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PHYSICS 2000/2001

Physics 2000/2001

by N8BEME9, Aug 2010

Subjects: 2000-2001 physics

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	1. acoustic shadowing distal to bone is due to: a. increased absorption of sound waves by the bone b. increased transmission c. refraction d. decreased velocity through bone	1. acoustic shadowing distal to bone is due to: a. increased absorption of sound waves by the bone b. increased transmission c. refraction d. decreased velocity through bone Answer: A – bone is the classic example of a substance the absorbs sound waves. Air (the lung) is the classic example of a substance that reflects sound waves.
	2. spin echo sequence in MR a. 90 followed by 180 degree pulse b. 180 followed by 90 c. 90 followed by 90 d. 90 followed by 180, followed by 180, followed by 180, followed by 180	2. spin echo sequence in MR a. 90 followed by 180 degree pulse b. 180 followed by 90 c. 90 followed by 90 d. 90 followed by 180, followed by 180, followed by 180, followed by 180 Answer: A – Option B describes inversion recovery and option D describes FSE. Option C describes a drunk tech or leprechaun or both.
	3. Echoplanar sequence is used for a. increased spatial resolution b. Increase contrast c. Increased SNR d. Very fast less than 100ms	3. Echoplanar sequence is used for a. increased spatial resolution b. Increase contrast c. Increased SNR d. Very fast less than 100ms Answer D – just remember that when a patient is moving, the tech does those fuzzy T2 looking images that say EPI on them (for Echo Planar Imaging).
	4. If the fractional standard deviation is 1% at count rate 50cps what is fractional standard deviation if count rate is increased to 100cps. a. 2% b. 1% c. 0.1% d. 0.5%	4. If the fractional standard deviation is 1% at count rate 50cps what is fractional standard deviation if count rate is increased to 100cps. a. 2% b. 1% c. 0.1% d. 0.5% Answer: C – another statistics question. The fractional SD is 1/√N. N is the count rate times the time. 1/√N = 0.01 √N = 1/0.01 = 100 N = 10,000 If you double the count rate, then you should double N. So plug 2N into the equation (Fractional SD = 1/√N) Fractional SD = 1/√20,000 = 0.001
	5. What is the minimal resolution in line pair/mm allowed if the line pair device is placed 4.5 cm on the table. a. 11 to 20 b. 21 to23 c. 30-34 d. 40-45	5. What is the minimal resolution in line pair/mm allowed if the line pair device is placed 4.5 cm on the table. a. 11 to 20 b. 21 to23 c. 30-34 d. 40-45 Answer: I'm not sure about this one either. Seems like some info is missing. The original recalls say the answer is A, but who knows...
	6. In DSA what is the post-processing method by which one can reduce the movement artifact (patient swallowed during carotid angio) a. increasing line pair b. change the mask image c. alter video display	6. In DSA what is the post-processing method by which one can reduce the movement artifact (patient swallowed during carotid angio) a. increasing line pair b. change the mask image c. alter video display Answer: B – on multiple old tests.
	7. What is the cause for the image quality of B compared to A? a. B has less counts than A b. A has a lower sensitivity collimator than B c. B has less quantum mottle than A d. Artifacts in B is from movement.	7. What is the cause for the image quality of B compared to A? a. B has less counts than A b. A has a lower sensitivity collimator than B c. B has less quantum mottle than A d. Artifacts in B is from movement. Answer: A – Image B has more quantum mottle, due to fewer counts. Possible causes include shorter imaging time and lower sensitivity collimator for Image B compared to Image A.
	8. Two Doppler study images were shown with the probe at different angulations A less than B. a. A has more velocity than B. b. B has more velocity than A. c. different perception of the vessel diameter	8. Two Doppler study images were shown with the probe at different angulations A less than B. a. A has more velocity than B. b. B has more velocity than A. c. different perception of the vessel diameter Answer: B As you increase the angle, the V max increases up to an angle of 90 degrees, when the cos 90 = 0. Also, the larger the angle, the larger the percent error. That is why you should measure Doppler shift at less than 60 degrees Vmax = (PRF * c)/(2fi * cosθ)
	9. . Which of the following CT window is best for cranial anatomy A. L: –1000 W: 1000 B. L: –20 W: 200 C. L: 0 W: 1000 D. L: –1000 W: 0 E. L: 30 W: 100	9. . Which of the following CT window is best for cranial anatomy A. L: –1000 W: 1000 B. L: –20 W: 200 C. L: 0 W: 1000 D. L: –1000 W: 0 E. L: 30 W: 100 Answer: E
	10. The binding energy of L and K shells are 30 keV and 40 keV, respectively, what would be the dynamic energy of the photoelectrons from L and K shells if the incident photon has an energy of 50 keV? A. 30 and 40 keV B. 20 and 10 keV C. 40 and 30 keV D. 50 and 50 keV	10. The binding energy of L and K shells are 30 keV and 40 keV, respectively, what would be the dynamic energy of the photoelectrons from L and K shells if the incident photon has an energy of 50 keV? A. 30 and 40 keV B. 20 and 10 keV C. 40 and 30 keV D. 50 and 50 keV Answer: B – Take the incident xray photon and subtract the binding energies. Also be familiar with what the energies of the characteristic xrays would be (not asked on this question).
	11. What is FDA’s guideline for MR during pregnancy? A. Specific Absorption Rate (SAR) less than 8 W/Kg peak in any 1 gram of tissue B. No MRI during pregnancy C. Only with scanner less than 2 tesla D. Should not stay in scanner longer than 3 hours	11. What is FDA’s guideline for MR during pregnancy? A. Specific Absorption Rate (SAR) less than 8 W/Kg peak in any 1 gram of tissue B. No MRI during pregnancy C. Only with scanner less than 2 tesla D. Should not stay in scanner longer than 3 hours Answer: A – MR is often done in pregnancy (for appendicitis for example).
	12. Reduced scatter of PET is due to A. Coincident detection B. Optimal energy level C. Better pulse height analyzer (PHA) D. Superior photomultiplier (PMT)	12. Reduced scatter of PET is due to A. Coincident detection B. Optimal energy level C. Better pulse height analyzer (PHA) D. Superior photomultiplier (PMT) Answer: bad question. The UC Davis syllabus states that coincident detection does not reduce scatter because scatter coincidence is true coincidences. PHA is used to reject some scatter, but is is not as useful as in SPECT and other nuc med images. Shortening the time window of detection will decrease RANDOM coincidence but not scatter coincidences. 2D acquisition is the only thing that reduces scatter fraction, but it also reduces efficiency. If I got this question on the test as remembered here, I don't know what I would pick...
	13. Given fixed focal spot size, how can we minimize focal spot blur? By A. increase the distance between subject to image (with fixed source to subject distance) B. decrease the distance between subject to image (with fixed source to subject distance) C. use faster film D. use slower screen E. use Beryllium filter instead of aluminum filter	13. Given fixed focal spot size, how can we minimize focal spot blur? By A. increase the distance between subject to image (with fixed source to subject distance) B. decrease the distance between subject to image (with fixed source to subject distance) C. use faster film D. use slower screen E. use Beryllium filter instead of aluminum filter Answer B – Option A describes moving the image (cassette) farther away from the subject, which will increase the penumbra.
	14. If there is a 20% window in a system using Tc99m, what is the window width? A. 126-154 B. 112-168 C. 140-160 D. 112-168	14. If there is a 20% window in a system using Tc99m, what is the window width? A. 126-154 B. 112-168 C. 140-160 D. 112-168 Answer: A – 140 x 0.2 = 28. So, the window is 140 +/- 14.
	. What is the role of the Automatic Brightness Control in a fluoro unit? A. Appropriate exposures B. Limiting scatter radiation C. Decreasing heat capacity of the anode D. Appropriate developer concentration in the processing unit. E. Proper aliment of th FOV and the cassette.	15. What is the role of the Automatic Brightness Control in a fluoro unit? A. Appropriate exposures B. Limiting scatter radiation C. Decreasing heat capacity of the anode D. Appropriate developer concentration in the processing unit. E. Proper aliment of th FOV and the cassette. Answer: A – also known as the Automatic Exposure control. Located just before the video camera (on the camera side of the aperture). Can adjust mA, kVp, or aperture size (last resort b/c image quality suffers from increased quantum mottle).
	16. In a mammographic unit with a Moly anode and a max 28 kVp, what is the max keV? A. 28 B. 17.5 C. 19.6 D. 20.2 E. 22.7	16. In a mammographic unit with a Moly anode and a max 28 kVp, what is the max keV? A. 28 B. 17.5 C. 19.6 D. 20.2 E. 22.7 Answer: A – the max keV is always the max kVp. The characteristic xrays for Moly are 17.5 and 19.6 keV, and the characteristic xrays for Rhodium are 20.2 and 22.7 keV. So those numbers should look familiar.
	17. A 1994 FDA rule concerning interventional fluoro units was concerned about what? A. Patient cancer rates B. Personnel cancer rates C. Faulty machinery D. Skin entrance doses to patients.	17. A 1994 FDA rule concerning interventional fluoro units was concerned about what? A. Patient cancer rates B. Personnel cancer rates C. Faulty machinery D. Skin entrance doses to patients. Answer: D – I added answer D to the recalls, but I think it is correct. See the following link if you want: http://www.fda.gov/cdrh/fluor.html
	18. Dark streaks in between 2 bones on a CT is due to what artifact? A. Beam hardening B. Aliasing C. Gibbs Artifact D. Volume averaging	18. Dark streaks in between 2 bones on a CT is due to what artifact? A. Beam hardening B. Aliasing C. Gibbs Artifact D. Volume averaging Answer: A
	19. Which of the following is FALSE concerning radiation dose to a fetus? A. The CNS risk is greater at 25–26 weeks vs. 8–15 weeks B. The risk of mental retardation is 40% per Sv C. Mental retardation is unlikely if exposure occurs during first week gestation D. Microcephaly can be seen with radiation induced mental retardation	19. Which of the following is FALSE concerning radiation dose to a fetus? A. The CNS risk is greater at 25–26 weeks vs. 8–15 weeks B. The risk of mental retardation is 40% per Sv C. Mental retardation is unlikely if exposure occurs during first week gestation D. Microcephaly can be seen with radiation induced mental retardation Answer: A
	20. The probability of decay in a radionuclide is given by what expression? A. 1-e-t B. et C. e-t	20. The probability of decay in a radionuclide is given by what expression? A. 1-e-t B. et C. e-t Answer: A – e-t is the probability of radionuclide remaining, and 1-e-t is the probability of decay.
	20. The probability of decay in a radionuclide is given by what expression? A. 1-e-t B. et C. e-t	20. The probability of decay in a radionuclide is given by what expression? A. 1-e-t B. et C. e-t Answer: A – e-t is the probability of radionuclide remaining, and 1-e-t is the probability of decay.
	21. In 2D Fourier transform in MR, extending k-space to higher values does what?	21. In 2D Fourier transform in MR, extending k-space to higher values does what? A. Improves spatial resolution.
	22. If a 3MHz transducer is used to image an object 5 cm deep which reflects 50%, what is the attenuation of the beam, in decibels (given 2 dB/cm/MHz)? A. 33 B. 63 C. 30 D. 50	22. If a 3MHz transducer is used to image an object 5 cm deep which reflects 50%, what is the attenuation of the beam, in decibels (given 2 dB/cm/MHz)? A. 33 B. 63 C. 30 D. 50 Answer: B. On the way to the object, there is 5cm x 3MHz x 2dB/cm/MHx = 30 dB. Then 50% is reflected (3dB loss) and then 30dB lost on the way back to the transducer. 30 + 3 + 30 = 63 dB.
	23. In a radiologist who develops erythema in his hands: A. Use of lead surgical gloves would decrease the dose by 90% B. Chronic cumulative doses of 3–4 Gy are often associated with erythema of the hands. C. The radiologist dose must have been greater than 2 Gy D. The radiologist will probabily experience moist desquamation in the coming weeks	23. In a radiologist who develops erythema in his hands: A. Use of lead surgical gloves would decrease the dose by 90% B. Chronic cumulative doses of 3–4 Gy are often associated with erythema of the hands. C. The radiologist dose must have been greater than 2 Gy D. The radiologist will probabily experience moist desquamation in the coming weeks Answer: C – See the small chart below. Note that option A is incorrect. Surgical lead gloves are considered to offer “minimal protection.” The lead is thin and the AEC increases the dose to penetrate the gloves, resulting in about the same dose to the hands.
	24. What is the typical range of effective dose for a CT of the chest? A. 20–30 mSv B. 5–7 mSv C. 0.6–2 mSv D. 30–80 Sv.	24. What is the typical range of effective dose for a CT of the chest? A. 20–30 mSv B. 5–7 mSv C. 0.6–2 mSv D. 30–80 Sv. Answer: B CT chest = 7 mSv CT A/P = 10 mSv
	25. There are 70 people in a study, and 50 people have the disease in question. If 40 people are correctly diagnosed as having the disease and the remaining 30 people are called normal, what is the sensitivity of the study? a. 4/5 b. 4/7 c. 5/7 d. 3/4	25. There are 70 people in a study, and 50 people have the disease in question. If 40 people are correctly diagnosed as having the disease and the remaining 30 people are called normal, what is the sensitivity of the study? a. 4/5 b. 4/7 c. 5/7 d. 3/4 Anwser: A – Know sensitivity and specificity. It is on every old test in some form or fashion. Sensitivity = TP/(TP+FN) Sensitivity = 40/(40+10) = 4/5 Specificity [TN/(TN+FP)] would be 100% since nobody without the disease had a positive test.
	26. For a 256x256 matrix with 256 shades of gray per pixel, how many megabytes per second are required for real-time transmission at 32 frames per second? a. 21 b. 16 c. 8 d. 2 e. 1	6. For a 256x256 matrix with 256 shades of gray per pixel, how many megabytes per second are required for real-time transmission at 32 frames per second? a. 21 b. 16 c. 8 d. 2 e. 1 Anwser: D – First figure out how many MB are needed for each frame. 256 x256 x 1 byte = 0.06MB/frame NOTE: 256 shades of gray equals 8 bits, which is 1 byte. Remember the number of shades of gray equals 2n, where n = #bits. Okay, so each frame needs 0.06MB, and they want 32 frames per second. 0.06MB/frame x 32 frames/sec = ~2MB/sec.
	27. If the computed tomography dose index (CTDI) is 20 mGy at the surface of the abdomen, what is the effective skin dose of 20 consecutive slices with 1 cm slice thickness? a. 20 mGy b. 40 mGy c. 100 mGy d. 200 mGy e. 400 mGy	27. If the computed tomography dose index (CTDI) is 20 mGy at the surface of the abdomen, what is the effective skin dose of 20 consecutive slices with 1 cm slice thickness? a. 20 mGy b. 40 mGy c. 100 mGy d. 200 mGy e. 400 mGy Answer: A – I don't know how to do this one, and I have no interest in figuring it out. I wonder if some info is missing (pitch, weighted CDTI) Maybe next year somebody can explain it.
	28. Which of the following is true for a radioactive sample in transient equilibrium? a. sample activity is constant b. parent activity is constant c. progeny activity is constant d. number of disintegrations per second is constant e. fraction of radionuclide undergoing decay per second is constant	28. Which of the following is true for a radioactive sample in transient equilibrium? a. sample activity is constant b. parent activity is constant c. progeny activity is constant d. number of disintegrations per second is constant e. fraction of radionuclide undergoing decay per second is constant Answer: E – the daughter concentration is higher than the parent, and the apparent daughter half-life equals the parent half-life.
	29. The best advise from the safety standpoint to a breast-feeding mother who is to undergo a radioiodine treatment for hyperthyroidism is: A) Stop breast-feeding indefinitely 2 weeks prior to treatment B) Stop breast-feeding indefinitely at the time of treatment C) Stop breast-feeding for 3 weeks starting 2 weeks prior to treatment D) Stop breast-feeding for 3 weeks starting at the time of treatment E) Stop breast-feeding for 5 days at the time of treatment	29. The best advise from the safety standpoint to a breast-feeding mother who is to undergo a radioiodine treatment for hyperthyroidism is: A) Stop breast-feeding indefinitely 2 weeks prior to treatment B) Stop breast-feeding indefinitely at the time of treatment C) Stop breast-feeding for 3 weeks starting 2 weeks prior to treatment D) Stop breast-feeding for 3 weeks starting at the time of treatment E) Stop breast-feeding for 5 days at the time of treatment Answer: A – I-131 is concentrated in breast milk. You want the woman to stop lactating before getting 1-131 not only to decrease dose to baby but also the mother's breast. In other words, you don't want I-131 hanging around in the woman's breast (bad, bad thing).
	30. Compared with the 3 MHz transducer crystal, a 5 MHz transducer is: A) greater in diameter B) smaller in diameter C) thinner D) thicker	30. Compared with the 3 MHz transducer crystal, a 5 MHz transducer is: A) greater in diameter B) smaller in diameter C) thinner D) thicker Answer: C – the wavelength is ½ the crystal diameter. Higher frequency transducer has a shorter wavelength and a thinner crystal.
	31. The automatic brightness control function in fluoroscopy is to ensure that: A) Same amount of light strikes the TV camera B) Same radiation exposure is used for every patient regardless of body habitus C) Same number of electrons strike the output phosphor D) Same number of electrons strike the input phosphor	31. The automatic brightness control function in fluoroscopy is to ensure that: A) Same amount of light strikes the TV camera B) Same radiation exposure is used for every patient regardless of body habitus C) Same number of electrons strike the output phosphor D) Same number of electrons strike the input phosphor Answer: A – the AEC is located on the backside of the aperture just before the TV camera. Can call for more kVp or mAs. If the radiation is maxed out, then the AEC will open the aperture (smaller f-number).
	32. The FDA regulation of 1995 dealing with fluoroscopy was concerned with: A. Injury risk to patients B. Injury risk to personnel C. Cancer risk in patients D. Cancer risk in personnel	32. The FDA regulation of 1995 dealing with fluoroscopy was concerned with: A. Injury risk to patients B. Injury risk to personnel C. Cancer risk in patients D. Cancer risk in personnel Answer: A – specificity skin injury
	33. The physical half-life of a radioisotope is 4 hours. If its effective half-life is 2 hours, what is its biological half-life? a. 0.5 hours b. 2 hours c. 4 hours d. 6 hours e. 8 hours	33. The physical half-life of a radioisotope is 4 hours. If its effective half-life is 2 hours, what is its biological half-life? a. 0.5 hours b. 2 hours c. 4 hours d. 6 hours e. 8 hours Answer: C – Here's how you do it: Teff = (Tb x Tp)/(Tb + Tp) 2 hr = 4Tb /(4+ Tb) 4 Tb = 8 hrs + 2 Tb 2 Tb= 8 hrs Tb = 4 hrs
	34. The percent standard deviation of pixel values in an abdominal CT will decrease by reducing: a. mAs b. kVp c. Scan time d. Slice thickness e. Patient size	34. The percent standard deviation of pixel values in an abdominal CT will decrease by reducing: a. mAs b. kVp c. Scan time d. Slice thickness e. Patient size Answer: E. The question is asking what will decrease relative noise (increase N). Reducing the other options decreases the number of photons reaching the detector. Just a point about test taking – if all of the options except one change the noise in one direction, then the answer has to be the option that changes noise in the opposite direction.
	35. In linear tomography, a streak artifact in the plane of the image is caused by: a. structure parallel to motion of x-ray tube b. structure perpendicular to motion c. moves in and out of plane d. at an obtuse angle e. at a right angle	35. In linear tomography, a streak artifact in the plane of the image is caused by: a. structure parallel to motion of x-ray tube b. structure perpendicular to motion c. moves in and out of plane d. at an obtuse angle e. at a right angle Answer: A – who cares? They might as well ask us about leech therapy for removing demons.
	36. The following artifact is caused by what occurring proximal to the artifact? a. high attenuating material b. low attenuating material c. two close interfaces d. side beam interference	36. The following artifact is caused by what occurring proximal to the artifact? a. high attenuating material b. low attenuating material c. two close interfaces d. side beam interference Answer: C – Ring down artifact is a form of reverberation.
	37. All other things equal, the SNR of a 1.5 T scanner (A) to a 0.5 T scanner (B) is: a. SNR A = 9 times SNR B b. SNR A = 3 times SNR B c. SNR A = SNR B d. SNR A times 3 = SNR B e. SNR A times 9 = SNR B	Answer: B – SNR is proportional to: B1.0 to B1.5 NEX0.5 Bandwidth-0.5 (decreasing BW causes more time and allows more signal) Volume
	38. An ROC curve without labels was given. The y and x axes represent: a. true positive fraction and false positive fraction b. true positive fraction and true negative fraction c. false positive and true positive d. sensitivity and specificity e. specificity and sensitivity	38. An ROC curve without labels was given. The y and x axes represent: a. true positive fraction and false positive fraction b. true positive fraction and true negative fraction c. false positive and true positive d. sensitivity and specificity e. specificity and sensitivity Answer: A – also sensitivity and 1-specificity. Here a chart of x and y axises to review:
	39. According to the quality control procedures of a nuclear medicine laboratory, a Co-57 source can be measured in a dose calibrator under the Tc-99m setting to ensure the instrument’s a. constancy b. spatial linearity c. count-rate linearity d. energy calibration e. dead time	39. According to the quality control procedures of a nuclear medicine laboratory, a Co-57 source can be measured in a dose calibrator under the Tc-99m setting to ensure the instrument’s a. constancy b. spatial linearity c. count-rate linearity d. energy calibration e. dead time Answer: A – constancy must be checked daily with a long lived radionuclide, such as Co-57. Basically the same test as accuracy except only one radionuclide is used and activity is compared to the value from the day instead of a national standard as in accuracy. Many places just do a daily accuracy test though. Usually Cesium is used for constancy and Cobalt for flood uniformity.

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