PET Course KFSH&RC General target theory, design and operation

1. 1
PET Isotope Targets
The Nucleus
•Composed of two nuclear particles (nucleons)
•Proton – positively charged
•Neutron – not charged
•The number of protons determines the element
•The total of neutrons and protons determines the mass
•For example
Name Number of
protons
Number of
neutrons
Total number of
nucleons
F-19 9 10 19
O-18 8 10 18
F-18 9 9 18

2. 2
PET Isotope Targets
Decay
•Most configurations of protons and neutrons are not stable, and decay
Forms of decay (transmutation)
•Electron capture (EC)
•An atomic electron is captured by the nucleus. A proton is
consumed, a neutron is generated, and a neutrino (very light, almost
undetectable particle) is emitted
• emissionα
•The nucleus spontaneously emits an particle (same thing as aα
helium nucleus, 2 protons and 2 neutrons)
• - emissionβ
•A neutron in the nucleus is converted into a proton, an electron and
an antineutrino

3. 3
PET Isotope Targets
• + (positron) emissionβ
•A proton in the nucleus is converted into a neutron, a positron
(antimatter electron) and a neutrino
•When F-18 (9 protons, 9 neutrons) decays, it goes to O-18 (8
protons, 10 neutrons)
•Also written 18
F
pn
+β
ν
There are other nuclear reactions, such as internal conversion or
gamma ray emission, that do not result in a change of element. These
are referred to as isomeric reactions

4. 4
PET Isotope Targets
Decay (cont.)
•The units of decay rate are Curies or Becquerels
•One Curie is 3.7 x 1010
decays per second
•The Becquerel is one decay per second
•One Curie is 37 Giga-Becquerels
•This is the typical way that radioactivity is quantified
•Note that this is the rate of decays, which is different from the total
number of activated nuclei

5. 5
PET Isotope Targets
Decay (cont.)
•There is a fixed probability of decay for each nucleus
•This results in a fixed fraction of the population decaying in a given
time period (exponential decay)
•The time it takes for the decay rate to drop to half its value is called the
half-life
•The decay rate at a given period of time can be calculated by
2/1693.0
0)( tt
eAtA ⋅−
⋅=
where:
•A0 = decay rate at t = 0
•t1/2 = half life (in same units as t)
•t = time since A0 was measured (same units as t1/2)

6. 6
PET Isotope Targets
Decay (cont.)
Decay of 1 Ci of F-18 over 8 hours
0
125
250
375
500
625
750
875
1000
1125
0 109.8 219.6 329.4 439.2
minutes
Activity(mCi)

7. 7
PET Isotope Targets
Decay (cont.)
•Each radioisotope has a unique half-life.
•The four commonly used isotopes in PET and their half lives are:
Name Half life
F-18 109.8 minutes
C-11 20.2 minutes
N-13 9.96 minutes
O-15 122 seconds

8. 8
PET Isotope Targets
Activation
•Stable nuclei can be bombarded with nucleons (neutrons or protons) or
other nuclei to form activated nuclei
•This process is referred to as a nuclear reaction
•The nomenclature for nuclear reactions is
target nucleus(bombarding particle, recoiling particle)product nucleus
For example
18
O(p,n)18
F
means 18
O bombarded by protons, results in 18
F and a neutron

9. 9
PET Isotope Targets
Production
•As a nuclear reaction takes place, the population of activated nuclei
grows, as does the decay rate
•At some point the decay rate and activation rate are equal
•The activity at this point is saturated. The saturation activity is an
indication of the production rate. Knowing the saturation activity one
can calculate the expected activity at any time using the following
equation:
( )2/1693.0
1)( tt
sat eAtA ⋅−
−=
where:
•Asat = Activity at saturation
•t1/2 = half life (in same units as t)
•t = time since beginning of production (same units as t1/2)

10. 10
PET Isotope Targets
Yield curve
•Saturation activity is an
indication of production rate
•Production rate is
proportional to current
•Production rate is sensitive
to energy
F-18 Thick target saturation yield vs. energy
0
50
100
150
200
250
300
0 2 4 6 8 10 12 14 16
Energy (MeV)
Yield(mCi/uAatsat.)

11. 11
PET Isotope Targets
Total number of F-18 nuclei (cont.)
For 1 Ci of F-18 (109.8 minute half life):
693.0
minsec60min8.109
sec107.3 1-
0
693.0
2/1 ⋅
⋅=⋅ eA
t
atomse1452.3= 352 trillion nuclei!

12. 12
PET Isotope Targets
Proton bombardment
•Protons have elemental charge = 1.6 x 10-19
Coulombs
•You can work out how many protons you have per microamp of beam
current
second
coulomb
amp =
second
coulomb
amp 6
101 −
=µ
coulomb
proton
second
coulomb
amp 19
6
106.1
101 −
−
×
⋅=µ
second
proton12
1025.6 ×=

13. 13
PET Isotope Targets
Exercise: calculate total number of protons required to produce
one F-18 nucleus
•Get number of F-18 nuclei produced per second from one microamp
•A good saturation activity for F-18 is 100 mCi/μA
•At saturation, decay and production are equal, so this is also an indication
of production rate
second
F
mCi
sec
A
mCi 18
9
17
107.3
107.3
100 ×=
×
⋅
−
µ
•We know number of protons per second in 1 microamp from previous
slide
F
proton
second
F
second
proton
1818
9
12
1689
107.3
1025.6
=
×
×

14. 14
PET Isotope Targets
Other Bombardment Processes
-Nuclear reactions (already looked at)
-Ionization heating
•Primary proton energy loss mechanism
•Secondary effects on material are the primary cause of stress to
targets
•Thermal expansion
•Pressure increases
•Temperature increases
-Hot atom chemistry
•Determines chemical form of the resulting nucleus and chemical
environment of the target

15. 15
PET Isotope Targets
Other Bombardment Processes – Ionization
•As it passes through matter, the energetic particle can be treated as a
bare nucleus with positive charge close to its atomic number.
•Protons (Hydrogen) have a charge of 1+
•Energetic F-18 would have a net charge of 9+
•As they pass atoms they pick up and shed atomic electrons many
times. This process is called ionization.
•The removal of the electron requires energy which is provided by the
bombarding particle.
•The bombarding loses energy, and the medium it passes through is
heated.
•Ionizations are much more frequent than nuclear reactions.

16. 16
PET Isotope Targets
Other Bombardment Processes – Ionization
Bethe Stopping Power equation





 ⋅⋅
⋅⋅⋅
⋅
⋅⋅⋅
=−
I
vm
ZN
vm
ze
dx
dE e
e
2
2
24
2
ln
4 π
Where:
N = number density of target atoms
Z = average atomic number of target material
z = charge of projectile
me = electron mass
v = projectile velocity

17. 17
Proton energy loss in water
0
2
4
6
8
10
12
0 0.05 0.1 0.15
Beam penetration, cm
Energy,MeV
0
100
200
300
400
500
600
700
800
900
1000
-dE/dx,MeV/cm
Energy
dE/ dx
PET Isotope Targets
Stopping power - the Bethe Equation

18. 18
PET Isotope Targets
Other Bombardment Processes – Ionization
How much heat?
•Total heat deposited is current times energy (μA* MeV)
•Example: 11 MeV * 40 μA = 440 watts
•Maximum heat density (in the Bragg peak)
32
000,68
5.0
40
850
cm
Watts
cm
A
cm
MeV
=⋅
µ

19. 19
PET Isotope Targets
Other Bombardment Processes – Pressure/expansion
•Heat causes expansion and pressure increases
•Gases undergo thermal expansion (ideal gas law)
nRTPV =
where
P = pressure
V = volume
N = number of gas molecules
R = gas constant
T = temperature

20. 20
PET Isotope Targets
Other Bombardment Processes – Hot atom chemistry
•The nuclear reaction must conserve momentum and energy
•Product radionuclide will have energy (on the order of a few MeV)
after the reaction, moving rapidly through the target medium, and will
be highly ionized
•The medium will also be highly ionized
•The final chemical form of the product radionuclide is sensitive to:
•Small contaminants
•Wall materials
•Dose rate (eV/molecule - minute)
•Dose (eV/molecule)

21. 21
PET Isotope Targets - Design
Target elements

22. 22
Eclipse High Power F- Target

23. 23
PET Isotope Targets - Design
Design constraints
•Temperature
•Pressure
•Medium and phase (liquid
water or hot gas)
•Chemistry
•Volume
•Lifetime
•Energy/yield

24. 24
High Power Target Body

25. 25
PET Isotope Targets - Design
Support equipment
•Small volumes, high pressures and low contamination
•Much of the equipment comes from HPLC
•Capillary tubing
•Chromatography valves (up to 1000 psig)
•Pressure transducers
•Syringe pumps
•Diaphragm pumps
•HPLC pumps

26. 26
PET Isotope Targets - Systems
Isotope: 18
F Reaction: 18
O(p,n)18
F
Chemical form: Fluoride ion (F-)
Target material: 18
O enriched water (argon overpressure)
Target volume: RD: 1.1 ml
HP: 2.3 ml
Target body: 99.99% pure silver or tantalum
Target window: .001” thick Havar (high strength Nickel-Cobalt
alloy)
Typical pressure: RD: 650 psig beam off, ~800 psig beam on
HP: 340 –350 psig beam off, ~600 psig beam on

27. 27
PET Isotope Targets - Systems
Isotope: 18
F Reaction: 18
O(p,n)18
F
Chemical form: Fluorine gas (F2)
Target material: 18
O enriched gas (Ar/5% F2 second shoot)
Target volume: RD: 7 ml (120 std. cc gas)
HP: no data
Target body: Aluminum
Target window: .001” thick Havar (high strength Nickel-Cobalt
alloy)
Typical pressure: 280 psig beam off, 900 psig beam on

28. 28
PET Isotope Targets - Systems
Isotope: 13
N Reaction: 16
O(p,α)13
N
Chemical form: Ammonium ion (NH4) in water
Target material: 5 mMol Ethanol in HPLC water
Target volume: RD: 2 ml
HP: No Data
Target body: Aluminum
Target window: .001” thick Titanium
Typical pressure: 320 psig

29. 29
PET Isotope Targets - Systems
Isotope: 15
O Reaction: 15
N(p,n)15
O
Chemical form: Oxygen gas (O2) in Nitrogen gas (N2)
Target material: 15
N enriched Nitrogen gas with 2.5% O2
Target volume: RD: 7 ml (120 std. cc)
HP: No Data
Target body: Aluminum
Target window: .001” thick Havar (high strength Nickel-Cobalt
alloy)
Typical pressure: 280 psig beam off, 900 psig beam on

30. 30
PET Isotope Targets - Systems
Isotope: 11
C Reaction: 14
N(p,α)11
C
Chemical form: Carbon dioxide gas (CO2) in Nitrogen gas (N2)
Target material: unenriched Nitrogen gas with 2.5% O2
Target volume: RD: 7 ml (120 std. cc)
HP: No Data
Target body: Aluminum
Target window: .001” thick Havar (high strength Nickel-Cobalt
alloy)
Typical pressure: 280 psig beam off, 900 psig beam on