Case Study 1: ${}^{18}\text{F}$-FDG PET — How Nuclear Physics Finds Cancer

The Patient

Maria, a 54-year-old schoolteacher, has been experiencing persistent fatigue and unexplained weight loss for three months. A CT scan reveals a 3 cm mass in her right lung. A biopsy confirms non-small-cell lung cancer. The oncologist orders a whole-body ${}^{18}\text{F}$-FDG PET/CT scan to stage the disease — to determine whether the cancer has spread to lymph nodes or distant organs — because the treatment strategy depends critically on the extent of the disease.

The Nuclear Physics, Step by Step

Step 1: Producing the Radionuclide (5:00 AM)

At a cyclotron facility 80 km from Maria's hospital, the day begins before dawn. A compact cyclotron (IBA Cyclone 18/9) accelerates protons to 18 MeV in a spiral path using a 1.9 T magnetic field. The proton beam, carrying a current of 40 $\mu$A, strikes a target of ${}^{18}\text{O}$-enriched water ($H_2{}^{18}\text{O}$, enrichment $>97\%$).

The nuclear reaction:

$${}^{18}\text{O} + p \to {}^{18}\text{F} + n \qquad Q = -2.44\,\text{MeV}$$

This is a $(p,n)$ reaction — the proton enters the oxygen-18 nucleus, and a neutron exits. The reaction is endothermic: the proton must supply at least 2.58 MeV in the lab frame (the threshold is slightly above $|Q|$ because of center-of-mass motion). At 18 MeV, the cross section is well above the peak, and the thick-target yield is approximately 9 GBq per $\mu$A$\cdot$hour.

After a 2-hour bombardment at 40 $\mu$A:

$$A_{\text{produced}} = 9 \times 40 \times 2 = 720\,\text{GBq} \approx 19.5\,\text{Ci}$$

This represents approximately $720 \times 10^9 / (0.693/6588) = 6.84 \times 10^{15}$ atoms of ${}^{18}\text{F}$ — a truly tiny mass: $6.84 \times 10^{15} \times 18 / (6.022 \times 10^{23}) = 0.20\,\mu\text{g}$. Nuclear medicine operates at the tracer level — the radioactive atoms are present in quantities too small to cause any pharmacological effect.

Step 2: Radiochemistry (5:30–6:30 AM)

The ${}^{18}\text{F}^-$ ions are extracted from the target water and fed into an automated radiosynthesis module. Through a nucleophilic substitution reaction, ${}^{18}\text{F}$ replaces the triflate leaving group at the 2-position of mannose triflate, followed by hydrolysis to yield ${}^{18}\text{F}$-fluorodeoxyglucose (FDG).

The chemistry must be fast — every minute costs 0.63% of the remaining ${}^{18}\text{F}$ activity. Total synthesis time: approximately 30 minutes. Quality control (radiochemical purity $>95\%$, radionuclidic purity $>99.5\%$, sterility, pyrogenicity, pH, residual solvents) adds another 30 minutes.

By 6:30 AM, the FDG is packaged in a lead-shielded vial and dispatched to hospitals.

Step 3: Decay During Transport

The courier drives 80 km to Maria's hospital, arriving at 8:00 AM — 90 minutes after the end of synthesis. During this time:

$$A(90\,\text{min}) = 720 \times e^{-0.693 \times 90/109.8} = 720 \times e^{-0.568} = 720 \times 0.567 = 408\,\text{GBq}$$

About 43% of the activity has been lost to radioactive decay during transport. This is why ${}^{18}\text{F}$'s half-life is ideal — long enough for this 90-minute journey, but much shorter-lived radionuclides (${}^{15}\text{O}$, $t_{1/2} = 2\,\text{min}$) would be completely useless by arrival.

Step 4: Injection and Biodistribution (9:00 AM)

Maria receives an intravenous injection of 370 MBq (10 mCi) of ${}^{18}\text{F}$-FDG. She is asked to rest quietly for 60 minutes in a dimly lit room (to minimize uptake in muscles and brain cortex) while the FDG distributes through her body.

FDG molecules enter cells via the same glucose transporters (GLUT1, GLUT3) that normal glucose uses. Inside the cell, hexokinase phosphorylates FDG to FDG-6-phosphate — but the fluorine atom at the 2-position blocks the next step of glycolysis. The molecule is trapped.

Cancer cells, with their dramatically upregulated glycolysis (the Warburg effect, first described by Otto Warburg in 1924), consume glucose at 5–10 times the rate of most normal tissues. The tumor in Maria's lung is a biochemical beacon, glowing with radioactive glucose.

Normal FDG distribution. Not all FDG uptake indicates cancer. The brain (which consumes $\sim 25\%$ of the body's glucose) always shows intense FDG uptake. The kidneys and bladder show activity because FDG is excreted in the urine. The liver shows moderate diffuse uptake. The heart shows variable uptake depending on the patient's metabolic state. The interpreting physician must distinguish pathological uptake (the tumor, the metastatic lymph node) from this normal physiological background — a skill that combines nuclear physics knowledge (understanding what produces the signal) with anatomic expertise.

Step 5: Radioactive Decay Inside the Patient

Each ${}^{18}\text{F}$ nucleus in Maria's body has a 50% chance of surviving each 110-minute interval. When it decays:

$${}^{18}\text{F} \to {}^{18}\text{O} + e^+ + \nu_e$$

The positron emerges with a kinetic energy drawn from the Fermi distribution (Chapter 14), ranging from 0 to 634 keV (mean $\approx$ 250 keV). In water-equivalent tissue, this positron loses energy through Coulomb collisions with atomic electrons — the same physics as the Bethe-Bloch formula for charged particles (Chapter 16) — and comes to rest after traveling an average of 0.6 mm.

At thermal energies, the positron captures an electron to form positronium. In the dense molecular environment of tissue, the positronium rapidly undergoes pick-off annihilation:

$$e^+ + e^- \to \gamma(511\,\text{keV}) + \gamma(511\,\text{keV})$$

The two photons emerge back-to-back (within $\pm 0.25°$, due to the residual center-of-mass momentum of the $e^+e^-$ pair).

Step 6: Detection (10:00 AM)

Maria lies in the PET/CT scanner — a donut-shaped ring of 23,000 lutetium-yttrium oxyorthosilicate (LYSO) scintillation crystals, each $4 \times 4 \times 20\,\text{mm}^3$, coupled to silicon photomultipliers (SiPMs). The ring diameter is 80 cm.

When a 511 keV photon strikes an LYSO crystal, it undergoes photoelectric absorption (dominant at this energy in the high-$Z$ lutetium-based crystal, $Z_{\text{Lu}} = 71$) or Compton scattering, producing a flash of scintillation light — approximately 26,000 visible-light photons per absorbed 511 keV photon. These scintillation photons are detected by a silicon photomultiplier (SiPM), which converts them to an electrical pulse with amplitude proportional to the deposited energy. Each detected event is time-stamped with a precision of $\sim 300\,\text{ps}$.

Why LYSO? The ideal PET detector crystal must have high density (to stop 511 keV photons efficiently in a thin crystal), high atomic number (to maximize photoelectric absorption over Compton scattering), high light output (for good energy resolution), and fast scintillation decay time (for good timing resolution and high count-rate capability). LYSO ($\text{Lu}_{1.8}\text{Y}_{0.2}\text{SiO}_5$:Ce) meets all these criteria: density 7.1 g/cm$^3$, effective $Z \approx 65$, light output $\sim 26,000$ photons/MeV, decay time $\sim 40$ ns. A 20 mm thick LYSO crystal stops approximately 80% of incident 511 keV photons.

Coincidence logic: The scanner's electronics continuously monitor all detector pairs. When two detectors register 511 keV photons within a coincidence time window of 4.5 ns, the system records a line of response (LOR) — a line connecting the two detectors along which the annihilation event must have occurred.

The coincidence timing also enables time-of-flight (TOF) PET: the slight difference in arrival times of the two photons ($\Delta t$) constrains the position of the annihilation event along the LOR to a segment of length $\Delta x = c \Delta t / 2$. With 300 ps timing resolution, $\Delta x \approx 3 \times 10^{10} \times 300 \times 10^{-12} / 2 = 4.5\,\text{cm}$ — not pinpoint, but enough to significantly improve signal-to-noise ratio in the reconstructed image.

Step 7: Corrections

The raw coincidence data must be corrected for several physical effects, all rooted in the radiation interaction physics of Chapter 16:

• Attenuation correction: 511 keV photons are attenuated as they traverse the body ($\mu \approx 0.096\,\text{cm}^{-1}$ in soft tissue, dominated by Compton scattering at this energy). A low-dose CT scan provides an attenuation map.

• Scatter correction: Some photons undergo Compton scattering in the body, changing direction and causing mispositioned LORs. Energy windowing (accepting only photons within $\pm 15\%$ of 511 keV) and model-based scatter estimation reduce this artifact.

• Random coincidences: Two photons from independent annihilation events may accidentally fall within the coincidence window. The random rate is $R = 2\tau S_1 S_2$, where $\tau$ is the coincidence window width and $S_1$, $S_2$ are the single-detector count rates.

• Dead-time correction: At high count rates, the electronics may miss events. Measured dead-time models correct for this.

Step 8: Reconstruction and Clinical Result

From $\sim 2 \times 10^8$ true coincidence events collected over a 20-minute whole-body scan (6–7 bed positions), iterative reconstruction algorithms (OSEM — ordered subset expectation maximization) produce a three-dimensional image of FDG distribution with spatial resolution $\sim 4\,\text{mm}$.

Maria's result: The PET/CT scan shows: 1. Intense FDG uptake in the known 3 cm right lung mass (SUV$_{\max} = 12.3$, where SUV — standardized uptake value — normalizes the tissue activity concentration to the injected dose per body weight). 2. A previously undetected 1.5 cm FDG-avid lymph node in the mediastinum (SUV$_{\max} = 8.1$). 3. No distant metastases.

The staging is revised from clinical stage I (surgery alone) to stage IIIA (requiring combined chemotherapy and radiation). The PET scan has changed the treatment plan — and potentially saved Maria from an inadequate surgical approach that would have missed the metastatic lymph node.

The Radiation Dose to the Patient

How much radiation does Maria receive from the scan? This is a critical question — a diagnostic procedure must not itself cause harm.

Dosimetry. The effective dose from a standard FDG PET scan (370 MBq) is approximately 7 mSv. For context: - The annual natural background radiation in the United States is approximately 3.1 mSv. - A chest CT delivers approximately 7 mSv. - The dose limit for occupational radiation workers is 20 mSv/year (ICRP recommendation).

The FDG dose is composed of contributions from the 511 keV annihilation photons (which irradiate the whole body as they traverse tissue) and the positrons (which deposit dose locally at the site of decay). The organs receiving the highest doses are the bladder (the primary excretion route for FDG) at approximately 16 mGy, and the brain (which has high glucose uptake) at approximately 7 mGy.

The additional dose from the CT component of the PET/CT scan (used for attenuation correction and anatomic localization) adds approximately 5–10 mSv, depending on the protocol. Modern low-dose CT protocols have reduced this considerably.

Risk-benefit analysis. The incremental cancer risk from 7 mSv is estimated at roughly 1 in 3,000 (using the conservative linear no-threshold model). For Maria, whose existing cancer is an immediate, life-threatening diagnosis, this minuscule additional risk is overwhelmingly outweighed by the benefit of accurate staging. The PET scan changes her treatment from surgery alone (which would have left residual metastatic disease) to combined modality therapy — a change that substantially improves her survival probability.

The Supply Chain: A Global Nuclear Physics Infrastructure

Maria's scan depended on an infrastructure that spans from nuclear physics research to hospital bedside:

1. Enrichment: ${}^{18}\text{O}$-enriched water ($>97\%$) is produced by distillation or electrolysis — a technically demanding separation of a stable isotope.
2. Cyclotron: A $\sim$\$3 million machine operated by specialized physicists and engineers.
3. Radiochemistry: Automated hot cells and synthesis modules, operated under Good Manufacturing Practice (GMP) regulations.
4. Transport: Time-sensitive delivery (${}^{18}\text{F}$ loses 0.63% of its activity per minute) in shielded containers by dedicated courier networks.
5. Quality control: Radiochemical purity, radionuclidic purity, sterility testing — all completed before the first dose is injected.

This infrastructure serves approximately 5 million PET scans per year worldwide, each requiring roughly 370 MBq of ${}^{18}\text{F}$. The total amount of ${}^{18}\text{F}$ consumed annually — across every hospital, every patient, every scan on Earth — is less than 1 mg. Nuclear medicine operates at the true tracer level: we detect individual nuclear decays in amounts so small they are invisible to chemistry.

The Nuclear Physics Score Card

Discussion Questions

1. Why is ${}^{18}\text{F}$ preferred over ${}^{15}\text{O}$ ($t_{1/2} = 2.04\,\text{min}$) for whole-body oncologic PET, despite the fact that ${}^{15}\text{O}$ can be incorporated into water molecules for perfusion imaging?

2. The positron range of ${}^{82}\text{Rb}$ ($E_{\max} = 3.35\,\text{MeV}$) is approximately 5 mm rms in tissue — nearly 10 times that of ${}^{18}\text{F}$. How does this affect spatial resolution? Why is ${}^{82}\text{Rb}$ still useful for cardiac PET?

3. The annual global demand for FDG PET scans is estimated at $\sim$5 million. Each scan requires $\sim$370 MBq. Estimate the total mass of ${}^{18}\text{F}$ produced globally per year. Comment on the extraordinary sensitivity of nuclear medicine — these scans collectively use a total mass of radioactive material measured in micrograms.

4. The 511 keV photon energy is a "fingerprint" of positron annihilation. Could a future detector system measure this energy with sufficient precision to distinguish annihilation photons from other sources of 511 keV radiation? What would be the advantage?