Question 12.10.Q1: Radionuclides produced in cyclotron with charged particle ac......
Radionuclides produced in cyclotron with charged particle activation are positron emitters used in positron emission tomography (PET) scanners for diagnostic imaging. PET scanning is considered a non-invasive imaging technique that provides a functional image of organs and tissues, in contrast to CT scanning and MRI scanning that provide anatomic images of organs and tissues. Carbon-11, a positron emitter decaying through β^+ decay with a halflife t_{1/2} of 20.4 minutes into boron-11, is the most stable artificial radioisotope of carbon and is one of four common positron emitters used in medicine for imaging with PET scanners.
A cyclotron operating at 40 µA and generating a 15 MeV proton beam is used to produce carbon-11 radionuclide through nuclear activation of a gaseous nitrogen-14 pressurized target. The principal nuclear activation reaction for production of carbon-11 is
{ }^{14} \mathrm{~N}+p \rightarrow{ }^{11} \mathrm{C}+\alpha . (12.321)
(a) Calculate the nuclear reaction energy Q for activation reaction (12.321). The appropriate rest energies and nuclear binding energies of components of (12.321) are provided in Appendix A. Use the three known methods for calculation of Q value (nuclear rest energy, atomic rest energy, and nuclear binding energy) and show that they give the same result.
(b) Calculate threshold kinetic energy \left(E_{\mathrm{K}}\right)_{\mathrm{thr}}^{\mathrm{p}} of the proton for the proton activation reaction (12.321) assuming that the target parent nucleus (nitrogen-14) is initially at rest.
(c) Determine the effective gaseous nitrogen-14 activation target thickness x_{eff} for the incident 15 MeV proton beam, if the pressure of the gas is 15 atm and its temperature is 20 °C. The density of nitrogen gas at STP (0 °C and 1 atm) is \rho_{\mathrm{STP}}=1.251 \times 10^{-3} \mathrm{~g} / \mathrm{cm}^3 .
(d) If the activity of the daughter is 3.52 Ci after an activation time t = \left(t_{1/2}\right)_{11_C}, calculate the mean activation cross section \bar{\sigma}_{\mathrm{P}} in the (12.321) activation reaction for the incident proton beam of kinetic energy \left(E_{\mathrm{K}}\right)_0^{\mathrm{p}}.
(a) Nuclear reaction energy Q, also known as reaction Q value for a nuclear reaction, provides the energy release or energy absorption during the nuclear reaction. We will determine Q value of the activation reaction (12.321) with the following three methods:
(1) Nuclear rest energy method: The sum of nuclear rest energies of the reaction products (i.e., total nuclear rest energy after reaction) is subtracted from the sum of nuclear rest energies of the reactants (i.e., total nuclear rest energy before reaction)
(2) Atomic rest energy method: The sum of atomic rest energies of the reaction products (i.e., total atomic rest energy after reaction) is subtracted from the sum of atomic rest energies of the reactants (i.e., total atomic rest energy before reaction)
(3) Binding energy method: The sum of nuclear binding energies of the reactants (i.e., total binding energy before reaction) is subtracted from the sum of nuclear binding energies of reaction products (i.e., total binding energy after reaction)
As expected, the three calculation methods of Q value (activation energy) give the same result equal to −2.92 MeV, indicating that the activation reaction is endothermic (endoergic) and that for the activation to happen, the proton must have a certain minimum kinetic energy called the threshold kinetic energy \left(E_{\mathrm{K}}\right)_{\mathrm{thr}}^{\mathrm{p}} thr that will be determined in (b).
(b) An exothermic reaction can occur spontaneously; an endothermic reaction cannot take place unless the projectile possesses total energy E exceeding the reaction threshold energy E_{thr}. This means that the projectile must possess kinetic energy exceeding threshold kinetic energy \left(E_K\right)_{thr} that can be determined from the relativistic invariant
\mathcal{J}=E^2-p^2 c^2=\mathrm{inv} (12.325)
where E stands for the total energy before the collision and total energy after the collision and p is the total momentum before collision and total momentum after the collision.
The invariant for conditions before the collision is written in laboratory coordinate system, the invariant for conditions after the collision is written in the centerof-mass coordinate system. For a general endothermic nuclear reaction A(a, b)B, where a is the projectile and A is the target, the relativistic invariant is given as follows
\mathcal{J}_{\text {before }}=\left(\sqrt{m_{\mathrm{a}}^2 c^4+p_{\mathrm{a}}^2 c^2}+m_{\mathrm{A}} c^2\right)^2-p_{\mathrm{a}}^2 c^2=\mathcal{J}_{\mathrm{after}}=\left(m_{\mathrm{B}} c^2+m_{\mathrm{b}} c^2\right)^2-0 (12.326)
Solving (12.326) for E_{\mathrm{thr}}^{\mathrm{a}}=\sqrt{m_{\mathrm{a}}^2 c^4+p_{\mathrm{a}}^2 c^2}, with p_a the total momentum before interaction, results in the following expression for the total threshold energy E_{\mathrm{thr}}^{\mathrm{a}} (T5.12) of the nuclear reaction
E_{\mathrm{thr}}^{\mathrm{a}}=\frac{\left(m_{\mathrm{B}} c^2+m_{\mathrm{b}} c^2\right)^2-\left(m_{\mathrm{a}}^2 c^4+m_{\mathrm{A}}^2 c^4\right)}{2 m_{\mathrm{A}} c^2} (12.327)
Noting that E_{\mathrm{thr}}^{\mathrm{a}}=\left(E_{\mathrm{K}}\right)_{\mathrm{thr}}^{\mathrm{a}}+m_{\mathrm{a}} c^2 \text {, where }\left(E_{\mathrm{K}}\right)_{\mathrm{thr}}^{\mathrm{a}} is the threshold kinetic energy of the projectile, we get the following expression for \left(E_{\mathrm{K}}\right)_{\mathrm{thj}}^{\mathrm{a}} (T5.13)
\left(E_{\mathrm{K}}\right)_{\mathrm{thr}}^{\mathrm{a}}=\frac{\left(m_{\mathrm{B}} c^2+m_{\mathrm{b}} c^2\right)^2-\left(m_{\mathrm{a}} c^2+m_{\mathrm{A}} c^2\right)^2}{2 m_{\mathrm{B}} c^2} (12.328)
The threshold kinetic energy \left(E_{\mathrm{K}}\right)_{\mathrm{thr}}^{\mathrm{a}} of the projectile given in (12.328) may now be written in terms of nuclear reaction Q value. First, we note that from the definition of Q value (T5.5) we can write the following expression linking Q value with rest masses of the projectile m_a, target m_A, and reaction products m_B\ and\ m_b.
\left(m_{\mathrm{B}} c^2+m_{\mathrm{b}} c^2\right)^2=\left(m_{\mathrm{a}} c^2+m_{\mathrm{A}} c^2\right)^2+Q^2-2 Q\left(m_{\mathrm{a}} c^2+m_{\mathrm{A}} c^2\right) (12.329)
Inserting the relationship (12.329) into (12.328) we obtain the following expression for \left(E_{\mathrm{K}}\right)_{\mathrm{thr}}^{\mathrm{a}} (T5.15)
\left(E_{\mathrm{K}}\right)_{\mathrm{thr}}^{\mathrm{a}}=-Q\left[\frac{m_{\mathrm{a}} c^2+m_{\mathrm{A}} c^2}{m_{\mathrm{A}} c^2}-\frac{Q}{2 m_{\mathrm{A}} c^2}\right] \approx-Q\left(1+\frac{m_{\mathrm{a}}}{m_{\mathrm{A}}}\right), (12.330)
where, since Q \ll m_{\mathrm{A}} c^2 \text {, we can ignore the } Q / 2 m_{\mathrm{A}} c^2 term in (12.330).
In (12.330) the threshold kinetic energy \left(E_{\mathrm{K}}\right)_{\mathrm{thr}}^{\mathrm{a}} of the projectile exceeds the |Q| value by a relatively small amount to account for conservation of both energy and momentum in the collision.
momentum in the collision.
We now use (12.327), (12.328), and (12.330) to calculate total threshold energy E_{\mathrm{thr}}^{\mathrm{p}} and threshold kinetic energy \left(E_{\mathrm{K}}\right)_{\mathrm{th}}^{\mathrm{p}} of the incident proton in the proton activation reaction { }^{14} \mathrm{~N}+p \rightarrow{ }^{11} \mathrm{C}+\alpha given in (12.321).
(1) Threshold energy E_{\mathrm{thr}}^{\mathrm{p}} of the proton using (12.327) is expressed as
resulting in threshold kinetic energy of 941.4055 MeV − 938.2720 MeV = 3.1335 MeV.
(2) Threshold energy E_{\mathrm{thr}}^{\mathrm{p}} of the proton using (12.328)
(3) Threshold kinetic energy \left(E_{\mathrm{K}}\right)_{\mathrm{thr}}^{\mathrm{p}} of the proton using (12.330)
From (12.331), (12.332), and (12.333) we note that the calculated threshold results are consistent, since
\left(E_{\mathrm{K}}\right)_{\mathrm{thr}}^{\mathrm{p}}=E_{\mathrm{thr}}^{\mathrm{p}}-m_{\mathrm{p}} c^2 \approx-Q\left(1+\frac{m_{\mathrm{p}} c^2}{M\left({ }_7^{14} \mathrm{~N}\right) c^2}\right)=3.13 \mathrm{MeV} (12.334)
In (a) we determined the activation energy |Q| = 2.92 MeV (12.322) for activation reaction (12.321) and in (12.334) we determined that the threshold proton kinetic energy \left(E_{\mathrm{K}}\right)_{\mathrm{thr}}^{\mathrm{p}}=3.13 \mathrm{MeV} \text { that is } \sim 7 \% higher than |Q| to account for the recoil kinetic energy of the target.
(c) Cyclotron targets are most commonly of the thick target variety, resulting in complete beam absorption in the target material. As the proton beam penetrates the target, its energy decreases as a result of Coulomb interactions with the orbital electrons of the target. Nuclear activation can only be produced in the target in layers where kinetic energy of protons is greater than or equal to the threshold kinetic energy of the projectile for the nuclear reaction to occur. We refer to this target thickness as the effective activation target thickness x_{eff} for a given incident kinetic energy \left(E_{\mathrm{K}}\right)_0^{\mathrm{p}} and determine it using the continuous-slowing-down-approximation range R_{CSDA} as follows
x_{\text {eff }}=\frac{R_{\mathrm{CSDA}}\left[\left(E_{\mathrm{K}}\right)_0^{\mathrm{p}}\right]-R_{\mathrm{CSDA}}\left[\left(E_{\mathrm{K}}\right)_{\mathrm{thr}}^{\mathrm{p}}\right]}{\rho}, (12.355)
where ρ is the mass density of the target and we accounted for threshold kinetic energy \left(E_{\mathrm{K}}\right)_{\mathrm{thr}}^{\mathrm{p}} of the proton by subtracting the proton CSDA range R_{\mathrm{CSDA}}\left[\left(E_{\mathrm{K}}\right)_{\mathrm{thr}}^{\mathrm{p}}\right] for the threshold kinetic energy from the proton range R_{\mathrm{CSDA}}\left[\left(E_{\mathrm{K}}\right)_0^{\mathrm{p}}\right] for the incident proton kinetic energy \left(E_{\mathrm{K}}\right)_0^{\mathrm{P}}.
The nitrogen-14 target of our example is pressurized to P = 15 and ran at a temperature T = 20 °C, resulting in the following target density ρ determined from the ideal gas formula
Looking up data for the proton CSDA range R_{CSDA} in nitrogen-14 from the NIST database (shown in Fig. 12.18) and using the density ρ calculated in (12.336), we get the following result for the required effective nitrogen target thickness x_{eff}.
where we used 15 MeV for the incident proton energy \left(E_{\mathrm{K}}\right)_0^{\mathrm{p}} and 3.13 MeV for threshold proton energy \left(E_{\mathrm{K}}\right)_0^{\mathrm{p}}, as determined in (12.334). Note: Proton data in the NIST table (physics.nist.gov/PhysRefData/Star/Text/PSTAR.html) are given for density at temperature of 20 °C and pressure of 101.325 kPa (1 atm).
(d) Discussions presented for neutron activation could in principle be generalized to charged particle activation; however, the following points should be considered:
(1) In neutron activation the target is immersed in a see of thermal neutrons and the neutron fluence is constant. In charged particle activation, beam attenuation in thick targets that are routinely used in production of medical positronemitting radionuclides complicates matters considerably. In the thick target the particle beam is completely stopped in the target or, at least, it is degraded in energy to a level below the threshold energy.
(2) As the charged particles traversing a thick target lose energy through Coulomb interactions with orbital electrons of the target, the activation yield is affected, since the cross section for activation depends on charged particle energy.
(3) The specific activities produced by charged particle activation are several orders of magnitude lower than those produced in neutron activation, so that in general parent nuclide depletion is not of concern in charged particle activation.
Charged particle activation is illustrated in Fig. 12.19. The charged particle beam is striking a target of cross sectional area A and activation takes place in the target from the surface to effective depth x_{eff}. Since the saturation model can be used to describe charged particle activation, we express the growth of activity d\mathcal{A}(t) in a slab thickness of dx as
\mathrm{d} \mathcal{A}(t)=\sigma_{\mathrm{P}}\left(E_{\mathrm{K}}\right) \dot{\varphi}\left(1-e^{-\lambda_{\mathrm{D}} t}\right) \mathrm{d} N_{\mathrm{P}}(0), (12.338)
where σ_P\left(E_K\right) is the activation cross section at proton kinetic energy E_K and dN_P(0) is the number of parent nuclei in the slab thickness dx.
For a beam current I the particle fluence rate \dot{φ} is given as
\dot{\varphi}=\frac{I}{q A}, (12.339)
where q is the charge of the incident charged particle and A is the cross sectional area of the beam. \mathrm{d} N_{\mathrm{P}}(0) can be expressed in terms of the density n^{\square} of the target nuclei (number of target nuclei per volume) as
\mathrm{d} N_{\mathrm{P}}(0)=n^{\square} A \mathrm{~d} x . (12.340)
Upon inserting (12.339) and (12.340) into (12.338) we get the following expression
\mathrm{d} \mathcal{A}(t)=n \frac{I}{q}\left(1-e^{-\lambda_{\mathrm{D}} t}\right) \sigma_{\mathrm{P}}\left(E_{\mathrm{K}}\right) \mathrm{d} x . (12.341)
The total daughter activity is obtained by integrating (12.341) as follows
\mathcal{A}(t)=\int_0^{x_{\mathrm{eff}}} \mathrm{d} \mathcal{A}(t)=n^{\square} \frac{I}{q}\left(1-e^{-\lambda_{\mathrm{D}} t}\right) \int_0^{x_{\mathrm{eff}}} \sigma_{\mathrm{P}}\left(E_{\mathrm{K}}\right) \mathrm{d} x (12.342)
The second integral in (12.341) can be expressed as
\int_0^{x_{\text {eff }}} \sigma_{\mathrm{P}}\left(E_{\mathrm{K}}\right) \mathrm{d} x=\int_{\left(E_{\mathrm{K}}\right)_0^{\mathrm{p}}}^{\left(E_{\mathrm{K}}\right)_{\text {thr }}^{\mathrm{p}}} \sigma_{\mathrm{P}}\left(E_{\mathrm{K}}\right) \frac{\mathrm{d} x}{\mathrm{~d} E_{\mathrm{K}}} \mathrm{d} E_{\mathrm{K}} \int_{\left(E_{\mathrm{K}}\right)_0^{\mathrm{p}}}^{\left(E_{\mathrm{K}}\right)_{\text {thr }}^{\mathrm{p}}} \frac{\sigma_{\mathrm{P}}\left(E_{\mathrm{K}}\right)}{\rho S_{\mathrm{col}}\left(E_{\mathrm{K}}\right)} \mathrm{d} E_{\mathrm{K}}, (12.343)
where S_{col}\left(E_K\right) is the collision stopping power for the charged particles. As shown in Fig. 12.20 for our example of proton activation of nitrogen-14 with protons of kinetic energy of 15 MeV, the activation cross section σ_P\left(E_K\right) is a complex function of the particle kinetic energy and (12.343) can be calculated by numerical methods. To simplify matters we express (12.343) as a product of thickness x_{eff} and a mean activation cross section \bar{\sigma}_{\mathrm{P}} to obtain the following simplified expression for (12.344)
\mathcal{A}(t)=n^{\square} \frac{I}{q} \bar{\sigma}_{\mathrm{P}} x_{\mathrm{eff}}\left(1-e^{-\lambda_{\mathrm{D}} t}\right)=n^{\square} \frac{I}{q} \bar{\sigma}_{\mathrm{P}} x_{\mathrm{eff}}\left(1-e^{-\frac{(\ln 2) t}{\left(t_{1 / 2}\right)_D}}\right), (12.344)
where we used the standard relationship between decay constant λ_D and half-life \left(t_{1/2}\right)_D for the activation product (daughter D) λ_D = (ln 2)/(t_{1/2})_D.
After solving (12.344) for \bar{σ}_P we get
\bar{\sigma}_{\mathrm{P}}=\frac{\mathcal{A}(t) q}{n^{\square} I x_{\mathrm{eff}}\left[1-e^{-\frac{\ln (2)}{\left(t_{1 / 2}\right) \mathrm{D}} t}\right]} . (12.345)
For the problem at hand, the nuclear density n^{\square} for the nitrogen-14 target is calculated as
n^{\square}=\rho \frac{N_{\mathrm{A}}}{A}=\left(1.748 \times 10^{-2} \mathrm{~g} / \mathrm{cm}^3\right) \times \frac{\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right)}{\left(14 \mathrm{~g} \cdot \mathrm{mol}^{-1}\right)}=7.459 \times 10^{20} \mathrm{~cm}^{-3}\quad (12.346)and (12.345) gives the following result for the mean activation cross section \bar{\sigma}_{\mathrm{N}-14} of nitrogen-14 with the following input data: x_{eff} = 15.37 cm, as determined in (12.337); activation time t = \left(t_{1/2}\right)_{C-11} = 20.4 min; activity A(t) at activation time t is 3.52 Ci = 1.30×10^{11} Bq; proton charge q = 1.602\times 10^{−19}\ C; and the proton beam current I = 4\times 10^{−5}\ A.
