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147 Final
| Question | Answer |
|---|---|
| K Shell Energy Level | 69.5 keV |
| L Shell Energy Level | 12.1 keV |
| Characteristic X-Ray's will always have ______ | 57.4 keV |
| An Increase in kVp will result in Decreased _______ | Contrast |
| An Increase in kVp will result in Increased _______ | Frequency |
| Coherent Scatter Energy | 10 keV or less |
| Pair Production Energy | 1.02 MeV |
| Photodisintegration Energy | Above 10 MeV |
| As _______, the proportion of Compton interactions relative to photoelectric interactions increases | kVp increases |
| Photons pass through the patient without interacting at all | Transmission |
| Reduction in number of x-rays due to partial or complete absorption | Attenuation |
| A Higher Atomic Number, Density, and amount of matter/tissue thickness increase the probability for _________ to occur | Scatter & PE Interaction Production |
| _________ is the basis for contrast of an image | Differential Absorption |
| _______ is the electric potential created by the generator | Voltage |
| Electron Cloud size depends on the _______ selected | mA |
| Projectile Electron (e-) needs a minimum of _______ to produce characteristic radiation | 70 kVp |
| Brems X-Rays have a _______ range | 20 – 150 kVp |
| kVp set between 70-100 will result in ______ of the beam being characteristic x-rays | 15% |
| Less than 70 kV means that _____ of x-ray beam is Bremmstrahlung | 100% |
| No X-rays can be produced with ________ | Alternating Current |
| During Half Wave Rectification (single phase, single pulse), X-Rays are produced _______ | Only during the positive half cycle |
| During Full Wave Rectification (single phase, single pulse), X-Rays are produced _______ | During both positive and negative halves of the cycle |
| A 3 phase 6 pulse generator has only ______ Ripple | 13% |
| A 3 phase 12 pulse generator has only ______ Ripple | 3% |
| A High Frequency generator has _____ Ripple | <1% |
| Less Ripple is ________ | More Efficient |
| Heat Units Formula | kVp x mA x seconds x c number |
| The _______ is used to determine the new mAs needed to maintain receptor exposure if the SID changes | Direct Square Law |
| Magnification Factor Formula | Image Size ______________ (Divided By) Actual Image Size |
| SOD + OID = | SID |
| SID - OID = | SOD |
| SID - SOD = | OID |
| For every change of 4 to 5 cm, _______ | Multiply or Divide the mAs by 2 |
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user-2016295