NUCLEAR MEDICINE FINAL EXAM
142 Actual Questions and Answers
Comprehensive Practice Review
2026/2027 NMTCB & ARRT Aligned
With Accurate Solutions | 2026 Update
April 2026
Aligned with NMTCB Certification, ARRT Nuclear Medicine,
SNMMI Procedure Guidelines, NRC Regulations (10 CFR 19, 20, 35),
ASNC Imaging Standards & Contemporary Nuclear Medicine Curriculum
, Nuclear Medicine Final Exam | 2026 Update
Section 1: Radiation Physics, Radiopharmaceuticals & Radiochemistry
Fundamentals (Q1-Q20)
Q1: A vial of Tc-99m MDP contains 80 mCi at 8:00 AM. The half-life of Tc-99m is 6.0 hours. What is the
remaining activity at 2:00 PM on the same day?
A. 10 mCi
B. 20 mCi
C. 40 mCi [CORRECT]
D. 60 mCi
Correct Answer: C
Rationale: After 6 hours (one half-life), activity halves: 80 mCi / 2 = 40 mCi. Option C is correct. Option A represents two
half-lives (16 hours); option B is a miscalculation; option D is only one-quarter decay.
Q2: A PET facility receives a dose of F-18 FDG calibrated at 370 MBq. What is this activity expressed in
millicuries? (1 mCi = 37 MBq)
A. 5.0 mCi
B. 10.0 mCi [CORRECT]
C. 15.0 mCi
D. 37.0 mCi
Correct Answer: B
Rationale: 370 MBq / 37 MBq per mCi = 10.0 mCi. Option B is correct. Option A divides by 74; option C is a rough
overestimate; option D is the denominator itself.
Q3: The physical half-life of I-131 is 8.04 days. What is the decay constant (lambda) in units of per day? (Use
the relationship lambda = ln(2) / T_half.)
A. 0.0582 per day
B. 0.0862 per day [CORRECT]
C. 0.1155 per day
D. 0.1570 per day
Correct Answer: B
Rationale: lambda = 0..04 = 0.0862 per day. Option B is correct. Option A inverts the calculation; option C uses
0.693/6; option D uses 0.693/4.41.
Q4: A sample of Lu-177-DOTATATE has an initial activity of 200 mCi. The physical half-life of Lu-177 is
6.71 days. Using the formula A = A0 * e^(-lambda*t), what is the approximate activity remaining after 3
days? (lambda for Lu-177 = 0.1033 per day)
A. 94 mCi
B. 112 mCi
C. 147 mCi [CORRECT]
D. 165 mCi
Correct Answer: C
Rationale: A = 200 * e^(-0.1033 * 3) = 200 * e^(-0.3099) = 200 * 0.733 = 146.6, approximately 147 mCi. Option C is
correct. Option A uses two half-lives; option B miscalculates the exponent; option D underestimates the decay.
Page 1
, Nuclear Medicine Final Exam | 2026 Update
Q5: A patient receives I-131 for thyroid ablation. The physical half-life is 8.04 days and the biological
half-life of iodine uptake in the thyroid is estimated at 18 days. What is the effective half-life?
A. 4.2 days
B. 5.6 days [CORRECT]
C. 9.4 days
D. 12.6 days
Correct Answer: B
Rationale: 1/T_eff = 1/T_phys + 1/T_bio = 1/8.04 + 1/18 = 0.1244 + 0.0556 = 0.180; T_eff = 1/0.180 = 5.56 days,
approximately 5.6 days. Option B is correct. Option A results from subtracting half-lives; option C is the arithmetic mean;
option D adds half-lives.
Q6: A Mo-99/Tc-99m generator was last eluted 24 hours ago. Mo-99 has a half-life of 66 hours and Tc-99m
has a half-life of 6 hours. Approximately what percentage of the maximum available Tc-99m has
reaccumulated?
A. 45%
B. 67%
C. 82%
D. 94% [CORRECT]
Correct Answer: D
Rationale: The transient equilibrium buildup reaches approximately 94% of maximum Tc-99m activity at about 23 hours
post-elution. Option D is correct. Option A corresponds to roughly 6 hours; option B to about 12 hours; option C to about
18 hours.
Q7: According to USP <825> and accepted radiopharmacy standards, what is the maximum acceptable
concentration of aluminum breakthrough in a Tc-99m eluate?
A. Less than 5 mcg/mL
B. Less than 10 mcg/mL [CORRECT]
C. Less than 15 mcg/mL
D. Less than 20 mcg/mL
Correct Answer: B
Rationale: The NRC and USP require that aluminum concentration in Tc-99m sodium pertechnetate be less than 10
mcg/mL. Option B is correct. Options A, C, and D are incorrect thresholds.
Q8: A Tc-99m eluate is tested for molybdenum breakthrough. The eluate contains 50 mCi of Tc-99m. The
acceptable limit for Mo-99 breakthrough is 0.15 mcg Mo per mCi Tc-99m at the time of elution. What is the
maximum allowable amount of Mo-99 in this eluate?
A. 2.5 mcg
B. 5.0 mcg
C. 7.5 mcg [CORRECT]
D. 10.0 mcg
Correct Answer: C
Rationale: Maximum Mo-99 = 0.15 mcg/mCi * 50 mCi = 7.5 mcg. Option C is correct. Option A uses 0.05 mcg/mCi;
option B uses 0.10 mcg/mCi; option D uses 0.20 mcg/mCi.
Page 2
, Nuclear Medicine Final Exam | 2026 Update
Q9: F-18 fluorodeoxyglucose (FDG) is produced in a cyclotron by bombarding O-18 enriched water with
protons. Which of the following correctly describes the decay characteristics of F-18?
A. Beta-minus emitter with 1.7 MeV max energy and 110-minute half-life
B. Positron emitter with 0.635 MeV max energy and 110-minute half-life [CORRECT]
C. Positron emitter with 0.635 MeV max energy and 65-minute half-life
D. Gamma emitter with 511 keV photons and 110-minute half-life
Correct Answer: B
Rationale: F-18 decays by positron emission (beta-plus) with a maximum positron energy of 0.635 MeV and a physical
half-life of 109.8 minutes (approximately 110 min). Option B is correct. Option A incorrectly states beta-minus; option C
has the wrong half-life; option D mislabels it as a direct gamma emitter rather than positron emitter.
Q10: A patient is scheduled for I-131 therapy for thyroid cancer. The prescribed dose is 150 mCi, to be
administered at noon. The dose is prepared at 6:00 AM and reads 150 mCi. The half-life of I-131 is 8.04 days.
Which of the following is the most appropriate action?
A. Administer the dose as prepared since the decay in 6 hours is negligible for an 8-day half-life
B. Recalibrate the dose at noon; the activity will have decayed slightly but remain therapeutically appropriate
[CORRECT]
C. Prepare a new dose at noon to ensure exact 150 mCi is delivered
D. Add 5 mCi to the prepared dose to compensate for decay
Correct Answer: B
Rationale: I-131 with an 8-day half-life decays only minimally in 6 hours (decay factor approximately 0.965, losing about
3.5%), so recalibrating at administration time is the standard practice. Option B is correct. Option A ignores the need to
verify activity; option C is unnecessary waste; option D introduces inaccuracy.
Q11: Lu-177-PSMA-617 is an FDA-approved radiopharmaceutical for the treatment of which condition?
A. Hepatocellular carcinoma
B. Neuroendocrine tumors
C. Metastatic castration-resistant prostate cancer [CORRECT]
D. Differentiated thyroid cancer
Correct Answer: C
Rationale: Lu-177-PSMA-617 (Pluvicto) is approved for the treatment of PSMA-positive metastatic castration-resistant
prostate cancer. Option C is correct. Option B describes Lu-177-DOTATATE (Lutathera); options A and D are incorrect
indications.
Q12: A technologist is preparing a Tc-99m MDP kit in the radiopharmacy. The kit label indicates that
reconstitution should occur with 5-10 mL of sterile sodium pertechnetate. The eluate has an activity
concentration of 25 mCi/mL. The department needs a total of 60 mCi. What volume of pertechnetate should
be added to the kit?
A. 2.4 mL using the highest concentration available
B. 5.0 mL to yield 125 mCi total in the kit vial
C. 8.0 mL to ensure the volume is within the 5-10 mL range and provides adequate activity [CORRECT]
D. 12.0 mL to maximize the total activity available
Correct Answer: C
Rationale: Adding 8 mL of 25 mCi/mL eluate yields 200 mCi total, well above the 60 mCi needed, and the 8 mL volume
falls within the kit's specified 5-10 mL range. Option C is correct. Option A yields only 60 mCi but 2.4 mL is below the
minimum 5 mL; option B yields 125 mCi but 5 mL is at the lower limit; option D exceeds the 10 mL maximum.
Page 3
142 Actual Questions and Answers
Comprehensive Practice Review
2026/2027 NMTCB & ARRT Aligned
With Accurate Solutions | 2026 Update
April 2026
Aligned with NMTCB Certification, ARRT Nuclear Medicine,
SNMMI Procedure Guidelines, NRC Regulations (10 CFR 19, 20, 35),
ASNC Imaging Standards & Contemporary Nuclear Medicine Curriculum
, Nuclear Medicine Final Exam | 2026 Update
Section 1: Radiation Physics, Radiopharmaceuticals & Radiochemistry
Fundamentals (Q1-Q20)
Q1: A vial of Tc-99m MDP contains 80 mCi at 8:00 AM. The half-life of Tc-99m is 6.0 hours. What is the
remaining activity at 2:00 PM on the same day?
A. 10 mCi
B. 20 mCi
C. 40 mCi [CORRECT]
D. 60 mCi
Correct Answer: C
Rationale: After 6 hours (one half-life), activity halves: 80 mCi / 2 = 40 mCi. Option C is correct. Option A represents two
half-lives (16 hours); option B is a miscalculation; option D is only one-quarter decay.
Q2: A PET facility receives a dose of F-18 FDG calibrated at 370 MBq. What is this activity expressed in
millicuries? (1 mCi = 37 MBq)
A. 5.0 mCi
B. 10.0 mCi [CORRECT]
C. 15.0 mCi
D. 37.0 mCi
Correct Answer: B
Rationale: 370 MBq / 37 MBq per mCi = 10.0 mCi. Option B is correct. Option A divides by 74; option C is a rough
overestimate; option D is the denominator itself.
Q3: The physical half-life of I-131 is 8.04 days. What is the decay constant (lambda) in units of per day? (Use
the relationship lambda = ln(2) / T_half.)
A. 0.0582 per day
B. 0.0862 per day [CORRECT]
C. 0.1155 per day
D. 0.1570 per day
Correct Answer: B
Rationale: lambda = 0..04 = 0.0862 per day. Option B is correct. Option A inverts the calculation; option C uses
0.693/6; option D uses 0.693/4.41.
Q4: A sample of Lu-177-DOTATATE has an initial activity of 200 mCi. The physical half-life of Lu-177 is
6.71 days. Using the formula A = A0 * e^(-lambda*t), what is the approximate activity remaining after 3
days? (lambda for Lu-177 = 0.1033 per day)
A. 94 mCi
B. 112 mCi
C. 147 mCi [CORRECT]
D. 165 mCi
Correct Answer: C
Rationale: A = 200 * e^(-0.1033 * 3) = 200 * e^(-0.3099) = 200 * 0.733 = 146.6, approximately 147 mCi. Option C is
correct. Option A uses two half-lives; option B miscalculates the exponent; option D underestimates the decay.
Page 1
, Nuclear Medicine Final Exam | 2026 Update
Q5: A patient receives I-131 for thyroid ablation. The physical half-life is 8.04 days and the biological
half-life of iodine uptake in the thyroid is estimated at 18 days. What is the effective half-life?
A. 4.2 days
B. 5.6 days [CORRECT]
C. 9.4 days
D. 12.6 days
Correct Answer: B
Rationale: 1/T_eff = 1/T_phys + 1/T_bio = 1/8.04 + 1/18 = 0.1244 + 0.0556 = 0.180; T_eff = 1/0.180 = 5.56 days,
approximately 5.6 days. Option B is correct. Option A results from subtracting half-lives; option C is the arithmetic mean;
option D adds half-lives.
Q6: A Mo-99/Tc-99m generator was last eluted 24 hours ago. Mo-99 has a half-life of 66 hours and Tc-99m
has a half-life of 6 hours. Approximately what percentage of the maximum available Tc-99m has
reaccumulated?
A. 45%
B. 67%
C. 82%
D. 94% [CORRECT]
Correct Answer: D
Rationale: The transient equilibrium buildup reaches approximately 94% of maximum Tc-99m activity at about 23 hours
post-elution. Option D is correct. Option A corresponds to roughly 6 hours; option B to about 12 hours; option C to about
18 hours.
Q7: According to USP <825> and accepted radiopharmacy standards, what is the maximum acceptable
concentration of aluminum breakthrough in a Tc-99m eluate?
A. Less than 5 mcg/mL
B. Less than 10 mcg/mL [CORRECT]
C. Less than 15 mcg/mL
D. Less than 20 mcg/mL
Correct Answer: B
Rationale: The NRC and USP require that aluminum concentration in Tc-99m sodium pertechnetate be less than 10
mcg/mL. Option B is correct. Options A, C, and D are incorrect thresholds.
Q8: A Tc-99m eluate is tested for molybdenum breakthrough. The eluate contains 50 mCi of Tc-99m. The
acceptable limit for Mo-99 breakthrough is 0.15 mcg Mo per mCi Tc-99m at the time of elution. What is the
maximum allowable amount of Mo-99 in this eluate?
A. 2.5 mcg
B. 5.0 mcg
C. 7.5 mcg [CORRECT]
D. 10.0 mcg
Correct Answer: C
Rationale: Maximum Mo-99 = 0.15 mcg/mCi * 50 mCi = 7.5 mcg. Option C is correct. Option A uses 0.05 mcg/mCi;
option B uses 0.10 mcg/mCi; option D uses 0.20 mcg/mCi.
Page 2
, Nuclear Medicine Final Exam | 2026 Update
Q9: F-18 fluorodeoxyglucose (FDG) is produced in a cyclotron by bombarding O-18 enriched water with
protons. Which of the following correctly describes the decay characteristics of F-18?
A. Beta-minus emitter with 1.7 MeV max energy and 110-minute half-life
B. Positron emitter with 0.635 MeV max energy and 110-minute half-life [CORRECT]
C. Positron emitter with 0.635 MeV max energy and 65-minute half-life
D. Gamma emitter with 511 keV photons and 110-minute half-life
Correct Answer: B
Rationale: F-18 decays by positron emission (beta-plus) with a maximum positron energy of 0.635 MeV and a physical
half-life of 109.8 minutes (approximately 110 min). Option B is correct. Option A incorrectly states beta-minus; option C
has the wrong half-life; option D mislabels it as a direct gamma emitter rather than positron emitter.
Q10: A patient is scheduled for I-131 therapy for thyroid cancer. The prescribed dose is 150 mCi, to be
administered at noon. The dose is prepared at 6:00 AM and reads 150 mCi. The half-life of I-131 is 8.04 days.
Which of the following is the most appropriate action?
A. Administer the dose as prepared since the decay in 6 hours is negligible for an 8-day half-life
B. Recalibrate the dose at noon; the activity will have decayed slightly but remain therapeutically appropriate
[CORRECT]
C. Prepare a new dose at noon to ensure exact 150 mCi is delivered
D. Add 5 mCi to the prepared dose to compensate for decay
Correct Answer: B
Rationale: I-131 with an 8-day half-life decays only minimally in 6 hours (decay factor approximately 0.965, losing about
3.5%), so recalibrating at administration time is the standard practice. Option B is correct. Option A ignores the need to
verify activity; option C is unnecessary waste; option D introduces inaccuracy.
Q11: Lu-177-PSMA-617 is an FDA-approved radiopharmaceutical for the treatment of which condition?
A. Hepatocellular carcinoma
B. Neuroendocrine tumors
C. Metastatic castration-resistant prostate cancer [CORRECT]
D. Differentiated thyroid cancer
Correct Answer: C
Rationale: Lu-177-PSMA-617 (Pluvicto) is approved for the treatment of PSMA-positive metastatic castration-resistant
prostate cancer. Option C is correct. Option B describes Lu-177-DOTATATE (Lutathera); options A and D are incorrect
indications.
Q12: A technologist is preparing a Tc-99m MDP kit in the radiopharmacy. The kit label indicates that
reconstitution should occur with 5-10 mL of sterile sodium pertechnetate. The eluate has an activity
concentration of 25 mCi/mL. The department needs a total of 60 mCi. What volume of pertechnetate should
be added to the kit?
A. 2.4 mL using the highest concentration available
B. 5.0 mL to yield 125 mCi total in the kit vial
C. 8.0 mL to ensure the volume is within the 5-10 mL range and provides adequate activity [CORRECT]
D. 12.0 mL to maximize the total activity available
Correct Answer: C
Rationale: Adding 8 mL of 25 mCi/mL eluate yields 200 mCi total, well above the 60 mCi needed, and the 8 mL volume
falls within the kit's specified 5-10 mL range. Option C is correct. Option A yields only 60 mCi but 2.4 mL is below the
minimum 5 mL; option B yields 125 mCi but 5 mL is at the lower limit; option D exceeds the 10 mL maximum.
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