Self-Assessment Quiz — Chapter 27

Test your understanding of the core concepts before moving on. Try to answer each question before checking the solutions at the end.

Q1. (Multiple Choice) In PET imaging, each positron annihilation event produces:

(a) A single 1.022 MeV photon (b) Two photons, each of energy 511 keV, emitted at 180 degrees (c) Two photons of variable energy emitted at random angles (d) Three photons sharing 1.022 MeV of energy

Q2. (Multiple Choice) The energy 511 keV corresponds to:

(a) The kinetic energy of the emitted positron (b) The rest mass energy of the electron (or positron): $m_e c^2$ (c) The binding energy of fluorine-18 (d) The gamma ray energy of ${}^{99\text{m}}\text{Tc}$

Q3. (True/False) PET imaging requires a physical collimator to determine the direction of the detected photons.

Q4. (Short Answer) Why does ${}^{18}\text{F}$-FDG accumulate preferentially in tumor tissue? Your answer should include the biological mechanism and the chemical reason FDG is trapped inside cells.

Q5. (Multiple Choice) The half-life of ${}^{99\text{m}}\text{Tc}$ is 6.01 hours. What fraction of the original activity remains after 24 hours?

(a) 50% (b) 25% (c) 12.5% (d) 6.25%

Q6. (Multiple Choice) The ${}^{99}\text{Mo}/{}^{99\text{m}}\text{Tc}$ generator works because:

(a) Molybdenum and technetium are chemically identical (b) The parent (${}^{99}\text{Mo}$) has a much longer half-life than the daughter (${}^{99\text{m}}\text{Tc}$), establishing secular/transient equilibrium (c) ${}^{99}\text{Mo}$ spontaneously fissions into ${}^{99\text{m}}\text{Tc}$ (d) The generator concentrates ${}^{99\text{m}}\text{Tc}$ from the environment

Q7. (Short Answer) Explain why the gamma ray energy of ${}^{99\text{m}}\text{Tc}$ (140.5 keV) is described as "ideal" for SPECT imaging. What would happen if the energy were much lower (30 keV) or much higher (500 keV)?

Q8. (Multiple Choice) The Bragg peak in a proton dose-depth curve occurs because:

(a) Protons are absorbed by nuclear reactions at a specific depth (b) The stopping power $(-dE/dx)$ increases as $1/v^2$ as the proton slows, causing maximum energy deposition just before the proton stops (c) Protons undergo pair production at a specific depth (d) The proton decays at the end of its range

Q9. (True/False) A 200 MeV proton beam has an exit dose — it deposits energy in tissue beyond the tumor, just like a photon beam.

Q10. (Multiple Choice) Carbon ion therapy has a higher relative biological effectiveness (RBE) than proton therapy primarily because:

(a) Carbon ions travel faster than protons (b) Carbon ions have higher linear energy transfer (LET), producing more densely ionizing tracks that cause irreparable clustered DNA damage (c) Carbon ions are heavier and therefore harder to stop (d) Carbon ions are radioactive

Q11. (Short Answer) Define relative biological effectiveness (RBE) in one equation and one sentence.

Q12. (Multiple Choice) Targeted alpha therapy (TAT) is effective at killing individual tumor cells because:

(a) Alpha particles have a range of several centimeters in tissue (b) Alpha particles have a range of $\sim 50$–$80\,\mu\text{m}$ (a few cell diameters) and deposit $\sim 100\,\text{keV}/\mu\text{m}$, causing lethal DNA damage within the targeted cell (c) Alpha particles are neutral and therefore penetrate deeply (d) Alpha particles stimulate the immune system

Q13. (True/False) In the ${}^{225}\text{Ac}$ decay chain, the recoil energy of the first daughter (${}^{221}\text{Fr}$) is sufficient to break it free from the targeting molecule.

Q14. (Short Answer) What is the "theranostic paradigm"? Give one specific clinical example with the diagnostic and therapeutic radionuclides named.

Q15. (Multiple Choice) In the MIRD dosimetry formalism, the absorbed dose to a target region is:

(a) $D = A_0 \times t_{1/2}$ (b) $D = \sum_{r_S} \tilde{A}(r_S) \times S(r_T \leftarrow r_S)$ (c) $D = E_\gamma / m$ (d) $D = \lambda \times N \times E$

Q16. (True/False) The effective half-life of a radionuclide in the body is always shorter than or equal to its physical half-life.

Q17. (Short Answer) Calculate the effective half-life of ${}^{131}\text{I}$ in the thyroid if the physical half-life is 8.02 days and the biological half-life in the thyroid is 80 days.

Q18. (Multiple Choice) ${}^{131}\text{I}$ therapy for thyroid cancer works because:

(a) The thyroid naturally concentrates iodine via the sodium-iodide symporter, and the $\beta^-$ particles from ${}^{131}\text{I}$ deposit their energy locally in thyroid tissue (b) ${}^{131}\text{I}$ emits alpha particles that destroy the thyroid (c) ${}^{131}\text{I}$ is chemically toxic to thyroid cells (d) The gamma rays from ${}^{131}\text{I}$ are focused on the thyroid by external magnets

Q19. (Short Answer) A patient is injected with 370 MBq of ${}^{18}\text{F}$-FDG. What activity remains after 5 half-lives ($\approx 9.2$ hours)?

Q20. (Multiple Choice) Which of the following is NOT a reason that ${}^{18}\text{F}$ is the most commonly used PET radionuclide?

(a) Its 110-minute half-life allows time for synthesis and transport (b) Its low positron energy gives good spatial resolution (c) It can be produced in a nuclear reactor (d) Fluorine forms a strong bond with carbon, enabling attachment to biomolecules

Solutions

Q1. (b) Two 511 keV photons at 180 degrees. Energy and momentum conservation in the nearly-at-rest $e^+e^-$ annihilation require two equal-energy photons emitted back-to-back.

Q2. (b) $m_e c^2 = 0.511\,\text{MeV} = 511\,\text{keV}$.

Q3. False. PET uses electronic collimation — coincidence detection of back-to-back 511 keV photon pairs defines the line of response without a physical collimator. This is why PET has higher sensitivity than SPECT.

Q4. Tumor cells have upregulated glucose metabolism (the Warburg effect) and overexpress glucose transporters (GLUT1/GLUT3). FDG enters cells via these transporters and is phosphorylated by hexokinase to FDG-6-phosphate. Because the 2-hydroxyl group is replaced by ${}^{18}\text{F}$, the molecule cannot proceed through glycolysis and is metabolically trapped, accumulating to high concentrations in tumors.

Q5. (d) $24/6.01 \approx 4$ half-lives. Fraction remaining $= (1/2)^4 = 1/16 = 6.25\%$.

Q6. (b) The long-lived parent ($t_{1/2} = 65.94\,\text{h}$) decays to the short-lived daughter ($t_{1/2} = 6.01\,\text{h}$), establishing transient (nearly secular) equilibrium. The daughter regrows after each elution.

Q7. At 140.5 keV, the photons are energetic enough to escape the body without excessive attenuation, but low enough to be efficiently detected by NaI(Tl) crystals and effectively collimated by lead collimators. At 30 keV, most photons would be absorbed before reaching the detector (high attenuation in tissue). At 500 keV, the collimator would need to be impractically thick, reducing sensitivity and spatial resolution.

Q8. (b) The $1/v^2$ dependence of the Bethe-Bloch stopping power means the proton deposits more energy per unit path length as it slows. The maximum energy deposition occurs just before the proton stops — this is the Bragg peak.

Q9. False. Protons have a well-defined range. Beyond the Bragg peak, the dose drops to essentially zero (ignoring the small contribution from nuclear fragmentation products). This is the fundamental advantage of proton therapy over photon therapy.

Q10. (b) Carbon ions have $z = 6$, so their LET ($\propto z^2$) is 36 times that of protons at the same velocity. The dense ionization track produces clustered, irreparable DNA damage.

Q11. $\text{RBE} = D_{\text{photon}} / D_{\text{particle}}$ at the same biological effect. RBE measures how much more effective a given type of radiation is at producing a specific biological endpoint compared to a reference photon beam.

Q12. (b) The short range ($\sim 50$–$80\,\mu\text{m}$) and high LET ($\sim 100\,\text{keV}/\mu\text{m}$) of alpha particles mean they deposit massive energy within a few cell diameters, producing lethal clustered DNA damage in the targeted cell while leaving cells beyond the range unharmed.

Q13. True. The recoil energy of ${}^{221}\text{Fr}$ is $\sim 106\,\text{keV}$, which is $\sim 10^4$–$10^5$ times the strength of any chemical bond ($\sim 1$–$5\,\text{eV}$). The daughter is ejected from its molecular carrier with certainty.

Q14. The theranostic paradigm uses the same molecular platform to both diagnose and treat disease: a diagnostic radionuclide (for imaging) identifies patients whose tumors express the target, and a therapeutic radionuclide (for treatment) is then delivered via the same molecule. Example: ${}^{68}\text{Ga}$-PSMA-11 (PET imaging) followed by ${}^{177}\text{Lu}$-PSMA-617 (therapy) for metastatic prostate cancer.

Q15. (b) $D = \sum_{r_S} \tilde{A}(r_S) \times S(r_T \leftarrow r_S)$, where $\tilde{A}$ is the cumulated activity and $S$ is the dose per unit cumulated activity.

Q16. True. $1/t_{1/2,\text{eff}} = 1/t_{1/2,\text{phys}} + 1/t_{1/2,\text{biol}}$, so $t_{1/2,\text{eff}} \leq t_{1/2,\text{phys}}$ (with equality only if there is no biological clearance, $t_{1/2,\text{biol}} \to \infty$).

Q17. $t_{1/2,\text{eff}} = (8.02 \times 80)/(8.02 + 80) = 641.6/88.02 = 7.29\,\text{days}$.

Q18. (a) The thyroid gland naturally concentrates iodine via the sodium-iodide symporter (NIS). ${}^{131}\text{I}$ is taken up like stable iodine, and its $\beta^-$ particles (mean range $\sim 0.4\,\text{mm}$) deposit energy locally, destroying the thyroid tissue.

Q19. $A = 370 \times (1/2)^5 = 370/32 = 11.6\,\text{MBq}$.

Q20. (c) ${}^{18}\text{F}$ is produced in a cyclotron via the ${}^{18}\text{O}(p,n){}^{18}\text{F}$ reaction, not in a nuclear reactor. Reactor production of ${}^{18}\text{F}$ is not practical.