Nuclear Medicine – NMTCB Practice Exam
2026/2027 | 300+ Questions with Answers |
Radiopharmaceuticals, Radiation Safety,
SPECT Imaging – Radiology Course
1. What is the half-life of Tc-99m?
A. 6 hours
B. 8 hours
C. 6.02 hours
D. 2.7 days
Answer: C
Rationale: Tc-99m has a physical half-life of 6.02 hours (approximately 6 hours). This is
the time required for the activity to decay to half its original value. This short half-life
allows for adequate imaging while minimizing patient radiation dose. Mo-99 has a half-
life of 66 hours .
2. Which of the following describes the photoelectric effect?
A. A photon interacts with an outer-shell electron, ejecting it and scattering with
reduced energy
B. A photon transfers all its energy to an inner-shell electron, ejecting it from the atom
C. A high-energy photon produces an electron-positron pair
D. A photon is scattered without energy loss
Answer: B
Rationale: In the photoelectric effect, an incident photon transfers all its energy to an
inner-shell electron, ejecting it from the atom. This is the primary interaction for
diagnostic imaging (photons in the 20-100 keV range). The ejected electron is called a
photoelectron. All the photon's energy is absorbed, providing the best image contrast .
,3. Which type of radiation interaction is primarily responsible for producing the
image in nuclear medicine?
A. Compton scatter
B. Photoelectric effect
C. Coherent scatter
D. Pair production
Answer: A
Rationale: Compton scatter is the primary photon interaction in nuclear medicine
imaging (energy range 100-200 keV). While it degrades image quality by creating
scatter radiation, it is the predominant interaction with tissue. The photoelectric effect
dominates at lower energies and contributes to patient dose .
4. What is the mode of decay for Tc-99m?
A. Beta decay
B. Positron emission
C. Isomeric transition
D. Alpha decay
Answer: C
Rationale: Tc-99m decays by isomeric transition (IT), emitting a 140 keV gamma photon.
The "m" indicates a metastable state; it transitions to Tc-99 without changing the
number of protons or neutrons. This is ideal for imaging due to the suitable energy and
lack of beta emissions .
5. A radioactive source has an activity of 10 mCi. What is the activity after one
half-life?
A. 2.5 mCi
B. 5 mCi
C. 7.5 mCi
D. 10 mCi
,Answer: B
Rationale: After one half-life, the activity is reduced to 50% of its original value.
Therefore, 10 mCi × 0.5 = 5 mCi. After two half-lives, it would be 2.5 mCi .
6. Which gas-filled detector operates in the saturation region and is used for dose
calibrators?
A. Geiger-Müller (GM) counter
B. Proportional counter
C. Ionization chamber
D. Scintillation counter
Answer: C
Rationale: Ionization chambers operate in the saturation region where all ion pairs
created are collected. This provides a current proportional to the activity, making them
ideal for dose calibrators. The output is independent of applied voltage in this region .
7. What is Bremsstrahlung radiation?
A. Radiation emitted during positron annihilation
B. Radiation produced by the deceleration of beta particles as they approach nuclei
C. Radiation emitted during isomeric transition
D. Characteristic x-rays from electron transitions
Answer: B
Rationale: Bremsstrahlung ("braking radiation") results from the deceleration of beta
particles as they approach the nuclei of atoms. This is important for high-energy beta
emitters like P-32 and Y-90, requiring thick shielding. It produces a continuous spectrum
of x-rays .
8. Which of the following is a reactor-produced radionuclide?
A. I-123
B. Tc-99m
, C. Mo-99
D. Tl-201
Answer: C
Rationale: Mo-99 is produced in a nuclear reactor by neutron bombardment of Mo-98
(or U-235 fission). Reactor-produced radionuclides include Mo-99/Tc-99m, I-131, I-125,
Xe-133, Sr-89, Sm-153, P-32, and Cr-51. I-123 and Tl-201 are accelerator-produced .
9. Which of the following is an accelerator-produced radionuclide?
A. Mo-99
B. I-131
C. I-123
D. Xe-133
Answer: C
Rationale: I-123 is produced in a cyclotron by proton bombardment of Xe-124 or other
targets. Accelerator-produced radionuclides include I-123, Tl-201, Ga-67, In-111, and all
positron emitters (F-18, C-11, N-13, O-15) .
10. What is the photopeak of Tc-99m?
A. 140 keV
B. 511 keV
C. 88 keV
D. 364 keV
Answer: A
Rationale: Tc-99m emits a 140 keV gamma photon, which is optimal for imaging with
NaI(Tl) detectors. This energy provides a good balance between tissue penetration and
detection efficiency. The 511 keV photopeak is from positron annihilation; 364 keV is
from I-131 .
11. What is the energy of the 511 keV annihilation photon?
2026/2027 | 300+ Questions with Answers |
Radiopharmaceuticals, Radiation Safety,
SPECT Imaging – Radiology Course
1. What is the half-life of Tc-99m?
A. 6 hours
B. 8 hours
C. 6.02 hours
D. 2.7 days
Answer: C
Rationale: Tc-99m has a physical half-life of 6.02 hours (approximately 6 hours). This is
the time required for the activity to decay to half its original value. This short half-life
allows for adequate imaging while minimizing patient radiation dose. Mo-99 has a half-
life of 66 hours .
2. Which of the following describes the photoelectric effect?
A. A photon interacts with an outer-shell electron, ejecting it and scattering with
reduced energy
B. A photon transfers all its energy to an inner-shell electron, ejecting it from the atom
C. A high-energy photon produces an electron-positron pair
D. A photon is scattered without energy loss
Answer: B
Rationale: In the photoelectric effect, an incident photon transfers all its energy to an
inner-shell electron, ejecting it from the atom. This is the primary interaction for
diagnostic imaging (photons in the 20-100 keV range). The ejected electron is called a
photoelectron. All the photon's energy is absorbed, providing the best image contrast .
,3. Which type of radiation interaction is primarily responsible for producing the
image in nuclear medicine?
A. Compton scatter
B. Photoelectric effect
C. Coherent scatter
D. Pair production
Answer: A
Rationale: Compton scatter is the primary photon interaction in nuclear medicine
imaging (energy range 100-200 keV). While it degrades image quality by creating
scatter radiation, it is the predominant interaction with tissue. The photoelectric effect
dominates at lower energies and contributes to patient dose .
4. What is the mode of decay for Tc-99m?
A. Beta decay
B. Positron emission
C. Isomeric transition
D. Alpha decay
Answer: C
Rationale: Tc-99m decays by isomeric transition (IT), emitting a 140 keV gamma photon.
The "m" indicates a metastable state; it transitions to Tc-99 without changing the
number of protons or neutrons. This is ideal for imaging due to the suitable energy and
lack of beta emissions .
5. A radioactive source has an activity of 10 mCi. What is the activity after one
half-life?
A. 2.5 mCi
B. 5 mCi
C. 7.5 mCi
D. 10 mCi
,Answer: B
Rationale: After one half-life, the activity is reduced to 50% of its original value.
Therefore, 10 mCi × 0.5 = 5 mCi. After two half-lives, it would be 2.5 mCi .
6. Which gas-filled detector operates in the saturation region and is used for dose
calibrators?
A. Geiger-Müller (GM) counter
B. Proportional counter
C. Ionization chamber
D. Scintillation counter
Answer: C
Rationale: Ionization chambers operate in the saturation region where all ion pairs
created are collected. This provides a current proportional to the activity, making them
ideal for dose calibrators. The output is independent of applied voltage in this region .
7. What is Bremsstrahlung radiation?
A. Radiation emitted during positron annihilation
B. Radiation produced by the deceleration of beta particles as they approach nuclei
C. Radiation emitted during isomeric transition
D. Characteristic x-rays from electron transitions
Answer: B
Rationale: Bremsstrahlung ("braking radiation") results from the deceleration of beta
particles as they approach the nuclei of atoms. This is important for high-energy beta
emitters like P-32 and Y-90, requiring thick shielding. It produces a continuous spectrum
of x-rays .
8. Which of the following is a reactor-produced radionuclide?
A. I-123
B. Tc-99m
, C. Mo-99
D. Tl-201
Answer: C
Rationale: Mo-99 is produced in a nuclear reactor by neutron bombardment of Mo-98
(or U-235 fission). Reactor-produced radionuclides include Mo-99/Tc-99m, I-131, I-125,
Xe-133, Sr-89, Sm-153, P-32, and Cr-51. I-123 and Tl-201 are accelerator-produced .
9. Which of the following is an accelerator-produced radionuclide?
A. Mo-99
B. I-131
C. I-123
D. Xe-133
Answer: C
Rationale: I-123 is produced in a cyclotron by proton bombardment of Xe-124 or other
targets. Accelerator-produced radionuclides include I-123, Tl-201, Ga-67, In-111, and all
positron emitters (F-18, C-11, N-13, O-15) .
10. What is the photopeak of Tc-99m?
A. 140 keV
B. 511 keV
C. 88 keV
D. 364 keV
Answer: A
Rationale: Tc-99m emits a 140 keV gamma photon, which is optimal for imaging with
NaI(Tl) detectors. This energy provides a good balance between tissue penetration and
detection efficiency. The 511 keV photopeak is from positron annihilation; 364 keV is
from I-131 .
11. What is the energy of the 511 keV annihilation photon?
