RAD112 Chapter 6 X-Production PDF
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Uploaded by DivineSkunk5245
Stellar Career College
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This document includes questions and answers on x-ray production, including topics such as characteristic radiation, bremsstrahlung radiation, beam quality, and half-value layer (HVL). The questions are likely from a radiation physics and radiobiology course at Stellar College.
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**RAD112 Chapter 6 X-Production** [RAD 0724\_RAD 112 Radiation Physics and Radiobiology I](https://lms.stellarcollege.edu/course/view.php?id=1811#section-7) Question **1** As the kVp increases beam energy and beam penetrability increases (As the kVp increases, the beam energy and penetrability...
**RAD112 Chapter 6 X-Production** [RAD 0724\_RAD 112 Radiation Physics and Radiobiology I](https://lms.stellarcollege.edu/course/view.php?id=1811#section-7) Question **1** As the kVp increases beam energy and beam penetrability increases (As the kVp increases, the beam energy and penetrability increases.) Question **2** During the bremsstrahlung interaction, the filament electron is attracted to the nucleus but not absorbed (During the bremsstrahlung interaction, the filament electron is attracted to the nucleus, causing it to slow down and change direction.) Question **3** Characteristic x-ray photons result when an inner-shell electron is knocked out and when outer-shell electrons fill the vacancy in an inner shell (To produce characteristic radiation, outer-shell electrons must drop into an inner-shell vacancy created by the filament electron knocking out the orbital electron.) Question **4** Filtration placed in the path of the x-ray beam absorbs low-energy photons (Beam filtration reduces beam quantity by absorbing low-energy photons.) Question **5** As the distance increases, the beam quantity reaching a specific area Decreases (Increasing the distance results in the number of photons in a specific area being decreased.) Question **6** For a tungsten target, the discrete line on the x-ray emission spectrum is approximately 69 keV (The discrete line is at 69 keV, based on a tungsten target.) Question **7** *Bremsstrahlung *means braking radiation (*Bremsstrahlung *is the German word for \"braking\" or \"slowing down\" radiation.) Question **8** A discrete emission spectrum is a graphic representation of characteristic radiation Question **9** A higher energy beam is said to be a \_\_\_\_\_\_ beam. hard Question **10** Changing from a high-frequency generator to a single-phase generator results in the x-ray emission spectrum- changing along the *y*-axis changing along the *x*-axis Question **11** Changes along the *x*-axis of the x-ray emission spectrum reflect changes in the \_\_\_\_\_ of the x-ray beam. quality Question **12** Half-value layer (HVL) is used to measure beam quality Question **13** A continuous emission spectrum is a graphic representation of bremsstrahlung radiation Question **14** Changing the \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ results in changes to the *x*-axis, *y*-axis, and location of the discrete line of the x-ray emission spectrum. target material Question **15** *Beam quality *refers to the energy level of the radiation Question **16** A filament electron removes a K shell electron and an M shell electron fills the vacancy. The K shell binding energy is 69.5 keV and the M shell binding energy is 2.8 keV. What is the energy of the K-characteristic photon produced?- The energy of the photon produced is determined by subtracting the binding energy of the outer shell electron from the binding energy of the inner shell electron (69.5 - 2.8 = 66.7) 66.7 keV. Question **17** At the anode target, how much of the energy from filament electrons is lost as heat and how much will result in x-ray production. 99% energy lost as heat; 1% result in x-rays Question **18** Adding filtration in the path of the beam results in the x-ray emission spectrum changing along the *y*-axis changing along the *x*-axis both of these Question **19** How many HVLs are needed to reduce beam intensity from 600 mR to 300 mR? One. Question **20** During x-ray production the electrons penetrate the target approximately 0.5 mm