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Question | Answer |
---|---|
When were X-rays discovered? | Novemeber 8, 1895 |
Who discovered X-rays? | Wilhelm Conrad Roentgen |
Largest radiologic science professional organization - | ASRT |
Establishes standards and examinations necessary to certify radiologic technologists - | ARRT |
National organization that formally conducts the accreditation of schools of radiologic technolgy - | JRCERT (Joint Review Committee on Education in Radiologic Technology |
LXMO | Limited X-ray Machine Operator - the scope of practice is limited compared to registered techonologists |
Physicians that specialize in interpreting radiographs and preform special imaging procedures - | Radiologist |
The source of the X-ray Beam - | X-ray Tube |
The x-ray tube is surrounded by a lead-lined ? | Tube Housing |
Cross section of the x-ray beam at the point where it is used is called ? | the radiation field |
An imaginary line in the center of the x-ray beam and perpendicular to the long axis of the X-ray beam - | Central Ray |
"Unseen" Image | Latenet Image |
Absorbtion of the X-ray Beam | Attenunation |
Radiation out side the body that has less engery but is hard to control - | Scatter radiation |
Causes unwanted exposure to the films and anyone in the room - | Scatter radiation |
Consists of the x-ray film and film holder | IR (Image Receptor)or Cassette |
A filmless system that uses digital format with computer technology | CR or Computed Radiography |
A box like device attached under the tube housing that allows variation in the size of the raditaion field. | Collimator |
A postition that allows the head to be lowered at least 15 degrees | Trandelenburg Position |
An oscilating grid under the table top | Bucky |
T or F The Primary X-ray Beam originates at a tiny point within the x-ray tube | True |
Unit that represents the rate at which x-rays are produced - | mA or Milliamperage |
The formula for determining mAs | mA x seconds = mAs |
Determine the mAs of an exposure made using 300 mA and 0.4 sec | 300 mA x .04 sec = 120 mAs |
Calculate the exposure time of 50 mA and 40 mAs - | 40 mAs / 50 mA = .8 secs |
Calculate the mA for the following exposure - 100 mAs and .5 sec | 100 mAs / .5 sec = 200 mA |
Distance in inches between the radiation source in the x-ray tube and the imaging plane | SID or Source to Image Distance |
T or F - Raditation intensity is inversly proportional to the square of the distance - | TRUE |
T or F - kVp is not a useful adjustment for relativly small variations from normal technique | False - kVp IS a useful adjustment for realativly small variations from normal, because large changes in kVp cause significant alterations in appreance. |
Beloe 85 kVp, and adjustment of ______ will compensate for small changes in part size - | 2 kVp/cm |
Above 85 kVp, and adjustment of ________ is necessary | 3 kVp/cm |
For a 2 cm increase in patient part size, increase the mAs by ________% | 30 (multiply the mAs by 1.3) |
For a 2cm decrease in patient part size, decrease the mAs by _________ % | 20 (multiple the mAs by 0.8) |
kVp is ________ to lengthen the scale of contrast - | increased |
Increasing kVp has these effects - | decreases contrast, increases latitude, and creates a GRAYER image |
kVp is ________ to shorten the scale of contrast - | decreased |
Decreasing kVp has this effect - | increased contrast, more black and white appearance |