chemicals from the film surface and in the emulsion
Prevents fixer chemicals from continuing to cause chemical changes in the emulsion after the finished product is produced
Insufficient washing causes artifacts as teh chemical continue to work

Over time film will degrade (mos. to yrs.)

Question 18

Developer

Answer 18

Activator - Sodium carbonate softens gelatin protective cover on film (preserves pH at 7

Reducing agents - (key ingredient) Hydroquinone or phenidione (metol or enol) reduces the exposed silver halide to black metallic silver

Preservative - Sodium sulfite

Restrainer - Potassium Bromide, acetic acid

Question 19

Fixer

Answer 19

(pH 4-4.5)
Clearing agent - Ammomium thiosulfate (key ingredient) Dissolves undeveloped silver halide and removes it from emulsion

Tanning agent - potassium alum

Activator - acetic Acid neutralizes the developer, Stops development and maintains pH

Question 20

Automatic developer system

Answer 20

Higher temps (92-96)
Stronger acids and bases than hand-tanking solutions
Constant agitation of chemicals
Rapid throughput of chemicals (90-80 secs)

Chemical require replenishment
Under-replenishment causes loss of contrast (grey) and loss of film density
Over-replenishment causes loss of contrast (increased fog) and wastes chemicals

Question 21

Causes of lowered contrast on a film

Answer 21

Safe light fog 
Room temperature or humidity too high 
Chemical fog from processor too hot
Chemical fog from chemicals too strong
Film is past expiration

Question 22

Every 1-2 months you must

Answer 22

Complete chemical changes and tank cleaning
Check replenisher rates
Check cycle time

Question 23

Sensitometer

Answer 23

Exposes film to a step wedge of known densities

Question 24

H & D Curve

Answer 24

Graphic measure of how a film records density over a range of exposures
Graph plot optical density (y-axis) vs. log relative exposure
Consists of the body, toe, and shoulder of the curve

Question 25

Toe of the curve

Answer 25

Starts above zero optical density even in the absence of any exposure because there is ALWAYS a small amount of light absorption by the film due to the base and emulsion plus a small amount of “auto exposure” that occurs with any emulsion of film

Question 26

Body of the curve

Answer 26

Straight line component of the curve Slope of teh curve determines the latitude of the film Short latitude film has a very High slope and is higher contrast Long latitude film has a lower slope and a lower contrast

Question 27

Shoulder of the curve

Answer 27

represents the D-max or maximum optical density the film will achieve no matter what

Question 28

Comparison of H and D surves

Answer 28

For two different types of films the line further to the left is faster. The one with a lower curve is a more latitude film with lower contrast

Question 29

Densitometer

Answer 29

Machine with the little arm on it that precisely measures the optical density of each step on a step wedge. This is reported on a logarithmic scale from 0 to about 4. Zero indicates zero absorption of light through the film- or the film transmits 100% of the incident light through the film. Optical density of 4 indicates 100% absorption of light- the film is entirely black and allows no transmission of light through it. Densitometer can tell us how much light transmits through any one of the steps on a step wedge in very precise ways

Question 30

Measurements we obtain to track performance of the processor over time:

Answer 30

Speed Index Contrast Index Gross fog index

Question 31

Speed index

Answer 31

Established by finding the step on the sensitometric step wedge strip that has the closest optical density to 1.0 (based on measures using a densitometer)

Question 32

Contrast index

Answer 32

Difference in optical density between the Speed Index step and the optical density reading (using the densitometer) of the step that is two steps darker on the sensitometric strip Should be about 1.4-1.7 for films used to obtain skeletal images (lower for films for chest images)

Question 33

Gross fog index

Answer 33

is film Base plus Fog It is the inherent optical density of the film when it has not been exposed at all and has just run through the processor blank Ideally it should be perfectly clear however there is always some stray radiation that causes silver bromide to break into silver and bromine and that then results in some elemental silver

Question 34

Usual Gross Fog Index

Answer 34

0.15 to 0.20 faster films have a higher gross fog index Length of the toe of teh H and D curve is related to the film's sensitivity to fog The longer the toe the less sensitive to fog

Question 35

Should not see reading greater than these values on serial (over months) recordings

Answer 35

+/- 0.15 for speed index +/- 0.15 for contrast index +/- 0.03 for gross fog index

Question 36

Spectral sensitivity of film

Answer 36

The bandwidth of light that it is most sensitive to during exposure Match the spectral emission of the screens that it is paired with

Question 37

Speed vs. detail film

Answer 37

Function of crystal size Slow speed with best detail: small crystals in thin layers Medium speed with medium detail: Medium sized crystals in a medium thick layer on the film Large crystals in a thin layer Small crystals in a thick layer Fast speed with low detail: large crystals in thick layers

Question 38

Latitude

Answer 38

Term used to describe the range of densities that can be recorded on the film Wide latitude film is best when you desire low contrast radiographs (i.e. chest radiograph) Short latitude film is best when you desire high contrast radiographs As contrast decreases… latitude increases (“Longest (widest) latitude”)

Question 39

Increasing film speed causes

Answer 39

More density per unit of exposure less mAs needed to create the film density that you want Faster films do not produce as much detail

Question 40

Fog may be caused by: TTT

Answer 40

Secondary radiation (scatter) Accidental radiation exposure Excessive or insufficient time, temperatures or chemical titration strength Exposure to darkroom chemical fumes

Question 41

Interactions of x-ray with matter

Answer 41

Coherent scattering: same energy of incident photon photon hits, spins, and is released without breaking apart ``` Compton Effect (key feature: Compton scatter is the most common mechanism of scatter production in human tissue): photon hits and scattered radiation is released ``` Photoelectric effect: Hits a positively charged atom, releases characteristic radiation and an excited photon

Question 42

Things that affect scatter production

Answer 42

The greater the tissue volume the greater the scatter produced (more than 10 cm and a grid is advised) Increasing the kvp Increases the energy of the beam and increases the number and energy of scattered photons. Can also lead to increased fog due to increased scatter (use high ratio grid to absorb high energy scatter)

Question 43

Minimize scatter production

Answer 43

Grid - place grid between patient and film to absorb scatter Air gap technique - Equal effect to an 8:1 grid Minimize exposed tissue through collimation May decrease kVP or mAs put this is risky

Question 44

Grid performance measures

Answer 44

Gird ratio: Height of lead strips: distance between lead strips 10:1 or 12:1 is best Grid frequency: Number of lead strips per square inch Bucky factor - Ratio of x-rays arriving at the grid to those transmitted through the grid Contrast improvement factor: the ratio of image contrast with a grid vs. image contrast without a grid

Question 45

Focused grids

Answer 45

Lead strips are aligned to the direction of the diverging primary beam Grid radius are the specific SID the grids are attuned to Grid focal range is the range of SIDs that will function with grid Grid cutoff: when the grid absorbs excessive amounts of remnant radiation resulting in visible and undesirable degradation of the image

Question 46

Cross hatch grid

Answer 46

lamination of 2 grids with lead strips at 90 degree angles to each other

Question 47

Stationary girds should

Answer 47

have at least 100 lines per inch

Question 48

Grid cutoff

Answer 48

Undesireable attenuation of the primary beam by grid | Oblique grids cause loss of image density on the side closer to the tube

Question 49

Sir gap technique

Answer 49

Increases OID 10 80-12 inches in order to reduce the volume of scatter that hits the film Must increase the SID to compensate for the magnification caused This in turn causes more mAs to accommodate for the increased distance Air gap will cause the scatter to miss the film so grid is not necessary

Question 50

Density

Answer 50

Level of overall darkness in the film 25-30% change is required for a visible effect on the film Film too dark: Over penetrated, too dense, overexposed, "midnight under a skillet" Film too light: Under penetrated, too light, Under exposed

Question 51

mAs and number of electrons

Answer 51

Relates to the number of electrons required required to make the beam QED. the number of photons in the beam

Question 52

mAs varies directly with the image density

Answer 52

``` Double = double the optical density of the image Half = cut the density in half ```

Question 53

When manipulating image density we try to use only ___

Answer 53

mAs It affects density in a linear fashion and is easily changed on control panel Few unintended consequences

Question 54

Occasionally, manipulating milliamperage (mA) separate from time (s) may be desirable because

Answer 54

Lengthening or shortening the exposure time to allow breathing during exposure or to stop motion in a tremor Can select a lower mA to get a smaller focal spot for greater detail and use a longer exposure time Using a lower mA setting and a long exposure to reduce the heat load build-up during large exposures (allows better heat dissipation during the exposure, thus lowering peak tube load)

Question 55

kVp effect on image density

Answer 55

As the energy of the beam increases a higher percentage of the photons in the beam will get through the target and hit the film Relationship is NOT linear Increasing the kVp 15% will double the density and lower the contrast Decreasing the kVp 15% will half the density and raise the contrast

Question 56

SID and image density

Answer 56

Increasing SID reduces beam intensity Decreasing SID increases beam intensity Effect is not linear - inverse square law Decreasign SID increases penumbra because the more divergent rays of the beam are being used

Question 57

Grids and image density

Answer 57

Grids selectively filter scatter and allows beam to pass through Scatter adds darkness in the form on non-image producing fog and this is always bad Grids also absorb some remnant radiation (the primary beam that would have made an image, this is BAD) The image will be underexposed unless one compensates with increased mAs higher contrast, less fog greater detail, higher radiation to patient

Question 58

Effects of a smaller focal spot

Answer 58

No magnification change Less penumbra Increased detail/sharpness

Question 59

Effects of a smaller OID

Answer 59

Less magnification Decreased penumbra Increased detail/sharpness

Question 60

Effects of Larger SID

Answer 60

Less magnification Decreased penumbra Increased detail

Question 61

Best sharpness with least magnification is achieved with

Answer 61

Smallest OID and Largest SID OID has a greater effect than SID

Question 62

Grid cutoff with grid characteristics

Answer 62

Off-level grid will cause visible cut-off on the side closest to the tube an inverted grid will acuse immediate and severe cut off beyond the central ray region

Question 63

CT

Answer 63

Modern scanners have multiple detector arrays that are stationary around the entire 360 degrees of the gantry rotations around the patient. The tube rotates around the patient in the axial plane Helical CT scanners acquire data continuously as the patient is moved within the gantry, creating a “spiral” acquisition of data Multi-slice scanners create the ability to acquire data in very thin slices creating what is referred to as volumetric acquisition. When the data is reformatted the data in total is as good as the axial data

Question 64

Reformatting (MPR= Multi Planar Reformatting)

Answer 64

Views of data from multiple orientations Helical CTs can produce an isotropic voxel (pixel with volume) which has the same depth width and height meaning that it can be viewed from any plane without losing resolution

Question 65

Hounsfield units

Answer 65

Used to describe the shade of gray that a tissue presents on a scan (-1000 to 3072) ``` water = 0 HU Air = -1000 HU Cancellous bone = 400 HU cortical bone = 700-2000 HU Metal = 2500 to 3000 HU ```

Question 66

Windowing

Answer 66

The process of emphasizing certain ranges in the HU scale Soft tissue windows allow fo rbetter viewing of the low level ends of the spectrum Bone values emphasize the higher end of the HU spectrum

Question 67

Creating the MRI image

Answer 67

Strong magnetic field (.3 to 3 teslas) is used to align the vector of spin of protons in the body A radiofrequency will be introduced to cause resonance of protons causing them to gain energy and spin away from their axis and wobble away from the dead center orientation at a higher energy when the radiofrequency is turned off the excess energy is expelled from each tissue as the precession returns to the aligned state of the magnet. The strength and character of the emitted energy is unique to the proton concentration of in that specific tissue and can be differentiated from other tissues Radiofrequency detectors pick up the emitted signals and are converted to a visible image

Question 68

T1 weighted image (MRI)

Answer 68

Fat is brightest Cord higher signal than CSF Medullary bone and spinal cord are higher signal water and cortical bone are lower signal

Question 69

T2 weighted image

Answer 69

``` Water and edema have brightest signal CSF higher than spinal cord Less precise detail Bright nucleolus pulposus fat loses signal Cortical bone is still dark ```

Question 70

STIR (Short Tau Inversion Recovery) images (AKA fat saturation sequences)

Answer 70

Suppresses fat signal (makes it appear black) so that edematous tissue and fluid can be better seen.

Question 71

Signal intensity and description

Answer 71

High signal = bright or light or white on the image  Low signal = dark or black on the image  Intermediate = anywhere in between

Question 72

Contraindications to MRI

Answer 72

Anything with batteries or ferrous metals (those including iron) Any patient who can’t hold still due to seizures, etc.  Patients who are clinically unstable and require electrical monitoring equipment in the MRI suite to ensure their medical stability Anything that has a battery or relies on radiofrequency signal to operate Insulin or infusion pump o Implanted drug infusion device Bone growth/fusion stimulator Cochlear, otologic, or ear implant

Question 73

CT vs. MRI

Answer 73

CT uses X-rays (ionizing radiation), MRI uses radiofrequency and magnets CT refers to densities in tissue; MRI refers to signal intensity CT provides exemplary detail; MRI provides exemplary tissue differentiation CT axial data can be reformatted in multiple planes data; MRI directly acquires multiplanar images- no reformatting o CT is very fast; MRI is not so fast CT can be done on patients with metal or battery implants; MRI can only be used if there are no batteries or ferrous metal in the patient CT can be done on clinically unstable patients; MRI is harder to use with clinically unstable patients

Question 74

Scintigraphy

Answer 74

Uses radioisotopes chelated to substances that are inert, but processed by various tissues in the body Technetium 99m is the most common Hot spot is an area of increased physiological activity Cold spot is an area lacking vascular supply

Question 75

Scintigraphy identifies ___

Answer 75

Physiological activity in organs and bones Characterizes areas of abnormally high or low metabolic activity High quality screening examination: Highly sensitive for areas of increased bone metabolism but poorly specific

Question 76

Types of sctintigraphy

Answer 76

2D SPECT - Single Proton Emission Computed Tomography Hybrid SPECT - Combining CT or MRI with scintigraphy to gain anatomic or hysioogical comparison

Question 77

Scintigraphy scan comprehensiveness

Answer 77

Whole body - entire skeletal system Limited - for specific region and usually for fractures Three phase - 3 phases - Flow, Blood pool, Delayed images.Used for infection, especially to determine osteomyelitis from cellulitis (extremely important to detect early infection in diabetics)

Question 78

Ultrasonography

Answer 78

Use of sound waves to create images Changes in tissue density will cause sound waves to “bounce back” to the ultrasound transducer at differing rates, creating the images we see on the scan. The transducer is a device that sends sound waves and has receiver microphones that detect echoes that come back… so it sends and hears sound waves. The typical diagnostic ultrasound wave frequencies range from 2-20 MHz (that’s 2-20 million Hz). Air will ruin image capture Because it is dynamic, it is highly user dependent Does not employ radiation

Question 79

Strengths of an ultrasound

Answer 79

Obstetric, gynecological and genitourinary evaluation Solid internal organ evaluation o Cardiac, abdominal and pelvic evaluation Large and medium blood vessel evaluation (arteries or veins) Evaluation of large and medium size joints- ligaments, menisci, tendons, labra, cartilage surfaces, fluid accumulations, intraarticular plica, etc. Muscles, tendons, ligaments and fascial bands, syndesmoses, compartment boundaries, etc. in the MSK system Spinal evaluation is limited to identifying perispinal structures and measuring muscle volume Intraoperative and interventional applications  Biopsies Needle or device placement guidance Localizing tissues at surgical depth