Set the Language

We weren't able to detect the audio language on your flashcards. Please select the correct language below.

Front

Back

Flashcards
»
PHYSICS 2002

Physics 2002

by N8BEME9, Aug 2010

Subjects: 2002 physics

Favorite

Add to folder

Flag

Shuffle
Toggle On

Toggle Off
Alphabetize
Toggle On

Toggle Off
Front First
Toggle On

Toggle Off
Both Sides
Toggle On

Toggle Off
Read
Toggle On

Toggle Off

Reading...

Front

Card Range To Study

through

Play button

Progress

1/96

Click to flip

Use LEFT and RIGHT arrow keys to navigate between flashcards;

Use UP and DOWN arrow keys to flip the card;

H to show hint;

A reads text to speech;

96 Cards in this Set

Front
Back

	1. PET detects: A. Positron particles B. Gamma rays C. Beta particles D. Annihilation photons	1. PET detects: A. Positron particles B. Gamma rays C. Beta particles D. Annihilation photons Answer: D
	2. Risk of malignancy in women is greatest at what age: A. <10 B. 10-20 C. 20-30 D. 30-40 E. 40-50	2. Risk of malignancy in women is greatest at what age: A. <10 B. 10-20 C. 20-30 D. 30-40 E. 40-50 Answer: E – Risk of malignancy in general is higher as you age, but younger women are more at risk for cancer due to the effects (stochastic) of radiation.
	3. Fat attenuates 3 dB/cm and you use a 5MHz transducer. If the lesion is 5 cm deep, what is the reduction in intensity at the transducer? A. 10 B. 13 C. 100 D. 130 E. 1000	3. Fat attenuates 3 dB/cm and you use a 5MHz transducer. If the lesion is 5 cm deep, what is the reduction in intensity at the transducer? A. 10 B. 13 C. 100 D. 130 E. 1000 Answer: E – 5cm there and 5cm back equals 10cm. 10cm x 3dB/cm = 30dB. Plug into equation: dB = 10 log I/I 30 = 10 log I/I 3 = log I/I I/I = 103 = 1000 A short cut is to divide the dB loss by ten and raise 10 to that number.
	4. The Larmour frequency of a proton with spin (1/2) is 100MHz when put in a magnetic field of 4T. What is the gyromagnetic ratio? A. 25 B. 100 C. 200 D. 4000	4. The Larmour frequency of a proton with spin (1/2) is 100MHz when put in a magnetic field of 4T. What is the gyromagnetic ratio? A. 25 B. 100 C. 200 D. 4000 Answer: A – 100MHz/4T = 25 MHz/T Any free system with a constant gyromagnetic ratio, such as a rigid system of charges, a nucleus, or an electron, when placed in an external magnetic field B (measured in teslas) that is not aligned with its magnetic moment, will precess at a frequency f (measured in hertz), that is proportional to the external field: . For this reason, values of γ/(2π), in units of hertz per tesla (Hz/T), are often quoted instead of γ.
	5. With DICOM which of the following is correct? .A. Increasing bandwidth allows faster transmission B. Refresh rate less than 5%	5. With DICOM which of the following is correct? .A. Increasing bandwidth allows faster transmission B. Refresh rate less than 5% Answer: Probably A.
	6. If an object transmits 5% of light, what is OD? A. 1 B. 1.3 C. 2 D. 3 E. 4	6. If an object transmits 5% of light, what is OD? A. 1 B. 1.3 C. 2 D. 3 E. 4 Answer: B – OD = log (I/Io). Transmittance is Io/I. So OD = log (1/T). OD = log (1/0.05) OD = log 20 = 1.3
	7. If you switch from parallel to pinhole collimation you increase A. Resolution B. Contrast C. Radiation D. Sensitivity	7. If you switch from parallel to pinhole collimation you increase A. Resolution B. Contrast C. Radiation D. Sensitivity Answer: A
	8. What is the MTF of a system given that the four components of the system are 1.0, 0.5, 0.8, and 0.7? A. 1.0 B. 0.50 C. 0.28 D. 0.14	8. What is the MTF of a system given that the four components of the system are 1.0, 0.5, 0.8, and 0.7? A. 1.0 B. 0.50 C. 0.28 D. 0.14 Answer: C – Remember MTFs multiply. ODs add.
	. Most tissue damage caused by ionizing radiation is through A. Electrons B. Neutrons C. Protons D. Pi Mesons	9. Most tissue damage caused by ionizing radiation is through A. Electrons B. Neutrons C. Protons D. Pi Mesons Answer: A – on multiple old tests.
	10. If a 10 microcurie dose of an agent is needed at 1:30 PM for a test, what dose should be prepared at 8 AM if the half life is 110 minutes? a. 150 b. 95 c. 80 d. 20	10. If a 10 microcurie dose of an agent is needed at 1:30 PM for a test, what dose should be prepared at 8 AM if the half life is 110 minutes? a. 150 b. 95 c. 80 d. 20 Answer: C – Here is how you do this one: Decay contant (λ) = 0.693/110min = 0.0063 min--1 10mCi = N e-(0.0063/min)(330min) N = ~80mCi
	12. A pregnant technologist should be monitored/restricted to assure that the fetal dose is for the entire pregnancy A. 5 mSv (500mrem) B. 50mSv (5000mrem) C. 0.5mSv (50mrem) D. 0.005mSv (5mrem)	12. A pregnant technologist should be monitored/restricted to assure that the fetal dose is for the entire pregnancy A. 5 mSv (500mrem) B. 50mSv (5000mrem) C. 0.5mSv (50mrem) D. 0.005mSv (5mrem) Answer: A – remember that the limit is 50mrem/mo as well. I remember the two numbers by thinking the pregnancy is really about 10 months (40wks), so: 50mrem/mo x 10 months = 500mrem/pregnancy
	13. Most of the energy deposited in biological tissue from diagnostic radiation is from: A. protons B. the photoelectric effect C. electrons	13. Most of the energy deposited in biological tissue from diagnostic radiation is from: A. protons B. the photoelectric effect C. electrons Answer: C – The bottom line is most of the energy deposited in tissues is from electrons, and the photoelectric effect deposits all its energy locally. Compton scatter electrons also deposit energy locally.
	14. If an ultrasound pulse returns to the transducer 20 microseconds after being emitted, how deep in the tissue is the object? (velocity not provided) A. 3.0 cm B. 1.5 cm C. 0.77 cm D. 7.7 cm E. 15 cm	14. If an ultrasound pulse returns to the transducer 20 microseconds after being emitted, how deep in the tissue is the object? (velocity not provided) A. 3.0 cm B. 1.5 cm C. 0.77 cm D. 7.7 cm E. 15 cm Answer: B – remember the simplified equation: Depth (cm) = 0.077 x time (microseconds) Depth (cm) = 0.077 x 20 microseconds = 1.5 cm
	15. A radiation POINT SOURCE is used to evaluate a SPECT machine with the collimator ON at a variety of angles. What function is being measured? A. system dead time B. something about linear response at a variety of angles C. Center of Rotation D. Flood view E. Intrinsic resolution	15. A radiation POINT SOURCE is used to evaluate a SPECT machine with the collimator ON at a variety of angles. What function is being measured? A. system dead time B. something about linear response at a variety of angles C. Center of Rotation D. Flood view E. Intrinsic resolution Answer: C – Center of Rotation is tested with a point source at varying angles with the collimator.
	16. Narrowing the optical aperture of a fluoroscopy system is likely to do which of the following? A. Increase exposure B. Reduce the field of view C. Decreases the f-number D. Decrease exposure	16. Narrowing the optical aperture of a fluoroscopy system is likely to do which of the following? A. Increase exposure B. Reduce the field of view C. Decreases the f-number D. Decrease exposure Answer: A – decreasing aperture lets less light through and the AEC bumps up radiation to make up the difference. The f-number is inversely related to the size of the whole. Narrowing the aperture will increase the f-number.
	17. In mammography, a source to image distance (SID) of 600 to 750 mm is used rather than the standard 1000 mm of other modalities why? A. To reduce the heel effect B. To allow for a stronger beam and a shorter exposure C. To accommodate the focal spot of the gird D. To improve geometric resolution throughout the breast E. To decrease magnification	17. In mammography, a source to image distance (SID) of 600 to 750 mm is used rather than the standard 1000 mm of other modalities why? A. To reduce the heel effect B. To allow for a stronger beam and a shorter exposure C. To accommodate the focal spot of the gird D. To improve geometric resolution throughout the breast E. To decrease magnification Answer: B – on multiple old tests.
	18. Which factors are most important for reducing image blur? The possible answers were multiple combinations of these terms: Screen thickness Emulsion thickness film-screen contact	18. Which factors are most important for reducing image blur? The possible answers were multiple combinations of these terms: Screen thickness Emulsion thickness film-screen contact Answer: I think they want screen thickness here.
	19. Which change will decrease the ratio of radiation entering to radiation exiting the patient? A. decreasing xray tube filtration B. decreasing the half value layer C. decreasing the kVp D. increasing the source to image distance while maintaining the subject to image distance	19. Which change will decrease the ratio of radiation entering to radiation exiting the patient? A. decreasing xray tube filtration B. decreasing the half value layer C. decreasing the kVp D. increasing the source to image distance while maintaining the subject to image distance Answer: D – you need to understand this type of question because it can be asked in a number of ways. The other options will make the beam “softer” and more radiation enters the patient that does not contribute to the image.
	20. A single exposure of 1 Gy (whole body) is likely to cause: a) skin erythema b) partial cataracts c) diarrhea in 50% of the population d) decreased absolute lymphocyte count	20. A single exposure of 1 Gy (whole body) is likely to cause: a) skin erythema b) partial cataracts c) diarrhea in 50% of the population d) decreased absolute lymphocyte count Answer: D
	21. Pertaining to the image receptor in mammography, which of the following is true: a) the screen moves to decrease grid lines showing b) the speed is 50 c) there are two screens in each cassette d) digital radiography is never used.	. Pertaining to the image receptor in mammography, which of the following is true: a) the screen moves to decrease grid lines showing b) the speed is 50 c) there are two screens in each cassette d) digital radiography is never used. Answer: B – Mammography cassettes use only a single thin screen, so the speed will be less than regular radiography. The grid lines have nothing to do with the receptor. The grid itself moves, not the screen. Read each option carefully!
	22. The Gibb’s phenomenon in 3rd generation CT scanners refers to the dark appearance around the brain directly adjacent to dense bone. This is caused by: a) beam hardening b) algorithm differences c) patient movement d) angry leprechaun	22. The Gibb’s phenomenon in 3rd generation CT scanners refers to the dark appearance around the brain directly adjacent to dense bone. This is caused by: a) beam hardening b) algorithm differences c) patient movement d) angry leprechaun Answer B -- be sure to read about Gibbs because they love to ask about it for some reason. It can be seen in CT or MRI and is due to algorithm difference (not enough PEG in MR, unsure about CT).
	23. Why is the radionuclide F-18 more difficult to use than I-131 or Tcm99? a) F-18 is more volatile and poses an inhalation risk b) F-18 causes positron emission c) F-18 is flammable	23. Why is the radionuclide F-18 more difficult to use than I-131 or Tcm99? a) F-18 is more volatile and poses an inhalation risk b) F-18 causes positron emission c) F-18 is flammable d) F-18 releases gamma rays that are 511keV and therefore shielding is difficult Answer: B – Well, because positrons annihilate with electrons and lead to 511 keV photons compared to about 360 something for I -131 and 140 for Tc99. Note option D is wrong b/c the photons are NOT gamma rays but annihilation photons. Read each option carefully.
	24. Horizontal resolution on video is improved by A. Increasing the number of scan lines B. Increasing refresh rate C. Increasing bandwidth D. Increasing the illuminence of the monitor	24. Horizontal resolution on video is improved by A. Increasing the number of scan lines B. Increasing refresh rate C. Increasing bandwidth D. Increasing the illuminence of the monitor Answer: C – The key word is horizontal. Horizontal resolution is limited by bandwidth; vertical resolution is limited by number of horizontal lines. Got it? Vote for Pedro.
	25. The linear attenuation coefficient (LAC) (of something?) is a) the product of the PE, mass attenuation coefficient (MAC), LAC and Compton scatter b) the product of PE, Compton Scatter, and LAC c) the sum of LAC, PE, and Compton scatter d) only PE e) only Compton scatter	25. The linear attenuation coefficient (LAC) (of something?) is a) the product of the PE, mass attenuation coefficient (MAC), LAC and Compton scatter b) the product of PE, Compton Scatter, and LAC c) the sum of LAC, PE, and Compton scatter d) only PE e) only Compton scatter 25. The linear attenuation coefficient (LAC) (of something?) is a) the product of the PE, mass attenuation coefficient (MAC), LAC and Compton scatter b) the product of PE, Compton Scatter, and LAC c) the sum of LAC, PE, and Compton scatter d) only PE e) only Compton scatter Answer: C – all the possible interactions are added together to make the LAC. See Bushberg syllabus. Eat it, Huda! The linear attenuation coefficient is the sum of the individual linear attenua tion coefficients for each type of interaction: LAC= Rayieigh+ Photoelectric Effect + Compton Scatter + pair production
	When using a scintillation camera, spatial resolution is affected most by a. collimator b. the crystal c. the photomultiplier tubes d. the ADC	26. When using a scintillation camera, spatial resolution is affected most by a. collimator b. the crystal c. the photomultiplier tubes d. the ADC Answer: A – the collimator resolution limits the system and is best at the collimator's surface. The intrinsic resolution (crystal) is not changed with increased distance.
	27. Recent analysis of data compiled from the affects of the atomic bomb in Japan have revealed that the likelihood of solid tumors and radiation dose has what relationship: a. a linear relationship with a threshold dose of…. b. A linear relationship with a threshold dose of…. c. A linear quadratic relationship with no threshold d. A linear quadratic relationship with a threshold dose of… e. linear relationship with no threshold	27. Recent analysis of data compiled from the affects of the atomic bomb in Japan have revealed that the likelihood of solid tumors and radiation dose has what relationship: a. a linear relationship with a threshold dose of…. b. A linear relationship with a threshold dose of…. c. A linear quadratic relationship with no threshold d. A linear quadratic relationship with a threshold dose of… e. linear relationship with no threshold Answer: E
	28. Calculate the change in minification gain with a 1.25 inch output phosphor and a change from 9-inch mode to 4.5-inch mode. a. 4x original b. 2x original c. ¼ original d. ½ original	28. Calculate the change in minification gain with a 1.25 inch output phosphor and a change from 9-inch mode to 4.5-inch mode. a. 4x original b. 2x original c. ¼ original d. ½ original Answer: C – Take the ratio of the input phosphor and square it. If you go from a larger screen to a smaller screen, make sure the bigger number is on the bottom (minification gain decreases) and visa versa if you go from smaller input to larger one.
	29. Chemical shift artifact in MRI results from a) T1 effects b) T2 effects c) Resonance frequency d) Patient motion	29. Chemical shift artifact in MRI results from a) T1 effects b) T2 effects c) Resonance frequency d) Patient motion Answer: C – KNOW THIS. It is on every recall ever made.
	30. What intervention will most reduce image unsharpness caused by patient motion? a. Nuclear medicine: Increase dose of radiopharmaceutical by 10% b. Radiography: Double mA and halve exposure time c. MR: Double the RF power (while keeping TR constant) d. US: Increase transducer frequency	30. What intervention will most reduce image unsharpness caused by patient motion? a. Nuclear medicine: Increase dose of radiopharmaceutical by 10% b. Radiography: Double mA and halve exposure time c. MR: Double the RF power (while keeping TR constant) d. US: Increase transducer frequency Answer: B
	. A well counter has a background count rate of 200 CPM. If a sample is placed in the counter for 2 minutes, how many counts will be 3 standard deviations above the mean? A. 212 B. 242 C. 400 D. 460 E. 475	31. A well counter has a background count rate of 200 CPM. If a sample is placed in the counter for 2 minutes, how many counts will be 3 standard deviations above the mean? A. 212 B. 242 C. 400 D. 460 E. 475 Answer: D – these statistics questions can be tricky. Hopefully there won't be more than 1 or 2 on the test. This one isn't that bad, just get the total number of counts (400) and take the square root to get the standard deviation (20). Then multiply the SD by 3, which gives you 60. Then add it to the total number of counts to get 460.
	32. A 60 keV photon is most likely to interact in soft tissue by: A. Compton scatter B. Photoelectric effect C. Pair production D. Coherent scatter	32. A 60 keV photon is most likely to interact in soft tissue by: A. Compton scatter B. Photoelectric effect C. Pair production D. Coherent scatter Answer: A – below 36keV is Photoelectric effect, above is Compton scatter.
	33. Compared to an ultrasound transducer with focal zone, a transducer with four zones will have increased: A. Axial resolution B. Lateral resolution C. Temporal resolution D. Contrast resolution	33. Compared to an ultrasound transducer with focal zone, a transducer with four zones will have increased: A. Axial resolution B. Lateral resolution C. Temporal resolution D. Contrast resolution Answer: B – four zones increases the length of the focal spot, and the lateral resolution is best at the focal spot (½ the beam diameter). Remember that axial resolution is independent of depth, and relies on the spatial pulse length (SPL).
	34. In computed radiography, a severely underexposed film will appear: A. Too dark B. Too light C. Contrast enhanced D. Contrast deficient	34. In computed radiography, a severely underexposed film will appear: A. Too dark B. Too light C. Contrast enhanced D. Contrast deficient Answer: D – an underexposed film will have increased noise b/c the number of xray photons used to make the image will be decreased. Increasing the noise will decrease the contrast. The wide dynamic range will allow the film to be “windowed” to the correct OD (won't be too dark or too light).
	35. Which of the following statements is FALSE regarding PET: A. It uses electronic collimation B. Random coincidences can produce noise in clinical images C. Typical dose with F-18 FDG is between 0.5 and 1.0 rem D. The ideal positron energy is slightly above 1.02MeV	35. Which of the following statements is FALSE regarding PET: A. It uses electronic collimation B. Random coincidences can produce noise in clinical images C. Typical dose with F-18 FDG is between 0.5 and 1.0 rem D. The ideal positron energy is slightly above 1.02MeV Answer: D – The energy of the annihilation photons add up to 1.02MeV, but their energy is a function of their all their mass turning into energy via Einstein's equation (E = mc2). If the positron energy was 1.02MeV, it would travel a far distance from the site of degradation and spatial resolution would be terrible.
	36. Increased collimation during fluoroscopy does not: A. Decrease patient integral dose B. Reduce dose to the fluoroscopist C. Improve contrast resolution D. Improve spatial resolution	36. Increased collimation during fluoroscopy does not: A. Decrease patient integral dose B. Reduce dose to the fluoroscopist C. Improve contrast resolution D. Improve spatial resolution Answer: D – the integral dose is the total dose to the body and is given in joules. The integral dose is NOT corrected for volume or mass. Absorbed dose is given in Gy or rad and is corrected for mass (joules/kg). So, the integral dose will decrease, but the absorbed dose will be unchanged with collimation. The contrast will be improved due to lower scatter, but the spatial resolution is unchanged.
	37. A 99Mo generator is eluted at 6AM and 12noon. Which is true? a. Tc activity is lower at 6am than 12noon b. Tc concentration is lower at 6am than 12noon c. Tc concentration is higher at 6am than 12noon d. The generator is in secular equilibrium at 12noon e. The generator is in transient equilibrium at 6am	37. A 99Mo generator is eluted at 6AM and 12noon. Which is true? a. Tc activity is lower at 6am than 12noon b. Tc concentration is lower at 6am than 12noon c. Tc concentration is higher at 6am than 12noon d. The generator is in secular equilibrium at 12noon e. The generator is in transient equilibrium at 6am Answer: C – the elution at 6am will be after 4 half-lives, and the elution at noon will be after only one half life.
	38. What’s the appropriate configuration for mammography? Four diagrams shown: a. Cathode against chest wall b. Anode against chest wall c. Cathode near, but not on, chest wall d. Anode near, but not on, chest wall	38. What’s the appropriate configuration for mammography? Four diagrams shown: a. Cathode against chest wall b. Anode against chest wall c. Cathode near, but not on, chest wall d. Anode near, but not on, chest wall Answer: A – the beam configuration of mammography has the cathode over the chest wall and anode over the nipple. It is a half beam geometry. In other words the center of the beam is at the chest wall. The “other half” of the beam is filtered out.
	39. A patient received 200 mCi I-131 for treatment of thyroid cancer. What is the best way to determine exposure around the patient? a. NaI uptake probe. b. Mobile scintillation camera c. Ionization chamber d. GM counter e. Dose calibrator	39. A patient received 200 mCi I-131 for treatment of thyroid cancer. What is the best way to determine exposure around the patient? a. NaI uptake probe. b. Mobile scintillation camera c. Ionization chamber d. GM counter e. Dose calibrator Answer: C – ionization chambers are good for measuring the gamma emission associated with I-131 (~339 keV). Ionization chambers are also good at finding the roentgen exposure, since roentgen is defined as ionizations (charge) in kg of air and ionization chambers are filled with air. So, if you see the word EXPOSURE, think IONIZATION CHAMBER!!
	40. 10 rad of any type of radiation is equal to what: A. 0.1 R B. 0.2 Sv C. 0.1 Gy D. 0.1 mGy	40. 10 rad of any type of radiation is equal to what: A. 0.1 R B. 0.2 Sv C. 0.1 Gy D. 0.1 mGy Answer: C – remember that 100 rad = 1 Gy (and 100 rem = 1 Sv)
	Human lethal LD 50/60 equals: A. 0.5 Gy B. 1.0 Gy C. 2.5 Gy D. 3.5 Gy	41. Human lethal LD 50/60 equals: A. 0.5 Gy B. 1.0 Gy C. 2.5 Gy D. 3.5 Gy Answer: D – I have also seen 3.0 Gy commonly quoted.

Share This Flashcard Set

Set the Language

Physics 2002

Add to Folders

Upgrade to Cram Premium

Card Range To Study

96 Cards in this Set