Limited Operator
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| When were X-rays discovered? | Novemeber 8, 1895
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| Who discovered X-rays? | Wilhelm Conrad Roentgen
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| Largest radiologic science professional organization - | ASRT
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| Establishes standards and examinations necessary to certify radiologic technologists - | ARRT
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| National organization that formally conducts the accreditation of schools of radiologic technolgy - | JRCERT (Joint Review Committee on Education in Radiologic Technology
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| LXMO | Limited X-ray Machine Operator - the scope of practice is limited compared to registered techonologists
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| Physicians that specialize in interpreting radiographs and preform special imaging procedures - | Radiologist
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| The source of the X-ray Beam - | X-ray Tube
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| The x-ray tube is surrounded by a lead-lined ? | Tube Housing
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| Cross section of the x-ray beam at the point where it is used is called ? | the radiation field
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| An imaginary line in the center of the x-ray beam and perpendicular to the long axis of the X-ray beam - | Central Ray
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| "Unseen" Image | Latenet Image
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| Absorbtion of the X-ray Beam | Attenunation
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| Radiation out side the body that has less engery but is hard to control - | Scatter radiation
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| Causes unwanted exposure to the films and anyone in the room - | Scatter radiation
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| Consists of the x-ray film and film holder | IR (Image Receptor)or Cassette
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| A filmless system that uses digital format with computer technology | CR or Computed Radiography
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| A box like device attached under the tube housing that allows variation in the size of the raditaion field. | Collimator
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| A postition that allows the head to be lowered at least 15 degrees | Trandelenburg Position
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| An oscilating grid under the table top | Bucky
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| T or F The Primary X-ray Beam originates at a tiny point within the x-ray tube | True
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| Unit that represents the rate at which x-rays are produced - | mA or Milliamperage
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| The formula for determining mAs | mA x seconds = mAs
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| Determine the mAs of an exposure made using 300 mA and 0.4 sec | 300 mA x .04 sec = 120 mAs
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| Calculate the exposure time of 50 mA and 40 mAs - | 40 mAs / 50 mA = .8 secs
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| Calculate the mA for the following exposure - 100 mAs and .5 sec | 100 mAs / .5 sec = 200 mA
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| Distance in inches between the radiation source in the x-ray tube and the imaging plane | SID or Source to Image Distance
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| T or F - Raditation intensity is inversly proportional to the square of the distance - | TRUE
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| 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.
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| Beloe 85 kVp, and adjustment of ______ will compensate for small changes in part size - | 2 kVp/cm
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| Above 85 kVp, and adjustment of ________ is necessary | 3 kVp/cm
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| For a 2 cm increase in patient part size, increase the mAs by ________% | 30 (multiply the mAs by 1.3)
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| For a 2cm decrease in patient part size, decrease the mAs by _________ % | 20 (multiple the mAs by 0.8)
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| kVp is ________ to lengthen the scale of contrast - | increased
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| Increasing kVp has these effects - | decreases contrast, increases latitude, and creates a GRAYER image
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| kVp is ________ to shorten the scale of contrast - | decreased
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| Decreasing kVp has this effect - | increased contrast, more black and white appearance
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