1f Detector A has an ionization energy of 10 eV and Detector B has an ionization

Question

1f Detector A has an ionization energy of 10 eV and Detector B has an ionization energy of 50 eV Detector A will exhibit higher energy resolution than Detector B B) Detector A will exhibit less dead time than Detector B C) Detector A will yield a photopeak with a more narrow FWHM (full-width at half-maximum) value D) A and B E) B and C F) A and C Relative to other scintillation/PMT detectors, sodium iodide detectors (Nal(TI)) are attractive because they A) are hygroscopic B) exhabit high efficiency produce visible light of a wavelength that is matched to the sensitivity of the photocathode A and B B and C A and C E) F) What property of typical semiconductor-based radiation detectors limits their use in medical imaging A) onization energy B) physical density C charge collection time D) hardness iv) The difference between effective dose and equivalent dose is A) effective dose takes into account the particle type, but equivalent dose does not. B) equivalent dose takes into account the particle type, but effective dose does not. C effective dose takes into account the tissue being irradiated, but equivalent dose does not D) equivalent dose takes into account the tissue being irradiated, but effective does does not The benefit of using a single-emulsion film versus a double-emulsion film for a radiographic procedure is that the radiograph produced using the single-emulsion film will ter A) exhibit higher spatial resolution 8) require a shorter exposure time exhibit higher image contrast D) result in reduced patient dose Consider a lower leg radiograph. Spot A on the radiograph represents transmission through 15 cm of soft tissue and Spot B represents transmission through 12 cm of soft tissue plus 3 cm of bone Due to attenuation by bone Spot B will exhibit transmittance and than Spot A optical density higher....higher...lower . C) lower...higher...lower D) higher...lower.. higher

Explanation / Answer

For (i) the answer is (A) because Detector A has less energy than Detector B

(ii) the answer is (C) because NaI counter will have more sensitive to the photo cathode in visible region.

(iii) the answer is (C)

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