NMTCB PET Certification Practice Test Bank 2026-2027: 350
Complex Exam Questions with Detailed Answers and
Rationales
SECTION 1: PET PHYSICS AND INSTRUMENTATION
Question 1. A PET technologist is calibrating the energy window using a
germanium-68 source. The energy resolution is noted to be 34 percent
between 425 and 600 keV. What is the primary purpose of this calibration?
A. To adjust the coincidence timing window
B. To set the lower and upper energy discriminator levels
C. To measure the dead time of the detector
D. To calculate the positron range
Answer: B
Rationale: Energy window calibration using a 68-Ge source sets the lower and
upper energy discriminator levels to accept 511 keV annihilation photons
while rejecting scattered events. The 34 percent energy resolution between
425 and 600 keV helps define these limits.
Question 2. A new PET scanner is installed with a spatial resolution of 4 to 5
mm. A physician asks why small lesions may still be difficult to detect. Which
factor most directly limits the detection of small lesions despite this
resolution?
A. High sensitivity of the detectors
B. Partial volume effect
C. Low random fraction
D. Short positron range
,Answer: B
Rationale: The partial volume effect causes small objects to appear with
falsely low intensity because the signal is averaged over the voxel volume.
Even with 4 to 5 mm resolution, lesions smaller than twice the resolution may
be underestimated.
Question 3. A PET technologist notices that the LSO detector block exhibits
intrinsic radioactivity. Which regulatory document addresses this issue?
A. NU-2-2007
B. 10 CFR 20
C. NUREG-1556
D. 21 CFR Part 212
Answer: A
Rationale: NU-2-2007 addresses intrinsic radioactivity in LSO and LYSO
crystals. This is due to the presence of lutetium-176, which emits beta
particles and gamma rays.
Question 4. During a PET/CT procedure, the technologist observes that the
photomultiplier tube is not amplifying the signal appropriately. What voltage
range should be applied to the PMT for proper operation?
A. 100 to 500 volts
B. 800 to 2000 volts
C. 3000 to 5000 volts
D. 50 to 100 volts
Answer: B
,Rationale: The voltage applied to the photomultiplier tube is typically 800 to
2000 volts. This high voltage is necessary for the dynodes to amplify the
electron cascade.
Question 5. A PET detector block contains 10,000 to 35,000 small crystals.
Each crystal is 4 mm on each side and 10 to 20 mm deep, with 4
photomultiplier tubes. What is the primary advantage of this configuration?
A. Increased dead time
B. Improved spatial resolution and light collection
C. Reduced sensitivity
D. Increased random events
Answer: B
Rationale: The small crystal size improves spatial resolution, while the depth
and multiple PMTs allow for efficient light collection and positioning of the
annihilation event.
Question 6. A technologist is reviewing the quality control results for a PET
scanner. The blank scan shows non-uniform sensitivity across the detector
modules. What is the most appropriate action?
A. Repeat the blank scan after adjusting the energy window
B. Perform normalization to correct for module sensitivity
C. Increase the coincidence timing window
D. Decrease the lower energy discriminator
Answer: B
Rationale: Normalization sets uniform sensitivity for detector modules and
creates a calibration sensitivity factor for each detector. The blank scan
provides the raw uniformity data used for this correction.
, Question 7. In PET imaging, the dominant interaction at 511 keV is Compton
scatter. Which factor does the scatter fraction depend on?
A. Only the source distribution
B. Only the size of the patient
C. Source distribution and size of the patient
D. Only the detector material
Answer: C
Rationale: The scatter fraction depends on both the source distribution and
the size of the patient. Larger patients produce more scatter, and the
distribution of activity affects the scatter profile.
Question 8. A patient is injected with 15 mCi of 18F-FDG. The technologist
notes that the exposure rate at 1 meter is 3 mR/hr. According to AAPM report
108, what is the significance of this measurement?
A. It is within normal limits
B. It exceeds the limit for a controlled area
C. It is below the limit for an uncontrolled area
D. It requires immediate evacuation
Answer: A
Rationale: AAPM report 108 suggests that the exposure rate from a patient
injected with 15 mCi is approximately 3 mR/hr at 1 meter. This is within
normal limits for a nuclear medicine patient.
Complex Exam Questions with Detailed Answers and
Rationales
SECTION 1: PET PHYSICS AND INSTRUMENTATION
Question 1. A PET technologist is calibrating the energy window using a
germanium-68 source. The energy resolution is noted to be 34 percent
between 425 and 600 keV. What is the primary purpose of this calibration?
A. To adjust the coincidence timing window
B. To set the lower and upper energy discriminator levels
C. To measure the dead time of the detector
D. To calculate the positron range
Answer: B
Rationale: Energy window calibration using a 68-Ge source sets the lower and
upper energy discriminator levels to accept 511 keV annihilation photons
while rejecting scattered events. The 34 percent energy resolution between
425 and 600 keV helps define these limits.
Question 2. A new PET scanner is installed with a spatial resolution of 4 to 5
mm. A physician asks why small lesions may still be difficult to detect. Which
factor most directly limits the detection of small lesions despite this
resolution?
A. High sensitivity of the detectors
B. Partial volume effect
C. Low random fraction
D. Short positron range
,Answer: B
Rationale: The partial volume effect causes small objects to appear with
falsely low intensity because the signal is averaged over the voxel volume.
Even with 4 to 5 mm resolution, lesions smaller than twice the resolution may
be underestimated.
Question 3. A PET technologist notices that the LSO detector block exhibits
intrinsic radioactivity. Which regulatory document addresses this issue?
A. NU-2-2007
B. 10 CFR 20
C. NUREG-1556
D. 21 CFR Part 212
Answer: A
Rationale: NU-2-2007 addresses intrinsic radioactivity in LSO and LYSO
crystals. This is due to the presence of lutetium-176, which emits beta
particles and gamma rays.
Question 4. During a PET/CT procedure, the technologist observes that the
photomultiplier tube is not amplifying the signal appropriately. What voltage
range should be applied to the PMT for proper operation?
A. 100 to 500 volts
B. 800 to 2000 volts
C. 3000 to 5000 volts
D. 50 to 100 volts
Answer: B
,Rationale: The voltage applied to the photomultiplier tube is typically 800 to
2000 volts. This high voltage is necessary for the dynodes to amplify the
electron cascade.
Question 5. A PET detector block contains 10,000 to 35,000 small crystals.
Each crystal is 4 mm on each side and 10 to 20 mm deep, with 4
photomultiplier tubes. What is the primary advantage of this configuration?
A. Increased dead time
B. Improved spatial resolution and light collection
C. Reduced sensitivity
D. Increased random events
Answer: B
Rationale: The small crystal size improves spatial resolution, while the depth
and multiple PMTs allow for efficient light collection and positioning of the
annihilation event.
Question 6. A technologist is reviewing the quality control results for a PET
scanner. The blank scan shows non-uniform sensitivity across the detector
modules. What is the most appropriate action?
A. Repeat the blank scan after adjusting the energy window
B. Perform normalization to correct for module sensitivity
C. Increase the coincidence timing window
D. Decrease the lower energy discriminator
Answer: B
Rationale: Normalization sets uniform sensitivity for detector modules and
creates a calibration sensitivity factor for each detector. The blank scan
provides the raw uniformity data used for this correction.
, Question 7. In PET imaging, the dominant interaction at 511 keV is Compton
scatter. Which factor does the scatter fraction depend on?
A. Only the source distribution
B. Only the size of the patient
C. Source distribution and size of the patient
D. Only the detector material
Answer: C
Rationale: The scatter fraction depends on both the source distribution and
the size of the patient. Larger patients produce more scatter, and the
distribution of activity affects the scatter profile.
Question 8. A patient is injected with 15 mCi of 18F-FDG. The technologist
notes that the exposure rate at 1 meter is 3 mR/hr. According to AAPM report
108, what is the significance of this measurement?
A. It is within normal limits
B. It exceeds the limit for a controlled area
C. It is below the limit for an uncontrolled area
D. It requires immediate evacuation
Answer: A
Rationale: AAPM report 108 suggests that the exposure rate from a patient
injected with 15 mCi is approximately 3 mR/hr at 1 meter. This is within
normal limits for a nuclear medicine patient.
