JPNM Physics
Some Fundamental Physics Considerations
Document created May 30, 1996; updated May 31, 1996
Instrumentation Consderations
All of the physics relating to colllimated F-18 imaging is found in any GOOD book on atomic physics. For example, The Atomic Nucleus by Evans is one of the most complete. Chapter 25 "Attenuation and Absorption of Electromagnetic Radiation" has excellent quantative data from which the tables below were compiled.
At 500 keV:
Mass Attenuation Coefficients
| cm2/g
|
|---|
| Material | Density | Photo | Compt scat | Compt abs | Tot abs | Tot atten
|
|---|
| Water | 1.0 | <<0.001 | 0.063 | 0.030 | 0.030 | 0.093
|
|---|
| NaI | 3.67 | 0.016 | 0.050 | 0.025 | 0.047 | 0.091
|
|---|
| Lead | 11.3 | 0.080 | 0.047 | 0.024 | 0.10 | 0.15
|
|---|
Linear Attenuation Coefficients
| cm-1
|
|---|
| Material | Density | Photo | Compt scat | Compt abs | Tot abs | Tot atten
|
|---|
| Water | 1.0 | <<0.001 | 0.063 | 0.030 | 0.030 | 0.093
|
|---|
| NaI | 3.67 | 0.059 | 0.18 | 0.092 | 0.17 | 0.33
|
|---|
| Lead | 11.3 | 0.90 | 0.53 | 0.27 | 1.1 | 1.7
|
|---|
Some important points to note are the following:
- In water, at this energy (511 kev), there is no photoelectric effect but there is significant Compton scattering.
- Lead has strong photoelectric effect AND very signifcant Compton scattering. This implies Compton scattering in the collimator will be significant.
- NaI has signifigant everything, i.e. photelectric effect and Compton scattering both happen to an appreciable degree. This demonstrates that NaI is a relatively poor detector at this energy.
- First order "shielding" calculations can be based on the half vaue layer:
At 500 keV: At 150 keV:
HVLH2O = 7.5 cm HVLH2O = 4.6 cm
HVLNaI = 2.1 cm HVLNaI = 0.32 cm
HVLPb = 0.41 cm HVLPb = 0.034 cm
- With F-18, for the same attenuation effect (first approximation), we need 1.6 times as much water, 6.6 times as much NaI and 12 times as much lead compared to Tc-99m.
Robert E. Zimmerman, zimmer@bwh.harvard.edu